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I am trying to find a non-constant function $f: \mathbb{R} \to [0,1]$ such that its derivative $f'$ is an even function and satisfies that $$ \mathbb{E}_{Z \sim \mathcal{N}(0,1)}[f'(Z)] =0. $$ Does anybody have suggestions? I was thinking of taking $f(\cdot)=\frac{1+\sin(\cdot)}{2}$ but $\mathbb{E}[\cos(Z)]=\frac{1}{\sqrt{e}}$.

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  • $\begingroup$ @achillehui: Can you give a reason? Notice that $Z$ is Gaussian. I already showed above why $\sin$ wouldn't work. $\endgroup$ – pikachuchameleon May 9 '17 at 20:14
  • $\begingroup$ I'm sorry, I misread it as uniform distribution. $\endgroup$ – achille hui May 9 '17 at 20:15
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How about $f(x) =1+ [\sin(x)-{e^{3/2}\over 2}\sin(2x)]/4$. As you note, $E[\cos(Z)]={1\over\sqrt{e}}$; also $E[\cos(2Z)]={1\over e^2}$. It follows that $E[f'(Z)]=0$. And you can check that $0\le f(x)\le 1$ for all $x$.

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Just take $f$ to be a constant function. Then its derivative is zero everywhere, so in particular $\mathbb{E}[f'(Z)] = 0$.

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