Sample Exam2

COURSE 3: MAY 2001 - 1 GO ON TO NEXT PAGE

May 2001Course 3

**BEGINNING OF EXAMINATION**

1. For a given life age 30, it is estimated that an impact of a medical breakthrough will be anincrease of 4 years in eo 30 , the complete expectation of life.

Prior to the medical breakthrough, s(x) followed de Moivres law with w = 100 as thelimiting age.

Assuming de Moivres law still applies after the medical breakthrough, calculate the newlimiting age.

(A) 104

(B) 105

(C) 106

(D) 107

(E) 108


2. On January 1, 2002, Pat, age 40, purchases a 5-payment, 10-year term insurance of100,000:

(i) Death benefits are payable at the moment of death.

(ii) Contract premiums of 4000 are payable annually at the beginning of each year for5 years.

(iii) i = 0.05

(iv) L is the loss random variable at time of issue.

Calculate the value of L if Pat dies on June 30, 2004.

(A) 77,100

(B) 80,700

(C) 82,700

(D) 85,900

(E) 88,000


3. Glen is practicing his simulation skills.

He generates 1000 values of the random variable X as follows:

(i) He generates the observed value l from the gamma distribution with a = 2 andq = 1 (hence with mean 2 and variance 2).

(ii) He then generates x from the Poisson distribution with mean l.

(iii) He repeats the process 999 more times: first generating a value l , thengenerating x from the Poisson distribution with mean l .

(iv) The repetitions are mutually independent.

Calculate the expected number of times that his simulated value of X is 3.

(A) 75

(B) 100

(C) 125

(D) 150

(E) 175


4. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins per minute.The denominations are randomly distributed:

(i) 60% of the coins are worth 1;

(ii) 20% of the coins are worth 5;

(iii) 20% of the coins are worth 10.

Calculate the variance of the value of the coins Tom finds during his one-hour walk towork.

(A) 379

(B) 487

(C) 566

(D) 670

(E) 768


5. For a fully discrete 20-payment whole life insurance of 1000 on (x), you are given:

(i) i = 0.06

(ii) qx+ =19 001254.

(iii) The level annual benefit premium is 13.72.

(iv) The benefit reserve at the end of year 19 is 342.03.

Calculate 1000 Px+20 , the level annual benefit premium for a fully discrete whole lifeinsurance of 1000 on (x+20).

(A) 27

(B) 29

(C) 31

(D) 33

(E) 35


6. For a multiple decrement model on (60):

(i) m601 0( ) ( ), ,t t follows the Illustrative Life Table.

(ii) m mt60 6012 0( ) ( )( ) ( ),t t t=

Calculate 10 60q( )t , the probability that decrement occurs during the 11th year.

(A) 0.03

(B) 0.04

(C) 0.05

(D) 0.06

(E) 0.07


7. A coach can give two types of training, light or heavy, to his sports team before agame. If the team wins the prior game, the next training is equally likely to be light orheavy. But, if the team loses the prior game, the next training is always heavy.

The probability that the team will win the game is 0.4 after light training and 0.8 afterheavy training.

Calculate the long run proportion of time that the coach will give heavy training to theteam.

(A) 0.61

(B) 0.64

(C) 0.67

(D) 0.70

(E) 0.73


8. For a simulation of the movement of a stocks price:

(i) The price follows geometric Brownian motion, with drift coefficient m = 0 01. and

variance parameter s 2 = 00004. .

(ii) The simulation projects the stock price in steps of time 1.

(iii) Simulated price movements are determined using the inverse transform method.

(iv) The price at t = 0 is 100.

(v) The random numbers, from the uniform distribution on 01, , for the first 2 stepsare 0.1587 and 0.9332, respectively.

(vi) F is the price at t = 1; G is the price at t = 2.

Calculate G F.

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5


9. (x) and (y) are two lives with identical expected mortality.

You are given:

P Px y= = 0.1Pxy = 006. , where Pxy is the annual benefit premium for a fully discrete

insurance of 1 on xyb g .d = 006.

Calculate the premium Pxy , the annual benefit premium for a fully discrete insurance of 1

on xyb g .

