Let $X_i$ be independent Poisson with mean $\lambda_i$. Show that

i)$\sum_1^\infty\lambda_i<\infty$ implies $\sum_1^\infty X_i$ converges almost surely to a finite limit.

ii)$\sum_1^\infty\lambda_i=\infty$ implies $\sum_1^\infty X_i=\infty$ almost surely

I could prove (i) by using Kolmogorov one series lemma just setting $Y_i=X_i-\lambda_i$, but I could not prove the second part. Any help/hint/solution for (ii) is appreciated. Thanks in advance

  • $\begingroup$ Have you tried Borel-Cantelli? $\endgroup$
    – Xi'an
    Feb 12, 2015 at 12:02
  • $\begingroup$ we could solve it by Kolmogorov 2-series theorem, i ll write answer soon $\endgroup$
    – hohoho
    Feb 12, 2015 at 15:16

1 Answer 1


Assume that $\sum_{i=1}^{+\infty}\lambda_i$ is infinite. Then we can construct inductively an increasing sequence of integers $\left(n_k\right)_{k\geqslant 1}$ such that for each $k\geqslant 1$, $$\tag{*}\sum_{i=n_k+1}^{n_{k+1}}\lambda_i\geqslant k.$$ This can be done by exploiting the fact that for each $k$, $\lim_{N\to +\infty}\sum_{i=n_k+1}^N\lambda_i=+\infty$.

Now define $Y_k:=\sum_{i=n_k+1}^{n_{k+1}}X_i$. In order to show that $\sum_{i=1}^{+\infty}X_i$ diverges almost surely, it suffices to prove that the sequence $\left(Y_k\right)_{k\geqslant 1}$ does not converge to $0$ almost surely. Define $A_k:=\left\{Y_k\geqslant 1\right\}$. Then the sequence $\left(A_k\right)_{k\geqslant 1}$ is independent and $$\sum_{k=1}^{+\infty}\mathbb P\left(A_k\right)=\sum_{k=1}^{+\infty}\left(1-\exp\left(-\sum_{i=n_k+1}^{n_{k+1}}\lambda_i\right)\right) \overset{\mbox{ by }(*)}{\geqslant}\sum_{k=1}^{+\infty}\left(1-\exp\left(-k\right)\right) =+\infty.$$


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