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Consider probability space W with pair of random variables having same distribution. On how much this variables distinct in terms of W symmetries? Namely, let's talk about automorphism as measure-preserving self-mapping of W defined almost everywhere. The following question must be well studied. When such random variables may be combined by some automorphism of W (up to measure-zero set, of course)? Sorry for bad english.

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I don't understand what you mean by "combined". Suppose we have our probability space $(W, \mathcal{F}, P)$, two identically distributed random variables $X,Y$. Your automorphism is (I think) a map $\Phi : W \to W$ such that $P(\Phi^{-1}(A)) = P(A)$ for all $A \in \mathcal{F}$. What is $\Phi$ supposed to do to $X$ and $Y$? –  Nate Eldredge Oct 28 '10 at 20:54
    
This reminds me of the mass transport problem (Monge-Kantorovich), but it's somewhat different. –  Florian Oct 28 '10 at 20:56
    
@Nate: As I understand it, $X\circ \Phi = Y$ almost everywhere. –  Florian Oct 28 '10 at 21:03
    
Florian rights, additionally i assume that P(A)=P(Φ(A)) (for a existence of inverse automorphisms) –  Bad English Oct 29 '10 at 6:27
    
@Bad English, I'm also uncertain about what you mean about combining the pair of identically distributed random variables. Do you simply mean that the distribution of $(X,Y)$ is equivalent to the distribution of $(Y,X)$ ? –  sleepless in beantown Oct 29 '10 at 6:49

2 Answers 2

up vote 3 down vote accepted

There are some obvious restrictions in the case when the base probability space $W$ is allowed to have atoms. For instance, if $W$ consists of an atom and a continuous part with equal masses $1/2$, then their indicator functions have the same distribution, but an automorphism in question clearly does not exist.

A slightly more involved counterexample is provided by the following pair of random variables defined on the unit interval endowed with the usual Lebesgue probability measure. One is the identity map, and the other one is $x\mapsto 2x$ (mod 1).

I will give a complete answer to your question in the situation when the base probability space $(W,P)$ is a Lebesgue (standard) probability space (which is the only reasonable generality in probability nowadays, although there is a lot of people here who are very fond of discussing various exotic if not outright pathological measure spaces). The signature of a measure space is the mass of its non-atomic part plus the non-increasing sequence of the weights of its atoms. Rohlin (see the above Wikipedia article for a reference to his 1949 article very appropriately called "On the fundamental ideas of measure theory") proved that, up to isomorphism, Lebesgue spaces are completely characterized by their signatures. In particular, there is only one purely non-atomic Lebesgue measure space, which is just the unit interval with the Lebesgue measure on it (whence the term).

It is less known that Rohlin also obtained a complete classification of homomorphisms of Lebesgue spaces (equivalently, of their measurable partitions, or of complete sub-$\sigma$-algebras). Signature of the quotient measure and signatures of the conditional measures associated with the homomorphism provide an obvious system of conjugacy invariants of such homomorphisms. Rohlin proved that this system is, in fact, a complete system of invariants. In the simplest purely non-atomic case it means that any homomorphism of Lebesgue spaces with a purely non-atomic quotient space and purely non-atomic conditional measures is conjugate to the coordinate projection of the unit square onto the unit interval (both being endowed with the canonical Lebesgue measures).

Applied to your original question, Rohlin's classification implies that two random variables with the same distribution on a Lebesgue space are equivalent in your sense if and only if for a.e. value taken by these variables the corresponding conditional measure spaces are isomorphic, i.e., have the same signature. In particular, the latter is the case if almost all conditional measures are purely non-atomic.

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R W: would you mind rewriting the question? I don't understand what is being asked. –  Kevin O'Bryant Nov 27 '10 at 4:18
    
Let $X$ and $Y$ be two random variables on the same probability space $(W,P)$ (the OP probably meant real valued random variables, but it does not matter). Call them equivalent if there exists an automorphism (measure preserving transformation) $\Phi$ of $(W,P)$ such that $Y=X(\Phi)$. Obviously, if $X$ and $Y$ are equivalent, then their distributions (i.e., the corresponding image measures on the target space) are the same. Question: conversely, are two random variables on the same probability space with the same distribution equivalent? –  R W Nov 27 '10 at 12:08

I think he means X and Y have same probability distribution functions.

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