# UMD Probability Qualifying Exams/Jan2006Probability

## Problem 1

 Let ${\displaystyle X_{1},X_{2},...}$ be i.i.d. r.v.'s such that ${\displaystyle E[X_{n}]=0}$ and ${\displaystyle |X_{n}|\leq 1}$ a.s., and let ${\displaystyle S_{n}=\sum _{k=1}^{n}X_{k}}$. (a) Find a number ${\displaystyle c}$ such that ${\displaystyle S_{n}^{2}-cn}$ is a martingale and justify the martingale property. (b) Define ${\displaystyle \tau _{m}=\inf\{n\geq 1:S_{n}>2m{\text{ or }}S_{n}<-m\}}$. Compute ${\displaystyle \lim _{m\to \infty }P(S_{\tau _{m}}>2m)}$. (c) Compute ${\displaystyle \lim _{m\to \infty }E[\tau _{m}]/m^{2}}$.

### Solution

#### (a)

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

{\displaystyle {\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 ${\displaystyle \sigma ^{2}}$ exists and is finite since each ${\displaystyle |X_{n}|\leq 1}$ almost surely. Therefore, in order to make ${\displaystyle S_{n}^{2}-cn}$ a martingale, we must have ${\displaystyle c=\sigma ^{2}}$.

## Problem 2

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

### Solution

First let us find the distribution of ${\displaystyle \tau _{r}}$:

${\displaystyle P(\tau _{r}

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

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

So then

{\displaystyle {\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 ${\displaystyle u=x^{r+1},dv=e^{-\lambda x}\,dx}$. We get:

{\displaystyle {\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 ${\displaystyle r}$ times and we get

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

## Problem 3

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

### Solution

#### (a)

${\displaystyle \varphi _{X_{kn}}(t)=(1-1/n)+1/ne^{itk^{2}}}$

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