I'm reading a paper and found an interesting expectation. I know the result the author found but I can't figure out the intermediary steps because the author provided none. My attempt is getting really ugly so I think I may be going in the wrong direction.
Suppose $(\epsilon,u,v)'\sim N(0,\Sigma)$ with $\Sigma = \begin{pmatrix} \sigma^2 & \psi_u \sigma & \psi_v \sigma \\ \psi_u \sigma & 1 & \rho \\ \psi_v \sigma & \rho & 1 \end{pmatrix}$, and $(-u,-v)$ are bivariate normal with correlation $\rho$.
Show: $\boxed{\mathbb{E}[\epsilon\cdot\mathbf{1}[u>-a, v>-b]]=\mathbb{E}[\sigma \psi_v \phi(-b)\Phi\left(\frac{a-\rho b}{\sqrt{1-\rho^2}}\right)+\sigma \psi_u \phi(-a)\Phi\left(\frac{b-\rho a}{\sqrt{1-\rho^2}}\right)]~~~~~}$
My attempt:
First, $\mathbb{E}[\epsilon\cdot\mathbf{1}[u>-a, v>-b]]=\mathbb{E}[\epsilon|u>-a, v>-b] \cdot \mathbb{P}(u>-a, v>-b)$
Then we know $\mathbb{E}(\epsilon|u, v) = \frac{\sigma}{(1-\rho^2)}[\psi_u-\rho\psi_v]u+\frac{\sigma}{(1-\rho^2)}[\psi_v-\rho\psi_u]v$
So
$\begin{split} \mathbb{E}[\epsilon|u>-a, v>-b]& = \mathbb{E}[\mathbb{E}(\epsilon|u, v)|u>-a, v>-b] \\ & = \mathbb{E}\left[\frac{\sigma}{(1-\rho^2)}(\psi_u-\rho\psi_v)u+\frac{\sigma}{(1-\rho^2)}(\psi_v-\rho\psi_u)v\bigg|u>-a, v>-b\right] \\ & = \frac{\sigma}{(1-\rho^2)}\left\{(\psi_u-\rho\psi_v)\mathbb{E}[u|u>-a, v>-b]+(\psi_v-\rho\psi_u)\mathbb{E}[v|u>-a, v>-b]\right\} \end{split}$
To evaluate those bivariate conditionals I was thinking of using this post: Expectation of truncated normal
For the probability $\mathbb{P}(u>-a, v>-b)$, one can write: $u \text{ and } z=\frac{v-\rho u}{\sqrt{1-\rho^2}}$. are independent.
\begin{split} \mathbb{P}\left(u>-a, z> \frac{-b-\rho u}{\sqrt{1-\rho^2}}\right) &= \mathbb{P}(u>-a)\cdot \mathbb{P}\left(z>\frac{-b-\rho u}{\sqrt{1-\rho^2}}\right) \\ & = \int_{-a}^{\infty}e^{\frac{-u^2}{2}} \int_{\frac{-b-\rho u}{\sqrt{1-\rho^2}}}^{\infty}e^{\frac{-z^2}{2}}dzdu \end{split}
I foundthis post which I think can be used to evaluate this probability: https://math.stackexchange.com/questions/414355/integral-involving-the-cdf-of-normal-distribution
Relevant other posts:
Truncated trivariate normal - conditional expectation
