Defining a uniform distribution of points in the plane

Question

I saw this question, and was wondering what is the best way to describe a random distribution of points in the plane such that the expected number of points in any region of unit area is $c$, where $c$ is a given positive constant, and such that the number of points in every pair of disjoint regions are independent. $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

I can mimic the derivation of the Poisson distribution in the following manner:

1. Take any $n \in \mathbb{N}_{>c}\,\,.$ Partition each unit square of the plane into an $n \times n$ grid of square cells.
2. With probability $\lfrac{c}{n^2}$ put a point into each cell, distributed uniformly within the cell, independently of other cells.
3. Then every region comprising $n^2$ cells has expected number of points $c$.
4. Also every pair of disjoint regions have independent numbers of points.
5. Now if $n \to \infty$, I presume that this will more and more accurately resemble the desired distribution of points.

But I have no idea how to take such a limit, nor what the limit distribution is!. I suspect it might be important that the cells are getting smaller in both dimensions, but I don't know. (If each square is partitioned into $n$ horizontal rectangles instead, each rectangle comprising $1 \times n$ cells, and the probability changed to $\lfrac{c}{n}$, the number of points in two vertical regions each comprising $n^2 \times 1$ cells will not be independent. This dependence might not affect the final limiting distribution, if such a limit exists at all, but a priori it is not clear to me.)

So how can we rigorously define such a distribution, and is it really the limit of the above concept, in which case how do we rigorously define this limit?

Answer 1

Thanks to Michael and Shalop for the core ideas, here is a rigorous definition and proof. $ \def\nn{\mathbb{N}} \def\zz{\mathbb{Z}} \def\rr{\mathbb{R}} \def\pp{\mathbb{P}} \def\ee{\mathbb{E}} \def\ii{\mathbf{1}} \def\wi{\subseteq} \def\none{\varnothing} \def\t{\text} \def\lsum{{\large\sum}} \def\lfrac#1#2{{\large\frac{#1}{#2}}} $

Definition

Let $S_{a,b} = [a,a+1) \times [b,b+1)$, for each $a,b \in \zz$.

Let $X_{a,b} \sim \t{Pois}(c)$ independently for each $a,b \in \zz$.

Let $P_{a,b}$ be a set of $X_{a,b}$ points distributed uniformly in $S_{a,b}$ independently for each $a,b \in \zz$.

Let $P = \bigcup_{a,b\in\zz} P_{a,b}$.

Theorem

$\#(P \cap U) \sim \t{Pois}(c|U|)$ for any $U \wi \rr^2$ with finite measure.

$\#(P \cap U),\#(P \cap V)$ are independent for any disjoint $U,V \wi \rr^2$ each with finite measure.

Proof

For each measurable $R \wi \rr^2$, let $N_R = \#( P \cap R )$.

[First we prove the 1-square case.]

Take any $a,b \in \zz$ and disjoint measurable $Q,R \wi S_{a,b}$.

Let $x = 1-|Q|-|R|$ and $y = |Q|$ and $z = |R|$.

Given any $m,n \in \nn$:

  $\pp( ⟨N_Q,N_R⟩ = ⟨m,n⟩ ) = \lsum_{k=0}^\infty e^{-c} \lfrac{c^{k+m+n}}{(k+m+n)!} \binom{k+m+n}{k,m,n} x^k y^m z^n$

  $\ = e^{-c} \lfrac{c^{m+n}}{m!n!} y^m z^n \lsum_{k=0}^\infty \lfrac{(cx)^k}{k!}$ $= e^{-c} \lfrac{c^{m+n}}{m!n!} y^m z^n e^{cx}$ $= e^{-cy} \lfrac{(cy)^m}{m!} \ e^{-cz} \lfrac{(cz)^n}{n!}$.

  $\pp(N_Q=m) = \pp(⟨N_Q,\none⟩ = ⟨m,0⟩) = e^{-cy} \lfrac{(cy)^m}{m!}$ and likewise $\pp(N_R=n) = e^{-cz} \lfrac{(cz)^n}{n!}$.

Therefore $N_Q,N_R$ are independent and $N_Q \sim \t{Pois}(c|Q|)$ and $N_R \sim \t{Pois}(c|R|)$.

[Now we prove the general case.]

Note that the 1-square case extends easily [by induction] to the finite-squares case.

For each measurable $R \wi \rr^2$, let $T_k = \bigcup_{a,b \in \{-k..k\}} S_{a,b}$ and $N_{R,k} = \#( P \cap R \cap T_k )$.

Take any disjoint measurable $U,V \wi \rr^2$ such that $|U|,|V|$ are finite.

