S OME PROBLEMS WITH SOLUTIONS FMSN25/MASM24 VALUATION OF D ERIVATIVE A SSETS

M AGNUS W IKTORSSON August 2015

CENTRUM SCIENTIARUM MATHEMATICARUM

Faculty of Engineering Centre for Mathematical Sciences Mathematical Statistics

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 )g e−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). pay−off function

Φ(x)

K

0 0

K x

2K

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:   ST ≤ K 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   σ2 ∂f (t, x) σ2 ∂ 2 f (t, x) ∂f (t, x) =0 + μ− + ∂t 2 ∂x 2 ∂x 2 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 consisting of 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 remain unchanged 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 provide a better approximation of the Black-Scholes model for a finite number of periods, TM is seldom used. The main reason for this it that TM gives an incomplete market. We are here going to study TM in the one period setting with u = 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 asset X where X is the derivative with pay-off S12 at time 1 and further suppose that the price of X at time zero is equal 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 the market price. This will then allow us to price other derivatives on the asset S. Show that the market is complete by 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:
dSt(1) = μ1 St(1) dt + St(1) σ11 dWt(1) + σ12 dWt(2) dSt(0)

=

rSt(0) 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 free of 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(ST(1) ) will have a price-formula that does not depend on the choice of martingale measure. (b) Show that the claim Y = 1W (2) >K , that is a contract which pays one unit of currency at time T if WT(2) > K , T 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: dSt(2) = μ2 St(2) dt + σ22 St(2) dWt(2) 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) + σ12 dW2 (t)) dS2 (t) = rS2 (t)dt + S2 (t)(σ21dW1 (t) + σ22 dW2 (t))

2

where W1 and W2 are two independent standard Q Brownian motions and where r, σ11 , σ12 , σ21 and σ22 are positive constants. 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 observed prices of Zero Coupon bonds (ZCB) as a discounting factor when pricing derivatives. The way to accomplish this is to express the dynamics of the underlying stock S under the forward measure QT , i.e. the martingale measure which has the ZCB as numeraire. Under the measure QT we have that the discounted stock process Z (t) = S(t)/p(t, T ) should be a martingale, where p(t, T ) is the price at time t of a ZCB with maturity T . Note that We assume the following model for Z (t) under QT : T dZ (t) = Z (t)v(t, T )dW Q (t), 0 ≤ t ≤ T , T

where W Q is a standard 2-dim QT Brownian motion and where v(t, T ) is a positive deterministic function (a 1 × 2 row 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 the stock, dP(t, T ) = dS(t) =

r(t)P(t, T )dt + P(t, T )(T − t)γdW1 (t), 0 ≤ t ≤ 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 when we 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 simplifies to 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 t hat either A or B will win and that the winning stock will increase over the coming T years. But you cannot make up your mind if you should buy stock A or stock B. Assume that you have the following model (under Q) for the stocks and the bank account. dB(t) = rB(t)dt, dSA (t) = rSA (t)dt + σA SA (t)dW1 (t), dSB (t) = rSB (t)dt + σB SB (t)(ρdW1 (t) +

p 1 − ρ2 dW2 (t)),

where r, σA , σB > 0 and −1 ≤ ρ < −0.5 with W1 and W2 being independent standard Brownian motions. The negative ρ should give the effect that one stock will go up and while the other stock goes down. For this setup to be realistic, one should assume that the initial prices of stock A and B are equal and that the volatilities also are roughly the 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 time t = 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 with maturity 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 cons with this approach compared to the derivatives described above. Look at aspects like initial price and possible final 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 two favourite stocks, SA and SB respectively, happened to have the same closing price on the previous day. To celebrate this they 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, neither Anna nor Belle will receive any money. Note that no money is paid until time T . Your task is now to investigate if this is a fair bet, by looking at the bet as a financial derivative. In order to do this we need a model for the stocks. Assume that the stocks under the martingale measure Q follow a two dimensional Black-Scholes model of the following form: dSA (t) = rSA (t)dt + σA SA (t)dW1 (t)   p dSB (t) = rSB (t)dt + σB SB (t) ρdW1 (t) + 1 − ρ2 dW2 (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 Anna and Belle decide to make the bet. Moreover explain how this is related to the value from Belle’s point of view. Is the 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 of view? (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 volatility on the SP500 index. The VIX is calculated using market prices of European put and call options on the SP500 index. Let ΠEP (t, t + τ, S(t), K ) and ΠEC (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 Ftt+t be the forward level at time t of the stock price at time t + τ, i.e. Ftt+t = EQ [S(t + τ)|Ft ]. Having access to a wide range of option prices over different strikes we can get an almost model-free estimate of the expected forward volatility. VIX (t, τ) =

NP−1 Ki+1 − Ki 2 X rt P e ΠE (t, t + τ, S(t), Ki ) I (Ki < Ftt+t ) τ i=1 Ki2

1/2 NC −1 Kj+1 − Kj 2 X rt C e ΠE (t, t + τ, S(t), Kj ) I (Kj > Ftt+t ) , + τ j=1 Kj2

where NP and NC are the number of available put and call options with maturity t + τ respectively1 . The empirically calculated VIX index squared VIX (t, τ)2 is an approximation of an integral ! Z ∞ Z Ftt+t 2 dK dK 2 rt C rt P VIX (t, τ) ≈ e ΠE (t, t + τ, S(t), K ) 2 e ΠE (t, t + τ, S(t), K ) 2 + τ K K Ftt+t 0 # "Z t+t Z ∞ Ft 2 Q dK dK = E (S(t + τ) − K )+ 2 | Ft (K − S(t + τ))+ 2 + t+t τ K K Ft 0     S(t + τ) 2 Q E − ln = | Ft . τ Ftt+t 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 model you shall use in the rest of the problem. 1 This

is a slightly simplified version of the real index.

