Let $\{{X_{j}}\}_{1}^{\infty}$ be independent r.v.s such that $\sum E( |X_{j}|) <\infty$. How to show that $\sum X_{j}$ converges almost surely. Can I argue simply that for every $\epsilon>0, \exists N$ such that $\forall j,k >N, E(|X{j}-X_{k}|)<\epsilon$. Then I proceed exactly as in

how to show convergence in probability imply convergence a.s. in this case?

  • $\begingroup$ can someone come up with a proof which does not involve Ottiavani's inequality..... $\endgroup$
    – user24367
    Feb 17, 2013 at 1:13
  • $\begingroup$ is it Levy's equivalence theorem? $\endgroup$
    – Yimin
    Feb 17, 2013 at 1:19
  • $\begingroup$ can you show its equivalent to Levy's equivalence theorem? $\endgroup$
    – user24367
    Feb 17, 2013 at 1:26

1 Answer 1


Here is a simple proof. By monotone convergence theorem: $$ \sum_j E|X_j| = E \big[ \sum_{j} |X_j| \big]. $$ It follows from the assumption that $E \big[ \sum_j |X_j| \big] < \infty$. Any random variable which has finite expectation should be finite almost surely. Thus, $\sum_j |X_j| < \infty$ almost surely. But absolute convergence for series implies convergence, hence $\sum_j X_j$ converges almost surely.

  • $\begingroup$ so $\{X_{j}\}$ need not be independent? $\endgroup$
    – user24367
    Feb 17, 2013 at 4:35
  • $\begingroup$ It doesn't seem so. $\endgroup$
    – passerby51
    Feb 17, 2013 at 4:48
  • $\begingroup$ in your argument where did you use the fact of independence $\endgroup$
    – user24367
    Feb 17, 2013 at 4:52
  • $\begingroup$ I meant you don't need independence. $\endgroup$
    – passerby51
    Feb 17, 2013 at 4:58
  • $\begingroup$ I have also made an edit. The last line should be $\sum_j X_j$ converges almost surely (not $\sum_j X_j < \infty$), since this is more precise. $\endgroup$
    – passerby51
    Feb 17, 2013 at 5:06

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