Math489/889 Stochastic Processes and Advanced Mathematical Finance Homework 9

Math489/889
Stochastic Processes and
Advanced Mathematical Finance
Homework 9

Steve Dunbar

Due Mon, November 22, 2010

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 $X (t) - X (s) = ρ (W_{1} (t) - W_{1} (s)) + \sqrt{1 - ρ^{2}} (W_{2} (t) - W_{1} (s)) .$
  Since $(W_{1} (t) - W_{1} (s)) \sim N (0, t - s)$ and $(W_{2} (t) - W_{2} (s)) \sim N (0, t - s)$ are independent normal random variables, $X (t) - X (s)$ is also a normal random variable with mean $0$ and variance $ρ^{2} (t - s) + {(\sqrt{1 - ρ^{2}})}^{2} (t - s) = (t - s)$ , that is, $X (t) - X (s) \sim N (0, t - s)$ .
2. For every pair of disjoint time intervals $[t_{1}, t_{2}]$ and $[t_{3}, t_{4}]$ , with $t_{1} < t_{2} \leq t_{3} < t_{4}$ , consider the increments $X (t_{4}) - X (t_{3}) = ρ (W_{1} (t_{4}) - W_{1} (t_{3})) + \sqrt{1 - ρ^{2}} (W_{2} (t_{4}) - W_{1} (t_{3}))$ and $X (t_{2}) - X (t_{1}) = ρ (W_{1} (t_{2}) - W_{1} (t_{1})) + \sqrt{1 - ρ^{2}} (W_{2} (t_{2}) - W_{1} (t_{1}))$ . For notational simplicity, let $H = ρ (W_{1} (t_{4}) - W_{1} (t_{3}))$ with p.d.f $f_{H} (x)$ , $J = \sqrt{1 - ρ^{2}} (W_{1} (t_{4}) - W_{1} (t_{3}))$ with p.d.f $f_{J} (x)$ , $K = ρ (W_{1} (t_{2}) - W_{1} (t_{1}))$ with p.d.f $f_{K} (x)$ and $L = \sqrt{1 - ρ^{2}} (W_{1} (t_{4}) - W_{1} (t_{3}))$ with p.d.f $f_{L} (x)$ . Then $H$ , $J$ , $K$ , and $L$ are all independent and the p.d.f. of $H + J$ is $\int_{- \infty}^{\infty} f_{H} (x - t) f_{J} (t) d t$ and the p.d.f of $K + L$ is $\int_{- \infty}^{\infty} f_{K} (x - t) f_{L} (t) d t$ . 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 $X (0) = ρ W_{1} (0) + \sqrt{1 - ρ^{2}} W_{2} (0) = 0$ .
4. As a linear combination of continuous functions, $X (t)$ is a continuous function.
1. Differentiate the c.d.f. of $T_{a}$ to obtain the expression for the p.d.f of $T_{a}$ .
2. Show that $E [T_{a}] = \infty$ 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 2π∫ a∕t∞ exp(−y2∕2)dy = 2 1 − Φ a t.
Differentiating with careful use of the chain rule and simplifying:
fTa(t) = 2 −Φ′ a t − a 2t3∕2 = a 2πt3∕2 exp(−a2∕2t).
Show that $E [T_{a}] = \infty$ for $a > 0$ .
Solution:
$\begin{array}{l} E [T_{a}] & = \int_{0}^{\infty} t \cdot \frac{a}{\sqrt{2 π} t^{3 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \\ = \int_{0}^{\infty} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \\ = \int_{0}^{\infty} \frac{1}{\sqrt{2 π}} exp (- 1 ∕ 2 u^{2}) d u \end{array}$
by the change of variables $u = \sqrt{t} ∕ a$ , with
$d u = \frac{1}{2 a \sqrt{t}} d t$
and $u (0) = 0$ and $u (\infty) = \infty$ . Since $exp (- 1 ∕ 2 u^{2}) \to 1$ as $u \to \infty$ , the integral does not converge, or more loosely, the integral and therefore the expectation is infinite.
Alternatively, write
$\begin{array}{l} E [T_{a}] & = \int_{0}^{\infty} t \cdot \frac{a}{\sqrt{2 π} t^{3 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \\ = \int_{0}^{1} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \\ + \int_{1}^{\infty} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \\ \geq \int_{0}^{1} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \\ + \int_{1}^{\infty} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2) d t \end{array}$
since $exp (- a^{2} ∕ 2 t) \geq exp (- a^{2} ∕ 2)$ for $t \geq 1$ . Since the integral
$\int_{1}^{\infty} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2) d t$
diverges like $\int_{0}^{\infty} 1 ∕ \sqrt{t}$ , and
$\int_{0}^{1} \frac{a}{\sqrt{2 π} t^{1 ∕ 2}} exp (- a^{2} ∕ 2 t) d t \geq 0$
then by the comparison test for improper integral divergence,
$\int_{0}^{\infty} t \cdot \frac{a}{\sqrt{2 π} t^{3 ∕ 2}} exp (- a^{2} ∕ 2 t) d t$
diverges.