4

(a) Calculate Ftt+t = EQ [S(t + τ)|Ft ]. (b) Evaluate the expectation     2 Q S(t + τ) | Ft . E − ln τ Ftt+t (c) Show that "Z Q E

0

Ftt+t

dK + (K − S(t + τ)) K2 +

#     S(t + τ) dK Q | Ft . | Ft = E − ln (S(t + τ) − K ) K2 Ftt+t Ftt+t

Z

∞

+

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, p p V (t)S(t)(ρdW1 (t) + 1 − ρ2 dW2 (t)), p dV (t) = κ(θ − V (t))dt + σv V (t)dW1 (t), dS(t) = rS(t)dt +

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     2 Q S(t + τ) | Ft , E − ln τ Ftt+t

where Ftt+t = EQ [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-Lee model for the short rate,  drs = Θ(s) ds + σdWs , for s ≥ t rt = r where Θ(s) = ce−s + σ2 s 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 the floating LIBOR rates   1 1 LTi−1 [Ti−1 , Ti ] = −1 Ti − Ti−1 p(Ti−1 , Ti )

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 value of 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 + σdWtQ , 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 numeraire, 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 and pay-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 correct pay-off X = 1 + (T2 − T1 )LT1 [T1 , T2 ] at time T2 where the rate is decided at time T1 , the system assigns the rate over the coming interval [T2 , T3 ], i.e. the pay-off is incorrectly set to Y = 1 + (T2 − T1 )LT2 [T2 , T3 ], where T2 − T1 = T3 − T2 = 0.25. The cash-flow is still at time T2 . For the correct pay-off X we have that the fair value at 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 the numeraire measure corresponding to the numeraire p(t, T3)) for the forward rate for T1 ≤ t ≤ u ≤ T3 as: df (t, u) = f (T1 , u) =

−σ(t)2(T3 − u)dt + σ(t)dW (t), 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 at time 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 the fair value in b)?

1.10

Dividend paying stocks

1.10.1 Continuous dividends Most real stock pay out dividends. One common model for a dividend paying stock is to assume that dividends are payed out in a continuous cash flow. More precisely, by holding a stock, S over the time interval [t, T ] we receive the following dividend cash flow Z T D(T ) − D(t) = qS(u)du, t

where the positive constant q is called dividend yield. The model for the stock and bank account B and dividends under Q is given by dS(u) = dB(u) =

αS(u)du + σS(u)dW Q (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 BS case (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 and T be to arbitrary times satisfying t < T . Buying the stock at time t and selling it at time T we get the following discounted cash flow Z T Z T B(t) B(t) B(t) B(t) S(T ) + dD(u) = S(T ) + q S(u)du. B(T ) B(u) B(T ) B(u) t t This cost us S(t) at time t so for no arbitrage we must have " # Z T B(t) B(t) S(t) = EQ S(T ) q + S(u)du|Ft , B(T ) B(u) t 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 arbitrage and 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 the stock 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 hold strategy will not work, since holding the stock you receive dividends whereas getting the stock at time T give you no dividends between t and T . A portfolio consisting of holdings in the stock and the bank account with value process 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 the result to the standard BS-formula. Conclusions? Remember that holding an European call option does not give you any dividends. (f ) Find a replicating portfolio for the contract in d). 1.10.2 Discrete dividends Most real stock pay out dividends. Another common model for a dividend paying stock is to assume that dividends are payed out at discrete time points. Let {tk }, k = 1, 2, · · · , be pre-specified ordered time points when dividends are payed out. More precisely, by holding a stock, S over the time interval [t, T ] we receive the following dividend cash flow ∞ X D(T ) − D(t) = δS(tk −)I (t < tk ≤ T ), k=1

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

Z

T

δS(u−)dN (u),

t

The model for the stock and bank account B and dividends under Q is given by dS(u) = dB(u) =

αS(u)du + σS(u)dW Q (u) − δS(u−)dN (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 model we need to use a generalised version of the Ito formula which gives us the following df (t, S(t)) = (ft ′ (t, S(t)) + αS(t)fs ′ (t, S(t)) + σ2 /2S(t)2fs ′′ s(t, S(t)))dt + σS(t)fs ′ (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 check this). (a) How should we choose α here? The key observation here is that we want an arbitrage free market. So let t and T be to arbitrary times satisfying t < T . Buying the stock at time t and selling it at time T we get the following discounted cash flow Z T Z T B(t) B(t) B(t) B(t) S(T ) + dD(u) = S(T ) + δ S(u−)dN (u). B(T ) B(u) B(T ) B(u) t t This cost us S(t) at time t so for no arbitrage we must have " # Z T B(t) B(t) Q S(t) = E S(T ) δ + S(u−)dN (u)|Ft , B(T ) B(u) t 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 arbitrage and 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 the stock 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 hold strategy will not work, since holding the stock you receive dividends whereas getting the stock at time T give you no dividends between t and T . A portfolio consisting of holdings in the stock and the bank account with value process 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 the result to the standard BS-formula. Conclusions? Remember that holding an European call option does not give you 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-dynamics dSt = rSt dt + σSt dWt and applying Itô’s formula to the process Xt = (St )g e−rt we get that dXt

1 = (−rXt + γrXt + σ2 γ(γ − 1)Xt )dt + γσXt dWt 2 1 = Xt (r(γ − 1) + σ2 γ(γ − 1))dt + γσXt dWt 2 1 = Xt (γ − 1)(r + σ2 γ)dt + γσXt dWt . 2

For Xt to be a martingale we need that the drift is zero which gives that γ should solve the equation 1 (γ − 1)(r + σ2 γ) = 0. 2 We see right away that it has the solutions γ = 1 and γ = − s2r2 and since we wanted a solution different from γ = 1 we see that the requested solution is γ = − s2r2 . We should also check that X satisfies the required moment conditions. Plugging in γ = − s2r2 we get the following dynamics for X: dXt = − which has the solution Xt = X0 exp( Now we have (assuming X0 = S0g > 0) E[|Xt |] = E[Xt ] = E[X0 exp(−

2r Xt dWt , σ

2r −2r 2 t − Wt ). 2 σ σ

2r 2 2r 2r 2 2r 2 t − Wt )] = X0 exp(− 2 t + 2 t) = X0 < ∞, 2 σ σ σ σ

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:   ∂f 1 ∂2f ∂f dt + 1 dWt + dXt = df (t, Wt ) = ∂t 2 ∂x 2 ∂x   1 t/2 1 = e cos(Wt ) − et/2 cos(Wt ) dt − et/2 sin(Wt )dWt 2 2 =

−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 Itô isometry, i.e. Z t Z t Z t E[( −es/2 sin(Ws )dWs )2 ] = E[es sin(Ws )2 ]ds ≤ E[es ]ds = et − 1 < ∞. 0

0

0

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 )es/2 )|Fs ] 2

es/2 cos(Ws )(e(t−s)/2 e−(t−s) /2 ) + sin(Ws )es/2 0 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.

