UMD Probability Qualifying Exams/Aug2006Probability

Problem 1

Consider a four state Markov chain with state space $\{1,2,3,4\}$, initial state $X_0=1$, and transition probability matrix $\left(\begin{array}{cccc}1/4& 1/4& 1/4& 1/4\\ 1/6& 1/3& 1/6& 1/3\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$ (a) Compute $\underset{n\to \mathrm{\infty }}{\lim }P(X_n=3)$. (b) Let $\tau =\inf \{n\ge 0:X_n\in \{3,4\}\}$. Compute $E(\tau )$.

Problem 2

If $X_1,...,X_n$ are independent uniformly distributed random variables on [0,1], then let $X_{(2),n}$ be the second smallest among these numbers. Find a nonrandom sequence $a_n$ such that $T_n=a_n-\log X_{(2),n}$ converges in distribution, and compute the limiting distribution.

Solution

$\begin{array}{cc}P(T_n\le x)& =P(e^{a_n-\log X_{(2),n}}\le e^x)=P(\frac{e^{a_n}}{X_{(2),n}}\le e^x)\\ & =P(X_{(2),n}\ge e^{a_n-x})=n(1-e^{a_n-x}{)}^{n-1}{e}^{a_n-x}+(1-e^{1_n-x}{)}^n\end{array}$

The two terms on the right hand side look like the limit definition of the exponential function. Can we choose $a_n$ appropriately so that it is?

Let $a_n=-\log n$. Then $\begin{array}{cc}P(T_n\le x)& =n(1-\frac{1}{n{e}^x}{)}^{n-1}\frac{1}{n{e}^x}+(1-\frac{1}{n{e}^x}{)}^n\\ & \to e^{-e^x}{e}^{-x}+e^{-e^x}\end{array}$

This is the distribution of $\underset{n\to \mathrm{\infty }}{\lim }{T}_n$.

Problem 3

Suppose that the real-valued random variables $\xi ,\eta$ are independent, that $\xi$ has a bounded density $p(x)$ (for $x\in ℝ$, with respect to Lebesgue measure), and that $\eta$ is integer valued. (a) Prove that $\zeta =\xi +\eta$ has a density. (b) Calculate the density of $\zeta$ in the case where $\xi \sim$ Uniform[0,1] and $\eta \sim$ Poisson(1).

Solution

(a)

$\begin{array}{cc}P(\zeta \le x)& =P(\xi \le x-\eta )=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}P(\eta =n)\underset{-\mathrm{\infty }}{\overset{x-n}{\int }}{p}_{\xi }(t)dt\\ & =\underset{N\to \mathrm{\infty }}{\lim }\sum _{n=-N}^{N}P(\eta =n)\underset{-\mathrm{\infty }}{\overset{x-n}{\int }}{p}_{\xi }(t)dt=\underset{N\to \mathrm{\infty }}{\lim }\sum _{n=-N}^{N}P(\eta =n)\underset{-\mathrm{\infty }}{\overset{x}{\int }}{p}_{\xi }(t-n)dt\\ & =\underset{N\to \mathrm{\infty }}{\lim }\underset{-\mathrm{\infty }}{\overset{x}{\int }}\sum _{n=-N}^{N}P(\eta =n)p_{\xi }(t-n)dt=\underset{-\mathrm{\infty }}{\overset{x}{\int }}\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}P(\eta =n)p_{\xi }(t-n)dt\end{array}$

where the last equality follows from Monotone Convergence Theorem.

Hence, we have shown explicitly that $\zeta$ has a density and it is given by $q_{\zeta }(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}P(\eta =n)p_{\xi }(t-n)$.

(b) When $\xi \sim$ Uniform[0,1] and $\eta \sim$ Poisson(1), we have $p_{\xi }(x)={\mathrm{X}}_{[0,1]}(x)$ and $p_{\eta }(k)=\frac{1}{k!e}$ with support on $k=0,1,2,...$.

Then from part (a), the density will be

$q_{\zeta }(x)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{p}_{\eta }(n)p_{\xi }(x-n)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{k!e}{\mathrm{X}}_{[0,1]}(x)$.

Problem 4

Let $(N(t),t\ge 0)$ be a Poisson process with unit rate, and let $W_{m,n}=\sum _{k=1}^{n}I\{N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\ge 2\}$ where $I(A)$ is the indicator of the event $A$. (a) Find a formula for $E(W_{m,n})$ in terms of $m,n$. (b) Show that if $m=n^{\alpha }$ with $\alpha >1/2$ a fixed constant, then $W_{m,n}\to \mathrm{\infty }$ in probability.

