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I've had no problems showing that

$$E[E[Y|X,Z]|Z]=E[Y|Z]$$

by the law of iterated expectation. For the latter I summed over $x$ for a certain value of $Z=z$: $$\begin{align} E[E[Y|X,Z]|Z] &= \sum_x E[Y|X=x,Z=z]\cdot P(X=x|Z=z)\\ &=\sum_{x,y} y\cdot P(Y=y|X=x,Z=z)P(X=x|Z=z)\\ &=\sum_{x,y} y\cdot\frac{P(Y=y,X=x,Z=z)}{P(X=x,Z=z)}\cdot\frac{P(X=x,Z=z)}{P(Z=z)}\\ &=\sum_{x,y} y\cdot\frac{P(Y=y,X=x,Z=z)}{P(Z=z)}\\ &=\sum_{x,y} y\cdot\frac{P(Y=y,X=x,Z=z)}{P(Z=z)}\\ &=\sum_{y} y\cdot\frac{P(Y=y,Z=z)}{P(Z=z)}\\ &=\sum_{y} y\cdot P(Y=y|Z=z)\\ &=E(Y|Z=z) \end{align}$$ However for $E[E[Y|X]|X,Z]=E[Y|X]$ i certainly have to go over $z$ for a certain value of $X=x$ which will be like: $$\begin{align} E[E[Y|X]|X,Z] &= \sum_z E[Y|X=x]\cdot P(Z=z|X=x)\\ &=\sum_{z,y} y\cdot P(Y=y|X=x)\cdot P(Z=z|X=x) \\ &=\sum_{z,y} y\cdot\frac{P(Y=y,X=x)}{P(X=x)}\cdot\frac{P(Z=z,X=x)}{P(X=x)} \end{align}$$ Now I'm kinda stuck....

Thanks in advance! Kind regards. Tim

/ed here the link where the statement comes from http://www.vwl.uni-mannheim.de/mammen/notes5.pdf (see page 4, Theorem 2.4, Section iii)

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    Are your random variables discrete ?2012-08-23
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    yes. for the discrete case is good enough for me :). if its easier to show with the continous case thats fine too tough.2012-08-23
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    I don't think that this holds. Can you explain why, in line 2 of your derivation, $x$ vanishes? How can the Information of $X$ be irrelevant? Are X, Y and Z somehow related or have special properties? Further more: what do you want to prove here? The claim is missing!2012-08-23
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    Hey. the missing x is indeed a mistake on my part which has been corrected (edited). The x was just not there :>. The reason why i want to show the equaltiy in E[E[Y|X] | X,Z] is that this is just a simplification of another problem i got. The problem im refering to is that if i calculate the conditional expectation of a conditional expectation that (in this case) the additional information about Z is no more needed since it is equal to the conditional expectation of y given the value of x.2012-08-23
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    in case this is still to abstract here the proper problem i got: Assume that E[e|x,z,d]=E[e|v,d]=E[e|c] (this is a assumption which i dont explain here but for the sake of this problem this is given). Know for y = xb + zd + E[e|c] + u we can show that E[u|x,z,d] = E[u|x,z,c] = 0. For this i need that E[E[e|c]|x,z,d]=E[E[e|c]|x,z,c]=E[e|c].2012-08-23

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