S K 0

)

−2K,0

max(S

−2m

ax(

S−

0

K

S

K,0

)

2K

The solid line correspond to the original pay-off and the dashed lines correspond to the stock, and one long European call option with strike 2K and two short European call options with strike K respectively (top to bottom). So we can replicate the pay-off using one stock, and one long European call option with strike 2K and two short European call option 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 S − 2 max(S − K , 0) + max(S − 2K , 0) = 2K − S   0

0≤S≤K K ≤ S ≤ 2K S ≥ 2K

and then move on to the right hand side

  S max(K − |S − K |, 0) = 2K − S   0

which shows the equivalence between the two pay-offs.

10

0≤S≤K K ≤ S ≤ 2K S ≥ 2K 

Sol: 1.2.2 Let Π (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 option and call option respectively both with maturity T and strike price K . The price Π (t) is given by any of the following equivalent 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.  S(T ) ≤ K  K 2K − S(T ) K ≤ S(T ) ≤ 2K  0 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 −

Xt

=

x.

σ2 )ds + σdWs , t ≤ s ≤ T 2

It is straightforward to see (at least it should be) that X is is a BM with drift μ − s2 and standard deviation σ starting at x at time t, i.e. σ2 XT = x + (μ − )(T − t) + σ(WT − Wt ). 2 2

Looking at exp(XT ) we see that it has exactly the same distribution as the geometric BM in the standard Black-Scholes model. We thus get that f (t, x) = =

i h s2 E[eXT |Xt = x] = E ex e(m− 2 )(T −t)+s(WT −Wt ) ex+(m−

s2 )(T −t)+ s2 (T −t) 2

2

= ex+m(T −t) .

We should also check the obtained solution fulfils the PDE and the boundary condition. We start with the last task f (T , x) = ex+0m = ex as prescribed. Finally we get that   ∂f (t, x) σ2 σ2 ∂f (t, x) σ2 ∂ 2 f (t, x) σ2 = −μf (t, x) + (μ − )f (t, x) + f (t, x) = 0, + μ− + 2 ∂t 2 ∂x 2 ∂x 2 2 which verifies that the obtained solution is correct.

Sol: 1.4 Sol: 1.4.1



Pricing of derivatives and hedging

(a) Under Q we have that ST = St e(r− 11

s2 )(T −t)+s(WT −Wt ) 2

.

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

= = = =

  Markov   e−r(T −t)EQ Φ(ST )|Ft = e−r(T −t) EQ Φ(ST )|St i h   e−r(T −t)EQ max(K − (ST )2 , 0)|St = e−r(T −t)EQ (K − (ST )2 )1ST ≤√K |St   s2 e−r(T −t)EQ (K − St2 e2(r− 2 )(T −t)+2s(WT −Wt ) )1 (r− s2 )(T −t)+s(WT −Wt ) √ |St 2 e ≤( K /St ) " # e−r(T −t)EQ (K − St2 e2(r−

s2 )(T −t)+2s√T −tZ 2

)1

Z≤

√ 2 log( K /St )−(r− s )(T −t) 2 √ s T −t

|St ,

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

Z d1 2s√T −tz− z2 z2 2 e− 2 e 2 (r−s2 )(T −t) √ dz − St e √ e K dz 2π 2π −∞ −∞ Z d1 − 1 (−4s√T −tz+z2 +4s2 (T −t)−4s2(T −t)) Z d1 − z 2 e 2 e 2 2 (r−s2 )(T −t) −r(T −t) √ dz − St e √ dz e K 2π 2π −∞ −∞ Z d1 − z 2 Z d1 − 1 (z−2s√T −t )2 e 2 e 2 −r(T −t) 2 (r+s2 )(T −t) √ dz − St e √ e K dz 2π 2π −∞ −∞ Z d1 −2s√T −t − 1 z2 Z d1 − z 2 e 2 e 2 2 (r+s2 )(T −t) −r(T −t) √ dz − St e √ dz e K 2π 2π −∞ −∞ √ −r(T −t) 2 (r+s2 )(T −t) e K N(d1 ) − St e N(d1 − 2σ T − t), −r(T −t)

Z

d1

where N is the standard normal distribution function and where √ 2 log( K /St ) − (r − s2 )(T − t) √ . d1 = 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-financing condition 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. The delta-hedge gives that we should choose hs (t) as hs (t) =

∂ F (t, St ) ∂St

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

∂ F (t, St ) F (t, St ) − St ∂S F (t, St ) − hs (t)St t = . Bt Bt

Using the F obtained in (a) we get that hs (t) = hb (t) =

√ 2 −2St e(r+s )(T −t) N(d1 − 2σ T − t)

√ 2 e−r(T −t)K N(d1 ) + St2 e(r+s )(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   σ2 St2 ∂ 2 ∂ ∂ F (t, St )dSt F (t, St ) + F (t, St ) dt + dΠt = 2 ∂t 2 ∂St ∂St   ∂ σ2 St2 ∂ 2 rBt ∂ F (t, St ) dt + F (t, St )dSt F (t, St ) + = 2 ∂t 2 ∂St rBt ∂St   ∂ σ2 St2 ∂ 2 1 ∂ dBt + F (t, St )dSt . F (t, St ) + F (t, S ) = t 2 ∂t 2 ∂St rBt ∂St 12

So if we choose hb (t) = hs (t) =

∂ σ2 St2 ∂ 2 F (t, St ) F (t, St ) + ∂t 2 ∂St2 ∂ F (t, St ) ∂St





1 rBt

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 ) − St ∂S F (t, St ) F (t, St ) − hs (t)St t = Bt 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 + q1 S0 + qu uS0 and that the risk neutral probabilities sum to one. (We must also have that the probabilities are positive and smaller than one.) We thus obtain the following linear system of equations for the risk neutral probabilities: 

