Suppose for two random variables $X_1$ and $X_2$ $\sigma(X_1) = \sigma(X_2)$. Why $E[Y | X_1] = E[Y | X_2]$ a.e. ?

the set where $X_1= X_1(\omega)$ can be different from the set $X_2= X_2(\omega) $. How can then conditional expectation conditioned on the event can be equal ?

  • $\begingroup$ conditional expectation doesn't depend on the values of $X$ whatsoever, it only depends on the sigma field generated by it. it seems quite hard to me to give a short and 'enlightening' argument without invoking the whole definition of conditional expectation, I suggest you just try to re-read it. as a trivial example you can consider $X_1 = X_2 + 4324$. you should see that the conditional expectations will be the same even though the sets on which these r.v. take values 4234 are different. also, consider any two constant r.v., for example $X_1 = 4$ and $X_2 = 6$. $\endgroup$ – mm-aops May 4 '14 at 14:09
  • $\begingroup$ @mm-aops: yes,from definition I can prove it. But, I am asking the intuition. $\endgroup$ – user144428 May 4 '14 at 15:25

Indeed, if $\sigma(X_1) = \sigma(X_2)$ then $E[Y | X_1] = E[Y | X_2]$ almost surely, thus $E[Y | X_1] = u_1(X_1)$ and $E[Y | X_2]=u_2(X_2)$ for some measurable functions $u_1$ and $u_2$ such that $u_1(X_1)=u_2(X_2)$ almost surely.

One can then define $E[Y | X_1=x]$ as $u_1(x)$ and $E[Y | X_2=x]$ as $u_2(x)$ for every $x$ in the target set of $X_1$ and $X_2$. This does not entail that $E[Y | X_1=x]$ and $E[Y | X_2=x]$ coincide since, in general, $u_1(x)\ne u_2(x)$.

To sum up, the condition that [$u_1(X_1)=u_2(X_2)$ almost surely] does not imply that [$u_1=u_2$].

Edit: Perhaps a simple example can help. Assume that $Y=6X_1=3X_2$, hence $X_2=2X_1$. Then $$E[Y | X_1]=E[Y | X_2]=Y=6X_1=3X_2,$$ hence, for every $x$, $E[Y | X_1=x]=6x$ and $E[Y | X_2=x]=3x$. If one selects some $\omega$ in $\Omega$ and one measures $X_1(\omega)=x_1$ and $X_2(\omega)=x_2$ then $x_2=2x_1$ hence $E[Y | X_1=x_1]=6x_1$ and $E[Y | X_2=x_2]=3x_2$, which implies that $$E[Y | X_1=x_1]=E[Y | X_2=x_2].$$ More generally, if $\sigma(X_1)=\sigma(X_2)$, there exists some invertible bimeasurable $v$ such that $X_2=v(X_1)$ almost surely hence $$ E[Y\mid X_2]=u_2(X_2)=u_2\circ v(X_1)=E[Y\mid X_1], $$ thus, for almost every $x_1$ (with respect to the distribution of $X_1$), $$ E[Y\mid X_2=v(x_1)]=u_2\circ v(x_1)=E[Y\mid X_1=x_1]. $$

  • $\begingroup$ I still did not get. Why is it true that for only a measure zero set $\{X_1=X_1(\omega)\}$ and $\{X_2=X_2(\omega)\}$ are disjoint ? $\endgroup$ – user144428 May 6 '14 at 5:02
  • $\begingroup$ I have no idea how your comment is related to my answer (maybe you can read my answer once more?) and the previous formulation of your post was clearer. Can you explain precisely what you call $\{X_1=X_1(\omega)\}$? $\endgroup$ – Did May 6 '14 at 6:01
  • $\begingroup$ Let $X_1(\omega) = r$ for some $\omega$. Now,consider the set $\{\omega: X_1(\omega)=r\}$. This I denote by $\{X_1 = X_1(\omega)\}$ for some $\omega$. Unless $\{X_1 = X_1(\omega)\}$ and $\{X_2 = X_2(\omega)\}$ are same conditional expectation conditioned on these events can't be same. But if they are not same they have to be disjoint as the sigma algebras are the same. Is it correct that conditional expectation means taking expectation on a reduced sample space (the space for which the condition is true) $\endgroup$ – user144428 May 6 '14 at 8:26
  • $\begingroup$ To sum up, in the general case (and as soon as one deals with continuous random variables), conditional expectations are NOT expectations on subsets of the sample space because, in general, $\{X=x\}$ have probability zero. One proceeds quite differently to define $E(Y\mid X=x)$ for some specific $x$--and this is what my answer describes: first one defines the random variable $E(Y\mid X)$, then one spots a function $u$ such that $E(Y\mid X)=u(X)$, finally one decides that $E(Y\mid X=x)=u(x)$. Other ways, pretending to be more intuitive, are just confusing and soon run into problems. $\endgroup$ – Did May 6 '14 at 8:50
  • $\begingroup$ yes, Radon- Nykodin derivative. $\endgroup$ – user144428 May 6 '14 at 9:44

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