SOME PROBLEMS WITH SOLUTIONS

FMSN25/MASM24 VALUATION OF DERIVATIVE ASSETS

MAGNUS WIKTORSSON

August 2015

Faculty of EngineeringCentre for Mathematical SciencesMathematical Statistics

CE

NT

RU

MSC

IEN

TIA

RU

MM

AT

HE

MA

TIC

AR

UM

Chapter 1

Problems

1.1 Martingales

1.1.1 Assume that the process {St}t≥0 follows the standard Black & Scholes model and that γ ∈ R. Find γ 6= 1 such that{(St)ge−rt}t≥0 will be a Q-martingale.

1.1.2 Show that the process X (t) = et/2 cos(Wt), where Wt is a standard Brownian motion, is a martingale for t ≥ 0.

1.2 Static replication of derivatives

1.2.1 We have a derivative with maturity T and pay-off

Φ(ST ) = max(K − |K − ST |, 0).

0 K 2K

0

K

x

Φ(x

)

pay−off function

Find a static hedge for this derivative using the asset S and European options on S.

1.2.2 Assume that X is a derivative with maturity T having the following pay- off function:

K ST ≤ K

2K − ST K ≤ ST ≤ 2K

0 ST ≥ 2K

Express the price for 0 < t < T of X using European put and call options, the stock S and the bank-account B.

1.3 PDE:s and Feynman-Kac formula

1.3.1 Solve the PDE

∂f (t, x)∂t

+

(

μ− σ2

2

)

∂f (t, x)∂x

+σ2

2∂2f (t, x)

∂x2= 0

f (T , x) = ex

for t ∈ [0, T ], x ∈ R using the Feynman-Kac representation, where μ and σ are real-valued constants.

1

1.4 Pricing of derivatives and hedging

1.4.1 Consider a derivative with pay-off Φ(ST ) = max(K − (ST )2, 0) in the Black-Scholes market.

(a) Find the price-formula for the derivative for 0 ≤ t ≤ T .

(b) Calculate the replicating portfolio (delta-hedge) for the derivative, that is find a self-financing portfolio consistingof the stock and the bank account that hedges the derivative.

1.5 Change of measure and completeness

1.5.1 A simple extension to the binomial model is the trinomial model (TM) where we also allow the stock price to remainunchanged with some probability. More precisely we have that

P(Sk+1 = uSk) = pu, P(Sk+1 = Sk) = p1, P(Sk+1 = dSk) = pd , k = 0, 1, · · · , (N − 1)

in the N -period case where u > 1 and d < 1 and a bank account B with Bk = exp(kr). Even though it can providea better approximation of the Black-Scholes model for a finite number of periods, TM is seldom used. The mainreason for this it that TM gives an incomplete market. We are here going to study TM in the one period setting withu = 6/5, d = 3/5, S0 = 1 and with zero interest rate.

(a) Show that TM is incomplete by showing that the risk-neutral probabilities qu, q1 and qd are not unique.

(b) If we add another asset to the market it is possible to make the market complete. Suppose that we add the assetX where X is the derivative with pay-off S2

1 at time 1 and further suppose that the price of X at time zero isequal to 26/25. If the price of X is given by the market we can view this as a calibration of our model to fit themarket price. This will then allow us to price other derivatives on the asset S. Show that the market is completeby finding the unique risk-neutral probabilities.

(c) Use the probabilities from (b) to price the derivative Y , with maturity T = 1 and pay-off max(S1 − 9/10, 0).Moreover find the replicating portfolio for this derivative using the assets S, X and B.

(d) Note that we cannot have an arbitrary price of X at time zero. Show that all prices outside the interval [1, 27/25]will give arbitrage in the sense that there are no risk-neutral probabilities compatible with those prices.

1.5.2 Assume that we have a market consisting of one risky asset S(1) and one bank account S(0) with the P-dynamics:

dS(1)t = μ1S(1)

t dt + S(1)t

(

σ11dW (1)t + σ12dW (2)

t

)

dS(0)t = rS(0)

t dt,

where W (1) and W (2) are two independent standard BM:s. Using the meta-theorem on this model we get that it is freeof arbitrage but incomplete, i.e. the martingale measure exists but is not unique.

(a) However, show that all simple claims X with maturity T of the form Φ(S(1)T ) will have a price-formula that does

not depend on the choice of martingale measure.

(b) Show that the claim Y = 1W (2)T >K , that is a contract which pays one unit of currency at time T if W (2)

T > K ,will have a price-formula that do depend on the choice of martingale measure.

(c) Show that if we add the asset S(2) with P dynamics:

dS(2)t = μ2S(2)

t dt + σ22S(2)t dW (2)

t

to the market, that the market will be free of arbitrage and complete by finding the unique Girsanov kernel(g1, g2).

(d) Price the derivative in (b) using this unique martingale measure.

1.6 Change of numeraires

1.6.1 Assume the following 2-dim Black-Scholes model (under Q) for the two stocks S1 and S2 ,

dS1(t) = rS1(t)dt + S1(t)(σ11dW1(t) + σ12dW2(t))

dS2(t) = rS2(t)dt + S2(t)(σ21dW1(t) + σ22dW2(t))

2

where W1 and W2 are two independent standard Q Brownian motions and where r, σ11, σ12, σ21 and σ22 are positiveconstants. Price the derivative with maturity T and pay-off:

Φ(S1(T ), S2(T )) = max(S2(T ) − S1(T ), 0),

for 0 < t < T . (Hint: the key is to find the right numeraires).

1.6.2 In a realistic situation the short interest rate r is not a deterministic constant. What one wants to do is to use observedprices of Zero Coupon bonds (ZCB) as a discounting factor when pricing derivatives. The way to accomplish this is toexpress the dynamics of the underlying stock S under the forward measure QT , i.e. the martingale measure which hasthe ZCB as numeraire. Under the measure QT we have that the discounted stock process Z (t) = S(t)/p(t, T ) shouldbe a martingale, where p(t, T ) is the price at time t of a ZCB with maturity T . Note that We assume the followingmodel for Z (t) under QT :

dZ (t) = Z (t)v(t, T )dW QT

(t), 0 ≤ t ≤ T ,

where W QT

is a standard 2-dim QT Brownian motion and where v(t, T ) is a positive deterministic function (a 1 × 2row vector). Note that by definition we have Z (T ) = S(T ) since p(T , T ) = 1.

(a) Now assume that under the usual martingale measure Q, we have the following model for the ZCB and thestock,

dP(t, T ) = r(t)P(t, T )dt + P(t, T )(T − t)γdW1(t), 0 ≤ t ≤ T ,

dS(t) = r(t)S(t)dt + σS(t)dW2(t),

where W1 and W2 are two independent standard Q Brownian motions and where σ and γ are positive constants.Calculate the volatility function v(t, T ) implied by this model. Recall that the volatilities do not change whenwe change the measure.

(b) Price a standard European call option with strike K and maturity T for this model, for t such that 0 < t < T .You should express the price in a Black-Scholes type of formula. Remember to check that your price simplifiesto the standard Black-Scholes formula if the interest rate is constant equal to r and |v(t, T )| ≡ σ.

1.6.3 You are thinking of investing in some stocks. Company A and B are competing on the same market. So you think that either A or B will win and that the winning stock will increase over the coming T years. But you cannot make upyour mind if you should buy stock A or stock B. Assume that you have the following model (under Q) for the stocksand the bank account.

dB(t) = rB(t)dt,

dSA(t) = rSA(t)dt + σASA(t)dW1(t),

dSB(t) = rSB(t)dt + σBSB(t)(ρdW1(t) +√

1 − ρ2dW2(t)),

where r,σA,σB > 0 and −1 ≤ ρ < −0.5 with W1 and W2 being independent standard Brownian motions. Thenegative ρ should give the effect that one stock will go up and while the other stock goes down. For this setup to berealistic, one should assume that the initial prices of stock A and B are equal and that the volatilities also are roughlythe same. You can use this assumption in the discussion of (d), but in (a)-(c) solve the general problem.

(a) Someone at your bank offers you a derivative with maturity T years and with pay-off

Φ(SA(T ), SB(T )) = max(SA(T ), SB(T )),

which take care of your problem to decide which stock to buy. What is the fair price of this contract at timet = 0 and 0 < t < T ?

(b) You talk to a friend who tell you to buy stock A now, keep it for T years, and to buy a spread option withmaturity T years and pay-off:

Φ(SA(T ), SB(T )) = (SB(T ) − SA(T ))+.

Verify that this setup is equivalent to the derivative offered by your bank.

(c) Find a hedge for the derivative in (a) (and therefore also for (b)) for 0 < t < T .

(d) Another possibility is of course to buy both the stocks now and then sell both after T years. Discuss pros and conswith this approach compared to the derivatives described above. Look at aspects like initial price and possiblefinal pay-off.

3

1.6.4 Two friends, we can call them Anna and Belle, meet at a café on a Saturday. They are amused to see that their twofavourite stocks, SA and SB respectively, happened to have the same closing price on the previous day. To celebrate thisthey decide to make a bet. If stock SA has a higher closing price than SB at time T Anna will receive SA(T ) from Belle,and if it is the other way around, Belle will receive SB(T ) from Anna. If the closing prices at time T are equal, neitherAnna nor Belle will receive any money. Note that no money is paid until time T . Your task is now to investigate if thisis a fair bet, by looking at the bet as a financial derivative. In order to do this we need a model for the stocks. Assumethat the stocks under the martingale measure Q follow a two dimensional Black-Scholes model of the following form:

dSA(t) = rSA(t)dt + σASA(t)dW1(t)

dSB(t) = rSB(t)dt + σBSB(t)(

ρdW1(t) +√

1 − ρ2dW2(t))

,

dB(t) = rB(t)dt,

where W1 and W2 are two independent Q-BM:s and where −1 < ρ < 1, σA, σB > 0 and r > 0.

(a) Calculate the value of the derivative (bet) from Anna’s point of view on the date of issue t = 0, i.e. the day Annaand Belle decide to make the bet. Moreover explain how this is related to the value from Belle’s point of view. Isthe bet fair?

(b) What is the value at time t for 0 < t < T for arbitrary positive values of SA(t) and SB(t) from Anna’s point ofview?

(c) Find a replicating portfolio for the bet from Anna’s and Belle’s points of view respectively.

(d) Do you think it is a good idea to hedge the bet? Motivate your answer properly.

1.7 Volatility

1.7.1 VIX: On Chicago Board Options Exchange (CBOE) there is an index called VIX, which measures expected volatilityon the SP500 index. The VIX is calculated using market prices of European put and call options on the SP500 index.LetΠ P

E (t, t +τ, S(t), K ) andΠ CE (t, t +τ, S(t), K ) be prices of a European put and a European call options respectively

given at time t with maturity t + τ with current stock price S(t) and with strike K. Let further F t+tt be the forward

level at time t of the stock price at time t + τ, i.e. F t+tt = E

Q[S(t + τ)|Ft]. Having access to a wide range of optionprices over different strikes we can get an almost model-free estimate of the expected forward volatility.

VIX (t, τ) =

(

2τ

NP−1∑

i=1

ertΠ PE (t, t + τ, S(t), Ki)

Ki+1 − Ki

K 2i

I (Ki < F t+tt )

+2τ

NC−1∑

j=1

ertΠ CE (t, t + τ, S(t), Kj)

Kj+1 − Kj

K 2j

I (Kj > F t+tt )

1/2

,

where NP and NC are the number of available put and call options with maturity t + τ respectively1. The empiricallycalculated VIX index squared VIX (t, τ)2 is an approximation of an integral

VIX (t, τ)2 ≈ 2τ

(

∫ F t+tt

0ertΠ P

E (t, t + τ, S(t), K )dK

K 2+

∫ ∞

F t+tt

ertΠ CE (t, t + τ, S(t), K )

dK

K 2

)

=2τE

Q

[

∫ F t+tt

0(K − S(t + τ))+

dK

K 2+

∫ ∞

F t+tt

(S(t + τ) − K )+dK

K 2| Ft

]

=2τE

Q

[

− ln

(

S(t + τ)

F t+tt

)

| Ft

]

.

Now assume that we have a market with the following Q-dynamics:

dB(t) = rB(t)dt,

dS(t) = rS(t)dt + σ(t)S(t)dW (t),

where r is the risk free rate, σ a deterministic function and W is a standard Q-Brownian motion. This is the modelyou shall use in the rest of the problem.

