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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) $?

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  • $\begingroup$ It's not clear to me what "$p[y \mid (z \mid x)]$" is intended to mean $\endgroup$
    – Glen_b
    Mar 9, 2016 at 3:53
  • $\begingroup$ My idea is to first calculate $ p(y \mid x) = p(y) $ and $ p(z \mid x) $ and then calculate the conditional of $ p(y) $ from its correlation with $ p(z \mid x) $, that's why I wrote $ p(y \mid (z \mid x)) $. I'm looking for the right notation. $\endgroup$ Mar 9, 2016 at 4:33
  • $\begingroup$ I'm no more enlightened, sorry -- your explanation didn't make sense to me either. $\endgroup$
    – Glen_b
    Mar 9, 2016 at 4:50
  • $\begingroup$ How about the first part of my question, is $ p(x \mid y, z) = p(x \mid z) $ ? $\endgroup$ Mar 9, 2016 at 5:29

1 Answer 1

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(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.)

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  • $\begingroup$ Thanks, makes sence. Then, since $ p(x \mid z) = \int_{-\infty}^{\infty} p(x \mid y=\epsilon, z) p(y=\epsilon) d\epsilon = \mathcal{N} \left(\Sigma_{xz}\Sigma_{zz}^{-1}(z), \Sigma_{xx} - \Sigma_{xz}\Sigma^{-1}_{zz}\Sigma_{xz} \right) $ ? $\endgroup$ Mar 9, 2016 at 5:52
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    $\begingroup$ I am not sure what you are asking here. I am simply saying that since the means are not the same, the distribution cannot be the same. $\endgroup$ Mar 9, 2016 at 5:55
  • $\begingroup$ I think this should be a separate question, it is about using both relations you expressed on the Bayes theorem. $\endgroup$ Mar 9, 2016 at 5:58

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