CS计算机代考程序代写 MATH3075/3975 Financial Mathematics

MATH3075/3975 Financial Mathematics

Tutorial 11: Solutions

Exercise 1 We consider the Black-Scholes model M = (B,S) with the initial stock price S0 = 9,
the continuously compounded interest rate r = 0.01 per annum and the stock price volatility σ = 0.1
per annum. Recall that dBt = rBt dt with B0 = 1 (equivalently, B(t, T ) = e

−r(T−t) and

dSt = St
(
r dt+ σ dWt

)
, S0 > 0,

where W is a standard Brownian motion under the martingale measure P̃.

(a) Using the Black-Scholes call option pricing formula

C0 = S0N
(
d+(S0, T )

)
−Ke−rTN

(
d−(S0, T )

)
we compute the price C0 of the European call option with strike price K = 10 and maturity
T = 5 years. We find that

d+(S0, T ) = −0.13578, d−(S0, T ) = −0.35938

and thus C0 = 0.59285.

(b) Using the Black-Scholes put option pricing formula

P0 = Ke
−rTN

(
− d−(S0, T )

)
− S0N

(
− d+(S0, T )

)
we find that the price P0 = 1.10514

(c) The put-call parity relationship holds since

C0 − P0 = 0.59285− 1.10514 = −0.51229 = 9− 10e−0.05 = S0 −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 = 0.416667 and thus

d+(S0, T ) = −1.53541, d−(S0, T ) = −1.59996.

Hence C0 = 0.015315 and P0 = 0.973735.

– We note that 5 days is equivalent to T = 0.013699 and thus

d+(S0, T ) = −8.98455, d−(S0, T ) = −8.99615.

Hence C0 = 1.49E − 21 and P0 = 0.99863.

(e) The call option (respectively, put option) price decreases to zero (respectively, increases to
K−S0 = 1) when the time to maturity tends to zero. This is related to the fact that S0 < 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 Exercise 2 Assume that the stock price S is governed under the martingale measure P̃ by the Black-Scholes stochastic differential equation dSt = St ( r dt+ σ dWt ) where σ > 0 is a constant volatility and r is a constant short-term interest rate. Let 0 < L < K be real numbers. We consider a path-independent contingent claim with the payoff X at maturity date T > 0 given as

X = min
(
|ST −K|, L

)
.

(a) It is easy to sketch the profile of the payoff X as the function of the stock price ST . The
decomposition of X in terms of the payoffs of standard call and put options reads

X = L− CT (K − L) + 2CT (K)− CT (K + L).

Note that other decompositions are possible.

(b) The arbitrage price πt(X) satisfies, for every t ∈ [0, T ],

πt(X) = Le
−r(T−t) − Ct(K − L) + 2Ct(K)− Ct(K + L).

(c) We will now find the limits of the arbitrage price limL→0 π0(X) and limL→∞ π0(X). We
observe the payoff X increases when L increases. Hence the price π0(X) is also an increasing
function of L. Moreover,

lim
L→0

π0(X) = −C0(K) + 2C0(K)− C0(K) = 0.

