5
$\begingroup$

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!

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.