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

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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 '14 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 '14 at 16:27
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.

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Half of the answer. – Did Mar 8 '14 at 16:29
@Did: indeed. It at least gives some potential candidates. – Davide Giraudo Mar 8 '14 at 16:33
The paragraph you added after my comment goes in the right direction. – Did Mar 8 '14 at 16:41

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