Conditional expectation as a random variable We have three random variables $x,y,z$. Is the condition "$y$ and $z$ are independent" enough to guarantee that "$\mathbb{E}(x\,|\,y)$ and $z$ are independent"? Would anyone give me a brief proof or counterexample? Thanks a lot!
My problem is that: By intuition I think it is not enough unless we also assume that $x$ is independent to $z$. However, by definition $\mathbb{E}(x\,|\,y)$ is $\sigma(y)$-measurable, which means that the $\sigma$-algebra generated by random variable $\mathbb{E}(x\,|\,y)$ is contained in $\sigma(y)$. Since we know that $\sigma(y)$ and $\sigma(z)$ are independent, it follows that $\mathbb{E}(x\,|\,y)$ and $z$ are independent. What is wrong in this argument?
 A: This requires carefully understanding the notation; $\mathbb{E}(X|Y)$ is a random variable that only depends on the value of random variable $Y$. It is independent of anything that $Y$ is independent of and dependent on anything that $Y$ is dependent on.
As mentioned in the comments above, consider any example where $X$ is not independent of $Z$, but $Y$ is independent of $Z$. In this case, $\mathbb{E}(X|Y)$ is a random variable which only depends on the distribution of $Y$. Any potential dependence on $Z$ has been 'integrated out'.
In the example given in comments on the op: $Y\sim U(0,1)$, $Z\sim U(0,1)$, $X\sim U(0,z)$, without any specification of $X$'s dependence on $Y$, I'll assume they are independent, so:
$$E(X|Y)=E(X)=\int_0^1\int_0^zx\frac{1}{z}dxdz=\frac{1}{4}.$$
In order to realize the dependence of $X$'s expectation on $Z$, one should write $\mathbb{E}(X|Z)$. In this example, $$\mathbb{E}(X|Z)=\frac{Z}{2}.$$
To be more explicit: $\mathbb{E}(X)$ (a number) is independent of $X$ (a random variable). The 'expectation' notation implies integration over the distribution of the random variable. Therefore if $\mathbb{E}(X)=\mu$, then $\mathbb{E}\left(\mathbb{E}(X)|X\right)=\mathbb{E}\left(\mu|X\right)=\mu=\mathbb{E}(\mu)=\mathbb{E}(\mathbb{E}(X))$. Hence $\mathbb{E}(X)$ is independent of $X$. This may seem silly, but it's still an important thing to realize.
