is a standard Brownian motion.

Transcription

1 Stochastic Calculus Final Examination Solutions June 7, 25 There are 2 problems and points each.. (Property of Brownian Bridge) Let Bt = {B t, t B = } be a Brownian bridge, and define dx t = Xt dt + db t t, with X =. a) Show that {B t, t } and {X t, t } have the same distribution. b) Show that Y t = ( + t)b t +t sol: (a) is a standard Brownian motion. ()Bt = B t tb, B t N(, t), B N(, ), tb N(, t 2 ) linear combination of normal distribution is still normal. Compute mean and variance. EB t =, EBs B t = s( t)ifs < t VarB t = t( t) Bt N(, t( t)) t. (2) According to the solution of OU process, X t = ( t) t db s s X t is the linear combination of N.D., therefore it is still a N.D. Compute mean and variance X t =, Ifs t : EX s X t = ( s)( t)e[ s = ( s)( t)e[ s = ( s)( t)e[( s db u u t db v v] db u u( s db v v + t db s v v)] db u u) 2 ] (independent increment) = ( s)( t)e[ s ( u )2 ds] (Ito s isometry) = ( s)( t) s s = s( t) VarX t = t( t) X t N(, t( t)) (b) B t +t N(, EX t = E( + t)(b t +t t t ), B s N(, s ), B +t +s +s N(, ) Y t N.D. B +t ) =

2 If s < t ( s < t ) : +s +t EX s X t = ( + s)( + t)[eb s B +s t t EB s B +t +t s EB +s +s t B + s +t VarY t = t Y t N(, t) ()Y = (2)Independent increment = t < t <... < t n = : Y t Y t, Y t2 Y t,..., Y tn Y tn Cov(Y ti Y ti, Y tj Y tj ) = (3)stationary increment E(Y t Y s ) = Var(Y t Y s ) = Var(Y t ) + Var(Y s ) 2Cov(Y t, Y s ) = t + s 2s = t s Y t Y s N(, t s) (4)Y t N(, t) (5)B t is continuous in t Y t is also continuous in t From () (5), Y t is a SBM. t +s +t EB2 ] = s 2. a) Solve the following stochastic differential equation dx t = ( αx t + β)dt + σdb t, where X = x and α >. b) Verify the solution can be written as X t = e αt (x + β α (eαt ) + σ t e αs db s ). c) Show that X t converges in distribution, as t, and find the limiting distribution. d) Find the covariance Cov(X s, X t ) for s < t. sol: (a) dx t + αx t dt = βdt + σdb t multiply e αt for both sides d(e αt X t ) = βe αt dt + σe αt db t e αs X s t = β t eαs ds + σ t eαs dbs 2

3 X t = e αt (x + β α (eαt ) + σ t eαs db s ) (b) X t t X t B t 2 X t Bt 2 = αx t + β = σ = σ By Ito s formula: dx t = ( αx t + β)dt + σdb t (c) ()EX t = x e αt + β ( α e αt ) β as t α (2)VarX t = σ 2 E t e 2α(t s) ds = σ2 ( 2α e 2αt ) σ2 as t 2α Therefore X D t N( β, σ2 ) as t α 2α (d) Cov(X s, X t ) = σ 2 e α(t+s) Cov( s eαu db u, t eαu db u ) = σ 2 e α(t+s) E s t e 2αu du = σ2 2α e α(t+s) (e 2α(s t) ) 3. a) Give a precise description of the simplest Girsanov theorem. b) Let L be the line given by the equation y = a + bt with a >. Define τ L = inf{t : B t = a + bt}. Use the simplest Girsanov theorem (Brownian motion with drift) to derive the probability density function f τl (t) of τ L. sol: (a) If the process {B t } is a P-BM and Q is the measure on C[, T ] induced by the process X t = B t + µt, then every bounded Borel measurable function W on the space C[, T ] satisfies E Q (W ) = E P (W M T ), where M t is the P-mtgle defined by M t = exp(µb t µ2 t 2 ) (b) 3

