TITLE: Help with question on joint Gaussian distribution
QUESTION [5 upvotes]: Does anyone know how to start this question?

Let random vectors $x,u,v$ have joint Gaussian distribution, and $u,v$ be independent. Show that
  $E(x|u,v)=E(x|u)+E(x|v)-E(x)$.

REPLY [3 votes]: Write $x=au+bv+w$ for some numbers $a$ and $b$, with $(u,v,w)$ gaussian and independent. 
Thus $y=E(x|u,v)$ is $y=au+bv+E(w)$. Since $(u,v)$ is independent, 
\begin{align}
& E(x|u) = E(y|u) = au+bE(v)+E(w),
\\
& E(x|v) = E(y|v) = aE(u)+bv+E(w)
\\
& E(x) = E(y) = aE(u)+bE(v)+E(w). 
\end{align}
This yields 
\begin{align}
E(x|u)+E(x|v) & = au+aE(u)+bv+bE(v)+2E(w)
\\
& = y+aE(u)+bE(v)+E(w)
\end{align}
hence 
$$
E(x|u)+E(x|v)=E(x|u,v)+E(x).
$$