Given any $m,n \in \nn$:

  For $k \in \nn$ as $k \to \infty$:

    $\pp( N_U \le m ) \approx \pp( N_{U,k} \le m )$   [by MCT for sets]

    $\ = \pp( \t{Pois}(c|U \cap T_k|) \le m )$   [by the finite-squares case and sum of Poisson r.v.]

    $\ \approx \pp( \t{Pois}(c|U|) \le m )$   [by MCT for sets]

    $\pp( N_U \le m \land N_V \le n )$

    $\ \approx \pp( N_{U,k} \le m \land N_{V,k} \le n )$   [by MCT for sets]

    $\ = \lsum_{i=0}^m \lsum_{j=0}^n \pp( N_{U,k} = i \land N_{V,k} = j )$

    $\ = \lsum_{i=0}^m \lsum_{j=0}^n \pp( N_{U,k} = i ) \ \pp( N_{V,k} = j )$   [by the finite-squares case]

    $\ = \pp( N_{U,k} \le m ) \ \pp( N_{V,k} \le n )$

    $\ \approx \pp( N_U \le m ) \ \pp( N_V \le n )$   [by MCT for sets].

  Therefore $\pp( N_U \le m ) = \pp( \t{Pois}(c|U|) \le m )$

  and $\pp( N_U \le m \land N_V \le n ) = \pp( N_U \le m ) \ \pp( N_V \le n )$.

Therefore $N_U \sim \t{Pois}(c|U|)$ and $N_U,N_V$ are independent.

Notes

At all the marked points we are implicitly using MCT for sets, which states the following:

Take any set $T$ such that $|T|$ is finite, and any sequence $(S_k)_{k\in\nn}$ of measurable sets such that $S_k \wi T$ for every $k \in \nn$ and $S_k \to S$ monotonically as $k \to \infty$. Then $|S_k| \to |S|$ as $k \to \infty$.

Answer 2

Say the expected number of sites in a given region is $c$; divide the region into $m$ parts. Approximate the desired distribution by saying you put a site into each part with probability $c/m$. This forbids the improbable event of more than one site in one of those $m$ parts, so it's not exact. We need a limit as $m\to\infty$.

The probability that the number (capital) $X$ of sites in the region is (lower-case) $x$ is \begin{align} \Pr(X=x) & = \binom m x \left( \frac c m \right)^x \left( 1 - \frac c m \right)^{m-x} \\[10pt] & = \frac{m(m-1)(m-2)(m-3)\cdots(m-x+1)}{x!} \cdot \frac{c^x}{m^x} \cdot \left( 1-\frac c m \right)^m \left( 1- \frac c m \right)^{-x} \\[10pt] & = \frac{m(m-1)(m-2)(m-3)\cdots(m-x+1)}{m^x} \cdot \frac{c^x}{x!} \cdot \left( 1 - \frac c m \right)^m \left( 1 - \frac c m \right)^{-x} \\[10pt] & = \underbrace{\frac{m(m-1)(m-2)(m-3)\cdots(m-x+1)}{m^x}}_{\text{This } \rightarrow \, 1} \cdot \frac{c^x}{x!} \cdot \underbrace{\left( 1 - \frac c m \right)^m}_{\text{This }\rightarrow\, e^{-c}}\ \underbrace{\left( 1 - \frac c m \right)^{-x}}_{\text{This }\rightarrow \,1} \\[15pt] & \to \frac{c^x e^{-c}}{x!} \qquad \text{as } m\to\infty. \end{align}

(Notice that the factor $\dfrac{c^x}{x!}$ does not change as $m\to\infty$, so its limit is not problematic (to say the least).)

In other words in the limit, $X$ has a Poisson distribution with expected value $c$.

Answer 3

There is one notable example of such a random collection of points in $\Bbb R^d$.

This example is the Poisson Point Process, also called the Spatial Poisson Process. This random set $N$ of points has the property that for any Borel set $B \subset \Bbb R^d$, we have that $N\cap B$ is distributed as a Poisson random variable with parameter $\lambda(B)$ where $\lambda$ is Lebesgue measure, and such that disjoint regions have independent numbers of points intersecting $N$.

If you're interested and want to know more about random point measures and Poisson Point Processes, you can see the Wiki article and its sources therein.

The construction of such a random cloud of points can be done in great generality, for instance on abstract measure spaces (not just $\Bbb R^n$ with Lebesgue measure). However, such a construction uses the Kolmogorov Extension Theorem and I won't be bothered to write out the details here (although you may want to try it as an exercise). In $\Bbb R^2$ there might be a much simpler construction without using such powerful abstract machinery.

See this PDF for a simple construction of the Poisson random measure using a method similar to yours. And see here for another (possibly unnecessarily convoluted, and rather poorly written) abstract construction using Kolmogorov Extension.