Professor

(a) What is the mean and variance of the price at time t = 4? SOLUTION: E(X(4)) = E(1 + 2B(4) + 20) = 21 + 2E(B(4)) = 21 + 0 = 21. V ar(X(4)) = V ar(1 + 2B(4) + 20) = V ar(2(B(4)) = 4V ar(B(4)) = 42 = 16. (b) What is the probability that at time t = 4, the price is > 14? SOLUTION: X(4) > 14 if and only if 21 + 2B(4) > 14. 2B(4) is N(0, 16) hence can be re-written as 4Z, where Z ∼ N (0, 1). Thus we want P (21 + 4Z > 14) = P (Z > −(7/4)) = P (Z > −1.75) = P (Z ≤ 1.75) = Φ(1.75) = 0.96. (c) Given that the price is 5.5 at time t = 6, what is the probability that the price is > 8 at time t = 7? SOLUTION: X(7) = X(6)+(X(7)−X(6)) = 5.5+(5+2(B(7)−B(6)) = 10.5+2(B(7)−B(6)). By stationary and independent increments, this has the same distribution as 10.5+2B(1) which has the same distribution as 10.5 + 2Z, where Z ∼ N (0, 1). So we want P (10.5 + 2Z > 8) = P (Z > −1.25) = P (Z ≤ 1.25) = Φ(1.25) = 0.89. (d) What is the probability the price goes up to 4 before down to 1/4? SOLUTION: X(t) = 1 + 2B(t) + 5t hits 4 before 1/4 if and only if 2B(t) + 5t hits a = 3 before −b = −3/4. With μ = 5 and σ = 2 we get 2μ/σ2 = (2× 5)/4 = 2.5.