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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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    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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One wants to show that $E(T\mid X,Z)=T$ with $T=E(Y\mid X)$. This holds true in full generality since (i) the random variable $T$ is $\sigma(X)$-measurable by definition hence $T$ is $\sigma(X,Z)$-measurable, and (ii) $E(U\mid X,Z)=U$ for every $\sigma(X,Z)$-measurable random variable $U$.

Recall that $E(U\mid V)$ is defined as the (almost surely) unique random variable $W$ such that (1.) $W$ is $\sigma(V)$-measurable, and (2.) $E(W\,\mathbf 1_A)=E(U\,\mathbf 1_A)$ for every $A$ in $\sigma(V)$. Additionally, (1.) is equivalent to (1'.) $W=\varphi(V)$ for some measurable function $\varphi$.

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    The fact that E(E(X|G)|H)=E(X|G) when G is a sigma-algebra included in the sigma-algebra H is true but this (true) fact is not called the tower property. I thought my previous comment made this clear.2012-08-26