Defining a uniform distribution of points in the plane 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: 


Take any $n \in \mathbb{N}_{&gt;c}\,\,.$ Partition each unit square of the plane into an $n \times n$ grid of square cells.
With probability $\lfrac{c}{n^2}$ put a point into each cell, distributed uniformly within the cell, independently of other cells.
Then every region comprising $n^2$ cells has expected number of points $c$. 
Also every pair of disjoint regions have independent numbers of points.
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?
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.
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) &amp; = \binom m x \left( \frac c m \right)^x \left( 1 - \frac c m \right)^{m-x} \\[10pt]
&amp; = \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]
&amp; = \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]
&amp; = \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]
&amp; \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 $c$.