1 1 3/5 1

1 6/5



   qd  q1  = 1 . 1 qu 

So we have two equations and three unknowns so there exists multiple (infinitely many) solutions. In principle we are done here since now we have that the risk neutral probabilities are not unique. We should check that at least 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 are positive 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 we obtain      1−θ 1 1 q1 = . 1 − 3/5θ qu 1 6/5 This has the solution [q1 , qu ] = [1 − 3θ, 2θ]. We here immediately see that 0 ≤ θ ≤ 1/3 for this to be probabilities. 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 market situations. (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 derivative match, i.e. we calibrate our model to fit the market price. We have that EQ [S12 |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 now we have three assets and three equations. So we should find hX , hS and hB (which are the portfolio weights for X = S 2 , S and B respectively) such that hX S12 + hS S1 + hB B1 = max(S1 − 9/10, 0), for all possible outcomes of S1 . This gives rise to following system of equations      9/25 3/5 1 0 hX  1 1 1   hS  =  1/10  36/25 6/5 1 hB 3/10 We easily solve this by Gauss elimination obtaining     5/4 hX  hS  =  −7/4  . 3/5 hB

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

EQ [S12 |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 the allowed upper and lower bounds for θ. This gives that 1 < EQ [S12 |S0 ] < 27/25, which was to be shown. So it is not possible to calibrate the model to prices outside this interval, since that would 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) = rSt(1) dt + St(1) σ11 dWt(1),Q + σ12 dWt(2),Q , where W (1),Q and W (2),Q are standard BM:s under Q. We therefore get that ST = St e(r−

s211 +s212 2

)(T −t)+

s11(WT(1),Q −Wt(1),Q )+s12 (WT(2),Q −Wt(2),Q ) ,

which has the same distribution as St e(r−

s211 +s212 2

)(T −t)+

√

s211 +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)

  − z2 √ 2 2√ s211 +s212 e 2 Φ St e(r− 2 )(T −t)+ s11 +s12 T −tz √ dz 2π −∞

Z

∞

regardless of the choice of Q, which was to be shown. (b) The dynamics of W (2) under Q is given by dWt(2) = dWt(2),Q − g2 (t)dt where (g1 (t), g2(t)) are any functions satisfying the equation μ1 − σ11 g1 (t) − σ12 g2 (t) = r, for, 0 ≤ t ≤ T and the Novikov condition

R T EP e 0 14

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

ds



< ∞.

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

e

−r(T −t)

and e

−r(T −t)

N

respectively.

N



Wt(2) − K √ T −t



! Wt(2) − K − ms1 −r (T − t) 12 √ T −t

(c) We get that (g1 (t), g2(t)) should solve the following system of linear equations −σ11 g1 (t) − σ12 g2 (t) = −σ22 g2 (t) = r − μ2

r − μ1

which has the unique solution (provided that σ11 σ22 6= 0) g2 (t) =

μ1 − r − σ12 ms2 −r μ2 − r 22 , g1 (t) = , σ22 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

! Wt(2) − K − ms2 −r (T − t) 22 √ . T −t 

Remark: Note that usually we say that a derivative or contingent claim should only be a function of the (trajectory of the) underlying asset (S1 ) up to maturity. The derivative in (a) is such a contract and there it does not matter 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 the reason why it is called a meta-theorem (see Åberg p. 111 and Björk p. 122) and not a theorem. The moral of this is perhaps that we should use a model description that only uses as many driving Brownian motions as we actually 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)EQ [max(S2 (T ) − S1 (T ), 0)|Ft ]. There are now some different approaches to calculate this expectation using change of numeraire techniques. The perhaps easiest approach is to use S1 as a numeraire this leads to     S2 (T ) 1 QS1 QS1 max( max(S2 (T ) − S1 (T ), 0)|Ft = S1 (t)E − 1, 0)|Ft , ΠX (t) = S1 (t)E S1 (T ) S1 (T ) where QS1 is the numeraire measure for S1 . This can now be seen as a European call option on the ratio S2 /S1 with strike 1. Moreover since S2 /S1 is a ratio between a traded asset and a numeraire it is automatically a martingale under QS1 . Using that the volatilities does not change when we change measure we can by calculate the volatilities under Q as     ∂ S2 ∂ S2 S1 (t)(σ11dW1 (t) + σ12 dW2 (t)) + S2 (t)(σ21dW1 (t) + σ22 dW2 (t)) ∂S1 S1 ∂S2 S1 S2 (t) = ((σ21 − σ11 )dW1 (t) + (σ22 − σ12 )dW2 (t)) S1 (t) 15

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

S1

S2 (s) S1 (s) S2 (t) S1 (t)

 S1 S1 S2 (s)  (σ21 − σ11 )dW1Q (s) + (σ22 − σ12 )dW2Q (s) , S1 (s) s2 (t) s1 (t)

= =

S1

where W1Q (s) and W2Q (s) are independent standard QS1 Brownian motions. We can now solve this SDE to obtain 
S1 S1 s2 (t) 1 S2 (T ) = exp − (σ21 − σ11 )2 + (σ22 − σ12 )2 (T − t) + (σ21 − σ11 )(W1Q (T ) − W1Q (t)) S1 (T ) s1 (t) 2  S1 S1 + (σ22 − σ12 )(W2Q (T ) − W2Q (t))

This has the same distribution as

  √ s2 (t) 1 exp − σ2 (T − t) + σ T − tG , s1 (t) 2 where G is a standard Gaussian random variable and where p σ = (σ21 − σ11 )2 + (σ22 − σ12 )2 .

We can therefore calculate the price ΠX (t) as     Z ∞ √ 1 exp(−g 2 /2) s2 (t) √ dg exp − σ2 (T − t) + σ T − tg − 1, 0 ΠX (t) = s1 (t) max s1 (t) 2 2π −∞    Z ∞ √ s2 (t) 1 exp(−g 2 /2) √ exp − σ2 (T − t) + σ T − tg − 1 = s1 (t) dg s1 (t) 2 2π −d    Z ∞ √ exp(−g 2 /2) 1 s2 (t) 1 √ exp − σ2 (T − t) + σ T − tg − g 2 /2 − √ dg = s1 (t) s1 (t) 2π 2 2π −d    Z ∞ √ 1 exp(−g 2 /2) s2 (t) 1 2 √ exp − (g − σ T − t) − √ dg = s1 (t) s1 (t) 2π 2 2π −d   Z Z ∞ ∞ √ exp(−g 2 /2) 1 1 √ exp − (g − σ T − t)2 dg − s1 (t) √ = s2 (t) dg 2 2π 2π −d −d √ = 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 S2

ΠX (t) = S2 (t)EQ This can now be rewritten as

QS2

ΠX (t) = S2 (t)E



  S1  I (S2 (T ) > S1 (T ))|Ft − S1 (t)EQ I (S2 (T ) > S1 (T ))|Ft .

