Wednesday, 25 December 2013

probability theory - Find the almost sure limit of Xn/n, where each random variable Xn has a Poisson distribution with parameter n




Xn independent and XnP(n) meaning that Xn has Poisson distributions with parameter n. What is the limnXnn 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 XnPois(n) for n=1,2,, then n1Xna.s.1. It suffices to show that for all ε>0,
P(lim infn|n1Xn1|<ε)=1.



We have {|n1Xn1|<ε}c={Xnn(1ε)}{Xnn(1+ε)},
so the Chernoff bounds yield P(Xnn(1ε))en(ne)n(1ε)(n(1ε))n(1ε)=(eε(1ε)1ε)n
and
P(Xnn(1+ε))en(ne)n(1+ε)(n(1+ε))n(1+ε)=(eε(1+ε)1+ε)n.



Now



eε(1ε)1ε=eε+(1ε)log(1ε)=eε22+O(ε3)>1
and
eε(1+ε)1+ε=eε+(1+ε)log(1+ε)=eε22+O(ε3)>1,
so n=1P(Xnn(1ε))n=1(eε(1ε)1ε)n< and
n=1P(Xnn(1+ε))n=1(eε(1+ε)1+ε)n<.
It follows from the first Borel-Cantelli lemma that P(lim supn{Xnn(1ε)})=P(lim supn{Xnn(1+ε)})=0,

and so we conclude.


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