Post on 07-Nov-2021
Research ArticleOptimal Control of Investment-ReinsuranceProblem for an Insurer with Jump-Diffusion Risk ProcessIndependence of Brownian Motions
De-Lei Sheng1 Ximin Rong12 and Hui Zhao1
1 Department of Mathematics Tianjin University Tianjin 300072 China2 Center for Applied Mathematics Tianjin University Tianjin 300072 China
Correspondence should be addressed to De-Lei Sheng tjhbsdl126com
Received 18 March 2014 Revised 20 June 2014 Accepted 25 June 2014 Published 24 July 2014
Academic Editor Simone Marsiglio
Copyright copy 2014 De-Lei Sheng et al This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited
This paper investigates the excess-of-loss reinsurance and investment problem for a compound Poisson jump-diffusion risk processwith the risk asset price modeled by a constant elasticity of variance (CEV) model It aims at obtaining the explicit optimal controlstrategy and the optimal value function Applying stochastic control technique of jump diffusion aHamilton-Jacobi-Bellman (HJB)equation is established Moreover we show that a closed-form solution for the HJB equation can be found by maximizing theinsurerrsquos exponential utility of terminal wealth with the independence of two Brownian motions 119882(119905) and 119882
1(119905) A verification
theorem is also proved to verify that the solution of HJB equation is indeed a solution of this optimal control problem Then wequantitatively analyze the effect of different parameter impacts on optimal control strategy and the optimal value function whichshow that optimal control strategy is decreasing with the initial wealth 119909 and decreasing with the volatility rate of risk asset priceHowever the optimal value function 119881(119905 119909 119904) is increasing with the appreciation rate 120583 of risk asset
1 Introduction
Bymeans of investment and reinsurance insurers can protectthemselves against potentially large losses or ensure theirearnings remain relatively stable Therefore many optimalinvestment and reinsurance problems have arisen in insur-ance risk management and have been extensively studied inthe literature
In the older forms reinsurance was often referred toas ldquoproportionalrdquo reinsurance few studies pay attentionto reinsurance Since Borch [1] studied the safety load-ing of reinsurance premiums a vast amount of literatureis particularly concerned about reinsurance However theexcess-of-loss reinsurance is a tool commonly employed inrisk management in the recent thirty years Tapiero andZuckerman [2] gave the optimum excess-loss reinsuranceunder a dynamic framework Taylor [3] studied reservingconsecutive layers of inwards excess-of-loss reinsurance Caoand Xu [4] investigated both proportional and excess-of-lossreinsurance under investment gains Gu et al [5] investigated
optimal control of excess-of-loss reinsurance and investmentfor insurers under a constant elasticity of variance modelbut without compound Poisson jump in their research Zhaoet al [6] studied optimal excess-of-loss reinsurance andinvestment problem for an insurer with jump-diffusion riskprocess under the Heston model
It is well known that the compound Poisson process is themost popular and useful model to describe claims processever since the classical Cramer-Lundberg model in risktheory However as a compound Poisson process perturbedby a standard Brownian motion jump diffusion has beenresearched extensively in the recent ten years Jump diffusioncan givemore practical description of claims than continuousmodels which widely used to describe dynamics of surplusprocess Yang and Zhang [7] investigated the problem ofoptimal investment for insurer with jump-diffusion riskprocess Li et al [8] studied the threshold dividend strategyfor renewal jump-diffusion process Ruan et al [9] studiedthe optimal portfolio and consumption with habit formationin a jump-diffusion market
Hindawi Publishing CorporationAbstract and Applied AnalysisVolume 2014 Article ID 194962 19 pageshttpdxdoiorg1011552014194962
2 Abstract and Applied Analysis
Not only does the insurer cede part of premiums forreinsurance but the insurer also invests in a financial marketconsisting of one risk-free asset and a risk asset The constantelasticity of variance (CEV) model describes the volatilityof risk asset that received widespread interests after beingproposed by Cox and Ross [10] for European option pricingpartly because the CEV model can capture the impliedvolatility skew and can explain the volatility smile as wellHsu et al [11] gave the integration and detailed derivationfor constant elasticity of variance (CEV) option pricingmodel Campi et al [12] investigated systematic equity-based credit risk for a CEV model with jump to defaultApplying Legendre transform and dual theory Xiao et al [13]investigated analytical strategies for annuity contracts underconstant elasticity of variance model The CEV model wasapplied to the optimal investment and reinsurance problemsin Gu et al [14] By maximizing the expected exponentialutility of terminal wealth Lin and Li [15] studied the optimalreinsurance and investment for a jump-diffusion risk processunder the CEV model
The goal of this paper is therefore to investigate theoptimal control strategy of excess-of-loss reinsurance andinvestment problem for an insurer with compound Poissonjump-diffusion risk process from the stochastic control pointof view Under the hypothesis that two Brownian motions119882(119905) and 119882
1(119905) are mutually independent we present
three main results We first give the optimal excess-of-lossreinsurance strategy and investment strategy respectivelyaccording to the first-order necessary condition ofmaximumWe also obtain the optimal value function by plugging areasonable conjecture into the HJB equation and solvinglinear second-order partial differential equations In the thirdpart we specialize in a verification theorem We also provethe verification theorem in detail inspired by the results ofTaksar and Zeng [16] Gu et al [5] and Zhao et al [6] At lastwe present some numerical examples to illustrate the impactsof model parameters
The paper is organized as follows Section 2 introducesthe risk asset price model under the CEV model and thewealth process with jump diffusion In Section 3 we con-sider the optimal excess-of-loss reinsurance and investmentstrategy and a verification theorem as well Section 4 containssome numerical simulations and theoretical results Section 5brings this paper to an end
2 The Model
Throughout this paper (ΩF 119875 F1199050le119905le119879
) denotes a com-plete probability space satisfying the usual condition wherea finite constant 119879 gt 0 represents the investment timehorizon two standard Brownianmotions119882
1(119905) and119882(119905) are
independent of each other 120574 is a Poisson random measureF
119905= F119882
119905or F
1198821
119905or F
120574
119905represents the minimal 120590-field
generated by F119882
119905 F119882
1
119905 and F
120574
119905 which contains all the
information available until time 119905 All stochastic processesinvolved in this paper are supposed to be F
119905adapted In
addition let 119909 and 119910 = min119909 119910 below
Following the same formulation of Zhao et al [6] thesurplus process of an insurer is described by the followingjump-diffusion model
d119877 (119905) = 119888119889119905 + 120590d119882(119905) minus d119862 (119905) (1)
where 119888 and 120590 are positive constants 119882(119905) is a standardBrownian motion and 120590119882(119905) describes the uncertaintyassociated with the surplus of the insurer at time 119905 Assume119862(119905) = sum
119873(119905)
119894=1119885
119894which represents the cumulative claims until
time 119905 where 119873(119905) is a homogeneous Poisson process withintensity 120582 and the claim sizes 119885
119894 119894 ge 1 are independent
and identically distributed positive random variables withcommon distribution 119865(119911) Denote the mean value 119864[119885
119894] =
120583infin
and
119863 = sup 119911 119865 (119911) le 1 lt +infin (2)
Suppose that 119865(0) = 0 0 lt 119865(119911) lt 1 if 119911 isin (0 119863) and 119865(119911) =1 if 119911 ge 119863 The premium according to the expected valueprinciple is 119888 = (1 + 120578)120582120583
infin where 120578 gt 0 is the safety loading
of the insurerAccording to the theory of Poisson random measure
(referring to Oslashksendal and Sulem [17]) we can rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894 120574 denotes the
Poisson random measure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(3)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (4)
so the compensated Poisson random measure of 120574 is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (5)
Insurance company can purchase excess-of-loss rein-surance to reduce the risk Supposing the insurerrsquos (fixed)retention level denoted by 119886 the corresponding reserveprocess is
d119877 (119905) = 119888(119886)119889119905 + 120590d119882(119905) minus d119862(119886)
(119905) (6)
and the reinsurerrsquos premium rate is
119888(119886)
= (1 + 120578) 120582120583infinminus (1 + 120579) 120582 120583
infinminus 119864 [119885
119894and 119886]
= (120578 minus 120579) 120582120583infin+ (1 + 120579) 120582119864 [119885
119894and 119886]
119864 [119885119894and 119886] = int
119886
0
119911d119865 (119911) + int
infin
119886
119886d119865 (119911)
= int
119886
0
(1 minus 119865 (119911)) d119911 ≜ int
119886
0
119865 (119911)d119911
(7)
Abstract and Applied Analysis 3
and 119862(119886)= sum
119873(119905)
119894=1(119885
119894and 119886) satisfying
119890int119905
0119890minus119903119904d119862(119886)(119904)
lt infin forall119905 lt infin (8)
In particular if the retention level 119886 = 119863 then
119888(119886)
= (1 + 120578) 120582120583infin= 119888
(119863)= 119888 (9)
Without loss of generality we always assume the reinsur-ance is not cheap that is 120579 gt 120578 and 120579 denotes the safetyloading of the reinsurer
Moreover the insurer also can invest in a financial marketconsisting of one risk-free asset and one risky assetThe priceprocess 119878
0(119905) of the risk-free asset is given by
d1198780(119905) = 119903119878
0(119905) d119905 (10)
where 119903 gt 0 is the risk-free interest rate The price process119878(119905) of the risky asset is described by the constant elasticity ofvariance (CEV) model
d119878 (119905) = 119878 (119905) [120583d119905 + 119896119878120573(119905) d119882
1(119905)] (11)
where 120583 and 119896 are positive constants 120583 gt 119903 is the expectedinstantaneous return rate of the risky asset 119896119878120573
(119905) is theinstantaneous volatility 120573 is the elasticity parameter and119882
1(119905) is a standard Brownian motion When 120573 = 0 a CEV
model degenerates into a GBM In this paper we assume thestandard Brownian motion119882
1(119905) is independent of119882(119905)
The control 120572 = (119886(119905) 120587(119905)) 119905 isin [0 119879] is a two-dimen-sional F(119905) adapted stochastic process where 119886(119905) isin [0 119863]
is the retention level of the excess-of-loss at time 119905 in which119886(119905) equiv 119863 means ldquono reinsurancerdquo and 119886(119905) equiv 0 means ldquofullreinsurancerdquo and 120587(119905) represents the proportion invested inthe risky asset Here short-selling is not allowed that is120587(119905) ge 0The amount invested in the risk-free asset is119883(119905)(1minus120587(119905)) where 119883(119905) is the wealth process for the insurer underthe strategy 120572 and the dynamics of119883(119905) is given by
d119883 (119905) = [(120578 minus 120579) 120582120583infin+ (1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120583120587 (119905)119883 (119905) + 119903 (1 minus 120587 (119905))119883 (119905) ] d119905
+ 120590d119882(119905) + 120587 (119905)119883 (119905) 119896119878120573(119905) d119882
1(119905) minus d119862(119886)
(119905)
119883 (0) = 1199090
(12)
A strategy 120572 is said to be admissible if forall119905 isin
[0 119879] (119886(119905) 120587(119905)) is F(119905) progressively measurable and119864[int
infin
0120587
2(119905)119883
2(119905)119878
2120573(119905)] lt infin 119886(119905) isin [0 119863] and 120587(119905) isin
[0 +infin) The set of all admissible strategies denoted byΛ and (12) has a unique (strong) solution Suppose theinsurer has a utility function 119880(119909) which is strictly concaveand continuously differentiable on (minusinfin +infin) and aims tomaximize the expected utility of hisher terminal wealth thatis
max120572isinΛ
119864 [119880 (119883 (119879))] (13)
3 Main Results
In this section we solve the excess-of-loss reinsurance andinvestment problem with independence of two Brownianmotions119882(119905) and119882
1(119905) By maximizing the expected utility
of terminal wealth the optimal strategy and value functionare given At the beginning let us give the following expo-nential utility function of an risk aversion insurer
119880 (119909) = minus1
119902119890minus119902119909
119902 gt 0 (14)
This utility function has a constant absolute risk aversionparameter 119902 and is the only utility function under the princi-ple of ldquozero utilityrdquo giving a fair premium that is independentof the level of reserves of insurers For an admissible strategy120572 = (119886(119905) 120587(119905)) we define the value function as
