# Does covariance matrix of conditional Gaussian random variable depend on $X_2$?

On Wikipedia, it states that the covariance matrix of the conditional Gaussian random variable, $f_{X_1|X_2}(x_1,x_2)$ is given by:

$${\displaystyle {\overline {\boldsymbol {\Sigma }}}={\boldsymbol {\Sigma }}_{11}-{\boldsymbol {\Sigma }}_{12}{\boldsymbol {\Sigma }}_{22}^{-1}{\boldsymbol {\Sigma }}_{21}.}$$

Does the conditional covariance matrix depend on $X_2$? I believe this is the case because, for instance, $\mathbf{\Sigma}_{22} = \mathbb{E}((X_2 - m_{X_2})(X_2 - m_{X_2}))$

However, I am not sure if I fully understood the meaning of "depending on $X_2$". Can someone chime in?

• The conditional covariance will not depend on $X_2$ if $X_1$ and $X_2$ are uncorrelated. That's why you have the cross-terms $\Sigma_{12/21}$ in that expression. Otherwise, in general, fixing $X_2$ to different values will produce different distributions of $X_1$ and hence different covariances. – Moss Murderer Feb 4 '18 at 7:33
• Do you mean that the expression $\overline \Sigma$ will not depend on $X_2$?. Is it because the product term $\Sigma_{12}....$ cancels out the $X_2$? Doesn't seem to be obvious – Detective Mooch Feb 4 '18 at 7:37
• I was specifically talking about the case of $\Sigma_{12}=\Sigma_{21}=0$, so that must be obvious. There may be other cases where those terms are non-zero, and yet there is no dependence on $X_2$. – Moss Murderer Feb 4 '18 at 7:42

Given that the variance-covariance matrix of the distribution of $X_1$ conditional on $X_2$ is $${\displaystyle {\overline {\boldsymbol {\Sigma }}}={\boldsymbol {\Sigma }}_{11}-{\boldsymbol {\Sigma }}_{12}{\boldsymbol {\Sigma }}_{22}^{-1}{\boldsymbol {\Sigma }}_{21}}$$ there are only constant terms in this expression and no function of the random variable $X_2$. The covariance matrix therefore does not depend on the realisation of $X_2$. Dependence can only be understood in this functional sense and the covariance matrix is not a random variable in this Gaussian case.
The conditional covariance of $X_1$ given $X_2$ does not depend on $X_2$. However, the conditional mean, i.e., the mean of $X_1$ conditional on $X_2$, let's call it $\overline{\mu}$, does depend on $X_2$. Therefore, the conditional distribution of $X_1$ given $X_2$ does depend on $X_2$.
Specifically, as stated at the same Wikipedia link, $\overline{\mu} = \mu_1 + \Sigma_{12}\Sigma_{22}^{-1}(a-\mu_2)$, where $\mu_1$ and $\mu_2$ are the unconditional means and $a$ is the value of $X_2$ being conditioned on.
Per the formulas, if that value $a$ being conditioned on happens to equal the unconditional mean of $X_2$, or if $X_1$ and $X_2$ are uncorrelated ($\Sigma_{12}$ = zero matrix), then the conditional mean of $X_1$ equals the unconditional mean of $X_1$, i.e., there is no adjustment to the unconditional mean to get the conditional mean. Otherwise, the conditional mean of $X_1$ is adjusted from its unconditional mean based on the realized value of $X_2$ per the formula shown.