Laplace Functionals and Campbell's Formula

From Counts to Functionals

A Poisson random measure describes random points through the numbers that fall in measurable sets. Many quantities of interest, however, depend on the points through a function: a total weight, a spatial cost, or a product of contributions. Two formulas connect these point-based quantities to the intensity measure. The Laplace functional gives the transform of a sum over points, while Campbell’s formula gives its expectation.

Work on a probability space \((\Omega,\mathcal{F},\mathbb{P})\), and let \((E,\mathcal{E})\) be a measurable space with a sigma-finite measure \(\nu\). A Poisson random measure \(N\) on \(E\) with intensity measure \(\nu\) has the following properties: for every \(A\in\mathcal{E}\) with \(\nu(A)<\infty\), \(N(A)\) has a Poisson distribution with mean \(\nu(A)\); and counts on any finite collection of pairwise disjoint measurable sets of finite measure are independent. Here \(N(A)\) counts points in \(A\), including multiplicities if points coincide.

For a nonnegative measurable function \(f:E\to[0,\infty]\), its integral against \(N\), written \(\int_E f(x)\,N(dx)\), is the sum of \(f\) over the points, with multiplicities. It is defined first for indicator functions, then for nonnegative simple functions, and then by increasing approximation for all nonnegative measurable functions. The integral may be infinite. An integral against \(\nu\) is defined in the same way.

The Laplace functional describes the distributional transform of a point sum, not just its mean. Campbell’s formula extracts the mean directly. Both apply to nonnegative measurable functions without requiring boundedness or finite support; when a right-hand side is infinite, the corresponding expectation is infinite.

Why the Formulas Hold

Begin with a measurable set \(A\) of finite intensity \(m=\nu(A)\). By the Poisson distribution, for \(a\ge0\),

This calculation also covers \(m=0\): then \(N(A)=0\) almost surely, and both sides equal \(1\). For the expectation, the Poisson mean gives \(\mathbb{E}N(A)=\nu(A)\), again including the zero-measure case.

Proof. First take a nonnegative simple function \(f=\sum_{\ell=1}^r a_\ell\mathbf{1}_{A_\ell}\), where the \(A_\ell\) are pairwise disjoint measurable sets of finite \(\nu\)-measure and \(a_\ell\ge0\). The random integral equals \(\sum_{\ell=1}^r a_\ell N(A_\ell)\). Independence of these counts factors the expectation of its exponential into \(r\) terms. Applying the Poisson calculation above to each term gives $$ \mathbb{E}\exp\!\left(-\int_E f\,dN\right) =\prod_{\ell=1}^r\exp\!\bigl(-\nu(A_\ell)(1-e^{-a_\ell})\bigr) =\exp\!\left(-\int_E(1-e^{-f})\,d\nu\right). $$ For a simple nonnegative \(g=\sum_{\ell=1}^r b_\ell\mathbf{1}_{A_\ell}\) on such sets, linearity of expectation and \(\mathbb{E}N(A_\ell)=\nu(A_\ell)\) give $$ \mathbb{E}\int_E g\,dN =\sum_{\ell=1}^r b_\ell\nu(A_\ell) =\int_E g\,d\nu. $$ To pass to general nonnegative measurable functions, use sigma-finiteness to choose measurable sets \(E_m\) of finite measure increasing to \(E\). Each \(f\) can be approximated from below by nonnegative simple functions supported on these sets; choose \(f_m\uparrow f\), and likewise \(g_m\uparrow g\). Then \(\int f_m\,dN\uparrow\int f\,dN\) and \(\int g_m\,dN\uparrow\int g\,dN\). The exponentials \(e^{-\int f_m\,dN}\) decrease to \(e^{-\int f\,dN}\), and are bounded by \(1\), so bounded convergence passes their expectations to the limit. Also \(1-e^{-f_m}\uparrow1-e^{-f}\), so monotone convergence passes the intensity integrals to the limit. Monotone convergence applied to \(g_m\) passes both sides of Campbell’s formula to the limit. Thus both identities hold, including when the nonnegative integrals are infinite. \(\square\)

A useful consequence is that if \(\int_E g\,d\nu<\infty\), then \(\int_E g\,dN\) is finite almost surely. Indeed, a nonnegative random variable that equals infinity with positive probability has infinite expectation. The formulas also specify what to do when integrability fails: Campbell’s identity remains valid, but may give an infinite expectation.

Worked Calculations

Where the Assumptions Matter

The independence condition concerns disjoint sets. Counts on overlapping sets are generally dependent, so multiplying their separate Laplace transforms can give a wrong answer. For a concrete calculation, suppose \(\nu(A\cap B)=\tfrac12\) and \(\nu(A\setminus B)=\nu(B\setminus A)=\tfrac12\). Then \(\nu(A)=\nu(B)=1\). For \(Y=(\log2)\{N(A)+N(B)\}\), incorrectly treating \(N(A)\) and \(N(B)\) as independent would give \(\mathbb{E}e^{-Y}=e^{-1}\cdot e^{-1}=e^{-2}\). The correction is to split the union into the three disjoint regions. On \(A\cap B\), the function value is \(2\log2\), and on each exclusive region it is \(\log2\). Thus the correct exponent in the Laplace functional is

Therefore the correct transform is \(e^{-7/8}\), not \(e^{-2}\). The overlap contributes to both counts, which is why the independent-count shortcut fails.

Another point to keep distinct is what each formula returns. The Laplace functional requires a nonnegative \(f\); Campbell’s formula requires a nonnegative \(g\). If \(\int g\,d\nu\) is finite, Campbell’s formula gives a finite expected sum. If it is infinite, the formula still holds in the extended sense and does not promise a finite sum. For functions that take both positive and negative values, one must separately control their positive and negative parts before using an expectation identity; the nonnegative version alone does not justify subtracting two infinite quantities.

These formulas are useful whenever a model records random points but the quantity of interest is built by adding contributions point by point. A simple function turns the calculation into a finite collection of Poisson counts. General nonnegative functions follow by increasing approximation, so spatial weights and other measurable contributions can be handled without requiring the points to lie in a finite set.