119867120572(119905 119909 119904) = 119864 [119880 (119883 (119879)) | 119883 (119905) = 119909 119878 (119905) = 119904] (15)
and the optimal value function is
119867(119905 119909 119904) = sup120572isinΛ
119867120572(119905 119909 119904) (16)
with the boundary condition
119867(119879 119909 119904) = minus1
119902119890minus119902119909
(17)
The objective of the insurer is to find an optimal strategy120572
lowast= (119886
lowast(119905) 120587
lowast(119905)) such that 119867(119905 119909 119904) = 119867
120572lowast
(119905 119909 119904) where119886
lowast(119905) is called the optimal reinsurance strategy and 120587
lowast(119905) is
called the optimal investment strategyFor any119867120572
isin 119862122
([0 119879]times119877+times119877
+) define the generator
A119867120572(119905 119909 119904)
= 119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909
+ 120583119904119867120572
119904+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+ (120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909+ 119896
2120587119909119904
2120573+1119867
120572
119909119904
+ 119867120572
119909(1 + 120579) 120582int
119886
0
119865(119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
(18)
where 119867120572
119905 119867120572
119909 119867120572
119904 119867120572
119909119909 and 119867
120572
119904119904denote the corresponding
first- and second-order partial derivatives of 119867120572(119905 119909 119904) with
respect to (wrt) the corresponding variables respectivelyApplying the classical tools of stochastic optimal control
we can derive the following Hamilton-Jacobi-Bellman (HJB)equation for problem (16)
sup120572isinΛ
A119867120572(119905 119909 119904) = 0 (19)
with the boundary condition (17)
4 Abstract and Applied Analysis
Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem
Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast
= (119886lowast 120587
lowast) in feedback form where
one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the
whole interval [0 119879] always is
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (20)
and the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(21)
The optimal value function is119881 (119905 119909 119904)
= minus1
119902exp minus 119902
times [119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909) d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
times 119865 (119909) d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
]
(22)
(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(23)
And the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(24)
The optimal value function is given by
119881 (119905 119909 119904)
= minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
=
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]
119892 (119905 119904)
= [(minus(120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573) 119890
2120573119903119905+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573
+ ((120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573)1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ (2120573 + 1) (120583 minus 119903)2
times (4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573) 119905 isin [0 119905
0)
119892 (119905 119904)
=
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
119905 isin [1199050 119879]
119891 (119905)
=(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [119905
0 119879]
Abstract and Applied Analysis 5
119891 (119905)
=(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
minus(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905
0)
+ (1 + 120579) 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [0 119905
0)
(25)Proof See Appendix A
Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= (120583 minus 119903) [1
119902minus
(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)
)
2119902119903]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587
lowast(119905) = ((120583 minus 119903)
1199091199021198962)119890
minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]
A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy
Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem
Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions
(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt
(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)
2minus 1205832120583)
then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast
= (119886lowast(119905) 120587
lowast(119905)) given in Theorem 1
Proof See Appendix B
Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump
4 Numerical Results
In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge
ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar
Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890
minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583
infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1
120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the
optimal strategy when 120573 ge 0
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= [
[
120583 minus 119903
119902minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
2119902119903
]
]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(27)
where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price
6 Abstract and Applied Analysis
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of x on the optimal investment strategy 120587
x = 5
x = 3
x = 1
Figure 1 The optimal investment strategy 120587 decreases with respectto 119909
Deriving 120587lowast with respect to 119909 we have
120597120587lowast
120597119909= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(28)
because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Remark 5 Obviously the optimal investment strategy 120587lowast
decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1
However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset
Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890
2119903119905120573 and 120583 gt (120583minus119903)we have
120597120587lowast
120597120583=
119890119903(119905minus119879(1+2120573))
119904minus2120573
[1198902119903119879120573
120583 minus 1198902119903119905120573
(120583 minus 119903)]
1198962119902119903119909gt 0 (29)
Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2
010 011 012 013 014 015 016
06
08
10
12
14
120583
120587
The effect of 120583 on the optimal investment strategy 120587
t = 8
t = 5
t = 1
Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset
Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt
119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)
Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions
120597120587lowast
120597120573= minus (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [1198902119903119905120573
(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
= (119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
(30)
Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573
gt 1198902119903119905120573
gt 0 and ln[119904] gt 0If 1199042 gt 119890
(120583minus119903)(119879minus119905) take logarithm on both sides
ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)
We have
1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905)
gt 1198902119903119905120573
119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0
(32)
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
2 Abstract and Applied Analysis
Not only does the insurer cede part of premiums forreinsurance but the insurer also invests in a financial marketconsisting of one risk-free asset and a risk asset The constantelasticity of variance (CEV) model describes the volatilityof risk asset that received widespread interests after beingproposed by Cox and Ross [10] for European option pricingpartly because the CEV model can capture the impliedvolatility skew and can explain the volatility smile as wellHsu et al [11] gave the integration and detailed derivationfor constant elasticity of variance (CEV) option pricingmodel Campi et al [12] investigated systematic equity-based credit risk for a CEV model with jump to defaultApplying Legendre transform and dual theory Xiao et al [13]investigated analytical strategies for annuity contracts underconstant elasticity of variance model The CEV model wasapplied to the optimal investment and reinsurance problemsin Gu et al [14] By maximizing the expected exponentialutility of terminal wealth Lin and Li [15] studied the optimalreinsurance and investment for a jump-diffusion risk processunder the CEV model
The goal of this paper is therefore to investigate theoptimal control strategy of excess-of-loss reinsurance andinvestment problem for an insurer with compound Poissonjump-diffusion risk process from the stochastic control pointof view Under the hypothesis that two Brownian motions119882(119905) and 119882
1(119905) are mutually independent we present
three main results We first give the optimal excess-of-lossreinsurance strategy and investment strategy respectivelyaccording to the first-order necessary condition ofmaximumWe also obtain the optimal value function by plugging areasonable conjecture into the HJB equation and solvinglinear second-order partial differential equations In the thirdpart we specialize in a verification theorem We also provethe verification theorem in detail inspired by the results ofTaksar and Zeng [16] Gu et al [5] and Zhao et al [6] At lastwe present some numerical examples to illustrate the impactsof model parameters
The paper is organized as follows Section 2 introducesthe risk asset price model under the CEV model and thewealth process with jump diffusion In Section 3 we con-sider the optimal excess-of-loss reinsurance and investmentstrategy and a verification theorem as well Section 4 containssome numerical simulations and theoretical results Section 5brings this paper to an end
2 The Model
Throughout this paper (ΩF 119875 F1199050le119905le119879
) denotes a com-plete probability space satisfying the usual condition wherea finite constant 119879 gt 0 represents the investment timehorizon two standard Brownianmotions119882
1(119905) and119882(119905) are
independent of each other 120574 is a Poisson random measureF
119905= F119882
119905or F
1198821
119905or F
120574
119905represents the minimal 120590-field
generated by F119882
119905 F119882
1
119905 and F
120574
119905 which contains all the
information available until time 119905 All stochastic processesinvolved in this paper are supposed to be F
119905adapted In
addition let 119909 and 119910 = min119909 119910 below
Following the same formulation of Zhao et al [6] thesurplus process of an insurer is described by the followingjump-diffusion model
d119877 (119905) = 119888119889119905 + 120590d119882(119905) minus d119862 (119905) (1)
where 119888 and 120590 are positive constants 119882(119905) is a standardBrownian motion and 120590119882(119905) describes the uncertaintyassociated with the surplus of the insurer at time 119905 Assume119862(119905) = sum
119873(119905)
119894=1119885
119894which represents the cumulative claims until
time 119905 where 119873(119905) is a homogeneous Poisson process withintensity 120582 and the claim sizes 119885
119894 119894 ge 1 are independent
and identically distributed positive random variables withcommon distribution 119865(119911) Denote the mean value 119864[119885
119894] =
120583infin
and
119863 = sup 119911 119865 (119911) le 1 lt +infin (2)
Suppose that 119865(0) = 0 0 lt 119865(119911) lt 1 if 119911 isin (0 119863) and 119865(119911) =1 if 119911 ge 119863 The premium according to the expected valueprinciple is 119888 = (1 + 120578)120582120583
infin where 120578 gt 0 is the safety loading
of the insurerAccording to the theory of Poisson random measure
(referring to Oslashksendal and Sulem [17]) we can rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894 120574 denotes the
Poisson random measure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(3)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (4)
so the compensated Poisson random measure of 120574 is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (5)
Insurance company can purchase excess-of-loss rein-surance to reduce the risk Supposing the insurerrsquos (fixed)retention level denoted by 119886 the corresponding reserveprocess is
d119877 (119905) = 119888(119886)119889119905 + 120590d119882(119905) minus d119862(119886)
(119905) (6)
and the reinsurerrsquos premium rate is
119888(119886)
= (1 + 120578) 120582120583infinminus (1 + 120579) 120582 120583
infinminus 119864 [119885
119894and 119886]
= (120578 minus 120579) 120582120583infin+ (1 + 120579) 120582119864 [119885
119894and 119886]
119864 [119885119894and 119886] = int
119886
0
119911d119865 (119911) + int
infin
119886
119886d119865 (119911)
= int
119886
0
(1 minus 119865 (119911)) d119911 ≜ int
119886
0
119865 (119911)d119911
(7)
Abstract and Applied Analysis 3
and 119862(119886)= sum
119873(119905)
119894=1(119885
119894and 119886) satisfying
119890int119905
0119890minus119903119904d119862(119886)(119904)
lt infin forall119905 lt infin (8)
In particular if the retention level 119886 = 119863 then
119888(119886)
= (1 + 120578) 120582120583infin= 119888
(119863)= 119888 (9)
Without loss of generality we always assume the reinsur-ance is not cheap that is 120579 gt 120578 and 120579 denotes the safetyloading of the reinsurer
