Xn independent and Xn∼P(n) meaning that Xn has Poisson distributions with parameter n. What is the limn→∞Xnn almost surely ?
I think we can write X(n)∼X(1)+X(1)+⋯+X(1) where the sum is taken on independent identical distribution then use the law of large number. But I am not sure that is it correct or not. Can anyone give me some hints? Thank you in advance!
Answer
If Xn∼Pois(n) for n=1,2,…, then n−1Xna.s.⟶1. It suffices to show that for all ε>0,
P(lim infn→∞|n−1Xn−1|<ε)=1.
We have {|n−1Xn−1|<ε}c={Xn⩽n(1−ε)}∪{Xn⩾n(1+ε)},
so the Chernoff bounds yield P(Xn⩽n(1−ε))⩽e−n(ne)n(1−ε)(n(1−ε))n(1−ε)=(eε(1−ε)1−ε)−n
and
P(Xn⩾n(1+ε))⩽e−n(ne)n(1+ε)(n(1+ε))n(1+ε)=(e−ε(1+ε)1+ε)−n.
Now
eε(1−ε)1−ε=eε+(1−ε)log(1−ε)=eε22+O(ε3)>1
and
e−ε(1+ε)1+ε=e−ε+(1+ε)log(1+ε)=eε22+O(ε3)>1,
so ∞∑n=1P(Xn⩽n(1−ε))⩽∞∑n=1(eε(1−ε)1−ε)−n<∞ and
∞∑n=1P(Xn⩾n(1+ε))⩽∞∑n=1(e−ε(1+ε)1+ε)−n<∞.
It follows from the first Borel-Cantelli lemma that P(lim supn→∞{Xn⩽n(1−ε)})=P(lim supn→∞{Xn⩾n(1+ε)})=0,
and so we conclude.
No comments:
Post a Comment