# Suppose $X_n$ are iid with a symmetric distribution. Then $\Sigma_n \frac{X_n}{n}<\infty ~\mathrm{a.s. iff }~\mathbb{E}|X_1|<\infty$

It seems to be solved by using Kolmogorov strong law of large numbers. Why $$X_n$$ have symmetric distribution?

• Do you really mean $\sum\limits_{n = 1}^\infty \dfrac{X_n}{n}$? If so, you are dealing with triangular arrays and I am not sure how to apply the SLLN then. – Will M. Dec 8 '18 at 6:12
• This might be killing a fly with a nuke, but Kolmogorov's three-series theorem would be helpful here. – Mike Earnest Dec 8 '18 at 6:17
• @MikeEarnest I think that OP is confused, I also thought of K3ST but that is for random series in general. Anyway, the exercise, as OP wrote it, is immediate from K3ST. – Will M. Dec 8 '18 at 6:19

As Mike's comment pointed out, by Kolmogorov's three-series theorem, the following three facts implies $$\sum_{j=1}^\infty \frac{X_j}{j} <\infty$$ almost surely:$$E|X_1|<\infty \Rightarrow \sum_{j\geq 1} P\left(|\frac{X_j}{j}|\geq 1\right) <\infty ,$$ $$\sum_j E\left[\frac{X_j}{j}1_{|X_j| \leq j}\right]=0,$$ which follows from symmetry of the distribution of $$X_j$$, and $$\begin{eqnarray} \sum_j E\left[\left(\frac{X_j}{j}\right)^21_{|X_j| \leq j}\right] &=& \sum_j \frac{1}{j^2}\int_{-j}^j x^2 dF(x) \\&=& \int_{-\infty}^\infty x^2 \left(\sum_{j\geq 1, j\geq |x|} \frac{1}{j^2}\right)dF(x)\\&\leq& C\int_{-\infty}^\infty |x|dF(x) <\infty, \end{eqnarray}$$ where $$F$$ is the distribution function of $$X_1$$.
Conversely, if $$E|X_1|=\infty$$, then we should have $$\sum_j P(|X_j|\geq j) = \infty.$$By Borel-Cantelli lemma, this implies that $$|\frac{X_j}{j}|\geq 1, \text{ i.o.}$$ almost surely. This makes the series $$\sum_j \frac{X_j}{j}$$ divergent almost surely. We can see this in another way: if $$\sum_{j=1}^\infty \frac{X_j}{j} <\infty$$ with positive probability $$p$$, then by Kronecker's lemma, we get $$\frac{1}{n}\sum_{j=1}^n X_j \to 0,$$ with probability $$p$$. However, if $$E|X_1|=\infty$$, by the converse of SLLN, we have that $$\limsup_{n\to \infty} \frac{1}{n}|\sum_{j=1}^n X_j| = \infty,$$ with probability $$1$$. This contradicts $$\sum_{j=1}^\infty \frac{X_j}{j} <\infty$$, with positive probability, thus the sum diverges almost surely.