Moreover the insurer also can invest in a financial marketconsisting of one risk-free asset and one risky assetThe priceprocess 119878
0(119905) of the risk-free asset is given by
d1198780(119905) = 119903119878
0(119905) d119905 (10)
where 119903 gt 0 is the risk-free interest rate The price process119878(119905) of the risky asset is described by the constant elasticity ofvariance (CEV) model
d119878 (119905) = 119878 (119905) [120583d119905 + 119896119878120573(119905) d119882
1(119905)] (11)
where 120583 and 119896 are positive constants 120583 gt 119903 is the expectedinstantaneous return rate of the risky asset 119896119878120573
(119905) is theinstantaneous volatility 120573 is the elasticity parameter and119882
1(119905) is a standard Brownian motion When 120573 = 0 a CEV
model degenerates into a GBM In this paper we assume thestandard Brownian motion119882
1(119905) is independent of119882(119905)
The control 120572 = (119886(119905) 120587(119905)) 119905 isin [0 119879] is a two-dimen-sional F(119905) adapted stochastic process where 119886(119905) isin [0 119863]
is the retention level of the excess-of-loss at time 119905 in which119886(119905) equiv 119863 means ldquono reinsurancerdquo and 119886(119905) equiv 0 means ldquofullreinsurancerdquo and 120587(119905) represents the proportion invested inthe risky asset Here short-selling is not allowed that is120587(119905) ge 0The amount invested in the risk-free asset is119883(119905)(1minus120587(119905)) where 119883(119905) is the wealth process for the insurer underthe strategy 120572 and the dynamics of119883(119905) is given by
d119883 (119905) = [(120578 minus 120579) 120582120583infin+ (1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120583120587 (119905)119883 (119905) + 119903 (1 minus 120587 (119905))119883 (119905) ] d119905
+ 120590d119882(119905) + 120587 (119905)119883 (119905) 119896119878120573(119905) d119882
1(119905) minus d119862(119886)
(119905)
119883 (0) = 1199090
(12)
A strategy 120572 is said to be admissible if forall119905 isin
[0 119879] (119886(119905) 120587(119905)) is F(119905) progressively measurable and119864[int
infin
0120587
2(119905)119883
2(119905)119878
2120573(119905)] lt infin 119886(119905) isin [0 119863] and 120587(119905) isin
[0 +infin) The set of all admissible strategies denoted byΛ and (12) has a unique (strong) solution Suppose theinsurer has a utility function 119880(119909) which is strictly concaveand continuously differentiable on (minusinfin +infin) and aims tomaximize the expected utility of hisher terminal wealth thatis
max120572isinΛ
119864 [119880 (119883 (119879))] (13)
3 Main Results
In this section we solve the excess-of-loss reinsurance andinvestment problem with independence of two Brownianmotions119882(119905) and119882
1(119905) By maximizing the expected utility
of terminal wealth the optimal strategy and value functionare given At the beginning let us give the following expo-nential utility function of an risk aversion insurer
119880 (119909) = minus1
119902119890minus119902119909
119902 gt 0 (14)
This utility function has a constant absolute risk aversionparameter 119902 and is the only utility function under the princi-ple of ldquozero utilityrdquo giving a fair premium that is independentof the level of reserves of insurers For an admissible strategy120572 = (119886(119905) 120587(119905)) we define the value function as
119867120572(119905 119909 119904) = 119864 [119880 (119883 (119879)) | 119883 (119905) = 119909 119878 (119905) = 119904] (15)
and the optimal value function is
119867(119905 119909 119904) = sup120572isinΛ
119867120572(119905 119909 119904) (16)
with the boundary condition
119867(119879 119909 119904) = minus1
119902119890minus119902119909
(17)
The objective of the insurer is to find an optimal strategy120572
lowast= (119886
lowast(119905) 120587
lowast(119905)) such that 119867(119905 119909 119904) = 119867
120572lowast
(119905 119909 119904) where119886
lowast(119905) is called the optimal reinsurance strategy and 120587
lowast(119905) is
called the optimal investment strategyFor any119867120572
isin 119862122
([0 119879]times119877+times119877
+) define the generator
A119867120572(119905 119909 119904)
= 119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909
+ 120583119904119867120572
119904+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+ (120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909+ 119896
2120587119909119904
2120573+1119867
120572
119909119904
+ 119867120572
119909(1 + 120579) 120582int
119886
0
119865(119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
(18)
where 119867120572
119905 119867120572
119909 119867120572
119904 119867120572
119909119909 and 119867
120572
119904119904denote the corresponding
first- and second-order partial derivatives of 119867120572(119905 119909 119904) with
respect to (wrt) the corresponding variables respectivelyApplying the classical tools of stochastic optimal control
we can derive the following Hamilton-Jacobi-Bellman (HJB)equation for problem (16)
sup120572isinΛ
A119867120572(119905 119909 119904) = 0 (19)
with the boundary condition (17)
4 Abstract and Applied Analysis
Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem
Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast
= (119886lowast 120587
lowast) in feedback form where
one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the
whole interval [0 119879] always is
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (20)
and the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(21)
The optimal value function is119881 (119905 119909 119904)
= minus1
119902exp minus 119902
times [119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909) d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
times 119865 (119909) d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
]
(22)
(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(23)
And the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(24)
The optimal value function is given by
119881 (119905 119909 119904)
= minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
=
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]
119892 (119905 119904)
= [(minus(120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573) 119890
2120573119903119905+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573
+ ((120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573)1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ (2120573 + 1) (120583 minus 119903)2
times (4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573) 119905 isin [0 119905
0)
119892 (119905 119904)
=
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
119905 isin [1199050 119879]
119891 (119905)
=(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [119905
0 119879]
Abstract and Applied Analysis 5
119891 (119905)
=(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
minus(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905
0)
+ (1 + 120579) 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [0 119905
0)
(25)Proof See Appendix A
Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= (120583 minus 119903) [1
119902minus
(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)
)
2119902119903]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587
lowast(119905) = ((120583 minus 119903)
1199091199021198962)119890
minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]
A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy
Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem
Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions
(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt
(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)
2minus 1205832120583)
then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast
= (119886lowast(119905) 120587
lowast(119905)) given in Theorem 1
Proof See Appendix B
Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump
4 Numerical Results
In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge
ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar
Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890
minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583
infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1
120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the
optimal strategy when 120573 ge 0
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= [
[
120583 minus 119903
119902minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
2119902119903
]
]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(27)
where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price
6 Abstract and Applied Analysis
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of x on the optimal investment strategy 120587
x = 5
x = 3
x = 1
Figure 1 The optimal investment strategy 120587 decreases with respectto 119909
Deriving 120587lowast with respect to 119909 we have
120597120587lowast
120597119909= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(28)
because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Remark 5 Obviously the optimal investment strategy 120587lowast
decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1
However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset
Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890
2119903119905120573 and 120583 gt (120583minus119903)we have
120597120587lowast
120597120583=
119890119903(119905minus119879(1+2120573))
119904minus2120573
[1198902119903119879120573
120583 minus 1198902119903119905120573
(120583 minus 119903)]
1198962119902119903119909gt 0 (29)
Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2
010 011 012 013 014 015 016
06
08
10
12
14
120583
120587
The effect of 120583 on the optimal investment strategy 120587
t = 8
t = 5
t = 1
Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset
Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt
119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)
Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions
120597120587lowast
120597120573= minus (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [1198902119903119905120573
(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
= (119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
(30)
Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573
gt 1198902119903119905120573
gt 0 and ln[119904] gt 0If 1199042 gt 119890
(120583minus119903)(119879minus119905) take logarithm on both sides
ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)
We have
1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905)
gt 1198902119903119905120573
119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0
(32)
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 3
and 119862(119886)= sum
119873(119905)
119894=1(119885
119894and 119886) satisfying
119890int119905
0119890minus119903119904d119862(119886)(119904)
lt infin forall119905 lt infin (8)
In particular if the retention level 119886 = 119863 then
119888(119886)
= (1 + 120578) 120582120583infin= 119888
(119863)= 119888 (9)
Without loss of generality we always assume the reinsur-ance is not cheap that is 120579 gt 120578 and 120579 denotes the safetyloading of the reinsurer
Moreover the insurer also can invest in a financial marketconsisting of one risk-free asset and one risky assetThe priceprocess 119878
0(119905) of the risk-free asset is given by
d1198780(119905) = 119903119878
0(119905) d119905 (10)
where 119903 gt 0 is the risk-free interest rate The price process119878(119905) of the risky asset is described by the constant elasticity ofvariance (CEV) model
d119878 (119905) = 119878 (119905) [120583d119905 + 119896119878120573(119905) d119882
1(119905)] (11)
where 120583 and 119896 are positive constants 120583 gt 119903 is the expectedinstantaneous return rate of the risky asset 119896119878120573
(119905) is theinstantaneous volatility 120573 is the elasticity parameter and119882
1(119905) is a standard Brownian motion When 120573 = 0 a CEV
model degenerates into a GBM In this paper we assume thestandard Brownian motion119882
1(119905) is independent of119882(119905)
The control 120572 = (119886(119905) 120587(119905)) 119905 isin [0 119879] is a two-dimen-sional F(119905) adapted stochastic process where 119886(119905) isin [0 119863]
is the retention level of the excess-of-loss at time 119905 in which119886(119905) equiv 119863 means ldquono reinsurancerdquo and 119886(119905) equiv 0 means ldquofullreinsurancerdquo and 120587(119905) represents the proportion invested inthe risky asset Here short-selling is not allowed that is120587(119905) ge 0The amount invested in the risk-free asset is119883(119905)(1minus120587(119905)) where 119883(119905) is the wealth process for the insurer underthe strategy 120572 and the dynamics of119883(119905) is given by
d119883 (119905) = [(120578 minus 120579) 120582120583infin+ (1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120583120587 (119905)119883 (119905) + 119903 (1 minus 120587 (119905))119883 (119905) ] d119905
+ 120590d119882(119905) + 120587 (119905)119883 (119905) 119896119878120573(119905) d119882
1(119905) minus d119862(119886)
(119905)
119883 (0) = 1199090
(12)
A strategy 120572 is said to be admissible if forall119905 isin
[0 119879] (119886(119905) 120587(119905)) is F(119905) progressively measurable and119864[int
infin
0120587
2(119905)119883
2(119905)119878
2120573(119905)] lt infin 119886(119905) isin [0 119863] and 120587(119905) isin
