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Let $\begin{bmatrix} Y_1 \\ Y_2 \\ Y_3 \end{bmatrix} \sim MVN\left( \begin{bmatrix} \mu_1 \\ \mu_2 \\ \mu_3 \end{bmatrix} , \begin{bmatrix} \Sigma_{11} & \Sigma_{12} & \Sigma_{13} \\ \Sigma_{21} & \Sigma_{22} & \Sigma_{23} \\ \Sigma_{31} & \Sigma_{32} & \Sigma_{33} \end{bmatrix}\right)$ where MVN denotes the multivariate normal distribution. I wish to calculate the conditional expectation $E[Y_3 | Y_1 = y_1]$, is it true then I can just ignore $Y_2$ basically and consider:

$$\begin{bmatrix} Y_1 \\ Y_3 \end{bmatrix} \sim MVN\left( \begin{bmatrix} \mu_1 \\ \mu_3 \end{bmatrix} , \begin{bmatrix} \Sigma_{11} & \Sigma_{13} \\ \Sigma_{31} & \Sigma_{33} \end{bmatrix}\right)$$. That is, $(Y_1, Y_3)$ are jointly bivariate normally distributed as above?

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Yes, you are correct.

If $X$ is a MVN random vector, then every component of $X$ is univariate Normal and every subvector of $X$ is also MVN, with appropriate partition of the the mean and covariance martrix. Further all conditional distributions are also normal.

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