The uniform distribution on a bounded interval $[a,b]$ has variance $(b-a)^2/12$. Consider any concave distribution on the same interval (concave in the sense that the graph of the pdf lies above any line segment joining two points of the graph). Does its variance $\sigma^2$ satisfy:

$$\sigma^2 \le \frac{(b-a)^2}{12}?$$

  • $\begingroup$ stats.stackexchange.com/q/142646/5536, stats.stackexchange.com/q/142653/5536 $\endgroup$
    – a06e
    Mar 20, 2015 at 18:59
  • $\begingroup$ I think you can prove this using the techniques from stochastic dominance theorems. $\endgroup$
    – Aksakal
    Mar 20, 2015 at 19:31
  • $\begingroup$ Is this work for some subject? $\endgroup$
    – Glen_b
    Mar 20, 2015 at 23:23
  • $\begingroup$ @Aksakal Do you have some references you can point out to me? $\endgroup$
    – a06e
    Sep 22, 2021 at 22:26

1 Answer 1


Yes, this is true.

First, note that any concave PDF must be unimodal. (This is an elementary consequence of the concavity.)

The answer to a related question shows that on the interval $[0,1]$, the variance of a distribution with mean $\mu$ cannot exceed $\mu(2-3\mu)/3$ (when $0 \le \mu \le 1/2$) or $(1-\mu)(3\mu-1)/3$ (when $1/2\le \mu \le 1$). Neither of these expressions exceeds $1/12$ (which is achieved solely when $\mu=1/2$.

By shifting and rescaling the interval $[0,1]$ to $[a,b]$, it immediately follows that $\sigma^2$ cannot exceed $(b-a)^2/12$, QED.

Evidently a stronger statement can be made when the mean $\mu$ is known: a tight upper bound is then

$$\sigma^2 \le \frac{(\mu-a)(2b+a-3\mu)}{3}$$

when $\mu \le (a+b)/2$. For $\mu \gt (a+b)/2$, switch $a$ and $b$ and replace $\mu$ by $b-\mu$.


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