(A) 0.14

(B) 0.16

(C) 0.18

(D) 0.20

(E) 0.22


10. For students entering a college, you are given the following from a multiple decrementmodel:

(i) 1000 students enter the college at t = 0 .

(ii) Students leave the college for failure 1b g or all other reasons 2b g .

(iii) m m1b gb gt = 0 4 tm 2 004b gb gt = . 0 4


11. You are using the inverse transform method to simulate Z, the present value randomvariable for a special two-year term insurance on (70). You are given:

(i) (70) is subject to only two causes of death, with

k k q701( )

k q702( )

0 0.10 0.10

1 0.10 0.50

(ii) Death benefits, payable at the end of the year of death, are:

During year Benefit for Cause 1 Benefit for Cause 2

1 1000 1100

2 1100 1200

(iii) i = 0 06.

(iv) For this trial your random number, from the uniform distribution on 01, , is 0.35.

(v) High random numbers correspond to high values of Z.

Calculate the simulated value of Z for this trial.

(A) 943

(B) 979

(C) 1000

(D) 1038

(E) 1068


12. You are simulating one year of death and surrender benefits for 3 policies. Mortalityfollows the Illustrative Life Table. The surrender rate, occurring at the end of the year, is15% for all ages. The simulation procedure is the inverse transform algorithm, with lowrandom numbers corresponding to the decrement occurring. You perform the followingsteps for each policy:

(1) Simulate if the policy is terminated by death. If not, go to Step 2; if yes, continuewith the next policy.

(2) Simulate if the policy is terminated by surrender.

The following values are successively generated from the uniform distribution on 01, :

0.3, 0.5, 0.1, 0.4, 0.8, 0.2, 0.3, 0.4, 0.6, 0.7,.

You are given:

Policy # Age DeathBenefitSurrenderBenefit

1 100 10 10

2 91 25 20

3 96 20 15

Calculate the total benefits generated by the simulation.

(A) 30

(B) 35

(C) 40

(D) 45

(E) 50


13. Mr. Ucci has only 3 hairs left on his head and he wont be growing any more.

(i) The future mortality of each hair follows

k xq k k= + =01 1 0 1 2 3. , , , ,b g and x is Mr. Uccis age

(ii) Hair loss follows the hyperbolic assumption at fractional ages.

(iii) The future lifetimes of the 3 hairs are independent.

Calculate the probability that Mr. Ucci is bald (has no hair left) at age x + 2.5 .

(A) 0.090

(B) 0.097

(C) 0.104

(D) 0.111

(E) 0.118


14. The following graph is related to current human mortality:

Which of the following functions of age does the graph most likely show?

(A) m xb g

(B) l xxmb g(C) l px x

(D) lx

(E) lx2

0 20 40 60 80 100

Age


15. An actuary for an automobile insurance company determines that the distribution of theannual number of claims for an insured chosen at random is modeled by the negativebinomial distribution with mean 0.2 and variance 0.4.

The number of claims for each individual insured has a Poisson distribution and themeans of these Poisson distributions are gamma distributed over the population ofinsureds.

Calculate the variance of this gamma distribution.

(A) 0.20

(B) 0.25

(C) 0.30

(D) 0.35

(E) 0.40


16. A dam is proposed for a river which is currently used for salmon breeding. You havemodeled:

(i) For each hour the dam is opened the number of salmon that will pass through andreach the breeding grounds has a distribution with mean 100 and variance 900.

(ii) The number of eggs released by each salmon has a distribution with mean of 5and variance of 5.

(iii) The number of salmon going through the dam each hour it is open and thenumbers of eggs released by the salmon are independent.

Using the normal approximation for the aggregate number of eggs released, determinethe least number of whole hours the dam should be left open so the probability that10,000 eggs will be released is greater than 95%.

(A) 20

(B) 23

(C) 26

(D) 29

(E) 32


17. For a special 3-year term insurance on ( )x , you are given:

(i) Z is the present-value random variable for the death benefits.

(ii) q kx k+ = +0 02 1. ( ) k = 0, 1, 2

(iii) The following death benefits, payable at the end of the year of death:

k bk+10 300,000

1 350,000

2 400,000

(iv) i = 0 06.

