UMD Probability Qualifying Exams/Jan2006Probability

Problem 1

Let $X_1,X_2,\mathrm{...}$ be i.i.d. r.v.'s such that $E[X_n]=0$ and $|X_n|\le 1$ a.s., and let $S_n=\sum _{k=1}^n{X}_k$. (a) Find a number $c$ such that $S_n^2-cn$ is a martingale and justify the martingale property. (b) Define ${\tau }_m=inf\{n\ge 1:S_n>2m\text{ or }{S}_n<-m\}$. Compute $\underset{m\to \mathrm{\infty }}{lim}P(S_{{\tau }_m}>2m)$. (c) Compute $\underset{m\to \mathrm{\infty }}{lim}E[{\tau }_m]/m^2$.

Solution

(a)

Each $S_n$ is clearly ${ℱ}_n$-measureable and finite a.s. (Hence $L^1$). Therefore we only need to verify the martingale property. That is, we want to show $S_n^2-cn=E[S_{n+1}^2-c(n+1_{|}{ℱ}_n]$

$\begin{array}{cc}{S}_n^2-cn=E[S_{n+1}^2-c(n+1_{|}{ℱ}_n]& =S_n^2+E[X_{n+1}^2+2\sum _{j=1}^n{X}_{n+1}{X}_j|{ℱ}_n]-cn-c\\ & =S_n^2-cn+E[X_{n+1}^2]+2\sum _{j=1}^n{X}_{j}E[X_{n+1}]\text{ by independence}\\ & =S_n^2-cn+Var[X_{n+1}]=S_n^2-cn+{\sigma }^2\end{array}$

We can assert that ${\sigma }^2$ exists and is finite since each $|X_n|\le 1$ almost surely. Therefore, in order to make $S_n^2-cn$ a martingale, we must have $c={\sigma }^2$.

Problem 2

Let $N_1(t),N_2(t)$ be independent Poisson processes with respective parameters $\lambda ,{\lambda }^2$, where $\lambda$ is an unspecified positive real number. For each $r\ge 1$, let ${\tau }_r=inf\{t>0:N_1(t)\ge r\}$. Show that ${\alpha }_r=E[N_2({\tau }_r^2)]$ does not depend on $\lambda$ and find ${\alpha }_r$ explicitly.

Solution

First let us find the distribution of ${\tau }_r$:

$P({\tau }_r<x)=P(N_1(x)\ge r)=\sum _{k=r}^{\mathrm{\infty }}\frac{(\lambda x{)}^k}{k!}{e}^{-\lambda x}$

Thus by the chain rule, our random variable ${\tau }_r$ has probability density function

$p_{{\tau }_r}(x)=\sum _{k=r}^{\mathrm{\infty }}\frac{k\lambda (\lambda x{)}^{k-1}}{k!}{e}^{-\lambda x}+\frac{(\lambda x{)}^k}{k!}(-\lambda )e^{-\lambda x}=\lambda \frac{(\lambda x{)}^{r-1}}{(r-1)!}{e}^{-\lambda x}$

So then

$\begin{array}{cc}E[N_2({\tau }_r^2)]=E[E[N_2(t)|{\tau }_r^2=t]]& ={\int }_0^{\mathrm{\infty }}{x}^2{\lambda }^2\lambda \frac{{\lambda }^{r-1}}{(r-1)!}{x}^{r-1}{e}^{-\lambda x}dx\\ & =\frac{{\lambda }^{r+2}}{(r-1)!}{\int }_0^{\mathrm{\infty }}{x}^{r+1}{e}^{-\lambda x}dx\end{array}$

Now integrate the remaining integral by parts letting $u=x^{r+1},dv=e^{-\lambda x}dx$. We get:

$\begin{array}{cc}E[N_2({\tau }_r^2)]& =\frac{{\lambda }^{r+2}}{(r-1)!}[-\frac{1}{\lambda }{e}^{-\lambda x}{x}^{r+1}{|}_0^{\mathrm{\infty }}+{\int }_0^{\mathrm{\infty }}{\frac{1}{\lambda }}^{r+1}{x}^r{e}^{-\lambda x}dx]\\ & =\frac{{\lambda }^{r+2}}{(r-1)!}[{\int }_0^{\mathrm{\infty }}{\frac{1}{\lambda }}^{r+1}{x}^r{e}^{-\lambda x}dx]\\ & =\frac{{\lambda }^{r+1}(r+1)}{(r-1)!}[{\int }_0^{\mathrm{\infty }}{\frac{1}{\lambda }}^{r+1}{x}^r{e}^{-\lambda x}dx]\\ \end{array}$

Repeat integration by parts another $r$ times and we get

$E[N_2({\tau }_r^2)]=(r+1)r\lambda {\int }_0^{\mathrm{\infty }}{e}^{-\lambda x}dx=\lambda (r+1)r\frac{-1}{\lambda }{e}^{-\lambda x}{|}_0^{\mathrm{\infty }}=(r+1)r$

Problem 3

Let $X_{kn},1\le k\le n$ be independent random variables such that $P(X_{kn}=0)=1-1/n,P(X_{kn}=k^2)=1/n$ (a) Find the characteristic function of $S-n=\sum _{k=1}^n{X}_{kn}$. (b) Show that $S_n/n^2$ converges in distribution to a non-degenerate random variable.

Solution

(a)

${\phi }_{X_{kn}}(t)=(1-1/n)+1/n{e}^{it{k}^2}$

Then by independence, we have ${\phi }_{X_n}(t)=\prod _{k=1}^n{\phi }_{X_{kn}}=\prod _{k=1}^n(1-1/n)+1/n{e}^{it{k}^2}$