0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What is the meaning of $1_{[0,\infty)}$. You probably mean $1_{[0,\infty)}(X)$. $\endgroup$ Mar 9, 2019 at 12:05
  • $\begingroup$ yes, that's what I mean $\endgroup$
    – max
    Mar 9, 2019 at 12:08

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks for the clear explanation. $\endgroup$
    – max
    Mar 9, 2019 at 12:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .