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.

Does the set of all random variables which generate the same $\sigma$-algebra form some nice set, such as a vector subspace? Thanks!

share|improve this question
1  
Did you try to prove that, given some random variable $X$, the random variables $Y$ such that $\sigma(Y)=\sigma(X)$ are exactly those such that there exists an invertible function $g$ such that $Y=g(X)$ almost surely? –  Did Mar 8 at 16:25
    
@Did, I thought of the reverse part, and then don't know why I quickly forgot it. I am not sure how you prove the other direction, i.e, find such $g$. –  Tim Mar 8 at 16:27

1 Answer 1

up vote 1 down vote accepted

It seems Doob-Dynkin theorem is the answer to your question. Let $X$ be a real valued random variable, $\sigma(X):=\{X^{-1}(B), B\in\mathcal B(\mathbb R)\}$. The random variable $Y$ is $\sigma(X)$-measurable if and only if there exists $f\colon\mathbb R\to\mathbb R$ Borel measurable such that $Y=f(X)$.

But there are restrictions on $f$ in order to guarantee that the inclusion $\sigma(Y)\subset\sigma(X)$ is not strict. For example, if $f$ is bijective it is OK.

share|improve this answer
    
Half of the answer. –  Did Mar 8 at 16:29
    
@Did: indeed. It at least gives some potential candidates. –  Davide Giraudo Mar 8 at 16:33
    
The paragraph you added after my comment goes in the right direction. –  Did Mar 8 at 16:41

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.