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.

  • $\begingroup$ 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 ? $\endgroup$ – Ahriman Aug 24 '12 at 11:16
  • $\begingroup$ @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. $\endgroup$ – Ilya Aug 24 '12 at 11:19
  • $\begingroup$ 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) $\endgroup$ – Ahriman Aug 24 '12 at 11:31
  • $\begingroup$ @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.. $\endgroup$ – Seyhmus Güngören Aug 24 '12 at 11:37
  • $\begingroup$ @SeyhmusGüngören: sure, I've updated - please tell me if the notation is clear. $\endgroup$ – Ilya Aug 24 '12 at 11:45