Assume that $X_n$ are independent (but not necessarily of the same distribution) and that $Var[X_n]>0$ for all $n$. We know that $\frac{X_n-E[X_n]}{n}\to 0 \textrm{ almost surely as $n\to\infty$},$ and that $E[X_n]>0$ for all $n$. We also know that $\sum_{n=0}^\infty \frac{Var[X_n]}{n^2}<\infty.$ How can we prove that $\sum_{i=1}^n X_i \to \infty\text{ almost surely as $n\to\infty$?}$ This seems to be intuitively clear, but a formal proof eludes me. But what can we say if $E[X_n] = 0$ for all $n$?
Almost sure convergence of random variables
0
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probability
statistics
probability-distributions
convergence-divergence
1 Answers
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You cannot: if $X_n = 2^{-n}$ almost surely then the expectations are also $2^{-n}$ and their sum is $1$.
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1@50c: have you tried to compute variances in the example I left you in the previous comment and see if it contradicts with your new condition? – 2012-05-31