Question:
Random Variable $Z$ follows a standard normal distribution with c.d.f. $\Phi$.
Find $E[Z^2\Phi(Z)]$ using the law of the unconscious statistician.
I am thinking of using some substitution to work it out, for example, using $\phi'(z)=-z\phi(z)$ can help to solve $E[Z\Phi(Z)]$, but I don’t know how to apply this trick for finding out $E[Z^2\Phi(Z)]$.
This is a tricky one! Consider
$$\int z^2 \Phi(z) \phi(z) \, dz = \int (z \Phi(z)) (z \phi(z)) \, dz \tag{1}$$ and choose $$u = z \Phi(z), \quad du = \Phi(z) + z \phi(z) \, dz, \\ dv = z \phi(z) \, dz, \quad v = -\phi(z). \tag{2}$$ Then $(1)$ becomes $$\begin{align} \int z^2 \Phi(z) \phi(z) \, dz &= -z \Phi(z) \phi (z) + \int \Phi(z) \phi(z) + z \phi(z)^2 \, dz \\ &= -z \Phi(z) \phi(z) + \frac{\Phi(z)^2}{2} - \frac{\phi(z \sqrt{2})}{2\sqrt{2\pi}} + C. \tag{3} \end{align}$$ All that remains is to evaluate the definite integral to obtain the expectation, noting that $\Phi(\infty) = 1$, and $\Phi(-\infty) = \phi(\pm\infty) = 0$: $$\operatorname{E}[Z^2 \Phi(Z)] = \frac{1}{2}.$$ I have left out some of the computational details, but these should be straightforward.