Conditional expectation of a joint normal distribution Let $X_1, X_2$ be jointly normal $N(\mu, \Sigma)$.
I know that in general, $\mathbb{E}[X_2|X_1]$ can be computed by integrating the conditional density, but in the case of jointly normal variables, it suffices to do a linear projection:
$\mathbb{E}[X_2 | \sigma(X_1)] = \mathbb{E}[X_2|\mathrm{span}(\mathbf{1}, X_1)] = \mu_2 + \frac{\mathrm{cov}(X_2, X_1)}{\mathrm{var}(X_1)} (X_1 - \mu_1) $
Is there a neat proof of this fact (one doesn't require doing any integrals)? Looking for references too. 
 A: Let $(X_1, X_2) \sim MVN (\mu, \Sigma)$, then recall that 
\begin{align}
f_{X_2|X_1}(x_2) = \frac{f_{X_1, X_2}(x_1, x_2)}{f_{X_1}(x_1)} \, , 
\end{align}
where 
$$
f_{X_1,X_2}(x_1,x_2)=\\
\frac{1}{2\pi \sigma_1\sigma_2 \sqrt{1-\rho}}\exp\left( - \frac{1}{2(1-\rho^2)}\left(\frac{(x_1 - \mu_1)^2}{\sigma^2} + \frac{(x_2 - \mu_2)^2}{\sigma^2} - \frac{2\rho(x_1-\mu_1)(x_2 - \mu_2)}{\sigma_1\sigma_2} \right) \right),
$$
and 
$$
f_{X_1}(x) = \frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left( - \frac{(x_1-\mu_1)^2}{2\sigma_1^2} \right).
$$
After some simple algebra and rearrengments you'll find that
$$
X_2|X_1 = x_1 \sim N\left( \mu_2 + \rho\frac{\sigma_2}{\sigma_1}(x_1 - \mu_1), (1- \rho^2)\sigma_2^2 \right).
$$ 
A: I've found an answer that I'm happy with:
$$ Y - \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X $$ is jointly normal with $X$ and uncorrelated, hence independent.
Therefore 
$$\mathbb{E}[Y|X] = \mathbb{E}[Y - \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X + \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X| X] \\
= \mathbb{E}[Y - \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} X] +\frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)}X \\
=  \mathbb{E} [Y] + \frac{\mathrm{cov}(X, Y)}{\mathrm{var}(X)} (X - \mathbb{E}[X])$$
