# If $(X_i)$ is i.i.d. with finite third moment, then $n^{-1/3}\cdot\max\limits_{1\leq i\leq n}|X_i|\to0$ almost surely

My question is to find a formal proof of the following statement or to prove it wrong. I have seen this statement in many literature.

Let $X_{1},...,X_{n}$ be independent and identically distributed random variables with $E(|X_{1}|^3)$ finite, then $\max\limits_{1\leq i \leq n} |X_{i}| =o(n^{1/3})$ almost surely.

Note that: the above statement involves little $o_p$. By virtue of the Borel Cantelli Lemma, since $E(X_{1}^3)<\infty$, we have $\sum_{n=1}^{\infty}Pr(|X_i|^3>n)<\infty.$ Thus, we conclude that there are finitely many such cases that $|X_i|^3>n$, and $\max\limits_{1\leq i \leq n} |X_{i}| =O_p(n^{1/3})$. But I do not have the result that $\max\limits_{1\leq i \leq n} |X_{i}| =o(n^{1/3})$ almost surely, which is little $o_p$.

Let $Y_n:=\max_{1\le i\le n}|X_i|$ For any $\epsilon>0$, $$\mathsf{P}(|X_n|>\epsilon n^{1/3} \text{ i.o.})=0$$ because $\mathsf{E}|X_1|^3<\infty$ .Thus $$\mathsf{P}(Y_n>\epsilon n^{1/3} \text{ i.o.})=0,$$ which implies that $\limsup Y_n/n^{1/3}=0$ a.s.
• Can you explain how you move from $|X_n|$ to $Y_n$? It seems that you jump many steps. – lzstat Sep 18 '17 at 23:58
• If $|X_n|\le\epsilon n^{1/3}$ eventually, the same is true for $Y_n$. Just fix $\omega$ from the first set and see what happens to $Y_n(\omega)$. – d.k.o. Sep 19 '17 at 0:03