1This is a slightly simplified version of the real index.

4

(a) Calculate F t+tt = E

Q[S(t + τ)|Ft].

(b) Evaluate the expectation2τE

Q

[

− ln

(

S(t + τ)

F t+tt

)

| Ft

]

.

(c) Show that

EQ

[

∫ F t+tt

0(K − S(t + τ))+

dK

K 2+

∫ ∞

F t+tt

(S(t + τ) − K )+dK

K 2| Ft

]

= EQ

[

− ln

(

S(t + τ)

F t+tt

)

| Ft

]

.

1.7.2 (VIX cont.) Assume that S follows the Heston model i.e. we have a market with the following Q-dynamics:

dB(t) = rB(t)dt,

dS(t) = rS(t)dt +√

V (t)S(t)(ρdW1(t) +√

1 − ρ2dW2(t)),

dV (t) = κ(θ− V (t))dt + σv

√

V (t)dW1(t),

where r is the risk free rate, σv, κ,θ are positive deterministic constants,−1 ≤ ρ ≤ 1 and W1 and W2 are independent standard Q-Brownian motions. Evaluate the expectation

2τE

Q

[

− ln

(

S(t + τ)

F t+tt

)

| Ft

]

,

where F t+tt = E

Q[S(t + τ)|Ft].

1.8 Simple interest rate contracts and Martingale models for the short rate

1.8.1 Calculate the price of a Zero Coupon bond with maturity T at time t where 0 < t < T in the following Ho-Leemodel for the short rate,

{

drs = Θ(s) ds + σdWs, for s ≥ trt = r

where Θ(s) = ce−s + σ2s and where σ, r and c are positive constants.

1.8.2 A floating rate bond with face value A is like a coupon bond with face value A which has coupon rates equal to thefloating LIBOR rates

LTi−1 [Ti−1, Ti] =1

Ti − Ti−1

(

1p(Ti−1, Ti)

− 1

)

over time intervals [Ti−1, Ti], i = 1, . . . , n. So this contracts pays out A at time Tn and the coupons ci = A(Ti −Ti−1)LTi−1[Ti−1, Ti], at times Ti, i = 1, . . . , n. The value of coupon i is not known until time Ti−1. Find the valueof this contract at time t = T0.

1.9 The HJM framework

1.9.1 Consider the following HJM model for the forward rate f (t, T ), T ≥ 0;

df (t, T ) = σ2(T − t)dt + σdW Qt , 0 ≤ t ≤ T

f (0, T ) = f ⋆(0, T ), T ≥ 0

where W is a standard BM and f ⋆(0, T ) is the observed forward rate on the e market.

(a) Find the ZCB price p(t, T ) for this model.

(b) Calculate the QS-dynamics, that is the dynamics under the numeraire measure where p(t, S) is used as a nu-meraire, for the process Xt = (1 + (S − T )Lt[T , S]) for 0 ≤ t ≤ T , where Lt [T , S] = (p(t, T )− p(t, S))/((S −T )p(t, S)).

(c) Use the dynamics obtained in (b) to price the Caplet for 0 ≤ t ≤ T , that is the derivative with maturity S andpay-off

(S − T ) max(LT [T , S] − K , 0),

where K is a positive constant.

5

1.9.2 You get an urgent phone call from a friend who has found a programming error in the bank’s accounting system.Instead of paying out a three month spot LIBOR-contract over the time interval [T1, T2] according to the correctpay-off X = 1 + (T2 − T1)LT1 [T1, T2] at time T2 where the rate is decided at time T1, the system assigns therate over the coming interval [T2, T3], i.e. the pay-off is incorrectly set to Y = 1 + (T2 − T1)LT2[T2, T3], whereT2 − T1 = T3 − T2 = 0.25. The cash-flow is still at time T2. For the correct pay-off X we have that the fair valueat time T1 is one regardless of which model we use i.e. πX (T1) = 1. You will however need at model to find Π Y (T1),i.e. the fair value of Y at time T1. So we now assume that we have the following QT3-dynamics (dynamics under thenumeraire measure corresponding to the numeraire p(t, T3)) for the forward rate for T1 ≤ t ≤ u ≤ T3 as:

df (t, u) = −σ(t)2(T3 − u)dt + σ(t)dW (t),

f (T1, u) = a + b(1 − e−(u−T1)),

where a, a + b > 0. Note that for arbitrary t, S1, S2 satisfying 0 ≤ t ≤ S1 < S2 we have

1 + (S2 − S1)Lt[S1, S2] = p(t, S1)/p(t, S2)

where p(t, S1) and p(t, S2) are ZCB values at time t for ZCB:s with maturity S1 and S2 respectively.

(a) Your task is now to find out the fair value of the pay-off Y at time T1 using the model above.

(b) You should now figure out if the fair value calculated in a) is larger or smaller than one (i.e. the fair value of X attime T1). To simplify things in b) we assume that a = 0.02, b = −0.01 and σ(t) ≡ σ where σ = 0.1.

(c) So would the bank on average win or loose on the programming error given the result you have obtain for thefair value in b)?

1.10 Dividend paying stocks

1.10.1 Continuous dividendsMost real stock pay out dividends. One common model for a dividend paying stock is to assume that dividends arepayed out in a continuous cash flow. More precisely, by holding a stock, S over the time interval [t, T ] we receive thefollowing dividend cash flow

D(T ) − D(t) =

∫ T

t

qS(u)du,

where the positive constant q is called dividend yield. The model for the stock and bank account B and dividendsunder Q is given by

dS(u) = αS(u)du + σS(u)dW Q(u),

dB(u) = rB(u)du, B(0) = 1

dD(u) = qS(u)du, D(0) = 0

where α is the drift, σ > 0 the volatility, W Q is a standard Q-BM, r the interest rate and q ≥ 0 is the dividend yield.So the stock behaves like a geometric BM as in the BS standard model. However we cannot have α = r as in the BScase (unless q = 0).

(a) How should we choose α here? The key observation here is that we want an arbitrage free market. So let t andT be to arbitrary times satisfying t < T . Buying the stock at time t and selling it at time T we get the followingdiscounted cash flow

S(T )B(t)B(T )

+

∫ T

t

B(t)B(u)

dD(u) = S(T )B(t)B(T )

+

∫ T

t

qB(t)B(u)

S(u)du.

This cost us S(t) at time t so for no arbitrage we must have

S(t) = EQ

[

S(T )B(t)B(T )

+

∫ T

t

qB(t)B(u)

S(u)du|Ft

]

,

for all 0 < t < T . No use this to find the correct value of α.

(b) Given the result in a), use the foundamental theorems of asset pricing to check if this market is free of arbitrageand complete.

6

(c) Now use the model with the α-value calculated in a) to find the value at time t of a contract where we receive thestock at time T , where T > t .

(d) Find a replicating portfolio for the contract in b). Notice that for dividend paying stocks a simple buy and holdstrategy will not work, since holding the stock you receive dividends whereas getting the stock at time T give youno dividends between t and T . A portfolio consisting of holdings in the stock and the bank account with valueprocess V (t) = a(t)S(t) + b(t)B(t) is self-financing if

dV (t) = a(t)dS(t) + a(t)dD(t) + b(t)dB(t).

Hint: Look at the delta-hedge and check that it will be self-financing.

(e) Now use the model with the α-value calculated in a) to price a standard European call option and compare theresult to the standard BS-formula. Conclusions? Remember that holding an European call option does not giveyou any dividends.

(f ) Find a replicating portfolio for the contract in d).

1.10.2 Discrete dividendsMost real stock pay out dividends. Another common model for a dividend paying stock is to assume that dividendsare payed out at discrete time points. Let {tk}, k = 1, 2, · · · , be pre-specified ordered time points when dividends arepayed out. More precisely, by holding a stock, S over the time interval [t, T ] we receive the following dividend cashflow

D(T ) − D(t) =

∞∑

k=1

δS(tk−)I (t < tk ≤ T ),

where the constant 0 < δ < 1 is called dividend fraction and S(tk−) is the value of the the stock just prior to thedividend pay-out. At the times {tk}, k = 1, 2, · · · , the stock S jumps downward with the jump size δS(tk−). LetN be a desterministic counter that counts the number dividends since time t = 0, i.e. N jumps up one unit at eachpay-out. We can then re-write the dividends stream as

D(T ) − D(t) =

∫ T

t

δS(u−)dN (u),

The model for the stock and bank account B and dividends under Q is given by

dS(u) = αS(u)du + σS(u)dW Q(u) − δS(u−)dN (u),

dB(u) = rB(u)du, B(0) = 1

dD(u) = δS(u−)dN (u), D(0) = 0

where α is the drift, σ > 0 the volatility, W Q is a standard Q-BM, r the interest rate and δ ≥ 0 is the dividend fraction.So the stock behaves like a geometric BM as in the BS standard model between the downward jumps. For this modelwe need to use a generalised version of the Ito formula which gives us the following

df (t, S(t)) = (f ′t (t, S(t)) + αS(t)f ′

s (t, S(t)) + σ2/2S(t)2f ′′s s(t, S(t)))dt + σS(t)f ′

s (t, S(t))dW Q(t)

+(f (t, S(t−)(1− δ)) − f (t, S(t−)))dN (t).

Applying this to the function f (t, S(t)) we of course get back the dynamics for S given above (but please do checkthis).

(a) How should we choose α here? The key observation here is that we want an arbitrage free market. So let t andT be to arbitrary times satisfying t < T . Buying the stock at time t and selling it at time T we get the followingdiscounted cash flow

S(T )B(t)B(T )

+

∫ T

t

B(t)B(u)

dD(u) = S(T )B(t)B(T )

+

∫ T

t

δB(t)B(u)

S(u−)dN (u).

This cost us S(t) at time t so for no arbitrage we must have

S(t) = EQ

[

S(T )B(t)B(T )

+

∫ T

t

δB(t)B(u)

S(u−)dN (u)|Ft

]

,

for all 0 < t < T . No use this to find the correct value of α.

7

(b) Given the result in a), use the foundamental theorems of asset pricing to check if this market is free of arbitrageand complete.

(c) Now use the model with the α-value calculated in a) to find the value at time t of a contract where we receive thestock at time T , where T > t .

(d) Find a replicating portfolio for the contract in b). Notice that for dividend paying stocks a simple buy and holdstrategy will not work, since holding the stock you receive dividends whereas getting the stock at time T give youno dividends between t and T . A portfolio consisting of holdings in the stock and the bank account with valueprocess V (t) = a(t)S(t) + b(t)B(t) is self-financing if

dV (t) = a(t)dS(t) + a(t)dD(t) + b(t)dB(t).

Hint: Look at the delta-hedge and check that it will be self-financing.

(e) Now use the model with the α-value calculated in a) to price a standard European call option and compare theresult to the standard BS-formula. Conclusions? Remember that holding an European call option does not giveyou any dividends.

(f ) Find a replicating portfolio for the contract in d).

8

Chapter 2

Solutions

Sol: 1.1 Martingales

Sol: 1.1.1 Using that St has Q-dynamicsdSt = rStdt + σStdWt

and applying Itô’s formula to the process Xt = (St)ge−rt we get that

dXt = (−rXt + γrXt +12σ2γ(γ− 1)Xt)dt + γσXtdWt

= Xt (r(γ− 1) +12σ2γ(γ− 1))dt + γσXtdWt

= Xt (γ− 1)(r +12σ2γ)dt + γσXtdWt .

For Xt to be a martingale we need that the drift is zero which gives that γ should solve the equation

(γ− 1)(r +12σ2γ) = 0.

We see right away that it has the solutions γ = 1 and γ = − 2rs2 and since we wanted a solution different from γ = 1we see that the requested solution is γ = − 2rs2 . We should also check that X satisfies the required moment conditions.Plugging in γ = − 2rs2 we get the following dynamics for X:

dXt = −2r

σXtdWt ,

which has the solution

Xt = X0 exp(−2r2

σ2t − 2r

σWt).

Now we have (assuming X0 = Sg0 > 0)

E[|Xt |] = E[Xt] = E[X0 exp(−2r2

σ2t − 2r

σWt )] = X0 exp(−2r2

σ2t +

2r2

σ2t) = X0 < ∞,

which shows that X is a martingale under Q.

