Math 3589 Introduction to Financial Mathematics Homework Assignment #8 Solutions Ohio State University

Introduction to Financial Mathematics Homework Assignment #8 Solutions Exercise 16. Consider the two period binomial model, with the stock price at time t = 0, S0 = 4, the “up factor” u = 2, “down factor” d = 1/2, and risk free interest rate r = 1/4 so that ˜p = 1/2. Assume in each period the probability P[H] = 3/4. Solve the two-period investors problem for ∆i , i = 0, 1, if U(x) = −2e −x and X0 = 10. Solution. Let f(x1, x2, x3, x4) = 9 16 −2e −x4
+ 3 16 −2e −xx
+ 3 16 −2e −x2
+ 1 16 −2e −x1
and g(x1, x2, x3, x4) = 16 25  1 4 x4 + 1 4 x3 + 1 4 x2 + 1 4 x1  − 10 = 0. For ¯x = (x1, x2, x3, x4), define the Lagrangian function L(¯x, λ) = f(¯x) − λg(¯x) = −18 16 e −x4 − 6 16 e −x3 − 6 16 e −x2 − 2 16 e −x1 − λ  16 100 (x4 + x3 + x2 + x1) − 10 . And we find the 5 equations    ∂ ∂x1 L(¯x, λ) = 2 16 e −x1 − 16 100λ = 0 ∂ ∂x2 L(¯x, λ) = 6 16 e −x2 − 16 100λ = 0 ∂ ∂x3 L(¯x, λ) = 6 16 e −x3 − 16 100λ = 0 ∂ ∂x4 L(¯x, λ) = 18 16 e −x4 − 16 100λ = 0 ∂ ∂λL(¯x, λ) = 16 100 (x4 + x3 + x2 + x1) − 10 = 0. (0.1) From here we see that x2 = x3, and setting the first and second equal, we see 2 16 e −x1 = 6 16 e −x2 =⇒ e −x1 = 3e −x2 =⇒ −x1 = ln(3) − x2 =⇒ x1 = x2 − ln(3). Similarly, 18 16 e −x4 = 6 16 e −x2 =⇒ 3e −x4 = e −x2 =⇒ x4 = x2 + ln(3). Plugging these into the last equation, we get 16 100 (4x2) = 10 =⇒ x2 = 12 Solving for x4, we find x4 ≈ 16.7236 and x1 ≈ 14.526. Now we find ∆1[H] = ln(3) 12 ≈ 0.091, ∆1[T] = ln(3) 3 ≈ 0.366 and ∆0 = X1[H] − X1[T] S1[H] − S1[T] = 4 5 1 2 125 8 + ln(3)
+