# $\mathbb E[ |X| ] < \infty \iff \forall \epsilon : \mathbb E[ |X / \epsilon | ] < \infty$

expected valueprobability theoryrandom variables

$$X$$ is a random variable.

$$\mathbb E ( | X | ) < \infty \implies \forall \epsilon > 0, \sum_n \mathbb P ( | X | \geq n \epsilon ) < \infty.$$

Any help to prove this?

(This amounts to prove that $$\mathbb E[ |X| ] < \infty \iff \forall \epsilon : \mathbb E[ |X / \epsilon | ] < \infty$$).

—————————————————- original post

Since I know that $$X \in L^1 \iff \sum \mathbb P ( | X | \geq n) < \infty$$ I was given the hint to try to prove the statement hereinafter:

Let us say that $$X$$ is a random variable.

$$\forall \epsilon > 0, \exists K>0, \forall n \in \mathbb N: \mathbb P( X \geq n \epsilon ) \leq K \mathbb P ( X \geq n),$$

is a useful fact that I need in some proof of my probability lecture, however I am quite unsure about how to prove such statement. Any idea?

—————————————————- comment

I guess that the original post is completly wrong because one cannot bound each term (cf. Ian comment), however bounding the whole sum is doable somehow.

$$\newcommand{\dd}{\mathop{}\!\mathrm{d}}$$Recall that $$\mathbb E|X|=\int_0^\infty\mathbb P(|X|>t)\dd t$$. Then since $$t\mapsto\mathbb P(|X|>t)$$ is a decreasing function, we can bound this integral via upper and lower Riemann sums as $$\epsilon\cdot\sum_{n=1}^\infty\mathbb P(|X|>n\epsilon)\leq\int_0^\infty\mathbb P(|X|>t)\dd t\leq\epsilon\cdot\sum_{n=0}^\infty\mathbb P(|X|>n\epsilon).$$ So it turns out that $$\mathbb E|X|<\infty\iff\sum\mathbb P(|X|>n\epsilon)<\infty$$.