UMD Probability Qualifying Exams/Jan2006Probability: Difference between revisions

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Problem 1

Let X1,X2,... be i.i.d. r.v.'s such that E[Xn]=0 and |Xn|1 a.s., and let Sn=k=1nXk.

(a) Find a number c such that Sn2cn is a martingale and justify the martingale property.

(b) Define τm=inf{n1:Sn>2m or Sn<m}. Compute limmP(Sτm>2m).

(c) Compute limmE[τm]/m2.


Solution

(a)

Each Sn is clearly n-measureable and finite a.s. (Hence L1). Therefore we only need to verify the martingale property. That is, we want to show Sn2cn=E[Sn+12c(n+1|n]

Sn2cn=E[Sn+12c(n+1|n]=Sn2+E[Xn+12+2j=1nXn+1Xj|n]cnc=Sn2cn+E[Xn+12]+2j=1nXjE[Xn+1] by independence=Sn2cn+Var[Xn+1]=Sn2cn+σ2

We can assert that σ2 exists and is finite since each |Xn|1 almost surely. Therefore, in order to make Sn2cn a martingale, we must have c=σ2.



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Problem 2

Let N1(t),N2(t) be independent Poisson processes with respective parameters λ,λ2, where λ is an unspecified positive real number. For each r1, let τr=inf{t>0:N1(t)r}. Show that αr=E[N2(τr2)] does not depend on λ and find αr explicitly.


Solution

First let us find the distribution of τr:

P(τr<x)=P(N1(x)r)=k=r(λx)kk!eλx

Thus by the chain rule, our random variable τr has probability density function

pτr(x)=k=rkλ(λx)k1k!eλx+(λx)kk!(λ)eλx=λ(λx)r1(r1)!eλx

So then

E[N2(τr2)]=E[E[N2(t)|τr2=t]]=0x2λ2λλr1(r1)!xr1eλxdx=λr+2(r1)!0xr+1eλxdx

Now integrate the remaining integral by parts letting u=xr+1,dv=eλxdx. We get:

E[N2(τr2)]=λr+2(r1)![1λeλxxr+1|0+01λr+1xreλxdx]=λr+2(r1)![01λr+1xreλxdx]=λr+1(r+1)(r1)![01λr+1xreλxdx]

Repeat integration by parts another r times and we get

E[N2(τr2)]=(r+1)rλ0eλxdx=λ(r+1)r1λeλx|0=(r+1)r

Problem 3


Let Xkn,1kn be independent random variables such that

P(Xkn=0)=11/n,P(Xkn=k2)=1/n

(a) Find the characteristic function of Sn=k=1nXkn.

(b) Show that Sn/n2 converges in distribution to a non-degenerate random variable.


Solution

(a)

φXkn(t)=(11/n)+1/neitk2

Then by independence, we have φXn(t)=k=1nφXkn=k=1n(11/n)+1/neitk2

Problem 4



Solution

Problem 5


Solution

Problem 6



Solution