Suppose $X_i$ are iid symmetric (about $0$) random variables. Then show that $\sum_n\dfrac{X_n}{n}$ converges almost surely if and only if $E|X_1|<\infty$.

I was thinking of applying the Kolmogorov 3-series theorem. It states that if $Y_n$ are independent then $\sum_n Y_n$ converges a.s. if and only if there exists a constant $c>0$ such that $\sum_n P(|Y_n|>c)<\infty,\sum_n Var(Y_n 1(|Y_n|\leq c))<\infty$ and $\sum_n E(Y_n 1(|Y_n|\leq c))$ converges.

Suppose $\sum_n X_n/n$ converges. Then there exists $c>0$ with $\sum_n P(|X_n|>nc)<\infty$ and since $X_i$ are iid, $\sum_n P(|X_1|>nc)<\infty$ implying $E|X_1|/c<\infty$ implying $E|X_1|<\infty$.

I am stuck in proving the converse. Suppose $E|X_1|<\infty$. This implies for any $c>0$, $\sum_n P(|X_n|>nc)<\infty$ and since $X_1$ is symmetric, $E(X_11(|X_1|\leq nc))=0$ for each $n$ and thus $\sum_n E(X_n1(|X_n|\leq nc))$ converges.

But I cannot show $\sum_n Var(X_11(|X_1|\leq nc))<\infty$.

  • $\begingroup$ Law of large numbers is fair game ? $\endgroup$ Apr 29 '18 at 13:34
  • $\begingroup$ No, you probably misinterpret it as $\sum_{1\leq k\leq n}X_k/n$ whereas I mean the infinite random sum $\sum_n X_n/n$. $\endgroup$ Apr 29 '18 at 14:06
  • $\begingroup$ Ah yes, sorry then. $\endgroup$ Apr 29 '18 at 14:16

We have to prove the convergence of the series $$\sum_n\color{red}{\frac 1{n^2}}\mathbb E\left[X_1^2\mathbf 1\left\{\left\lvert X_1\right\rvert \leqslant n\right\}\right].$$ Letting $a_k:=\mathbb E\left[X_1^2\mathbf 1\left\{k-1\lt \left\lvert X_1\right\rvert \leqslant k\right\}\right]$, we have to prove the convergence of $$\sum_{n=1}^{\infty}\frac 1{n^2} \sum_{k=1}^na_k, $$ which can be done by summing first over $n\geqslant k$ and then over $k$.

  • $\begingroup$ Sorry but I fail to see why is that convergent? $\endgroup$ Apr 29 '18 at 14:52
  • $\begingroup$ I fail to see what you fail to see in your attempt for proving the convergence. $\endgroup$ Apr 29 '18 at 14:55
  • $\begingroup$ Yes, got it now. Thanks! BTW you mean sum first over n and then over k. $\endgroup$ Apr 29 '18 at 14:57

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