If (X,Y) are jointly normal $(\mu,\Sigma)$, How to calculate $E(X|Y>0)$?
I have a solution using $E(X|Y>0)=E\{E(X|Y)|Y>0\}$, since we know $E(X|Y)=\mu_x+\rho\sigma_x/\sigma_y(Y-\mu_y)$. After a very long algebra I get the final solution is $\mu_x+\rho \sigma_x \phi(m)/\Phi(m)$ where $m=\mu_y/\sigma_y$. Do you think it's correct?
This is an example of an expectation of the form $\mathbb{E}[X|A]$, where $A$ is an event with a non zero probability. Its value is given by
$$ \mathbb{E}[X|A] = \frac{\mathbb{E}[X\cdot 1_A]}{\mathbb{P}(A)} $$
In this case we need to compute the following:
$$ \mathbb{E}[X|Y>0] = \frac{\int_0^{\infty}\int_\mathbb{R}xf(x,y)dxdy}{\int_0^{\infty}\int_\mathbb{R}xf(x,y)dxdy} $$
with $f(x,y)$ the joint distribution of a normal random variable.