Solution

(a)

We know that $N(\frac{mk}{n})-N(\frac{m(k-1)}{n})$ is distributed as Poisson with parameter $m/n$. So

$P(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\ge 2)=1-P(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})=1\text{ or }2)=1-e^{-m/n}-\frac{m}{n}{e}^{-m/n}$

Then $\begin{array}{cc}E(W_{m,n})& =E[\sum _{k=1}^{n}I(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\ge 2)\\ & =\sum _{k=1}^{n}E(I(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\ge 2))\text{ by linearity}\\ & =\sum _{k=1}^{n}P(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\ge 2)=n(1-e^{-m/n}-\frac{m}{n}{e}^{-m/n})=n-e^{-m/n}(n+m)\end{array}$

(b)

If $\underset{n\to \mathrm{\infty }}{\lim }{W}_{n,n^{\alpha }}=\mathrm{\infty }$ then we must have $I(N(\frac{mk}{n})-N(\frac{m(k-1)}{n})\ge 2)=0$ only finitely often. The probability of this even (from part a) is $e^{-n^{\alpha -1}}(1+n^{\alpha -1})$.

This decays to 0 for $\alpha >1$. Then clearly, we see that the probability of $\underset{n\to \mathrm{\infty }}{\lim }{W}_{n,n^{\alpha }}=\mathrm{\infty }$ is equal to 1.

I don't know how to show the result for $1/2<\alpha <1$...

Problem 5

Let $X_0=0$ and for $n\ge 1,X_n=\sum _{j=1}^n{\xi }_j$ where the r.v.'s ${\xi }_j$ are i.i.d. with $P({\xi }_j=-2)=1/4,P({\xi }_j=1)=3/4$. (a) Prove that there exist constants $a,b$ such that $Y_n=X_n-an$ and $Z_n=\exp (b{X}_n)$ are martingales. (b) If $\tau =\inf \{n\ge 1:X_n=3\}$, then prove that $\tau <\mathrm{\infty }$ almost surely and find $E(\tau )$. (c) Prove that $\exp (b{X}_n)$ is not a uniformly integrable martingale.

Solution

(a)

We want $Y_n=E[Y_{n+1}|{ℱ}_n]$. We can compute both sides of this equation explicitly.

$X_n-an=E[X_{n+1}-a(n+1)|X_n]=(X_n-2)\frac{1}{4}+(X_n+1)\frac{3}{4}-a(n+1)=X_n+1/4-an-a$

Thus if we want this equality to hold we must have $a=1/4$.

Similarly, if we want $Z_n=E[Z_{n+1}|{ℱ}_n]$ then

$e^{b{X}_n}=E[e^{b{X}_{n+1}}|X_n]=\frac{1}{4}{e}^{b(X_n-2)}+\frac{3}{4}{e}^{b(X_n+1)}=e^{b{X}_n}(\frac{1}{4}{e}^{-2b}+\frac{3}{4}{e}^b)$

We can easily check that $b=0$ gives a trivial solution to the equation. Using the substitution $x=e^b$ we can find another solution for $b$. We should get $b=\log (1+\sqrt{13})-\log (6)<0$.

(b)

We've just shown that $Y_n=X_n-\frac{1}{4}n$ is a martingale. Thus, $E[Y_n|Y_0]=E[Y_0]=0$. Then since each $\xi$ is i.i.d., we can apply the Strong Law of Large Numbers to say $1/n(X_n-n/4)\to 0$ almost surely. In other words, $X_n\to \mathrm{\infty }$ almost surely and so certainly $\tau <\mathrm{\infty }$ almost surely.

Now to calculate $E[\tau ]$. We introduce new notation: let ${\tau }_k(x)=\inf \{n\ge 1:X_n=k,X_0=x\}$. Then $\begin{array}{cc}E[{\tau }_n(0)]& =\frac{3}{4}(1+E[{\tau }_n(1)])+\frac{1}{4}(1+E[{\tau }_n(-2)])\\ & =\frac{3}{4}(1+E[{\tau }_{n-1}(0)])+\frac{1}{4}(1+E[{\tau }_{n+2}(0)])=1+\frac{3}{4}{\tau }_{n-1}(0)+\frac{1}{4}{\tau }_{n+2}(0)\end{array}$

by a symmetry argument.

So we can write $E[{\tau }_1(0)]=1+\frac{3}{4}0+\frac{1}{4}{\tau }_3(0)$. But I don't know how to calculate $E[{\tau }_1(0)]$....

(c)

Recall from part (a) that the nontrivial solution for $b$ must be some negative number. Then $\underset{n\to \mathrm{\infty }}{\lim }{Z}_n\to 1$ almost surely by part (a) as well.

However, $Z_n=e^{b{X}_n}\ne 1=E[Z_{\mathrm{\infty }}|{ℱ}_n]$. This by the definition, means the martingale is not right closable. A martingale is right-closable iff uniformly integrable. Thus, we're done.