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 $0\leq X_n\leq 1$

  • $\begingroup$ The $X_n$ are exponential variables, not Poisson - the title should be changed $\endgroup$
    – Alex
    Dec 2, 2022 at 14:10
  • $\begingroup$ Kolmogorov three-series theorem makes quick work of this. $\endgroup$ Dec 3, 2022 at 16:53
  • $\begingroup$ @MikeEarnest But $X_n$ is unbounded,then how to use Kolmogorov three-series theorem? $\endgroup$
    – qmww987
    Dec 4, 2022 at 9:26

1 Answer 1


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.

  • $\begingroup$ Why does $\sum X_n < \infty$ a.e imply E(exp($-s\sum X_n$)) converges a.e? $\endgroup$
    – qmww987
    Dec 4, 2022 at 9:23
  • $\begingroup$ @qmww987 Did you actually read my answer? I went on to explain that in detail... $\endgroup$
    – Math1000
    Dec 4, 2022 at 14:35
  • $\begingroup$ 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? $\endgroup$
    – qmww987
    Dec 4, 2022 at 16:11
  • $\begingroup$ If $\sum_{n=1}^\infty X_n$ is infinite then $e^{-s\sum_{n=1}^\infty X_n}\to 0$ as $n\to\infty$.... $\endgroup$
    – Math1000
    Dec 5, 2022 at 18:25
  • $\begingroup$ 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. $\endgroup$
    – qmww987
    Dec 6, 2022 at 2:31

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