[0 +infin) The set of all admissible strategies denoted byΛ and (12) has a unique (strong) solution Suppose theinsurer has a utility function 119880(119909) which is strictly concaveand continuously differentiable on (minusinfin +infin) and aims tomaximize the expected utility of hisher terminal wealth thatis
max120572isinΛ
119864 [119880 (119883 (119879))] (13)
3 Main Results
In this section we solve the excess-of-loss reinsurance andinvestment problem with independence of two Brownianmotions119882(119905) and119882
1(119905) By maximizing the expected utility
of terminal wealth the optimal strategy and value functionare given At the beginning let us give the following expo-nential utility function of an risk aversion insurer
119880 (119909) = minus1
119902119890minus119902119909
119902 gt 0 (14)
This utility function has a constant absolute risk aversionparameter 119902 and is the only utility function under the princi-ple of ldquozero utilityrdquo giving a fair premium that is independentof the level of reserves of insurers For an admissible strategy120572 = (119886(119905) 120587(119905)) we define the value function as
119867120572(119905 119909 119904) = 119864 [119880 (119883 (119879)) | 119883 (119905) = 119909 119878 (119905) = 119904] (15)
and the optimal value function is
119867(119905 119909 119904) = sup120572isinΛ
119867120572(119905 119909 119904) (16)
with the boundary condition
119867(119879 119909 119904) = minus1
119902119890minus119902119909
(17)
The objective of the insurer is to find an optimal strategy120572
lowast= (119886
lowast(119905) 120587
lowast(119905)) such that 119867(119905 119909 119904) = 119867
120572lowast
(119905 119909 119904) where119886
lowast(119905) is called the optimal reinsurance strategy and 120587
lowast(119905) is
called the optimal investment strategyFor any119867120572
isin 119862122
([0 119879]times119877+times119877
+) define the generator
A119867120572(119905 119909 119904)
= 119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909
+ 120583119904119867120572
119904+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+ (120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909+ 119896
2120587119909119904
2120573+1119867
120572
119909119904
+ 119867120572
119909(1 + 120579) 120582int
119886
0
119865(119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
(18)
where 119867120572
119905 119867120572
119909 119867120572
119904 119867120572
119909119909 and 119867
120572
119904119904denote the corresponding
first- and second-order partial derivatives of 119867120572(119905 119909 119904) with
respect to (wrt) the corresponding variables respectivelyApplying the classical tools of stochastic optimal control
we can derive the following Hamilton-Jacobi-Bellman (HJB)equation for problem (16)
sup120572isinΛ
A119867120572(119905 119909 119904) = 0 (19)
with the boundary condition (17)
4 Abstract and Applied Analysis
Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem
Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast
= (119886lowast 120587
lowast) in feedback form where
one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the
whole interval [0 119879] always is
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (20)
and the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(21)
The optimal value function is119881 (119905 119909 119904)
= minus1
119902exp minus 119902
times [119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909) d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
times 119865 (119909) d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
]
(22)
(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(23)
And the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(24)
The optimal value function is given by
119881 (119905 119909 119904)
= minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
=
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]
119892 (119905 119904)
= [(minus(120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573) 119890
2120573119903119905+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573
+ ((120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573)1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ (2120573 + 1) (120583 minus 119903)2
times (4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573) 119905 isin [0 119905
0)
119892 (119905 119904)
=
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
119905 isin [1199050 119879]
119891 (119905)
=(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [119905
0 119879]
Abstract and Applied Analysis 5
119891 (119905)
=(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
minus(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905
0)
+ (1 + 120579) 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [0 119905
0)
(25)Proof See Appendix A
Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= (120583 minus 119903) [1
119902minus
(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)
)
2119902119903]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587
lowast(119905) = ((120583 minus 119903)
1199091199021198962)119890
minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]
A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy
Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem
Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions
(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt
(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)
2minus 1205832120583)
then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast
= (119886lowast(119905) 120587
lowast(119905)) given in Theorem 1
Proof See Appendix B
Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump
4 Numerical Results
In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge
ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar
Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890
minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583
infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1
120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the
optimal strategy when 120573 ge 0
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= [
[
120583 minus 119903
119902minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
2119902119903
]
]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(27)
where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price
6 Abstract and Applied Analysis
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of x on the optimal investment strategy 120587
x = 5
x = 3
x = 1
Figure 1 The optimal investment strategy 120587 decreases with respectto 119909
Deriving 120587lowast with respect to 119909 we have
120597120587lowast
120597119909= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(28)
because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Remark 5 Obviously the optimal investment strategy 120587lowast
decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1
However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset
Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890
2119903119905120573 and 120583 gt (120583minus119903)we have
120597120587lowast
120597120583=
119890119903(119905minus119879(1+2120573))
119904minus2120573
[1198902119903119879120573
120583 minus 1198902119903119905120573
(120583 minus 119903)]
1198962119902119903119909gt 0 (29)
Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2
010 011 012 013 014 015 016
06
08
10
12
14
120583
120587
The effect of 120583 on the optimal investment strategy 120587
t = 8
t = 5
t = 1
Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset
Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt
119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)
Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions
120597120587lowast
120597120573= minus (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [1198902119903119905120573
(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
= (119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
(30)
Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573
gt 1198902119903119905120573
gt 0 and ln[119904] gt 0If 1199042 gt 119890
(120583minus119903)(119879minus119905) take logarithm on both sides
ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)
We have
1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905)
gt 1198902119903119905120573
119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0
(32)
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
4 Abstract and Applied Analysis
Standard results (eg Fleming and Soner [18]) tell us(19) admits the unique strong solution which togetherwith the fact that the value function is twice-continuouslydifferentiable gives the following theorem
Theorem 1 For the optimal excess-of-loss reinsurance andinvestment problem with jump-diffusion risk process under theCEV model a solution to HJB equation (19) with boundarycondition (17) is given by 119881(119905 119909 119904) and the correspondingmaximizer is given by 120572lowast
= (119886lowast 120587
lowast) in feedback form where
one has the following(1) If 119863 ge ln(1 + 120579)119902 the optimal retention level on the
whole interval [0 119879] always is
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (20)
and the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(21)
The optimal value function is119881 (119905 119909 119904)
= minus1
119902exp minus 119902
times [119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909) d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
times 119865 (119909) d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
]
(22)
(2) If119863 lt ln(1 + 120579)119902 the optimal retention level is
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(23)
And the optimal investment strategy is given by
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
(24)
The optimal value function is given by
119881 (119905 119909 119904)
= minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
=
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [0 1199050)
minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119892 (119905 119904) + 119891 (119905)] 119905 isin [1199050 119879]
119892 (119905 119904)
= [(minus(120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573) 119890
2120573119903119905+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573
+ ((120583 minus 119903)
2
119890minus2119903119879120573
41199021199031198962120573)1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ (2120573 + 1) (120583 minus 119903)2
times (4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573) 119905 isin [0 119905
0)
119892 (119905 119904)
=
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
119905 isin [1199050 119879]
119891 (119905)
=(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [119905
0 119879]
Abstract and Applied Analysis 5
119891 (119905)
=(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
minus(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905
0)
+ (1 + 120579) 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [0 119905
0)
(25)Proof See Appendix A
Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= (120583 minus 119903) [1
119902minus
(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)
)
2119902119903]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587
lowast(119905) = ((120583 minus 119903)
1199091199021198962)119890
minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]
A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy
Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem
Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions
(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt
(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)
2minus 1205832120583)
then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast
= (119886lowast(119905) 120587
lowast(119905)) given in Theorem 1
Proof See Appendix B
Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump
4 Numerical Results
In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge
ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar
Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890
minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583
infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1
120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the
optimal strategy when 120573 ge 0
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= [
[
120583 minus 119903
119902minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
2119902119903
]
]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(27)
where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price
6 Abstract and Applied Analysis
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of x on the optimal investment strategy 120587
x = 5
x = 3
x = 1
Figure 1 The optimal investment strategy 120587 decreases with respectto 119909
Deriving 120587lowast with respect to 119909 we have
120597120587lowast
120597119909= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(28)
because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Remark 5 Obviously the optimal investment strategy 120587lowast
decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1
However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset
Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890
2119903119905120573 and 120583 gt (120583minus119903)we have
120597120587lowast
120597120583=
119890119903(119905minus119879(1+2120573))
119904minus2120573
