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

I want to prove the following:

Let $X\in L^1(\Omega,\mathcal{F},\mathbb{P})$ be a random variable and suppose $\mathcal{G,M}\subset\mathcal{F}$ are sub–$\sigma$–algebras such that $\mathcal{G}$ is independent of $X$ and $\mathcal{M}$. Prove that $$\mathbb{E}(X\mid\sigma\,(\mathcal{G},\mathcal{M}))=\mathbb{E}(X\mid\mathcal{M}).$$

Following hints from my textbook, I have proved that for all $G\in\mathcal{G}$ and $M\in\mathcal{M}$ $$\int_{G\cap M}\mathbb{E}(X\mid\mathcal{M})\,\text{d}\mathbb{P}=\int_{G\cap M}X\,\text{d}\mathbb{P}.$$ However, there's number of issues I do not understand and I hope you could clarify them to me.

I know that $\mathbb{E}(X\mid\sigma\,(\mathcal{G},\mathcal{M}))$ is the only r.v. $\eta$ such that
(i) $\eta$ is $\sigma\,(\mathcal{G},\mathcal{M})$–measurable,
(ii) $\forall_{A\in\sigma\,(\mathcal{G},\mathcal{M})} \int_A\eta=\int_AX.$
By showing that RHS of the equality in the problem satisfies (i) and (ii), I'm done. However, my questions are:

$1)$ Why is $\mathbb{E}(X\mid\mathcal{M})$ a $\sigma\,(\mathcal{G},\mathcal{M})$–measurable r.v.?
$2)$ So far, I proved (ii) for $A$'s of the form $A_1\cap A_2$ where $A_1\in\mathcal{G}$, $A_2\in\mathcal{M}$. Does the claim follows from that? If so, why?

Thanks for help!

Edit: Ad question 2): I've just found that it is sufficient to verify (ii) for every set $A$ in some $\pi$–system that contains $\Omega$ and generates $\sigma\,(\mathcal{G},\mathcal{M})$. Is it true that $\{G\cap M\mid G\in\mathcal{G}, M\in\mathcal{M}\}$ is a generator of $\sigma\,(\mathcal{G},\mathcal{M})$? I will strongly appreciate any stories that may possibly improve my understanding of $\sigma$–algebras.

share|cite|improve this question… for a counterexample if independence assumption is weakened. – Jisang Yoo May 31 '14 at 15:57
up vote 2 down vote accepted
  1. Because $E[X\mid \cal M]$ is by definition $\cal M$-measurable and $\cal M\subset \sigma(\cal G,\cal M)$.
  2. The idea which consists to use $\pi$-system is good. Let $\mathcal{P}:=\{G\cap M,G\in\mathcal{G},M\in\cal M\}$. Then $\cal P\supset \cal M,\cal G$ as $\Omega\in\cal M\cap\cal G$. If $G\in \cal G$ and $M\in\cal M$, then $G\in\sigma(\cal G,\cal M)$ and $M\in\sigma(\cal G,\cal M)$, and their intersection still will be in $\sigma(\cal G,\cal M)$, proving that $\sigma(\cal P)=\sigma(\cal G,\cal M)$.

Now we conclude showing that the collection of subsets $S$ of $\Omega$ satisfying $$\int_SE[X\mid \mathcal M]\mathrm{d}\Bbb P=\int_SX\mathrm{d}\Bbb P$$ is a $\lambda$-system.

share|cite|improve this answer

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.