# Partitioned Multivariate Gaussian

My question is on partitioned multivariate Gaussians which are shown as

$$f(x;\mu,\Sigma) = \frac{ 1}{(2\pi)^{(p+q)/2} \det(\Sigma)^{1/2}} \exp \bigg\{ -\frac{ 1}{2} \left[\begin{array}{r} x_1 - \mu_1\\ x_2 - \mu_2 \end{array}\right]^T \left[\begin{array}{rr} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{array}\right]^{-1} \left[\begin{array}{r} x_1 - \mu_1\\ x_2 - \mu_2 \end{array}\right] \bigg\}$$

and

$$\Sigma = \left[\begin{array}{rr} \Sigma_{11} & \Sigma_{12}\\ \Sigma_{21} & \Sigma_{22} \end{array}\right]$$

In order to derive marginal and conditional densities for pieces of this Gaussian, in book Introduction to Graphical Models chapter 12, Jordan shows

These are the final steps of this algebra, right after Schur's complement is used which was used to obtain the pieces for the inverse. However in the last step, the transition from 2nd to 3rd equation, I am not able to follow the derivation, for example in the last equation $-\Sigma_{22}^{-1}\Sigma_{21}$ seems to have disappeared, but it is there in the second equation. Does anyone know how? Is it because transpose of $\Sigma_{12}$ is $\Sigma_{21}$, he had to pull out transpose operator to a larger group of symbols, therefore had to use $(\Sigma_{21}^T)^T$, and inside the paranthesis it is $\Sigma_{12}$.. maybe?

Thanks,

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I am talking about the entire third equation, how it was obtained, but the situation of $\Sigma_{12}$ was most perplexing. – BB_ML Feb 24 '12 at 13:33
I don't know the context, but: if it is assumed that $\Sigma_{12} = \Sigma^T_{21}$, and $x_1, x_2$ etc are vectors, then yes, what you surmised is correct. – Willie Wong Feb 24 '12 at 13:42
Have you tried to look at how the term is defined $\Sigma$? Adapting their notation to the notation of this link you will discover through these intricate algebraic operations the answer. I think that $\Sigma=$ [Shur complement][1]. [1]: en.wikipedia.org/wiki/Schur_complement – MathOverview Feb 24 '12 at 13:42
I updated the question so it shows $\Sigma$. And $\Sigma/\Sigma_{22}$ is Schur's complement. – BB_ML Feb 24 '12 at 14:35

Have you tried to look at how the term is defined $\Sigma/\Sigma)^{-1}=$? Adapting their notation to the notation of this link you will discover through these intricate algebraic operations the answer. I think that $\Sigma/\Sigma)^{-1}=$ Shur complement.
And with this notation you can see that this operation is legitimate only if $\Sigma=\Sigma^T$ i.e. $\Sigma_{12} = \Sigma_{21}^T$, $\Sigma_{11}=\Sigma_{11}^T$ and $\Sigma_{22}=\Sigma_{22}^T$ are symmetric. See more here.. In this case: $-\Sigma_{22}^{-1}\Sigma_{21}=-\Sigma_{12}\Sigma_{22}^{-1}$