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Let $X_1, X_2, X_3$ be jointly normal with means $\mu_1,\mu_2,\mu_3$ and covariances $\sigma_{ij}$. I know that $$ \mathbb{E}[X_1\mid X_2=x_2]=\mu_1+\frac{\sigma_{12}}{\sigma_2^2}(x_2-\mu_2) $$ and $$ \mathbb{E}[X_1\mid X_2<x_2]=\mu_1-\sigma_{12}\frac{\phi(\frac{x_2-\mu_2}{\sigma_2})}{\Phi(\frac{x_2-\mu_2}{\sigma_2})} $$ where $\phi$ and $\Phi$ are the density and probability distribution functions for the standard normal distribution. Is there a similar expression for $$ \mathbb{E}[X_1\mid X_2=x_2,X_3<x_3]~? $$ Thank you in advance.

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    $\begingroup$ Did you derive the first 2 expectations or were they given? $\endgroup$ Commented Jun 4, 2018 at 0:30
  • $\begingroup$ $$ \operatorname E(X_1\mid X_2=x_2) = \mu_1 + \frac{\sigma_{12}}{\sigma_2^2} (x_2-\mu_2), $$ with a "plus" sign. Note that if $\sigma_{12}>0$ then this conditional expectation should increase as $x_2$ increases. $\endgroup$ Commented Jun 4, 2018 at 2:10
  • $\begingroup$ Thanks Michael for pointing out the typo in my derivation, I've corrected it. $\endgroup$
    – Jason V
    Commented Jun 4, 2018 at 2:16
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    $\begingroup$ $(X_1, X_3)|X_2 = x_2$ has a bivariate normal distribution. Then you can apply your second formula. $\endgroup$
    – BGM
    Commented Jun 4, 2018 at 3:39
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    $\begingroup$ How did you get your conditional expectation given $X_2<x_2$? You've written an expression that does not depend on $\mu_2. \qquad$ $\endgroup$ Commented Jun 4, 2018 at 3:48

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For $\operatorname E(X_1\mid X_3<x_3)$ I'm getting $$ \operatorname E(X_1\mid X_3<x_3) = \mu_1 - \frac{\sigma_{13}}{\sigma_3} \cdot \frac{\varphi\left( \dfrac{x_3-\mu_3}{\sigma_3} \right)}{\Phi\left( \dfrac{x_3-\mu_3}{\sigma_3} \right)}. \tag 1 $$ Once you're given $X_2=x_2,$ you have a certain conditional probability distribution of $(X_1,X_3),$ which is bivariate normal with expected value given by \begin{align} & \operatorname E(X_1\mid X_2=x_2) = \mu_1 + \frac{\sigma_{12}}{\sigma_2^2} ( x_2-\mu_2 ), \\[10pt] & \operatorname E(X_3\mid X_2=x_2) = \mu_3 + \frac{\sigma_{32}}{\sigma_2^2} ( x_2-\mu_2 ), \end{align} and variance given by \begin{align} & \operatorname{var}(X_1\mid X_2=x_2) = \sigma_1^2 - \frac{\sigma_{12}^2}{\sigma_2^2}, \\[10pt] & \operatorname{var}(X_3\mid X_2=x_2) = \sigma_3^2 - \frac{\sigma_{23}^2}{\sigma_2^2}, \\[10pt] & \operatorname{cov}(X_1,X_3\mid X_2=x_2) = \sigma_{13} - \frac{\sigma_{12} \sigma_{23}}{\sigma_2^2}. \end{align} Now apply line $(1)$ above but with these conditional expectations and conditional variances. Where you had $\sigma_{13}$ in line $(1),$ you'll have $\sqrt{\sigma_{13} - \frac{\sigma_{12} \sigma_{23}}{\sigma_2^2}},$ etc.

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  • $\begingroup$ Your answer makes sense, thanks a lot Michael! I have corrected my second formula as well. After conditioning $(1)$ on $X_2=x_2$ as you suggested, I was able to obtain the final answer. $\endgroup$
    – Jason V
    Commented Jun 4, 2018 at 6:14
  • $\begingroup$ @JasonV : Recall that you can "accept" this answer. $\endgroup$ Commented Jun 4, 2018 at 14:34
  • $\begingroup$ I did not notice that convenient feature, your answer has been accepted. $\endgroup$
    – Jason V
    Commented Jun 4, 2018 at 14:56
  • $\begingroup$ @JasonV : Thank you. $\endgroup$ Commented Jun 4, 2018 at 15:51
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    $\begingroup$ @Michael Hardy, The (1) should be: $E[X_1|X_3<x_3]=\mu_1-\cdots$. (Consider the case of $X_1=X_3$ and $x_3=\mu_1$, $E[X_1|X_1<\mu_1]<\mu_1$). $\endgroup$
    – JGWang
    Commented Jun 6, 2018 at 8:40

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