TITLE: $\lim_n \frac{1}{n} E(\max_{1\le j\le n} |X_j|) = 0$
QUESTION [6 upvotes]: If $\{X_n\}$ is a sequence of identically distributed r.v.'s with finite mean, then
$$\lim_n \frac{1}{n} E(\max_{1\le j\le n} |X_j|) = 0$$
The inequality
$$\frac{1}{n}E(\max_{1\le j\le n} |X_j|) \le \frac{1}{n}E(|X_1| + \cdots + |X_n|)=E(|X_1|)$$
suggests that the result has something to do with the "graphs" of the $|X_j|$ overlapping so that $E(|X_j|)=\int_0^\infty |X_j| P(d\omega)$ isn't fully counted $n$ times in the value of $E(\max_{1\le j\le n} |X_j|)$. But how do we make this rigorous? Notice there is no mention of the $X_j$ being independent.

REPLY [3 votes]: Let $u(x)=P[|X_1|\geqslant x]$ and $M_n=\max\limits_{1\leqslant k\leqslant n}|X_k|$, then $[M_n\geqslant x]=\bigcup\limits_{k=1}^n[|X_k|\geqslant x]$ hence
$$
n^{-1}E[M_n]=\int_0^{+\infty} u_n(x)\mathrm dx,
$$
with
$$
u_n(x)=n^{-1}P[M_n\geqslant x]\leqslant u(x).
$$
Now, $0\leqslant u_n\leqslant n^{-1}$ hence $u_n\to0$ pointwise. Furthermore, $u$ is integrable since
$$
\int_0^{+\infty}u(x)\mathrm dx=E[|X_1|],
$$
hence, by dominated convergence,
$$
\int_0^{+\infty}u_n(x)\mathrm dx\to0.
$$
Thus,
$$
n^{-1}E[M_n]\to0.
$$