�

Sol: 1.1.2 Let Xt = f (t, Wt) = et/2 cos(Wt). We now calculate dXt with Itôs formula:

dXt = df (t, Wt) =

{

∂f

∂t+

12

∂2f

∂x2

}

dt + 1∂f

∂xdWt

=

{

12

et/2 cos(Wt) −12

et/2 cos(Wt )

}

dt − et/2 sin(Wt)dWt

= −et/2 sin(Wt)dWt

9

This process has no drift term moreover we have that the diffusion part has finite second moment using the Itôisometry, i.e.

E[(∫ t

0−es/2 sin(Ws)dWs)

2] =

∫ t

0E[es sin(Ws)

2]ds ≤∫ t

0E[es]ds = et − 1 < ∞.

This gives that X (t) is a Martingale.

Alternative solution:We can directly calculate E[Xt |Fs] for s < t as

E[Xt |Fs] = E[et/2 cos(Wt)|Fs] = E[e(t−s)/2 cos(Wt − Ws + Ws)es/2|Fs]

= E[e(t−s)/2(cos(Wt − Ws) cos(Ws)es/2 − sin(Wt − Ws) sin(Ws)e

s/2)|Fs]

= es/2 cos(Ws)(e(t−s)/2e−(t−s)2/2) + sin(Ws)e

s/20

= es/2 cos(Ws) = Xs.

Finally we need to establish that E[|Xt |] < ∞ which is easily done since |Xt | < et/2 < ∞. �

Sol: 1.2 Static replication of derivatives

Sol: 1.2.1 The easiest way to find a hedge for the pay-off is to draw a picture.

0 K 2K

0K

Smax(S−2K,0)

−2max(S−K,0)

S

The solid line correspond to the original pay-off and the dashed lines correspond to the stock, and one long Europeancall option with strike 2K and two short European call options with strike K respectively (top to bottom). So we canreplicate the pay-off using one stock, and one long European call option with strike 2K and two short European calloption with strike K . To do things properly we should check that

S − 2 max(S − K , 0) + max(S − 2K , 0) = max(K − |S − K |, 0)

for all S ≥ 0. We start with the left hand side

S − 2 max(S − K , 0) + max(S − 2K , 0) =

S 0 ≤ S ≤ K

2K − S K ≤ S ≤ 2K

0 S ≥ 2K

and then move on to the right hand side

max(K − |S − K |, 0) =

S 0 ≤ S ≤ K

2K − S K ≤ S ≤ 2K

0 S ≥ 2K

which shows the equivalence between the two pay-offs. �

10

Sol: 1.2.2 LetΠ (t) be the price of the derivative X for t ≤ T , moreover let PK (t) and CK (t) be the price at time t of a put optionand call option respectively both with maturity T and strike price K . The price Π (t) is given by any of the followingequivalent portfolios (there are more possible equivalent portfolios).

Π (t) = P2K (t) − PK (t)

= K B(t)/B(T )− CK (t) + C2K (t)

= S(t) + PK (t) − 2CK (t) + C2K (t)

Looking at the payoffs at time T show that the portfolios have the same payoff as the contract X , i.e.

K S(T ) ≤ K2K − S(T ) K ≤ S(T ) ≤ 2K0 S(T ) ≥ 2K

�

Sol: 1.3 PDE:s and Feynman-Kac formula

Sol: 1.3.1 According to Feynman-Kac’s representation theorem the PDE is solved by

f (t, x) = E[eXT |Xt = x],

where X has the dynamics

dXs = (μ− σ2

2)ds + σdWs, t ≤ s ≤ T

Xt = x.

It is straightforward to see (at least it should be) that X is is a BM with drift μ− s2

2 and standard deviation σ startingat x at time t , i.e.

XT = x + (μ− σ2

2)(T − t) + σ(WT − Wt).

Looking at exp(XT ) we see that it has exactly the same distribution as the geometric BM in the standard Black-Scholesmodel. We thus get that

f (t, x) = E[eXT |Xt = x] = E

[

exe(m− s2

2 )(T−t)+s(WT−Wt )]

= ex+(m− s2

2 )(T−t)+s2

2 (T−t)= ex+m(T−t).

We should also check the obtained solution fulfils the PDE and the boundary condition. We start with the last taskf (T , x) = ex+0m = ex as prescribed. Finally we get that

∂f (t, x)∂t

+

(

μ− σ2

2

)

∂f (t, x)∂x

+σ2

2∂2f (t, x)

∂x2= −μf (t, x) + (μ− σ

2

2)f (t, x) +

σ2

2f (t, x) = 0,

which verifies that the obtained solution is correct. �

Sol: 1.4 Pricing of derivatives and hedging

Sol: 1.4.1 (a) Under Q we have that

ST = Ste(r−s2

2 )(T−t)+s(WT−Wt ).

11

According to the risk-neutral-valuation-formula we have that the price of the derivative, Πt = F (t, St), is givenby

F (t, St) = e−r(T−t)E

Q[

Φ(ST )|Ft

] Markov= e−r(T−t)

EQ[

Φ(ST )|St

]

= e−r(T−t)E

Q[

max(K − (ST )2, 0)|St

]

= e−r(T−t)E

Q[

(K − (ST )2)1ST≤√

K |St

]

= e−r(T−t)E

Q

[

(K − S2t e2(r− s2

2 )(T−t)+2s(WT−Wt ))1e(r− s2

2 )(T−t)+s(WT −Wt )≤(√

K /St )|St

]

= e−r(T−t)E

Q

[

(K − S2t e2(r− s2

2 )(T−t)+2s√T−tZ )1Z≤

log(√

K /St )−(r− s22 )(T−t)s√T−t

|St

]

,

where Z ∈ N(0, 1). Expressing the expectation as an integral we get

F (t, St) = e−r(T−t)K

∫ d1

−∞

e−z2

2

√2π

dz − S2t e(r−s2)(T−t)

∫ d1

−∞

e2s√T−tz− z2

2

√2π

dz

= e−r(T−t)K

∫ d1

−∞

e−z2

2

√2π

dz − S2t e(r−s2)(T−t)

∫ d1

−∞

e−12 (−4s√T−tz+z2

+4s2(T−t)−4s2(T−t))√

2πdz

= e−r(T−t)K

∫ d1

−∞

e−z2

2

√2π

dz − S2t e(r+s2)(T−t)

∫ d1

−∞

e−12 (z−2s√T−t)2

√2π

dz

= e−r(T−t)K

∫ d1

−∞

e−z2

2

√2π

dz − S2t e(r+s2)(T−t)

∫ d1−2s√T−t

−∞

e−12 z2

√2π

dz

= e−r(T−t)K N(d1) − S2t e(r+s2)(T−t)N(d1 − 2σ

√T − t),

where N is the standard normal distribution function and where

d1 =log(

√K /St ) − (r − s2

2 )(T − t)

σ√

T − t.

(b) Let h = (hb, hs) be a self-financing portfolio with value process V (t) = hb(t)Bt + hs(t)St . By the self-financingcondition we get that the dynamics of V is given by

dV (t) = hb(t)dBt + hs(t)dSt .

We now want to choose hb and hs such V and the derivative with price process Π has the same dynamics. Thedelta-hedge gives that we should choose hs(t) as

hs(t) =∂

∂StF (t, St)

and since we should have Vt = F (t, St) we get that

hb(t) =F (t, St) − hs(t)St

Bt=

F (t, St) − St∂

∂StF (t, St)

Bt.

Using the F obtained in (a) we get that

hs(t) = −2Ste(r+s2)(T−t)N(d1 − 2σ

√T − t)

hb(t) =e−r(T−t)K N(d1) + S2

t e(r+s2)(T−t)N(d1 − 2σ√

T − t)ert

.

Alternative derivation: If you have forgotten the delta-hedge you should look at following to refresh your memory.Applying Itô’s formula to Πt = F (t, St) we get that

dΠt =

(

∂

∂tF (t, St) +

σ2S2t

2∂2

∂S2t

F (t, St)

)

dt +∂

∂StF (t, St)dSt

=

(

∂

∂tF (t, St) +

σ2S2t

2∂2

∂S2t

F (t, St)

)

rBt

rBtdt +

∂

∂StF (t, St)dSt

=

(

∂

∂tF (t, St) +

σ2S2t

2∂2

∂S2t

F (t, St)

)

1rBt

dBt +∂

∂StF (t, St)dSt .

12

So if we choose

hb(t) =

(

∂

∂tF (t, St) +

σ2S2t

2∂2

∂S2t

F (t, St)

)

1rBt

hs(t) =∂

∂StF (t, St)

we get that V and Π will have the same dynamics. We can simplify this further by using that Vt = Πt = F (t, St)(which also follows from that F satisfies the Black-Scholes PDE) so that we get that

hb(t) =F (t, St) − hs(t)St

Bt=

F (t, St) − St∂

∂StF (t, St)

Bt

�

Sol: 1.5 Change of measure and completeness

Sol: 1.5.1 The mathematics in this solution is only simple linear algebra but the financial implications are still noteworthy.

(a) The condition for the risk neutral probabilities are that

S0 = EQ[S1|S0] = qd dS0 + q1S0 + quuS0

and that the risk neutral probabilities sum to one. (We must also have that the probabilities are positive andsmaller than one.) We thus obtain the following linear system of equations for the risk neutral probabilities:

[

1 1 13/5 1 6/5

]

qd

q1

qu

=

[

11

]

.

So we have two equations and three unknowns so there exists multiple (infinitely many) solutions. In principlewe are done here since now we have that the risk neutral probabilities are not unique. We should check that atleast two solutions satisfy that they are non-negative and smaller than one. We however need the full solution in(d) so we solve it now. We start by finding all solutions and then we make restrictions so that the probabilities arepositive and smaller than one. Three unknowns and two equations gives us a one parametric family of solutions.We put qd = θ and aim to express the other probabilities using θ. Plugging this into the our equations above weobtain

[

1 11 6/5

] [

q1

qu

]

=

[

1 − θ1 − 3/5θ

]

.

This has the solution [q1, qu] = [1 − 3θ, 2θ]. We here immediately see that 0 ≤ θ ≤ 1/3 for this to beprobabilities. For financial reasons we do not want the cases θ = 0 (implies S = B with prob 1) and θ = 1/3(which gives us back the binomial model with prob 1). Taking this into consideration gives us the solutions[qd , q1, qu] = [θ, 1 − 3θ, 2θ], 0 < θ < 1/3. We can thus by changing θ calibrate to different marketsituations.

(b) Since we all ready have the full solution all we need to do here is to find θ so that the price of the derivativematch, i.e. we calibrate our model to fit the market price. We have that E

Q[S21 |S0 = 1] = 26/25 which gives

that

θ(3/5)2+ (1 − 3θ)12

+ 2θ(6/5)2= 26/25,

1 + θ(9 − 75 + 72)/25 = 26/25,

1 + 6θ/25 = 26/25

6θ = 1

θ = 1/6.

So we have a unique risk neutral measure with [qd , q1, qu] = [1/6, 1/2, 1/3].

13

(c) We now use the probabilities from (b) to price the derivative

EQ[max(S1 − 9/10, 0)|S0 = 1] = 0 · 1/6 + (1 − 9/10) · 1/2 + (6/5 − 9/10) · 1/3

= 1/20 + 1/10 = 3/20.

To find the hedge portfolio we must solve system of linear equations just as in the binomial model but nowwe have three assets and three equations. So we should find hX , hS and hB (which are the portfolio weights forX = S2, S and B respectively) such that

hX S21 + hSS1 + hBB1 = max(S1 − 9/10, 0),

for all possible outcomes of S1. This gives rise to following system of equations

9/25 3/5 11 1 1

36/25 6/5 1

hX

hS

hB

=

01/103/10

We easily solve this by Gauss elimination obtaining

hX

hS

hB

=

5/4−7/4

3/5

.

(d) From (a) and (b) we get that

EQ[S2

1 |S0 = 1] = 1 + 6θ/25, 0 < θ < 1/3.

This is an increasing function of θ thus we get the upper and lower bounds for the price by plugging in theallowed upper and lower bounds for θ. This gives that

1 < EQ[S2

1 |S0] < 27/25,

which was to be shown. So it is not possible to calibrate the model to prices outside this interval, since thatwould imply negative probabilities and or probabilities larger than one.

