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:

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

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:

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

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

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

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:

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

Proof via the Bhatia–Davis inequality

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

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

Thus,

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

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

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