$(X,Y)$ follows bivariate normal distribution with parameters $(\mu_1, \mu_2, \sigma_1^2, \sigma_2^2, \rho)$. Find the conditional expectation $E(Y|X>x)$.
I know that $Y|x$ follows Normal distribution with parameters $(\mu_2+\frac{\rho\sigma_2}{\sigma_1}(x-\mu_1), \sigma_2^2(1-\rho^2))$. So It is easy to find $E(Y|x)$. But I cannot understand how to find $E(Y|X>x)$.
Any help will be appreciated. Thanks in advance.
Probability that $X>x_0$ is given by $c=1-\Phi(\frac{x_0-\mu_1}{\sigma_1})$, so desired expectation is $$\int_{-\infty}^\infty \int_{x_0}^\infty \frac y cf(x,y)dxdy$$ with $f(x, y)$ being the bivariate normal density. $f(x,y); x_0<x$ is the part of the bivariate normal distribution that has $X>x_0$; when we divide by $c$ we get $\frac{f(x,y)}{c}=\frac{f(x,y)}{P(X>x_0)}; x_0<x$, a reweighted part of the bivariate density corresponding only to the desired part that is a pdf: $f(x, y|X>x_0)$. The inner integral marginalizes out $x$ from the density, giving $g(y|X>x_0)$.
The outer integral computes the expectation, with a multiplication by $y$ (as it is constant in the inner integral).
For $\mu_1=1, \mu_2=2, \sigma_1^2=3, \sigma_2^2=4, \rho=1$, a simulation of 100,000 bivariate normal random variables gives an approximation of $E(Y|X>1)\approx 2.798615$. The value given by computing the integral above is $2.79788$.