Why $E(X|Y)=E(E(X|Z,Y)|Y)$?
I know that $E(U)=E(E(U|V))$. So, $E(U|W)$ should be $E(E(U|V,W))$. But the latter expression is free from $W$ so it is not possible.
I can get a intuitive idea that right hand side of the above expression should be a function of $Y$, so we are conditioning two times. But how to prove it rigorously?
On the right hand side you condition $X$ first on the sigma algebra $\sigma(Y,Z)$ and then on the coarser sigma algebra $\sigma(Y)$. Since $\sigma(Y)\subset\sigma(Y,Z)$ the first conditioning has no effect on the right hand side, thus the equality. If you want to prove it rigorously, go back to the definition of conditional expectation with respect to sigma algebras.
