I have the following question:
Let $X\sim\mathcal{N}(0,1)$ and $\mathbb{E}[Xf(X)\mathbb{1}_{[0,\infty)}(X)]=0$. Clearly it is sufficient that $f(x)=0$ for all values of the domain. Is it also a necessary condition?
So I separated the function $f$ into its positive and negative parts $f=f^{+}-f^{-}$. It follows that $\mathbb{E}[Xf^{+}(X)\mathbb{1}_A]>0$ for $A:=\{\omega: g(X(\omega))>0\}$. Now I need to show that $\mathbb{P}(A)=0$ but I'm not sure how to proceed. Do I need to use some sort of 0-1 law? The help would be much appreciated.
It does not follow that $f$ is $0$. Example: take $f(x)=a$ for $0<x<1$ and $f(x)=b$ for $1 <x<\infty$. Then the hypothesis becomes $a\int_0^{1}x\phi(x)dx+b\int_1^{\infty} x \phi(x)dx=0$ where $\phi$ is the standard normal density function. It is clear that you can choose non-zero $a$ and $b$ satisfying this equation.