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When talking about a single random variable, knowing only its distribution, the construction of a probability space is quite easy. Namely, let $(X,\mathscr A)$ be a measurable space and let $\mathsf Q$ be some probability measure over this space which we refer to as a distribution of some random variable. The usual definition states that there is some probability space $(\Omega,\mathscr F,\mathsf P)$, the random variable is $$ \xi:(\Omega,\mathscr F)\to(X,\mathscr A) $$ i.e. it is a measurable map, and its distribution is a pushforward measure: $$ \mathsf Q:=\xi_*(\mathsf P) $$ i.e. $\mathsf Q(A) = \mathsf P(\xi^{-1}(A))$ for any $A\in \mathscr A$.

Clearly, given $(X,\mathscr A,\mathsf Q)$ for a single random variable there is no reason to come up with a new sample space and we can take $(\Omega,\mathscr F,\mathsf P) = (X,\mathscr A,\mathsf Q)$ and $\xi:=\mathrm{id}_X$.

Let us stick to this latter case. It may happen, that there is a map $$ \eta:(X,\mathscr A)\to(X,\mathscr A) $$ such that $\eta\neq\mathrm{id}_X$ but still it holds that $\mathsf Q = \eta_*(\mathsf Q)$. I wonder if the existence of this other maps is studied somewhere.

The brief statement of the problem is thus the following: given a probability space $(X,\mathscr A,\mathsf Q)$ if the identity map $\mathrm{id}_X$ is the unique solution of the equation $$ \mathsf Q = \xi_*(\mathsf Q) \tag{1} $$ where the variable $\xi$ is any measurable map from $(X,\mathscr A)$ to itself. As far as I am not mistaken, the space of solutions of $(1)$ is a monoid as it is closed under the composition of maps.

Also, if $\xi$ is a bijection which solves $(1)$ then $\xi^{-1}$ solves it as well: $$ \xi^{-1}_*(\mathsf Q)(A) = \mathsf Q(\xi(A)) = \mathsf Q(\xi^{-1}(\xi(A))) = \mathsf Q(A). $$

Hence, solutions of $(1)$ which are bijection form a group - which may be thought of a group of "symmetries" of $\mathsf Q$, apparently.


A small example just to add some clarity to the problem statement.

If $X = \{a,b\}$, $\mathscr A = 2^X$ and $\mathsf Q(a) = 0.4$ then the solution is unique. However, if $\mathsf Q(a) = 0.5$ there are exactly two solutions. Indeed, there are exactly $4$ maps $\xi:X\to X$ namely $$ \begin{align} \xi^1:(a,b) \mapsto (a,a) & &\xi^2:(a,b) \mapsto (a,b) \\ \xi^3:(a,b) \mapsto (b,a) & &\xi^4:(a,b) \mapsto (b,b) \end{align} $$ In the first case, the pushforwards are $$ \begin{align} \xi^1_*(\mathsf Q) &= &(1,0) & &\xi^2_*(\mathsf Q) &= &(0.4,0.6) \\ \xi^3_*(\mathsf Q) &= &(0.6,0.4) & &\xi^4_*(\mathsf Q) &= &(0,1) \end{align} $$ and in the second case: $$ \begin{align} \xi^1_*(\mathsf Q) &= &(1,0) & &\xi^2_*(\mathsf Q) &= &(0.5,0.5) \\ \xi^3_*(\mathsf Q) &= &(0.5,0.5) & &\xi^4_*(\mathsf Q) &= &(0,1) \end{align} $$ hence in the first case $(1)$ has the only solution $\xi^2$ which is of course $\mathrm{id}_X$, but in the second case both $\xi^2$ and $\xi^3$ solve the problem.

Due to this reason, I expect a non-uniqueness of the solution to reflect some kind of symmetries in the distribution $\mathsf Q$.


I posted the same question on MO.

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In dynamical systems, given a measurable transformation of a measurable space, one is often looking for interesting probability measures which are invariant under the transformation. You're considering the converse problem, in some sense : the probability measure is given, and you're looking for transformations which leave it invariant, don't you ? –  Ahriman Aug 24 '12 at 11:16
    
@Ahriman: exactly - this is can be considered as a sort of a dual problem to the invariant measures. Though in my case, the solution does always exist, but it can be non-unique. One can also say, that the question is a uniqueness of the representation of the distribution via the random variable over the fixed sample space. –  Ilya Aug 24 '12 at 11:19
    
This is known as the inverse Perron-Frobenius problem, at least when the probability is absolutely continuous w.r.t. a "natural" one. (for instance, Lebesgue measure on an interval) –  Ahriman Aug 24 '12 at 11:31
    
@Ilya As I am trying to learn more I read your question. Didnt understand completely. In your example what is the unique solution and what are exactly two solutions? I am not knowledgeble person in this area as I said and I wanna understand the things first. If I can finally manage then I will start thinking.. –  Seyhmus Güngören Aug 24 '12 at 11:37
    
@SeyhmusGüngören: sure, I've updated - please tell me if the notation is clear. –  Ilya Aug 24 '12 at 11:45
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