Calculate E Zb g .

(A) 36,800

(B) 39,100

(C) 41,400

(D) 43,700

(E) 46,000


18. For a special fully discrete 20-year endowment insurance on (55):

(i) Death benefits in year k are given by b kk = -21b g, k = 1, 2, , 20.(ii) The maturity benefit is 1.

(iii) Annual benefit premiums are level.

(iv) kV denotes the benefit reserve at the end of year k, k = 1, 2,, 20.

(v) 10 = 5.0V

(vi) 19 = 0.6V

(vii) q65 0.10=

(viii) i = 0.08

Calculate 11 .V

(A) 4.5

(B) 4.6

(C) 4.8

(D) 5.1

(E) 5.3


19. For a stop-loss insurance on a three person group:

(i) Loss amounts are independent.

(ii) The distribution of loss amount for each person is:

Loss Amount Probability

0 0.4

1 0.3

2 0.2

3 0.1

(iii) The stop-loss insurance has a deductible of 1 for the group.

Calculate the net stop-loss premium.

(A) 2.00

(B) 2.03

(C) 2.06

(D) 2.09

(E) 2.12


20. An insurers claims follow a compound Poisson claims process with two claims expectedper period. Claim amounts can be only 1, 2, or 3 and these are equal in probability.

Calculate the continuous premium rate that should be charged each period so that theadjustment coefficient will be 0.5.

(A) 4.8

(B) 5.9

(C) 7.8

(D) 8.9

(E) 11.8


21-22. Use the following information for questions 21 and 22.

The Simple Insurance Company starts at time t = 0 with a surplus of S = 3. At thebeginning of every year, it collects a premium of P = 2. Every year, it pays a randomclaim amount:

Claim Amount Probability of Claim Amount

0 0.15

1 0.25

2 0.50

4 0.10

Claim amounts are mutually independent.

If, at the end of the year, Simples surplus is more than 3, it pays a dividend equal to theamount of surplus in excess of 3. If Simple is unable to pay its claims, or if its surplusdrops to 0, it goes out of business. Simple has no administrative expenses and its interestincome is 0.

21. Determine the probability that Simple will ultimately go out of business.

(A) 0.00

(B) 0.01

(C) 0.44

(D) 0.56

(E) 1.00


21-22. (Repeated for convenience) Use the following information for questions 21 and 22.

The Simple Insurance Company starts at time t = 0 with a surplus of S = 3. At thebeginning of every year, it collects a premium of P = 2. Every year, it pays a randomclaim amount:

Claim Amount Probability of Claim Amount

0 0.15

1 0.25

2 0.50

4 0.10

Claim amounts are mutually independent.

If, at the end of the year, Simples surplus is more than 3, it pays a dividend equal to theamount of surplus in excess of 3. If Simple is unable to pay its claims, or if its surplusdrops to 0, it goes out of business. Simple has no administrative expenses and its interestincome is 0.

22. Calculate the expected dividend at the end of the third year.

(A) 0.115

(B) 0.350

(C) 0.414

(D) 0.458

(E) 0.550


23. A continuous two-life annuity pays:

100 while both (30) and (40) are alive;70 while (30) is alive but (40) is dead; and50 while (40) is alive but (30) is dead.

The actuarial present value of this annuity is 1180. Continuous single life annuitiespaying 100 per year are available for (30) and (40) with actuarial present values of 1200and 1000, respectively.

Calculate the actuarial present value of a two-life continuous annuity that pays 100 whileat least one of them is alive.

(A) 1400

(B) 1500

(C) 1600

(D) 1700

(E) 1800


24. For a disability insurance claim:

(i) The claimant will receive payments at the rate of 20,000 per year, payablecontinuously as long as she remains disabled.

(ii) The length of the payment period in years is a random variable with the gammadistribution with parameters a q= =2 1and .

(iii) Payments begin immediately.

(iv) d = 005.

Calculate the actuarial present value of the disability payments at the time of disability.

(A) 36,400

(B) 37,200

(C) 38,100

(D) 39,200

(E) 40,000


25. For a discrete probability distribution, you are given the recursion relation

p kk

p kb g b g= -2 1* , k = 1, 2,.

