Solution for Assignment 3 Question 1 a) Determine the distribution of Z1 (t). Solution: It is easy to see Z1 (t) can be written as the Riemann sum of normal random variables. Using the property that the linear combination of normal random variables is still a normal random variable, Z1 (t) is a normal random variable with mean 0. To obtain the variance, we compute it as following Z tZ t Z tZ t Z tZ t min(u, v) du dv E(Wu Wv ) du dv = Wu Wv du dv) = V ar(Z1 (t)) = E( 0 = Z t ( 0 Z v u du) dv + 0 0 0 0 0 Z t ( 0 Z t v du) dv = v 0 t3 . 3 t3 ) 3 Hence Z1 (t) ∼ N (0, b) Determine the covariance Cov[Wt , Z1 (t)]. Z t Z t Z t t2 s ds = E(Wt Ws ) ds = Cov[Wt , Z1 (t)] = E(Wt Ws ds) = 2 0 0 0 c) Determine the variance of Z2 (t). Consider g(Wt ) = Wt2 /2, we get 1 t dg = Ws dWs + dt ⇒ Z2 (t) = Wt2 /2 − . 2 2 Thus we have 1 V ar[Z2 (t)] = V ar(Wt2 /2) = t2 . 2 d) Is Z3 (t) as a process a martingale? To see whether it is a martingale or not, we consider the SDE Z3 (t) satisfies, 1 dZ3 (t) = (−dt + 2Wt dWt + dt) = −Wt dWt . 2 Since the coefficient of the dt term is 0, we conclude that Z3 (t) is a martingale. 1 Question 2 Let X(t) = [at + bt2 + ctWt + γWt2 ]eWt − 2 t . Find the value of a, b, c, d to make it a martingale. Solution: Apply Ito’s lemma to 1 g(x, t) = [at + bt2 + ctx + γx2 ]ex− 2 t . 1 Thus X(t) = g(t, Wt ) and hence 1 1 dX(t) = [(2γ + c)Wt + (a + γ) + (c + 2b)t]eWt − 2 t dt + [(a + c)t + bt2 + (2γ + ct)Wt + γWt2 ]eWt − 2 t dWt . Zero drift implies c + 2b = 0, 2γ + c = 0, a + γ = 0 ⇒ γ = b, a = −γ, c = −2γ. For example, take a = 1, b = −1, c = 2, γ = −1. Question 3 Solution: 1. By solving the SDE dSt = e−0.2t dt + 0.3e−0.6t dWt , St We have d log(St ) = (e−0.2t − 0.045e−1.2t ) dt + 0.3e−0.6t dWt . Z t −0.2t −1.2t e−0.6s dWs ). ⇒ St = S0 exp(5(1 − e ) − 0.0375(1 − e ) + 0.3 0 2. For any fixed t, get the distribution of St , and its mean and variance. Obviously, St follows a log-normal distribution. To obtain its mean and variance, we just need to focus on stochastic part. Obviously, Rt exp(0.3 0 e−0.6s dWs ) follows a log-normal distribution with mean e0.0375(1−e −1.2t ) and variance (e0.0375(1−e −1.2t ) − 1)e0.075(1−e −1.2t ) . Hence, E(St ) = S0 exp(5(1 − e−0.2t )), and V ar(St ) = S02 exp(10(1 − e−0.2t ))(e0.0375(1−e −1.2t ) − 1). 3. The probability that stock have higher return than money market, we get P (e5(1−e −0.1 )−0.0375(1−e−0.6 )+0.3 R 0.5 0 e−0.6s dWs 2 > e0.015 ) ⇒ P (Z ≥ −2.413) = 0.992