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 lim almost surely ?





I think we can write X(n) \sim X(1)+X(1)+\cdots+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 X_n\sim\mathsf{Pois}(n) for n=1,2,\ldots, then n^{-1}X_n\stackrel{\mathrm{a.s.}}\longrightarrow 1. It suffices to show that for all \varepsilon>0,
\mathbb P\left(\liminf_{n\to\infty}\left|n^{-1}X_n-1\right|<\varepsilon\right)=1.



We have \left\{\left|n^{-1}X_n-1\right|<\varepsilon\right\}^c = \left\{X_n\leqslant n(1-\varepsilon) \right\}\cup\left\{X_n\geqslant n(1+\varepsilon) \right\},
so the Chernoff bounds yield \mathbb P(X_n\leqslant n(1-\varepsilon)) \leqslant \frac{e^{-n}(ne)^{n(1-\varepsilon)}}{(n(1-\varepsilon))^{n(1-\varepsilon)}}= \left(e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon} \right)^{-n}
and
\mathbb P(X_n\geqslant n(1+\varepsilon)) \leqslant \frac{e^{-n}(ne)^{n(1+\varepsilon)}}{(n(1+\varepsilon))^{n(1+\varepsilon)}}= \left(e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} \right)^{-n} .



Now



e^\varepsilon(1-\varepsilon)^{1-\varepsilon} = e^{\varepsilon + (1-\varepsilon)\log(1-\varepsilon)}=e^{\frac{\varepsilon^2}2+O(\varepsilon^3)}>1
and
e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} = e^{-\varepsilon+(1+\varepsilon)\log(1+\varepsilon)}=e^{\frac{\varepsilon^2}2+O(\varepsilon^3)}>1,
so \sum_{n=1}^\infty \mathbb P(X_n\leqslant n(1-\varepsilon))\leqslant\sum_{n=1}^\infty \left(e^{\varepsilon}(1-\varepsilon)^{1-\varepsilon} \right)^{-n}<\infty and
\sum_{n=1}^\infty \mathbb P(X_n\geqslant n(1+\varepsilon))\leqslant\sum_{n=1}^\infty \left(e^{-\varepsilon}(1+\varepsilon)^{1+\varepsilon} \right)^{-n}<\infty.
It follows from the first Borel-Cantelli lemma that \mathbb P\left(\limsup_{n\to\infty} \{X_n\leqslant n(1-\varepsilon)\}\right)=\mathbb P\left(\limsup_{n\to\infty} \{X_n\geqslant n(1+\varepsilon)\}\right)=0,

and so we conclude.


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