two random variables and conditional expectation Consider two random variables $X$ and $Y$. Let $Z_1,Z_2$ be two random variables, measurable with respect to the $\sigma$-field generated by $X,Y$ such that 
$$\mathbb E (X\mid Z_1)=E (X\mid Z_2)$$
$$\mathbb E (Y\mid Z_1)=E (Y\mid Z_2)$$
What can I say about $Z_1$ and $Z_2$? Do they generate the same $\sigma$-algebra? Does there exist a one-to-one function $f$ such that $Z_1=f(Z_2)$?
Thanks!

No.  Let $X$ and $Y$ be iid $U(-1,1)$ rvs.  Let $Z_1$ be the indicator rv of the event  $[|X|+|Y|<1]$ and let $Z_2$ be the indicator of the event $[X^2+Y^2<1]$.  Clearly $E(X\mid Z_i) = E(Y\mid Z_i) = 0$  for $i=1,2$, but $Z_1$ and $Z_2$ generate different $\sigma$-algebras, and $Z_1$ is not a function of $Z_2$, etc.
Another example of this sort:  $X$ and $Y$ as above, $Z_1=|X+Y|$ and $Z_2=|X-Y|$. Identifying the sample space $\Omega$ with $[-1,1]^2$,  all events in $\sigma(Z_1)$ are center-symmetric subsets of  $[-1,1]^2$ and hence have centers of mass (barycenters) equal to $(0,0)$, and similarly for $\sigma(Z_2)$. This  yields $E(X\mid Z_1)=0$, and so on, as before.
