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This question was originally asked on the Mathematics StackExchange by User smcc

Consider a continuous random variable $V$ with cumulative distribution function $F$ and density function $f$. Suppose that $f$ is compactly supported on $[a,b]\subset\mathbb R_{\geq 0}$. Let $\mu$ denote the expected value of $V$.

Conjecture. Then it holds that \begin{equation}\tag 1\label 1 \mu F(\mu)^2\geq\int_{\mu}^{b}F(x)[1-F(x)] \,\mathrm dx. \end{equation}

Question. Is this conjecture true? How can we (dis)-prove it?

Remark. In the original question, we had the additional assumptions that $f$ should be differentiable on $]a,b[$ and that $\ln\circ f$ should be concave on $[a,b]$. I don't think that these assumptions are necessary though.

Notation. For $x\in[a,b]$, let $$\phi(x):=x F(x)^2-\int_{x}^{b}F(y)\cdot(1-F(y))\,\mathrm dy.$$

Then \eqref{1} is $\phi(\mu)\geq0$.

Partial results. (Refer to the original question for the derivations)

  • We have $\phi(x)=g(x)-h(x)$, where $$g(x)=\int_{x}^{b}yf(y)\,\mathrm dy+\mu F(x), \qquad \text{and} \qquad h(x)=\int_{x}^{b}y\cdot(2f(y)\cdot F(y))\,\mathrm dy.$$ In particular, $\phi(a)<0$ and $\phi(b)>0$. Additionally, $\phi$ is strictly increasing (as $\phi'>0$ on $]a,b[$) and $g$ achieves its maximum at $x=\mu$.
  • Inequality \eqref{1} holds if [$\mu$ is greater or equal than the median of $V$] and $\mu\geq \frac b2$. In particular, if $f$ is flip-symmetrical to $\frac{a+b}2$, then \eqref{1} holds.
  • Inequality \eqref{1} holds if [$f$ is increasing on $[a,b]$] and $F(\mu)\geq\frac12$.
  • We have $$\phi(\mu)\geq \mu F(\mu)^2-\int_{a}^{\mu}F(x)\,\mathrm dx.$$ However, the right-hand side of the last equation can be negative. For example, for $F(x)=\ln(1+(e-1)\cdot x)$ on $[0,1]$, the right-hand side is $\approx-0.0008$. (Remark: The according density function to my $F$ is not $\ln$-concave. I wasn't able to find a $\ln$-concave counter-example).
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This conjecture is false. E.g., let $V$ take values $0,\frac54,\frac64$ with probabilities $\frac14,\frac24,\frac14$, respectively. Then the left-hand side of your inequality is $\frac2{32}$ and its right-hand side is $\frac3{32}$, so that the inequality fails to hold.

(Note that here $\mu=1$, which differs from each of the values $0,\frac54,\frac64$ of $V$, so that the cumulative distribution function $F$ of $V$ is continuous at $\mu$. Thus, if you wish, you can tweak this distribution of $V$ slightly -- say by convolving it with the uniform distribution on the interval $[0,h]$ for small enough $h>0$ -- so that the distribution become absolutely continuous and yet your inequality continue to fail to hold.)

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  • $\begingroup$ 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). $\endgroup$
    – smcc
    Commented Oct 24, 2019 at 11:15
  • $\begingroup$ @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 $x$. $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2019 at 12:35

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