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:^[1]

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

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:^[2]

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

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

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

where μ is the expectation of the random variable.^[3]

In the case of an independent sample of n observations from a bounded probability distribution, the von Szokefalvi Nagy inequality^[4] gives a lower bound to the variance of the sample mean:

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

Proof via the Bhatia–Davis inequality[edit]

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

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

Thus,

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

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

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

References[edit]

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