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I've read that if $\Phi$ is a Poisson point process (on $\mathbb{R}^d$, say), then conditional on there being $k$ points in some $A \subseteq \mathbb{R}^d$, the positions $X_1,\ldots,X_k$ of these points are uniformly distributed in $A$.

I'm having trouble making sense of what this means. "Conditional on $\Phi(A)=k$ I guess means consider the process $\Phi 1_{\Phi(A)=k}$ and then divide probabilities by $P(\Phi(A)=k)$. But, probabilities of what exactly? How am I labeling the points $X_1,\ldots,X_k$? In $\mathbb{R}$ If I did so by $X_1< X_2 < \cdots < X_k$ then clearly they are not uniformly distributed, so clearly the way that I label them matters. Hence my question, what is meant by saying $X_1,\ldots,X_k$ are uniformly distributed?

nullUser
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  • That, for every $k$, the set ${X_1,X_2,\ldots,X_k}$ is conditionally distributed like the set ${U_1,U_2,\ldots,U_k}$ , where the sequence $(U_n)_n$ is i.i.d. uniform on $A$. For example, in $\mathbb R$, $X_1$ is conditionally distributed as $\min(U_1,U_2,\ldots,U_n)$, $X_n$ is conditionally distributed as $\max(U_1,U_2,\ldots,U_n)$, and so on. – Did Nov 14 '15 at 22:58
  • I don't follow. How can $X_1$ and $X_2$ be conditionally independent if $X_1<X_2$ is guaranteed? Your two statements, that ${X_1,\ldots,X_k}$ is conditionally distributed like ${U_1,\ldots,U_k}$ and $X_1$ is conditionally distributed as $\min(U_1,\ldots,U_k)$ seem to contradict eachother. – nullUser Nov 14 '15 at 23:11
  • Read again: the (unordered) sets are identically distributed (indeed, $X_1$ is not conditionally distributed like $U_1$, as I explain just afterwards). – Did Nov 14 '15 at 23:26
  • How does one define the distribution of these unordered sets? Sum over all orderings and then divide by $k!$? – nullUser Nov 14 '15 at 23:32
  • Or just thinking of the map $\omega\mapsto \mathrm{supp}(\Phi) \cap A$ from $\Omega$ to the set of closed subsests or $\mathbb{R}^d$, and then looking at this as a random closed set? – nullUser Nov 14 '15 at 23:57
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    The most usual way to describe the distribution of a random set $A$, at least when $A$ is almost surely locally finite, is to specify the distribution of $(#(A\cap B_i))$ for every finite collection of Borel subsets $B_i$. – Did Nov 17 '15 at 08:23

2 Answers2

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For every measurable set $A\subseteq\mathbb R^d$ of finite measure, and every measurable set $B\subseteq A$, let $p$ be the conditional probability that the number of sites in $B$ is $\ell$, given that the number of of sites in $A$ is $k$.

Suppose $X_1,\ldots,X_k\sim\text{i.i.d. Uniform}(A)$. Let $q$ be the probability that $|\{ X_1,\ldots,X_k \} \cap B| = \ell$.

Then, regardless of which sets are $A$ and $B$ and which numbers are $k$ and $\ell$, we have $p=q$.

In other words, the probability distribution of the number of points falling in $B$ given the number in $A$, is always the same in either of those two scenarios.

Michael Hardy
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  • This makes much more sense. Just to help me clarify, would it be equivalent to say that the capacity functional $T_Z(K):= P(Z \cap K \neq \emptyset)$ of the random closed set $Z:={X_1,\ldots,X_k}$, where $X_i$ are iid uniform in $A$ is the same as the conditional capacity functional $T'_Z(K) := P(\mathrm{supp}(\Phi) \cap A \cap K | \Phi(A)=k)$? – nullUser Nov 15 '15 at 02:00
  • I'm not sure exactly what you mean by a probabliity of $\operatorname{supp}(\Phi)\cap A\cap K$, since $\operatorname{supp}(\Phi)\cap A\cap K$ does not appear to be an event. Might you have meant $\operatorname{supp}(\Phi)\cap A\cap K=\varnothing$? By $\operatorname{supp}(\Phi)$ do you mean the random set of sites distributed according to the Poisson process? (BTW, notice the typographical difference between $P(\mathrm{supp}(\Phi) \cap A \cap K | \Phi(A)=k)$ and $P(\mathrm{supp}(\Phi) \cap A \cap K \mid \Phi(A)=k)$. The latter uses \mid and has proper spacing.) ${}\qquad{}$ – Michael Hardy Nov 15 '15 at 02:14
  • Yes I meant $P(\mathrm{supp}(\Phi)\cap A \cap K \neq \emptyset \mid \Phi(A) = k)$, and by $\mathrm{supp}(\Phi)$ I mean the set of atoms of $\Phi$, which form a random closed set. – nullUser Nov 15 '15 at 05:08
  • In that case your two capacity functionals would be equal. – Michael Hardy Nov 15 '15 at 16:32
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I've read that if $\Phi$ is a Poisson point process (on $\mathbb{R}^d$, say), then conditional on there being $k$ points in some $A \subseteq \mathbb{R}^d$, the positions $X_1,\ldots,X_k$ of these points are uniformly distributed in $A$.

$\Phi$ is a process which generates iid distributed points in $\mathbb{R}^d$, then if $A \subseteq \mathbb{R}^d$ with $k$ points, then the $k$ points within $A$ are uniformly distributed in $A$.

MaxW
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  • Could you define what you mean by "generates iid distributed points" and what you mean by "If $A\subseteq \mathbb{R}^d$ with $k$ points". These are precisely the statements that I am asking about in my question. – nullUser Nov 14 '15 at 23:53
  • Think of 3D space. The points are chosen random over the volume of the 3D region. So random in each of x, y, and z. I choose some region of some plane in X-Y that has k points. The k points will then be evenly distributed in the chosen X-Y planar region. But the plane doesn't have to be in X-Y. It could be any plane. So within that plane I have X' and Y' coordinates. Then the points will be iid with respect to X' and Y' too. – MaxW Nov 15 '15 at 00:08
  • What you are describing sounds like the distribution of $(X,Y)$, not the distribution of ${X,Y}$ – nullUser Nov 15 '15 at 01:49