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As a sort of a follow-up and a generalization from a previous question, suppose that we have two independent, identically distributed random variables $X, Y$ and a third random variable $W$. Is it true that $\mathbb E[X \mid W]=\mathbb E[Y \mid W]$?

More in general, what conditions are to be imposed in order to have equality? When $W=X+Y$ the statement is true. But supposed there is no relation, a priori between $W$ and $X, Y$. What can we say in this case?

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  • If $W$ is a symmetric function in $X$ and $Y$, then the answer is yes. – Batominovski Jun 30 '16 at 10:07
  • Did you read this comment on a previous comment of yours? – Did Jun 30 '16 at 10:23
  • @Did, yes, if I understand correctly you are saying the following: If $X, Y$ are i.i.d., then the pairs $(X, X+Y), (Y, X+Y)$ have the same distribution. To be honest, I still do not see why. My background here is insufficient. Could you please suggest me a book where this statement is proved exhaustively? That would help me a lot. Thanks. – RandomGuy Jun 30 '16 at 10:36
  • "To be honest, I still do not see why" Note that $(X,X+Y)=g(X,Y)$ and $(Y,X+Y)=g(Y,X)$ for some function $g$ that you can probably define yourself, and that $(X,Y)$ and $(Y,X)$ are identically distributed, hence $g(X,Y)$ and $g(Y,X)$ are identically distributed, right? – Did Jun 30 '16 at 11:58

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Suppose $W = X$. Then $E[X\mid W] = X$ whereas $E[Y\mid W] = E[Y]$. There is too much freedom in the choice of $W$. You have to control for some property to arrive at a useful conclusion. Otherwise, any conclusion is possible as the example above shows.

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