I think I got the definition of the conditional expectation now, but I'm still having some problems with actual calculations...
Let $(X,Y,Z)$ be a real gaussian vector. X and Y centered and independent. I need to show that
$\mathbb{E}(Z|X,Y)=\mathbb{E}(Z|X)+\mathbb{E}(Z|Y)-\mathbb{E}(Z)$
what sounds completely intuitive for me, and it works for all the special cases where Z is either independend from X and/or Y. But I don't know how to proof that. At the proof using the def I'm stuck at how the $\mathbb{1}_{A}$ for $A\in\sigma(X,Y)$ looks like.... If you have any good tips on how to calculate/proof such equations, pls tell me :)
EDIT: Ok, I guess if i can assume that $A=\{X\in B,Y\in C\}$ for $B,C$ Borel sets on $\mathbb{R}$ I can proof it. But is it sufficient to show def of the conditional exp. only for a generator of the sigma algebra?
Thanks in advise
Since $(X,Y,Z)$ is a Gaussian vector, it is known that also $Z$ given $(X,Y)$ (or actually any combination) is also Gaussian, and that $$ E(Z\mid X,Y) = E(Z)+\left( \text{cov}(X,Z)\ \ \text{cov}(Z,Y)\right)\left( \begin{matrix} \text{var}(X) & \text{cov}(X,Y) \\ \text{cov}(X,Y) & \text{var}(Y) \end{matrix}\right)^{-1}\left(\begin{matrix} X-EX \\ Y-EY \end{matrix}\right) $$ namely the conditional expectation has a linear form in $X,Y$. Now, simply use that fact that $EX=EY=0$, and that $X,Y$ are independent, to obtain the result.
EDIT: See here, here, and here.