What is E[X|X+Y < z] with X, Y independent Normals?

Question

Let $X\sim N(\mu_X,\sigma_X^2)$ and $Y\sim N(\mu_Y,\sigma_Y^2)$ and $Cov(X,Y)=\sigma_{XY}$. Define $Z=X+Y$.

I know that $E[X|Z=z]=\mu_X + \frac{\sigma_X^2+\sigma_{XY}}{\sigma_X^2+2\sigma_{XY}+\sigma_Y^2}(z-\mu_X-\mu_Y)$.

But what is $E[X|Z\leq z]$?

Can I just integrate over all possible realizations of $Z\leq z$ and use linearity to do the following?

$$ \begin{align*}E[X|Z\leq z] & = \mu_X + \frac{\sigma_X^2+\sigma_{XY}}{\sigma_X^2+2\sigma_{XY}+\sigma_Y^2}(E[Z|Z\leq z]-\mu_Z)\\ & = \mu_X + \frac{\sigma_X^2+\sigma_{XY}}{\sqrt{\sigma_X^2+2\sigma_{XY}+\sigma_Y^2}}\frac{\phi(\frac{z-\mu_Z}{\sigma_Z})}{\Phi(\frac{z-\mu_Z}{\sigma_Z})} \end{align*}$$

Answer 1

Yes, using the law of total expectation, \begin{align} E(X|Z\le z) &=E(E(X|Z)|Z\le z) \\&=E(a + bZ|Z\le z) \\&=a + bE(Z|Z\le z). \end{align}