10
$\begingroup$

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?

$\endgroup$
3
  • $\begingroup$ can someone come up with a proof which does not involve Ottiavani's inequality..... $\endgroup$
    – user24367
    Feb 17 '13 at 1:13
  • $\begingroup$ is it Levy's equivalence theorem? $\endgroup$
    – Yimin
    Feb 17 '13 at 1:19
  • $\begingroup$ can you show its equivalent to Levy's equivalence theorem? $\endgroup$
    – user24367
    Feb 17 '13 at 1:26
10
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ so $\{X_{j}\}$ need not be independent? $\endgroup$
    – user24367
    Feb 17 '13 at 4:35
  • $\begingroup$ It doesn't seem so. $\endgroup$
    – passerby51
    Feb 17 '13 at 4:48
  • $\begingroup$ in your argument where did you use the fact of independence $\endgroup$
    – user24367
    Feb 17 '13 at 4:52
  • $\begingroup$ I meant you don't need independence. $\endgroup$
    – passerby51
    Feb 17 '13 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 '13 at 5:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.