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:
σ 2 ≤ 1 4 ( M − m ) 2 . {\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:
σ 2 + ( μ 3 2 σ 2 ) 2 ≤ 1 4 ( M − m ) 2 , {\displaystyle {\sigma ^{2}+\left({\frac {\mu _{3}}{2\sigma ^{2}}}\right)^{2}}\leq {\frac {1}{4}}(M-m)^{2},}
Where μ 3 {\displaystyle \mu _{3}} is the third central moment. If one additionally assumes knowledge of the expectation, then the stronger Bhatia–Davis inequality holds
σ 2 ≤ ( M − μ ) ( μ − m ) {\displaystyle \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:
σ 2 ≥ ( M − m ) 2 2 n . {\displaystyle \sigma ^{2}\geq {\frac {(M-m)^{2}}{2n}}.}
Proof via the Bhatia–Davis inequality Let A {\displaystyle A} be a random variable with mean μ {\displaystyle \mu } , variance σ 2 {\displaystyle \sigma ^{2}} , and Pr ( m ≤ A ≤ M ) = 1 {\displaystyle \Pr(m\leq A\leq M)=1} . Then, since m ≤ A ≤ M {\displaystyle m\leq A\leq M} ,
0 ≤ E [ ( M − A ) ( A − m ) ] = − E [ A 2 ] − m M + ( m + M ) μ {\displaystyle 0\leq \mathbb {E} [(M-A)(A-m)]=-\mathbb {E} [A^{2}]-mM+(m+M)\mu } . Thus,
σ 2 = E [ A 2 ] − μ 2 ≤ − m M + ( m + M ) μ − μ 2 = ( M − μ ) ( μ − m ) {\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, a b ≤ ( a + b 2 ) 2 {\displaystyle ab\leq \left({\frac {a+b}{2}}\right)^{2}} , with a = M − μ {\displaystyle a=M-\mu } and b = μ − m {\displaystyle b=\mu -m} , yields the desired result:
σ 2 ≤ ( M − μ ) ( μ − m ) ≤ ( M − m ) 2 4 {\displaystyle \sigma ^{2}\leq (M-\mu )(\mu -m)\leq {\frac {\left(M-m\right)^{2}}{4}}} .
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