Having shown the Bhatia-Davis inequality in our previous post, we find that the Wikipedia article also mentions a curious lower bound on the variance \(\sigma^2\) of a bounded random variable: \[(M-\mu)(\mu-m) - \frac{(M-m)^3}{6} \leq \sigma^2.\] Alas, it is incorrect.
A clear counterexample is the uniform random variable on \([0, \frac{1}{2}].\) Obviously its \(M\) is \(\frac{1}{2}\), \(m = 0\), and \(\mu = \frac{1}{4}\), so the LHS is \(\frac{1}{24}\); on the other hand \(\sigma^2 = \frac{1}{48}\).
It is necessary for a correct lower bound to modify the second term slightly as follows.
Let \(f\) be a probability density function for a bounded random variable \(X\) with maximum \(M\) and minimum \(m\). Assume that \(f \in L^\infty\) i.e. \(\lvert f(x) \rvert \leq \lVert f \rVert_\infty\) for almost every \(x.\) Then \[(M-\mu)(\mu-m) - \frac{(M-m)^3}{6} \lVert f \rVert_\infty \leq \sigma^2.\]
Proof. We can assume that \(m = 0\) without loss of generality. \begin{align}\sigma^2 - (M-\mu)\mu + \frac{M^3}{6} \lVert f \rVert_\infty &= (\mathbb{E}[X^2] - \mu^2) - M\mu + \mu^2 + \frac{M^3}{6} \lVert f \rVert_\infty\\ &= \mathbb{E}[X^2] - M \mathbb{E}[X] + \frac{M^3}{6} \lVert f \rVert_\infty\\ &= \int_0^M (x-M) x f(x) dx + \int_0^M (M-x) x \lVert f \rVert_\infty dx\\ &= \int_0^M (M-x) x (\lVert f \rVert_\infty - f(x)) dx\\ &\geq 0\end{align} as \(M-x, x,\) and \(\lVert f \rVert_\infty - f(x)\) are nonnegative for almost every \(x\) in \([0, M].\) QED.