5
$\begingroup$

Probabilist often work on Polish spaces. Does somebody know an ("non-exotic") example, for which it is not possible to work on a Polish space, but instead one has to work on a general measurable space? By non-exotic example I mean something like a stochastic process, which is really used in applications, and cannot be defined on a Polish space...(I posted this question also here).

$\endgroup$
  • $\begingroup$ I'm no probabilist, but shouldn't something like $[0,1]^{\mathbf R_+}$ naturally arise in probabilistic contexts? It's completely regular, but certainly not Polish. $\endgroup$ – tomasz Jul 10 '12 at 12:44
  • 1
    $\begingroup$ There is a good link on MO, with examples of what can go wrong when the assumption Polish is being dropped...this might help in constructing an example... $\endgroup$ – Andy Teich Jul 10 '12 at 12:48
1
$\begingroup$

There re a number of constructions that do not work for Polish spaces, but a certain class of probability spaces, variously known as super-atomless, saturated, nowhere countably generated and a number of other names. A nice overview can be found here.

A probability space $(\Omega,\Sigma,\mu)$ is saturated if for every two Poilsh spaces $X$ and $Y$, every probability measure $\nu$ on $X\times Y$ and every random variable $f:\Omega\to X$ such that its distribution $\mu f^{-1}$ equals the marginal of $\nu$ on $X$, there is a random variable $g:\Omega\to Y$ such that the joint distribution of $(f,g)$ is $\nu$.

The following definition is conceptually different, but can be shown to be equivalent:

A probability space $(\Omega,\Sigma,\mu)$ is super-atomless if there is no $A\in\Sigma$ satisfying $\mu(A)>0$, such that the pseudo-metric space obtained by endowing the trace $\sigma$-algebra on $A$ with the pseudo-metric $d(A,B)=\mu(A\triangle B)$ is separable.

$\endgroup$
  • $\begingroup$ Could you explain a bit more? The definition of "saturated probability space" in their article makes no sense, likely because of faulty notations... $\endgroup$ – D. Thomine Jul 10 '12 at 16:46
  • $\begingroup$ @D.Thomine: I hope it is clearer now. $\endgroup$ – Michael Greinecker Jul 10 '12 at 17:16
  • $\begingroup$ Yes, thank you. Now I still have to understand this theory, but that's another problem :) $\endgroup$ – D. Thomine Jul 10 '12 at 17:22
  • $\begingroup$ @MichaelGreinecker: The paper you are referring to seems to give only examples of set-valued functions...are there also examples for single-valued functions or does that already follow from this paper? $\endgroup$ – Andy Teich Jul 11 '12 at 9:04
  • $\begingroup$ @Andy: Section 4 is about the existence of Nash equilibria, which are certain functions, even though the proofs make use of set-valued functions. Here is a paper that uses these kind of spaces to construct extended product spaces which are used in the modeling of ideosyncratic risk in large populations in economics. $\endgroup$ – Michael Greinecker Jul 11 '12 at 9:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.