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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 $\eta\neq \rm id_X$ 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 $\rm id_X$ 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, bijective solutions of $(1)$ form a group - which may seem to be thought of a group of "symmetries" of $\mathsf Q$. For example, the standard normal distribution over reals $\mathsf Q = \mathscr N(0,1)$ admits at least two representations $\xi(\omega) = \omega$ and $\xi(\omega) = -\omega$. As well as any Haar measure over a group admits representation via $\xi(\omega) = \alpha \omega$ where $\alpha$ is any element of the group.

I've asked this question on MSE, but I have not received any answers.


Edited: To clarify (as requested), my question is exactly as follows: are such groups of symmetries of measures studied somewhere in the literature - may be, providing some interesting results for measures exhibiting such symmetries. I have studied the Lie groups of ODE/PDE symmetries, and I wonder if there is anything similar known for measures.

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    $\begingroup$ What exactly is the question here? Are you just asking whether such symmetries have been studied? $\endgroup$ Sep 4, 2012 at 13:40
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    $\begingroup$ Incidentally, it's easy to give a lot more symmetries of the normal distribution. Start by representing a uniform random variable $U$ in $[0,1]$ as $\sum_{n=1}^\infty 2^{-n} \xi_n$, where the $\xi_n$ are i.i.d. Bernoulli random variables. Then every permutation of $\mathbb{N}$ induces a symmetry of the distribution of $U$. If $F$ is the c.d.f. of the standard normal distribution, then $F^{-1}(U)$ is a standard normal random variable, and those permutations yield symmetries of the standard normal distribution. $\endgroup$ Sep 4, 2012 at 13:51
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    $\begingroup$ I haven't thought through the details, but at least any probability distribution on $\mathbb{R}$ with strictly positive density on a bounded or unbounded interval can be shown in this way to possess infinitely many symmetries. $\endgroup$ Sep 4, 2012 at 13:53
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    $\begingroup$ Of course Noah's answer, typed simultaneously with my comments, gives a probably simpler approach. $\endgroup$ Sep 4, 2012 at 13:55
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    $\begingroup$ I don't know how much work has gone into describing the automorphism group of a probability space $(X,\mu)$, but it is likely to be fairly large for some natural examples. Any amenable group $G$ acting on a compact Hausdorff space will preserve some Radon probability measure. If that measure turns out to be isomorphic to the one you started with, then you have an embedding of $G$ into $\mathrm{Aut}(X,\mu)$. $\endgroup$
    – Colin Reid
    Sep 5, 2012 at 3:05

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This is a fascinating topic. One impressive systematic study of symmetries is in the book by Olav Kallenberg (2005)

In there, though, the measurable space has to have some structure to get the most out of the results.

I don't know of any systematic applications of Lie groups to probability theory. However, there are here and there some interesting results. For instance, this book contains a study of measures that are invariant under O(n).

There is also plenty of results and applications of discrete symmetries (among others) in here:

Maybe one should ask a community wiki question where everyone tries to list the results they know. That would be a very interesting list!

Edit: I recently came across this book that is a quite relevant reference for studying symmetries of probability measures:

It has an extensive discussion on Lie groups ans Lie algebras.

Edit 2: Another book with an extensive discussion on Lie groups in Probability and Statistics!

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Maps such as $\eta$ and $\xi$ are called measure-preserving and are studied in ergodic theory. In particular ergodic theory views these as dynamical systems, because the maps can be iterated. One then studies properties of such maps, such as Poincaré Recurrence.

For, say, Lebesgue space there are many such transformations. Perhaps the simplest such maps defined on $[0,1)$ with Lebesgue measure are the maps $x\mapsto nx+\alpha\mod 1$ for fixed $\alpha\in\mathbb{R}$ and $n\in\mathbb{N}$ (invertible iff $n=1$). See e.g. Silva's textbook for a variety of more intricate examples.

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  • $\begingroup$ Thanks a lot for the answer, though as far as I faced ergodic theory, there was considered some (fixed) measure-preserving map and nice results were indeed derived for the asymptotic properties of the iterations of this map. I've added clarification for what I mean with my question: I am more interested in studies on the group of symmetries of a given measure (or, perhaps, a family of measure). I cannot check out Silva's book at the moment, maybe it concerns the topic I've mentioned as well. $\endgroup$
    – SBF
    Sep 4, 2012 at 14:14
  • $\begingroup$ @Ilya: one might fit ergodic theory into the framework of your question by saying it deals with cyclic subgroups (or cyclic subsemigroups) of the group of symmetries. $\endgroup$ Sep 4, 2012 at 15:14
  • $\begingroup$ @MarK: indeed, but my hope was that there is something beyond it. $\endgroup$
    – SBF
    Sep 4, 2012 at 15:16
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Usually, such symmetries have been either studied in the context of Lebesgue spaces or studied in the context of homogenous measure algebras, where autmorphisms are easy to study. Every automorphism of a probability space gives rise to an automorphism of the corresponding measure algebra.

The easiest case are Lebesgue spaces, or even simpler, studying the uniform distribution on $[0,1]$. This is of course essentially the case of a Haar measure. A nice property is that one can take any automorphism of the measure algebra and find an automorphism of the probability space inducing it. Moreover, two automorphisms of the probability space giving rise to the same automorphism of the measure algebra can differ only on a set of measure zero. Every, homogenous, atomless, separable measure algebra can be represented as a Lebesgue space.

If one starts with a homogenous measure algebra, one may look for probability spaces representing the measure algebra. The two most prominent representations are by the Stone space of the measure algebra or in the form of a product of coinflips $\{0,1\}^\kappa$ with $\kappa$ infinite, which can represent every atomless (normed) homogenous measure algebra by Maharam's theorem. In the case of the Stone space, the automorphisms of the measure algebra correspond essentially to the automorphisms of the representing probability space. In the coin-flipping representation, every automorphism of the measure algebra is induced by an automorphism of the probability space. But very different automorphisms may give rise to the same automorphism of the measure algebra. Actually, there exists an automorphism of $\{0,1\}^\mathfrak{c}$ that induces the identity on the measure algebra but has no fixed point.

The discussion so far is largely adapted from the introduction to Ergodic theory on homogeneous measure algebras. by Choksi and Prasad. The book this has appeared in is likely to be available somewhere on the internet...

One can also study the case of rigid probability spaces, where there is essentially no automorphism but the identity. It is actually possible to find a countably generated and countably separated measurable space in which all automorphisms differ from the identity only on a countable set. This is done in the booklet Borel Spaces by Rao and Rao in Proposition 4. There also is an example of an atomless, countably generated probability space with no autmorphism but the identity (up to a countable set) in Section 48 of Values of non-atomic games by Aumann and Shapley.

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I don't know of any systematic study of such symmetries in any great generality. On the other hand, as in most (if not all) fields of mathematics, probability theory happily exploits symmetries to help solve more concrete problems. For example, if $X$ is a random variable and $\xi$ is a bijective solution of your (1), then $X' = \xi(X)$ is a new random variable with the same distribution as $X$, coupled to $X$ in a nontrivial way, which can be a helpful technical tool. In particular, if $\xi\circ \xi = \mathrm{id}$, then $(X,X')$ is an exchangeable pair, which can be used together with Stein's method to prove distributional approximation theorems for $X$.

In a similar vein, your example for Haar measure is essentially the definition of Haar measure, and as such can of course be used (frequently quite directly) to prove many things about Haar measures.

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