Sol: 1.5.2 (a) Regardless of the choice of martingale measure Q we have that S(1) will have the Q-dynamics

dS(1)= rS(1)

t dt + S(1)t

(

σ11dW (1),Qt + σ12dW (2),Q

t

)

,

where W (1),Q and W (2),Q are standard BM:s under Q. We therefore get that

ST = Ste(r−s2

11+s212

2 )(T−t)+s11(W (1),QT −W (1),Q

t )+s12(W (2),QT −W (2),Q

t ),

which has the same distribution as

Ste(r−s2

11+s212

2 )(T−t)+√s2

11+s212

√T−tZ , where, Z ∈ N(0, 1).

Thus the price at time t , Πt , of a simple claim with maturity T and pay-off Φ(ST ) is given by

Πt = e−r(T−t)∫ ∞

−∞Φ

(

Ste(r−s2

11+s212

2 )(T−t)+√s2

11+s212

√T−tz

)

e−z2

2

√2π

dz

regardless of the choice of Q, which was to be shown.

(b) The dynamics of W (2) under Q is given by

dW (2)t = dW (2),Q

t − g2(t)dt

where (g1(t), g2(t)) are any functions satisfying the equation

μ1 − σ11g1(t) − σ12g2(t) = r, for, 0 ≤ t ≤ T

and the Novikov condition

EP

[

eR T

0g1(s)2+g2(s)2

2 ds

]

< ∞.

14

Two possible choices are e.g. (g1(t), g2(t)) = ((μ1 − r)/σ11, 0) and (g1(t), g2(t)) = (0, (μ1 − r)/σ12). These twochoices will give different prices to the derivative 1W (2)

T >K , more precisely

e−r(T−t)N

(

W (2)t − K√T − t

)

and

e−r(T−t)N

(

W (2)t − K − m1−rs12

(T − t)√

T − t

)

respectively.

(c) We get that (g1(t), g2(t)) should solve the following system of linear equations

−σ11g1(t) − σ12g2(t) = r − μ1

−σ22g2(t) = r − μ2

which has the unique solution (provided that σ11σ22 6= 0)

g2(t) =μ2 − r

σ22, g1(t) =

μ1 − r − σ12m2−rs22

σ11,

and since the Girsanov kernel is unique we get that the market is free of arbitrage and complete.

(d) Using the result from (c) we get that the price of the derivative is given as

e−r(T−t)N

(

W (2)t − K − m2−rs22

(T − t)√

T − t

)

.

�

Remark: Note that usually we say that a derivative or contingent claim should only be a function of the (trajec-tory of the) underlying asset (S1) up to maturity. The derivative in (a) is such a contract and there it does notmatter that we have added an extra Brownian motion. However the contract in (b) is not a function of S1 only,here we need more information to know the value at maturity. This ambiguity regarding completeness is thereason why it is called a meta-theorem (see Åberg p. 111 and Björk p. 122) and not a theorem. The moral ofthis is perhaps that we should use a model description that only uses as many driving Brownian motions as weactually need to represent the model.

Sol: 1.6 Change of numeraires

Sol: 1.6.1 We want to price the derivative X with maturity T and pay-off:

Φ(S1(T ), S2(T )) = max(S2(T ) − S1(T ), 0).

According to the risk-neutral valuation formula the price at time t , ΠX (t), is given by

ΠX (t) = e−r(T−t)E

Q[max(S2(T ) − S1(T ), 0)|Ft].

There are now some different approaches to calculate this expectation using change of numeraire techniques. Theperhaps easiest approach is to use S1 as a numeraire this leads to

ΠX (t) = S1(t)EQS1

[

1S1(T )

max(S2(T ) − S1(T ), 0)|Ft

]

= S1(t)EQS1

[

max(S2(T )S1(T )

− 1, 0)|Ft

]

,

where QS1 is the numeraire measure for S1. This can now be seen as a European call option on the ratio S2/S1 withstrike 1. Moreover since S2/S1 is a ratio between a traded asset and a numeraire it is automatically a martingale underQS1 . Using that the volatilities does not change when we change measure we can by calculate the volatilities under Q

as(

∂

∂S1

S2

S1

)

S1(t)(σ11dW1(t) + σ12dW2(t)) +

(

∂

∂S2

S2

S1

)

S2(t)(σ21dW1(t) + σ22dW2(t))

=S2(t)S1(t)

((σ21 − σ11)dW1(t) + (σ22 − σ12)dW2(t))

15

This gives that the dynamics for S2/S1 under QS1 for s ≥ t is given by

dS2(s)S1(s)

=S2(s)S1(s)

(

(σ21 − σ11)dW QS1

1 (s) + (σ22 − σ12)dW QS1

2 (s))

,

S2(t)S1(t)

=s2(t)s1(t)

where W QS1

1 (s) and W QS1

2 (s) are independent standard QS1 Brownian motions. We can now solve this SDE to obtain

S2(T )S1(T )

=s2(t)s1(t)

exp

(

−12

(

(σ21 − σ11)2+ (σ22 − σ12)2

)

(T − t) + (σ21 − σ11)(W QS1

1 (T ) − W QS1

1 (t))

+ (σ22 − σ12)(W QS1

2 (T ) − W QS1

2 (t)))

This has the same distribution as

s2(t)s1(t)

exp

(

−12σ2(T − t) + σ

√T − tG

)

,

where G is a standard Gaussian random variable and where

σ =√

(σ21 − σ11)2 + (σ22 − σ12)2.

We can therefore calculate the price ΠX (t) as

ΠX (t) = s1(t)∫ ∞

−∞max

(

s2(t)s1(t)

exp

(

−12σ2(T − t) + σ

√T − tg

)

− 1, 0

)

exp(−g2/2)√2π

dg

= s1(t)∫ ∞

−d

(

s2(t)s1(t)

exp

(

−12σ2(T − t) + σ

√T − tg

)

− 1

)

exp(−g2/2)√2π

dg

= s1(t)∫ ∞

−d

(

s2(t)s1(t)

1√2π

exp

(

−12σ2(T − t) + σ

√T − tg − g2/2

)

− exp(−g2/2)√2π

)

dg

= s1(t)∫ ∞

−d

(

s2(t)s1(t)

1√2π

exp

(

−12

(g − σ√

T − t)2

)

− exp(−g2/2)√2π

)

dg

= s2(t)∫ ∞

−d

1√2π

exp

(

−12

(g − σ√

T − t)2

)

dg − s1(t)∫ ∞

−d

exp(−g2/2)√2π

dg

= s2(t)(1 − N (−d − σ√

T − t)) − s1(t)(1 − N (−d))

= s2(t)N (d + σ√

T − t) − s1(t)N (d),

where

d =ln(s2(t)/s1(t)) − σ2(T − t)/2

σ√

T − t,

and where N is the distribution function of the standard Gaussian distribution. Alternative solution:We can also use both S1 and S2 as numeraires which gives that

ΠX (t) = S2(t)EQS2 [

I (S2(T ) > S1(T ))|Ft

]

− S1(t)EQS1 [

I (S2(T ) > S1(T ))|Ft

]

.

This can now be rewritten as

ΠX (t) = S2(t)EQS2

[

I

(

S1(T )S2(T )

< 1

)

|Ft

]

− S1(t)EQS1

[

I

(

S2(T )S1(T )

> 1

)

|Ft

]

.

Now S1/S2 is a QS2 -martingale and S2/S1 is a QS1 -martingale. By using the same type of argument and calculationsas in the first solution we obtain that S1(T )/S2(T ) under QS2 has the same distribution as

s1(t)s2(t)

exp

(

−12σ2(T − t) + σ

√T − tG

)

,

and that S2(T )/S1(T ) under QS1 has the same distribution as

s2(t)s1(t)

exp

(

−12σ2(T − t) + σ

√T − tG

)

,

16

where σ and G are as defined in the first solution. Straightforward calculation now give that

ΠX (t) = s2(t)N (d + σ√

T − t) − s1(t)N (d),

where d are defined as in the first solution.Alternative solution 2:We can also calculate the dynamics for S1 and S2 under both QS1 and QS2 . For this we need to find two two-dimensional Girsanov kernels, where we also need to look at dynamics of the bank-account to get the right Girsanovkernels. This is however as seen from the calculations above an unnecessary detour. For the sake of completeness wesupply the appropriate Girsanov kernels:

gS11 = −σ11, gS1

2 = −σ12

gS21 = −σ21, gS2

2 = −σ22.

By the same type of calculations as in the first two solutions we finally arrive at the same answer. �

Sol: 1.6.2 (a) By using that Z (t) = S(t)/P(t, T ), for 0 ≤ t ≤ T , is a QT martingale and that volatilities do note change whenwe change measure, we see that we can calculate the volatility function v(t, T ) using the diffusion part of theQ-dynamics for S(t)/P(t, T ). This gives that

Z (t)v(t, T )

[

dW1(t)dW2(t)

]

=

(

∂

∂P

S

P

)

P(t, T )γ(T − t)dW1(t) +

(

∂

∂S

S

P

)

S(t)σdW2(t)

= −Z (t)γ(T − t)dW1(t) + Z (t)σdW2(t)

= Z (t)[

−γ(T − t) σ]

[

dW1(t)dW2(t)

]

.

This gives that v(t, T ) =[

−γ(T − t) σ]

.

(b) Using that S(T ) = Z (T ) we can view the contract as written on the process Z instead of S. So the price of theEuropean call option can thus be calculated as

Π (t) = p(t, T )EQT [

max(Z (T ) − K , 0)|Ft

]

.

To calculate this we must find the distribution of Z (T ) under QT . Solving the SDE for Z under QT gives that

Z (T ) = Z (t) exp

(

−12

∫ T

t

|v(s, T )|2ds +

∫ T

t

v(s, T )dW QT

(s)

)

.

This now has the same distribution as

Z (t) exp

(

−12Σ(t, T )2(T − t) + Σ(t, T )

√T − tG

)

,

where

Σ(t, T ) =

√

∫ T

t γ2(T − s)2 + σ2ds

T − t=

√

γ2(T − t)3/3 + σ2(T − t)T − t

,

and where G is standard Gaussian random variable. By exactly the same calculations as in the derivation of thestandard Black-Scholes formula but with σ replaced by Σ(t, T ) and e−r(T−t) replace by p(t, T ) we obtain

Π (t) = p(t, T )Z (t)N (d + Σ(t, T )√

T − t) − p(t, T )KN (d)

= S(t)N (d + Σ(t, T )√

T − t) − p(t, T )KN (d),

where

d =ln(Z (t)/K ) − Σ(t, T )2(T − t)/2

Σ(t, T )√

T − t=

ln(S(t)/K ) − ln(P(t, T )) − Σ(t, T )2(T − t)/2

Σ(t, T )√

T − t,

and where N is the distribution function of the standard Gaussian distribution.

17

To check that we get back the original formula if r(t) ≡ r and |v(t, T )| ≡ σ , we notice that this imply thatγ = 0 and p(t, T ) = exp(−r(T − t)). This gives that Σ(t, T ) ≡ σ and − ln(p(t, T )) = r(T − t). Now pluggingthis into our price gives

Π (t) = S(t)N (d + σ√

T − t) − e−r(T−t)KN (d),

where

d =ln(S(t)/K ) + r(T − t) − σ2(T − t)/2

σ√

T − t,

which we recognize as the ordinary Black-Scholes formula. �

Sol: 1.6.3 (a) We start by examining the pay-off:

max(SA(T ), SB(T )) = SA(T )I (SA(T ) ≥ SB(T )) + SB(T )I (SB(T ) > SA(T ))

= SA(T )I

(

SB(T )SA(T )

≤ 1

)

+ SB(T )I

(

SA(T )SB(T )

< 1

)

.

After this rewriting we are ready to attack the problem with RNVF. Let Π (t, SA(t), SB(t)) be the value of t hecontract at time t . Thus we have

Π (t, SA(t), SB(t)) = EQ[

B(t)B(T )

max(SA(T ), SB(T ))|Ft]

= EQ

[

B(t)B(T )

SA(T )I

(

SB(T )SA(T )

≤ 1

)

|Ft

]

+EQ

[

B(t)B(T )

SB(T )I

(

SA(T )SB(T )

< 1

)

|Ft

]

.

