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I am having trouble with the following problem regarding conditional expectation. Consider the random variables $X,Y$ and $Z$ with mean and dispersion matrix

$$\mu=\begin{bmatrix}1 \\ 1\\ 1\end{bmatrix},\ \Sigma=\begin{bmatrix}1 & \rho & \rho \\ \rho & 1 & \rho\\ \rho & \rho & 1\end{bmatrix}$$

Furthermore, consider the random variables $U=X+Y+Z$ and $V=2X-Y-Z$.

I am now asked to calculate $\text{E}\left(\begin{bmatrix}X\\ Y\end{bmatrix}\large{|}Z=z\right)$ and from my lecture notes I am told that computing this is done in the following way:

$$\text{E}(X_1|X_2=x_2)\mu_1+\Sigma_{12}\Sigma_{22}^{-1}(x_2-\mu_2)$$

where the normally distributed random variable $X$ has been partitioned into the following:

$$X=\begin{bmatrix}X_1\\X_2\end{bmatrix},\ \mu=\begin{bmatrix}\mu_1\\\mu_2\end{bmatrix},\ \Sigma=\begin{bmatrix}\Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22}\end{bmatrix}$$

I am completely confused with how I am supposed to use this with the formula given above. Can someone help me figure out how to determine what $\Sigma_{12}$ and $\Sigma_{22}$ is? I can't connect this to the given setting introduced in the problem.

Thanks in advance.

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    $\begingroup$ Is there an unstated but crucial assumption that $X$, $Y$, and $Z$ are jointly gaussian? $\endgroup$ – kimchi lover Feb 22 at 16:28
  • $\begingroup$ The random variable $X$ is normally distributed. My bad, I forgot to mention that. $\endgroup$ – James Feb 22 at 16:56
  • $\begingroup$ What about $Y$ and $Z$ ? $\endgroup$ – Bertrand Feb 22 at 17:13
  • $\begingroup$ They are also normally distributed. $\endgroup$ – James Feb 22 at 17:49
  • $\begingroup$ What do $U$ and $V$ have to do with the problem statement? $\endgroup$ – kimchi lover Feb 22 at 21:13

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