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I'm trying to solve the following exercise.

Show that if $(X_n)_{n =1}^\infty$ is a sequence of independent real-valued random variables with mean zero and uniformly bounded fourth moment, $n^{-1}\sum_{j=1}^n X_j \to 0$ almost surely. (Hint: Think about a fourth moment tail bound.)

Here's what I did. Without loss, by rescaling $X_i$, we may assume that $\mathbf{E} X_i^4 \leq 1$. Note that this means $\mathbf{E} X_i^2 = \|X_i\|_2^2 \leq \|X_i\|_4^2 = (\mathbf{E} X_i^4)^{1/2} \leq 1$. Put $S_n = \sum_{j=1}^n X_j$, observe that $\mathbf{E}S_n^4 \leq n + (n^2 - n) = n^2$, simply because the other cross terms vanish due to independence.

Following the hint, notice that by Markov's inequality, $$ \mathbf{P}(|n^{-1}S_n|\geq n^{-1/8}) = \mathbf{P}(S_n^4 \geq n^{3.5}) \leq n^{-3.5} \mathbf{E}S_n^4 \leq n^{-1.5}. $$ The upshot of this is that with $A_n = \{|n^{-1}S_n| \geq n^{-1/8}\}$, $\mathbf{P}(A_n)$ is summable, and hence $\mathbf{P}(A_n~\mathrm{i.o.}) = 0$ by Borel-Cantelli Lemma 1.

If $\omega$ is such that $|n^{-1}S_n(\omega)| \geq n^{-1/8}$ for finitely many terms, then $\limsup_n |n^{-1}S_n(\omega)| = 0$, and hence $n^{-1}S_n(\omega) \to 0$. But the set of such $\omega$ is almost sure (indeed, the complement of $A_n~\mathrm{i.o.}$), proving the claim.

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This is almost correct, except for the crucial part: it is not necessarily true that the expected value of $S_n^4$ will not be greater than $n$. You fail to take into account the $\sim n^2$ terms $\mathbb{E}[X_i^2X_j^2]$.

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  • $\begingroup$ Ah, @Mindlack, good point. I think, however since $u\mapsto u^4$ is convex, we have that $(1/n \sum_{j=1}^n X_j(\omega))^4 \leq 1/n \sum_{j=1}^n X_j(\omega)^4$, in which case we actually have that $\mathbf{E} (S_n/n)^4 \leq 1$. $\endgroup$ – Drew Brady Dec 12 '18 at 0:22
  • $\begingroup$ Ah, but this bound is not strong enough to carry forth the argument. $\endgroup$ – Drew Brady Dec 12 '18 at 0:25
  • $\begingroup$ @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$. $\endgroup$ – Mindlack Dec 12 '18 at 0:26
  • $\begingroup$ Thanks for the edits. Can you take a look at my revised argument. Tell me if you think the details look right now. I forgot that $L^p$ is monotone on probability spaces! $\endgroup$ – Drew Brady Dec 12 '18 at 0:48
  • $\begingroup$ The edited solution works now! $\endgroup$ – Mindlack Dec 12 '18 at 12:00

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