Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

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


share|cite|improve this question
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
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
up vote 2 down vote accepted

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 $X:=\chi_{\{b\}}$.

share|cite|improve this answer
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

This is not true in general. Consider what happens if $\mathcal{G} = \mathcal{F}$ and $\mathcal{H} = \{Y, \emptyset\}$ is the trivial $\sigma$-field.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.