UMD Probability Qualifying Exams/Aug2008Probability

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Problem 1

Let (ξ1,ξ2) be a Gaussian vector with zero mean and covariance matrix D with entries D11=D22=1,D12=D21=1/2. Find E(ξ12ξ2|2ξ1ξ2)


Solution

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Problem 2

Let Xn,n0 be a Markov chain on the state space X={1,2} having transition matrix P with elements P11=1/3,P12=2/3,P21=P22=1/2. Let f:X be the function with f(1)=1 and f(2)=4. Find a function g:X such that

Yn=f(Xn)f(X0)i=0n1g(Xi),n1,

is a martingale relative to the filtration nX generated by the process Xn.


Solution

Notice that since f,g are measurable functions, then Yn is composed of linear combinations of n-measurable functions and hence Yn is n-adapted. Furthermore, for any n, Yn is finite everywhere, hence is L1.

Therefore, we only need to check the conditional martingale property, i.e. we want to show Yn=E(Yn+1|n.

That is, we want

f(Xn)f(X0)i=1n1g(Xi)=E[f(Xn+1)f(X0)i=1ng(Xi)|n]f(Xn)=E[f(Xn+1)|n]g(Xn)

Therefore, if Yn is to be a martingale, we must have

g(Xn)=E[f(Xn+1)|n]f(Xn).

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

g(1)=E[f(Xn+1)|Xn=1]f(1)=(11/3+42/3)1=2

g(2)=E[f(Xn+1)|Xn=2]f(2)=(11/2+41/2)=32

This explicitly defines the function g and verifies that Yn is a martingale.

Problem 3

Let ξn be independent identically distributed random variables with uniform distribution on [0,1]. For which values of α>0 does the series

n=1(ξn+nα)(nα+1)

converge almost surely?



Solution

Problem 4



Solution

Problem 5


Solution

Problem 6



Solution