Math489/889
Stochastic Processes and
Advanced Mathematical Finance
Homework 9

Steve Dunbar

Due Mon, November 22, 2010
  1. Let W1(t) be a Brownian motion and W2(t) be another independent Brownian motion, and ρ is a constant between 1 and 1. Then consider the process X(t) = ρW1(t) + 1 ρ2W2(t). Is this X(t) a Brownian motion?

    Solution: Check each of the four criteria in the definition of Brownian motion:

    1. Consider the increment
      X(t) X(s) = ρ W1(t) W1(s) + 1 ρ2 W2(t) W1(s) .

      Since W1(t) W1(s) N(0,t s) and W2(t) W2(s) N(0,t s) are independent normal random variables, X(t) X(s) is also a normal random variable with mean 0 and variance ρ2(t s) + (1 ρ2)2(t s) = (t s), that is, X(t) X(s) N(0,t s).

    2. For every pair of disjoint time intervals [t1,t2] and [t3,t4], with t1 < t2 t3 < t4, consider the increments X(t4) X(t3) = ρ W1(t4) W1(t3) + 1 ρ2 W2(t4) W1(t3) and X(t2) X(t1) = ρ W1(t2) W1(t1) + 1 ρ2 W2(t2) W1(t1). For notational simplicity, let H = ρ W1(t4) W1(t3) with p.d.f fH(x), J = 1 ρ2 W1(t4) W1(t3) with p.d.f fJ(x), K = ρ W1(t2) W1(t1) with p.d.f fK(x) and L = 1 ρ2 W1(t4) W1(t3) with p.d.f fL(x). Then H, J, K, and L are all independent and the p.d.f. of H + J is f H(x t)fJ(t)dt and the p.d.f of K + L is f K(x t)fL(t)dt. Then the p.d.f. of the joint distribution of H + J and K + L is the product of the two distributions, and the two increments are independent.
    3. Easily X(0) = ρW1(0) + 1 ρ2W2(0) = 0.
    4. As a linear combination of continuous functions, X(t) is a continuous function.
    1. Differentiate the c.d.f. of Ta to obtain the expression for the p.d.f of Ta.
    2. Show that E Ta = for a > 0. (Hint: use the Integral Comparison Test, see any calculus book in the section on Improper Integrals.)

    Solution: The c.d.f. is:

    FTa(t) = 2 2πat exp(y22)dy = 2 1 Φ a t.

    Differentiating with careful use of the chain rule and simplifying:

    fTa(t) = 2 Φ a t a 2t32 = a 2πt32 exp(a22t).
  2. Show that E Ta = for a > 0.

    Solution:

    E Ta =0t a 2πt32 exp(a22t)dt =0 a 2πt12 exp(a22t)dt =0 1 2π exp(12u2)du

    by the change of variables u = ta, with

    du = 1 2atdt

    and u(0) = 0 and u() = . Since exp(12u2) 1 as u , the integral does not converge, or more loosely, the integral and therefore the expectation is infinite.

    Alternatively, write

    E Ta =0t a 2πt32 exp(a22t)dt =01 a 2πt12 exp(a22t)dt +1 a 2πt12 exp(a22t)dt 01 a 2πt12 exp(a22t)dt +1 a 2πt12 exp(a22)dt

    since exp(a22t) exp(a22) for t 1. Since the integral

    1 a 2πt12 exp(a22)dt

    diverges like 01t, and

    01 a 2πt12 exp(a22t)dt 0

    then by the comparison test for improper integral divergence,

    0t a 2πt32 exp(a22t)dt

    diverges.