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Problem 1

Let XnN(0,1+1/n) and XN(0,1). Find a joint distribution of (Xn,X) so that XnX in probability.

Say(Xn,X) is bivariate normal with μXn=μX=0, σXn=1+1/n, σX=1 and cor(Xn,X)=1, then XnX has a normal distribution with

E[XnX]=0var(XnX)=var(Xn)+var(X)2cov(Xn,X)=(1+1/n)2+122(1+1/n)×1×1=1+2/n+1/n2+122/n=1/n2 and so

P(|XnX|<ϵ)=2FXnX(ϵ)1=2Φ(nϵ)12×11=1 as n because then nϵ.

Problem 2

Let XU[0,1] and Xn|X=xU[x,x+1/n]. Show that XnX in probability.

We have X<Xn<X+1/n, or 0<XnX<1/n. Now let nϵ=1/ϵ, so that for all n>nϵ we have ϵ>1/n and therefore

P(|XnX|<ϵ)=P(XnX<ϵ)P(XnX<1/n)=1

Problem 3

Let Xn be a random variable with P(Xn=0)=1/n,P(Xn=1)=12/n,P(Xn=2)=1/n. Show that Xn1 in probability

  1. using the definition 3.2.6

if ϵ<1 we have

P(|XnX|<ϵ)=P(|Xn1|<ϵ)=P(Xn=1)=12/n1

  1. using theorem 3.2.9a

E[X]=0×1/n+1×(12/n)+2×1/n=1E[X2]=02×1/n+12×(12/n)+22×1/n=1+2/nvar(Xn)=1+2/n12=2/n0 so Xn1 in quadratic mean, and therefore by theorem 3.2.9a also in probability.