Borel-Catelli - convergence

Question

Let $(X_n)$ be an sequence of independent and identically distributed random variables and $E|X_1|=\infty$. Show that $\limsup_{n\to\infty}\frac{Xn}{n} $$ = \infty$ almost surely.

I think that I have to use Borel-Cantelli.

Answer 1

This statement is clearly wrong for non-positive $X_n$ (or $X_n$ with bounded positive part).

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If the $X_n$ are non-negative, however, the claim follows from $$ P(X_n\ge nC \text{ for infinitely many $n\in\mathbb N$})=1 \quad \text{for all $C>0$.}$$ This statement can in fact be proved via Borel-Cantelli by showing that $$ \sum_{n=1}^\infty P(X_n\ge nC) = \infty,$$ but $P(X_n\ge nC)=P(X_1/C\ge n)$, so this sum converges if and only if $E(X_1/C)<\infty$.

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We have thus effectively proved that under your general assumptions $$ \limsup_{n\to\infty}\frac{|X_n|}{n}=\infty \quad \text{$P$-almost surely}.$$

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By the way: Writing $S_n:=\sum_{k=1}^nX_n$, it is easy to deduce from this that also $$ \limsup_{n\to\infty}\frac{|S_n|}{n}=\infty \quad \text{$P$-almost surely},$$ since $$ \frac{|X_n|}{n}=\frac{|S_n-S_{n-1}|}{n}\le \frac{|S_n|}{n}+\frac{n-1}{n}\frac{|S_{n-1}|}{n-1}.$$