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For $G=\mathbb{R}^d$ I know that a stationary point process $X$ either has 0 or infinitely many points, a.s. Daley and Vere-Jones refer to this as the 0-Infinity dichotomy. They hint that this fact is known in a more general setting. What is the most general setting for which the answer is known? Does it hold for all (locally compact second countable hausdorff topological) groups $G$? References appreciated.

Edit: I'm really interested in when the dichotomy DOES hold. If you have some example where it doesn't hold, please post as a comment unless you feel that your example really does answer my question.

Edit: Obviously $G$ must be non-compact for the dichotomy to hold, else a Poisson process will give a counterexample.

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This question is actually a very good illustration of the lamentable degree of understanding of the measure theory in the probabilistic community. The claim is that, given a locally compact non-compact group $G$, there are no translation invariant probability measures on the space $\mathcal F(G)$ of non-empty finite subsets of $G$.

Indeed, let $\pi$ be the natural map which assigns to any subset $A\in\mathcal F(G)$ the uniform probability measure on $A$. Then the $\pi$-image of any translation invariant probability measure on $\mathcal F(G)$ is a translation invariant probability measure on $G$, which contradicts non-compactness of $G$.

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  • $\begingroup$ 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). $\endgroup$ Jul 15, 2016 at 22:23
  • $\begingroup$ 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? $\mu(B) = P(\pi^{-1}(??))$. $\endgroup$
    – nullUser
    Jul 15, 2016 at 22:50
  • $\begingroup$ @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. $\endgroup$
    – R W
    Jul 15, 2016 at 23:08
  • $\begingroup$ @RW I am very much interested in the measure theoretic approach, may we please resume discussion thereof? $\endgroup$
    – nullUser
    Jul 15, 2016 at 23:15
  • $\begingroup$ @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. $\endgroup$
    – R W
    Jul 15, 2016 at 23:18
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Edit 2: There is in fact a very simple proof that works for any locally compact group. Suppose $Z$ is an invariant point process on a locally compact group that $0<|Z|<\infty$ with positive probability. By conditioning, we can assume $0<|Z|<\infty$ almost surely. Let $X \in G$ be the random variable defined by choosing one of the points of $Z$ uniformly at random. The stationarity of $Z$ implies that the law of $X$ must be Haar measure on $G$. If $G$ is not compact, then its Haar measure is not a probability measure, and we get a contradiction.


To prove a theorem of this kind in a general setting, it is very useful to use the so-called mass-transport principle. Edit 2: Even in light of the above simple argument, I think the argument below is still instructive; the mass-transport principle has many further uses.

Edit 3: In particular, the mass-transport principle can be used to show that a point-process can't have 'too few' points: in particular, there can be no way of assigning disjoint subsets of the group to each point in such a way that some point gets assigned a set of infinite volume. I don't believe this can be done without the MTP (also, this only works with the true, unimodular, version of the MTP, it is false for nonunimodular groups).

A good reference is Probability on Trees and Networks by Lyons and Peres.

I will do things in the finitely generated case since I usually think in terms of graphs. The continuous case is quite similar, although I don't know the optimal conditions off the top of my head.

(Edit: The mass-transport principle relies on the group being unimodular, i.e., the left and right Haar measures being equal. This is automatically satisfied in the discrete case, since there both the measures are just counting measure. However, there is a biased version of the mass-transport principle for nonunimodular groups, involving the so-called modular function. This version of mass-transport is generally much less powerful than the unimodular version, but it does still suffice to prove what you want to here.)

Claim: Let $G$ be an infinite, finitely generated group, and let $Z$ be a point process on $G$ whose law is invariant under the action of $G$. Then $Z$ has either no points, or infinitely many points almost surely.

Proof: Let $F:G^2\times \{0,1\}^G\to [0,\infty]$ be a Borel function that takes as input a pair of points $x,y \in G$ and a set $\omega\subseteq G$, which represents a possible configuration of the point process. The mass-transport principle says that if $F$ is diagonally invariant, meaning that $$F(x,y,\omega) = F(\gamma x, \gamma y, \gamma \omega)$$ for every $\gamma \in G$ and every $(x,y,\omega) \in G^2\times \{0,1\}^G$, then $$\mathbb{E}\left[\sum_{x \in G}F(e,x,Z)\right]=\mathbb{E}\left[\sum_{x \in G}F(x,e,Z)\right],$$ where $e$ is the identity element of $G$. We think of as a rule for sending a positive amount of mass from $x$ to $y$ that depends on $\omega$, and the mass-transport principle tells us that the expected mass received by the identity is equal to the expected mass sent out by the identity.

For each $x \in G$ and $\omega \subseteq G$, let $\omega_x$ be the set of points in $\omega$ that are of minimal distance to $x$. Define $F$ by setting
$$F(x,y,\omega)=\frac{\mathbb{1}(y \in \omega_x)}{|\omega_x|},$$ letting $F(x,y,\omega)=0$ for all $y$ if $\omega = \emptyset$.

Then we have $$\mathbb{E}\left[\sum_{x \in G}F(e,x,Z)\right] = \mathbb{P}(\omega \neq \emptyset) \leq 1.$$

Suppose for contradiction that, with positive probability, $Z$ has a finite, non-zero, number of points. On this event, there must exist a point $y \in Z$ such that $y \in \omega_x$ for infinitely many $x \in G$. By stationarity, we deduce that $e \in \omega_x$ for infinitely many $x$ with positive probability. But then we have $$\mathbb{E}\left[\sum_{x \in G}F(x,e,Z)\right] \geq \mathbb{E}\left[\sum_{x \in G}\frac{\mathbb{1}(e \in \omega_x)}{|Z|}\right]=\infty$$ since the quantity we are taking the expectation of is infinite with positive probability. Thus, we have contradicted the mass-transport principle. $\square$

The proof of the mass-transport principle is very easy (see Lyons & Peres), but it turns out to be extremely useful. Essentially, it allows one to do 'averaging' arguments that work even in non-amenable settings where the ergodic theorem is not available. See Lyons and Peres for references to many papers in which it is used.

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  • $\begingroup$ Forgive me but this proof seems too good. Why is the proof in D&VJ such a nightmare if this simple proof suffices? Just to nitpick, conditioning on $0<|Z|< \infty$ preserves stationarity? And $X$ is defined by its conditional distribution given $Z$? How do we know we can actually construct such an $X$? $\endgroup$
    – nullUser
    Jul 15, 2016 at 17:10
  • $\begingroup$ And do you have a book I can cite for the full locally compact case? $\endgroup$
    – nullUser
    Jul 15, 2016 at 17:10
  • $\begingroup$ I don't, but it is rather obvious as R W said below. The mass-transport principle can give some much stronger results, as I started to indicate in my third edit. See Lyons and Peres for examples where this is very useful in e.g. percolation. $\endgroup$
    – tmh
    Jul 17, 2016 at 8:08
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    $\begingroup$ Also, conditioning preserves stationarity because $\{0<|Z|<\infty\}$ is an invariant event. $\endgroup$
    – tmh
    Jul 17, 2016 at 8:11

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