# fixedpoint.jp

## A lower bound of the Bhatia-Davis inequality (2018-10-14)

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.