# $X_n$ are independent exponential variables, if $\sum_{n=1}^\infty λ_n^{-1}=\infty$,show that $\sum_{n=1}^\infty X_n=\infty$ a.e.

Let $$X_1,X_2,\cdots,X_n$$ be independent nonnegative random variables, with $$X_n$$ having density $$λ_n\exp(-λ_n x),x\geq 0,λ_n\geq 0$$,if $$\sum_{n=1}^\infty λ_n^{-1}=\infty$$, show that $$\sum_{n=1}^\infty X_n=\infty$$ a.e.

Attempts:$$\sum_{n=1}^\infty λ_n^{-1}=\infty$$ implies $$\sum_n E(X_n)=E(\sum_n X_n)=\infty$$,also I get a hint to try to consider $$\exp(-\sum_nX_n)$$,and from $$X_n$$ are independent,$$E(\exp(-\sum_nX_n))=\Pi_n E(\exp(-X_n))$$, but I don't know what to do next?

A very similar question:Let $X_1, X_2, \ldots$ be independent r.v.'s with $0 \leq X_n \leq 1$ and $\sum_n E(X_n) = \infty$. Show $\sum_n X_n = \infty$ with probability 1? but differs in $$0\leq X_n\leq 1$$

• The $X_n$ are exponential variables, not Poisson - the title should be changed
– Alex
Dec 2, 2022 at 14:10
• Kolmogorov three-series theorem makes quick work of this. Dec 3, 2022 at 16:53
• @MikeEarnest But $X_n$ is unbounded,then how to use Kolmogorov three-series theorem? Dec 4, 2022 at 9:26

The sum $$\sum_{n=1}^\infty X_n$$ converges if and only if $$\sum_{n=1}^\infty \lambda_n^{-1}$$ does. If $$\sum_{x=1}^\infty X_n<\infty$$ then for all $$s>0$$ we have $$0<\mathbb E\left[e^{-s\sum_{n=1}^\infty X_n}\right] = \prod_{n=1}^\infty \mathbb E[e^{-sX_n}] = \prod_{n=1}^\infty \frac{\lambda_n}{\lambda_n+s}.$$ This infinite product converges if and only if $$\sum_{n=1}^\infty \left(1-\frac{\lambda_n}{\lambda_n+s}\right)$$ does (as can be shown by considering the convergence of $$\sum_{n=1}^\infty -\log\left(\frac{\lambda_n}{\lambda_n+s}\right)$$ ), and $$\sum_{n=1}^\infty \left(1-\frac{\lambda_n}{\lambda_n+s}\right) = \sum_{n=1}^\infty \frac s{s+\lambda_n}$$ converges if and only if $$\sum_{n=1}^\infty \lambda_n^{-1}$$ does.
Conversely, if $$\sum_{n=1}^\infty \lambda_n^{-1}<\infty$$, then by monotone convergence, $$\mathbb \sum_{n=1}^\infty \mathbb E[X_n] = \mathbb E\left[\sum_{n=1}^\infty X_n\right]<\infty,$$ and hence $$\sum_{n=1}^\infty X_n<\infty$$ as the $$X_n$$ are almost surely nonnegative.
• Why does $\sum X_n < \infty$ a.e imply E(exp($-s\sum X_n$)) converges a.e? Dec 4, 2022 at 9:23
• Sorry but if I hadn't misunderstand your answer,you state that $∑X_n<∞$ a.e implies $E(\exp(−s\sum X_n))$ converges a.e,and you show that $E(\exp(−s\sum X_n))$ converges a.e iff $\sum_n\lambda_n^{-1}$ converges(which are in detail), but you didn't explain why $∑X_n<∞$ a.e implies $E(\exp(−s\sum X_n))$ converges a.e? Dec 4, 2022 at 16:11
• If $\sum_{n=1}^\infty X_n$ is infinite then $e^{-s\sum_{n=1}^\infty X_n}\to 0$ as $n\to\infty$.... Dec 5, 2022 at 18:25
• I see.But it should be: $E(\exp(−s\sum X_n))$ doesn't converge to 0 a.e instead of $E(\exp(−s\sum X_n))$ doesn't converge a.e,actually $E(\exp(−s\sum X_n))$ always converges. Dec 6, 2022 at 2:31