TITLE: Conditional Expectation of Gaussian Random Vector of length n
QUESTION [5 upvotes]: I am trying to prove the following:
Let $(X_1,\dots,X_n)$ be a Gaussian vector with mean 0 and covariance matrix B.  Find the distribution of $E(X_1\mid X_2,\dots,X_n).$
I know in general for two Gaussian r.v. $X_1$ and $X_2$ we can show that $f_{X_1|X_2} (x_1|x_2) = \frac{1}{\sigma_{X_1} \sqrt{2 \pi (1-\rho^2)}}\exp\frac{(-(x_{1}- \rho(\sigma_{X_1}/ \sigma_{X_2})x_2)^2)}{2 \sigma^{2}_{X_1}(1- \rho^{2})}$ where $\rho$ is the correlation and
$E(X_{1}|X_{2})=\int x_{1} f_{X_1|X_2} (x_1|x_2)dx_{1}.  $
How can I generalize this for a gaussian random vector of size $n$?  Can I conclude on the distribution of $E(X_1|X_2,...X_n)$ based on the form of $f_{X_1|X_2,\dots,X_n} (x_1|x_2,\dots,x_n)$?  For example, in the case of 2 Gaussian r.v we produce a normal r.v with variance $2 \sigma^{2}_{X_1}(1- \rho^{2})$ and mean $\rho(\sigma_{X_1}/ \sigma_{X_2})x_2$.

REPLY [7 votes]: Partition the column vector $X:=(X_1, X_2,\ldots, X_n)^T$ into subvectors $X_a$ and $X_b$:
$$
X = \left(\begin{matrix}X_a\\X_b\end{matrix}\right)
$$
(in your case $X_a=X_1$ is univariate) and correspondingly partition the mean vector $\mu$ and covariance matrix $\Sigma$ of $X$:
$$
\mu = \left(\begin{matrix}\mu_a\\ \mu_b\end{matrix}\right)
$$
$$
\Sigma=\left(\begin{matrix}\Sigma_{a,a}&\Sigma_{a,b}\\\Sigma_{b,a}&\Sigma_{b,b}\end{matrix}\right)$$
Assume that $\Sigma_{b,b}$ is invertible. To find the conditional expectation $E(X_a\mid X_b)$, first find a matrix $C$ of constants such that
$Z:=X_a- C X_b$
is uncorrelated with $X_b$. For this to be true we demand
$$
0= \operatorname{cov} (Z, X_b)=\operatorname{cov} (X_a - CX_b, X_b)=\Sigma_{a,b}-C\Sigma_{b,b},
$$
which yields
$$
C=\Sigma_{a,b}\Sigma_{b,b}^{-1}.
$$
Therefore
$$
\begin{align}
E(X_a\mid X_b)&=E(Z + C X_b\mid X_b)\\
&=E(Z\mid X_b) + CX_b\\
&\stackrel{(*)}=E(Z) + CX_b\\
&= E(X_a) + C(X_b - E(X_b))\\
&= \mu_a + \Sigma_{a,b}\Sigma_{b,b}^{-1}(X_b - \mu_b),\tag1
\end{align}
$$
where in (*) we use the fact that $Z$ and $X_b$ are uncorrelated multivariate Gaussian random vectors and therefore are independent. From (1) we see that $E(X_a\mid X_b)$ is a linear combination of the elements of the Gaussian vector $X_b$, and therefore also is Gaussian.