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 think, that I need something like the following, but do not find it anywhere in textbooks. I am not even sure if it makes sense.

If you recognize it, please provvide some pointers.

  • Two measurable spaces $(\Omega_i,\mathcal{F}_i)$, with $i\in\{1,2\}$
  • A product space with probability $(\Omega=\Omega_1\times\Omega_2,\mathcal{F}=\mathcal{F}_1\times\mathcal{F}_2,P)$
  • Projection(better name?): $\pi_1(A\in\mathcal{F}_1):=A\times \Omega_2$ and $\pi_2(A\in\mathcal{F}_2):=\Omega_1 \times A$
  • Also possible: $\pi_1(\mathcal{F_1})=\{A\times\Omega_2|A\in\mathcal{F}_1\}$

Now you can have something like marginal probabilities? $P_i$ is a probability measure on $\mathcal{F}_i$:


In the classical two random variable setting, one has expectation conditional on one of the two. I would generalize this as conditional expectation with respect to the sigma-algebra $\pi_i(\mathcal{F}_i)$, e.g.


This should integrate out any dependence on states of the $i$-th space?

share|cite|improve this question
a) If $\mathcal{F}_i$ is a sigma-algebra, should'nt $\pi_i(\mathcal{F}_i)\subset \mathcal{F}$ also be one? b) Where can I read about marginals of $P$? c) I can always take an expectation restricted on any set $\pi_1(A), A\in\mathcal{F}_1$. The set of all $\pi_1(A)$ is $\pi_1(\mathcal{F}_1)$, so I thought this could be captured by a conditional expectation. d) In this setting I do not have random variables, but in the two variable setting, there is a marginal expectation. I am looking for something similar for my situation. – JSG Jun 24 '12 at 21:35

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.