UMD Probability Qualifying Exams/Aug2008Probability

Problem 1

Let $(\xi _{1},\xi _{2})$ be a Gaussian vector with zero mean and covariance matrix $D$ with entries $D_{11}=D_{22}=1,D_{12}=D_{21}=1/2$ . Find $E(\xi _{1}^{2}\xi _{2}|2\xi _{1}-\xi _{2})$

Solution

Problem 2

Let $X_{n},n\geq 0$ be a Markov chain on the state space $\mathrm {X} =\{1,2\}$ having transition matrix $P$ with elements $P_{11}=1/3,P_{12}=2/3,P_{21}=P_{22}=1/2$ . Let $f:\mathrm {X} \to \mathbb {R}$ be the function with $f(1)=1$ and $f(2)=4$ . Find a function $g:\mathrm {X} \to \mathbb {R}$ such that

$Y_{n}=f(X_{n})-f(X_{0})-\sum _{i=0}^{n-1}g(X_{i}),n\geq 1,$

is a martingale relative to the filtration ${\mathcal {F}}_{n}^{X}$ generated by the process $X_{n}$ .

Solution

Notice that since $f,g$ are measurable functions, then $Y_{n}$ is composed of linear combinations of ${\mathcal {F}}_{n}$ -measurable functions and hence $Y_{n}$ is ${\mathcal {F}}_{n}$ -adapted. Furthermore, for any $n$ , $Y_{n}$ is finite everywhere, hence is $L^{1}$ .

Therefore, we only need to check the conditional martingale property, i.e. we want to show $Y_{n}=E(Y_{n+1}|{\mathcal {F}}_{n}$ .

That is, we want

${\begin{aligned}f(X_{n})-f(X_{0})-\sum _{i=1}^{n-1}g(X_{i})&=E[f(X_{n+1})-f(X_{0})-\sum _{i=1}^{n}g(X_{i})|{\mathcal {F}}_{n}]\\f(X_{n})&=E[f(X_{n+1})|{\mathcal {F}}_{n}]-g(X_{n})\end{aligned}}$

Therefore, if $Y_{n}$ is to be a martingale, we must have

$g(X_{n})=E[f(X_{n+1})|{\mathcal {F}}_{n}]-f(X_{n})$ .

Since $\mathrm {X} =\{1,2\}$ , we can compute the right hand side without too much work.

$g(1)=E[f(X_{n+1})|X_{n}=1]-f(1)=(1\cdot 1/3+4\cdot 2/3)-1=2$

$g(2)=E[f(X_{n+1})|X_{n}=2]-f(2)=(1\cdot 1/2+4\cdot 1/2)=-{\frac {3}{2}}$

This explicitly defines the function $g$ and verifies that $Y_{n}$ is a martingale.

Problem 3

Let $\xi _{n}$ be independent identically distributed random variables with uniform distribution on [0,1]. For which values of $\alpha >0$ does the series

$\sum _{n=1}^{\infty }(\xi _{n}+n^{-\alpha })^{(n^{\alpha +1})}$

converge almost surely?

Solution

Problem 4

Solution

Problem 5

Solution

Problem 6

Solution

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