# Independent Poisson distributed RV, limsup

Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of independent Poisson-distributed random variables with parameter 1, i.e. I have the probability mass function given by $$P(k)=\mathbb{P}(X=k)={e^{-1}\over k!}$$

I would like to prove that $$\limsup\limits_{n\to\infty}{\log \log n \over \log n}X_n=1.$$

I found a proposal for solution but I cannot follow well:

1) Let $a_n$ be the integer part of ${\delta \log n \over \log\log n}$ for a $\delta >0$. Then for $\lambda >0$: $$P(X=a_n)={e^{-\lambda}\lambda^{a_n}\over a_n!}=e^{-\lambda}e^{a_n \log \lambda}e^{-\sum_{j=1}^{a_n}\log j}=e^{-a_n\log a_n (1+o(1))}=n^{-\delta +o(1)}.$$ 2) Then, $\sum_{n=1}^{\infty} P(X_n\geq a_n)<\infty$ or $=\infty$ depending upon $\delta >1$ or $\delta<1$. This implies, by the Borel-Cantelli lemma, $$\limsup{X_n\over \log n / \log\log n}=1 \ \ a.s.$$

First is I don't understand the last equation in (1). And, the Borel-Cantelli lemma says, that if $A_1,A_2,A_3,\dotsc$ are independent with $\sum_{n=1}^{\infty}P(A_n)=\infty$, then $P(\limsup_{n\to\infty}A_n)=1$. What is $A_n$ in this case?

Thank you!

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Yes: Borel-Cantelli lemmas. – Did Dec 19 '12 at 16:11