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


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

Mr. Barrrington
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1 Answers1


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.

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  • at first thanks for the formula, never seen it before but it seems quite useful!! Can you (or someone else) give me some reference? or a hint for the proof of my concrete problem? – Mr. Barrrington Oct 04 '13 at 16:03
  • thanks again, just to clarify, by any-combination you really mean ANY (mb) combination between them? So that i can for example also calculate $\mathbb{E}(X|XY)$ with this formula? – Mr. Barrrington Oct 05 '13 at 09:24
  • I meant that since $X,Y,Z$ are jointly Gaussian, the various possible conditional probabilities formed by them of them are also Gaussian (e.g. $X$ given $Y,Z$, $Y$ given $X,Z$, $X,Y$ given $Z$, and so on). If you meant $\mathbb{E}(X\mid X,Y)$ (and not the product $XY$) then the result is of course $\mathbb{E}(X\mid X,Y)=X$, and the result in the solution will still holds true - $X$ given $X,Y$ is still Gaussian. – user91011 Oct 05 '13 at 09:49
  • Ok, thanks again! But I was actually really talking about the product :) can I still use this forumla for $\mathbb{E}(X|X*Y)$? Or even conditioned on a measurable function of X and Y? – Mr. Barrrington Oct 05 '13 at 14:58
  • Well, in this case the above formula is not true. The product $XY$ is not even Gaussian (see http://math.stackexchange.com/questions/101062/is-the-product-of-two-gaussian-random-variables-also-a-gaussian). The direct way to find the conditional expectation is using the definition + Bayes theorem. – user91011 Oct 05 '13 at 15:02
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    I'll give it a try, thanks a lot for all your help! – Mr. Barrrington Oct 06 '13 at 09:04