# UMD Probability Qualifying Exams/Aug2006Probability

## Problem 1

 Consider a four state Markov chain with state space ${\displaystyle \{1,2,3,4\}}$ , initial state ${\displaystyle X_{0}=1}$ , and transition probability matrix ${\displaystyle {\begin{pmatrix}1/4&1/4&1/4&1/4\\1/6&1/3&1/6&1/3\\0&0&1&0\\0&0&0&1\end{pmatrix}}}$  (a) Compute ${\displaystyle \lim _{n\to \infty }P(X_{n}=3)}$ . (b) Let ${\displaystyle \tau =\inf\{n\geq 0:X_{n}\in \{3,4\}\}}$ . Compute ${\displaystyle E(\tau )}$ .

## Problem 2

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

### Solution

{\displaystyle {\begin{aligned}P(T_{n}\leq x)&=P(e^{a_{n}-\log X_{(2),n}}\leq e^{x})=P({\frac {e^{a_{n}}}{X_{(2),n}}}\leq e^{x})\\&=P(X_{(2),n}\geq e^{a_{n}-x})=n(1-e^{a_{n}-x})^{n-1}e^{a_{n}-x}+(1-e^{1_{n}-x})^{n}\end{aligned}}}

The two terms on the right hand side look like the limit definition of the exponential function. Can we choose ${\displaystyle a_{n}}$  appropriately so that it is?

Let ${\displaystyle a_{n}=-\log n}$ . Then {\displaystyle {\begin{aligned}P(T_{n}\leq x)&=n(1-{\frac {1}{ne^{x}}})^{n-1}{\frac {1}{ne^{x}}}+(1-{\frac {1}{ne^{x}}})^{n}\\&\to e^{-e^{x}}e^{-x}+e^{-e^{x}}\end{aligned}}}

This is the distribution of ${\displaystyle \lim _{n\to \infty }T_{n}}$ .

## Problem 3

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

### Solution

(a)

{\displaystyle {\begin{aligned}P(\zeta \leq x)&=P(\xi \leq x-\eta )=\sum _{n=-\infty }^{\infty }P(\eta =n)\int _{-\infty }^{x-n}p_{\xi }(t)\,dt\\&=\lim _{N\to \infty }\sum _{n=-N}^{N}P(\eta =n)\int _{-\infty }^{x-n}p_{\xi }(t)\,dt=\lim _{N\to \infty }\sum _{n=-N}^{N}P(\eta =n)\int _{-\infty }^{x}p_{\xi }(t-n)\,dt\\&=\lim _{N\to \infty }\int _{-\infty }^{x}\sum _{n=-N}^{N}P(\eta =n)p_{\xi }(t-n)\,dt=\int _{-\infty }^{x}\sum _{n=-\infty }^{\infty }P(\eta =n)p_{\xi }(t-n)\,dt\end{aligned}}}

where the last equality follows from Monotone Convergence Theorem.

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

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

Then from part (a), the density will be

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

## Problem 4

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

### Solution

#### (a)

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

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

Then {\displaystyle {\begin{aligned}E(W_{m,n})&=E[\sum _{k=1}^{n}I(N({\frac {mk}{n}})-N({\frac {m(k-1)}{n}})\geq 2)\\&=\sum _{k=1}^{n}E(I(N({\frac {mk}{n}})-N({\frac {m(k-1)}{n}})\geq 2)){\text{ by linearity}}\\&=\sum _{k=1}^{n}P(N({\frac {mk}{n}})-N({\frac {m(k-1)}{n}})\geq 2)=n\left(1-e^{-m/n}-{\frac {m}{n}}e^{-m/n}\right)=n-e^{-m/n}(n+m)\end{aligned}}}

#### (b)

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

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

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

## Problem 5

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

### Solution

#### (a)

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

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

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

Similarly, if we want ${\displaystyle Z_{n}=E[Z_{n+1}|{\mathcal {F}}_{n}]}$  then

${\displaystyle e^{bX_{n}}=E[e^{bX_{n+1}}|X_{n}]={\frac {1}{4}}e^{b(X_{n}-2)}+{\frac {3}{4}}e^{b(X_{n}+1)}=e^{bX_{n}}({\frac {1}{4}}e^{-2b}+{\frac {3}{4}}e^{b})}$

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

#### (b)

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

Now to calculate ${\displaystyle E[\tau ]}$ . We introduce new notation: let ${\displaystyle \tau _{k}(x)=\inf\{n\geq 1:X_{n}=k,X_{0}=x\}}$ . Then {\displaystyle {\begin{aligned}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{aligned}}}

by a symmetry argument.

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

#### (c)

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

However, ${\displaystyle Z_{n}=e^{bX_{n}}\neq 1=E[Z_{\infty }|{\mathcal {F}}_{n}]}$ . This by the definition, means the martingale is not right closable. A martingale is right-closable iff uniformly integrable. Thus, we're done.