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).

1 Answer 1


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.)

  • $\begingroup$ 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 $x$. $\endgroup$ Commented Oct 24, 2019 at 12:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.