A proof of the Bhatia-Davis inequality (2018-10-10)

The Bhatia–Davis inequality gives an upper bound on the variance of a bounded random variable. It has the following simple proof.

Let \(X\) be a bounded random variable with maximum \(M\) and minimum \(m.\) Also, let \(\mu\) be \(X\)'s expected value. Without loss of generality we can assume that \(m = 0.\) (Otherwise, another random variable \(Y := X-m\) has the same variance as \(X\)'s, and \((M-\mu)(\mu-m) = \{(M-m)-(\mu-m)\}\{(\mu-m)-0\}.\)) Then\[(M-\mu)(\mu-m)-\mathbb{E}[(X-\mu)^2] = \{(M+m)\mu+\mu^2-Mm\}-(\mathbb{E}[X^2]-\mu^2) = (M+m)\mu-Mm-\mathbb{E}[X^2] = M\mu-\mathbb{E}[X^2] \geq M\mu - M\mu = 0.\]QED.

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