How to find $E[X \Phi(X)]$ with $X$ normally distributed?

Question

Let $X$ be distributed as a standard normal variable, with CDF $\Phi(x)$.

How do I prove that $E[X \Phi(X)] = \frac{1}{2\sqrt{\pi}}$?

That is, how to show that

$$\int^\infty_{-\infty} X \frac{1}{\sqrt{2\pi}}e^{-0.5X^2}\int^{X}_{-\infty} \frac{1}{\sqrt{2\pi}}e^{-0.5Z^2}dZdX = \frac{1}{2\sqrt{\pi}} $$

Answer 1

Using Stein's lemma, we have

$$E[g(X) X] = E[g'(X)]$$

Using $g(x) = \Phi(x), g'(x) = \phi(x)$,

$$\begin{align*}E[\Phi(X)X] = E[\phi(X)] &= \frac{1}{\sqrt{2\pi}}\int^{\infty}_{-\infty} \frac{1}{\sqrt{2\pi}}e^{-X^2}dX \\ &= \frac{1}{\sqrt{2\pi}} \times \frac{1}{\sqrt{2}} \\ &= \frac{1}{2 \sqrt{\pi}} \end{align*}$$