4 τ L = inf{t : B t = a + bt}, τ a = inf{t : X t = a}, wherex t = B t bt P (τ L t) = Q(τ a t) = E Q ((τ a t)) = E P ((τ a t)m T )(simplest Girsanov theorem) wherem t = exp{ bb T b2 2 T } = E P ((τ a t)m t τa ) ({τ a t} is F t τa measurable) = E P ((τ a t) exp( ab b2 τ 2 a)) = t exp( ab b2 s) a φ( a 2 s 2 3 s )ds f τa (t) = P (τ t L t) = a φ( a+bt t 3/2 t ) for t. 4. a) State the Martingale representation theorem. b) State the Feynman-Kac representation theorem for Brownian motion. Sol: (a) X t is an {F t }-martingale, where {F t } is the standard Brownian filtration, if a and T, such that E(XT 2) <, and! φ(s, ω) H 2 [, T ], such that X t = t φ(s, ω)db s, t T. (b) Suppose that the function q : R R is bounded. Consider { ut (t, x) = 2 u xx(t, x) + q(x)u(t, x) u(, x) = f(x) where f : R R is also bounded. If u(t, x) is the unique bounded solution of the function, then 5. The Black-Scholes model is assumed to be u(t, x) = E[f(x + B t )exp( t q(x + B s)ds)] ds t = µs t dt + σs t db t, dβ t = rβ t dt. a) By using arbitrage theory to show the procedure of deriving the Black-Scholes PDE for European call option. 4

5 b) Show how to transform this PDE into a heat equation. 6. (continued) c) Use Fourier transform technique to derive the solution of the above heat equation. d) Derive Black-Scholes formula for European call option via Feynman-Kac representation. Sol: (a) Hedging price Y t = β t B t + γ t S t. Ito s formula: And dy = ( Y t + µs Y S + 2 σ2 S 2 2 Y S 2 )dt + σs Y S dw t dy = β t db t + γ t ds t = (rβ t B t + µγ t S t )dt + σγ t S t dw t Matching coefficient: Y t = rb t ( Y t + 2 σ2 S 2 2 Y S 2 )B t + Y S S t { Y { γt = Y S β t = rb t ( Y t + 2 σ2 S 2 2 Y S 2 ) (T, S t t) + 2 σ2 (T, S t )ST 2 2 Y (T, S S 2 t ) = Y (T, S) = f(t, S) = (S K) + (b) Let θ = σ 2 (T t) z = ln S + (r σ2 )(T t) 2 Set u(θ, z) = e r(t t)y (t,st), then { u = 2 u θ 2 z 2 u(, z) = f(z) = (e z K) + 5

6 (c) Take Fourier transform: û(θ, w) = ˆf(w)e 2 w2 θ Inverse Fourier transform: { tû(w, θ) = 2 (iw)2 û(w, θ) = 2û(w, θ) û(, w) = ˆf(w) u(θ, z) = 2π ˆf(w)e 2 w2θ e iwz dw By transform of Gaussian, we have F(e z2 2θ )(w) = θe 2 w2 θ e z2 2θ Therefore, = F( θ e z2 2θ )(w) F(g θ (z))(w) u(θ, z) = f g θ (z) = 2π f(y)g θ(z y)dy f(y) 2π θ e (z y)2 2θ dy = = (z y)2 f(y)e 2θ 2πθ dy 2πθ (ey K) + e (z y)2 2θ dy = (d) y = W theta + z dy = dw θ, where {W θ } is the standard Brownian motion. E(e W θ+z K) + = E(e z e θw K) + = E(e z+ θ 2 e θw θ 2 K) + = e z+ θ z ln K+θ 2 Φ( θ ) Ke z+ θ 2 Φ( z ln K θ ) = S Φ( ln( S K )+(r+ σ2 )(T t) 2 ) Ke rt Φ( ln( S σ T t K )+(r σ2 )(T t) 2 σ T t 7. Derive the pricing formula for European put option by using the technique of change of numeraire and Girsanov theorem. payoff X = (K S(T )) + p = Ẽ[e rt (K S(T )) + ] = Ẽ[e rt (K S(T ))I(k S(T )] = e rt K P (K S(T )) e rt Ẽ[S(T )I(K S(T ))] = () (2) 6 )

