# Show that $\frac{S_n}{n} \to 0$ almost surely

For the below question:

Suppose that $$X_i$$ are independent random variables such that $$\mathbb{E}(X_i)=0$$ and $$\mathbb{E}(X_i^2) \leq c \sqrt{i}$$ for some fixed but positive constant $$c$$. Show that $$\displaystyle{\frac{S_n}{n} = \frac{X_1 + \cdots X_n}{n}}$$ converges to $$0$$ a.s.

My first thought was to use Chebychev's Inequality to bound the probability, then apply Borel-Cantelli Lemma. However, it seems in this case that I cannot find a convergent series to bound the probability. Any help would be thankful.

• Do you know the theorem that if $X_i$ are independent, $E(X_i)=0$ and $\sum_i \operatorname{Var}(X_i)<\infty,$ then $\sum_iX_i$ converges a.s.? If so, apply it to the sequence $X_n/n,$ and infer from that that $S_n/n\to 0$ a.s. Feb 3, 2023 at 2:44
• Please notice following fact: if $\{X_n,n\ge 1\}$ are independent r.v.s with $\mathsf{E}X_n=0$, $\mathsf{E}X_n^2=\sigma^2_n$, $S_n=\sum\limits_{i=1}^n X_i$, $s_n^2=\sum\limits_{i=1}^n\sigma_i^2\to\infty$, then $\sum_n\frac{X_n}{s_n^{\alpha}}$ convergence a.s. and $S_n/s_n^{\alpha} \stackrel{\text{a.s.}}\longrightarrow 0$ for $\alpha>\frac12$. Feb 3, 2023 at 8:15
• @spaceisdarkgreen can you give a reference? I am often also guilty of not doing what I am now preaching, but it seems to me that you should make your comment into an answer. Thanks! Feb 4, 2023 at 1:51
• @JGWang same comment that I made to spaceisdarkgreen: "... I am often also guilty of not doing what I am now preaching, but it seems to me that you should make your comment into an answer." Thanks! Feb 4, 2023 at 1:52
• @peterag Thank you! That is also what I want to recommend to people who answer this question since the answer is be very nontrivial to me. Feb 4, 2023 at 10:21

We have $$\sum_n \operatorname{Var}(X_n/n) \le \sum_n \frac{c}{n^{3/2}} <\infty,$$ so by Kolmogorov's two-series theorem, $$\sum_n X_n/n$$ converges almost surely, so by Kronecker's lemma, $$S_n/n\to 0.$$
On a side note, in addition to the proof in the link above, the two-series theorem is immediate from the martingale convergence theorem, since if $$E(Z_n)=0$$ and $$\sum_n \operatorname{Var}(Z_n)<\infty,$$ then $$M_n = \sum_{k\le n} Z_k$$ is an $$L^2$$-bounded martingale.
• Instead of $n$ , we can take any monotonic sequence $a_{n}\to\infty$ . Then if $\sum_{n}\text{Var}(\frac{X_{n}}{a_{n}})<\infty$ then $\frac{S_{n}-E(S_{n})}{a_{n}}\xrightarrow{a.s} 0$ . Feb 4, 2023 at 18:36
• @Mr.GandalfSauron Yes, the question is suboptimal in the sense that we can just as easily show, e.g. $S_n/n^a\to 0$ a.s. for $a>3/4.$ Feb 4, 2023 at 19:06