[1198902119903119879120573
120583 minus 1198902119903119905120573
(120583 minus 119903)]
1198962119902119903119909gt 0 (29)
Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2
010 011 012 013 014 015 016
06
08
10
12
14
120583
120587
The effect of 120583 on the optimal investment strategy 120587
t = 8
t = 5
t = 1
Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset
Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt
119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)
Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions
120597120587lowast
120597120573= minus (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [1198902119903119905120573
(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
= (119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
(30)
Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573
gt 1198902119903119905120573
gt 0 and ln[119904] gt 0If 1199042 gt 119890
(120583minus119903)(119879minus119905) take logarithm on both sides
ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)
We have
1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905)
gt 1198902119903119905120573
119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0
(32)
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 5
119891 (119905)
=(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
minus(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905
0)
+ (1 + 120579) 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909) d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909) d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903 119905 isin [0 119905
0)
(25)Proof See Appendix A
Remark 2 FromTheorem 1 we find that the optimal invest-ment strategy is a function of (120583 minus 119903) 119902 119909 119896119904120573 119903 and 119905which fully reflects the influence of various factors on theinvestment strategy 119909 is the initial wealth 119896119904120573 representsthe volatility of risk asset price and 120583 minus 119903 is the profit ofrisk asset appreciation higher than risk-free interest rate Bysimple deformation
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= (120583 minus 119903) [1
119902minus
(120583 minus 119903) (1 minus 119890minus2119903120573(119879minus119905)
)
2119902119903]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(26)Obviously the optimal investment strategy decreases withrespect to the initial wealth 119909 and the volatility of risk assetprice 119896119904120573 In particular if 120573 = 0 then 120587
lowast(119905) = ((120583 minus 119903)
1199091199021198962)119890
minus119903(119879minus119905) which is the same as that in Theorem 1 in Guet al [5]
A surprising finding is that the optimal reinsurancestrategy has nothing to do with the initial wealth and riskasset However insurerrsquos safe load 120579 and the parameter 119902 inthe utility function play an important role in determiningreinsurance strategy
Motivated by the results of Taksar and Zeng [16] Gu et al[5] and Zhao et al [6] we obtain the following verificationtheorem
Theorem 3 (verification theorem) For the optimal excess-of-loss reinsurance and investment problem with jump-diffusionrisk process under the CEV model the wealth process 119883(119905)is associated with an admissible strategy (119886(119905) 120587(119905)) If thesolution to HJB equation (19) with boundary condition (17) isgiven by 119881(119905 119909 119904) and the parameters 120583 gt 119903 gt 0 satisfy eitherof the following conditions
(I) 119903 gt (1 minus 1radic6)120583(II) 119903 lt (1 minus 1radic6)120583 119879 lt
(1120573radic6(120583 minus 119903)2minus 1205832) arctan(minusradic6(120583 minus 119903)
2minus 1205832120583)
then the optimal value function is 119867(119905 119909 119904) = 119881(119905 119909 119904) andthe optimal strategy is 120572lowast
= (119886lowast(119905) 120587
lowast(119905)) given in Theorem 1
Proof See Appendix B
Remark 4 In contrast with Gu et al [5] they also consideredthe excess-of-loss with a compound Poisson jump under theCEVmodel however they simplified the compound Poissonjump by diffusion approximation process which leads theproblem back to the case without jump
4 Numerical Results
In this section we analyze the impacts of some parameterson the optimal strategies and the value function Theoreticalanalysis and some corresponding numerical examples aregiven to illustrate the influences of model parameters on theoptimal strategy and the optimal value function when 119863 ge
ln(1 + 120579)119902 The analysis for the case of 119863 lt ln(1 + 120579)119902 issimilar
Throughout this section we always assume 120573 ge 0 theclaim sizes follow exponential distribution 119865(119909) = 1 minus 119890
minus119909119898119909 gt 0 and if 120573 lt 0 some of the conclusions will be differentThe parameters are given by 119898 = 2 119902 = 05 119879 = 10 120590 = 1120583
infin= 05 120579 = 02 119903 = 005 120583 = 012 119896 = 02 120582 = 2 119909 = 1
120573 = 05 and 119904 = 1(1) First let us pay attention to the expression of the
optimal strategy when 120573 ge 0
120587lowast(119905) = [
[
120583 minus 119903
1199091199021198962minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
21199021199091199031198962
]
]
119890minus119903(119879minus119905)
119904minus2120573
= [
[
120583 minus 119903
119902minus
(120583 minus 119903)2
(1198902119903120573(119905minus119879)
minus 1)
2119902119903
]
]
sdot 119890minus119903(119879minus119905)
sdot1
119909sdot
1
(119896119904120573)2
(27)
where 119909 is the initial wealth and 119896119904120573 represents the volatilityof risk asset price
6 Abstract and Applied Analysis
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of x on the optimal investment strategy 120587
x = 5
x = 3
x = 1
Figure 1 The optimal investment strategy 120587 decreases with respectto 119909
Deriving 120587lowast with respect to 119909 we have
120597120587lowast
120597119909= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(28)
because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Remark 5 Obviously the optimal investment strategy 120587lowast
decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1
However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset
Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890
2119903119905120573 and 120583 gt (120583minus119903)we have
120597120587lowast
120597120583=
119890119903(119905minus119879(1+2120573))
119904minus2120573
[1198902119903119879120573
120583 minus 1198902119903119905120573
(120583 minus 119903)]
1198962119902119903119909gt 0 (29)
Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2
010 011 012 013 014 015 016
06
08
10
12
14
120583
120587
The effect of 120583 on the optimal investment strategy 120587
t = 8
t = 5
t = 1
Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset
Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt
119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)
Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions
120597120587lowast
120597120573= minus (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [1198902119903119905120573
(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
= (119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
(30)
Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573
gt 1198902119903119905120573
gt 0 and ln[119904] gt 0If 1199042 gt 119890
(120583minus119903)(119879minus119905) take logarithm on both sides
ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)
We have
1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905)
gt 1198902119903119905120573
119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0
(32)
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
6 Abstract and Applied Analysis
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of x on the optimal investment strategy 120587
x = 5
x = 3
x = 1
Figure 1 The optimal investment strategy 120587 decreases with respectto 119909
Deriving 120587lowast with respect to 119909 we have
120597120587lowast
120597119909= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(28)
because 120583 gt 119903 gt 0 119879 gt 119905 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Remark 5 Obviously the optimal investment strategy 120587lowast
decreases with respect to the initial wealth 119909 which tells usthat for a risk aversion insurer with the initial fund biggerand bigger the investment proportion on risk asset must bereduced tomake the actual amount invested on risk asset stayat an appropriate level to avoid the potential risk of beingunbearable See Figure 1
However the optimal strategy increases with respect toappreciation rate 120583 of the risky asset
Since119879 gt 119905 120583 gt 119903 gt 0 so that 1198902119903119879120573gt 119890
2119903119905120573 and 120583 gt (120583minus119903)we have
120597120587lowast
120597120583=
119890119903(119905minus119879(1+2120573))
119904minus2120573
[1198902119903119879120573
120583 minus 1198902119903119905120573
(120583 minus 119903)]
1198962119902119903119909gt 0 (29)
Figure 2 shows that the optimal strategy increases withrespect to the return rate of the risky asset 120583 which isobviously established The higher appreciation rate 120583 of riskyasset will attract more money invested on the risk asset andforce the investment proportion 120587 increase See Figure 2
010 011 012 013 014 015 016
06
08
10
12
14
120583
120587
The effect of 120583 on the optimal investment strategy 120587
t = 8
t = 5
t = 1
Figure 2 The optimal investment strategy increases with theappreciation rate 120583 of risky asset
Proposition 6 Suppose 120573 ge 0 is the volatility parameter inCEV model 120587lowast is the optimal investment strategy If 1199042 gt
119890(120583minus119903)(119879minus119905) 119904 gt 1 120583 gt 119903 gt 0 119879 gt 119905 then the optimalstrategy 120587lowast decreases with respect to 119896 120573 and 119904 In fact theoptimal investment strategy decreases with the volatility rate ofrisk asset price denoted by 119896119904120573(119905)
Proof The optimal strategy with respect to 120573 may raisevarious possible cases under different conditions
120597120587lowast
120597120573= minus (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [1198902119903119905120573
(119903 minus 120583) (119903 (119905 minus 119879) minus ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
= (119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) (119903 (119879 minus 119905) + ln [119904])
minus 1198902119903119879120573
(119903 + 120583) ln [119904]]) times (1198962119902119903119909)
minus1
(30)
Since 119904 gt 1 120583 gt 119903 and 119879 gt 119905 then 120583 minus 119903 gt 0 119879 minus 119905 gt 01198902119903119879120573
gt 1198902119903119905120573
gt 0 and ln[119904] gt 0If 1199042 gt 119890
(120583minus119903)(119879minus119905) take logarithm on both sides
ln 1199042 gt (120583 minus 119903) (119879 minus 119905) ie 2 ln 119904 gt (120583 minus 119903) (119879 minus 119905) (31)
We have
1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905)
gt 1198902119903119905120573
119903 (2 ln [119904] minus (120583 minus 119903) (119879 minus 119905)) gt 0
(32)
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 7
so that120597120587
lowast
120597120573= (119896
2119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [1198902119903119905120573
(120583 minus 119903) ln [119904]
+ 1198902119903119905120573
(120583 minus 119903) 119903 (119879 minus 119905) minus 1198902119903119879120573
times (120583 minus 119903) ln [119904] minus 2119903 ln [119904] 1198902119903119879120573]
= (1198962119902119903119909)
minus1
119890119903(119905minus119879(1+2120573))
119904minus2120573
(120583 minus 119903)
times [ minus (1198902119903119879120573
minus 1198902119903119905120573
) (120583 minus 119903) ln [119904]
minus (1198902119903119879120573
2119903 ln [119904] minus 1198902119903119905120573
119903 (120583 minus 119903) (119879 minus 119905))] lt 0
(33)
which means that the optimal strategy decreases with respectto 120573 (see Figure 3(c))
120597120587lowast
120597119904= (119890
119903(119905minus119879(1+2120573))119904minus2120573
(119903 minus 120583)
times [(1198902119903119905120573
(119903 minus 120583) + 1198902119903119879120573
(119903 + 120583))]) times (21198962119902119903119909
2)minus1
= minus ((120583 minus 119903) 119904minus2120573
119890119903(119905minus119879(1+2120573))
times [119903 (1198902119903119879120573
+ 1198902119903119905120573
) + 120583 (1198902119903119879120573
minus 1198902119903119905120573
)] )
times (21198962119902119903119909
2)minus1
lt 0
(34)
because 120583 minus 119903 gt 0 120573 gt 0 119904minus2120573gt 0 119890119903(119905minus119879(1+2120573))
gt 0 and120583(119890
2119903119879120573minus 119890
2119903119905120573) gt 0
Consider
120597120587lowast
120597119896= minus
119890minus119903(119879minus119905)
119904minus2120573
(120583 minus 119903) (120583 + 119903 minus 119890minus2119903(119879minus119905)120573
(120583 minus 119903))
1198963119902119903119909lt 0
(35)
Thus the optimal strategy decreases with respect to 119904 and119896 respectively See Figures 3(a) and 3(b)
All the calculations above show that the optimal strategydecreases with respect to 119896 119904 and 120573 simultaneously So itcomes to a conclusion that the optimal investment strategydecreases with the volatility rate 119896119904120573(119905) of risk asset price