Determine p 4b g.

(A) 0.07

(B) 0.08

(C) 0.09

(D) 0.10

(E) 0.11


26. A company insures a fleet of vehicles. Aggregate losses have a compound Poissondistribution. The expected number of losses is 20. Loss amounts, regardless of vehicletype, have exponential distribution with q = 200.

In order to reduce the cost of the insurance, two modifications are to be made:

(i) a certain type of vehicle will not be insured. It is estimated that this willreduce loss frequency by 20%.

(ii) a deductible of 100 per loss will be imposed.

Calculate the expected aggregate amount paid by the insurer after the modifications.

(A) 1600

(B) 1940

(C) 2520

(D) 3200

(E) 3880


27. An actuary is modeling the mortality of a group of 1000 people, each age 95, for the nextthree years.

The actuary starts by calculating the expected number of survivors at each integral age by

l p kk k95 951000+ = =, 1, 2, 3

The actuary subsequently calculates the expected number of survivors at the middle of eachyear using the assumption that deaths are uniformly distributed over each year of age.

This is the result of the actuarys model:

Age Survivors

95 1000

95.5 800

96 600

96.5 480

97 --

97.5 288

98 --

The actuary decides to change his assumption for mortality at fractional ages to the constantforce assumption. He retains his original assumption for each k p95 .

Calculate the revised expected number of survivors at age 97.5.

(A) 270

(B) 273

(C) 276

(D) 279

(E) 282


28. For a population of individuals, you are given:

(i) Each individual has a constant force of mortality.

(ii) The forces of mortality are uniformly distributed over the interval (0,2).

Calculate the probability that an individual drawn at random from this population dieswithin one year.

(A) 0.37

(B) 0.43

(C) 0.50

(D) 0.57

(E) 0.63


29-30. Use the following information for questions 29 and 30.

You are the producer of a television quiz show that gives cash prizes. The number ofprizes, N, and prize amounts, X, have the following distributions:

n Pr N n=b g x Pr X x=b g1 0.8 0 0.2

2 0.2 100 0.7

1000 0.1

29. Your budget for prizes equals the expected prizes plus the standard deviation of prizes.

Calculate your budget.

(A) 306

(B) 316

(C) 416

(D) 510

(E) 518


29-30. (Repeated for convenience) Use the following information for questions 29 and 30.

You are the producer of a television quiz show that gives cash prizes. The number ofprizes, N, and prize amounts, X, have the following distributions:

n Pr N n=b g x Pr X x=b g1 0.8 0 0.2

2 0.2 100 0.7

1000 0.1

30. You buy stop-loss insurance for prizes with a deductible of 200. The cost of insuranceincludes a 175% relative security load.

Calculate the cost of the insurance.

(A) 204

(B) 227

(C) 245

(D) 273

(E) 357


31. For a special fully discrete 3-year term insurance on xb g :(i) Level benefit premiums are paid at the beginning of each year.

(ii) k bk+1 qx k+0 200,000 0.03

1 150,000 0.06

2 100,000 0.09

(iii) i = 0.06

Calculate the initial benefit reserve for year 2.

(A) 6,500

(B) 7,500

(C) 8,100

(D) 9,400

(E) 10,300


32. For a special fully continuous whole life insurance on (x):

(i) The level premium is determined using the equivalence principle.

(ii) Death benefits are given by b itt= +1b g where i is the interest rate.

(iii) L is the loss random variable at t = 0 for the insurance.

(iv) T is the future lifetime random variable of (x).

Which of the following expressions is equal to L ?

(A)n T x

x

A

A

-

-c hc h1

(B) n T x xA A- +c hc h1

(C)n T x

x

A

A

-

+c hc h1

(D) n T x xA A- -c hc h1

(E)v A

A

Tx

x

+

+d ic h1


33. For a 4-year college, you are given the following probabilities for dropout from allcauses:

q

q

q

q

0

1

2

3

015

010

005

001

====

.

.

.

.

Dropouts are uniformly distributed over each year.

Compute the temporary 1.5-year complete expected college lifetime of a student entering

the second year, eo11 5: . .