Now we apply the change of numeraire technique to simply the calculations:

Π (t, SA(t), SB(t)) = EQSA

[

SA(t)SA(T )

SA(T )I

(

SB(T )SA(T )

≤ 1

)

|Ft

]

+EQSB

[

SB(t)SB(T )

SB(T )I

(

SA(T )SB(T )

< 1

)

|Ft

]

= SA(t)EQSA

[

I

(

SB(T )SA(T )

≤ 1

)

|Ft

]

+ SB(t)EQSB

[

I

(

SA(T )SB(T )

< 1

)

|Ft

]

= SA(t)QSA

(

SB(T )SA(T )

≤ 1|Ft

)

+ SB(t)QSB

(

SA(T )SB(T )

< 1|Ft

)

.

We can now use that SB(u)/SA(u) is a QSA martingale and that SA(u)/SB(u) is a QSB martingale, since they areboth ratios of traded assets and numeraires. We then get the following QSA -dynamics for SB(u)/SA(u) (using Itoand the MG-property)

dSB(u)SA(u)

=SB(u)SA(u)

(

(σBρ− σA)dW QSA

1 (t) + σB

√

1 − ρ2dW QSA

2 (t))

d=

SB(u)SA(u)

√

(σBρ− σA)2 + σ2BdW QSA

(t)

=SB(u)SA(u)

σdW QSA(t),

where σ =√

(σBρ− σA)2 + σ2B and W QSA is a standard QSA BM. So then we get that

SB(T )SA(T )

d=

SB(t)SA(t)

e−s2

2 (T−t)+s√T−tG ,

where G is standard Gaussian random variable. Using the same type of arguments we get the following distribu-tion for SA(u)/SB(u) under QSB .

SA(T )SB(T )

d=

SA(t)SB(t)

e−s2

2 (T−t)+s√T−tG ,

18

where G is standard Gaussian random variable and where σ =√

(σBρ− σA)2 + σ2B as before. Putting this

together we obtain

Π (t, SA(t), SB(t)) = SA(t)QSA

(

SB(t)SA(t)

e−s2

2 (T−t)+s√T−tG ≤ 1|Ft

)

+SB(t)QSB

(

SA(t)SB(t)

e−s2

2 (T−t)+s√T−tG < 1|Ft

)

= SA(t)QSA

G ≤ln(

SA(t)SB (t)

)

+s2

2 (T − t)

σ√

T − t|Ft

+SB(t)QSB

G <ln(

SB(t)SA(t)

)

+s2

2 (T − t)

σ√

T − t|Ft

= SA(t)N

ln(

SA(t)SB (t)

)

+s2

2 (T − t)

σ√

T − t

+ SB(t)N

ln(

SB (t)SA(t)

)

+s2

2 (T − t)

σ√

T − t

= SA(t)N(d1) + SB(t)N(d2).

As a preparation for (d) we also look explicitly at the price at time zero

Π (0, SA(0), SB(0)) = SA(0)N

ln(

SA(0)SB (0)

)

+s2

2 T

σ√

T

+ SB(0)N

ln(

SB (0)SA(0)

)

+s2

2 T

σ√

T

.

(b) The easiest way of showing this is to re-write the value at time T for (b)

Φ(b)(SA(T ), SB(T )) = SA(T ) + (SB(T ) − SA(T ))+

= SA(T ) + (SB(T ) − SA(T ))I (SA(T ) < SB(T ))

= SA(T )I (SA(T ) ≥ SB(T )) + SB(T )I (SA(T ) < SB(T ))

= max(SA(T ), SB(T )).

This was exactly what we needed to show since now the values in (a) and (b) coincide for all possible outcomesat maturity and thus they must also coincide for all previous times to avoid arbitrage.

(c) The Black-Scholes like market in this problem is complete so we can hedge all contingent claims. To find thehedge we use the standardΔ-hedge approach.

hB(t) =1

B(t)(Π (t, SA(t), SB(t)) − hSA (t)SA(t) − hSB (t)SB(t))

hSA (t) =∂

∂SAΠ (t, SA(t), SB(t))

hSB (t) =∂

∂SBΠ (t, SA(t), SB(t))

We start with hSA (t)

hSA (t) =∂

∂SA(SA(t)N(d1) + SB(t)N(d2))

= N(d1) + SA(t)n(d1)∂

∂SA(d1) + SB(t)n(d2)

∂

∂SA(d2)

= N(d1) +1

σ√

T − t

(

SA(t)SA(t)

n(d1) − SB(t)SA(t)

n(d2)

)

= N(d1) +1

σ√

T − t

(

n(d1) − SB(t)SA(t)

n(d2)

)

where n(x) = (d/dx)N (x) = e−x2/2/√

2π,Using almost similar calculations we obtain that

hSB (t) = N (d2) +1

σ√

T − t

(

−SA(t)SB(t)

n(d1) + n(d2)

)

.

19

Using this we finally obtain

hB(t) =1

B(t)(Π (t, SA(t), SB(t)) − hSA (t)SA(t) − hSB (t)SB(t))

=1

B(t)(Π (t, SA(t), SB(t)) − SA(t)N (d1) − SB(t)N (d2)

− 1

σ√

T − t(SA(t)n(d1) − SB(t)n(d2) − SA(t)n(d1) + SB(t)n(d2))

)

=1

B(t)(Π (t, SA(t), SB(t)) −Π (t, SA(t), SB(t))) = 0,

hSA (t) = N (d1) +1

σ√

T − t

(

n(d1) − SB(t)SA(t)

n(d2)

)

,

hSB (t) = N (d2) +1

σ√

T − t

(

−SA(t)SB(t)

n(d1) + n(d2)

)

.

(d) From (a) we get that the price of derivative at time zero is

Π (0, SA(0), SB(0)) = SA(0)N

ln(

SA(0)SB (0)

)

+s2

2 T

σ√

T

+ SB(0)N

ln(

SB (0)SA(0)

)

+s2

2 T

σ√

T

.

Now using that SA(0) = SB(0) = S and that σA = σB = σ we get

Π (0, SA(0), SB(0)) = 2SN

(

σ√

1 − ρ√2

√T

)

.

We can then compare this to buying both stocks which costs 2S. So buying both stocks are always more expen-sive. Now looking at the potential pay-off we have that the derivative gives the best of the two stocks. We havethat the correlation is negative so if one stocks goes up the other goes down compared to the risk free rate. Inmost cases one of the stocks will be worth considerably more than the other so most of the sum of both stockswill come from the best performing stock. So the derivative will in most cases give almost as high pay off asthe sum of the two stocks but with less initial investment. So if our assumption about the correlation is true weshould prefer the derivative to the sum of the two stocks. �

Sol: 1.6.4 In this problem we first observe that we can view the bet as a derivative written on two assets. The key step will be tofind a clever choice of numeraires to avoid calculating two dimensional integrals. According to the general risk neutralvaluation formula we have that the price at time t of a derivative X with maturity T is given as

Π (t, X ) = EQN

[

N (t)N (T )

X |Ft

]

,

where N is a numeraire and QN is the corresponding numeraire measure.

(a) We now take a look at the pay off from Anna’s point of view (pow)

Φ(SA(T ), SB(T )) =

SA(T ) SB(T ) < SA(T ),

0 SB(T ) = SA(T ),

−SB(T ) SA(T ) < SB(T ),

This can be re-written as

Φ(SA(T ), SB(T )) =

SA(T ) SB(T )/SA(T ) < 1,

0 SB(T )/SA(T ) = 1,

SB(T ) SA(T )/SB(T ) < 1,

and we further see that it can also be written as

Φ(SA(T ), SB(T )) = SA(T )I (SB(T )/SA(T ) ≤ 1) − SB(T )I (SA(T )/SB(T ) ≤ 1).

20

Plugging this into the general RNVF gives

Π (t, X ) = EQN

[

N (t)N (T )

Φ(SA(T ), SB(T ))|Ft

]

= EQ

N

[

N (t)N (T )

SA(T )I (SB(T )/SA(T ) ≤ 1)|Ft

]

−EQN

[

N (t)N (T )

SB(T )I (SA(T )/SB(T ) ≤ 1)|Ft

]

So if we choose SA as numeraire in the first expectation and SB as numeraire in the second expectation we get thefollowing simplification

Π (t, X ) = SA(t)ESA[

I (SB(T )/SA(T ) ≤ 1)|Ft

]

− SB(t)ESB[

I (SA(T )/SB(T ) ≤ 1)|Ft

]

. (∗)

Now we have under SA that SB/SA is the ratio of a traded asset and the numeraire and under SB that SA/SB is theratio of a traded asset and the numeraire. This further simplifies our calculations since both ratios are martingalesunder their respective numeraire measure, i.e. SA and SB. So when we calculate the dynamics we only need toconsider the diffusion parts of the dynamics since we know that the drift parts should be zero. First we considerSB/SA under SA

dSB(u)/SA(u) = −(SB(u)/SA(u)2)SA(u)σAdW SA1 (u)

+(1/SA(u))SB(u)(ρσBdW SA1 (u) + σB

√

1 − ρ2dW SA2 (u))

= (SB(u)/SA(u))(ρσB − σA)dW SA1 (u) +

√

1 − ρ2σBdW SA2 (u)).

We then get that

(SB(T )/SA(T )) = (SB(t)/SA(t)) exp{

−(T − t)/2(

(ρσB − σA)2+ (√

1 − ρ2σB)2)

+(ρσB − σA)(W SA1 (T ) − W SA

1 (t)) +√

1 − ρ2σB(W SA2 (T ) − W SA

2 (t))}

d= (SB(t)/SA(t)) exp(−Σ2(T − t)/2 + Σ

√T − tG1),

where Σ2 = (ρσB − σA)2 + (√

1 − ρ2σB)2 = σ2A − 2ρσAσb + σ2

B and where G1 is a standard Gaussian randomvariable.

Using the same type of calculations we get that SA/SB under SB has the distribution

(SA(T )/SB(T ))d= (SA(t)/SB(t)) exp(−Σ2(T − t)/2 + Σ

√T − tG2),

where Σ2 = σ2A − 2ρσAσb + σ2

B and where G2 is a standard Gaussian random variable. We can now plug thisinto the pricing equ ation (∗) to obtain

Π (t, X ) = SA(t)∫ ∞

−∞I ((SB(t)/SA(t))e−S2(T−t)/2+Sx ≤ 1)

e−x2

2

√2π

dx

−SB(t)∫ ∞

−∞I ((SA(t)/SB(t))e−S2(T−t)/2+Sx ≤ 1)

e−x2

2

√2π

dx

= SA(t)∫ d1

−∞

e−x2

2

√2π

dx − SB(t)∫ d2

−∞

e−x2

2

√2π

dx (∗∗)

= SA(t)N(d1) − SB(t)N(d2),

where

d1 =ln(SA(t)/SB(t)) + Σ2(T − t)/2

Σ√

T − t, d2 =

ln(SB(t)/SA(t)) + Σ2(T − t)/2

Σ√

T − t

and where N is the distribution function of a standard Gaussian random variable. The value from Belles pow isjust minus the value from Anna’s pow. Using that SA(0) = SB(0) we obtain that the value at time zero from bothAnna’s and Belle’s pow is zero. So the value at time zero for both participants is zero and they both also make azero investment at time zero so we can see the bet as fair.

(b) See the calculations in (a).

21

(c) To find a replicating portfolio we want to find a self-financing portfolio consisting of the assets in our market thathas the same dynamics as the derivative. Let V be the value of the self-financing portfolio V (t) = a0(t)B(t) +

a1(t)SA(t) + a2(t)SB(t). The self-financing condition gives that the dynamics of V is

dV (t) = a0(t)dB(t) + a1(t)dSA(t) + a2(t)dSB(t)

We now look at the dynamics of Π and it is given by

dΠ (t, X ) = Πt (t, X )dt +ΠSA (t, X )dSA(t) +ΠSB (t, X )dSB(t) +ΠSASA (t, X )(dSA(t))2/2

+ΠSASB (t, X )(dSA(t))(dSB(t)) +ΠSBSB (t, X )(dSB(t))2/2, (∗ ∗ ∗)

where

Πt (t, X ) =∂

∂tΠ (t, X )

ΠSA (t, X ) =∂

∂SAΠ (t, X )

ΠSB (t, X ) =∂

∂SBΠ (t, X )

ΠSASA (t, X ) =∂2

∂S2A

Π (t, X )

ΠSASB (t, X ) =∂2

∂SA∂SBΠ (t, X )

ΠSBSB (t, X ) =∂2

∂S2B

Π (t, X )

Using that Π satisfies the Black-Scholes equation we obtain that

Πt (t, X )dt +ΠSASA (t, X )(dSA(t))2/2 +ΠSASB (t, X )(dSA(t))(dSB(t)) +ΠSBSB (t, X )(dSB(t))2/2

= r(Π (t, x) − SA(t)ΠSA(t, X ) − SB(t)ΠSB (t, X ))dt

Plugging this into (***) gives

dΠ (t, X ) = r(Π (t, X ) − SA(t)ΠSA(t, X ) − SB(t)ΠSB (t, X ))dt +ΠSA (t, X )dSA(t) +ΠSB (t, X )dSB(t).