        S2 (T ) S1 (T ) QS1 I I < 1 |Ft − S1 (t)E > 1 |Ft . S2 (T ) S1 (T )

Now S1 /S2 is a QS2 -martingale and S2 /S1 is a QS1 -martingale. By using the same type of argument and calculations as in the first solution we obtain that S1 (T )/S2 (T ) under QS2 has the same distribution as   √ s1 (t) 1 exp − σ2 (T − t) + σ T − tG , s2 (t) 2 and that S2 (T )/S1(T ) under QS1 has the same distribution as   √ s2 (t) 1 exp − σ2 (T − t) + σ T − tG , s1 (t) 2 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 twodimensional Girsanov kernels, where we also need to look at dynamics of the bank-account to get the right Girsanov kernels. This is however as seen from the calculations above an unnecessary detour. For the sake of completeness we supply the appropriate Girsanov kernels: g1S1 = −σ11 , g1S2 = −σ21 ,

g2S1 = −σ12 g2S2 = −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 when we change measure, we see that we can calculate the volatility function v(t, T ) using the diffusion part of the Q-dynamics for S(t)/P(t, T ). This gives that       ∂ S ∂ S dW1 (t) Z (t)v(t, T ) = P(t, T )γ(T − t)dW1 (t) + S(t)σdW2 (t) dW2 (t) ∂P P ∂S P = −Z (t)γ(T − t)dW1 (t) + Z (t)σdW2 (t)     dW1 (t) . = Z (t) −γ(T − t) σ 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 the European call option can thus be calculated as T

Π (t) = p(t, T )EQ



 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 1 T Z (T ) = Z (t) exp − v(s, T )dW Q (s) . |v(s, T )|2 ds + 2 t t This now has the same distribution as   √ 1 2 Z (t) exp − Σ(t, T ) (T − t) + Σ(t, T ) T − tG , 2 where Σ(t, T ) =

s

RT t

γ2 (T − s)2 + σ2 ds = T −t

r

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 the standard 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 ln(S(t)/K ) − ln(P(t, T )) − Σ(t, T )2(T − t)/2 √ √ = , Σ(t, T ) T − t 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 that γ = 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 plugging this 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 ))     SB (T ) SA (T ) SA (T )I ≤ 1 + SB (T )I SB ) + SB I (SB > SA ) + SA − SB = SA I (SA < SB ) − SB I (SB < SA ) this positionhas less 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 Sol: 1.7.1

Volatility

(a) Using that we get that

e−r(t+t−t) EQ [S(t + τ)|Ft ] = S(t) Ftt,t+t = EQ [S(t + τ)|Ft ] = er t S(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 −

Z

t+

t

σ2 (u)/2du +

t

Z

t+

t

σ(u)dW (u).

t

Using that the last term is a martingale we obtain Q

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

Z

t+

t

σ2 (u)/2du.

t

This together with the result in (a) then gives that     Z 2 1 t+t 2 2 = − σ (u)du − ln(S(t)) − rτ ln(S(t)) + rτ − EQ − ln(S(t + τ)/Ftt,t+t)|Ft τ τ 2 t Z t+t 1 σ2 (u)du. = τ t 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 = Ftt+t . We start by direct calculation of the integrals Z ∞ Z F Z S Z F dK dK dK dK (S − K )+ 2 = (K − S) 2 I (S < F ) + (S − K ) 2 I (F < S) (K − S)+ 2 + K K K K F S F 0 Z F Z S 1 1 S S = − − 2 dKI (S < F ) + dKI (F < S) 2 K K K K S F F F   S S I (S < F ) + − − ln(K ) I (F < S) = ln(K ) + K S K S S = (ln(F ) − ln(S) + − 1)(I (S < F ) + I (F < s)) F S S = ln(F ) − ln(S) + − 1 = − ln(S/F ) + − 1. F F Taking conditional expectation and using the definition of Ftt+t we obtain EQ [− ln(S(t+τ)/Ftt+t)+

S(t + τ Ftt+t Q t+t −1|F ] = E [− ln(S(t+τ)/F −1 = EQ [− ln(S(t+τ)/Ftt+t)|Ft ], )|F ]+ t t t Ftt+t Ftt+t

which shows that # "Z t+t     Z ∞ Ft S(t + τ) Q Q + dK + dK − ln = E + | F E (S(t + τ) − K ) (K − S(t + τ)) | F t t . K2 K2 Ftt+t Ftt+t 0  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 + We solve this by direct integration to obtain Z ln(S(t + τ)) = ln(S(t)) + rτ −

t+

t

p p V (u)(dW1 (u) + 1 − ρ2 dW2 (u)).

V (u)/2du +

t

Z

t

Using that the last term is a martingale we obtain

t+

tp

EQ [ln(S(t + τ))|Ft ] = ln(S(t)) + rτ − EQ 24

V (u)(dW1 (u) +

Z

t

t+

t

p 1 − ρ2 dW2 (u)).

 V (u)/2du|Ft .

This together with the that Ftt+t = er t S(t) then gives that       Z t+t 2 2 1 EQ − ln(S(t + τ)/Ftt,t+t )|Ft V (u)du|Ft − ln(S(t)) − rτ = − ln(S(t)) + rτ − EQ τ τ 2 t  Z t+t 1 Q V (u)du|Ft . = E τ t Now to calculate this we need to use the dynamics of V . Using direct integration we see that we can represent V (u), u > t as Z u Z up V (s)σv dW1 (s). V (u) = V (t) + κ(θ − V (s))dss + t

t

Taking conditional expectation on both sides using that the last term is a martingale we obtain Z u    Z u   V (s)ds|Ft . κ(θ − V (s))ds|Ft = V (t) + κθ(u − t) − κEQ EQ V (u)|Ft = EQ V (t) + t

t

Q

  Now put m(u) = E V (u)|Ft we then get

m(u) = V (t) +

Z

u

κ(θ − m(s))ds.

t

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

˜ = (θ − V (t))e−k(u−t) m(u)

and then we get that

m(u) = θ − (θ − V (t))e−k(u−t).

