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

Question

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

Answer 1

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.