(A) 1.25

(B) 1.30

(C) 1.35

(D) 1.40

(E) 1.45


34. Lee, age 63, considers the purchase of a single premium whole life insurance of 10,000with death benefit payable at the end of the year of death.

The company calculates benefit premiums using:

(i) mortality based on the Illustrative Life Table,

(ii) i = 0.05

The company calculates contract premiums as 112% of benefit premiums.

The single contract premium at age 63 is 5233.

Lee decides to delay the purchase for two years and invests the 5233.

Calculate the minimum annual rate of return that the investment must earn to accumulateto an amount equal to the single contract premium at age 65.

(A) 0.030

(B) 0.035

(C) 0.040

(D) 0.045

(E) 0.050


35. You have calculated the actuarial present value of a last-survivor whole life insurance of1 on (x) and (y). You assumed:

(i) The death benefit is payable at the moment of death.

(ii) The future lifetimes of (x) and (y) are independent, and each life has a constantforce of mortality with m = 0 06. .

(iii) d = 005.

Your supervisor points out that these are not independent future lifetimes. Eachmortality assumption is correct, but each includes a common shock component withconstant force 0.02.

Calculate the increase in the actuarial present value over what you originally calculated.

(A) 0.020

(B) 0.039

(C) 0.093

(D) 0.109

(E) 0.163


36. The number of accidents follows a Poisson distribution with mean 12. Each accidentgenerates 1, 2, or 3 claimants with probabilities 12 , 13 , 16 , respectively.

Calculate the variance in the total number of claimants.

(A) 20

(B) 25

(C) 30

(D) 35

(E) 40


37. For a claims process, you are given:

(i) The number of claims N t tb gm r, 0 is a nonhomogeneous Poisson process withintensity function:

l( ),

,

,

t

t

t

t

= < =08 015 03. .

total Death and Surrender Benefits = 10+20+0 = 30

Test Question: 13 Key: E

2

3

1 01 0 2 07

0 7 03 04

p

px

x

= - - =

= - =

. . .

. . .

Use lx = 1 (arbitrary, doesnt affect solution)

so l lx x+ += =2 307 04. .

By hyperbolic 1

51

51

2 5 2 3l l lx x x+ + += +

.

. .

= + =..

..

.57

54

19643

l px x+ = =2 5 2 505091. ..

2 5 1 05091 04909. . .qx = - =

Prob (all 3 failed) = =04909 01183. .b g

Test Question: 14 Key: B

This is a graph of l xxmb g.m xb g would be increasing in the interval 80 100,b g .The graphs of l px x , lx and lx

2 would be decreasing everywhere.The graph shown is comparable to Figure 3.3.2 on page 65 of Actuarial Mathematics

Test Question: 15 Key: A

Using the conditional mean and variance formulas:

E N E N= L Lc h

Var N Var E N E Var N= +L LL Lc hd j c hd jSince N, given lambda, is just a Poisson distribution, this simplifies to:

E N E= L Lb gVar N Var E= +L LL Lb g b gWe are given that E N = 0 2. and Var N = 04. , subtraction gives Var Lb g = 0.2


N = number of salmonX = eggs from one salmonS = total eggs.E(N) = 100tVar(N) = 900t

E S E N E X tb g b g b g= = 500Var S E N Var X E X Var N t t tb g b g b g b g b g= + = + =2 100 5 25 900 23 000,P S P

S tt

tt

> =-

>-F

HGIKJ = 10 000

500

23 000

10 000 500

23 00095,

,

,

,.b g

10 000 500 1645 23000 250

40 2

2 40 0

1 1 320

44 73

22 4

2

, .

.

.

- = - = -

- = -

- - =

= +

=

=

t t t

t t

t t

t

t

d i

round up to 23


A P V (xs benefits) = + +=

+v b p qk k k xk

x k1

10

2

= + +

=

1000 300 0 02 350 098 004 400 098 0 96 006

36 829

2 3v v v. . . . . .

,

b g b gb g b gb gb g


p denotes benefit premium19V APV= future benefits - APV future premiums

0 61

1080326.

