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We know that for two real numbers $a,b$ and two random variables $X,Y$ we have that $E(a X + b Y ) = a E(X) + b E(Y)$. Under what conditions is it also true that for any three random variables $X,Y,Z$, we have that $E\bigl(X (Y + Z)\bigr) = E(X Y + X Z) = E(X Y) + E(X Z)$?

In this equation you are allowed to assume that there exist joint distributions for $X,Y$ and $X,Z$ but not necessarily for $X,Y,Z$. So, as per Michael's answer below, the question would seem to reduce to when a joint distribution exists given distributions for $X,Y$ and $X,Z$.

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    That's distributivity, not associativity.2012-09-10
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    For real numbers, you have $X(Y+Z) = XY + XZ$. Hence you have $E(X(Y+Z)) = E(XY + XZ)$. Expectation is linear, so you have $ E(XY + XZ) = E XY + E XZ$.2012-09-10
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    Thanks mjqxxxx, editted. I think it can't always be true because the violation of this property is needed to arrive at the violation of Bell inequalities in quantum mechanics.2012-09-10
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    Your equation $E(aU+bV)=aE(U)+bE(V)$ answers the question.2012-09-10
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    Thanks to everyone who tried to answer my question. I have now found a partial answer to my question. Unfortunately there is no guarantee that given marginal or conditional distributions for $X,Y$ and $X,Z$, there exists a joint distribution for $X,Y,Z$. This is why the distributive law does not always hold. The paper dealing with conditional distributions can be seen [here](http://www.jstor.org/stable/2289858?seq=1). However, if anyone can lucidate the matter with a simple condition, the question is still open. I have also editted the question to make it clearer what I want.2012-09-11

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