Factorial Moment Measures and Void Probabilities

From Point Configurations to Factorial Moments

A point process records random locations, but many questions concern groups of distinct points rather than individual counts. How many ordered pairs fall into specified regions? What is the probability that a region is empty? Factorial moment measures answer these questions by averaging over ordered tuples of distinct points. For a Poisson random measure, they have a particularly simple form: the $k$th factorial moment measure is the $k$-fold product of the intensity measure.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, let $(E,\mathcal{E})$ be a measurable space of locations, and let $N$ be a Poisson random measure on $E$ with a $\sigma$-finite intensity measure $\mu$. Thus, for disjoint measurable sets of finite intensity, their counts under $N$ are independent, and $N(A)$ has a Poisson distribution with mean $\mu(A)$. We allow $E$ to be a spatial domain such as $\mathbb{R}^2$. The results below concern measurable sets of finite intensity unless stated otherwise.

Definition: For an integer $k\ge1$, the $k$th factorial moment measure $\alpha^{(k)}$ is the measure on $E^k$ whose integral against a nonnegative measurable function $f$ is the expected sum of $f$ over all ordered $k$-tuples of distinct points counted by $N$. Distinctness refers to point occurrences: if multiple points occur at the same location, they are still distinct occurrences. In particular, $\alpha^{(2)}$ describes ordered pairs of distinct points.

For a set $A$, the number of ordered $r$-tuples of distinct points in $A$ is the falling factorial $(N(A))_r=N(A)(N(A)-1)\cdots(N(A)-r+1)$, with $(N(A))_0=1$. Hence $\alpha^{(r)}(A^r)=\mathbb{E}[(N(A))_r]$. This differs from $\mathbb{E}[N(A)^r]$: the latter also counts tuples that reuse the same point occurrence.

Theorem (Factorial moment measures of a Poisson random measure): For every integer $k\ge1$, $\alpha^{(k)}=\mu^{\otimes k}$, the $k$-fold product measure. Consequently, for measurable sets $A_1,\ldots,A_k$ of finite intensity, $$ \mathbb{E}\!\left[\sum_{\text{distinct }x_1,\ldots,x_k\text{ counted by }N} \prod_{\ell=1}^k\mathbf{1}_{A_\ell}(x_\ell)\right] =\prod_{\ell=1}^k\mu(A_\ell). $$

Proof. First, if $Z$ has the Poisson distribution with mean $a\ge0$, then for $r\ge1$, $$ \mathbb{E}[(Z)_r]=\sum_{n=r}^{\infty}\frac{n!}{(n-r)!}e^{-a}\frac{a^n}{n!} =e^{-a}a^r\sum_{m=0}^{\infty}\frac{a^m}{m!}=a^r. $$ For $a=0$, $Z=0$ almost surely and the same identity holds for $r\ge1$; for $r=0$, both sides are $1$.

Now take a finite collection $A_1,\ldots,A_k$. Partition their union into the finitely many disjoint measurable cells formed by membership or non-membership in each $A_\ell$, discarding empty cells. Assign each coordinate $x_\ell$ to one cell contained in $A_\ell$. For any fixed assignment of cells to the $k$ coordinates, suppose cell $C$ is assigned $r_C$ times. The number of ways to choose distinct point occurrences for those coordinates in $C$ is $(N(C))_{r_C}$. Counts in distinct cells are independent, and the Poisson factorial-moment calculation gives expectation $\prod_C\mu(C)^{r_C}$. Summing this expression over every allowable cell assignment expands to $\prod_{\ell=1}^k\mu(A_\ell)$. Thus the two measures agree on measurable rectangles $A_1\times\cdots\times A_k$, which determine the product measure; hence $\alpha^{(k)}=\mu^{\otimes k}$. $\square$

Joint Counts on Disjoint Regions

For disjoint measurable sets $A_1,\ldots,A_m$ of finite intensity, put $N_i=N(A_i)$ and $\mu_i=\mu(A_i)$. Applying the theorem to tuples whose coordinates lie in each $A_i$ gives, for nonnegative integers $r_1,\ldots,r_m$,

\mathbb{E}\!\left[\prod_{i=1}^m(N_i)_{r_i}\right] =\prod_{i=1}^m\mu_i^{r_i}.

