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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.

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    $\begingroup$ 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. $\endgroup$ Feb 3, 2023 at 2:44
  • $\begingroup$ 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 $. $\endgroup$
    – JGWang
    Feb 3, 2023 at 8:15
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    $\begingroup$ @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! $\endgroup$
    – peter a g
    Feb 4, 2023 at 1:51
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    $\begingroup$ @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! $\endgroup$
    – peter a g
    Feb 4, 2023 at 1:52
  • $\begingroup$ @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. $\endgroup$ Feb 4, 2023 at 10:21

1 Answer 1

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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.

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  • $\begingroup$ +1, just learned about this theorem in the past week! the frequency illusion at work (was wondering whether the harmonic series with terms randomly set to positive or negative would converge) $\endgroup$ Feb 4, 2023 at 18:27
  • $\begingroup$ 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 $\frac{S_{n}-E(S_{n})}{a_{n}}\xrightarrow{a.s} 0$ . $\endgroup$ Feb 4, 2023 at 18:36
  • $\begingroup$ @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 $a>3/4.$ $\endgroup$ Feb 4, 2023 at 19:06

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