Is $E[E[X|Y,Z]|Y]=E[X|Y]$ for independence between X and Z only? Does $E[X|Y,Z]=E[X|Y]$ hold only for independence between X and Z or in general?
Thanks!


The independence of $X$ and $Z$ does not guarantee that $E[X\mid Y,Z]=E[X\mid Y]$. 

For an example, consider $Z=XY$ with $(X,Y)$ uniform on $\{-1,+1\}^2$ (equivalently, $X$ and $Y$ are independent symmetric Bernoulli random variables). Then $E[X\mid Y,Z]=X$ because $X=YZ$, while $E[X\mid Y]=E[X]=0$ because $X$ and $Y$ are independent and $X$ is symmetric.
Conversely:

The fact that $E[X\mid Y,Z]=E[X\mid Y]$ does not guarantee the independence of $X$ and $Z$. 

For an example, assume that $Z$ is $Y$-measurable, then $E[X\mid Y,Z]=E[X\mid Y]$ without any assumption on the dependence structure of $(X,Z)$.
Edit:

The independence of $(X,Y)$ and $Z$ guarantees that $E[X\mid Y,Z]=E[X\mid Y]$.

To see this, note that $T=E[X\mid Y]$ is measurable with respect to $(Y,Z)$ hence, to check that $T=E[X\mid Y,Z]$, one just has to check that $E[T;(Y,Z)\in B]=E[X;(Y,Z)\in B]$ for every Borel set $B$, and that in fact, it suffices to check that $E[T;Y\in B,Z\in C]=E[X;Y\in B,Z\in C]$ for every Borel sets $B$ and $C$. 
To show this last statement, first note that $T$ is $Y$-measurable and that $Y$ and $Z$ are independent hence $E[T;Y\in B,Z\in C]=E[T;Y\in B]\cdot P[Z\in C]$. Furthermore, $T=E[X\mid Y]$ hence $E[T;Y\in B]=E[X;Y\in B]$. On the other hand, $(X,Y)$ and $Z$ are independent hence $E[X;Y\in B,Z\in C]=E[X;Y\in B]\cdot P[Z\in C]$. This proves the claim.
