$X$ 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 $\mathbb E[ |X| ] < \infty \iff \forall \epsilon : \mathbb E[ |X / \epsilon | ] < \infty$).
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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.
Best Answer
$\newcommand{\dd}{\mathop{}\!\mathrm{d}}$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 $\mathbb E|X|<\infty\iff\sum\mathbb P(|X|>n\epsilon)<\infty$.