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Let $(X, Y)$ be distributed as a multivariate normal with parameters

$$ \mu = \begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix} \qquad \Sigma = \begin{bmatrix} \sigma_X^2 & \sigma_{XY} \\ \sigma_{XY} & \sigma_Y^2 \end{bmatrix}. $$

I would like to calculate $E(X | y_1 < Y < y_2)$, where $y_1$ and $y_2$ are constants.

From Wikipedia, I have managed to work out that $$ E(X | Y > y_1) = \mu_X + \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}{1 - \Phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}\right]\\ E(X | Y < y_2) = \mu_X - \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}{\Phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}\right] ,$$ where $\phi(\cdot)$ and $\Phi(\cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 < Y < y_2)$, though.

I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them:

  1. Conditional expectation in the multivariate normal distribution
  2. Expectation of conditional normal distribution
  3. Conditional expectation of bivariate normal
  4. https://math.stackexchange.com/q/2807096/83294
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  • 1
    $\begingroup$ See stats.stackexchange.com/questions/356023/…. $\endgroup$ – StubbornAtom Jan 3 at 10:49
  • $\begingroup$ @StubbornAtom, thank you. I am studying the answers there and will post one here (since my question is different and the solution to it is but a step used in the answers posted there) ASAP. $\endgroup$ – Waldir Leoncio Jan 3 at 11:27
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you should use the same approach in point 3. the conditional expectation $$E(X|Y)=\mu_X+\sigma_{XY}\frac{Y-\mu_Y}{\sigma_Y^2}$$ Then take the expectation of the RHS of this expression given $y_1<Y<y_2$. The only random variable is $Y$ and this has conditional expectation of $$E(Y|y_1<Y<y_2)=\mu_Y+\sigma_Y\left[\frac{\phi\left(\frac{y_1-\mu_Y}{\sigma_Y}\right)-\phi\left(\frac{y_2-\mu_Y}{\sigma_Y}\right)}{\Phi\left(\frac{y_2-\mu_Y}{\sigma_Y}\right)-\Phi\left(\frac{y_1-\mu_Y}{\sigma_Y}\right)}\right] $$ Plugging this in gives you $$E(X|y_1<Y<y_2)=\mu_X+\frac{\sigma_{XY}}{\sigma_Y}\left[\frac{\phi\left(\frac{y_1-\mu_Y}{\sigma_Y}\right)-\phi\left(\frac{y_2-\mu_Y}{\sigma_Y}\right)}{\Phi\left(\frac{y_2-\mu_Y}{\sigma_Y}\right)-\Phi\left(\frac{y_1-\mu_Y}{\sigma_Y}\right)}\right] $$

this also contains your two answers as special cases $y_1\to-\infty$ and $y_2\to\infty$

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  • $\begingroup$ Thank you for your insightful answer, it helped me with a subsequential issue I was having over here. $\endgroup$ – Waldir Leoncio Jan 3 at 11:59

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