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Given a probability space, a random variable $X$ and a sub sigma algebra $\mathcal G$. What is the sigma algebra generated by the conditional expectation $E (X\mid\mathcal G)$?

Is it the one generated by the intersection between the sigma algebra of $X$ and $\mathcal G$? Perhaps not.

It must be contained in $\mathcal G$, since $E (X\mid\mathcal G)$ is measurable wrt $\mathcal G$.

Thanks and regards!

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  • $\begingroup$ I changed $E(X|\mathcal{G})$ to $E(X\mid\mathcal{G})$, coded as E(X\mid\mathcal{G}). They look different: $E(X|\mathcal{G})$ versus $E(X\mid\mathcal{G})$. The latter is standard usage. $\endgroup$ – Michael Hardy Mar 2 '13 at 16:01
  • $\begingroup$ @Tim "contain" should be "be contained in" $\endgroup$ – user940 Mar 2 '13 at 17:23
  • $\begingroup$ @ByronSchmuland: Right. Thanks! $\endgroup$ – Tim Mar 2 '13 at 17:24
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Let $\cal H$ be the $\sigma$-algebra generated by ${\Bbb E} (X\mid\mathcal G)$. As you say, since ${\Bbb E} (X\mid\mathcal G)$ is $\mathcal G$-measurable, $\cal H$ is contained in $\mathcal G$. It doesn't have to be contained in $\sigma(X)$. For a simple counterexample, let the probability space be $\Omega=\{1,2,3,4\}$ with all subsets measurable and the uniform probability measure, and let $${\mathcal G}=\{\emptyset,\{1,2\},\{3\},\{4\},\{3,4\},\{1,2,3\},\{1,2,4\},\Omega\},$$ $$X(1)=1, X(2)=X(3)=X(4)=0.$$ Then $$ {\Bbb E} (X\mid\mathcal G)(1)={\Bbb E} (X\mid\mathcal G)(2)=\frac12, \ \ \ \ {\Bbb E} (X\mid\mathcal G)(3)={\Bbb E} (X\mid\mathcal G)(4)=0$$ so ${\cal H}=\{\emptyset,\{1,2\},\{3,4\},\Omega\}$ is not contained in $\sigma(X)=\{\emptyset,\{1\},\{2,3,4\},\Omega\}$.

Since $\cal H$ does not have to be contained in $\sigma(X)$, it need not equal ${\cal G}\cap\sigma(X)$. The example above shows that it also does not have to equal $\cal G$.

Addendum: As Byron Schmuland says below, if $X$ is an ${\cal F}/{\cal B}({\Bbb R})$-measurable real-valued function, then $\cal H$ is countably generated as it's the pullback of the countably generated $\sigma$-algebra ${\cal B}(\Bbb R)$ under ${\Bbb E} (X\mid\mathcal G)^ {-1}$. Also, any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\cal H$ for some real-valued $X$. (${\Bbb E} (X\mid\mathcal G)$ is only defined up to equality on a set of measure $1$, so to be precise you should say that any countably generated $\sigma$-algebra which is contained in $\cal G$ can occur as a value of $\sigma({\Bbb E}(X\mid\mathcal G))$ for some real-valued $X$ and some version of ${\Bbb E}(X\mid \mathcal G)$.)

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  • $\begingroup$ Thanks, can it be any subsigma algebra of $\mathcal G$? $\endgroup$ – Tim Mar 3 '13 at 1:38
  • $\begingroup$ @Tim The $\sigma$-algebra generated by $\mathbb{E}(X\mid {\cal G})$, like any $\sigma$-algebra generated by a random variable, must be countably generated. So it cannot be any sub $\sigma$-algebra of ${\cal G}$. $\endgroup$ – user940 Mar 3 '13 at 1:50
  • $\begingroup$ @ByronSchmuland: it is countably generated, because a random variable takes value in $\mathbb R$ with the Borel sigma algebra? $\endgroup$ – Tim Mar 3 '13 at 2:08
  • $\begingroup$ @Tim Yes, that is right. $\endgroup$ – user940 Mar 3 '13 at 3:08
  • $\begingroup$ Can anyone tell me what E(E(X | G)) would be? $\endgroup$ – Branden Keck Jul 3 '18 at 3:00

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