Conditional Probability Distribution of Multivariate Gaussian

I've been working on the following question:

Let the vector $(X,Y,Z)$ be a multivariate Gaussian random variables with mean vector and covariance matrix:

$\mu = \begin{pmatrix} 0 \\ 3 \\ -1 \end{pmatrix}$ and $\Sigma = \begin{pmatrix} 16 & 8 & 0 \\ 8 & 9 & -2 \\ 0 & -2 & 1 \end{pmatrix}$

Derive the distribution of $(X,Z)$ given $Y=0$

I know (from my lecture notes) that the conditional distribution of $X_1$ given $X_2=x_2$ is Gaussian with mean: $\mu_1 + \Sigma_{12}\Sigma_{22}^{-1}(x_2-\mu_2)$ and variance-covariance matrix: $\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}$

So I rewrote $\Sigma$ as $\Sigma = \begin{pmatrix} 16 & 0 & 8 \\ 0 & 1 & -2 \\ 8 & -2 & 9 \end{pmatrix}$

And then I let $X_1=\begin{pmatrix} X \\ Z \end{pmatrix}$ and $Y_2=(Y)$. So $E(X_1)=\begin{pmatrix} 0 \\-1 \end{pmatrix}$ and $E(X_2)=(3)$

And $Cov(\begin{pmatrix} Y_1 \\ Y_2 \end{pmatrix}) = \begin{pmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{pmatrix}$ where $\Sigma_{11}=\begin{pmatrix} 16 & 0 \\ 0 & 1 \end{pmatrix}, \Sigma_{12}=\begin{pmatrix} 8 \\ -2 \end{pmatrix} = \Sigma_{21}^T, \Sigma_{22}=(9)$

Meaning that $E(Y_1 \mid Y_2 =0) = \begin{pmatrix} 0 \\ -1 \end{pmatrix} + \begin{pmatrix} 8 \\ -2 \end{pmatrix}(\frac{1}{9})(0-3) = \begin{pmatrix} \frac{-8}{3} \\ \frac{1}{3}\end{pmatrix}$

and $Cov(Y_1 \mid Y_2=0) = \begin{pmatrix} 16 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} 8 \\ -2 \end{pmatrix}(\frac{1}{9})\begin{pmatrix} 8 & -2 \end{pmatrix} = \begin{pmatrix} 16 & 0 \\ 0 & 1 \end{pmatrix} - \begin{pmatrix} \frac{64}{9} & -\frac{16}{9} \\ -\frac{16}{9} & \frac{4}{9} \end{pmatrix} = \frac{1}{9}\begin{pmatrix} 80 & 16 \\ 16 & 5 \end{pmatrix}$

Am I correct in this approach? Is there something I'm fundamentally missing here? The subsequent question is "What is the distribution of $(2X-Z,3Y+Z)$?" How might I approach this one?

Thanks for any help/suggestions/feedback.

• You might want to add the self-study tag and read it's wiki. May 12, 2018 at 8:09

As for the distribution of $(2X−Z,3Y+Z)$, viewed as a 2 element column vector. Consider$(X.Y,Z)$ as a 3 element column vector. You need to determine the matrix $A$ such that $A*(X,Y,Z) = (2X−Z,3Y+Z)$ . Hint: what dimensions must $A$ have to transform a 3 by 1 vector into a 2 by 1 vector? Then use the result $\text{Cov} (A*(X,Y,Z)) = A* \text{Cov}(X,Y,Z)*A^T$ combined with the trivial calculation of the mean, and your knowledge of the type of distribution which a linear transformation of a Multivariate Gaussian has.