Suppose $(X_1, X_2, X_3)^T$ is multivariate normal.

What is the conditional expectation $E(X_1 \mid X_2, I(X_3 > 0))$?

Here, $I(X_3>0)$ is the random variable that takes the value one when $X_3 > 0$ and zero otherwise. I know that, without using any properties of the normal distribution, $$ E(X_1 \mid I(X_3>0)) = I(X_3 > 0)\frac{E[ X_1I(X_3 > 0)]}{P(X_3 > 0)}+ I(X_3 \leq 0)\frac{E[ X_1I(X_3 \leq 0)]}{P(X_3 \leq 0)}. $$

Based on this, I'm tempted to infer that,

$$ E(X_1\mid X_2, I(X_3> 0)) = I(X_3 > 0)\frac{E[ X_1I(X_3 > 0) \mid X_2]}{P(X_3 > 0\mid X_2)} + I(X_3 \leq 0)\frac{E[ X_1I(X_3 \leq 0)\mid X_2]}{P(X_3 \leq 0 \mid X_2)}, $$

but am unable to confirm or refute this. If this is true, one could proceed using properties of conditional distributions for multivariate normal vectors to get more specific expressions.

  • $\begingroup$ The joint distribution of $(X_1,X_2,Z_3=\mathbb I_{X_3>0})$ has density $$p(x_1,x_2|\mathbf \mu,\mathbf \Sigma)\Phi(-\mathbb E[X_3|x_1,x_2]/\text{sd}(X_3|x_1,x_2))^{z_3}\Phi(\mathbb E[X_3|x_1,x_2]/\text{sd}(X_3|x_1,x_2))^{1-z_3}$$ $\endgroup$ – Xi'an Nov 1 '19 at 15:58

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