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I have the following normal distribution $$\begin{pmatrix}X_1\\X_2\\X_3\\\end{pmatrix} \sim N\left[\begin{pmatrix}\mu_1 \\\mu_2 \\\mu_3\\ \end{pmatrix},\begin{pmatrix}\sigma_1^2 & \ a & b \\ \ c & \sigma_2^2 & e\\ \ f & g & \sigma_3^2 \end{pmatrix} \right]$$

How do I calculate: $E(X_1 X_2 | X_3)$

by definition $E(X_1 X_2 | X_3) = E(X_1 X_2) + Cov(X_1 X_2 ,X_3)(Var(X_3))^{-1}(X_1 X_2-E(X_3| X_1 X_2 ))$

$E(X_1 X_2) = Cov(X_1, X_2) + E(X_1)E(X_2) = a +\mu_1 \mu_2$

$Cov(X_1 X_2 ,X_3)=E(X_1 X_2 X_3)-E(X_1X_2)E(X_3)$

How to get $E(X_1 X_2 X_3)$ ?

or is there another way of calculating $Cov(X_1 X_2 ,X_3)$

MC1325
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  • By regressing $(X_1,X_2)$ against $X_3$ (as the explanatory variable) you find that the conditional distribution is bivariate Normal, thereby reducing your question to finding the expectation of the product of bivariate Normal variables. The product can be expressed as a linear combination of two squared Normal variables (the sum and difference), so all that's left to do is look up a formula for the expectation of a squared Normal variable (which, by definition, has a [non-central chi-squared distribution](https://stats.stackexchange.com/questions/188626)). – whuber Jun 27 '20 at 12:57

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