For i.i.d. $\{X_n\}$ proving $\frac1n\max\limits_{1\leqslant k\leqslant n}|X_k|\xrightarrow{\mathrm{a.s.}}0\Leftrightarrow E(|X_1|)<+∞$

Question

$\def\d{\mathrm{d}}$Given that $X_1, X_2, \cdots$ are i.i.d., prove that$$ \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \xrightarrow{\mathrm{a.s.}} 0 \Longleftrightarrow E(|X_1|) < +\infty. $$

My try:

Because $\varliminf\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ and $\varlimsup\limits_{n \to \infty} \dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ are both measurable with respect to the exchangeable σ-algebra of $X_1, X_2, \cdots$, they are both almost surely constants by the Hewitt-Savage 0-1 law, i.e.$$ \varliminf_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_1, \quad \text{a.s.} \tag{1}\\ \varlimsup_{n \to \infty} \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| = c_2, \quad \text{a.s.} $$ where $c_1, c_2 \in [0, +\infty]$. Also, if the distribution function of $|X_1|$ is $F(x)$, then the distribution function of $\dfrac{1}{n} \max\limits_{1 \leqslant k \leqslant n} |X_k|$ is$$ F_n(x) = P\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \leqslant x \right) = \prod_{k = 1}^n P(|x_k| \leqslant nx) = (F(nx))^n, $$ which implies\begin{align*} E\left( \frac{1}{n} \max_{1 \leqslant k \leqslant n} |X_k| \right) &= \int_0^{+\infty} x \,\d((F(nx))^n) = \frac{1}{n} \int_0^{+\infty} y \,\d((F(y))^n)\\ &= \frac{1}{n} \int_0^{+\infty} y·n (F(y))^{n - 1} \,\d F(y) = \int_0^{+\infty} y (F(y))^{n - 1} \,\d F(y)\\ &\leqslant \int_0^{+\infty} y \,\d F(y) = E(|X_1|). \end{align*} Also for any $c > 0$,$$ \sum_{k = 1}^\infty P(|X_1| \geqslant ck) < \infty \Longleftrightarrow E(|X_1|) < +\infty. \tag{2} $$ However, I failed to connect (1) and (2) to prove the proposition.

Any ideas are appreciated. Thanks in advance.

Answer 1

Hints:

1. Show that $$\frac{1}{n} \max_{1 \leq k \leq n} |X_k(\omega)| \to 0 \tag{3}$$ if, and only if, $$\forall \epsilon>0 \, \, \exists N \in \mathbb{N} \, \, \forall n \geq N: \quad |X_n(\omega)| \leq n \epsilon. \tag{4}$$
2. Use the Borel cantelli lemma to show that $(4)$ (and hence $(3)$) holds almost surely if, and only if, $$\forall \epsilon>0: \quad \sum_{n \geq 1} \underbrace{\mathbb{P}(|X_n| \geq n \epsilon)}_{\mathbb{P}(|X_1| \geq n \epsilon)} < \infty.$$
3. Combine Step 2 with $(2)$ to prove the assertion.

Remark: Step 1 is actually a purely deterministic result. For any sequence $(a_n)_{n \in \mathbb{N}} \subseteq \mathbb{R}$ of real numbers the following equivalence holds: $$\frac{a_n}n \xrightarrow[]{n \to \infty} 0 \iff \frac1n\max_{k \leq n} |a_k| \xrightarrow[]{n \to \infty} 0.$$