7 σ2 (r K S e 2 )T +σ T Z z σ [ln( S T K where d 2 = σ [ln( S σ2 ) + (r )T ] T K 2 () = Ke rt P (z d2 ) = Ke rt Φ( d 2 ) (2) = S Ẽ[e rt S(T ) I(K S(T ))] S() (Let d P d P = e rt S(T ) = S() eσ W T σ 2 2 T Let W t = W t σt σ2 (r k S e 2 )T +σ T Z σ2 (r+ = S e 2 )T +σ W T z d ) ) + (r σ2 2 )T ] d 2, = S P (z d ) = S Φ( d ) Therefore, p = Ke rt Φ( d 2 ) S Φ( d ) 8. Barrier option pricing: Find the arbitrage price of a contingent claim that pays M if the stock price gets to a level K or higher during the time period [, T ] and that pays zero otherwise. In other words, find a formula for the arbitrage price of the claim sol: X = MI { sup S t>k}. t [,T ] sup S t > K sup B t + σ2 (r )t ln K σ 2 σ S t [,T ] t [,T ] Let a = ln K σ S, b = σ2 (r )t σ 2 τ = inf{b t = a + bt} f τ = a (a+bt)2 e 2t 2πt 3 u = e rt Ẽ[M( sup t [,T ] S t > K)] = e rt M P ( sup t [,T ] S t > K) = e rt MP (τ T ) (Law(µ + σb t, t T P T ) = Law(r + σb t, t T P T )) = e rt MP (τ L T ) = e rt M[ Φ( ln K (r σ2 S 2 )T σ ) + K τ S e 2 σ 2 (r σ2 2 ) Φ( ln K (r σ 2 S 2 )T σ T )] 9. a) State the Poisson process martingale property theorem. b) Let N(t) be a Poisson process with intensity λ. Show that exp{ln( u)n(t) + uλt}, < u <, 7

8 is a martingale. Sol: (a) Let N(t) be a Poisson process with intensity λ. Then M(t) = N(t) λt is a martingale. (b) E[exp{ln( u)n(t + s)} F t ] = E[exp{ln( u)n(t) + ln( u)(n(t + s) N(t))} F t ] = exp{ln( u)n(t)}e[exp{ln( u)(n(t + s) N(t))} F t ] = exp{ln( u)n(t)}e[exp{ln( u)(n(t + s) N(t))}] = exp{ln( u)n(t)} exp{ uλs} Therefore, E[exp{ln( u)n(t + s) + uλ(t + s)} F t ] = exp{ln( u)n(t) + uλt}. Consider the stock and bond model given by ds t = µ(t, S t )dt + σ(t, S t )db t and dβ t = r(t, S t )β t dt, where all of the model coefficients µ(t, S t ), σ(t, S t ), and r(t, S t ) are given by explicit functions of the current time and the current stock price. Use the coefficient matching method to show that arbitrage price at time t of a European option with terminal time T and payout h(s T ) is given by f(t, S t ), where f(t, S t ) is the solution of the terminal value problem Sol: f t (t, S t ) = 2 σ2 (t, x)f xx (t, x) r(t, x)xf x (t, x) + r(t, x)f(t, x), f(t, x) = h(x). (a) For European call option, the payout is h(s t ) = (S T K) +. Consider the replicating portfolio at time t, V t Let a t : the number of units of stock. b t : the number of units of the bond. Total value of the portfolio at time t is V t = a t S t + b t β t We require that the restructuing of the portfolio be self-financing. So, the requirement is dv t = a t ds t + b t dβ t 8

