I'm having trouble computing the expectation:
$$\mathbb E [Z_1 \mid Z_1 > Z_2]$$
even in the simple case where $Z_1, Z_2$ are iid standard normal rvs. It's clear to me that, at least in some cases, conditioning on the event $Z_1>Z_2$ does give us some information about $Z_1$: consider the case where $Z_2$ is a constant.
My typical approach for this kind of problem would be to condition on $Z_2$ and try take advantage of the independence assumption:
$$\mathbb E [Z_1 \mid Z_1 > Z_2] =\mathbb E \left [\mathbb E [Z_1 \mid Z_1 > Z_2, Z_2] \mid Z_1 > Z_2\right ]$$
But that does not help much since I cannot simplify the inner expectation.
At first glance I would guess the expectation should be easy to compute, but I am having trouble simplifying it using properties of expectation (or by trying to compute a marginal distribution which I can use to express the expectation). How can I solve this problem?
Let $f(\mu,\sigma)=\mathbb E(Z_1\mid Z_1>Z_2)$ when the $Z$'s are $\mathcal N(\mu,\sigma^2)$'s. It is intuitively clear that $f(\mu,\sigma^2)=\sigma f(0,1)+\mu$, so we only need to calculate $f(0,1)$.
It is also intuitively clear that $\mathbb P(Z_1>Z_2)=\frac12$, so the pdf of $Z_1$ given $Z_1>Z_2$ is $$f_{Z_1\mid Z_1>Z_2}(x)=\left(\frac1{2\pi}\int_{-\infty}^xe^{-(x^2+y^2)/2}\,dy\right)\bigg/\frac12$$ $$=\frac1\pi\int_{-\infty}^xe^{-(x^2+y^2)/2}\,dy$$ Thus $$f(0,1)=\int_{-\infty}^\infty\frac x\pi\int_{-\infty}^xe^{-(x^2+y^2)/2}\,dy\,dx$$ $$=\frac1\pi\int_{-\infty}^\infty e^{-y^2/2}\int_y^\infty xe^{-x^2/2}\,dx\,dy$$ $$=\frac1\pi\int_{-\infty}^\infty e^{-y^2/2}\int_{y^2/2}^\infty e^{-u}\,du\,dy$$ $$=\frac1\pi\int_{-\infty}^\infty e^{-y^2}\,dy=\frac1{\sqrt\pi}$$ and $f(\mu,\sigma)=\frac\sigma{\sqrt\pi}+\mu$.