Comparing with the dynamics for V we obtain

a0(t) = (Π (t, x) − SA(t)ΠSA(t, X ) − SB(t)ΠSB (t, X ))/B(t),

a1(t) = ΠSA (t, X ),

a2(t) = ΠSB (t, X ).

This is the general formula for hedging a derivative written on two assets for the model in this problem. Thisis in fact true for any complete market consisting of two risky assets and a bank account.This is just the usua ldelta-hedge in two dimensions and could have been considered as given. he derivation of the formula was donemostly in order for the solution to be complete. We can now calculate the exact portfolio weights using theformula for Π . We right up the hedge from Belle’s pow which is t he portfolio that replicates the pay-off fromAnna’s pow. The hedge from Anna’s pow is just the reversed positions. We start with ΠSA (t, X )

ΠSA (t, X ) =∂

∂SAΠ (t, X )

=∂

∂SA(SA(t)N(d1) − SB(t)N(d2))

= N(d1) + SA(t)n(d1)∂

∂SA(d1) − SB(t)n(d2)

∂

∂SA(d2)

= N(d1) + 1/(Σ√

T − t)(

SA(t)n(d1)/SA(t) + SB(t)n(d2)/SA(t))

where n(x) = (d/dx)N (x) = e−x2/2/√

2π,

22

We now plug this into the previous equation and using the formula for d1 and d2 we obtain

ΠSA (t, X ) = N(d1)

+1

Σ√

T − texp

−

ln(

SA(t)SB(t) )

)2+ (Σ2(T − t)/2)2

2Σ2(T − t)

(

exp

(

− ln

(

SA(t)SB(t)

)

/2

)

+SB(t)SA(t)

exp

(

− ln

(

SB(t)SA(t)

)

/2

))

= N (d1)

+2

Σ√

T − texp

−

ln(

SA(t)SB(t) )

)2+ (Σ2(T − t)/2)2

2Σ2(T − t)

exp

(

− ln

(

SA(t)SB(t)

)

/2

)

= N (d1) +2

Σ√

T − tn(d1).

Using almost similar calculations we obtain that

ΠSB (t, X ) = −N (d2) − n(d2)

Σ√

T − t− SA(t)

SB(t)n(d1)

Σ√

T − t

= −N (d2) − SA(t)SB(t)

2

Σ√

T − tn(d1).

Using this we finally obtain

a0(t) =

(

Π (t, x) − SA(t)N (d1) + SB(t)N (d2) − 2

Σ√

T − t

(

SA(t)n(d1) − SB(t)SA(t)SB(t)

n(d1)

))

/B(t)

= (Π (t, X ) −Π (t, X ))/B(t) = 0,

a1(t) = N (d1) +2

Σ√

T − tn(d1),

a2(t) = −N (d2) − SA(t)SB(t)

2

Σ√

T − tn(d1).

The fact that we here always should hold a zero amount in the bank account is due to the specific nature of thecontract. We also see that price and the hedge do not depend on the interest rate.

(d) If one should hedge or not has multiple aspects. The first is that among friends it is not considered “commes ilfaut” (appropriate) to hedge a bet. Moreover it would not be much of a bet if you neither could gain nor loseanything. In addition to this I guess you would not enter into the bet without believing in your own stock. Evenif we disregard these aspects there is one problem with the hedge defined in (c). If t is close to T and SA(t) isclose to SB(t) the hedge weights will assume very large values. The hedge will be long in SA and short in SB withreally large positions. This problem comes from that the pay-off is discontinous at SA = SB . This lead to largerisks if there is any trouble with liquidity in either SA or SB . One way out of this is to use a partial static hedge.Belle could for instance buy SA and sell SB which is a zero intial investment since SA(0) = SB(0). At maturity shewill then have −SAI (SA > SB) + SBI (SB > SA) + SA − SB = SAI (SA < SB) − SBI (SB < SA) this positionhasless risk than the original bet. It is however somewhat strange from Belle’s pow if she really believes in here stock.The bottomline is that you should probably not hedge the bet.

�

Sol: 1.7 Volatility

Sol: 1.7.1 (a) Using thate−r(t+t−t)

EQ[S(t + τ)|Ft] = S(t)

we get thatF t,t+t

t = EQ[S(t + τ)|Ft] = ertS(t).

23

(b) By applying Ito’s formula to ln(S(u)) we get that

d ln(S(u)) = (r − σ2(u)/2)du + σ(u)dW (u).

We solve this by direct integration to obtain

ln(S(t + τ)) = ln(S(t)) + rτ−∫ t+t

t

σ2(u)/2du +

∫ t+tt

σ(u)dW (u).

Using that the last term is a martingale we obtain

EQ[ln(S(t + τ))|Ft] = ln(S(t)) + rτ−

∫ t+tt

σ2(u)/2du.

This together with the result in (a) then gives that

EQ

[

−2τ

ln(S(t + τ)/F t,t+tt )|Ft

]

= −2τ

(

ln(S(t)) + rτ− 12

∫ t+tt

σ2(u)du − ln(S(t)) − rτ

)

=1τ

∫ t+tt

σ2(u)du.

So here we see that the VIX-index squared really is an approximation of the average squared future volatility.

(c) To ease the notational complexity we put S = S(t + τ) and F = F t+tt . We start by direct calculation of the

integrals

∫ F

0(K − S)+

dK

K 2+

∫ ∞

F

(S − K )+dK

K 2=

∫ F

S

(K − S)dK

K 2I (S < F ) +

∫ S

F

(S − K )dK

K 2I (F < S)

=

∫ F

S

1K

− S

K 2dKI (S < F ) +

∫ S

F

S

K 2− 1

KdKI (F < S)

=

[

ln(K ) +S

K

]F

S

I (S < F ) +

[

− S

K− ln(K )

]F

S

I (F < S)

= (ln(F ) − ln(S) +S

F− 1)(I (S < F ) + I (F < s))

= ln(F ) − ln(S) +S

F− 1 = − ln(S/F ) +

S

F− 1.

Taking conditional expectation and using the definition of F t+tt we obtain

EQ[− ln(S(t+τ)/F t+t

t )+S(t + τ

F t+tt

−1|Ft] = EQ[− ln(S(t+τ)/F t+t

t )|Ft]+F t+t

t

F t+tt

−1 = EQ[− ln(S(t+τ)/F t+t

t )|Ft],

which shows that

EQ

[

∫ F t+tt

0(K − S(t + τ))+

dK

K 2+

∫ ∞

F t+tt

(S(t + τ) − K )+dK

K 2| Ft

]

= EQ

[

− ln

(

S(t + τ)

F t+tt

)

| Ft

]

.

�

Sol: 1.7.2 By applying Ito’s formula to ln(S(u)) we get that

d ln(S(u)) = (r − V (u)/2)du +√

V (u)(dW1(u) +√

1 − ρ2dW2(u)).

We solve this by direct integration to obtain

ln(S(t + τ)) = ln(S(t)) + rτ−∫ t+t

t

V (u)/2du +

∫ t+tt

√

V (u)(dW1(u) +√

1 − ρ2dW2(u)).

Using that the last term is a martingale we obtain

EQ[ln(S(t + τ))|Ft] = ln(S(t)) + rτ− E

Q

[∫ t+tt

V (u)/2du|Ft

]

.

24

This together with the that F t+tt = ertS(t) then gives that

EQ

[

−2τ

ln(S(t + τ)/F t,t+tt )|Ft

]

= −2τ

(

ln(S(t)) + rτ− EQ

[

12

∫ t+tt

V (u)du|Ft

]

− ln(S(t))− rτ

)

=1τE

Q

[∫ t+tt

V (u)du|Ft

]

.

Now to calculate this we need to use the dynamics of V . Using direct integration we see that we can representV (u), u > t as

V (u) = V (t) +

∫ u

t

κ(θ− V (s))dss +

∫ u

t

√

V (s)σvdW1(s).

Taking conditional expectation on both sides using that the last term is a martingale we obtain

EQ[

V (u)|Ft

]

= EQ

[

V (t) +

∫ u

t

κ(θ− V (s))ds|Ft

]

= V (t) + κθ(u − t) − κEQ

[∫ u

t

V (s)ds|Ft

]

.

Now put m(u) = EQ[

V (u)|Ft

]

we then get

m(u) = V (t) +

∫ u

t

κ(θ− m(s))ds.

Taking derivatives w.r.t u of both sides we obtain the ODE

m(u) = κ(θ− m(u))

m(t) = V (t).

Making the change of variables m(u) = θ− m(u) we obtain a standard linear ODE

˙m(u) = −κm(u))

m(t) = θ− V (t).

We solve this straight away yielding˜m(u) = (θ− V (t))e−k(u−t)

and then we get thatm(u) = θ− (θ− V (t))e−k(u−t).

We can now finally calculate

1τE

Q

[∫ t+tt

V (u)du|Ft

]

=1τ

∫ t+tt

m(u)du

=1τ

∫ t+tt

θ− (θ− V (t))e−k(u−t)du

= θ− θ− V (t)κτ

(1 − e−kt)= V (t)

(1 − e−kt)κτ

+ θe−kt − 1 + κτ

κτ

Note that this expression tends to V (t) as τ tends to zero and it tends to θ as τ tends to infinity. This makes sense sinceV (t) is the current squared volatility and θ is the long term mean of squared volatility.

Sol: 1.8 Simple interest rate contracts and Martingale models for the short rate

Sol: 1.8.1 The price of the ZCB, p(t, T ), is given by

p(t, T ) = E

[

e−R T

tr(s)ds]

.

The short rate r(s) for s > t is given by

r(s) = r +

∫ s

t

Θ(u)du +

∫ s

t

σdW (u) = r − c(e−s − e−t) +σ2

2(s2 − t2) + σ

∫ s

t

dW (u) .

25

The integral of the short rate r(s) between t and T is then given by

−∫ T

t

r(s)ds = −∫ T

t

(

r − c(e−s − e−t) +σ2

2(s2 − t2) + σ

∫ s

t

dW (u)

)

ds

= −r(T − t) − c(e−T − e−t ) − ce−t (T − t) − σ2

6(T 3 − t3) +

σ2

2t2(T − t) − σ

∫ T

t

∫ s

t

dW (u)ds .

The expectation of the stochastic term is

E

[

∫ T

t

∫ s

t

dW (u)ds

]

= 0 ,

and the variance is

V

[

−∫ T

t

∫ s

t

dW (u)ds

]

= E

(

−∫ T

t

∫ s

t

dW (u)ds

)2

= E

(

∫ T

t

∫ T

u

dsdW (u)

)2

= E

(

∫ T

t

(T − u)dW (u)

)2

=

∫ T

t

(T − u)2du =

[

− (T − u)3

3

]T

t

=(T − t)3

3.

We see that the stochastic term has a normal distribution, and therefore

−σ∫ T

t

∫ s

t

dW (u)ds ∼ N

(

0,σ2 (T − t)3

3

)

.

Moreover we have that

E

[

e−s R

Tt

R

st

dW (u)ds]

= es2

6 (T−t)3

.

This gives that p(t, T ) is given by

p(t, T ) = exp{

−r(T − t) − c(e−T − e−t) − ce−t (T − t)

− σ2

6(T 3 − t3) +

σ2t2

2(T − t) +

σ2

6(T − t)3

}

.

