A Note on an Inequality Involving the Normal Distribution

Abstract

The following inequality is useful in studying a variation of the classical isoperimetric problem. Let $X$ be normally distributed with mean 0 and variance 1. If $g$ is absolutely continuous and $g(X)$ has finite variance, then $E \{\lbrack g'(X)\rbrack^2\} \geq \operatorname{Var}\lbrack g(X)\rbrack$ with equality if and only if $g(X)$ is linear in $X$. The proof involves expanding $g(X)$ in Hermite polynomials.