Proof verification: Strong Law of Large Numbers Under Fourth Moment Control

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.

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 $$\mathbb{E}[X_i^2X_j^2]$$.
• 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 $\mathbf{E} (S_n/n)^4 \leq 1$. – Drew Brady Dec 12 '18 at 0:22
• @Drew Brady: Yes indeed. However, the terms mentioned in my answer are the only ones that matter, so the expected value is less than $n^2$. – Mindlack Dec 12 '18 at 0:26
• 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! – Drew Brady Dec 12 '18 at 0:48