Timeline for When does a stationary point process on group $G$ have $0$ or $\infty$ many points a.s.?
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| Jul 17, 2016 at 13:18 | comment | added | Pablo Lessa | @RW I just had a look, and yes that's the one I remember the most (especially the example of sine and cosine on $[0,2\pi]$ being two uncorrelated random variables which aren't independent). Concerning the same issue Doob also wrote a two page "commentary on probability" in a book "A century of mathematics", and a short chapter "Probability vs Measure" in a book honoring Halmos. | |
| Jul 17, 2016 at 11:34 | comment | added | R W | @Pablo Lessa - Thank you for mentioning Doob - do you mean "The development of rigor in mathematical probability"? | |
| Jul 17, 2016 at 2:38 | comment | added | Pablo Lessa | @RW I agree there are some probabilist who eschew measure theory. I think we only disagree on how many of them do. I would guess that only those doing things in a finite setting can avoid going deeply into measure theory. Your complaint reminds me of a very interesting article by Doob in which he related seeing the attitude towards measure theory in probability change throughout the decades. I would like to think at this point most of the resistance to Kolmogorov's foundational approach has dissipated. But then again, maybe I'm wrong... | |
| Jul 16, 2016 at 2:30 | comment | added | nullUser | Ahh I think I see the lost step. Consider the space $(\mathcal{M}(G), \pi(P))$. Define the intensity measure $\Lambda$ on $G$ by $\Lambda(A) := \int_{\mathbb{M}(G)}\phi(A)\,\pi(P)(d\phi)$. Then $\Lambda$ is shift invariant, and a prob measure since $\phi(G) = 1$ for $\pi(P)$-a.e. $\phi$. Now $\Lambda$ is a Haar probability measure on a noncompact space, contradiction. | |
| Jul 16, 2016 at 2:14 | comment | added | nullUser | Okay so lets think of $\pi: \mathcal{F}(G) \to \mathbb{M}(G)$ with codomain the space of Radon measures on $G$. Measurability aside, $\pi$ induces a natural mapping of measures $\pi(\mu)(B) := \mu(\pi^{-1}(B))$ for $\mu$ a measure on $\mathcal{F}(G)$ and $B\subset \mathbb{M}(G)$ measurable. So if $P$ is our translation invariant probability measure on $\mathcal{F}(G)$, then $\pi(P)(B) := P(\pi^{-1}(B))$ is a shift invariant prob measure on $\mathbb{M}(G)$, not on $G$. How do we proceed from here? | |
| Jul 15, 2016 at 23:18 | comment | added | R W | @nullUser Yes, any measurable map $X\to Y$ naturally extends to a map between spaces of measures on $X$ and $Y$. As for the natural $\sigma$-algebra on $\mathcal F(G)$, one can present $\mathcal F(G)$ as a disjoint union of spaces $\mathcal F_n(G)$ of $n$-point subsets, and each $\mathcal F_n(G)$ as an image of a measurable subset of the product space $G^n$. Of course, formally one should also verify measurability of $\pi$. However, in questions of this kind usually there are "natural" $\sigma$-algebras, and measurability with respect to them is then almost automatic. | |
| Jul 15, 2016 at 23:15 | comment | added | nullUser | @RW I am very much interested in the measure theoretic approach, may we please resume discussion thereof? | |
| Jul 15, 2016 at 23:08 | comment | added | R W | @Pablo Lessa - Well, it's very much opinion based of course. But what I want to say is that most probabilists avoid using the very notion of measure, and examples of this are readily provided by this very discussion. Look, for instance, at the formulation of the "mass transfer principle" above (which, by the way, has nothing to do with the original question) or at the proof of the "zero-infinity dichotomy" in Daley and Vere-Jones. | |
| Jul 15, 2016 at 22:50 | comment | added | nullUser | I don't follow your suggestion. We consider the space $\mathcal{F}(G)$ with what $\sigma$-algebra? Then $\pi : \mathcal{F}(G) \to \mathrm{Prob}(G)$ is defined by $\pi(A) := \mathrm{Unif}(A)$. Now suppose $P$ is a translation invariant probability measure on $\mathcal{F}(G)$. What is meant by its $\pi$-image? $\mu(B) = P(\pi^{-1}(??))$. | |
| Jul 15, 2016 at 22:23 | comment | added | Pablo Lessa | I think you went way overboard on that first phrase. The probabilistic community thrives on abstract measure theory (seriously, who else is actually calculating Lebesgue integrals on abstract measure spaces, using disintegration into conditional measures daily, or is worried about different types of filtrations and sigma algebras). | |
| Jul 15, 2016 at 21:08 | history | answered | R W | CC BY-SA 3.0 |