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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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    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 ?2012-08-24
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    @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.2012-08-24
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    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)2012-08-24
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    @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..2012-08-24
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    @SeyhmusGüngören: sure, I've updated - please tell me if the notation is clear.2012-08-24
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    @Ilya For your last question, yes, if $\xi$ is invertible, and if it is measurable, as well as its inverse, then $\xi$ solves $(1)$ iff $\xi^{-1}$ solves it. It suffices to write it.2012-08-24
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    @Ilya I think you write the questions in a very good way. It is more related to me as I have weak knowledge. It seems that this property happens only when $P(x_i\in X)=c\forall i$?2012-08-24
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    @SeyhmusGüngören: that's nice to know, thanks! I didn't get, what does your latter expression mean, though.2012-08-24
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    @Ahriman: yeah, I've even updated it before you answered - needed to take $\xi^{-1}$ at the right place.2012-08-24
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    $(0.5,0.5)$ or $(0.2,0.2,0.2,0.2)$ if the sample space has $5$ symbols and you will have again more than one solution but iff $(0.2,0.2,0.2,0.2)$. For other cases it is again unique. Or?2012-08-24
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    @SeyhmusGüngören: or. Consider $(0.1,0.1,0.1,0.1,0.6)$. All the maps that just permute first $4$ states and leave the $5$-th on its place are solutions.2012-08-24
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    @Ilya yes you are right. Apperantly there are $N-k$ permuations with $k$ equal probabilities such that the equation that you mentioned is satisfied. Accordingly we have $(N-k)!$ such $\xi$. In such cases $\xi$ is not identity mapping. Then we still have a random variable unique to its distribution function right?2012-08-26
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    @SeyhmusGüngören: well, *by definition* we are looking for all such random variables $\xi:X\to X$ which have the same *fixed* distribution $\mathsf Q$, aren't we?2012-08-26
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    @Ilya ah ok so you mean we have a single distribution but quite many $\xi$ which are mapping to the same distribution which means that the mapping from a distribution to a random variable is not unique. From what I see it seems correct but this happens for (N-k)! special cases if there are. So what is the implication of this fact?2012-08-26
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    @SeyhmusGüngören: indeed, this is what I mean. I am not aware of any implications, but solutions of $(1)$ seems to serve as different representations of the same distribution. Moreover, *bijective* solutions of $(1)$ form something similar to a group of symmetries of a measure - but I didn't find such notion used anywhere.2012-08-26
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    @Ilya though I find it interesting. Perhaps it doesnt change anything when one uses for example not $\xi_2$ but say $\xi_3$ as the probabilities are the same at the same indexes although indexes themselves (the mapping procedure) are changing.2012-08-26
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    This question is put on-hold due to a duplicate being posted to MathOverflow, with answer accepted.2014-05-27
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    @WillieWong: hey, no problem with closing this question - but perhaps another reason rather than off-topic would be more appropriate? Also, maybe in the closing message you'd be able to put an explicit link to the MO version of the question?2014-06-04
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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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