Timeline for Does the inequality $\mu F(\mu)^2\geq \int_{\mu}^{b}F(x)\cdot(1-F(x))\,\mathrm dx$ hold for compactly supported continuous random variables?

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Oct 24, 2019 at 12:35	comment	added	Iosif Pinelis		@smcc : Of course one can make $F$ differentiable and even infinitely smooth, say by convolving the discrete distribution with a distribution supported on the interval $[0,h]$ with an infinitely smooth density $p$ -- if, again, $h>0$ is small enough. Such a density may be given by the formula $p(x)=\frac ch\,\exp\{-\frac{h^2}{(h-x)x}\}1_{0<x<h}$ for a certain suitable universal positive real constant $c$ and all real $x$.
Oct 24, 2019 at 11:15	comment	added	smcc		Thanks for your answer. The context of the original problem requires that a density exists (and that it is differentiable and log-concave), and so I had not been looking for a discrete counterexample, but it is useful to know there is one. You say this example could be tweaked to make $F$ absolutely continuous. Could it be tweaked to ensure a differentiable density? Clearly this example could not be tweaked to make the density log-concave (as it would not even be unimodal).
Oct 23, 2019 at 15:57	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 82 characters in body
Oct 23, 2019 at 0:46	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 14 characters in body
Oct 23, 2019 at 0:39	history	edited	Iosif Pinelis	CC BY-SA 4.0	added 274 characters in body
Oct 23, 2019 at 0:32	history	answered	Iosif Pinelis	CC BY-SA 4.0