1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

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

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 $\mathbb{E}[\tilde{x}_2 = x_2 | \tilde{x}_2 > a ]$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .