## Math489/889 Stochastic Processes and Advanced Mathematical Finance Homework 9

Due Mon, November 22, 2010
1. Let ${W}_{1}\left(t\right)$ be a Brownian motion and ${W}_{2}\left(t\right)$ be another independent Brownian motion, and $\rho$ is a constant between $-1$ and $1$. Then consider the process $X\left(t\right)=\rho {W}_{1}\left(t\right)+\sqrt{1-{\rho }^{2}}{W}_{2}\left(t\right)$. Is this $X\left(t\right)$ a Brownian motion?

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

1. Consider the increment
$X\left(t\right)-X\left(s\right)=\rho \left({W}_{1}\left(t\right)-{W}_{1}\left(s\right)\right)+\sqrt{1-{\rho }^{2}}\left({W}_{2}\left(t\right)-{W}_{1}\left(s\right)\right).$

Since $\left({W}_{1}\left(t\right)-{W}_{1}\left(s\right)\right)\sim N\left(0,t-s\right)$ and $\left({W}_{2}\left(t\right)-{W}_{2}\left(s\right)\right)\sim N\left(0,t-s\right)$ are independent normal random variables, $X\left(t\right)-X\left(s\right)$ is also a normal random variable with mean $0$ and variance ${\rho }^{2}\left(t-s\right)+{\left(\sqrt{1-{\rho }^{2}}\right)}^{2}\left(t-s\right)=\left(t-s\right)$, that is, $X\left(t\right)-X\left(s\right)\sim N\left(0,t-s\right)$.

2. For every pair of disjoint time intervals $\left[{t}_{1},{t}_{2}\right]$ and $\left[{t}_{3},{t}_{4}\right]$, with ${t}_{1}<{t}_{2}\le {t}_{3}<{t}_{4}$, consider the increments $X\left({t}_{4}\right)-X\left({t}_{3}\right)=\rho \left({W}_{1}\left({t}_{4}\right)-{W}_{1}\left({t}_{3}\right)\right)+\sqrt{1-{\rho }^{2}}\left({W}_{2}\left({t}_{4}\right)-{W}_{1}\left({t}_{3}\right)\right)$ and $X\left({t}_{2}\right)-X\left({t}_{1}\right)=\rho \left({W}_{1}\left({t}_{2}\right)-{W}_{1}\left({t}_{1}\right)\right)+\sqrt{1-{\rho }^{2}}\left({W}_{2}\left({t}_{2}\right)-{W}_{1}\left({t}_{1}\right)\right)$. For notational simplicity, let $H=\rho \left({W}_{1}\left({t}_{4}\right)-{W}_{1}\left({t}_{3}\right)\right)$ with p.d.f ${f}_{H}\left(x\right)$, $J=\sqrt{1-{\rho }^{2}}\left({W}_{1}\left({t}_{4}\right)-{W}_{1}\left({t}_{3}\right)\right)$ with p.d.f ${f}_{J}\left(x\right)$, $K=\rho \left({W}_{1}\left({t}_{2}\right)-{W}_{1}\left({t}_{1}\right)\right)$ with p.d.f ${f}_{K}\left(x\right)$ and $L=\sqrt{1-{\rho }^{2}}\left({W}_{1}\left({t}_{4}\right)-{W}_{1}\left({t}_{3}\right)\right)$ with p.d.f ${f}_{L}\left(x\right)$. Then $H$, $J$, $K$, and $L$ are all independent and the p.d.f. of $H+J$ is ${\int }_{-\infty }^{\infty }{f}_{H}\left(x-t\right){f}_{J}\left(t\right)\phantom{\rule{0em}{0ex}}dt$ and the p.d.f of $K+L$ is ${\int }_{-\infty }^{\infty }{f}_{K}\left(x-t\right){f}_{L}\left(t\right)\phantom{\rule{0em}{0ex}}dt$. 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 $X\left(0\right)=\rho {W}_{1}\left(0\right)+\sqrt{1-{\rho }^{2}}{W}_{2}\left(0\right)=0$.
4. As a linear combination of continuous functions, $X\left(t\right)$ 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 $E\left[{T}_{a}\right]=\infty$ 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:

$\begin{array}{llll}\hfill {F}_{{T}_{a}}\left(t\right)& =\frac{2}{\sqrt{2\pi }}{\int }_{a∕\sqrt{t}}^{\infty }exp\left(-{y}^{2}∕2\right)\phantom{\rule{0em}{0ex}}dy\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\left(1-\Phi \left(\frac{a}{\sqrt{t}}\right)\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Differentiating with careful use of the chain rule and simplifying:

$\begin{array}{llll}\hfill {f}_{{T}_{a}}\left(t\right)& =2\left(-{\Phi }^{\prime }\left(\frac{a}{\sqrt{t}}\right)\right)\frac{-a}{2{t}^{3∕2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{a}{\sqrt{2\pi }{t}^{3∕2}}exp\left(-{a}^{2}∕2t\right).\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$
2. Show that $E\left[{T}_{a}\right]=\infty$ for $a>0$.

Solution:

$\begin{array}{llll}\hfill E\left[{T}_{a}\right]& ={\int }_{0}^{\infty }t\cdot \frac{a}{\sqrt{2\pi }{t}^{3∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{0}^{\infty }\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{0}^{\infty }\frac{1}{\sqrt{2\pi }}exp\left(-1∕2{u}^{2}\right)\phantom{\rule{0em}{0ex}}du\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

by the change of variables $u=\sqrt{t}∕a$, with

$du=\frac{1}{2a\sqrt{t}}\phantom{\rule{3.26288pt}{0ex}}dt$

and $u\left(0\right)=0$ and $u\left(\infty \right)=\infty$. Since $exp\left(-1∕2{u}^{2}\right)\to 1$ as $u\to \infty$, the integral does not converge, or more loosely, the integral and therefore the expectation is infinite.

Alternatively, write

$\begin{array}{llll}\hfill E\left[{T}_{a}\right]& ={\int }_{0}^{\infty }t\cdot \frac{a}{\sqrt{2\pi }{t}^{3∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={\int }_{0}^{1}\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+{\int }_{1}^{\infty }\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \ge {\int }_{0}^{1}\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \phantom{\rule{2em}{0ex}}+{\int }_{1}^{\infty }\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2\right)\phantom{\rule{0em}{0ex}}dt\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

since $exp\left(-{a}^{2}∕2t\right)\ge exp\left(-{a}^{2}∕2\right)$ for $t\ge 1$. Since the integral

${\int }_{1}^{\infty }\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2\right)\phantom{\rule{0em}{0ex}}dt$

diverges like ${\int }_{0}^{\infty }1∕\sqrt{t}$, and

${\int }_{0}^{1}\frac{a}{\sqrt{2\pi }{t}^{1∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt\ge 0$

then by the comparison test for improper integral divergence,

${\int }_{0}^{\infty }t\cdot \frac{a}{\sqrt{2\pi }{t}^{3∕2}}exp\left(-{a}^{2}∕2t\right)\phantom{\rule{0em}{0ex}}dt$

diverges.