�

Sol: 1.8.2 Let X (T0) denote the value of the floating rate bond at time To. We start by noting that the value of the floating ratebond is just all expected cash flows discounted back to time T0. This gives that

X (T0) = EQ

[

B(To)B(Tn)

A +

n∑

i=1

B(To)B(Ti)

A(Ti − Ti−1)LTi−1 [Ti−1, Ti]|FT0

]

= EQ

[

B(To)B(Tn)

A|FT0

]

+

n∑

i=1

EQ

[

B(To)B(Ti)

A(Ti − Ti−1)LTi−1[Ti−1, Ti]|FT0

]

= EQ

[

B(To)B(Tn)

A|FT0

]

+

n∑

i=1

EQ

[

B(To)B(Ti)

A

(

1p(Ti−1, Ti)

− 1

)

|FT0

]

Now we change numeraires so that each cash flow is evaluated under the corresponding forward measure, that is a cashflow at time Ti is evaluated using the measure QTi with numeraire p(t, Ti). This then gives

X (T0) = p(T0, Tn)EQTn [

A|FT0

]

+

n∑

i=1

p(T0, Ti)EQTi

[

A

(

1p(Ti−1, Ti)

− 1

)

|FT0

]

.

26

Now using that 1p(Ti−1,Ti)

=P(Ti−1,Ti−1)

p(Ti−1,Ti)is martingale under QTi we get

X (T0) = Ap(T0, Tn) + A

n∑

i=1

p(T0, Ti)

(

p(To, Ti−1)p(T0, Ti)

− 1

)

= Ap(T0, Tn) + A

n∑

i=1

p(To, Ti−1) − p(T0, Ti)

telescoping sum= Ap(T0, Tn) + A(p(T0, T0) − p(T0, Tn)) = Ap(T0, T0) = A.

So this floating rate bond always has a value equal to the face value A at time T0 for all arbitrage free models. �

Sol: 1.9 The HJM framework

Sol: 1.9.1 (a) Using that

f (t, u) = f (0, u)⋆ +

∫ t

0σ2(u − s)ds +

∫ t

0σdW Q

s

= f ⋆(0, u) − σ2

2((u − t)2 − u2) + σW Q

t

and thatp(t, T ) = e−

R Tt

f (t,u)du

we get that

p(t, T ) = e−R T

tf ⋆(0,u)du+ s2

2

R Tt

(u−t)2−u2du−s R Tt

W Qt du

= e−R

Tt

f ⋆(0,u)du+ s2

2(T−t)3−T 3+t3

3 −(T−t)sW Qt

= e−R T

tf ⋆(0,u)du− s2

2 tT (T−t)−(T−t)sWQt .

Note that p⋆(0, T )/p⋆(0, t) = exp(−∫ T

t f ⋆(0, u)du) and that r(t) = f (t, t) = f ⋆(0, t) + t2σ2/2 + σW Qt , so we

can further simplify the expression as

p(t, T ) =p⋆(0, T )p⋆(0, t)

exp

(

(T − t)f ⋆(0, t) − σ2

2t(T − t)2 − (T − t)r(t)

)

,

which is the Ho-Lee model’s ZCB-price when calibrated to the initial forward curve. This form is however, not so wellsuited for further calculations.

(b) Using the definition of Xt and Lt [T , S] we see that

Xt = 1 + (S − T )p(t, T )− p(t, S)(S − T )p(t, S)

=p(t, T )p(t, S)

.

If we now use the result from (a) we get that

Xt = e−R

Tt

f ⋆(0,u)du−s2

2 tT (T−t)−(T−t)sW Qt +

R

St

f ⋆(0,u)du s2

2 tS(S−t)+(S−t)sW Qt

= eR S

Tf ⋆(0,u)du+ s2

2 t(S(S−t)−T (T−t))+(S−T )sW Qt

= eR S

Tf ⋆(0,u)du− s2

2 t(S−T )(S+T−t)+(S−T )sW Qt

To find the dynamics under QS we note that Xt is the ratio of the traded asset p(t, T ) and the numeraire p(t, S)therefore we must have that Xt is a martingale under QS . Therefore we only need to calculate the diffusion partof the dynamics since we know that the drift part must be zero, doing this we obtain that Xt has the followingdynamics under QS

dXt = σ(S − T )XtdW QS

t ,

where W QS

t is a standard BM under QS . This gives that

XT = Xte− s2(S−T )2

2 (T−t)+(S−T )s(W QS

T −W QS

t )

27

(c) We start by observing that we can express the pay-off in terms of Xt instead of Lt [T , S] giving that

(S − T ) max(LT [T , S] − K , 0) = (S − T ) max

(

XT − 1S − T

− K , 0

)

= max(XT − (1 + (S − T )K ), 0).

This can now be seen as a standard European call option on XT with strike level (1 + (S − T )K ), except that wewill not get paid at until time S as opposed to time T in the standard case. Using the RNVF for the numerairemeasure QS we get that the price of the Caplet at time t Πt for 0 ≤ t ≤ T is given by

Πt = p(t, S)EQS [

max(XT − (1 + (S − T )K ), 0)|Ft

]

.

Using the result from (b) we get that

Πt = p(t, S)XtN(d1) − p(t, S)(1 + (S − T )K )N(d2) = p(t, T )N(d1) − p(t, S)(1 + (S − T )K )N(d2),

where d1 = (log(

Xt/(1 + (S − T )K ))

+ (S − T )2σ2(T − t)/2)/((S − T )σ√

T − t) andd2 = d1 − (S − T )σ

√T − t.

�

Sol: 1.9.2 (a) We should calculate the fair value of Y = 1 + (T2 − T1)LT2 [T2, T3] at time T1. Using the formula

1 + (S2 − S1)Lt [S1, S2] = p(t, S1)/p(t, S2)

where we put S1 = T2, S2 = T3 and t = T2 we can express Y in terms of ZCB values as

Y = 1 + (T2 − T1)LT2 [T2, T3] =p(T2, T2)p(T2, T3)

.

To find the fair value at time T1 we use the change of trick numeraire trick. So we get that

Π Y (T1) = EQN

[

N (T1)N (T2)

Y |FT1

]

= EQN

[

N (T1)N (T2)

p(T2, T2)p(T2, T3)

|FT1

]

.

A reasonable choice of numeraire is to use P(t, T3), which gives that we use the numeraire measure QT3 . Thisalso is the measure under which the dynamics for the forward rate is given in the problem. We thus obtain

Π Y (T1) = EQT3

[

p(T1, T3)p(T2, T3)

p(T2, T2)p(T2, T3)

|FT1

]

= p(T1, T3)EQT3

[

1p(T2, T3)2

|FT1

]

.

We now re-express this using forward rates

Π Y (T1) = e−R T3

T1f (T1,u)du

EQT3

[

e2R T3

T2f (T2,u)du|FT1

]

= e−

R T3T1

f (T1,u)duE

QT3

[

e2

R T3T2

f (T1,u)+R T2

T1ds f (s,u)du|FT1

]

= e−R T3

T1f (T1,u)du+2

R T3T2

f (T1,u)duE

QT3

[

e2R T3

T2

R T2T1

ds f (s,u)du|FT1

]

= eR T3

T2f (T1,u)du−

R T2T1

f (T1,u)duE

QT3

[

e2

R T3T2

R T2T1

ds f (s,u)du|FT1

]

.

So far the calculations hold for all models. We now plug in the dynamics given in the problem and use that

28

T3 − T2 = T2 − T1. We then get that

Π Y (T1) = eR T3

T2a+b(1−e−(u−T1))du−

R T2T1

b(1−e−(u−T1))duE

QT3

[

e2R T3

T2

R T2T1

−s(s)2(T3−u)ds+R T2

T1s(s)dW (s)du|FT1

]

=e−R T3

T2be−(u−T1)du+

R T2T1

be−(u−T1)duE

QT3

[

e−2

R T2T1

“

R T3T2

(T3−u)du”s(s)2ds+2

“

R T3T2

du”

R T2T1

s(s)dW (s)|FT1

]

= eb(e−(T3−T1)−e−(T2−T1))+b(1−e−(T2−T1))E

QT3[

e−R T2

T1(T3−T2)2s(s)2ds+2(T3−T2)

R T2T1

s(s)dW (s)|FT1

]

T3−T2=T2−T1= eb(e−2(T2−T1)−2e−(T2−T1)

+1)E

QT3[

e−(T2−T1)2 R T2T1

s(s)2ds+2(T2−T1)R T2

T1s(s)dW (s)|FT1

]

= eb(1−e−(T2−T1))2

EQT3[

e−

R T2T1

(T2−T1)2s(s)2ds+2(T2−T1)R T2

T1s(s)dW (s)|FT1

]

= eb(1−e−(T2−T1))2

EQT3

[

e−(T2−T1)2 R T2

T1s(s)2ds+2(T2−T1)

q

R T2T1

s(s)2dsG |FT1

]

, G ∈ N(0, 1)

= eb(1−e−(T2−T1))2+(T2−T1)2(

R T2T1

s(s)2ds)(−1+2)

= eb(1−e−(T2−T1))2

+(T2−T1)2(R T2

T1s(s)2ds)

So we get that

Π Y (T1) = eb(1−e−(T2−T1))2

+(T2−T1)2(R T2

T1s(s)2ds)

.

(b) Plugging that that b = −0.01, σ(t) = σ = 0.1 and that T2 − T1 = 0.25 we get that

Π Y (T1) = e−0.01(1−e−0.25)2+(0.25)3·0.01 ≈ 0.9997.

So the value is slightly less than one.

(c) Since the value of Y is slightly less than one the bank will on average pay out less than the contract X which hasvalue one. This mean that the bank will on average make a small profit on the error. The gain is due to thatthe term structure for the forward rate is decreasing. Testing the stability of the result by slightly changing themodel parameters reveal that the result is not very stable. If we instead had b = 0.01 then the value of Y wouldbe ≈ 1.0006 and the bank would on average lose instead. Keeping b at -0.01 and increasing σ to 0.2 would give≈ 1.0001 so the bank would also here lose. We thus see that it is quite a delicate matter if the bank would winor lose on the error.

�

Sol: 1.10 Dividend paying stocks

Sol: 1.10.1 (a) Since S is a GBM, straight forward calculations give

S(u) = S(t) exp((α−σ2/2)(u−t)+σ(W (u)−W (t)))d= S(t) exp((α−σ2/2)(u−t)+σ

√u − tG), G ∈ N(0, 1), u > t.

Using this we obtain

EQ

[

S(T )B(t)B(T )

+

∫ T

t

qB(t)B(u)

S(u)du|Ft

]

posintegrand= E

Q[

S(t) exp((α− r − σ2/2)(T − t) + σ√

T − tG)|Ft

]

+

∫ T

t

EQ[

qS(t) exp((α− r − σ2/2)(u − t) + σ√

u − tG)|Ft

]

du

= S(t) exp((α− r)(T − t)) +

∫ T

t

qS(t) exp((α− r)(u − t))du

= S(t)

(

exp((α− r)(T − t))(1 +q

α− r) − q

α− r

)

.

This expression should equal S(t) for all 0 < t < T . This is possible only if

q

α− r= 1,

29

which gives thatα = r − q.

Alternative solution: We can using the Ito formula express S(T )B(t)/B(T ) as

S(T )B(t)B(T )

= S(t) +

∫ T

t

d

(

S(u)B(t)B(u)

)

= S(t) +

∫ T

t

B(t)B(u)

dS(u) + S(u)dB(t)B(u)

= S(t) +

∫ T

t

B(t)B(u)αS(u)du +

B(t)B(u)σS(u)dW Q(u) − rS(u)

B(t)B(u)

du

= S(t) +

∫ T

t

B(t)B(u)

(α− r)S(u)du +B(t)B(u)σS(u)dW Q(u).

This gives that

S(T )B(t)B(T )

+

∫ T

t

B(t)B(u)

dD(u)

= S(t) +

∫ T

t

B(t)B(u)

(α− r)S(u)du +B(t)B(u)σS(u)dW Q(u) +

∫ T

t

B(t)B(u)

qS(u)du

= S(t) +

∫ T

t

B(t)B(u)

(α− r + q)S(u)du +B(t)B(u)σS(u)dW Q(u).

