Let $X_1,X_2,\dots$ be i.i.d. random variables. I want to show that $P(|X_n| \ge n \ i.o.) = 0$ if and only if $E[|X_1|] < \infty$.
My thoughts: One direction is the straightforward application of the Borel-Cantelli Lemma, i.e. $E[|X_1|] < \infty \Rightarrow \sum_{n=1}^\infty P(|X_n| \ge n) < \infty \Rightarrow P(|X_n| \ge n \ i.o.) = 0$. But I am having trouble figuring out the other direction. I am thinking along the line of proof by contradiction. Assume $P(|X_n| \ge n \ i.o. ) > 0$, and use the fact that the $X_i$'s are i.i.d. to derive $E[|X_i|] = \infty$. But I don't know what theorem I can use since Borel-Cantelli only guarantees convergence of the tail of the series to $0$. I am not the sure whether the reverse direction is true. Can someone give me a hint? Thanks in advance!
Best Answer
Assume that $E[|X_1|] = +\infty$. Then, by Fubini's theorem, $\sum_{n \ge 1} P(|X_n| \ge n) = \sum_{n \ge 1} P(|X_1| \ge n) \ge E[|X_1|-1] = +\infty$.
Since $X_1, X_2, \cdots$ are independent, the events $|X_n| \ge n$ are also independent. Now use the Borel-Cantelli second lemma to conclude $P(\{|X_n| \ge n, i.o.\}) = 1$.