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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?

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

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.

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so $\{X_{j}\}$ need not be independent? –  user24367 Feb 17 '13 at 4:35
    
It doesn't seem so. –  passerby51 Feb 17 '13 at 4:48
    
in your argument where did you use the fact of independence –  user24367 Feb 17 '13 at 4:52
    
I meant you don't need independence. –  passerby51 Feb 17 '13 at 4:58
    
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. –  passerby51 Feb 17 '13 at 5:06
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