## Math489/889 Stochastic Processes and Advanced Mathematical Finance Homework 7

Due Wed, October 20, 2010
1. Suppose $X$ is a continuous random variable with mean and variance both equal to $20$. What can be said about $ℙ\left[0\le X\le 40\right]$?

Solution: $ℙ\left[0\le X\le 40\right]=ℙ\left[|X-20|\le 20\right]=ℙ\left[|X-\mu |\le 20\right]=1-ℙ\left[|X-\mu |>20\right]\ge 1-20∕2{0}^{2}=19∕20$.

1. Look up the distribution of a Poisson random variable with parameter $\lambda$, state it and use that to calculate $ℙ\left[X\ge 1\right]$, and $ℙ\left[X\ge 2\right]$ where $X$ is a Poisson random variable with parameter $1$.
2. Given that the m.g.f. ${\varphi }_{X}\left(t\right)$ of a Poisson random variable with parameter $\lambda$ is ${e}^{\lambda \left({e}^{t}-1\right)}$, show that the sum of independent Poisson random variables ${X}_{1}$ with parameter ${\lambda }_{1}$ and ${X}_{2}$ with parameter ${\lambda }_{2}$ is again Poisson with parameter ${\lambda }_{1}+{\lambda }_{2}$.
3. Using the fact that the sum of two independent Poisson random variables with means ${\lambda }_{1}$ and ${\lambda }_{2}$ is again Poisson with mean ${\lambda }_{1}+{\lambda }_{2}$ find the exact probability that $ℙ\left[{X}_{1}+\cdots +{X}_{10}>15\right]$. Take ${\lambda }_{i}=1$ where each ${X}_{i}$ is a Poisson random variable with parameter $1$.
4. Use the Markov Inequality to get a bound on $ℙ\left[{X}_{1}+\cdots +{X}_{10}>15\right]$ where each ${X}_{i}$ is a Poisson random variable with parameter $1$.

Solution:

1. $ℙ\left[X\ge 1\right]=1-ℙ\left[X=0\right]=1-\frac{{e}^{-1}{\left(1\right)}^{0}}{0!}=1-{e}^{-1}$

and

$ℙ\left[X\ge 2\right]=1-ℙ\left[X=0,1\right]=1-\frac{{e}^{-1}{\left(1\right)}^{0}}{0!}-\frac{{e}^{-1}{\left(1\right)}^{1}}{1!}=1-2{e}^{-1}$

2. The mgf of ${X}_{1}+{X}_{2}$ is the product of the respective mgfs of ${X}_{1}$ and ${X}_{2}$ that is ${e}^{{\lambda }_{1}\left({e}^{t}-1\right)}\cdot {e}^{{\lambda }_{2}\left({e}^{t}-1\right)}={e}^{\left({\lambda }_{1}+{\lambda }_{2}\right)\left({e}^{t}-1\right)}$ which is the mgf of a Poisson random variable with parameter ${\lambda }_{1}+{\lambda }_{2}$.
3. By part (b), ${X}_{1}+\cdots +{X}_{10}$ is Poisson with parameter 10, so the probability is
$ℙ\left[{X}_{1}+\dots {X}_{10}>15\right]\approx 0.0487404033.$

4. ${X}_{1},{X}_{2},\dots ,{X}_{10}$ are all positive (and integer-valued) random variables and $E\left[{X}_{1}+\cdots +{X}_{10}\right]=10$. Then by the Markov inequality bound, $ℙ\left[{X}_{1}+\cdots +{X}_{10}>15\right]=ℙ\left[{X}_{1}+\cdots +{X}_{10}\ge 16\right]=10∕16=0.625$. The Markov bound is not sharp, although it is true.
2. A first simple assumption is that the daily change of a company’s stock on the stock market is a random variable with mean $0$ and variance ${\sigma }^{2}$. That is, if ${S}_{n}$ represents the price of the stock on day $n$ with ${S}_{0}$ given, then
${S}_{n}={S}_{n-1}+{X}_{n},n\ge 1$

where ${X}_{1},{X}_{2},\dots$ are independent, identically distributed continuous random variables with mean $0$ and variance ${\sigma }^{2}$. (Note that this is an additive assumption about the change in a stock price. In the binomial tree models, we assumed that a stock’s price changes by a multiplicative factor up or down. We will have more to say about these two distinct models later.) Suppose that a stock’s price today is $100$. If ${\sigma }^{2}=1$, what can you say about the probability that after $10$ days, the stock’s price will be between $95$ and $105$ on the tenth day?

Solution: Let $X={\sum }_{i=1}^{10}{X}_{i}$, so $E\left[X\right]=0$ and $Var\left[X\right]=10$ because of the independence. Then by Chebyshev’s inequality

$\begin{array}{llll}\hfill ℙ\left[-5<{X}_{1}+\dots {X}_{10}<5\right]& =ℙ\left[|X|<5\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =1-ℙ\left[|X|\ge 5\right]\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \ge 1-10∕{5}^{2}=1-10∕25=3∕5.\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Note that the problem does not assume that the daily changes ${X}_{i}$ are normal, so we cannot use the normal distribution for the sum. With only $10$ summands, the Central Limit Theorem cannot be reliably used.

3. Find the moment generating function ${\varphi }_{X}\left(t\right)=E\left[exp\left(tX\right)\right]$ of the random variable $X$ which takes values $1$ with probability $1∕2$ and $-1$ with probability $1∕2$. Show directly (that is, without using Taylor polynomial approximations) that ${\varphi }_{X}{\left(t∕\sqrt{n}\right)}^{n}\to exp\left({t}^{2}∕2\right)$. (Hint: Use L’Hopital’s Theorem to evaluate the limit, after taking logarithms of both sides.)

Solutions: The m.g.f. is

${\varphi }_{X}\left(t\right)=\frac{{e}^{-t}+{e}^{t}}{2}=sinh\left(t\right).$

Then

${\varphi }_{X}{\left(t∕\sqrt{n}\right)}^{n}={\left(\frac{{e}^{-t∕\sqrt{n}}+{e}^{t∕\sqrt{n}}}{2}\right)}^{n}$

Then let

$\begin{array}{llll}\hfill L\left(t,n\right)& =log\left({\left(\frac{{e}^{-t∕\sqrt{n}}+{e}^{t∕\sqrt{n}}}{2}\right)}^{n}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =nlog\left(\left(\frac{{e}^{-t∕\sqrt{n}}+{e}^{t∕\sqrt{n}}}{2}\right)\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =log\left(\left(\frac{{e}^{-t∕\sqrt{n}}+{e}^{t∕\sqrt{n}}}{2}\right)\right)∕\left(1∕n\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Now apply L’Hopital’s Rule twice to evaluate ${lim}_{n\to \infty }L\left(t,n\right)$. (Details are omitted.) The result is:

$\underset{n\to \infty }{lim}L\left(t,n\right)=\frac{{t}^{2}}{2}.$

Therefore ${\varphi }_{X}\left(t∕\sqrt{n}\right)\to exp\left({t}^{2}∕2\right)$ as $n\to \infty$.