Poisson Random Measures and Intensity Measures

Random Measures as Counts on Sets

A Poisson process on a time interval records how many arrivals occur by each time. A Poisson random measure extends this idea: it assigns a random count to each measurable region of a space, which might represent time, location, or another measurable attribute. The measure controlling the expected counts is called the intensity measure. Its values determine the count law, while disjoint regions have independent counts.

Let \((E,\mathcal{E})\) be a measurable space, and let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space. A measure \(\nu\) on \((E,\mathcal{E})\) is sigma-finite if \(E\) is a countable union of measurable sets of finite \(\nu\)-measure. A random counting measure \(N\) assigns to each \(A\in\mathcal{E}\) a random variable \(N(A)\) taking values in \(\{0,1,2,\ldots,\infty\}\). For almost every outcome, \(N(\varnothing)=0\) and \(N\) is countably additive over disjoint measurable sets. The value \(N(A)\) is the number of points in \(A\), counted with multiplicity.

The intensity measure is the mean measure: for any set with finite intensity, \(\mathbb{E}[N(A)]=\nu(A)\). Indeed, the mean of a Poisson random variable with parameter \(\nu(A)\) is \(\nu(A)\). The definition gives the count law directly, but its independence clause initially concerns only sets of finite intensity. We will show that independence extends to disjoint measurable sets even when some have infinite intensity.

Sigma-finiteness is the condition that allows an infinite-intensity space to be handled in finite-intensity pieces. Choose measurable sets \(C_1,C_2,\ldots\) of finite intensity whose union is \(E\), and set \(E_n=C_1\cup\cdots\cup C_n\). Then \(E_n\) increases to \(E\), and \(\nu(E_n)<\infty\). Such a sequence is called a finite-measure exhaustion.

Count Laws and Disjoint-Set Independence

Proof. Take a finite-measure exhaustion \(E_n\) and define \(A_i^{(n)}=A_i\cap E_n\). Each \(A_i^{(n)}\) has finite intensity, and for fixed \(n\) these sets are pairwise disjoint. Therefore their counts are independent. For fixed nonnegative integers \(k_1,\ldots,k_m\), the events \(\{N(A_i^{(n)})\le k_i\text{ for every }i\}\) decrease as \(n\) increases. Their intersection is exactly \(\{N(A_i)\le k_i\text{ for every }i\}\): the counts increase to \(N(A_i)\), so the full count is at most \(k_i\) precisely when every truncated count is at most \(k_i\). Independence at each \(n\) gives \[ \mathbb{P}\bigl(N(A_i^{(n)})\le k_i\text{ for every }i\bigr) =\prod_{i=1}^m\mathbb{P}\bigl(N(A_i^{(n)})\le k_i\bigr). \] By continuity of probability for decreasing events, taking limits gives the same factorization for the full counts. These joint distribution functions determine the joint law of the integer-valued counts, so the counts are independent. If \(\nu(A_i)=\infty\), then \(\nu(A_i^{(n)})\) increases to infinity. For any fixed integer \(K\ge0\), \[ \mathbb{P}(N(A_i)\le K) \le \mathbb{P}(N(A_i^{(n)})\le K) =e^{-\nu(A_i^{(n)})}\sum_{r=0}^{K}\frac{\nu(A_i^{(n)})^r}{r!}\longrightarrow 0. \] The final limit is zero because the sum has finitely many terms, each a polynomial in \(\nu(A_i^{(n)})\) multiplied by its negative exponential. Thus \(\mathbb{P}(N(A_i)\le K)=0\) for every \(K\), and \(N(A_i)=\infty\) almost surely. \(\square\)

The decreasing events in this proof are events that each truncated count is at most its threshold. They are not events that each truncated count equals its full-count value: points outside \(E_n\) can still contribute to the full count. The threshold events work because an increasing sequence of counts has limit at most \(k\) exactly when every term is at most \(k\).

For disjoint finite-intensity sets, the joint count probabilities are products of Poisson probabilities. For instance, if \(\nu(A)=a<\infty\) and \(\nu(B)=b<\infty\), with \(A\cap B=\varnothing\), then \[ \mathbb{P}(N(A)=r,N(B)=s) =e^{-a}\frac{a^r}{r!}\,e^{-b}\frac{b^s}{s!}. \] The sum \(N(A)+N(B)\) is Poisson with parameter \(a+b\). To verify this, for \(k\ge0\) sum over all \(r+s=k\): \[ \mathbb{P}(N(A)+N(B)=k) =e^{-(a+b)}\sum_{r=0}^{k}\frac{a^r b^{k-r}}{r!(k-r)!} =e^{-(a+b)}\frac{(a+b)^k}{k!}, \] where the last equality is the binomial theorem. For more disjoint sets, the same argument gives the sum of their intensities as the Poisson parameter.

Finite and Infinite Intensity

A finite intensity measure means \(\nu(E)<\infty\). In that case, the total number of points \(N(E)\) is finite almost surely and has a Poisson distribution with parameter \(\nu(E)\). A sigma-finite measure can instead assign infinite intensity to some sets. Such a set does not have a Poisson distribution with an infinite parameter; its count is infinite almost surely, as proved above. Finite-intensity subsets within it still have ordinary Poisson counts.

A Poisson random measure can be constructed by splitting a sigma-finite space into disjoint sets \(C_r\) of finite measure. On a piece with \(\nu(C_r)=\lambda_r>0\), first draw a Poisson count \(Z_r\) with parameter \(\lambda_r\), then place \(Z_r\) independent points with distribution \(\nu(\,\cdot\,\cap C_r)/\lambda_r\). Use independent constructions on different pieces. To check the count law within one piece, partition it into disjoint measurable subsets with intensities \(\lambda_1,\ldots,\lambda_q\), including the leftover subset if necessary. Conditional on \(Z_r\), the subset counts have a multinomial distribution. Multiplying that multinomial probability by the Poisson probability of \(Z_r\) gives, for counts \(k_1,\ldots,k_q\), the product \(\prod_{\ell=1}^q e^{-\lambda_\ell}\lambda_\ell^{k_\ell}/k_\ell!\). Thus these subset counts are independent Poisson variables. Combining independent pieces gives the stated count laws on finite-intensity sets; increasing unions handle the sigma-finite space.

A Pitfall: Overlapping Sets Do Not Have Independent Counts

Independence is tied to disjointness, not merely to having different sets. Here is a concrete calculation. Take Lebesgue intensity on \([0,2]\), \(A=[0,1]\), and \(B=[\tfrac12,\tfrac32]\). Both sets have intensity \(1\), but their overlap \([\tfrac12,1]\) has intensity \(\tfrac12\). Decompose their union into \(L=[0,\tfrac12)\), \(C=[\tfrac12,1]\), and \(R=(1,\tfrac32]\). These pieces are disjoint and each has intensity \(\tfrac12\). The event that both \(N(A)\) and \(N(B)\) are zero is exactly the event that all three piece counts are zero. By independence on disjoint sets, its probability is \[ \mathbb{P}(N(A)=0,N(B)=0) =e^{-1/2}e^{-1/2}e^{-1/2}=e^{-3/2}. \] Treating the two overlapping counts as independent would instead give \(e^{-1}e^{-1}=e^{-2}\), which is incorrect. The shared region contributes to both counts, so the two counts are dependent.

For a quick check when applying the theory, identify the measurable sets first, compute their \(\nu\)-measures, and verify disjointness before multiplying probabilities. If a set has finite intensity, use its finite-parameter Poisson law. If it has infinite intensity, use the almost-surely infinite count result rather than substituting infinity into a Poisson probability formula.