Let $\Phi$ be the distribution function of $Z \sim N(0,1)$.Find $E[Z \Phi(Z)]$ and $E[Z^2 \Phi(Z)]$.

Question

Let $\Phi$ be the distribution function of $Z \sim N(0,1)$.Find $E[Z \Phi(Z)]$ and $E[Z^2 \Phi(Z)]$. In the first case, I am struggling with the integral $\int_{-\infty}^{\infty} z^2 \Phi(z) \phi(z) dz$ How to evaluate this?

Answer 1

On base of $\Phi(z)+\Phi(-z)=1$ we find:$$Z^2=Z^2\Phi(Z)+(-Z)^2\Phi(-Z)\tag1$$

$Z$ and $-Z$ have the same distribution, so also $Z^2\Phi(Z)$ and $(-Z)^2\Phi(-Z)$ have the same distribution.

Then taking expectation on both sides of $(1)$ we find:$$1=\mathsf EZ^2=2\mathsf EZ^2\Phi(Z)$$

So: $$\mathsf EZ^2\Phi(Z)=\frac12$$

Answer 2

We have $$(-z)^2 \Phi(-z) \phi(-z) = z^2 (1-\Phi(z)) \phi(z)$$ and so

$$\begin{align*}\int_{-\infty}^{0} z^2 \Phi(z) \phi(z) dz &= \int_{0}^{\infty} (-z)^2 \Phi(-z) \phi(-z) dz\\ &= \int_{0}^{\infty} z^2 \phi(z) dz - \int_{0}^{\infty} z^2 \Phi(z) \phi(z) dz\end{align*}$$

And we get:

$$\begin{align*}\int_{-\infty}^{\infty} z^2 \Phi(z) \phi(z) dz &= \int_{-\infty}^{0} z^2 \Phi(z) \phi(z) dz + \int_{0}^{\infty} z^2 \Phi(z) \phi(z) dz \\ &= \left(\int_{0}^{\infty} z^2 \phi(z) dz - \int_{0}^{\infty} z^2 \Phi(z) \phi(z) dz\right) + \int_{0}^{\infty} z^2 \Phi(z) \phi(z) dz\\ &= \int_{0}^{\infty} z^2 \phi(z) dz \\&= \frac{1}{2}E[Z^2] = \frac{1}{2}\end{align*}$$

But actually you can have this much easier just use the expectation: $$\begin{align*}E[Z^2 \Phi(Z)] &= E[Z^2 \Phi(Z) 1_{Z\le 0}] + E[Z^2 \Phi(Z) 1_{Z\ge 0}] \\ &= E[Z^2 \Phi(-Z) 1_{Z\ge 0}] + E[Z^2 \Phi(Z) 1_{Z\ge 0}] \\ &= E[Z^2 1_{Z \ge 0}] = \frac{1}{2}\end{align*}$$ The same you can do with $E[Z \Phi(Z)]$