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I am reading about multivariate Gaussian distributions and ran into the following equation about condtional distributions which says that any conditional distribution is also Gaussian.

$$ p(x,y) = \mathcal{N} \Bigg([\mu_x~\mu_y],\begin{bmatrix}\Sigma_x & \Sigma_{xy} \\ \Sigma_{xy}^T & \Sigma_y\end{bmatrix}\Bigg) $$

$$ p(x|y) = \mathcal{N}\bigg( \mu_x + \Sigma_{xy}\Sigma_y^{-1}(y-\mu_y), \Sigma_x - \Sigma_{xy}\Sigma_y^{-1}\Sigma_{xy}^T \bigg) $$

I am having a difficult time convincing myself that the variance of this conditional distribution at a particular $y$ is independent of $y$.

To visualize this I am trying to imagine a simple case (without much loss of generality) where $\Sigma_{xy} = 0$ so that there is no correlation between the random variables, which gives

$$ p(x|y) = \mathcal{N}(\mu_x, \Sigma_x) $$

I am thinking of this conditional distribution as a cross section of the bivariate at that particular $y$. How can the variance be a constant($\Sigma_x$) at any cross section of the bivariate? The reasoning seems obvious when I think of $y$ that lies far away from $[\mu_x~\mu_y]$. Could someone please help me understand the fault in my reasoning?

Thank you

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  • $\begingroup$ It is a conditional distribution, not a marginal distribution.What you are referring to is the homoscedasticity of the distribution of $X\mid Y$. $\endgroup$ – StubbornAtom Aug 14 '18 at 16:21
  • $\begingroup$ My bad, its the conditional distribution. I'll edit the question. $\endgroup$ – krishna Aug 14 '18 at 16:29
  • $\begingroup$ "I am having a difficult time convincing myself that the variance of this conditional distribution at a particular y is independent of y" ?? The conditional variance is $$\Sigma_x-\Sigma_{xy}\Sigma_y^{-1}\Sigma_{xy}^T$$ This matrix does not depend on $y$, QED. Is this your question? $\endgroup$ – Did Aug 14 '18 at 16:39
  • $\begingroup$ Yes. Shouldn't the variance increase the farther we move from $[\mu_x~\mu_y]$ because the height of the peak is reducing? $\endgroup$ – krishna Aug 14 '18 at 16:59

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