Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Conditional probability and conditional independence are unique almost surely, but relative to what: the conditioning field or the underlying field?

More precisely, consider the case of conditional independence. Let $\left(\Omega, \mathcal{A}, P\right)$ be a probability space, let $\mathcal{B}$ be a sub-$\sigma$-algebra of $\mathcal{A}$ and let $D,E\in\mathcal{A}$. Then by definition (see, e.g. Kallenberg (1995) p. 86) $D,E$ are conditionally independent given $\mathcal{B}$ iff $$P\left(\left.D\cap E\right|\mathcal{B}\right)=P\left(\left.D\right|\mathcal{B}\right)P\left(\left.E\right|\mathcal{B}\right)\space\space\mathrm{a.s}$$ But does "a.s." mean "up to a null set $F\in\mathcal{A}$" or "up to a null set $F\in\mathcal{B}$"?

share|improve this question
1  
I answered this on the other page. The answer is: BOTH, since $P(D\cap E\mid\mathcal B)$ and $P(D\mid\mathcal B)P(E\mid\mathcal B)$ are $\mathcal B$-measurable. –  Did Mar 10 '13 at 21:55
    
@Did: I see now. Thanks. –  Evan Aad Mar 10 '13 at 22:28

1 Answer 1

up vote 0 down vote accepted

See Did's comment to this thread as well as the one he posted in reply to my comment on this thread.

share|improve this answer

Your Answer

 
discard

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.