By analysing the payoff X when L tends to infinity (obviously, we no longer assume here that
the inequality L < K holds since K is fixed and L tends to infinity), we obtain lim L→∞ min ( |ST −K|, L ) = |ST −K| = (K − ST )+ + (ST −K)+ = PT (K) + CT (K) and thus lim L→∞ π0(X) = P0(K) + C0(K). (d) To find the limit limσ→∞ π0(X), we observe that lim σ→∞ d+(S0, T ) =∞, lim σ→∞ d−(S0, T ) = −∞, so that lim σ→∞ N ( d+(S0, T ) ) = 1, lim σ→∞ N ( d−(S0, T ) ) = 0. Hence the price of the call option satisfies, for all strikes K ∈ R+, lim σ→∞ C0(K) = S0. This in turn implies that limσ→∞ π0(X) = Le −rT = π0(L). 2 Exercise 3 We denote by v the Black-Scholes call option pricing, that is, the function v : R+ × [0, T ]→ R such that Ct = v(St, t) for all t ∈ [0, T ]. (a) We need to show that, for every s ∈ R+, lim t→T v(s, t) = (s−K)+ 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, N(d+(s,K)) and N(d−(s,K)) tend to 1 (respectively, 0) when t→ T and s > K (respectively, s < K). This in turn implies that v(s, 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 and to check that limt→T v(s, t) = 0 when s = K. (b) (MATH3975) Observe that v(s, t) = c(s, T − t) where the function c is such that Ct = c(St, T − t). Our goal is to check that the pricing function of the European call option satisfies the Black-Scholes partial differential equation (PDE) ∂v ∂t + 1 2 σ2s2 ∂2v ∂s2 + rs ∂v ∂s − rv = 0, ∀ (s, t) ∈ (0,∞)× (0, T ), (1) with the terminal condition v(s, T ) = (s−K)+. Equivalently, the function c satisfies − ∂c ∂t + 1 2 σ2s2 ∂2c ∂s2 + rs ∂c ∂s − rc = 0, ∀ (s, t) ∈ (0,∞)× (0, T ), with the initial condition c(s, 0) = (s−K)+. From the Black-Scholes theorem, we know that v is given by the following expression v(s, t) = sN(d+(s, T − t))−Ke−r(T−t)N(d−(s, T − t)). (2) Straightforward computations show that the partial derivatives are: vs(s, t) = N(d+(s, T − t)), vss(s, t) = n(d+(s, T − t)) σs √ T − t , vt(s, t) = − σs 2 √ T − t n(d+(s, T − t))−Kre−r(T−t)N(d−(s, T − t)) where n(x) is the density function of the standard normal distribution. Hence − sσ 2 √ T − t n(d+(s, T − t))−Kre−r(T−t)N(d−(s, T − t)) + 1 2 σ2s2 n(d+(s, T − t)) sσ √ T − t + rsN(d+(s, T − t))− rv(s, t) = 0 where we have also used the equality (2). It is worth noting that the pricing function w(s, t) = p(s, T − t) for the put option also satisfies the Black-Scholes PDE but with the terminal condition w(s, T ) = (K − s)+. This can be checked either by computing directly the partial derivatives or by combining already established PDE (1) with the put-call parity relationship, which reads v(s, t)− w(s, t) = s−Ke−r(T−t). 3 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 −rtSt is a martingale with respect to the filtration F = (Ft)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, EP̃ ( Ŝt ∣∣Fs) = EP̃ (Ŝs eσ(Wt−Ws− 12σ2(t−s)) ∣∣Fs) = Ŝs e − 1 2 σ2(t−s) EP̃ ( eσ(Wt−Ws) | Fs ) = Ŝs e − 1 2 σ2(t−s) EP̃ ( eσ(Wt−Ws) | Fs ) = Ŝs e − 1 2 σ2(t−s) EP̃ ( eσ(Wt−Ws) ) where in the last equality we used the independence of increments of the Wiener process. Recall also that Wt −Ws = √ t− sZ where Z ∼ N(0, 1), and thus EP̃ ( Ŝt | Fs ) = Ŝs e − 1 2 σ2(t−s) EP̃ ( eσ √ t−sZ). It is known (and easy to check by integration) that if Z ∼ N(0, 1) then for any real number a EP̃ ( eaZ ) = e 1 2 a2 . (3) By setting a = σ √ t− s, we obtain EP̃ ( Ŝt | 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 P̃. (b) To compute the expectation EP̃(St), we observe that EP̃(St) = e rt EP̃(Ŝt) = e rt EP̃(Ŝ0) = e rtŜ0 = e rtS0. To compute the variance Var P̃(St), we recall that Var P̃(St) = EP̃(S 2 t )− [ EP̃(St) ]2 where in turn EP̃(S 2 t ) = S 2 0e 2rt EP̃ [ e2σWt−σ 2t ] = S20e 2rteσ 2t EP̃ [ e2σWt− 1 2 (2σ √ t)2 ] = S20e 2rteσ 2t EP̃ [ eaZ− 1 2 a2 ] where we denote a = 2σ √ t and Z ∼ N(0, 1). Since (see (3)) EP̃ [ eaZ− 1 2 a2 ] = 1 we conclude that EP̃(S 2 t ) = S 2 0e 2rteσ 2t and thus Var P̃(St) = S 2 0e 2rt ( eσ 2t − 1 ) . 4