Popoviciu's inequality on variances

In probability theory, Popoviciu's inequality, named after Tiberiu Popoviciu, is an upper bound on the variance σ² of any bounded probability distribution. Let M and m be upper and lower bounds on the values of any random variable with a particular probability distribution. Then Popoviciu's inequality states:

${\sigma }^2\le \frac{1}{4}(M-m{)}^2.$

This equality holds precisely when half of the probability is concentrated at each of the two bounds.

Sharma et al. have sharpened Popoviciu's inequality:

${\sigma }^2+{\left(\frac{\text{Third central moment}}{2{\sigma }^2}\right)}^2\le \frac{1}{4}(M-m{)}^2.$

If one additionally assumes knowledge of the expectation, then the stronger Bhatia–Davis inequality holds

${\sigma }^2\le (M-\mu )(\mu -m)$

where μ is the expectation of the random variable.

In the case of an independent sample of n observations from a bounded probability distribution, the von Szokefalvi Nagy inequality gives a lower bound to the variance of the sample mean:

${\sigma }^2\ge \frac{(M-m{)}^2}{2n}.$

Proof via the Bhatia–Davis inequality

Let $A$ be a random variable with mean $\mu$, variance ${\sigma }^2$, and $\mathrm{Pr}(m\le A\le M)=1$. Then, since $m\le A\le M$,

$0\le \mathbb{𝔼}[(M-A)(A-m)]=-\mathbb{𝔼}[A^2]-mM+(m+M)\mu$.

Thus,

${\sigma }^2=\mathbb{𝔼}[A^2]-{\mu }^2\le -mM+(m+M)\mu -{\mu }^2=(M-\mu )(\mu -m)$.

Now, applying the Inequality of arithmetic and geometric means, $ab\le {\left(\frac{a+b}{2}\right)}^2$, with $a=M-\mu$ and $b=\mu -m$, yields the desired result:

${\sigma }^2\le (M-\mu )(\mu -m)\le \frac{{(M-m)}^2}{4}$.

References

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