..= - =p p

1110 65

65

108 10V

V qp

=+ -pb gb g b gb g.

=+ -

-5 0 0326 108 010 10

1 010

. . . .

.b gb g b gb g

=5.28

Test Question: 19 Key: C

X = losses on one lifeE X = + +

=03 1 0 2 2 01 3

1

. . .b gb g b gb g b gb g

S = total lossesE S E X

E S E S F

E S f

s

s

= =

- = - -

= - -

= - -

= -=

+

3 3

1 1 1 0

1 1 0

3 1 1 04

3 0 936

2 064

3

b g b gc hb g b gc h

b gd i..

.


M r E e

e e e

Me e e

xrx

r r r

x

b g

b g

=

= + +

= + + =

2 3

0 5 1 5

3

053

2 95. .. .

p E X11 2 3

32= = + + =

l M r crxb g - =1Since l = 2 and r = 0.5,2 05 1 05

2 2 95 1 05

39 05

7 8

M c

c

c

c

x . .

. .

. .

.

b gb g

- =

- === = premium rate per period


Simples surplus at the end of each year follows a Markov process with four states:State 0: out of businessState 1: ending surplus 1State 2: ending surplus 2State 3: ending surplus 3 (after dividend, if any)

State 0 is absorbing (recurrent). All the other states are transient states.Thus eventually Simple must reach state 0.

Test Question: 22 Key: D

(See solution to problem 21 for definition of states).

t = 0

0 0 0 1

10 0 0 000 000

01 05 025 015

01 0 0 050 0 40

0 0 01 000 090

00 01 00 09 1

. . . .

. . . .

. . . .

. . . .

. . . .

L

N

MMMM

O

Q

PPPP= =at t

t = 1

00 01 0 0 09

10 00 0 00 000

01 05 0 25 015

01 00 050 0 40

00 01 0 00 090

0 01 014 0 025 0825 2. . . .

. . . .

. . . .

. . . .

. . . .

. . . .

L

N

MMMM

O

Q

PPPP= =at t

Expected dividend at the end of the third year =

k=

0

3

(probability in state k at t = 2) (expected dividend if in state k)

0.01*0 + 0.14*0 + 0.025(0*0.85 + 1*0.15) + 0.825*(0*0.6 +1*0.25 + 2*0.15) = 0.4575


1180 70 50 20

1180 70 12 50 10 20

8

12 10 8 14

100 1400

30 40 30 40

30 40

30 40

30 40 30 40 30 40

30 40

= + -= + -== + - = + - =

=

a a a

a

a

a a a a

a

:

:

:

: :

:

b gb g b gb g


a a f t dtt

= zo b g = -

- -z 1 0 05 120 05e

te dtt

t.

. Gb go

= -- -z10 05 105. .te t tte dtb io

= - + + +FHGIKJ

LNM

OQP

- -

1

0 051

105

1

10521 05

0. . ..t e

tet tb g

= - FHGIKJ

LNMM

OQPP

=10 05

11

105185941

2

. ..

20 000 185941 37 188, . , =


p kk

p k

kp k

b g b g

b g

= -

= +LNMOQP -

21

02

1

Thus an (a, b, 0) distribution with a = 0, b = 2.

Thus Poisson with l = 2 .

pe

42

40 09

2 4

b g ==

-

!.


By the memoryless property, the distribution of amounts paid in excess of 100 is stillexponential with mean 200.

With the deductible, the probability that the amount paid is 0 is F e100 1 0 393100 200b g = - =- / . .Thus the average amount paid per loss is (0.393) (0) + (0.607) (200) = 121.4

The expected number of losses is (20) (0.8) = 16.

The expected amount paid is (16) (121.4) = 1942.


From UDD ll l

96 596 97

2.= +

480600

236097 97=

+ =l l

Likewise, from l l97 97 5360 288= =and . ,we get l98 216=

For constant force,

ell

p e

l l

-

-

= = =

= = =

= = =

m

m

98

97

0 5 975

97 5 97

216360

0 6

0 6 07746

07746 0 7746 360 27886

12

.

. .

. . .

..

.

b gb g b gb g


Let M = the force of mortality of an individual drawn at random; and T = future lifetime of theindividual.