We can now finally calculate 1 Q E τ

Z

t+

t

V (u)du|Ft

t



= =

1 τ 1 τ

Z

t+

t

Z

t

t+

t t

m(u)du θ − (θ − V (t))e−k(u−t)du

θ − V (t) (1 − e−kt ) κτ e−kt − 1 + κτ (1 − e−kt ) +θ = V (t) κτ κτ

= θ−

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 since V (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 i h RT p(t, T ) = E e− t r(s)ds .

The short rate r(s) for s > t is given by Z s Z s Z s σ2 dW (u) . r(s) = r + Θ(u)du + σdW (u) = r − c(e−s − e−t ) + (s2 − t 2 ) + σ 2 t t t 25

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

Z

t

T

r(s)ds = −

T

Z

t



r − c(e

−s

σ2 − e ) + (s2 − t 2 ) + σ 2 −t

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

s

Z



dW (u) ds

t

σ2 3 σ2 (T − t 3 ) + t 2 (T − t) − σ 6 2

Z

t

T

Z

s

dW (u)ds . t

The expectation of the stochastic term is E

"Z

T

t

Z

#

s

dW (u)ds = 0 ,

t

and the variance is " Z V −

T

t

Z

s

dW (u)ds

t

#



E −

=



E

=

Z

=

t

T

Z

Z

T

T

Z

t

t

Z

s

t

T u

!2  dW (u)ds 

 !2  Z dsdW (u)  = E 

T

t



(T − u)2 du = −

(T − u) 3

 3 T

=

t

!2  (T − u)dW (u) 

(T − t)3 . 3

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

T

Z

t

Z

s

dW (u)ds

∼

t

Moreover we have that

N

  (T − t)3 0, σ2 . 3

h i RT Rs 3 s2 E e−s t t dW (u)ds = e 6 (T −t) .

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 t 2 σ2 σ2 3 (T − t 3 ) + (T − t) + (T − t)3 6 2 6



. 

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 rate bond is just all expected cash flows discounted back to time T0 . This gives that X (T0 )

=

=

=

"

# n X B(To ) B(To ) E A+ A(Ti − Ti−1 )LTi−1 [Ti−1 , Ti ]|FT0 B(Tn ) B(Ti ) i=1     X n B(To ) B(To ) EQ EQ A|FT0 + A(Ti − Ti−1 )LTi−1 [Ti−1 , Ti ]|FT0 B(Tn ) B(Ti ) i=1    X    n 1 Q B(To ) Q B(To ) E E A|FT0 + A − 1 |FT0 B(Tn ) B(Ti ) p(Ti−1 , Ti ) i=1 Q

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

X (T0 ) = p(T0 , Tn )EQ



  n  X Ti A|FT0 + p(T0 , Ti )EQ A i=1

26

1 p(Ti−1 , Ti )

  − 1 |FT0 .

Now using that

1 p(Ti−1 ,Ti )

=

P(Ti−1 ,Ti−1 ) p(Ti−1 ,Ti )

X (T0 )

is martingale under QTi we get Ap(T0 , Tn ) + A

=

n X

p(T0 , Ti )

i=1

Ap(T0 , Tn ) + A

=

n X i=1

telescoping sum



p(To , Ti−1 ) −1 p(T0 , Ti )



p(To , Ti−1 ) − p(T0 , Ti )

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 Sol: 1.9.1



The HJM framework

(a) Using that f (t, u) =

f (0, u)⋆ +

Z

0

=

t

σ2 (u − s)ds +

t

Z

0

σdWsQ

σ2 f (0, u) − ((u − t)2 − u2 ) + σWtQ 2 ⋆

and that p(t, T ) = e−

RT t

f (t,u)du

we get that p(t, T ) = e−

RT

f ⋆ (0,u)du+ s2

= e−

RT

f ⋆ (0,u)du+ s2

t

t

2

RT t

(u−t)2 −u2 du−

2 (T −t)3 −T 3 +t 3 3

s RtT WtQ du s

−(T −t) WtQ

tT (T −t)−(T −t)sWt = e− t f . R T Note that p⋆ (0, T )/p⋆(0, t) = exp(− t f ⋆ (0, u)du) and that r(t) = f (t, t) = f ⋆ (0, t) + t 2 σ2 /2 + σWtQ , so we can further simplify the expression as   σ2 p⋆ (0, T ) ⋆ 2 exp (T − t)f (0, t) − t(T − t) − (T − t)r(t) , p(t, T ) = ⋆ p (0, t) 2 RT

2 ⋆ (0,u)du− s2

Q

which is the Ho-Lee model’s ZCB-price when calibrated to the initial forward curve. This form is however, not so well suited 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, T ) − p(t, S) = . (S − T )p(t, S) p(t, S)

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

RT

=

e−

=

e

RS

e

RS

=

T

T

t

s

f ⋆ (0,u)du− s2 tT (T −t)−(T −t) WtQ + 2

RS t

s

f ⋆ (0,u)du s2 tS(S−t)+(S−t) WtQ 2

s

f ⋆ (0,u)du+ s2 t(S(S−t)−T (T −t))+(S−T ) WtQ 2

2 f (0,u)du− s2 t(S−T )(S+T −t)+(S−T )sWtQ ⋆

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 part of the dynamics since we know that the drift part must be zero, doing this we obtain that Xt has the following dynamics under QS S dXt = σ(S − T )Xt dWtQ , S

where WtQ is a standard BM under QS . This gives that XT = Xt e −

s2 (S−T )2 (T −t)+(S−T )s(W QS −W QS ) 2

27

T

t

(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 − 1 − K,0 S−T



= 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 we will not get paid at until time S as opposed to time T in the standard case. Using the RNVF for the numeraire measure QS we get that the price of the Caplet at time t Πt for 0 ≤ t ≤ T is given by  S  Πt = p(t, S)EQ max(XT − (1 + (S − T )K ), 0)|Ft .

