MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider a planar point process $X$ and call $N_A = \text{Card}\big( X \cap A\big)$ the number of points inside the subset $A \subset \mathbb{R}^2$. If one knows the law of $(N_{A_1}, \ldots, N_{A_r})$ for any sets $A_1, \ldots, A_r$, then the process is completely characterized. I recently learned that it in fact suffices to know $f(A)=P(N_A=0)$ (called the void-probability function) for any set $A$ in order to completely characterize the law of $X$.

Intuitively, I do not understand why such a result is true. Indeed, the knowledge of the function $f$ brings some information in the correlation structure of the process $X$: nevertheless, I still fail to understand how the function $f$ can encode the whole correlation structure of the process. Any thoughts on this ?

share|cite|improve this question
up vote 7 down vote accepted

This is only true for simple point processes (no duplicate points).

By the inclusion-exclusion principle, $f$ determines the joint distribution of several (disjoint) sets being empty or occupied. If the process is simple this allows recovering the law of $(N_{A_1},\dots,N_{A_r})$ as a limit over finer partitions.

share|cite|improve this answer
Hi Omer! Welcome to MO. – Louigi Addario-Berry Sep 21 '10 at 16:08
indeed! Thank you very much. – Alekk Sep 23 '10 at 9:54
Very nice. I'd known only one really nice example of inclusion-exclusion that could be presented in an introductory probability course with calculus as a prerequisite: the probability that a random permutation is a derangement approaches $1/e$ as the number of things being permuted grows. – Michael Hardy Dec 8 '10 at 16:25

To add to Omer's concise explanation, the general result is known as Choquet's capacity theorem. It says that the void probabilities characterise any random closed set. Simple point processes are an example of random closed sets.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.