Remark 7 The conclusion of this proposition is very naturalRisk averse investors always adopt a relatively conservativeinvestment strategy to avoid risk With the increase of riskasset price volatility rate 119896119904120573(119905)whichmeans risk increase therisk averse insurer must reduce the investment proportion 120587on risk asset to keep the companyrsquos risk at an appropriate levelSo the optimal investment strategy 120587lowast decreases with the riskasset price volatility rate 119896119904120573(119905) See Figure 3
(2) Next we are concerned with the expression of thevalue function for the special case when 120578 = 120579 = 0 which
makes the value function simplified completely Without lossof generality we also suppose 120573 ge 0 the case of 120573 lt 0 issimilar Hence
119881 (119905 119909 119904)
= minus1
119902exp
times minus119902[119909119890119903(119879minus119905)
minus120590
2119902119890
2119903(119879minus119905)
4119903
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573)) (120583 minus 119903)2
)
times (81199021205731199032)minus1
+120590
2119902
4119903]
(36)
Proposition 8 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 119896 is the volatility coefficient
of risk asset price in CEVmodel If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and119879 gt 119905 then the optimal value function119881(119905 119909 119904) decreases withrespect to 119896 and 119904 respectively
Proof If 119896 gt 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905 then
(21198963119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
gt 0 (37)
so that
120597119881
120597119896= minus(2119896
3119902119903120573)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573) )] ]
lt 0
(38)
which means the value function decreases with respect to 119896See Figure 4(b)
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
8 Abstract and Applied Analysis
05 10 15 20 25 30 35 4000
01
02
03
k
120587
t = 7
The optimal investment strategy 120587 with k
t = 4
t = 1
(a)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
s = 15
The effect of s on the optimal investment strategy 120587
s = 10
s = 5
(b)
0 2 4 6 8 1000
01
02
03
04
05
06
07
t
120587
The effect of 120573 on the optimal investment strategy 120587
120573 = 15
120573 = 1
120573 = 05
(c)
Figure 3 (a) The optimal investment strategy decreases with respect to 119896 (b) The optimal investment strategy 120587 decreases with respect to 119904(c) The optimal investment strategy 120587 decreases with respect to 120573
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 9
0 2 4 6 8 10
minus25
minus20
minus15
minus10
minus05
00
V
t = 9
t = 4
t = 1
s
The optimal value function V varies with s
(a)
01 02 03 04 05
minus20
minus15
minus10
minus05
00
k
V
t = 7
t = 3
The optimal value function V decreases with k
t = 1
(b)
Figure 4 (a) The optimal value function 119881 decreases with 119904 (b) The value function 119881 decreases with 119896
Completely similar to the analysis in (1) we also have
120597119881
120597119904= minus(2119896
2119902119903)
minus1
(1198902119903119879120573
minus 1198902119903119905120573
) 119904minus1minus2120573
(119903 minus 120583)2
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
lt 0
(39)
which shows that the value function decreases with respect to119904 See Figure 4(a)
Let
119863120573(119905) ≜ 119896
21199042120573
times (1198902119903119879120573
(1 + 4119903 (119879 minus 119905) 1205732)
minus 1198902119903119905120573
(1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])120597119881
120597120573= (8119896
2119902119903
2120573
2)minus1
119904minus2120573
(119903 minus 120583)2
sdot 11989621199042120573(119890
2119903119879120573(1 + 4119903 (119879 minus 119905) 120573
2) minus 119890
2119903119905120573
times (1 + 2119903 (119879 minus 119905) 120573 (1 + 2120573)))
+ 21198902119903119905120573
119903 (1 + 2119903 (119879 minus 119905) 120573 + 2120573 ln [119904])
minus 21198902119903119879120573
119903 (1 + 2120573 ln [119904])
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
10 Abstract and Applied Analysis
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
times [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
times (1 + 2119903 (119905 minus 119879) 120573) ) ] ]
(40)
Obviously 119863120573(119879) = 0 Deriving 119863120573 defined above with
respect to 119905 we have
120597119863120573
120597119905= 4119903120573
2minus119890
211990311987912057311989621199042120573+ 2119890
2119903119905120573119903 ln [119904]
+ 1198902119903119905120573
(21199032(minus119905 + 119879) + 119896
21199042120573
times (1 + 119903 (119905 minus 119879) (1 + 2120573)) )
(41)
Remark 9 If 119905 is big enough relative to 119879 then 120597119863120573120597119905 gt 0
which tells us that 119863120573(119905) rarr 0 increases as time 119905 closes to
119879 So 119863120573(119905) lt 0 as 119905 rarr 119879 The sign of 120597119881120597120573 is the same as
119863120573(119905) that is 120597119881120597120573 lt 0when 119905 is large enough relative to 119879
which means the value function decreases with respect to 120573
when 119905 is large enough Else if 119905 is small enough relative to 119879the value function may increase with respect to 120573
In conclusion the volatility 119896119904120573 of risk asset has a signifi-cant negative influence on the expected utility of insurancecompany through varying parameters 119896 119904 and 120573 someapproximate rule can be found
Proposition 10 119881(119905 119909 119904) is the optimal value function underthe optimal strategy (119886lowast
120587lowast) and 120583 is the expected return of risk
asset price in CEVmodel If 120573 ge 0 119904 gt 0 120583 gt 119903 gt 0 and 119879 gt 119905then the optimal value function119881(119905 119909 119904) increases with respectto the appreciation rate of risk asset 120583
Proof Letting
119863120583(119905) ≜ [119896
21199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
(42)
obviously119863120583(119879) = 0 and
120597119863120583
120597119905= [119896
21199042120573(1 + 2120573) minus 2119903] (minus2119903120573119890
2119903119905120573)
+ 211990312057311989621199042120573(1 + 2120573) 119890
2119903119879120573
= minus21199031205731198902119903119905120573
11989621199042120573(1 + 2120573)
+ 41199032120573119890
2119903119905120573+ 2119903120573119896
21199042120573(1 + 2120573) 119890
2119903119879120573
= 211990312057311989621199042120573(1 + 2120573) (119890
2119903119879120573minus 119890
2119903119905120573)
+ 41199032120573119890
2119903119905120573gt 0
(43)
which shows that 119863120583(119905) increases to 119863120583
(119879) = 0 with respectto time 119905 Thus
[11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573le 0
(44)
So
120597119881
120597120583= minus(4119896
2119902119903
2120573)
minus1
119904minus2120573
(120583 minus 119903)
sdot [11989621199042120573(1 + 2120573) minus 2119903] (119890
2119903119879120573minus 119890
2119903119905120573)
minus 2119903 (119879 minus 119905) 12057311989621199042120573(1 + 2120573) 119890
2119903119879120573
sdot exp minus 2119903119879120573 minus1
8119902
times [8119890119903(minus119905+119879)
119909 minus
2 (minus1 + 1198902119903(minus119905+119879)
) 1199021205902
119903
minus (1198962119902119903
2120573)
minus1
119890minus2119903119879120573
119904minus2120573
(119903 minus 120583)2
sdot [2 (1198902119903119905120573
minus 1198902119903119879120573
) 119903 + 11989621199042120573(1 + 2120573)
times (minus1198902119903119905120573
+ 1198902119903119879120573
(1 + 2119903 (119905 minus 119879) 120573))] ]
ge 0
(45)
The value function 119881 increases with 120583 referring toFigure 5 which shows that the higher appreciation rate ofrisk asset will lead to relatively more greater expected wealthutility As a matter of fact appreciation rate always is themain driving factor to make the wealth increase Of coursethe insurerrsquos utility maximization can be better realized for alarger appreciation rate
5 Conclusion
In this paper we describe the dynamics of the risky assetsrsquoprices with the CEVmodel under which the optimal excess-of-loss reinsurance and investment problem with jump-diffusion risk process is investigated by maximizing theinsurerrsquos exponential utility of terminal wealth Applying
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 11
010 015 020 025 030
minus090
minus085
minus080
minus075
minus070
minus065
minus060
120583
V
The value function V varies with 120583
Figure 5 The value function 119881 varies with 120583
stochastic control of jump diffusion a Hamilton-Jacobi-Bellman (HJB) equation associatedwith a compound Poissonjump risk process is established The explicit solution forthe exponential utility function is given In the last part ofthe text some numerical examples are given to illustratethe effects of model parameters on the optimal strategyand the optimal value function Meanwhile we give somepropositions and remarks which enriched the innovation ofthe paper
Appendices
A Proof of Theorem 1
This appendix collects the proofs of the main results statedin Section 3 Other necessary lemmas are also presented andproved in this appendix
Proof Differentiating (19) with respect to 120587 gives the optimalinvestment policy
120587lowast= minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A1)
Putting (A1) into (19) after simplification we have
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A2)
That is
119867120572
119905+ [(120578 minus 120579) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904+1
2120590
2119867
120572
119909119909
+1
211989621199042120573+2
119867120572
119904119904minus(120583 minus 119903)
2
(119867120572
119909)2
211989621199042120573119867120572
119909119909
minus11989621199042120573+2
(119867120572
119909119904)2
2119867120572
119909119909
minus(120583 minus 119903) 119904119867
120572
119909119867
120572
119909119904
119867120572
119909119909
+max119886
119867120572
119909(1 + 120579) 120582int
119886
0
119865 (119909)d119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 0
(A3)
We try to conjecture (A2) having a solution 119881(119905 119909 119904) ofthis form
119881 (119905 119909 119904) = minus1
119902119890minus119902[119909119890
119903(119879minus119905)+119891(119905)+119892(119905119904)]
119905 isin [0 119879]
119891 (119879) = 0
119892 (119879 119904) = 0
119881 (119879 119909 119904) = minus1
119902119890minus119902119909
(A4)
Differentiating the solution (A4) with respect to the corre-sponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 119891
1015840(119905) + 119892
119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
1198922
119904+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119886) 119904) minus 119867
120572(119905 119909 119904)]
= 119881119902119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
(A5)
Plugging the above derivatives (A5) into (A1) gives
120587lowast=
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A6)
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
12 Abstract and Applied Analysis
and substituting the above derivatives into (A2) yields
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
+ 119903119904119892119904
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+max119886
(1 + 120579) 120582119890119903(119879minus119905)
int
119886
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
119886
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A7)
Differentiating with respect to 119886 in (A7) we find that theminimizer 1198860
lowast(119905) satisfies
[(1 + 120579) minus 1198901199021198860(119905)119890119903(119879minus119905)
] 119865 (1198860(119905)) = 0 (A8)
If 1198860(119905) isin [0 119863) then 0 lt 119865(1198860(119905)) le 1 and (1 + 120579) minus
1198901199021198860(119905)119890119903(119879minus119905)
= 0 which gives
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
(A9)
To ensure that 0 le 1198860
lowast(119905) lt 119863 that is 0 le (ln(1 + 120579)
119902)119890minus119903(119879minus119905)
lt 119863 it must be ensured that
119905 lt 119879 +1
119903ln [
119902119863
ln (1 + 120579)] ≜ 119905
0 (A10)
which leads to two different cases
Case 1 If119863 ge (ln(1 + 120579)119902) then 1199050 ge 119879 and
1198860
lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905) (A11)
always is optimal retention level on the whole interval [0 119879]Inserting (A5) into (A1) the optimal investment policy
is given by
120587lowast(119905) =
120583 minus 119903
11990911990211989621199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905)(A12)
and 119886lowast= 119886
0(119905) = (ln(1 + 120579)119902)119890
minus119903(119879minus119905) The correspondingHJB equation (A7) becomes
1198911015840(119905) + 119892
119905+ 119903119904119892
119904+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)+1
211989621199042120573+2
119892119904119904
+(120583 minus 119903)
2
211990211989621199042120573+ 119890
119903(119879minus119905)(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909 minus 120582119890119903(119879minus119905)
times int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A13)
To solve (A13) we decompose (A13) into two equations
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119890119903(119879minus119905)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909
= 0
(A14)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