Pr T 1 = E T MPr 1n s= =

z Pr T M f dM10 m m mb g= -zz m mme dt dt 120102= - = + - = +- - -z 1 12 12 2 1 12 12 202 e du e emd i d i d i= 056767.


E N

E N

E X

= + =

= + =

= - =

= + =

08 1 02 2 12

08 1 0 2 4 16

16 12 016

70 100 170

2

2

. . .

. . .

. . .

b gb g b gb gb g b gb g

b gVar N

Var Xb g b g= - = + - =E X E X2 2 27000 100 000 170 78100, ,E S E N E X= = =12 170 204. b gVar S Var X Var Nb g b g b g b g b g= + = + =E N E X 2 212 78100 170 016 98 344. , . ,

Std dev Sb g = =98 344 3136, .So B = 204 + 314 = 518


f

f

f

s

s

s

1000 08 01 0 2 2 02 01 0088

1100 02 2 07 01 0028

2000 02 01 0 0022

b g b gb g b gb gb gb gb g b gb gb gb gb g b gb g

= + =

= =

= =

. . . . . .

. . . .

. . .

E S - = + +

=+200 0 088 800 0028 900 0002 1800

99 2

b g b gb g b gb g b gb g. . ..

With 175% relative security loading, cost = (2.75) (99.2) = 272.8

Alternatively,

f F

f

F

s s

s

s

0 0 08 0 2 02 02 0168

100 08 07 02 2 02 07 0 616

100 0168 0616 0 784

2b g b g b gb g b gb gb g b gb g b gb gb gb gb g

= = + =

= + =

= + =

. . . . .

. . . . . .

. . .

E S = 204 [from problem 29]

E S E S F

E S F F

s

s s

- = - - -

= - - - -

= - - - -=

+ +200 100 100 1 100

100 1 0 100 1 100

204 100 1 0168 100 1 0 784

99 2

b g b g b g b gc hb g b gc h b g b gc hb gb g b gb g. .

.cost = (2.75) (99.2) = 272.8


Let p = benefit premium

Actuarial present value of benefits =

= + += + +=

003 200 000 0 97 006 150 000 097 094 0 09 100 000

566038 7769 67 689008

20 32013

2 3. , . . , . . . ,

. . .

, .

b gb g b gb gb g b gb gb gb gv v v

Actuarial present value of benefit premiums=

= + +

=

= =

&&

. . .

.

, .

..

:a

v v

x 3

21 0 97 097 0 94

2 7266

20 32013

2 7266745255

p

p

p

p

b gb g

1745255 106 200 000 0 03

1 003195846

V =-

-=

. . , .

..

b gb g b gb g

Initial reserve, year 2 = 1V + p = 1958.56 + 7452.55 = 9411.01


Let p denote the premium.

L b v a i v a

a

TT

TT T

T

T

= - = + -

= -

p p

p

1

1

b g

E L ax ax= - = =1 0 1p p

= - = - =- -

=- -

= --

L aa

a

a v

a

v a

av A

A

TT

x

xT

xT

x

x

Tx

x

1 11

1

1

pd

d

dd

d i

b g


1 1

2 1

1 01 0 9

09 1 005 0855

p

p

= - == - =

( . ) .

. . .b gb g

since uniform, 1 5 1 09 0855 2

08775. . . /

.

p = +=b g

e t to 1 1 5 0 15

1 0 9

21

0 9 08775

205

0 95 0 444

1394

: . .

. . ..

. .

.

= = =

= +FHGIKJ +

+FHG

IKJ

= +=

Area between and

b g b g

Alternatively,

e p dt

p dt p p dx

t dt x dx

t x

t

t x

t x

o11 5 10

1 5

10

1

1 1 20

0 5

0

1

0

0 5

0 12

0

10 05

20

0 5

1 01 09 1 0 05

09

095 0 444 1394

2 2

: ..

.

.

. ..

. . .

.

. . .

