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Let $x \in \mathbb R$ and $f(x) = x^2$ if $|x| \le 1$ and $f(x) = |x|$ otherwise and $(X_n)$ a sequence of symmetric and independent random variables. Why does it follow from $\sum{ \mathbb E [f(X_n)]} < \inf $ that $ \sum{X_n} $ converges almost surely?

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  • $\begingroup$ sorry, didn't remark this. From now on, I do :) Thank you :) $\endgroup$
    – JohnD
    Apr 20, 2013 at 23:07

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You know that both $\sum X_n 1 \{ 0 < X_n \leq 1 \}$ and $\sum X_n 1 \{-1 \leq X_n < 0 \}$ converge almost surely because they are monotonic and bounded in $\mathbb L^2$. The same argument applies to $\sum X_n 1 \{X_n > 1 \}$ and $\sum X_n 1 \{ X_n < -1\}$ which are bounded in $\mathbb L^1$.

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  • $\begingroup$ I don't understand your proof. Where do you use the expectation E? Would be very nice if you could give some more details! thx! ;) $\endgroup$
    – user72898
    Apr 16, 2013 at 18:17
  • $\begingroup$ Ok. $Y_N = \sum_{n =1}^N X_n 1 \{ 0 < X_n < 1 \}$ increases with $N$, so the limit $Y$ exists. Moreover, $Y$ is finite almost surely because $\lim_{N \to \infty} \mathbb E(Y_N^2) \leq \sum_{n=1}^\infty \mathbb E(f(X_n)) < \infty$. So $Y_N$ converges almost surely toward $Y$. Now write that $$ \sum_{n=1}^N X_n = \sum_{n =1}^N X_n 1 \{ 0 < X_n < 1 \} + \sum_{n =1}^N X_n 1 \{ -1 < X_n < 0 \} + \sum_{n =1}^N X_n 1 \{X_n \leq -1 \} + \sum_{n =1}^N X_n 1 \{ X_n \geq 1 \} $$ where each part can be shown to cv almost surely by the same argument as before. $\endgroup$
    – roger
    Apr 16, 2013 at 18:44

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