proving this equality of expectation of a random variable times a function of that r.v.

Question

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be differentiable and let $X$~$N(0,\sigma^2)$. I want to show that:

$\mathbb{E}[Xf(X)]=\sigma^2\mathbb{E}[f'(X)]$

My attempt:

I understand I can apply the by-parts rule of integration, but am not sure. quite how

$$\mathbb{E}[Xf(X)]=\int_{\Omega}{X(\omega)f(\omega)d\mathbb{P}(\omega)}$$

I am not sure how to deal with this integral. Is my definition of expectation correct?

Answer 1

This requires some additional hypothesis. For example if $f(x)=e^{x^{2}}$ then $Ef'(X)$ does not exist.

For 'well behaved' $f$ you can prove this by an integration by parts.

If $\phi (x)=\frac 1 {\sqrt {2 \pi}} e^{-x^{2}/2\sigma ^{2}}$ then $Ef'(X)=\int f'(x)\phi (x) dx=f(x)\phi (x) |_{-\infty} ^{\infty}+\int x f(x) \phi (x) dx$. This gives the answer provided $f(x)\phi (x) |_{-\infty} ^{\infty}=0$. [This would be true if $f$ is bounded or has polynomial growth].