Suppose $X$ and $Y$ are correlated with correlation coefficient $\rho$. They are jointly normal with means $\mu_X$ and $\mu_Y$ respectively. Then what is $E[X | Y \geq T]$? Feel free to add additional assumptions if necessary.
In the case where $X = Y$, we can use the formula here: https://en.wikipedia.org/wiki/Truncated_normal_distribution#One_sided_truncation_(of_lower_tail)[4].
The same expectation can be written in terms of standard normal RVs, with the same correlation coefficient: $$\mathbb E[X|Y\geq t] = \sigma_x \mathbb E\left[\frac{X-\mu_x}{\sigma_x} \bigg| \frac{Y-\mu_y}{\sigma_y} \geq \frac{t-\mu_y}{\sigma_y}\right] + \mu_x = \sigma_x\mathbb E[X'|Y'>t']+\mu_x$$
Then, we can use the formula given in the wikipedia:
$$\mathbb E[X' | Y' > t']=\mathbb E[X' | Y'\geq t']=\rho \frac{\phi(t')}{1-\Phi(t')}$$
and, plug in. $\phi(z)$ and $\Phi(z)$ are the PDF and CDF of standard normal RV, respectively.
As a closed form solution has already been proposed, here is a relatively simple numerical algorithm to arrive at an answer as well.
Generate a random uniform deviate between 0 and 1.
Use this value in a software supplied Normal random deviate inversion algorithm to create a Normal deviate, say $Y_i$, for a Normal distribution with associated mean = $μ_y$ and sigma = $σ_y$.
If $Y_i$ ≥ t, keep the sample point, otherwise return to Step 1.
Compute $E[X|Y = Y_i ]$ per the well known formula $\mathbb E[X | Y_i ] = \mu_x + ρ\frac{\sigma_x}{\sigma_y}(Y_i -\mu_y) $. Source see, for example, this education reference.
Repeat starting at Step 1 for a large number of iterations.
Finally, calculate the average of all values obtained in Step 4.
Note, one can easily adjust Step 3 in this Monte Carlo simulation approach to handle more complex cases (like a range).
