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If there are two sub-sigma algebras $\mathcal{G}$ and $\mathcal{H}$ of $\mathcal{F}$, neither a subset of the other from a probability space $(Y,\mathcal{F},P)$ and a random variable $X$ which is not measurable with respect to either $\mathcal{G}$ or $\mathcal{H}$, can I apply double expectation on the conditional expectation of $X|\mathcal{G}$ like this:

$$ E[E[X|\mathcal{G}]|\mathcal{H}] = E[X|\mathcal{G}] $$

Thanks.

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    What is $H$ in the property you want? Are $\cal G$ and $\cal H$ _sub_$\sigma$-algebras of $\cal F$?2012-11-18
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    Did you mean to condition on $\mathcal{H}$ rather than $\mathcal{F}$?2012-11-18
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    Yeah my mistake... G and H are sub-algs of F and the conditional is E[E[X|G]|H]... Edited above. Thanks.2012-11-18
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    If this were the case, then by definition of conditional expectation, $E[X\mid \mathcal{G}]$ has to be $\mathcal{H}$-measurable. But we already know that it is $\mathcal{G}$-measurable, so if $\mathcal{G}\subseteq \mathcal{H}$ doesn't hold, there is lots of examples where this is not true (see e.g. Davide's example).2012-11-18

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The property doesn't hold when $\Omega=\{a,b,c\}$, $\cal F:= 2^\Omega$, $\cal G:=\{\emptyset,\{a,b\},\{c\},\Omega\}$, $\cal H:=\{\emptyset,\{a\},\{b,c\},\Omega\}$ and $X:=\chi_{\{b\}}$.

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    This is exactly the sort of thing I was thinking of... Does it not hold because the information from G {a,b} and H {b,c} just ends up being omega? Is there any way to simplify this when the tower property doesn't apply?2012-11-18
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    With the imposed condition, a random variable is measure with respect to both $\cal G$ and $\cal H$ if and only if it's constant.2012-11-18
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    And then would the condition also be equal to E[E[X|H]|G]?2012-11-18
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    Of which case are you talking about?2012-11-18
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    Using the sigma algebras you described since X is not measurable wrt to F or G does the order of the conditioning matter? Does E[E[X|H]|G] =E[E[X|G]|H]?2012-11-18
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    As each term is $\cal G$ and $\cal H$ measurable, it's constant. Now you have to check whether it's the same value.2012-11-18