I want to ask the following probability inequality:


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


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



Inequality involving the weak second moment
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.


