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$\mathcal{F}$ and $\mathcal{G}$ are two sigma algebras on the same space, neither a subset of the other.

If $X$ is $\mathcal{F}$-measurable but not $\mathcal{G}$-measurable is:

$$ E[E[X\mid \mathcal{G}]\mid \mathcal{F}] = X? $$

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1 Answer 1

up vote 3 down vote accepted

Let $\mathcal{G}=\{\Omega,\emptyset\}$ and $X$ be not almost surely constant. Then $E[X|\mathcal{G}]$ is simply the expectation and constant almost surely. So the answer is no.

In the following example, the two $\sigma$-algebras are not comparable:

Let the underlying probability space be $([0,1],\mathcal{B},\mu)$, with $\mathcal{B}$ being the Borel $\sigma$-algebra and $\mu$ being the uniform distribution (Lebesgue measure). Let $\mathcal{G}$ consists of all subsets that are countable or a have countable complement. Let $\mathcal{F}=\{\emptyset,\Omega,[0,1/2],(1/2,1]\}$. Let $X$ be a random-variable that is $\mathcal{F}$-measurable, but not $\mathcal{G}$-measurable. Then $E[x|\mathcal{G}]$ is equal to a constant function random variable almost surely. This is easily shown. For each positive $n$, there is a closed subinterval of $\mathbb{R}$ with length at most $1/n$ and probability $1$ under the distribution $\mu\circ X^{-1}$ since all sets in $\mathcal{G}$ have measure $0$ or $1$. Pick such an interval for each $n$. Their intersection contains a single point $r$, so $\mu\circ X^{-1}\{r\}=1.$

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Thanks, but in that scenario G is a subset of F... I'm wondering in the case where it is not a subset. –  Dirk Calloway Nov 18 '12 at 18:36
    
Does that make a difference? –  Dirk Calloway Nov 18 '12 at 18:36
    
@Dirk I've added a counterexample satisfying your restriction. –  Michael Greinecker Nov 18 '12 at 19:18

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