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:


 * $$ \sigma^2 \le \frac14 ( M - m )^2. $$

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:


 * $$ {\sigma^2 + \left( \frac \text {Third central moment} {2\sigma^2} \right)^2} \le \frac14 (M - m)^2. $$

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 $$\Pr(m \leq A \leq M) = 1$$. Then, since $$m \leq A \leq M$$,

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

Thus,

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

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:

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