What is the $\mathbb E[(X-\mu)g(X)]$?
Let $X\sim N(\mu , \sigma ^2)$ and $g:R \Rightarrow R$ and $g$ is bounded and differentiable.
I need to show $E[(X-\mu)g(X)]=\sigma ^2 \mathbb E[g'(X)]$
My attempt
I don't know if I can split the expectations into 2, are they independent? Well I assumed I could.
I got $\mathbb E[(X-\mu)g(X)]=\mathbb E[(X-\mu)\mathbb Eg(X)$
Then I know that $\mathbb E[(X-\mu)=\mathbb E[X]-\mathbb E[\mu]=0$
This shows that infact, they can't be independent and I'm not sure why.
Then I can rewrite $\mathbb E[(X-\mu)g(X)]=\mathbb E[Xg(x)]-\mathbb E[\mu g(X)]$
I attmpted to use integration by parts, but I can't see where the derivative can come from. Any help is very much appreciated.