211990211989621199042120573= 0 (A15)
By simple transposition and integration for (A14) with119891(119879) = 0 we have
119891 (119905) =(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
(A16)
To solve (A20) we conjecture a solution with the followingform with 119860(119879) = 0 and 119861(119879) = 0
119892 (119905 119904) = 119860 (119905) 119904minus2120573
+ 119861 (119905) (A17)
Differentiating (A17) with respect to the correspondingvariables respectively
119892119905= 119860
1015840(119905) 119904
minus2120573+ 119861
1015840(119905) 119892
119904= minus2120573119904
minus2120573minus1119860 (119905)
119892119904119904= 2120573 (2120573 + 1) 119904
minus2120573minus2119860 (119905)
(A18)
Substituting the above derivatives into (A20) yields
1198601015840(119905) 119904
minus2120573+ 119861
1015840(119905) minus 2119903120573119904
minus2120573119860 (119905)
+ 1198962120573 (2120573 + 1)119860 (119905) +
(120583 minus 119903)2
21199021198962119904minus2120573
= 0
(A19)
To solve (A20) we decompose (A20) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0 (A20)
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0 (A21)
Obviously 119904minus2120573gt 0 (A20) is equivalent to
1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962= 0 (A22)
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 13
Equation (A22) with 119860(119879) = 0 is easy to be solved and thesolution is
119860 (119905) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962 (A23)
Inserting (A23) into (A21) with 119861(119879) = 0 we have
119861 (119905) = (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A24)
Substituting (A23) and (A24) back into (A17) we obtain
119892 (119905 119904) =
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
(A25)
Substituting (A16) and (A25) back into (A4) the optimalvalue function is given by
119881 (119905 119909 119904)
= minus1
119902exp
times
minus 119902
times [
[
119909119890119903(119879minus119905)
+(120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
+ int
119879
119905
(119890119903(119879minus119910)
(1 + 120579) 120582
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119865 (119909)d119909) d119910
minus int
119879
119905
(120582119890119903(119879minus119910)
times int
(ln(1+120579)119902)119890minus119903(119879minus119910)
0
119890119902119909119890119903(119879minus119910)
119865 (119909)d119909) d119910
minus(120578 minus 120579) 120582120583
infin
119903+120590
2119902
4119903
+ (119890minus2119903120573119879
(1 + 2120573)
times (1198902119903120573119905
+ 1198902119903120573119879
(minus1 + 2119903 (119879 minus 119905) 120573))
times (120583 minus 119903)2
) times (81199021205731199032)minus1
+
119890minus2119903120573119879
(minus1198902119903120573119905
+ 1198902119903120573119879
) (120583 minus 119903)2
41199021199031205731198962119904minus2120573]
]
119905 isin [0 119879]
(A26)
Case 2 If 119863 lt ln(1 + 120579)119902 then 1199050lt 119879 the whole interval
[0 119879] is divided into two parts [0 1199050) and [1199050 119879] 1198860
(119905) =
(ln(1 + 120579)119902)119890minus119903(119879minus119905) is the optimal retention level only on
[0 1199050) however 1198860
(119905) ge 119863 on [1199050 119879] we can only take 119863 as
the retention level naturally Therefore the optimal retentionlevel is denoted by
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050)
119863 119905 isin [1199050 119879]
(A27)
The solving process depends on two different time intervalscorresponding to the different optimal retention level respec-tively
(1) For
119886lowast(119905) = 119863 119905 isin [119905
0 119879] (A28)
the corresponding HJB equation is
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
+max120587
(120583 minus 119903) 120587119909119867120572
119909+1
2120587
2119909
211989621199042120573119867
120572
119909119909
+1198962120587119909119904
2120573+1119867
120572
119909119904
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A29)
The first-order maximizing condition for the optimal invest-ment strategy is
120587lowast(119905) = minus
(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904
11990911989621199042120573119867120572
119909119909
(A30)
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
14 Abstract and Applied Analysis
Inserting (A30) into (A29) and simplifying the HJB equa-tion becomes
119867120572
119905+ [(1 + 120578) 120582120583
infin+ 119903119909]119867
120572
119909+ 120583119904119867
120572
119904
+1
2120590
2119867
120572
119909119909+1
211989621199042120573+2
119867120572
119904119904
minus
[(120583 minus 119903)119867120572
119909+ 119896
21199042120573+1
119867120572
119909119904]2
211989621199042120573119867120572
119909119909
+ 120582119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)] = 0
(A31)
We conjecture (A31) having a solution 119881(119905 119909 119904) as fol-lows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [1199050 119879]
119881 (119879 119909 119904) = 119880 (119909)
119891 (119879) = 0
119892 (119879 119904) = 0
(A32)
Differentiating the conjecture (A32) with respect to thecorresponding variables
119881119905= 119881 (minus119902) [minus119903119909119890
119903(119879minus119905)+ 1198911015840
(119905) + 119892119905]
119881119909= 119881 (minus119902) 119890
119903(119879minus119905) 119881
119909119909= 119881(minus119902)
2
1198902119903(119879minus119905)
119881119904= 119881 (minus119902) 119892
119904 119881
119904119904= 119881 [(minus119902)
2
119892119904
2+ (minus119902) 119892
119904119904]
119881119909119904= 119881(minus119902)
2
119890119903(119879minus119905)
119892119904
119864 [119867120572(119905 119909 minus (119885
119894and 119863) 119904) minus 119867
120572(119905 119909 119904)]
= 119881int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) minus 119881
(A33)
Substituting the above derivatives (A33) into (A31) and(A30) after simplifying it gives
120587lowast(119905) =
(120583 minus 119903)
11990211989621199091199042120573119890119903(119879minus119905)minus
119904119892119904
119909119890119903(119879minus119905) (A34)
1198911015840(119905) + 119892
119905+ (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ 119903119904119892119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A35)
Decompose (A35) into two equations
1198911015840(119905) + (1 + 120578) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
minus120582
119902int
119863
0
119890119902119909119890119903(119879minus119905)
d119865 (119909) + 120582
119902= 0
(A36)
119892119905+ 119903119904119892
119904+1
211989621199042120573+2
119892119904119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A37)
By transposition and integration for (A36) with 119891(119879) = 0
119891 (119905) =(1 + 120578) 120582120583
infin119890119903(119879minus119905)
119903minus120590
2119902119890
2119903(119879minus119905)
4119903
minus int
119879
119905
[120582
119902int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus120582 (119905 minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A38)
For (A37) we try to find a solution as follows
119892 (119905 119904) = 119860 (119905)119904minus2120573
+ 119861 (119905) (A39)
satisfying 119860(119879) = 0 and 119861(119879) = 0 Equation (A37) turns into
1198601015840(119905)119904
minus2120573+ 1198611015840
(119905) minus 2119903120573119860 (119905)119904minus2120573
+ 1198962120573 (2120573 + 1)119860 (119905)
+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0
(A40)
Decompose (A40) into two equations
[1198601015840(119905) minus 2119903120573119860 (119905) +
(120583 minus 119903)2
21199021198962] 119904
minus2120573= 0
1198611015840(119905) + 119896
2120573 (2120573 + 1)119860 (119905) = 0
(A41)
Solving the above simple ordinary differential equationswe get
119860 (119905) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573
119861 (119905) =
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
(A42)
So
119892 (119905 119904) =
(1 minus 1198902119903120573(119905minus119879)
) (120583 minus 119903)2
41199021198962119903120573119904minus2120573
+
(1 + 2120573) [1198902119903120573(119905minus119879)
minus 1 + 2119903 (119879 minus 119905) 120573] (120583 minus 119903)2
81199021199032120573
120587lowast(119905) =
2119903 (120583 minus 119903) + (120583 minus 119903)2
(1 minus 1198902119903120573(119905minus119879)
)
211990211989621199031199091199042120573119890119903(119879minus119905)
(A43)
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 15
(2) For
119886lowast(119905) =
ln (1 + 120579)
119902119890minus119903(119879minus119905)
119905 isin [0 1199050) (A44)
the corresponding HJB equation is the same as (A29) whichcorresponds to [0 1199050) having a solution 119881(119905 119909 119904) as follows
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891 (119905) + 119892 (119905 119904)]
119905 isin [0 1199050]
119891(1199050) =
119891(1199050)
119892(1199050 119904) =
119892(1199050 119904)
(A45)
After the same calculation as (1) we obtain the followingresults
1198911015840(119905) + 119892
119905+ (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A46)
Splitting (A46) into two equations
119892119905+1
211989621199042120573+2
119892119904119904+ 119903119904119892
119904+(120583 minus 119903)
2
21199021198962119904minus2120573
= 0 (A47)
The equation (A47) is a linear second-order partial differ-ential equation with variable coefficients which exhibits aunique solution according to the appendix of Badaoui andFernandez [19] or is ensured by the theoremof Friedman [20]
1198911015840(119905) + (120578 minus 120579) 120582120583
infin119890119903(119879minus119905)
minus1
2120590
2119902119890
2119903(119879minus119905)
+ (1 + 120579) 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119865 (119909)d119909
minus 120582119890119903(119879minus119905)
int
(ln(1+120579)119902)119890minus119903(119879minus119905)
0
119890119902119909119890119903(119879minus119905)
119865 (119909)d119909 = 0
(A48)
By transposition and integration to (A48) the result isgiven by
119891 (119905) =(120578 minus 120579) 120582120583
infin
119903119890119903(119879minus119905)
minus120590
2119902
41199031198902119903(119879minus119905)
minus int
119905
0
[(1 + 120579) 120582119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
+ 120582int
119905
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+ 1198623
(A49)
where 119908 can be determined by 119891(1199050) = 119891(1199050) Consider
1198623= minus
(120578 minus 120579) 120582120583infin
119903119890119903(119879minus119905
0)+ (1 + 120579) 120582
times int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119865 (119909)d119909] d120591
minus 120582int
1199050
0
[119890119903(119879minus120591)
int
(ln(1+120579)119902)119890minus119903(119879minus120591)
0
119890119902119909119890119903(119879minus120591)
119865 (119909)d119909] d120591
+(1 + 120578) 120582120583
infin
119903119890119903(119879minus119905
0)
minus120582
119902int
119879
1199050
[int
119863
0
119890119902119909119890119903(119879minus120591)
d119865 (119909)] d120591
minus
120582 (1199050minus 119879)
119902minus(1 + 120578) 120582120583
infin
119903+120590
2119902
4119903
(A50)
So
119891lowast(119905) =
119891 (119905) 119905 isin [0 1199050)
119891 (119905) 119905 isin [1199050 119879]
(A51)
With the same calculations as before we obtain
119892 (119905 119904) = [11986211198902120573119903119905
+(120583 minus 119903)
2
41199021198962120573119903] 119904
minus2120573minus 119862
1
1198962(2120573 + 1)
21199031198902120573119903119905
minus(2120573 + 1)
119902119903(120583 minus 119903)
2
+ 1198622
(A52)
here the undetermined constants 1198621and 119862
2will be deter-
mined by using the continuity of 119881(119905 119909 119904)By the continuity of 119881(119905 119909 119904) at 119905 = 119905
0 we have
119892 (1199050 119904) =119892(1199050 119904) (A53)
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
16 Abstract and Applied Analysis
Comparing the items and the coefficients on both sides of(A53) we obtain the undetermined constants 119862
1and 119862
2
1198621= minus
(120583 minus 119903)2
119890minus2119903119879120573
41199021199031198962120573
1198622= (2120573 + 1) (120583 minus 119903)
2
(4 + 119879 minus 119905
0
4119902119903minus
1
81199021199032120573)
(A54)
So
119892lowast(119905 119904) =
119892 (119905 119904) 119905 isin [0 1199050)
119892 (119905 119904) 119905 isin [1199050 119879]
(A55)
And the optimal value function is
119881 (119905 119909 119904) = minus1
119902exp minus119902 [119909119890119903(119879minus119905)
+ 119891lowast(119905) + 119892
lowast(119905 119904)]
(A56)
B Proof of Theorem 3
A new lemma is necessary to prove Theorem 3 For conve-nience denote 119876 = [0 +infin) times [0 +infin)
Lemma B1 Take a sequence of bounded open sets 1198761 119876
2
1198763 with 119876
119894sub 119876
119894+1sub 119876 119894 = 1 2 and 119876 = cup
119894119876
119894 Let 120591
119894
denote the exit time of (119883120572lowast
(119905) 119878(119905)) from 119876119894 If the conditions
inTheorem 3 hold then one has119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and
119879)) | 119883(119905) = 119909 119878(119905) = 119904] lt infin for 119894 = 1 2
Proof According to Oslashksendal and Sulem [17] we rewrite thecompound Poisson process 119862(119905) = sum
119873(119905)
119894=1119885
119894in terms of a
Poisson random measure Suppose 120574 is the Poisson randommeasure then
