Let $\{X_1,X_2,\dots\}$ be a sequence of i.i.d. random variables. Define $M_n=\sup\limits_{k\leq n} |X_k|$. Suppose $X_1$ is $p-$th integrable i.e. $E[|X_1|^p]<\infty$ for some $p\in(0,\infty)$. Show that

$\frac{1}{n^{1/p}}M_n\to 0$

with probability 1.

I have already proven that for an integrable random variable $X$,

$\sum\limits_{n=1}^\infty P(|X|\geq \epsilon n)<\infty$

Hence to prove the statement, let $\epsilon >0$. Define $A_n=\{|X_n|\geq \epsilon n^{1/p}\}$.

Then $\sum\limits_{n=1}^\infty P(A_n)=\sum\limits_{n=1}^\infty P(|X_n|\geq \epsilon n^{1/p})\leq \sum\limits_{n=1}^\infty P(M_n\geq \epsilon n^{1/p})$

I want to show that $\sum\limits_{n=1}^\infty P(A_n)<\infty$ and then use the Borel-Cantelli lemma to show $P(\limsup A_n)=0$.

I don't know how to show $\sum\limits_{n=1}^\infty P(M_n\geq \epsilon n^{1/p})<\infty$. I don't know how to use the condition that $X_1$ is p-th integrable.


Since the $X_j$ are identically distributed, $\mathbb{P}(|X_n| \geq \epsilon n^{1/p}) = \mathbb{P}(|X_1| \geq \epsilon n^{1/p}) = \mathbb{P}(|X_1|^p \geq \epsilon^pn)$. Then as $X_1$ is $p$-th integrable from what you've already shown $\sum_{n=1}^{\infty} \mathbb{P}(|X_n|^p \geq \epsilon^pn) = \sum_{n=1}^{\infty} \mathbb{P}(|X_1|^p \geq \epsilon^pn) < \infty$ so, as you noted, by Borel-Cantelli $\mathbb{P}(\lim \sup A_n) = 0$.

Now we argue that the event $\{\lim_{n \to \infty} \frac{1}{n^{1/p}}M_n < 2 \epsilon \}$ contains the complement of $\lim \sup A_n$ and in particular has probability $1$.

Suppose $\omega \not \in \lim \sup A_n$. Then $|X_n(\omega)| < \epsilon n^{1/p}$ eventually, say for $n \geq N$. So $\frac{1}{n^{1/p}}M_n(\omega) < \frac{1}{n^{1/p}}\max_{k < N} |X_k(\omega)| + \epsilon$ for any $n \geq N$. But since the $\max$ is over finitely many terms the first part of the right hand side tends to $0$ as $n \to \infty$ so there is an $M$ such that for $n \geq M$, $\frac{1}{n^{1/p}}\max_{k < N} |X_k(\omega)| < \epsilon$. Hence for $n \geq \max\{M,N\}$ we have that $\frac{1}{n^{1/p}}M_n(\omega) < 2 \epsilon$ so $\omega \in \{\lim_{n \to \infty} \frac{1}{n^{1/p}}M_n < 2 \epsilon \}$.

To conclude, let $B_k = \{\lim_{n \to \infty} \frac{1}{n^{1/p}}M_n < \frac{2}{k} \}$. By taking $\epsilon = \frac{1}{k}$ in the above $\mathbb{P}(B_k) = 1$ for each $k$ so $\mathbb{P}(\bigcap B_k) = 1$. But since $M_n \geq 0$, $\bigcap B_k = \{\lim_{n \to \infty} \frac{1}{n^{1/p}}M_n = 0 \}$

Edit: I've realised that a problem with the original answer that I give above is that it presupposes the existence of the limit in the third paragraph. An easy fix for this is to just look at the set $\{\lim \sup_{n \to \infty} \frac{1}{n^{1/p}}M_n < 2 \epsilon \}$ instead and then all of the above argument goes through.

  • $\begingroup$ Note that $\frac{1}{n^{1/p}} |X_j(\omega)| < ...$ for each $j$ then just take the $\sup$ on the left hand side. $\endgroup$ – Rhys Steele May 6 '17 at 15:25

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