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## probability inequality

I want to ask the following probability inequality:

Is it true that for any random variable $X\ge 0$, we have

$$\sup_{t>0}(t\mathbb E(X\mathbf 1_{X\ge t})) \le 2\sup_{t>0}(t^2 \mathbb P(X \ge t))?$$

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 You should tell us why you want to know; why you think it might be true etc. – Anthony Quas Dec 9 at 16:27 It's from a big proof of some theorem. The right hand side of this inequality is a weaker version of the second moment. This inequality gets a lower bound of this weak second moment. – honglangwang Dec 9 at 19:37

This is an interesting inequality.

Let $f$ be the density of $|X|$. Let $$g(t)=t\int_t^\infty xf(x)dx.$$ Then our inequality is $$\sup_t g(t)\leq \sup_t t^2\int_t^\infty f(x)dx.$$ Let us assume that $\sup$ of the LHS is attained at some point $t=a$, so that $g(a)$ be the maximal value of $g$.

(It will be easy to get rid of this assumption, as well as of the assumption about existence of the density in the end). Our $g$ is bounded. Then the RHS evaluated at the same point $a$ is $$-2a^2\int_a^\infty\left(\frac{g(t)}{t}\right)^\prime\frac{dt}{t}.$$ I integrate this by parts and obtain $$2a^2\left( g(a)a^{-2}-\int_a^\infty\frac{g(t)}{t^3} dt\right).$$ Now, using that $a$ is the maximum of $g$, I estimate this from below as $$2g(a)-2g(a)a^2\int_a^\infty\frac{dt}{t^3}\geq g(a),$$ which completes the proof.

Getting rid of the assumptions made before is routine: just approximate your distribution by a distribution satisfying those assumptions.

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