Let $N\sim \mathcal N(0,\sigma^2)$ be a normal RV. Let $g$ be continuous. Can we bound $E[g(N)]$ in terms of $\sigma^2$?

Question

Let $N\sim \mathcal N(0,\sigma^2)$ be a normal RV. Let $g$ be continuous. Can we bound $E[g(N)]$ in terms of $\sigma^2$?

This seems fairly easy but it might be wrong. What if we assume additional structure on $g$? For example what if assume that $g$ is continuous convex? This is something that came up in research.

Answer 1

If, in addition, $g$ is Lipschitz with $g(0)=0$, then \begin{align} |\mathsf{E}g(N)|&\le |g(0)|+\mathsf{E}|g(N)-g(0)| \\ &\le |g(0)|+\operatorname{Lip}(g)\mathsf{E}|N|=\operatorname{Lip}(g)\sigma\sqrt{\frac{2}{\pi}}. \end{align}

Answer 2

Let $X\sim \mathcal{N}(0,\sigma^2)$ and suppose that $g$ is twice differentiable with a bounded second derivative.

We will now consider the first order taylor approximation $t(x) = g(0) + g'(0)x$ around $x=0$, and we note that $\mathbb{E}[t(X)]=g(0)$. This means that

\begin{align*} \Big|\mathbb{E}[g(X)]-g(0) \Big| &= \Big|\mathbb{E}[g(X)-t(X)] \Big| \\ &\leq \mathbb{E}\big[ \big|g(X)-t(X)|] \end{align*} But if the second derivative is bounded by some constant $C>0$, then taylors formula gives that $$|g(x)-t(x)| \leq \frac{C}{2}x^2$$ and thus we get that $$\Big|\mathbb{E}[g(X)]-g(0) \Big| \leq \frac{C}{2}\mathbb{E}[X^2] = \frac{C\sigma^2}{2} $$