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 the this question also here).
$\begingroup$
$\endgroup$
6
-
1$\begingroup$ I run the risk of incurring the wrath of some hardcore probabilists here, but for problems that are truly applied, I doubt even measure theory is that useful. As Hamming said, "Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether, say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane." $\endgroup$– Simon LyonsCommented Jul 10, 2012 at 17:12
-
$\begingroup$ @Simon: The real advantage of the Lebesgue integral is that it is meaningful for functions over other spaces than the Euclidean space, say, the symbolic space $\lbrace 0,1\rbrace^\mathbb{Z}$. $\endgroup$– AlgernonCommented Jul 10, 2012 at 18:41
-
$\begingroup$ @AndyTeich In general the sample paths of a continuous-time stochastic process do not lie in a Polish space, do they ? $\endgroup$– Stéphane LaurentCommented Jul 11, 2012 at 5:20
-
$\begingroup$ @Stéphane Laurent: As an example, the Skorokhod space (see wikipedia), i.e. the space of cadlag functions, is a Polish space under a certain metric...of course one could give it a metric so that it is not a Polish space, but this would be "exotic"... – Andy Teich 0 secs ago $\endgroup$– Andy TeichCommented Jul 11, 2012 at 7:30
-
$\begingroup$ @AndyTeich The addition is not continuous under the Skorokhod topology, isn't it ? Hence a bigger topology could be more appropriate without being so exotic. $\endgroup$– Stéphane LaurentCommented Jul 11, 2012 at 20:07
|
Show 1 more comment
1 Answer
$\begingroup$
$\endgroup$
2
This answer is identical to the one I gave when the same question was posted at M.SE.:
There are 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.
answered Jul 10, 2012 at 19:14
-
$\begingroup$ I don't really see any example in these links. $\endgroup$ Commented Jul 12, 2012 at 18:15
-
$\begingroup$ I'm not sure wha you want. The paper shows that certain properties needed in applications do not hold on polish spaces. It also shows that one can do them on the space of uncountable coin-flips, a non-Polish space. $\endgroup$ Commented Jul 12, 2012 at 20:05