# 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?

### 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 $$$$\tag 1\label 1 \mu F(\mu)^2\geq\int_{\mu}^{b}F(x)[1-F(x)] \,\mathrm dx.$$$$

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).

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.)
• 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).
• @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 $x$. Commented Oct 24, 2019 at 12:35