These joint factorial moments determine the full joint count law. To see this, expand $z_i^{N_i}=(1+(z_i-1))^{N_i}$. For $0\le z_i\le1$, the expansion can be averaged term by term: the sum of the expectations of the absolute terms is at most $\exp(\sum_i\mu_i|z_i-1|)$, which is finite by the factorial-moment formula. Therefore the joint probability generating function is $$ \mathbb{E}\!\left[\prod_{i=1}^m z_i^{N_i}\right] =\sum_{r_1,\ldots,r_m\ge0}\prod_{i=1}^m \frac{\mu_i^{r_i}(z_i-1)^{r_i}}{r_i!} =\prod_{i=1}^m e^{\mu_i(z_i-1)}. $$ This is the product of the probability generating functions of Poisson counts with means $\mu_i$; thus the joint counts are independent Poisson random variables.

Worked Example: Joint Counts in Two Disjoint Rectangles

Let $N$ be a homogeneous Poisson random measure on $\mathbb{R}^2$ with intensity density $\lambda=\tfrac12$ per unit area. Take disjoint rectangles $A_1=[0,2]\times[0,1]$ and $A_2=[3,4]\times[0,3]$. Find the probability that $A_1$ contains exactly one point and $A_2$ contains exactly two.

The first rectangle has area $2\cdot1=2$, so $\mu_1=\tfrac12\cdot2=1$. The second has area $1\cdot3=3$, so $\mu_2=\tfrac12\cdot3=\tfrac32$. Their joint factorial moments are $\mathbb{E}[(N_1)_{r_1}(N_2)_{r_2}]=1^{r_1}(\tfrac32)^{r_2}$. By the joint count law just derived,

\mathbb{P}(N_1=1,N_2=2) =\left(e^{-1}\frac{1^1}{1!}\right) \left(e^{-3/2}\frac{(3/2)^2}{2!}\right) =\frac98e^{-5/2}.

Check: the means used in the two Poisson probabilities are exactly $1$ and $\tfrac32$, and the joint probability factors because the rectangles are disjoint. The associated mixed factorial moment for one point in $A_1$ and two distinct points in $A_2$ is $1\cdot(\tfrac32)^2=\tfrac94$, as the product-intensity formula requires.

Empty Regions and Void Probabilities

A void event is the event that a region contains no points. The factorial-moment identity gives its probability directly. The result includes the degenerate case $\mu(B)=0$: then $N(B)=0$ almost surely.

Theorem (Poisson void probability): If $B\in\mathcal{E}$ and $\mu(B)<\infty$, then $$ \mathbb{P}(N(B)=0)=e^{-\mu(B)}. $$

Proof. Write $Z=N(B)$ and $a=\mu(B)$. For every nonnegative integer $n$, the binomial identity gives $\mathbf{1}_{\{n=0\}}=\sum_{r=0}^{n}(-1)^r\binom{n}{r}$. Thus $\mathbf{1}_{\{Z=0\}}=\sum_{r=0}^{Z}(-1)^r(Z)_r/r!$. Taking expectations term by term is justified because the expected sum of the absolute terms is $$ \sum_{r=0}^{\infty}\frac{\mathbb{E}[(Z)_r]}{r!} =\sum_{r=0}^{\infty}\frac{a^r}{r!}=e^a<\infty. $$ The factorial-moment calculation also holds for $r=0$, so $$ \mathbb{P}(Z=0)=\sum_{r=0}^{\infty}\frac{(-1)^r a^r}{r!}=e^{-a}. $$ If $a=0$, this calculation gives $1$, consistent with $Z=0$ almost surely. $\square$

Worked Example: An Empty Disk

Suppose $N$ is a homogeneous Poisson random measure on $\mathbb{R}^2$ with intensity density $\lambda=\tfrac1\pi$. Find the probability that the disk of radius $2$ centered at the origin is empty, and the probability that it contains at least one point.

The disk has area $\pi(2)^2=4\pi$. Its intensity measure is $\mu(B)=\tfrac1\pi(4\pi)=4$. Therefore $$ \mathbb{P}(N(B)=0)=e^{-4},\qquad \mathbb{P}(N(B)\ge1)=1-e^{-4}. $$

Check: summing the Poisson count probabilities gives $\sum_{n=1}^{\infty}e^{-4}4^n/n!=e^{-4}(e^4-1)=1-e^{-4}$, the same probability of at least one point. The radius enters through the area: changing the radius changes $\mu(B)$, not the form of the void formula.