Using the result from (b) we get that

Πt = p(t, S)Xt N(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 )2 σ2 (T − t)/2)/((S − T )σ T − t) and √ d2 = 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 N

Π Y (T1 ) = EQ



   N N (T1 ) p(T2 , T2 ) N (T1 ) Y |FT1 = EQ |FT1 . N (T2 ) N (T2 ) p(T2 , T3 )

A reasonable choice of numeraire is to use P(t, T3 ), which gives that we use the numeraire measure QT3 . This also is the measure under which the dynamics for the forward rate is given in the problem. We thus obtain Π Y (T1 )

 p(T1 , T3 ) p(T2 , T2 ) |FT1 p(T2 , T3 ) p(T2 , T3 )   T3 1 = p(T1 , T3 )EQ |F T1 . p(T2 , T3 )2 T3

= EQ



We now re-express this using forward rates Π Y (T1 )

= e−

R T3

f (T1 ,u)du QT3

−

R T3

f (T1 ,u)du QT3

= e−

R T3

f (T1 ,u)du+2

= e

= e

T1

T1

R T3 T2

T1

E

E

f (T1 ,u)du−

R T2 T1





R T3 T2

e2

R T3

f (T2 ,u)du

2

R T3

f (T1 ,u)+

e

T2

T2

|FT1

R T2 T1



ds f (s,u)du

|FT1



  RT R T2 3 e2 T2 T1 ds f (s,u)du |FT1   RT R T2 d f (s,u)du 2 3 |FT1 . e T2 T1 s

f (T1 ,u)du QT3

E

f (T1 ,u)du QT3

E

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 Y

Π (T1 ) = e −

= e

R T3 T2

R T3 T2

a+b(1−e −(u−T1 ) )du−

be −(u−T1 ) du+

R T2 T1

R T2 T1

b(1−e −(u−T1 ) )du QT3

E

be −(u−T1 ) du QT3

E



e

−2



e

R T2 “R T3 T1

T2

2

R T3 R T2 T1

T2

(T3 −u)du

s

− (s)2 (T3 −u)ds+

s(s)2 ds+2

”

“R T

3 T2

R T2

du

T1

s(s)dW (s)du |F

”R

T2 T1

T1

s(s)dW (s) |F



T1



i R T2 R T2 2 2 e− T1 (T3 −T2 ) s(s) ds+2(T3 −T2 ) T1 s(s)dW (s) |FT1 h i R R T2 2 T2 2 T3 −T2 =T2 −T1 b(e −2(T2 −T1 ) −2e −(T2 −T1 ) +1) QT3 e−(T2 −T1 ) T1 s(s) ds+2(T2 −T1 ) T1 s(s)dW (s) |FT1 = e E i h R T2 RT −(T2 −T1 ) 2 (T −T )2 s(s)2 ds+2(T2 −T1 ) T 2 s(s)dW (s) − ) QT3 1 |FT1 e T1 2 1 = eb(1−e E   qR RT T2 −(T2 −T1 )2 T 2 s(s)2 ds+2(T2 −T1 ) s (s)2 dsG b(1−e −(T2 −T1 ) )2 QT3 T 1 1 e =e E |FT1 , G ∈ N(0, 1) = eb(e

−(T3 −T1 ) −e −(T2 −T1 ) )+b(1−e −(T2 −T1 ) )

(

R T2

s(s)2 ds)(−1+2)

b(1−e −(T2 −T1 ) )2 +(T2 −T1 )2 (

R T2

s(s)2 ds)

= eb(1−e =e

h

T3

EQ

−(T2 −T1 ) 2 ) +(T

2 −T1 )

2

T1

T1

So we get that Π Y (T1 ) = e

b(1−e −(T2 −T1 ) )2 +(T2 −T1 )2 (

R T2 T1

s(s)2 ds) .

(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 has value one. This mean that the bank will on average make a small profit on the error. The gain is due to that the term structure for the forward rate is decreasing. Testing the stability of the result by slightly changing the model parameters reveal that the result is not very stable. If we instead had b = 0.01 then the value of Y would be ≈ 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 win or lose on the error. 

Sol: 1.10 Sol: 1.10.1

Dividend paying stocks

(a) Since S is a GBM, straight forward calculations give √ d S(u) = S(t) exp((α−σ2/2)(u−t)+σ(W (u)−W (t))) = S(t) exp((α−σ2 /2)(u−t)+σ u − tG), G ∈ N(0, 1), u > t. Using this we obtain # Z T B(t) B(t) + q S(u)du|Ft E S(T ) B(T ) B(u) t i h √ posintegrand Q = E S(t) exp((α − r − σ2 /2)(T − t) + σ T − tG)|Ft Z T   √ + EQ qS(t) exp((α − r − σ2 /2)(u − t) + σ u − tG)|Ft du Q

"

t

Z

T

= S(t) exp((α − r)(T − t)) + qS(t) exp((α − r)(u − t))du t   q q = S(t) exp((α − r)(T − t))(1 + . )− α−r α−r

This expression should equal S(t) for all 0 < t < T . This is possible only if q = 1, α−r 29

which gives that α = r − q. Alternative solution: We can using the Ito formula express S(T )B(t)/B(T ) as B(t) S(T ) B(T )

=

S(t) +

Z

T

t

=

S(t) +

Z

T

t

=

S(t) +

Z

=

S(t) +

t

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

T

B(t) B(t) B(t) αS(u)du + σS(u)dW Q (u) − rS(u) du B(u) B(u) B(u)

T

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

t

Z

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

This gives that Z T B(t) B(t) S(T ) + dD(u) B(T ) B(u) t Z T Z T B(t) B(t) B(t) (α − r)S(u)du + σS(u)dW Q (u) + qS(u)du = S(t) + B(u) B(u) B(u) t t Z T B(t) B(t) = S(t) + (α − r + q)S(u)du + σS(u)dW Q (u). B(u) B(u) t 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 straight forward see that Z T B(t) σS(u)dW Q (u) B(u) t

is martingale (using the Ito isometry, since GBM:s have finite moments of all orders) so that # " Z T B(t) B(t) Q + q S(u)du|Ft E S(T ) B(T ) B(u) t " # Z T B(t) Q Q = E S(t) + σS(u)dW (u)|Ft = S(t). B(u) t

(b) According to the fundamental theorems of asset pricing we have: A model is free of arbitrage if and only if there 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 unique solution in a) the market is free of arbitrage and complete. (c) Using the result in a) we get   B(t) Π (t) = EQ S(T ) |Ft B(T )

= S(t) exp((r − q − r)(T − t)) = S(t) exp(−q(T − t))

(d) Following the hint we get a(t) = b(t) =

∂ Π = exp(−q(T − t)) ∂S(t) S(t) exp(−q(T − t)) − S(t) exp(−q(T − t)) Π (t) − a(t)S(t) = = 0. B(t) B(t)

