MATH3075/3975 Financial Mathematics


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1 MATH3075/3975 Financial Mathematics Week 11: Solutions Exercise 1 We consider the BlackScholes model M = B, S with the initial stock price S 0 = 9, the continuously compounded interest rate r = 0.01 per annum and the stock price volatility σ = 0.1 per annum. a Using the BlackScholes call option pricing formula C 0 = S 0 N d + S 0, T Ke rt N d S 0, T we compute the price C 0 of the European call option with strike price K = 10 and maturity T = 5 years. We find that and thus C 0 = d + S 0, T = , d S 0, T = b Using the BlackScholes put option pricing formula P 0 = Ke rt N d S 0, T S 0 N d + S 0, T we find that the price P 0 = c The putcall parity relationship holds since C 0 P 0 = = = 9 10e 0.05 = S 0 Ke rt. d We now recompute the prices of call and put options for modified maturities T = 5 months and T = 5 days. We note that 5 months is equivalent to T = and thus d + S 0, T = , Hence C 0 = and P 0 = d S 0, T = We note that 5 days is equivalent to T = and thus d + S 0, T = , Hence C 0 = 1.49E 21 and P 0 = d S 0, T = e The call option respectively, put option price decreases to zero respectively, increases to K S 0 = 1 when the time to maturity tends to zero. This is related to the fact that S 0 < K and thus for short maturities it is unlikely respectively, very likely that the call option respectively, put option will be exercised at expiration. 1
2 Exercise 2 Assume that the stock price S is governed under the martingale measure P by the BlackScholes stochastic differential equation ds t = S t r dt + σ dwt where σ > 0 is a constant volatility and r is a constant shortterm interest rate. Let 0 < L < K be real numbers. Consider the contingent claim with the payoff X at maturity date T > 0 given as X = min S T K, L. a It is easy to sketch the profile of the payoff X as the function of the stock price S T. The decomposition of X in terms of the payoffs of standard call and put options reads X = L C T K L + 2C T K C T K + L. Note that other decompositions are possible. b The arbitrage price π t X satisfies, for every t [0, T ], π t X = Le rt t C t K L + 2C t K C t K + L. c We will now find the limits of the arbitrage price lim L 0 π 0 X and lim L π 0 X. We observe the payoff X increases when L increases. Hence the price π 0 X is also an increasing function of L. Moreover, lim π 0X = C 0 K + 2C 0 K C 0 K = 0. L 0 By analysing the payoff X when L tends to infinity we no longer assume that L < K, we obtain lim π 0X = P 0 K + C 0 K. L d To find the limit lim π 0 X, we observe that so that lim d +S 0, T =, lim N d + S 0, T = 1, lim d S 0, T =, lim N d S 0, T = 0. Hence the price of the call option satisfies, for all strikes K R +, lim C 0K = S 0. This in turn implies that lim π 0 X = Le rt = π 0 L. 2
3 Exercise 3 We denote by v the BlackScholes call option pricing, that is, the function v : R + [0, T ] R such that C t = vs t, t for all t [0, T ]. a Our goal is to check that the pricing function of the European call option satisfies the BlackScholes PDE v t σ2 s 2 2 v s + rs v rv = 0, s, t 0, 0, T, 2 s with the terminal condition vs, T = s K +. Note that vs, t = cs, T t where the function c is such that C t = cs t, T t. Hence vs, t = snd + s, T t Ke rt t Nd s, T t. 1 Straightforward, but a bit lengthy, computations show that the partial derivatives are see Section 5.5 in the course notes v s s, t = Nd + s, T t, v ss s, t = nd +s, T t σs, T t v t s, t = σs 2 T t nd +s, T t Kre rt t Nd s, T t, where nx is the density function of the standard normal distribution. Hence sσ 2 T t nd +s, T t Kre rt t Nd s, T t σ2 s 2 nd +s, T t sσ T t where we have also used the equality 1. b We need to show that, for every s R +, + rsnd + s, T t rvs, t = 0 lim vs, t = s K+ t T For this purpose, we observe that d + s, K and d s, K tend to respectively, when t T and s > K respectively, s < K. Consequently, Nd + s, K and Nd s, K tend to 1 respectively, 0 when t T and s > K respectively, s < K. This in turn implies that vs, T tends to either s K or 0 depending on whether s > K or s < K. The case when s = K is also easy to analyse. 3
4 Exercise 4 MATH3975 We consider the stock price process S given by the Black and Scholes model. a We will first show that Ŝt = e rt S t is a martingale with respect to the filtration F = F t t 0 generated by the stock price process S. We observe that this filtration is also generated by W. Using the properties of the conditional expectation, we obtain, for all s t, Ŝs e σ W t W s 1 2 σ2 t s Fs E PŜt Fs = E P = Ŝs e 1 2 σ2 t s E P e σw t W s F s = Ŝs e 1 2 σ2 t s E P e σw t W s F s = Ŝs e 1 2 σ2 t s E P e σw t W s where in the last equality we used the independence of increments of the Wiener process. Recall also that W t W s = t s Z where Z N0, 1, and thus E PŜt F s = Ŝ s e 1 2 σ2 t s E P e σ t sz. Let us finally recall that if Z N0, 1 then for any real number a By setting a = σ t s, we obtain E P e az = e 1 2 a2. E PŜt Ŝu, u s = Ŝs e 1 2 σ2 t s e 1 2 σ2 t s = Ŝs, which shows that Ŝ is a martingale under P. b To compute the expectation E PS t, we observe that E PS t = e rt E PŜt = e rt E PŜ0 = e rt Ŝ 0 = e rt S 0. To compute the variance Var PS t, we recall that where in turn Var PS t = E PS 2 t [ E PS t ] 2 E PS 2 t = S 2 0e 2rt E P [ e 2σW t σ 2 t = S0e 2 2rt e σ2t E P = S0e 2 2rt e σ2t E P [e az 1 a2] 2 = S 2 0e 2rt e σ2 t [ ] e 2σW t 1 2 2σ2 t where we denote a = σ t. Hence Var PS t = S 2 0e 2rt e σ2t 1. ] 4
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