=

= +

= - + -

= - + -

= + =

zz zz zb g b g

tp1

t1 2

(1.5, 0.8775)

(2, 0.885)

(0, 1) (1, 0.9)


10 000 112 523363, .A b g =A

AA i q

p

A

A

xx x

x

63

1

64

65

0 4672

1

04672 105 001788

1 0 0178804813

04813 105 0 01952

1 00195204955

=

=+ -

=-

-=

=-

-=

+

.

. . .

..

. . .

..

b g

b gb g

b gb g

Single contract premium at 65 = (1.12) (10,000) (0.4955) = 5550

15550

5233

5550

52331 0 02984

2+ = = - =i ib g .


Original Calculation (assuming independence):

mm

m

x

y

xy

==

= + =

006

006

0 06 006 012

.

.

. . .

A

A

xx

x

yy

y

=+

=+

=

=+

=+

=

mm d

mm d

006

0 06 0 05054545

0 06006 0 05

054545

.

. ..

.. .

.

A

A A A A

xyxy

xy

xy x y xy

=+

=+

=

= + - = + - =

mm d

012

012 0050 70588

054545 054545 070588 038502

.

. ..

. . . .

Revised Calculation (common shock model):

m m

m mx x

T x

y yT y

= =

= =

006 0 04

006 004

. , .

. , .

*

*

b g

b g

m m m mxy xT x

yT y Z= + + + + + =* * . . . .b g b g 004 0 04 0 02 010

A

A

xx

x

yy

y

=+

=+

=

= =+

=+

mm dm

m d

006

0 06 0050 54545

0 06006 005

054545

.

. ..

.. .

.

A

A A A A

xyxy

xy

xy x y xy

=+

=+

=

= + - = + - =

mm d

010

010 0050 66667

054545 054545 066667 042423

.

. ..

. . . .

Difference = - =0 42423 038502 003921. . .


Treat as three independent Poisson variables, corresponding to 1, 2 or 3 claimants.

rate

rate

rate2

3

1 61

212

4

2

= = LNMOQP

==

Var

Var

Var

1

2

3

=

= =

=

6

16 4 2

18

2

total Var = 6 16 18 40+ + = , since independent.

Alternatively,

E X 22 2 212

23

36

103

d i = + + =

For compound Poisson, Var S E N E X= 2

= FHGIKJ =12

10

340b g


l lt dt Nb g b g= =z 6 3 603 so is Poisson with .P is Poisson with mean 3 (with mean 3 since Prob yi < =500 05b g g.P and Q are independent, so the mean of P is 3, no matter what the value of Q is.


At age x :

Actuarial Present value (APV) of future benefits = 1

51000Ax

FHG

IKJ

APV of future premiums = 4

5&&ax

FHG

IKJ p

10005

4525 25

A a= p && by equivalence principle

10004

14

816516 2242

12582525

Aa&&

..

.= = =p p

10V = APV (Future benefits) APV (Future benefit premiums)

= -

= -

=

1000

5

4

515

128 7245

1258 153926

1025

35 35A ap &&

. . .

.

b g b gb g


Let Y = present value random variable for payments on one lifeS Y= = present value random variable for all payments

E Y a= =10 14816640&& .

Var YA A

d=

-

= -

=

10

100 004863 016132 106 006

70555

22

40 402

2

2 2

d i

d ib g. . . / ..

E S E Y

S Y

= =

= =

100 14 816 6

100 70 555

, .

,Var Var

Standard deviation S = =70 555 26562, .

By normal approximation, needE [S] + 1.645 Standard deviations = 14,816.6 + (1.645) (265.62)

= 15,254


Initial Benefit Prem =-

-

5 4

5 4

30 30 201

30 35 30 20

A A

a a:

: :&& &&

e j

=

--

= --

= =

5 010248 4 002933

5 14 835 4 11959

05124 011732

74175 47 836

039508

263390015

. .

. .

. .

. .

.

..

b g b gb g b g

Where

A A A30 201

30 20 30 201 032307 029374 002933: : : . . .= - = - =e j

and

&&...

.::a

A

d30 2030 201 1 0 32307

0 06106

11959=-

= -FHG

IKJ

=

Comment: the numerator could equally well have been calculated as A E A30 20 30 504+= 0.10248 + (4) (0.29374) (0.24905)= 0.39510

Sample Exam2

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