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.

  • $\begingroup$ What if in addition, we also know, say, $\mathbb E(Y\mid Z_1)=Z_1=\mathbb E(Y\mid Z_2)$? Is there anything I can say? Thanks $\endgroup$ – Ryan Oct 30 '17 at 3:42
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    $\begingroup$ I don't know, in part because I don't have an intuitive feel for what kind of result you are seeking. I suppose it's something like, if two explanations do equally well at predicting $X$ and $Y$, they must in some sense be equivalent. I certainly don't like my $EX\mid Z_1 = 0$ equations: they make my examples look like the cooked-up one-purpose counterexamples they are! $\endgroup$ – kimchi lover Oct 30 '17 at 13:41

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