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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 $\cal F$? –  Davide Giraudo Nov 18 '12 at 12:59
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Did you mean to condition on $\mathcal{H}$ rather than $\mathcal{F}$? –  Stefan Hansen Nov 18 '12 at 12:59
    
Yeah my mistake... G and H are sub-algs of F and the conditional is E[E[X|G]|H]... Edited above. Thanks. –  Dirk Calloway Nov 18 '12 at 13:21
    
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). –  Stefan Hansen Nov 18 '12 at 13:53
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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? –  Dirk Calloway Nov 18 '12 at 14:14
    
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. –  Davide Giraudo Nov 18 '12 at 14:33
    
And then would the condition also be equal to E[E[X|H]|G]? –  Dirk Calloway Nov 18 '12 at 14:36
    
Of which case are you talking about? –  Davide Giraudo Nov 18 '12 at 14:46
    
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]? –  Dirk Calloway Nov 18 '12 at 15:04
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This is not true in general. Consider what happens if $\mathcal{G} = \mathcal{F}$ and $\mathcal{H} = \{Y, \emptyset\}$ is the trivial $\sigma$-field.

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But in this cawe $\cal G\supset \cal H$, and it's not the conditions the OP wants. –  Davide Giraudo Nov 18 '12 at 13:34
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