9 Hence put () and (2) into (4) dv t = µdt + σ(t, S t )db t = [a t u(t, S t ) + b t r(t, S t )β t ]dt + a t σ(t, S t )db t V t = f(t, S t ) and Ito formula for geometric Brownian Motion. dv t = f t (t, S t )dt + (/2)f xx (t, S t )ds t ds t + f x (t, S t )ds t = [f t (t, S t ) + (/2)f xx (t, S t σ 2 )(t, S t ) + f x (t, S t µ(t, S t ))]dt + f x (t, S t )σ(t, S t )db t Compare () and (2) a t = f x (t, S t ) b t = r(t,s t)β t (f t (t, S t ) + /2f xx (t, S t )σ 2 (t, S t )) V t = f(t, S t ) = a t S t + b t β t = f x (t, S t )S t + r(t,s t)β t (f t (t, S t ) + /2f xx (t, S t )σ 2 (t, S t )) The arbitrage price at time t of European option with terminal time T is f(t, x) which is the solution of the terminal value problem. (b)from (a) { ft (t, x) = 2 σ2 (t, x)f xx (t, x) r(t, x)xf x (t, x) + r(t, x)f(t, x) f(t, x) = h(x) Hence they replicate h(s T ). a t = f x (t, S t ) b t = r(t,s t)β t (f t (t, S t ) + /2f xx (t, S t )σ 2 (t, S t )) V t = f(t, S t ) = a t S t + b t β t. The price of a security U is defined as the product of two asset prices R and S, i.e., U t = R t S t. Suppose R and S are Ito processes given by dr t = 2( R t )dt + S t db t, ds t = R t S t dt + S t db t. 9

10 Show that {tu t, t } is also an Ito process, and find its expected instantaneous rate of return at time if R =.5 and S = 35. Sol: (a). du t = U R dr + U S ds + U RS drds = S t (2( R t )dt + S t db t ) + R t (R t S t dt + S t db t ) + St 2dt = (2S t 2S t R t + Rt 2 S t + St 2 )dt + (St 2 + R t S t )db t Let f = tu t, by Ito formula dtu t = f t dt + f U du + f 2 UU(dU) 2 = U t dt + tdu t = U t dt + (2tS t 2tU t + tu t R t + tst 2 )dt + (tst 2 + tr t S t )db t (b). E( dtut dt ) = U tdt + 2tS t 2tU t + tu t R t + ts 2 t = U + (t ) = R S =.5 35 = Assume that X follows a geometric Brownian motion with drift α and volatility σ. The economy is risk-neutral, and the risk-free rate of interest is r. A machine prints a certificate worth X(t) at random time t generated by a Poisson arrival process with intensity λ. What is the value of the machine? Hint: By how much should the asset value change at the time the certificate is printed? Assume that V is linear in X. sol: dx = αxdt + σxdw To find the value V of the machine, which pays a cash flow of X(t) determined by the Poisson process: V = V (X) By Ito s lemma: dv = V X dx + 2 V XX(dX) 2 = [αxv X + 2 σ2 X 2 V XX ]dt + σxv X dw Expected cash gain (ECG) = E[dV ] = [αxv X 2 σ2 X 2 V XX ]dt Expected cash flow (ECF) = Xλdt (a cash flow paid with probability λdt) Total Revenue (TR)=ECG+ECF=[αXV X + 2 σ2 X 2 V XX + Xλ]dt Arbitrage theory: [αxv X + 2 σ2 X 2 V XX + Xλ]dt = rv dt

11 αxv X + 2 σ2 X 2 V XX + Xλ = rv Let V = AX + B, V X = A, V XX = r(ax + B) = αxa + + Xλ ra = αa + λrb = A = λ, B = r α Hence, V = λx r α