# Sum of independent Possion random variables tends to its expectation almost surely?

This is Exercise 2.3.3 of Durrett's book "Probability: Theory and examples":

Let $X_n$ be independent Poission r.v's with $EX_n=\lambda_n$, and let $S_n=X_1+\cdots+X_n.$ Show that if $\sum\lambda_n=\infty,$ then $$\frac{S_n}{\mathbb ES_n}\overset{a.s.}\longrightarrow 1$$

By Chebyshev's inequality, it is not hard to see that $S_n/\mathbb ES_n\to 1$ in probability. And our aim is to show $$P\left(|S_n-\mathbb ES_n|>\delta\mathbb ES_n \text{ i.o.}\right)=0$$ for all $\delta>0$. By Chebyshev's inequality, I can find a subsequence of $S_n$ satisfying this property. But I don't know how to extend it to the whole sequence.

• you mean subsequence of $S_n$? – Momo Nov 6 '16 at 1:01
• Yes, sorry for the typo. – Connor Nov 6 '16 at 1:05
• See the proof of Theorem 2.3.8 in Durrett's book. – John Dawkins Nov 6 '16 at 1:24
• @JohnDawkins Since in that theorem, we know $EX_m\le 1$, which we need for extending the property of subsequence to the whole sequence. But in this problem, $EX_m =\lambda_m$, which is not bounded. I don't know which subsequence I should pick. – Connor Nov 6 '16 at 1:28
• You can overcome the boundness problem by chopping your Poisson r.v.'s into multiple Poisson r.v.'s with $\lambda<1$. So you embed your $S_n$ as a subsequence of another sequence $R_n$ for which you can apply the aforementioned theorem. I explained about it in the answer below. – Momo Nov 6 '16 at 4:41

You can decompose your random variables $X_i$ into independent Poisson random variables with smaller $\lambda$. So you can construct another sequence which has bounded parameters. You apply your result on the modified sequence. The sum of variances will be the same like the sum of variances on the original sequence, so the divergence of the sum of variances is maintained.