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Let (\Omega, F, P) be a probability space, which may have atoms (important), S be a set of measure-preserving transformations T:\Omega\to\Omega, that is, such that preimage T^{-1}(A) is measurable whenever A is measurable, and P(A)=P(T^{-1}(A)). Then, obviously, random variables X and Y given by Y(\omega)=X(T(\omega)), T\in S, have the same distribution. I need the fact that the converse also true: if X and Y have the same distribution, then Y(\omega)=X(T(\omega)), for some T\in S. I am sure that this fact should be known, therefore do not want to reprove it. In the worst case, it should be a corollary from some known theorems, with 2-3 line proof. The question is: is it indeed known, and if so, where to refer?

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1 Answer 1

It is not true, at least as stated here.

For example, if $\Omega = [0,1]$ with Lebesgue measure, then $X(\omega)=\omega$ and $Y(\omega)=2\omega\text{ mod }1$ have the same distribution, and there exists a transformation that maps $X$ to $Y$, but it is not invertible, and there is no measure-preserving transformation from $Y$ to $X$.

Indeed, if $X(\omega) = Y(T(\omega))$, then $T(\omega)$ should select either $\omega/2$ or $(\omega+1)/2$ in a measure-preserving way. $P[T^{-1}([0,1/2])] = 1/2$, but note the "squeezing" map $\omega \mapsto \omega/2$ from $T^{-1}([0,1/2])$ to $[0,1/2]$ obviously does not preserve Lebesgue measure.

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Thabk you! I will update the question. –  bogdan Aug 23 '12 at 10:43
    
I don't understand the question now. What is this $S$ for? And I believe that my counterexample applies here for trivial reasons: if there is no transformation at all, then surely there is no transformation in $S$, right? And by the way, if you place $'s around your LaTeX code, it would be easier to read. –  Alexander Shamov Aug 23 '12 at 16:10

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