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
119911120574 (d119911 d119906)
119862119886(119905) =
119873(119905)
sum
119894=1
119885119894= int
119905
0
int119877+
(119911 and 119886) 120574 (d119911 d119906)
(B1)
The compensator ] of the random measure 120574 is
] (d119911 d119905) = 120582d119865 (119911) d119905 (B2)
so the compensated Poisson random measure of 119862(119905) is
120574 (d119911 d119905) = 120574 (d119911 d119905) minus 120582d119865 (119911) d119905 (B3)
Set (119905) = 119881(119905 119883120572lowast
(119905) 119878(119905)) Applying Itorsquos formula to
2(119905)
d2(119905)
= 2 (119905) 119881119909120590d119882(119905) + 119896119878
120573(119905)
times [119881119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905) +A (119905)
+ [1198812
119909120590
2+ 119881
2
119909120587
2119883(119905)
211989621198782120573(119905) + 119881
2
11990411989621198782120573+2
(119905)
+ 2119881119909119881
119904120587 (119905)119883 (119905) 119896
21198782120573+1
(119905)] d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905)) minus 2119881 (119905 119883 (119905minus) 119878 (119905))
sdot 119881119909sdot (119911 and 119886) ] 120582d119865 (119911) d119905
+ int119877+
[1198812(119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus 1198812(119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B4)
Since 120572lowast is the optimal strategy for (19) thenA120572lowast
(119905) =
0 Plugging the expression of119881119909119881
119904 and 119886 lowast (119905) into (B4) we
obtain the following equation by simple calculations
d2(119905)
2(119905)
=
minus 2119902120590119890119903(119879minus119905)d119882(119905) minus
2 (120583 minus 119903) 119904minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903] d119882
1(119905)
+
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119905)
+
1199022120590
21198902119903(119879minus119905)
+(120583 minus 119903)
2
119904minus2120573
1198962[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
2
+
(1 minus 119890minus2119903120573(119879minus119905)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962
minus
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times[1 +
(1 minus 119890minus2119903120573(119879minus119905)
) (120583 minus 119903)
2119903]
d119905
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 17
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119905)minus 1] 120582d119865 (119911) d119905
+ int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119905)
minus 1] 120574 (d119911 d119905)
(B5)
The solution of (B5) is
2(119905)
2(0)
= exp
int
119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906)
minus1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
+ int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1] 120582d119865 (119911) d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906)
(B6)
where (1199090 119904
0) is the state at time 0 Applying Itorsquos formula to
(11) we can derive
d119878minus2120573(119905) = (120573 (2120573 + 1) 119896
2minus 2120573119903119878
minus2120573(119905)) d119905
minus 2120573119896radic119878minus2120573(119905)d119882
1(119905)
(B7)
According to the results of Taksar and Zeng [16] and Guet al [5] and other existing literature 120583 gt 119903 gt 0 we know that
expint119905
0
minus2119902120590119890119903(119879minus119906)d119882(119906) minus
1
2int
119905
0
41199022120590
21198902119903(119879minus119906)d119906
exp
int
119905
0
minus2 (120583 minus 119903) 119904
minus120573
119896
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119882
1(119906)
minus1
2int
119905
0
4(120583 minus 119903)2
119904minus2120573
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
exp
int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)2
119904minus120573
119903119896d119882
1(119906)
minus1
2int
119905
0
(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
11990321198962d119906
(B8)
are martingales Since 119911 isin [0 119863] 119863 = sup119911 119865(119911) le 1 lt
+infin and
119864[int
119879
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1]
2
120582d119865 (119911) d119906] lt infin (B9)
then the process
int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
minus 1] 120574 (d119911 d119906) (B10)
is also a martingaleAccording to Taksar and Zeng [16]
119864[expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906] lt infin (B11)
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
18 Abstract and Applied Analysis
when either of conditions (I) and (II) in Theorem 3 issatisfied Taking expectation from both sides of (B6) yields
119864 [2(119905)]
= 2(0) 119864
times [
[
exp
int
119905
0
31199022120590
21198902119903(119879minus119906)d119906
+ int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
+ int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
+ int
119905
0
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903] d119906
+ int
119905
0
int119877+
[119890minus2119902(119911and119886
lowast)119890119903(119879minus119906)
+ 2119902 (119911 and 119886lowast) 119890
119903(119879minus119906)minus 1]
times 120582d119865 (119911) d119906
]
]
(B12)
If 120573 ge 0 then 0 le (1 minus 119890minus2119903120573(119879minus119906)
) lt 1 For 120583 gt 119903 gt 0 and119879 gt 0 are constants we have
exp
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
2
d119906
lt expint119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962[1 +
(120583 minus 119903)
2119903]
2
d119906
= exp[1 +(120583 minus 119903)
2119903]
2
int
119905
0
3(120583 minus 119903)2
119904minus2120573
(119906)
1198962d119906
exp
int
119905
0
3(1 minus 119890minus2119903120573(119879minus119906)
)2
(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
lt expint119905
0
3(120583 minus 119903)4
119904minus2120573
(119906)
411990321198962d119906
exp
int
119905
0
10038161003816100381610038161003816100381610038161003816100381610038161003816
minus
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)3
119904minus2120573
(119906)
1199031198962
times [1 +
(1 minus 119890minus2119903120573(119879minus119906)
) (120583 minus 119903)
2119903]
10038161003816100381610038161003816100381610038161003816100381610038161003816
d119906
lt expint119905
0
(120583 minus 119903)3
119904minus2120573
(119906)
1199031198962[1 +
(120583 minus 119903)
2119903] d119906
(B13)
For the case of 120573 lt 0 |1 minus 119890minus2119903120573(119879minus119906)
| lt 119890minus2119903120573119879 which is
similar After simple calculation and analysis we obtain
119864 [1198812(119905 119883
120572lowast
(119905) 119878 (119905))] = 119864 [2(119905)] lt infin (B14)
So119864[1198812(120591
119894and119879119883
120572lowast
(120591119894and119879) 119878(120591
119894and119879))] lt infin for 119894 = 1 2
We use Lemma B1 to prove the verificationTheorem 3 asfollows
Proof By Itorsquos formula for Ito-Levy process we have
d119881 (119905 119883120572(119905) 119878 (119905))
= A120572119881 (119905 119883
120572(119905) 119878 (119905)) + 119881
119909120590d119882(119905)
+ 119896119878120573(119905) [119881
119909120587 (119905)119883 (119905) + 119881
119904119878 (119905)] d119882
1(119905)
+ int119877+
[119881 (119905 119883 (119905minus) + (119911 and 119886) 119878 (119905))
minus119881 (119905 119883 (119905minus) 119878 (119905))] 120574 (d119911 d119905)
(B15)
Take a sequence of bounded open sets1198761119876
2119876
3 with
119876119894sub 119876
119894+1sub 119876 119894 = 1 2 For (119909 119904) isin 119876
1 let 120591
119894denote the
exit time of (119909 119904) from 119876119894 Then 120591
119894and 119879 rarr 119879 when 119894 rarr infin
Integrating both sides of the above equation
119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
= 119881 (119905 119909 119904) + int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
+ int
120591119894and119879
119905
119881119909120590d119882(119906)
+ int
120591119894and119879
119905
119896119878120573(119906) [119881
119909120587119909 + 119881
119904119878 (119906)] d119882
1(119906)
+ int
120591119894and119879
119905
int119877+
[119881 (119906119883 (119906minus) + (119911 and 119886) 119878 (119906))
minus119881 (119906119883 (119906minus) 119878 (119906))] 120574 (d119911 d119906)
(B16)
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975
Abstract and Applied Analysis 19
Since sup120572isinΛ
A120572119881(119905 119909 119904) = 0 then A120572
119881(119905 119909 119904) lt 0The last three terms of (B16) are square-integrable martin-gales with zero expectation taking conditional expectationgiven (119905 119909 119904) on two sides of the above formula
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))
| 119883120572(119905) = 119909 119878 (119905) = 119904]
= 119881 (119905 119909 119904)
+ 119864 [int
120591119894and119879
119905
A120572119881 (119906119883
120572(119906) 119878 (119906)) d119906
| 119883120572(119905) = 119909 119878 (119905) = 119904]
le 119881 (119905 119909 119904)
(B17)
From Lemma B1 119881(120591119894and 119879119883
120572(120591
119894and 119879) 119878(120591
119894and 119879)) 119894 =
1 2 are uniformly integrable Thus we have
119867(119905 119909 119904)
= sup120572isinΛ
119864 [119880 (119883 (119879)) | 119883120572lowast
(119905) = 119909 119878 (119905) = 119904]
= lim119905rarrinfin
119864 [119881 (120591119894and 119879119883
120572(120591
119894and 119879) 119878 (120591
119894and 119879))]
le 119881 (119905 119909 119904)
(B18)
When 120572 = 120572lowast the inequality in the above formula
becomes an equality that is119867(119905 119909 119904) = 119881(119905 119909 119904) The proofis finished
Conflict of Interests
The authors declare that they have no conflict of interestsregarding the publication of this paper
Acknowledgments
This research was supported by the National Natural ScienceFoundation of China (Grant nos 11201335 and 11301376)and Humanity and Social Science Foundation of Ministry ofEducation of China (Grant no 11YJC910007)
References
[1] K Borch ldquoThe safety loading of reinsurance premiumsrdquo Scan-dinavian Actuarial Journal vol 1960 no 3-4 pp 163ndash184 1960
[2] C S Tapiero and D Zuckerman ldquoOptimum excess-loss rein-surance a dynamic frameworkrdquo Stochastic Processes and TheirApplications vol 12 no 1 pp 85ndash96 1982
[3] G Taylor ldquoReserving consecutive layers of inwards excess-of-loss reinsurancerdquo Insurance Mathematics amp Economics vol 20no 3 pp 225ndash242 1997
[4] Y Cao and J Xu ldquoProportional and excess-of-loss reinsuranceunder investment gainsrdquo Applied Mathematics and Computa-tion vol 217 no 6 pp 2546ndash2550 2010
[5] A Gu X Guo Z Li and Y Zeng ldquoOptimal control of excess-of-loss reinsurance and investment for insurers under a CEVmodelrdquo Insurance Mathematics amp Economics vol 51 no 3 pp674ndash684 2012
[6] H Zhao X Rong and Y Zhao ldquoOptimal excess-of-loss rein-surance and investment problem for an insurer with jump-diffusion risk process under the Heston modelrdquo InsuranceMathematics amp Economics vol 53 no 3 pp 504ndash514 2013
[7] H L Yang and L H Zhang ldquoOptimal investment for insurerwith jump-diffusion risk processrdquo Insurance Mathematics ampEconomics vol 37 no 3 pp 615ndash634 2005
[8] B Li R Wu and M Song ldquoA renewal jump-diffusion processwith threshold dividend strategyrdquo Journal of Computational andApplied Mathematics vol 228 no 1 pp 41ndash55 2009
[9] X Ruan W Zhu J Hu and J Huang ldquoOptimal portfolio andconsumption with habit formation in a jump diffusion marketrdquoApplied Mathematics and Computation vol 222 pp 391ndash4012013
[10] J C Cox and S A Ross ldquoThe valuation of options for alternativestochastic processesrdquo Journal of Financial Economics vol 3 no1-2 pp 145ndash166 1976
[11] Y L Hsu T I Lin and C F Lee ldquoConstant elasticity of vari-ance (CEV) option pricing model integration and detailedderivationrdquoMathematics and Computers in Simulation vol 79no 1 pp 60ndash71 2008
[12] L Campi S Polbennikov and A Sbuelz ldquoSystematic equity-based credit risk a CEVmodel with jump to defaultrdquo Journal ofEconomic Dynamics amp Control vol 33 no 1 pp 93ndash108 2009
[13] J W Xiao S H Yin and C L Qin ldquoConstant elasticity ofvariance model and analytical strategies for annuity contractsrdquoApplied Mathematics and Mechanics vol 27 no 11 pp 1499ndash1506 2006
[14] M Gu Y Yang S Li and J Zhang ldquoConstant elasticity ofvariance model for proportional reinsurance and investmentstrategiesrdquo Insurance Mathematics amp Economics vol 46 no 3pp 580ndash587 2010
[15] X Lin and Y Li ldquoOptimal reinsurance and investment fora jump diffusion risk process under the CEV modelrdquo NorthAmerican Actuarial Journal vol 15 no 3 pp 417ndash431 2011
[16] M Taksar and X D Zeng ldquoA general stochastic volatility modeland optimal portfolio with explicit solutionsrdquo Working Paper2009
[17] B Oslashksendal and A Sulem Applied Stochastic Control of JumpDiffusions Springer Berlin Germany 2009
[18] W H Fleming and H M Soner Controlled Markov Processesand Viscosity Solutions Springer Berlin Germany 1993
[19] M Badaoui and B Fernandez ldquoAn optimal investment strategywith maximal risk aversion and its ruin probability in thepresence of stochastic volatility on investmentsrdquo InsuranceMathematics and Economics vol 53 no 1 pp 1ndash13 2013
[20] A Friedman Stochastic Differential Equations and ApplicationsAcademic Press New York NY USA 1975