Second-Order Spatial Summaries

The second factorial moment measure gives expected ordered pair counts. In particular, for finite-intensity sets $A,B$, it gives the expected number of ordered pairs whose first point is in $A$ and second point is in $B$, counting distinct point occurrences. When $A$ and $B$ are disjoint, this is $\mathbb{E}[N(A)N(B)]=\mu(A)\mu(B)$, so the covariance of the two counts is zero. For a single set $A$, it gives $\mathbb{E}[N(A)(N(A)-1)]=\mu(A)^2$. Since $N(A)^2=N(A)(N(A)-1)+N(A)$, it follows that $\operatorname{Var}(N(A))=\mu(A)$.

If $\mu$ has a density $\lambda(x)$ with respect to spatial volume, then the second factorial moment measure has density $\lambda(x)\lambda(y)$ on pairs $(x,y)$. Where $\lambda(x)\lambda(y)>0$, the pair correlation—the second factorial density divided by the product of the first-order densities—is therefore $1$. This describes the absence of excess or reduced pair counts relative to the product-intensity benchmark. The factorial measure concerns distinct point occurrences; it does not count a point paired with itself.

Worked Example: Pair Counts Under a Spatially Varying Intensity

On $E=[0,1]^2$, let $N$ be a Poisson random measure with intensity density $\lambda(x,y)=1+x$ relative to area. Let $A=[0,\tfrac12)\times[0,1]$ and $C=[\tfrac12,1]\times[0,1]$. Find the expected ordered pair count with one point in each region, and the expected number of ordered pairs of distinct points in the whole square.

The mean count in $A$ is $$ \mu(A)=\int_0^1\int_0^{1/2}(1+x)\,dx\,dy =\frac12+\frac18=\frac58. $$ In $C$, it is $$ \mu(C)=\int_0^1\int_{1/2}^{1}(1+x)\,dx\,dy =\frac12+\frac38=\frac78. $$ The regions are disjoint, so the expected ordered pair count with one point in each is $\mu(A)\mu(C)=\tfrac58\cdot\tfrac78=\tfrac{35}{64}$. The whole square has mean $\mu(E)=\tfrac12+1=\tfrac32$, so its expected ordered pair count is $$ \mathbb{E}[N(E)(N(E)-1)]=\mu(E)^2=\left(\frac32\right)^2=\frac94. $$

Check: the two region means add to $\tfrac58+\tfrac78=\tfrac{12}{8}=\tfrac32$, the whole-square mean. Also, the second factorial density is $(1+x)(1+x')$ for locations $(x,y)$ and $(x',y')$; dividing by the product of the first-order densities gives pair correlation $1$.

A Common Counting Pitfall

A frequent error is to treat $N(A)^2$ as the number of ordered pairs of distinct points. It is not: $N(A)^2$ also includes the $N(A)$ diagonal pairs in which a point occurrence is paired with itself. For example, if $\mu(A)=\tfrac32$, the expected distinct ordered-pair count is $\mathbb{E}[N(A)(N(A)-1)]=\tfrac94$, whereas $$ \mathbb{E}[N(A)^2] =\mathbb{E}[N(A)(N(A)-1)]+\mathbb{E}[N(A)] =\frac94+\frac32=\frac{15}{4}. $$ Using $\tfrac{15}{4}$ as the factorial pair count would overcount by $\tfrac32$, the expected number of points paired with themselves. For pair-based spatial summaries, first decide whether the statistic includes such diagonal terms; factorial moment measures exclude them by definition.

Factorial moment measures are useful because they convert configuration-level questions into integrals against $\mu^{\otimes k}$. Joint counts on disjoint regions, empty-region probabilities, and second-order summaries are all consequences of that identity. The finite-intensity condition matters: it ensures the count and the void probability formulas above are defined in the stated form.

Check Your Understanding

For each question, distinguish counts of point occurrences from counts of distinct ordered tuples.

If $\mu(A)=2$, what are $\mathbb{E}[N(A)(N(A)-1)]$ and $\operatorname{Var}(N(A))$?
Two disjoint regions have intensities $\tfrac12$ and $2$. What is the expected number of ordered pairs with the first point in the first region and the second in the second?
A finite-intensity region $B$ has $\mu(B)=0$. What is its void probability, and why is this not an exceptional case of the formula?
For a spatial Poisson random measure with intensity density $\lambda(x)$, write the second factorial moment density and the pair correlation where the density is positive.
Explain why $\mathbb{E}[N(A)^2]$ is not the second factorial moment, and state the term that separates the two.