Conditional second moment normal distribution

Question

Can someone help me find the expression of

$$ E(Z^2 | Z \leq 0) \quad when \quad Z\sim N(0,1) $$

I'm aware that $$ E(Z | Z \leq 0) = \frac{2}{\sqrt{2\pi}} \quad , \,\, Z \sim N(0,1) $$ using the Inverse Mills Ratio, but I don't know how to apply it to the second moment.

Answer 1

Does this method work? \begin{align*} E(Z^2) &= E(Z^2 | Z \leq 0) \mathbb P(Z\leq 0)+E(Z^2 | Z \geq 0) \mathbb P(Z\geq 0)\\ &= E(Z^2 | Z \leq 0) \frac{1}{2}+E(Z^2 | Z \geq 0) \frac{1}{2}\\ &= E(Z^2 | Z \leq 0) \frac{1}{2}+E(Z^2 | Z \leq 0) \frac{1}{2}\\ &= E(Z^2 | Z \leq 0) \end{align*} where the first equality is by iterated expectation (total probability), the second equality is by property of $\mathcal N (0,1)$, the third equality is by symmetry of Z around 0.

But $E(Z^2) = 1 +(E[Z])^2 =1$. So $E(Z^2 | Z \leq 0) =1$