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Moderator's Note: This question has been put on hold due to the version over at MathOverflow having received better attention and produced an accepted answer. Interested readers are advised to visit the above link.

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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    @Ilya: "off-topic" is the only one that can handle "custom" reasons like the one I entered in the comments above. None of the others are better fits. And please read the moderator message that I entered _waaaay_ at the top of the question.2014-06-04

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