I'm trying to prove that if [itex]\{X_n\}[/itex] is independent and [itex]E(X_n)=0[/itex] for all n, and [itex]\sum_{n}E(X_n^2) <\infty[/itex], then [itex]\sum_{n}X_n[/itex] converges almost surely.(adsbygoogle = window.adsbygoogle || []).push({});

What I've got so far is the following: Denote the partial sums by [itex]\{S_n\}[/itex], then proving almost sure convergence is equivalent to showing that there exists a random variable, X, such that for all [itex]\epsilon > 0[/itex], [itex]\sum_{n}P(|S_n-X|>\epsilon)<\infty[/itex]. Using the Markov inequality gives [itex]\sum_{n}P(|S_n-X|>\epsilon)\leq\sum_{n}E(|S_n-X|)/\epsilon[/itex]. Then we just need to show that this sum converges. But I can't find anyway to do this.

It's easy to show that [itex]\sum_{n}X_n[/itex] converges in the [itex]L^2[/itex] sense, so it also converges in the [itex]L^1[/itex] sense, which means that [itex]\lim_{n\to\infty}E(|S_n-X|)=0[/itex], but this isn't strong enough to imply that [itex]\sum_{n}E(|S_n-X|)<\infty[/itex], which is what I need.

Can anyone help me with this?

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# Almost sure convergance of sum of rv

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