Show that $E\{ZΦ(Z)\} = 1 / \left( 2\sqrt{\pi} \right)$ I need to show that $E\{ZΦ(Z)\} = 1 / \left( 2\sqrt{\pi} \right)$. Let $Z$ be a standard normal random variable with density $ϕ$ and distribution function $Φ$
I don't know how to start. 
 A: This is S. Catterall's hint:
\begin{align*}
E\{ZΦ(Z)\} &= \int_{\mathbb{R}} z \Phi(z) \phi(z) dz.
\end{align*}
And an extra hint: let $u=\Phi(z)$ and $v' = z\phi(z)$.
One more hint: $v = \int_{-\infty}^z \phi(s)sds = - \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}}$.
A: I'm going to offer an alternative that uses a special property of the Gaussian distribution - but it really comes from applying integration by parts, let $f$ be a function (I won't say much about what this function class is) then it is the case that if $Z$ is a standard Gaussian random variable we have
$$
\begin{align}
\mathbb{E} \left[ Z f(Z) \right] = \mathbb{E} \left[ f^{\prime}(Z)\right],
\end{align}
$$
so applying that to this problem we get
$$
\begin{align}
\mathbb{E}\left[ Z \Phi(Z) \right] &= \mathbb{E}\left[ \phi(Z) \right] \\
&= \frac{1}{2\pi} \int e^{-z^2}dz \\ &= \frac{\sqrt{\pi}}{2\pi} = \frac{1}{2\sqrt{\pi}}.
\end{align}
$$
The identity used above is the starting point of what is called Stein's method in probability and statistics.
