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Let $X_1,X_2,\ldots$ be iid random variables. I want to show that $\mathbb{P}(|X_n|\geq n \text{ infinitely often})=0$ if and only if $\mathbb{E}(|X_1|)<\infty$.

I think I can use Borel Cantelli to show at least one direction, but then I need to show that if $\mathbb{E}(|X_1|)=\int_0^\infty \mathbb{P}(|X_1|\geq n)<\infty$, then $\sum_{n=1}^\infty \mathbb{P}(|X_n|\geq n)=\sum_{n=1}^\infty \mathbb{P}(|X_1|\geq n)<\infty$. Can I simply say that $\sum_{n=1}^\infty \mathbb{P}(|X_1|\geq n)\leq \int_0^\infty \mathbb{P}(|X_1|\geq n)<\infty$?

Also, I am not sure how to get the converse.

mathim1881
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1 Answers1

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I think you can argue both directions with the two versions of Borel-Cantelli and the integral comparison test, which in this case tells us that $∫ \mathbb{P}(|X_1| > x) \ \mathrm{d}x $ converges iff $ \sum_{n \geq 1} \mathbb{P}(|X_1| > n) $ converges

jro
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