Doubt on conditional probability of correlated normal variables I have a system of the form
$ \begin{bmatrix} x \\ y\\ z\end{bmatrix} \text{~} \mathcal{N}\left(\begin{bmatrix} 0 \\ 0\\ 0\end{bmatrix}, \begin{bmatrix} \Sigma_{xx} & 0 & \Sigma_{xz} \\ 0 & \Sigma_{yy} & \Sigma_{xz} \\ \Sigma_{xz}^{T} & \Sigma_{yz}^{T} & \Sigma_{zz} \end{bmatrix}\right)$,
thus x and y with each other are independent but z is dependent in both x and y.
In this scenario is $p\left(x \mid y, z \right) = p(x\mid z)$? 
And what would be the correct notation for $ p[y \mid (z \mid x)]$? $ p(y\mid x,z) $?
 A: (I am assuming all entries in the matrix are scalar.)
Lets see if $p(x|y,z) = p(x|z)$. Using the laws of conditional distribution for the normal distribution as mentioned here,
$$p(x|y,z) \sim N\left(0 + [0 \, \, \, \, \,  \Sigma_{xz}]\left[ \begin{array}{cc}\Sigma_{yy} & \Sigma_{yz} \\ \Sigma_{yz} & \Sigma_{zz}  \end{array}  \right]^{-1} \left[ \begin{array}{c} y \\ z \end{array} \right], \Sigma_{xx} - [0 \, \, \, \, \,  \Sigma_{xz}]\left[ \begin{array}{cc}\Sigma_{yy} & \Sigma_{yz} \\ \Sigma_{yz} & \Sigma_{zz}  \end{array}  \right]^{-1}  \left[ \begin{array}{c} 0 \\ \Sigma_{xz} \end{array} \right] \right)$$
and 
$$p(x|z) = N\left(0 + \Sigma_{xz}\Sigma_{zz}^{-1}(z), \Sigma_{xx} - \Sigma_{xz}\Sigma^{-1}_{zz}\Sigma_{xz} \right) $$
Now
\begin{align*}
[0 \, \, \, \, \,  \Sigma_{xz}]\left[ \begin{array}{cc}\Sigma_{yy} & \Sigma_{yz} \\ \Sigma_{yz}^T & \Sigma_{zz}  \end{array}  \right]^{-1} \left[ \begin{array}{c} y \\ z \end{array} \right] & = [0 \, \, \, \, \,  \Sigma_{xz}]\dfrac{1}{\Sigma_{yy}\Sigma_{zz} - \Sigma_{yz}^2}\left[ \begin{array}{cc}\Sigma_{zz} & -\Sigma_{yz} \\ -\Sigma_{yz} & \Sigma_{yy}  \end{array}  \right] \left[ \begin{array}{c} y \\ z \end{array} \right]\\
& = \dfrac{1}{\Sigma_{yy}\Sigma_{zz} - \Sigma_{yz}^2} \left[\begin{array}{cc}-\Sigma_{xz}\Sigma_{yz} &  \Sigma_{xz}\Sigma_{yy}\end{array} \right]\left[ \begin{array}{c} y \\ z \end{array} \right]\\
& \ne \Sigma_{xz}\Sigma_{zz}^{-1}(z).
\end{align*}
(Hopefully, I haven't made any typos.)
