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Let $X$ be a non-negative random variable then $\mathbb{E}(|X|) <\infty$ if and only if $\sum_{n}^{\infty}\mathbb{P}(|X| > \epsilon n) < \infty $ for every $\epsilon > 0$.

I tried in the following way,

let $\epsilon = 1$

Now $$ \mathbb{E}(|X|) = \int_0^{\infty}\mathbb{P}(|X|>t)dt\\ = \sum_{n=1}^{\infty} \int_{n-1}^{n}\mathbb{P}(|X|>t)dt\\ $$ Since $\mathbb{P}(|X|>t)$ is non-increasing function then $$ \sum_{n=1}^{\infty}\mathbb{P}(|X|>n)\leq\mathbb{E}(|X|)\leq \sum_{n=1}^{\infty}\mathbb{P}(|X|>n-1)\\ \Rightarrow \sum_{n=1}^{\infty}\mathbb{P}(|X|>n)\leq\mathbb{E}(|X|)\leq 1 +\sum_{n=1}^{\infty}\mathbb{P}(|X|>n) $$ Hence the result follow by comparison test. How can I prove this for every $\epsilon >0$.

Thanks in advance

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What you have done is correct if you repalce $i$ by $n$ throughout.

Note that $E|X|<\infty$ iff $E|\frac X {\epsilon} | <\infty$ for any $\epsilon >0$. Apply what you have proved to the random variable $\frac X {\epsilon}$.

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  • $\begingroup$ Yeah, sry for that mistake.. I edited.. and thanks for answer. $\endgroup$
    – annie_lee
    May 9 '20 at 6:51

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