B8.3 Mathematical Models for Financial Derivatives. Hilary Term Solution Sheet 2

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1 B8.3 Mathematical Models for Financial Derivatives Hilary Term 18 Solution Sheet In the following W t ) t denotes a standard Brownian motion and t > denotes time. A partition π of the interval, t is a sequence of points = t < t 1 < t < < t n = t and π = max k t k+1 t k ). On a given partition W k W tk, δw k W k+1 W k, δt k t k+1 t k and if f is a function on, t, f k ft k ) and δf k f k+1 f k. 1. Show that if t, s then EW s W t = mins, t). If s = t then we have EW s W t = EW t = t. If not, assume s < t and write W t = W s + W t W s. Then EW s W t = E W s W s + W s Wt W s ) = EW s W s + E W s Wt W s ) = EW s = s = mins, t).. Suppose we define the following two stochastic integrals, the anti-itô integral fw s, s) dw s = lim and the Stratonovich integral π k= fw k+1, t k+1 )W k+1 W k ), fw s, s) dw s = lim π k= 1 fwk+1, t k+1 ) + fw k, t k ) ) W k+1 W k ). Show that W s dw s = Wt + t, E W s dw s = Wt, E W s dw s W s dw s = t, = t 1

2 W s dw s = lim π k= π k= ) W k+1 Wk+1 W k = lim W k+1 Wk ) ) + lim Wk+1 W k = W n W + lim π k= π k= ) Wk+1 W k = Wt + t. Since we know E Wt = t, we see that E W s dw s = E Wt + t = t. Similarly we have and so W s dw s = lim π k= = lim π k= = W n W = W t, ) ) Wk+1 + W k Wk+1 W k W k+1 Wk ) E W s dw s = E Wt = t. 3. Assuming that both the integral and its variance exist, show that var fw s, s) dw s = E fw s, s) ds. Note: if the integral and its variance exist then it is legitimate to interchange the order of expectation and integration. If the integral exists then its expected value is zero so ) var fw s, s) dw s = E fw s, s) dw s. Set f k = fw k, t k ) and consider ) f k δw k = k= = f j f k δw j δw k k= j= fk δw k) + f j f k δw j δw k. k= k=1 j<k

3 We can eliminate the double sum using the tower law, E f j f k δw j δw k = E E tk fj f k δw j δw k = E f j f k δw j E tk δwk since t j < t k, which means f j and δw j are known at time t k, as is f k = fw k, t k ). As E tk δwk =, this term vanishes, leaving ) f k δw k E k= = E fk δw k) = E = E k= E tk f k δw k ) k= fk E t k δwk ) k= = E fk δt k = E fk δtk k= k= and as we refine the partition, π, we see lim π k= E fk δtk E fw s, s) ds. 4. Use the differential version of Itô s lemma to show that a) W s ds = t W t We have which integrates to show Rearranging gives s dw s = t s) dw s, d t W t ) = Wt dt + t dw t t W t = W s ds = t W t W s ds + s dw s = s dw s. t s) dw s. 3

4 b) W s dw s = 1 3 W 3 t W s ds, This time d Wt 3 ) = 3 W t dw t + 3 W t dt which integrates to give W 3 t = 3 W s dw s + 3 Dividing by 3 and rearranging gives W s dw s = 1 3 W 3 t W s ds. W s ds. c) E e awt a t/ = 1. Set fw, t) = expaw 1 a t) and f t = fw t, t). Itô s lemma gives df t = 1 a f t dt + a f t dw t + 1 a f t dt = a f t dw t. This integrates to give f t f = a exp aw t 1 a t ) = 1 + a Taking expectations gives E exp aw t 1 a t ) = 1 + a E = 1. f s dw s or, in full, exp aw s 1 a s ) dw s exp aw s 1 a s ) dw s 5. Define X t to be the area under a Brownian motion, X = and X t = W u du for t >. Show that X t is normally distributed with From Question 4a) we have E X t =, E X t = 1 3 t3. W s ds = t s) dw s. 4

5 As t s does not depend on W s, the integral is normally distributed with E W s ds = E t s) dw s = and var W s ds = var t s) dw s = = 1 3 t3. t s) ds Note that X t is continuously differentiable for t >, with Ẋt = W t recall W t is continuous in t). Now define Y t as the average area under a Brownian motion, { if t =, Y t = X t /t if t >. Show that Y t has EY t =, E Yt = t/3 and that Yt is continuous for all t. Is 3 Y t a Brownian motion? Give reasons for you answer. For any t >, Y t is normal because X t is, with and E Y t = E Xt /t = var Y t = var Xt /t = var X t /t = 1 3 t. For t > we have Y t = X t /t which is the ratio of two differentiable functions and as the denominator is never zero it follows that Y t is differentiable for t >, which implies continuous. Moreover, it means we can use l Hopital s rule to show lim Y W t t = lim t + t + 1 =, which shows that Y t is continuous at t =. The function 3 Y t is not only continuous but differentiable for t > and therefore it can t be a Brownian motion. The hard way to do this part of the question is to show that the increments over disjoint intervals are not independent.) 6. Show that if ds t S t = µ dt + σ dw t, 5

