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I'd appreciate any hint for this problem. I've been thinking a lot about it and I don't see how to proceed. Let $X_n$ a sequence of independent positive random variables. Is it true that $$\sum_{k=1}^{\infty}E(X_i)<\infty$$ when $\sum_{k=1}^{\infty}X_i<\infty$ almost surely? I reviwed many questions in the forum but none seem to be specific about it. In fact the question Expected value of infinite sum cites some references for an additional condition for the result, but none of these references suposses the random variables are independent. Any hint or reference or counterexample? Thank you for your help.

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I think I found a counter example:

Let $X_n$ be independent random variables where $$P(X_n = 2^n) = 2^{-n}$$ and $$P(X_n=0) = 1 - 2^{-n}.$$ Obviously, $E(X_n) = 1$ and $\sum_{n=1}^\infty E(X_n) = \infty.$ Now $\sum_{n=1}^\infty X_n(\omega) = \infty$ for some outcome $\omega$ if and only if $X_n(\omega) = 2^n$ infinitely often. Let $E_n$ denote the event that $X_n$ is not zero. Then $$\sum_{n=1}^\infty P(E_n) = \sum_{n=1}^\infty 2^{-n} =1< \infty$$ and so by the Borel-Cantelli Lemma the probability of infinitely many $E_n$ occuring is $0$. Furthermore, $$P\left(\sum_{n=1}^\infty X_n = \infty\right) = P(\{ E_n: \text{infinitely often}\}) = 0$$ so $\sum_{n=1}^\infty X_n < \infty$ almost surely.

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  • $\begingroup$ +1 You do not even need the Borel-Cantelli Lemma, since $P(\sum X_i \ge 2^n) =2^{-n}+(1-2^{-n})2^{-1-n}+(1-2^{-n})(1-2^{-1-n})2^{-2-n}+\ldots\le 2^{1-n}$ and so tends to zero as $n$ increases $\endgroup$ – Henry Jun 15 '17 at 23:19
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Assuming by positive you are still allowing the value $0$, this is actually equivalent to asking "is $E[Y]<\infty$ for any positive random variable $Y$?" which clearly has the answer "no". Indeed, put $Y=\sum X_i$. By the monotone convergence theorem (since each $X_i$ is positive), $E[Y]=\sum E[X_i]$. On the other hand, given a positive $Y$ we can define $X_1=Y$ and $X_i=0$ for $i\ge1$; then $\{X_i\}$ are independent and $Y=\sum X_i$.

For a complete characterization of when $\sum X_i$ converges for independent $X_i$, see the Kolmogorov three series theorem. Notably, the convergence of $\sum EX_i$ is not necessary.

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