# UMD Probability Qualifying Exams/Jan2006Probability

## Problem 1

 Let $X_{1},X_{2},...$ be i.i.d. r.v.'s such that $E[X_{n}]=0$ and $|X_{n}|\leq 1$ 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 $\tau _{m}=\inf\{n\geq 1:S_{n}>2m{\text{ or }}S_{n}<-m\}$ . Compute $\lim _{m\to \infty }P(S_{\tau _{m}}>2m)$ . (c) Compute $\lim _{m\to \infty }E[\tau _{m}]/m^{2}$ .

### Solution

#### (a)

Each $S_{n}$ is clearly ${\mathcal {F}}_{n}$ -measureable and finite a.s. (Hence $L^{1}$ ). Therefore we only need to verify the martingale property. That is, we want to show $S_{n}^{2}-cn=E[S_{n+1}^{2}-c(n+1_{|}{\mathcal {F}}_{n}]$ {\begin{aligned}S_{n}^{2}-cn=E[S_{n+1}^{2}-c(n+1_{|}{\mathcal {F}}_{n}]&=S_{n}^{2}+E[X_{n+1}^{2}+2\sum _{j=1}^{n}X_{n+1}X_{j}|{\mathcal {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{aligned}} We can assert that $\sigma ^{2}$ exists and is finite since each $|X_{n}|\leq 1$ 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 $r\geq 1$ , let $\tau _{r}=\inf\{t>0:N_{1}(t)\geq r\}$ . 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} Thus by the chain rule, our random variable $\tau _{r}$ has probability density function

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

{\begin{aligned}E[N_{2}(\tau _{r}^{2})]=E[E[N_{2}(t)|\tau _{r}^{2}=t]]&=\int _{0}^{\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}^{\infty }x^{r+1}e^{-\lambda x}\,dx\end{aligned}} Now integrate the remaining integral by parts letting $u=x^{r+1},dv=e^{-\lambda x}\,dx$ . We get:

{\begin{aligned}E[N_{2}(\tau _{r}^{2})]&={\frac {\lambda ^{r+2}}{(r-1)!}}[-{\frac {1}{\lambda }}e^{-\lambda x}x^{r+1}|_{0}^{\infty }+\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\&={\frac {\lambda ^{r+2}}{(r-1)!}}[\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\&={\frac {\lambda ^{r+1}(r+1)}{(r-1)!}}[\int _{0}^{\infty }{\frac {1}{\lambda }}^{r+1}x^{r}e^{-\lambda x}\,dx]\\\end{aligned}} Repeat integration by parts another $r$ times and we get

$E[N_{2}(\tau _{r}^{2})]=(r+1)r\lambda \int _{0}^{\infty }e^{-\lambda x}\,dx=\lambda (r+1)r{\frac {-1}{\lambda }}e^{-\lambda x}|_{0}^{\infty }=(r+1)r$ ## Problem 3

 Let $X_{kn},1\leq k\leq n$ 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)

$\varphi _{X_{kn}}(t)=(1-1/n)+1/ne^{itk^{2}}$ Then by independence, we have $\varphi _{X_{n}}(t)=\prod _{k=1}^{n}\varphi _{X_{kn}}=\prod _{k=1}^{n}(1-1/n)+1/ne^{itk^{2}}$ 