6 and f t = fs t ) = S 3 t, then Write the SDE for S t as df t f t = 3 µ + σ ) dt + 3 σ dw t. ds t = µ S t dt + σ S t dw t so Itô s lemma for a general F t = F S t, t) is df t = F t S t, t) + 1 σ S t F ) S S t, t) dt + F S S t, t) ds t. With fs) = S 3, so f S) = 3 S and f S) = 6 S, this becomes df t = 3 σ St 3 dt + 3 St ds t = 3 σ f t dt + St ) ) µ St dt + σ S t dw t = 3 f t µ + σ ) ) dt + σ dw t which we could write as df t f t = 3 µ + σ ) dt + 3 σ dw t. 7. Find solutions of the Black-Scholes terminal value problem when + r y) S r V =, S >, t < T, S S V S, T ) = fs), S >, a) fs) = C, where C is a constant; In this case it makes sense to look for solutions which don t depend on S, so / S = and V/ S = identically. Then we have V = V t) and the partial differential equation reduces to the ODE dv dt r V =, V T ) = C which has the solution V = C e rt t). b) fs) = S α, where α is a constant. Hint: you don t need the Feynman-Kǎc formula to do either of these. Look for simple functional forms of the solution. 6

7 In this case, let s look for separable solutions We find that t = Sα ft), V S, t) = S α ft), ft ) = 1. S S = α Sα ft), S V S = αα 1) Sα ft) and the Black-Scholes equation becomes ) S α ft) + 1 σ αα 1) ft) + r y) α ft) r ft) =. As this has to hold for all S > we have an ordinary differential equation for ft) which we can write as where The solution is and ft) = λ ft), ft ) = 1, λ = r r y) α 1 σ αα 1). λt t) ft) = e V S, t) = e λt t) S α. 7

8 Optional questions 8. The Black-Scholes equation from a binomial method. One step of the Cox, Ross & Rubinstein binomial method can be written as V S, t) = e rδt q V u + 1 q) V d) where V u = V S u, t + δt ), V d = V S d, t + δt ), δt S u = S e σ δt, S d = S e σ δt, q = erδt e σ e σ δt e, σ δt σ > is the volatility, r is the risk-free rate and δt > is the length of the time-step. Supposing this is true for all S > and that V S, t) may be expanded in a Taylor series in both S and t, show that as δt q = 1 + r 1 σ δt + Oδt), σ V u = V + δt σ S S + δt t + 1 σ S V d = V δt σ S S + δt S + S V )) S + Oδt 3/ ), t + 1 σ S S + S V )) S + Oδt 3/ ), where V and all its partial derivatives are evaluated at S, t). Hence show that in the limit δt the option price satisfies the zero dividend-yield) Black-Scholes equation As δt we can expand S + r S S S r V =. e σ δt = 1 + σ δt + 1 σ δt + Oδt 3/ ), from which it follows that e σ δt = 1 σ δt + 1 σ δt + Oδt 3/ ), e rδt = 1 + rδt + Oδt ), q = σ δt + r 1 σ )δt + Oδt 3/ ) σ δt + Oδt 3/ ) = 1 + r 1 σ δt + Oδt), σ 8

9 and that V S e σ δt, t + δt) = V S, t) + δt t S, t) + σ δt + 1 σ δt ) S S, t) S + 1 σ δt S V S S, t) + Oδt3/ ) = V + δt σ S S + δt V S e σ δt, t + δt) Therefore t + 1 σ S )) S + S V S + Oδt 3/ ) = V δt σ S S + δt t + 1 σ S )) S + S V S + Oδt 3/ ). q V S e σ δt, t + δt) + 1 q) V S e σ δt, t + δt) = V + δt S + 1 σ S S = V + δt S + r S S ) + r 1 σ ) δt S S + Oδt3/ ) ) + Oδt 3/ ) and hence ) e rδt q V S e σ δt, t + δt) + 1 q) V S e σ δt, t + δt) = 1 rδt) V + δt S + r S S = V + δt S + r S S r V )) + Oδt 3/ ) ) + Oδt 3/ ). Substituting this back into the binomial equation gives V = V + δt S + r S ) S r V + Oδt 3/ ) and cancelling V and dividing by δt then gives S + r S ) S r V + Oδt 1/ ) =. Taking the limit δt gives the Black-Scholes equation, S + r S S r V =. 9

10 9. The variation of a function, or stochastic process, over, t, is f t = lim π k= f k+1 f k. If f t is finite on, t we say f has bounded variation on, t. Show that: a) if f is C 1, t then f t = f t) dt < ; If f is C 1 on, t then the intermediate value theorem asserts that f k+1 f k = f ξ k )t k+1 t k ) for some ξ k t k, t k+1. Since δt k = t k+1 t k ) >, f t = lim π k= f ξ k ) δt k = f s) ds. If f s) is continuous on, t then so too is f s) and as, t is a closed and bounded set, f s) must be bounded on, t. Therefore the integral must be finite and hence f has bounded variation on, t. b) if f is a continuous function with f t < then its quadratic variation is zero, f t = ; We have ) f t = lim fk+1 f k and we also have π k= ) ) fk+1 f k max fj+1 f j fk+1 f k ). k= j If f has bounded variation then the sum on the right-hand side is finite as π. If f is continuous then the maximum value of f j+1 f j as π. Thus the product of the two vanishes as π. c) Brownian motion does not have bounded variation; Brownian motion is continuous in t and has non-zero quadratic variation. Therefore it must have unbounded variation by the previous part of the question. d) the arc length of the graph of a Brownian motion is infinite for any t >. Hint: if y = X t has an arc length s then ds = dy + dt dy = dy. k= 1

11 Assume that we can define arc length for a Brownian motion. Then it is given by s = where if y = X t then ds is given by Now observe that ds ds = dt + dy, ds = dt + dy. ds = dt + dy dy = dy. If the integral of dy existed it would equal the variation of W t, but we know that W t doesn t have bounded variation. This implies that the length of the graph of a Brownian motion is infinite for any t > ). 11