# the conditional expectation of truncated multivariate normal distribution

I want to calculate the conditional expectation of truncated multivariate normal distribution. Specifically, $$\mathbb{E} (\tilde{y} \mid \tilde{x}_1 = x_1, \tilde{x}_2 \geq a),$$ where $$\tilde{Y} = \begin{pmatrix} \tilde{y} \\ \tilde{x}_1 \\ \tilde{x}_2 \end{pmatrix} \sim N(\mu, \Sigma)$$ with $$\mu = \begin{pmatrix} b \\ b \\ b \end{pmatrix}$$ and $$\Sigma = \begin{pmatrix} \sigma^2_y & \sigma^2_y & \sigma^2_y \\ \sigma^2_y & \sigma^2_x + \sigma^2_y & \sigma^2_y \\ \sigma^2_y & \sigma^2_y & \sigma^2_x + \sigma^2_y \end{pmatrix}.$$

My attempt is as follows: $$\mathbb{E} (\tilde{y} \mid \tilde{x}_1 = x_1, \tilde{x}_2 \geq a) = \mathbb{E} \left[ \mathbb{E} (\tilde{y} \mid \tilde{x}_1 = x_1, \tilde{x}_2) \mid \tilde{x}_2 \geq a \right] = \mathbb{E} \left[ \frac{b \sigma_x^2}{\sigma_x^2 + 2\sigma_y^2} + \frac{ (x_1 + \tilde{x}_2) \sigma_y^2}{\sigma_x^2 + 2\sigma_y^2} \mid \tilde{x}_2 \geq a \right].$$ For the last term, it seems to me that since $$x_1$$ is a realization but $$\tilde{x}_2$$ is still a random variable, the term with $$x_1$$ can go outside of expectation term. Then, it will be $$\frac{b \sigma_x^2}{\sigma_x^2 + 2\sigma_y^2} + \frac{ x_1 \sigma_y^2}{\sigma_x^2 + 2\sigma_y^2} + \frac{\sigma_y^2}{\sigma_x^2 + 2\sigma_y^2} \mathbb{E}(\tilde{x}_2 \mid \tilde{x}_2 \geq a),$$ where the last expectation is nothing but $$b + \sigma_x\frac{\phi(Z)}{1-\Phi(Z)}$$ with $$Z = \frac{a-b}{\sigma_x}.$$

Do you think that the calculations are correct?

The answer to your question is available here:

Conditional expectation of multivariate normal distribution with inequality condition

When following the steps, just replace the formula

$$\mathrm{E}\left(X_{1} \mid X_{3}

with

$$\mathrm{E}\left(X_{1} \mid X_{3}>x_{3}\right)=\mu_{1}+\frac{\sigma_{13}}{\sigma_{3}} \cdot \frac{\varphi\left(\frac{x_{3}-\mu_{3}}{\sigma_{3}}\right)}{1 - \Phi\left(\frac{x_{3}-\mu_{3}}{\sigma_{3}}\right)}$$

and you should be good to go.

The reason your approach is not quite right is that when you use the Law of Total Expectation, you must still condition on $$\tilde{x}_1 = x_1$$. Your last formula will depend on $$\mathbb{E}[\tilde{x}_2 = x_2 | \tilde{x}_2 > a , \tilde{x}_1 = x_1]$$ instead of $$\mathbb{E}[\tilde{x}_2 = x_2 | \tilde{x}_2 > a ]$$.