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)