TITLE: Conditional distributions of the multivariate normal
QUESTION [7 upvotes]: Wikipedia gives details on the conditional distribution of the multivariate normal:

If $\mu$ and $\Sigma$ are partitioned as follows
$\boldsymbol\mu = \begin{bmatrix}  \boldsymbol\mu_1 \\  \boldsymbol\mu_2 \end{bmatrix}$
$\boldsymbol\Sigma = \begin{bmatrix}  \boldsymbol\Sigma_{11} & \boldsymbol\Sigma_{12} \\  \boldsymbol\Sigma_{21} & \boldsymbol\Sigma_{22} \end{bmatrix} \quad$ 
then, the distribution of $x_1$ conditional on $x_2= a$ is
  multivariate normal $(x_1|x_2=a) \sim N(\bar{\mu}, \bar{\Sigma})$
  where
$\bar{\boldsymbol\mu} = \boldsymbol\mu_1 + \boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \left(  \mathbf{a} - \boldsymbol\mu_2 \right) $
and covariance matrix
$\overline{\boldsymbol\Sigma} = \boldsymbol\Sigma_{11} - \boldsymbol\Sigma_{12} \boldsymbol\Sigma_{22}^{-1} \boldsymbol\Sigma_{21}.  $

How can I prove this result? Wikipedia cites Eaton, Morris L. (1983). Multivariate Statistics: a Vector Space Approach. John Wiley and Sons. pp. 116–117., but I don't have this book handy...

REPLY [10 votes]: Consider the following transformation 
$$\begin{align} Y_1 &=X_1-\Sigma_{12}\Sigma_{22}^{-1}X_2 \\ Y_2 &=X_2\end{align}$$
Then $$\begin{pmatrix}
        Y_1 \\
        Y_2 
        \end{pmatrix} \sim\mathcal{N_p}\left[\begin{pmatrix}
        \mu_1-\Sigma_{12}\Sigma_{22}^{-1}\mu_2 \\
        \mu_2 
        \end{pmatrix} ,\begin{pmatrix}
        \bar \Sigma & 0 \\
        0 & \Sigma_{22}
        \end{pmatrix} \right]
$$
Hence the PDF of 
$\begin{pmatrix}
        Y_1 \\
        Y_2 
        \end{pmatrix}
$ is 
$$\begin{align}n(y_1,y_2) &= n(y_1|\mu_1-\Sigma_{12}\Sigma_{22}^{-1}\mu_2, \bar \Sigma)\cdot n(y_2| \mu_2,\Sigma_{22}) \end{align}$$
Since $Y_1 \sim \mathcal{N_{p1}}(\mu_1-\Sigma_{12}\Sigma_{22}^{-1}\mu_2, \bar \Sigma)$ and $Y_2 \sim \mathcal{N_{p-p1}}(\mu_2,\Sigma_{22})$ independently, as $Cov(Y_1,Y_2)=0$.
The PDF of  $\begin{pmatrix}
        X_1 \\
        X_2 
        \end{pmatrix}
$ will be obtained by replacing $y_1$ by $x_1-\Sigma_{12}\Sigma_{22}^{-1}x_2$ and $y_2$ by $x_2$. Note that here jacobian of the transformation is unity.
Hence the PDF of 
$\begin{pmatrix}
        X_1 \\
        X_2 
        \end{pmatrix}
$ is 
$$\begin{align}n(x_1,x_2) &= n((x_1-\Sigma_{12}\Sigma_{22}^{-1}x_2)|\mu_1-\Sigma_{12}\Sigma_{22}^{-1}\mu_2, \bar \Sigma)\cdot n(x_2| \mu_2,\Sigma_{22}) \end{align}$$
Hence the conditional PDF of $X_1$ given $X_2=a$ is 
$$\begin{align}f_{X_1|X_2=a}(x_1) &= \dfrac{n(x_1,a)}{n(a|\mu_2,\Sigma_{22})} \\ &=n(x_1-\Sigma_{12}\Sigma_{22}^{-1}a|\mu_2-\Sigma_{12}\Sigma_{22}^{-1}\mu_2, \bar \Sigma) \\ &= \dfrac{1}{(2 \pi)^{p/2}\sqrt{|\bar \Sigma|}}\exp\left( \frac{1}{2}\left[ (x_1-\mu_1- \Sigma_{12}\Sigma_{22}^{-1}(a-\mu_2))'\bar \Sigma^{-1}(x_1-\mu_1- \Sigma_{12}\Sigma_{22}^{-1}(a-\mu_2)) \right]\right) \end{align}$$
Hence $X_1|X_2=a \sim \mathcal{N_{p1}}(\bar \mu, \bar \Sigma)$.