We when immediately see that the integral will be a good candidate for being a martingale if α − r + q = 0,which is equivalent to α = r − q. Plugging this α into the SDE for S and observing that S is a GBM we straightforward see that

∫ T

t

B(t)B(u)σS(u)dW Q(u)

is martingale (using the Ito isometry, since GBM:s have finite moments of all orders) so that

EQ

[

S(T )B(t)B(T )

+

∫ T

t

qB(t)B(u)

S(u)du|Ft

]

= EQ

[

S(t) +

∫ T

t

B(t)B(u)σS(u)dW Q(u)|Ft

]

= S(t).

(b) According to the fundamental theorems of asset pricing we have: A model is free of arbitrage if and only ifthere exists at least one probability measure Q such that any discounted traded asset is a martingale under Q.Moreover: If a model is free of arbitrage then it is complete if and only if Q is unique. Since we have a uniquesolution in a) the market is free of arbitrage and complete.

(c) Using the result in a) we get

Π (t) = EQ

[

S(T )B(t)B(T )

|Ft

]

= S(t) exp((r − q − r)(T − t))

= S(t) exp(−q(T − t))

(d) Following the hint we get

a(t) =∂

∂S(t)Π = exp(−q(T − t))

b(t) =Π (t) − a(t)S(t)

B(t)=

S(t) exp(−q(T − t)) − S(t) exp(−q(T − t))B(t)

= 0.

This now gives us the portfolio (a(t), b(t)) = (exp(−q(T − t)), 0) with value process

V (t) = a(t)S(t) + b(t)B(t) = S(t) exp(−q(T − t)).

30

We can now check the self-financing condition

dV (t) = a(t)dS(t) + a(t)dD(t) + b(t)dB(t).

We start with the LHS

dV (t) = exp(−q(T − t))dS(t) + q exp(−q(T − t))S(t)dt.

We then calculate the RHS

a(t)dS(t) + a(t)dD(t) + b(t)dB(t) = exp(−q(T − t))dS(t) + exp(−q(T − t))qS(t)dt + 0

= exp(−q(T − t))dS(t) + exp(−q(T − t))qS(t)dt.

Thus we have LHS=RHS which was to be shown.

(e) So we should here find the valueΠ cE of a standard European call on stock paying continuous dividends. Accord-

ing to the RNVF we have

Π cE (t) = E

Q

[

B(t)B(T )

max(S(T ) − K , 0)|Ft

]

= EQ

[

B(t)B(T )

max(S(t) exp((r − q − σ2/2)(T − t) + σ√

T − tG) − K , 0)|Ft

]

,

G ∈ N(0, 1), T > t

=

∫ ∞

−∞

B(t)B(T )

max(S(t) exp((r − q − σ2/2)(T − t) + σ√

T − tx) − K , 0)exp(−x2/2)√

2πdx

=

∫ ∞

−∞S(t) exp(−q(T − t))

exp(

− 12 (σ2(t − t) − 2σ

√T − tx + x2)

)

)√2π

I (x > −d)dx

−∫ ∞

−∞K exp(−r(T − t))

exp(−x2/2)√2π

I (x > −d)dx

=

∫ ∞

−d−s√T−t

S(t) exp(−q(T − t))exp(−x2/2)√

2πdx

−∫ ∞

−d

K exp(−r(T − t))exp(−x2/2)√

2πdx

= S(t) exp(−q(T − t))(1 − N (−d − σ√

T − t)) − K exp(−r(T − t))(1 − N (−d))symmetri

= S(t) exp(−q(T − t))N (d + σ√

T − t)) − K exp(−r(T − t))N (d),

where

d =ln(

S(t)K

)

+ (r − q − s2

2 )(T − t)

σ√

T − t

= =

ln(

S(t) exp(−q(T−t))K

)

+ (r − s2

2 )(T − t)

σ√

T − t

and where N is the cumulative distribution function of the standard Gaussian disytribution. Comparing thiswith the standard BS formula we see that putting q = 0 the formula reduces to the standard BS formula.Moreover we see that by replacing S(t) by S(t) exp(−q(T − t)) in the original BS formula (this observation is infact true for all simple claims, but it is in general not true for path dependend options), we obtain the formulawe calculated above and since the original BS formula is montonically increarsing in S we get that the Europeancall options always are cheaper with dividends than without.

(f ) So we want to find a self-financing portfolio (a(t), b(t)) with value process V (t) = a(t)S(t) + b(t)B(t) whichreplicates the call option. Using the result from e) that by replacing S(t) by S(t) exp(−q(T − t)) in the originalBS formula we obtain the formula we calculated above, the hedge calculation in the standard BS case and thechain rule we immediately obtain

a(t) =∂

∂S(t)Π c

E (t) = exp(−q(T − t))N (d + σ√

T − t))

b(t) =Π c

E (t) − a(t)S(t)B(t)

= −K exp(−rT )N (d).

�

31

Sol: 1.10.2 (a) We can using the Ito formula express S(T )B(t)/B(T ) as

S(T )B(t)B(T )

= S(t) +

∫ T

t

d

(

S(u)B(t)B(u)

)

= S(t) +

∫ T

t

B(t)B(u)

dS(u) + S(u)dB(t)B(u)

= S(t) +

∫ T

t

B(t)B(u)αS(u)du +

B(t)B(u)σS(u)dW Q(u) − δS(u−)dN (u) − rS(u)

B(t)B(u)

du

= S(t) +

∫ T

t

B(t)B(u)

(α− r)S(u)du +B(t)B(u)σS(u)dW Q(u) − δS(u−)dN (u).

This gives that

S(T )B(t)B(T )

+

∫ T

t

B(t)B(u)

dD(u)

= S(t) +

∫ T

t

B(t)B(u)

(α− r)S(u)du +B(t)B(u)σS(u)dW Q(u) − B(t)

B(u)δS(u−)dN (u) +

∫ T

t

B(t)B(u)δS(u−)dN (u)

= S(t) +

∫ T

t

B(t)B(u)

(α− r)S(u)du +B(t)B(u)σS(u)dW Q(u).

We when immediately see that the integral will be a good candidate for being a martingale if α− r = 0, which isequivalent to α = r. Since S is positive and only jumps downwards, looking at another process Y which behaveslike S but without the jumps, we find will make Y is always larger than S. Now Y is a standard GBM with allmoments finite so then S most also have all moments finite. Using this we straight forward see that

∫ T

t

B(t)B(u)σS(u)dW Q(u)

is martingale (using the Ito isometry) so that

EQ

[

S(T )B(t)B(T )

+

∫ T

t

δB(t)B(u)

S(u−)dN (u)|Ft

]

= EQ

[

S(t) +

∫ T

t

B(t)B(u)σS(u)dW Q(u)|Ft

]

= S(t).

(b) According to the fundamental theorems of asset pricing we have: A model is free of arbitrage if and only ifthere exists at least one probability measure Q such that any discounted traded asset is a martingale under Q.Moreover: If a model is free of arbitrage then it is complete if and only if Q is unique. Since we have a uniquesolution in a) the market is free of arbitrage and complete.

(c) To get out a formula for the stock process we apply the Ito formula to ln(S(u)). This gives us

d ln(S(u)) = r − σ2/2du + σdW Q(u) + (ln(S(u−)(1 − δ)) − ln(S(u−))dN (u)

= r − σ2/2du + σdW Q(u) + ln(1 − δ)dN (u).

Integrating this from t to T we get

ln(S(T )) = ln(S(u)) +

∫ T

t

d ln(S(u))

= ln(S(u)) + (r − σ2/2)(T − t) + σ(W Q(T ) − W Q(t)) + ln(1 − δ)(N (T ) − N (t)).

Taking the exponential of both sides we obtain

S(T ) = S(t) exp(ln(1 − δ)(N (T ) − N (t))) exp((r − σ2/2)(T − t) + σ(W Q(T ) − W Q(t)).

Using this result we get

Π (t) = EQ

[

S(T )B(t)B(T )

|Ft

]

= S(t) exp(ln(1 − δ)(N (T ) − N (t)))exp((r − r)(T − t))

= S(t) exp(ln(1 − δ)(N (T ) − N (t))).

32

(d) Following the hint we get

a(t) =∂

∂S(t)Π = exp(ln(1 − δ)(N (T ) − N (t)))

b(t) =Π (t) − a(t)S(t)

B(t)

=S(t)(exp(ln(1 − δ)(N (T ) − N (t))) − exp(ln(1 − δ)(N (T ) − N (t))))

B(t)= 0.

This now gives us the portfolio (a(t), b(t)) = (exp(ln(1 − δ)(N (T ) − N (t)))), 0) with value process

V (t) = a(t)S(t) + b(t)B(t) = S(t) exp(ln(1 − δ)(N (T ) − N (t)))).

We can now check the self-financing condition

dV (t) = a(t−)dS(t) + a(t−)dD(t) + b(t)dB(t).

We start with the LHS

dV (t) = a(t−)dS(t) + S(t−)da(t) + dS(t)da(t)

= a(t−)rS(t)dt + a(t−)σS(t)dW (t)

−a(t−)δS(t−)dN (t) + a(t−)S(t−)(exp(ln(1 − δ)) − 1)dN (t)

−a(t−)S(t−)δ(exp(ln(1 − δ)) − 1)dN (t)

= a(t−)rS(t)dt + a(t−)σS(t)dW (t)

+a(t−)S(t−)(−δ+ 11 − δ − 1 − δ( 1

1 − δ − 1))dN (t)

= a(t−)rS(t)dt + a(t−)σS(t)dW (t)

+a(t−)S(t−)(1

1− δ − 1 +−δ

1 − δ )dN (t)

= a(t−)rS(t)dt + a(t−)σS(t)dW (t)

+a(t−)S(t−)(1− δ1− δ − 1)dN (t)

= a(t−)rS(t)dt + a(t−)σS(t)dW (t)

We then calculate the RHS

a(t−)dS(t) + a(t−)dD(t) + b(t)dB(t)

= a(t−)dS(t) + a(t−)δS(t−)dN (t) + 0

= a(t−)rS(t)dt + a(t−)σS(t)dW (t) + a(t−)S(t−)(−δ+ δ)dN (t)

= a(t−)rS(t)dt + a(t−)σS(t)dW (t).

Thus we have LHS=RHS which was to be shown.

(e) So we should here find the value Π cE of a standard European call on stock paying discrete dividends. According

to the RNVF we have

Π cE (t) = E

Q

[

B(t)B(T )

max(S(T ) − K , 0)|Ft

]

= EQ

[

B(t)B(T )

max(S(t) exp(ln(1 − δ)(N (T ) − N (t)))) exp((r − σ2/2)(T − t) + σ√

T − tG) − K , 0)|Ft

]

, G ∈ N(0, 1), T > t

We here immediately see that this would be equivalent to calculating the standard BS formula with S(t) replacedby S(t) exp(ln(1− δ)(N (T )−N (t))) (this observation is in fact true for all simple claims, but it is in general nottrue for path dependend options). We thus obtain

Π cE (t) = S(t) exp(ln(1 − δ)(N (T ) − N (t)))Φ(d + σ

√T − t)) − K exp(−r(T − t))Φ(d),

where

d =

ln(

S(t) exp(ln(1−d)(N (T )−N (t)))K

)

+ (r − s2

2 )(T − t)

σ√

T − t,

33

and whereΦ is the cumulative distribution function of the standard Gaussian disytribution. Comparing this withthe standard BS formula we see that putting q = 0 the formula reduces to the standard BS formula. Moreoveras noted above we see that by replacing S(t) by S(t) exp(−q(T − t)) in the original BS formula we obtain theformula we calculated above and since the original BS formula is montonically increarsing in S we get that theEuropean call options always are cheaper with dividends than without.

(f ) So we want to find a self-financing portfolio (a(t), b(t)) with value process V (t) = a(t)S(t) + b(t)B(t) whichreplicates the call option. Using the result from e) that by replacing S(t) by S(t) exp(ln(1 − δ)(N (T ) − N (t)))in the original BS formula we obtain the formula we calculated above, the hedge calculation in the standard BScase and the chain rule we immediately obtain

a(t) =∂

∂S(t)Π c

E (t) = exp(ln(1 − δ)(N (T ) − N (t)))Φ(d + σ√

T − t))

b(t) =Π c

E (t) − a(t)S(t)B(t)

= −K exp(−rT )Φ(d).

34

August 2015

Mathematical Statistics

Centre for Mathematical SciencesLund University

Box 118, SE-221 00 Lund, Sweden

http://www.maths.lth.se/