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 value ΠEc of a standard European call on stock paying continuous dividends. According to the RNVF we have   B(t) Q c ΠE (t) = E max(S(T ) − K , 0)|Ft B(T )   √ B(t) Q 2 max(S(t) exp((r − q − σ /2)(T − t) + σ T − tG) − K , 0)|Ft , = E B(T ) G ∈ N(0, 1), T > t Z ∞ √ B(t) exp(−x 2 /2) = max(S(t) exp((r − q − σ2 /2)(T − t) + σ T − tx) − K , 0) √ dx 2π −∞ B(T ) √
Z ∞ exp − 12 (σ2 (t − t) − 2σ T − tx + x 2 ) ) √ = S(t) exp(−q(T − t)) I (x > −d)dx 2π −∞ Z ∞ exp(−x 2 /2) I (x > −d)dx − K exp(−r(T − t)) √ 2π −∞ Z ∞ exp(−x 2 /2) √ S(t) exp(−q(T − t)) dx = √ 2π −d −s T −t Z ∞ exp(−x 2 /2) − K exp(−r(T − t)) √ dx 2π −d √ = 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 )(T − t) √ σ T −t 2

ln = =



S(t) exp(−q(T −t)) K



2 + (r − s2 )(T − t)

√ σ T −t

and where N is the cumulative distribution function of the standard Gaussian disytribution. Comparing this with 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 in fact true for all simple claims, but it is in general not true for path dependend options), we obtain the formula we calculated above and since the original BS formula is montonically increarsing in S we get that the European 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) which replicates the call option. Using the result from e) that by replacing S(t) by S(t) exp(−q(T − t)) in the original BS formula we obtain the formula we calculated above, the hedge calculation in the standard BS case and the chain rule we immediately obtain √ ∂ ΠEc (t) = exp(−q(T − t))N (d + σ T − t)) a(t) = ∂S(t) ΠEc (t) − a(t)S(t) = −K exp(−rT )N (d). b(t) = B(t) 

31

Sol: 1.10.2

(a) We can using the Ito formula express S(T )B(t)/B(T ) as  Z T  B(t) B(t) S(T ) = S(t) + d S(u) B(T ) B(u) t Z T B(t) B(t) dS(u) + S(u)d = S(t) + B(u) B(u) t Z T B(t) B(t) B(t) = S(t) + αS(u)du + σS(u)dW Q (u) − δS(u−)dN (u) − rS(u) du B(u) B(u) B(u) t Z T B(t) B(t) (α − r)S(u)du + σS(u)dW Q (u) − δS(u−)dN (u). = S(t) + B(u) B(u) t This gives that

Z T B(t) B(t) S(T ) + dD(u) B(T ) B(u) t Z T B(t) (α − r)S(u)du + = S(t) + B(u) t Z T B(t) (α − r)S(u)du + = S(t) + B(u) t

B(t) B(t) σS(u)dW Q (u) − δS(u−)dN (u) + B(u) B(u) B(t) σS(u)dW Q (u). B(u)

Z

T t

B(t) δS(u−)dN (u) B(u)

We when immediately see that the integral will be a good candidate for being a martingale if α − r = 0, which is equivalent to α = r. Since S is positive and only jumps downwards, looking at another process Y which behaves like S but without the jumps, we find will make Y is always larger than S. Now Y is a standard GBM with all moments finite so then S most also have all moments finite. Using this we straight forward see that Z T B(t) σS(u)dW Q (u) B(u) t is martingale (using the Ito isometry) so that # " Z T B(t) B(t) Q + δ S(u−)dN (u)|Ft E S(T ) B(T ) B(u) t " # Z T B(t) Q Q = E S(t) + σS(u)dW (u)|Ft = S(t). B(u) t

(b) According to the fundamental theorems of asset pricing we have: A model is free of arbitrage if and only if there 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 unique solution 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 Z ln(S(T )) = ln(S(u)) +

T

d ln(S(u))

t

=

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   B(t) |Ft Π (t) = EQ S(T ) B(T )

= 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) = b(t) = =

∂ Π = exp(ln(1 − δ)(N (T ) − N (t))) ∂S(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)))) = 0. B(t)

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) 1 1 − 1 − δ( − 1))dN (t) +a(t−)S(t−)(−δ + 1−δ 1−δ a(t−)rS(t)dt + a(t−)σS(t)dW (t) 1 −δ +a(t−)S(t−)( −1+ )dN (t) 1− δ 1−δ a(t−)rS(t)dt + a(t−)σS(t)dW (t) 1− δ − 1)dN (t) +a(t−)S(t−)( 1− δ 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 ΠEc of a standard European call on stock paying discrete dividends. According to the RNVF we have   B(t) Q c max(S(T ) − K , 0)|Ft ΠE (t) = E B(T )   √ B(t) Q 2 = E max(S(t) exp(ln(1 − δ)(N (T ) − N (t)))) exp((r − σ /2)(T − t) + σ T − tG) − K , 0)|Ft B(T ) , G ∈ N(0, 1), T > t We here immediately see that this would be equivalent to calculating the standard BS formula with S(t) replaced by 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 not true for path dependend options). We thus obtain √ ΠEc (t) = S(t) exp(ln(1 − δ)(N (T ) − N (t)))Φ(d + σ T − t)) − K exp(−r(T − t))Φ(d), where ln d=



d

S(t) exp(ln(1− )(N (T )−N (t))) K



√ σ T −t

33

2 + (r − s2 )(T − t)

,

and where Φ is the cumulative distribution function of the standard Gaussian disytribution. Comparing this with the standard BS formula we see that putting q = 0 the formula reduces to the standard BS formula. Moreover as noted above we see that by replacing S(t) by S(t) exp(−q(T − t)) in the original BS formula we obtain the formula we calculated above and since the original BS formula is montonically increarsing in S we get that the European 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) which replicates 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 BS case and the chain rule we immediately obtain a(t) = b(t) =

√ ∂ ΠEc (t) = exp(ln(1 − δ)(N (T ) − N (t)))Φ(d + σ T − t)) ∂S(t) ΠEc (t) − a(t)S(t) = −K exp(−rT )Φ(d). B(t)

34

August 2015 Mathematical Statistics Centre for Mathematical Sciences Lund University Box 118, SE-221 00 Lund, Sweden http://www.maths.lth.se/