Let Xn∼N(0,1+1/n) and X∼N(0,1). Find a joint distribution of (Xn,X) so that Xn→X in probability.
Say(Xn,X) is bivariate normal with μXn=μX=0, σXn=1+1/n, σX=1 and cor(Xn,X)=1, then Xn−X has a normal distribution with
E[Xn−X]=0var(Xn−X)=var(Xn)+var(X)−2cov(Xn,X)=(1+1/n)2+12−2(1+1/n)×1×1=1+2/n+1/n2+1−2−2/n=1/n2 and so
P(|Xn−X|<ϵ)=2FXn−X(ϵ)−1=2Φ(nϵ)−1→2×1−1=1 as n→∞ because then nϵ→∞.
Let X∼U[0,1] and Xn|X=x∼U[x,x+1/n]. Show that Xn→X in probability.
We have X<Xn<X+1/n, or 0<Xn−X<1/n. Now let nϵ=⌈1/ϵ⌉, so that for all n>nϵ we have ϵ>1/n and therefore
P(|Xn−X|<ϵ)=P(Xn−X<ϵ)≥P(Xn−X<1/n)=1
Let Xn be a random variable with P(Xn=0)=1/n,P(Xn=1)=1−2/n,P(Xn=2)=1/n. Show that Xn→1 in probability
if ϵ<1 we have
P(|Xn−X|<ϵ)=P(|Xn−1|<ϵ)=P(Xn=1)=1−2/n→1
E[X]=0×1/n+1×(1−2/n)+2×1/n=1E[X2]=02×1/n+12×(1−2/n)+22×1/n=1+2/nvar(Xn)=1+2/n−12=2/n→0 so Xn→1 in quadratic mean, and therefore by theorem 3.2.9a also in probability.