When is $E[f(X)]=0$ for even functions

Question

Let $X$ be standard normal and $f$ a function that satisfies

1. $f(0)=0$
2. $f$ is even
3. $(x-{\rm sign}(x) \cdot a) \cdot x \le f(x) \le (x+{\rm sign}(x) \cdot a) \cdot x$, for all $x$, and some fixed $a>0$. Moreover, these bounds are asymptotcily tight.

I am either trying to find and example of $f(x)$ such that \begin{align} E[f(X)]=0 \end{align} or that to show that this can not happen.

Thanks.

Answer 1

If $a < \sqrt{\dfrac{\pi}{2}}$, then for any $f$ satisfying the given conditions, we have:

$\mathbb{E}[f(X)]$ $= \displaystyle\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}f(x)\,dx$ $= \displaystyle\int_{0}^{\infty}\sqrt{\dfrac{2}{\pi}}e^{-x^2/2}f(x)\,dx$

$\ge\displaystyle\int_{0}^{\infty}\sqrt{\dfrac{2}{\pi}}e^{-x^2/2}(x^2-ax)\,dx$ $= 1 - a\sqrt{\dfrac{2}{\pi}} > 0$.

So, no such function exists if $a < \sqrt{\dfrac{\pi}{2}}$.

If $a \ge \sqrt{\dfrac{\pi}{2}}$, then we can pick $f(x) = \begin{cases}x^2-\sqrt{\dfrac{\pi}{2}}x & \text{if} \ x \ge 0 \\ x^2+\sqrt{\dfrac{\pi}{2}}x & \text{if} \ x < 0\end{cases}$.

This $f$ satisfies the given conditions, and

$\mathbb{E}[f(X)]$ $= \displaystyle\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}f(x)\,dx$ $= \displaystyle\int_{0}^{\infty}\sqrt{\dfrac{2}{\pi}}e^{-x^2/2}f(x)\,dx$

$= \displaystyle\int_{0}^{\infty}\sqrt{\dfrac{2}{\pi}}e^{-x^2/2}\left(x^2-\sqrt{\dfrac{\pi}{2}}x\right)\,dx = 0$.

Answer 2

Such $f$ always exist if $a$ can be selected based on $f$. Take $f(x) = \begin{cases}x &\text{if }|x| \leq c \\ -1 &\text{if } c < |x| \leq 10 \\ x^2 &\text{else} \end{cases}$. For $c$ near $0$ you obtain $\mathbb{E}(f(x))=-1$ while for $c=10$ you obtain $\mathbb{E}(f(x))>0$. By continuity in $c$ and the intermediate value theorem, there is a $c$ strictly between $0$ and $10$ such that $\mathbb{E}(f(x))=0$. For such $c$, you can find an $a$ such that your condition holds.