From Spatial Points to a Random Field
A spatial Poisson model describes random locations, but many applications are interested in the combined effect of all those locations at an observation point. Each point may contribute a signal, a dose, or a local influence that decreases with distance. Adding these contributions produces a shot-noise field. Its mean and variance summarize its scale and variability, while its Laplace transform can describe its distribution even when its mean is infinite.
Let \((\Omega,\mathcal{F},\mathbb{P})\) be a probability space, let \(E\) be a spatial domain with a sigma-finite measure \(\Lambda\), and let \(N\) be a Poisson random measure on \(E\) with intensity measure \(\Lambda\). Thus, for measurable \(A\subseteq E\) with \(\Lambda(A)<\infty\), \(N(A)\) has a Poisson distribution with mean \(\Lambda(A)\), and counts on disjoint sets are independent. For example, for homogeneous Poisson points in \(\mathbb{R}^d\) with intensity \(\lambda>0\), \(\Lambda(du)=\lambda\,du\), where \(du\) is \(d\)-dimensional volume.
The response may vanish outside a finite region, decay with distance, or include a random mark such as signal strength. A sum over infinitely many points is defined by increasing finite or simple-function approximations. The conditions below say when its expectation and variance are finite; they are not needed merely to define the nonnegative extended sum.
Mean, Variance, and Laplace Transform
Proof. First take \(h=\sum_{\ell=1}^r c_\ell\mathbf{1}_{A_\ell}\), where \(c_\ell\ge0\) and the disjoint measurable sets \(A_\ell\) have finite intensity. Then \(X=\sum_\ell c_\ell N(A_\ell)\), a sum of independent variables. A Poisson variable with mean \(a\) has mean and variance \(a\), and its Laplace transform is \(\mathbb{E}e^{-s cN}=\sum_{k\ge0}e^{-sck}e^{-a}a^k/k!=\exp\{a(e^{-sc}-1)\}\). Independence therefore gives \(\mathbb{E}X=\sum_\ell c_\ell\Lambda(A_\ell)=\int h\,d\Lambda\), \(\operatorname{Var}(X)=\sum_\ell c_\ell^2\Lambda(A_\ell)=\int h^2\,d\Lambda\), and \(\mathbb{E}e^{-sX}=\exp\{-\sum_\ell(1-e^{-sc_\ell})\Lambda(A_\ell)\}\). For general nonnegative measurable \(h\), sigma-finiteness permits an increasing sequence of nonnegative simple functions \(h_n\uparrow h\), supported on sets of finite intensity. Their integrals \(X_n=\int h_n\,dN\) increase to \(X\). Monotone convergence gives \(\mathbb{E}X=\lim_n\mathbb{E}X_n=\int h\,d\Lambda\), and also gives the Laplace formula, since \(e^{-sX_n}\downarrow e^{-sX}\) and \(1-e^{-sh_n}\uparrow1-e^{-sh}\). If \(\int h\,d\Lambda<\infty\), then \(\mathbb{P}(X=\infty)=0\), since a nonnegative random variable that is infinite with positive probability has infinite expectation. If also \(\int h^2\,d\Lambda<\infty\), monotone convergence applied to \(X_n^2\uparrow X^2\) gives \(\mathbb{E}X^2=\lim_n\{(\int h_n\,d\Lambda)^2+\int h_n^2\,d\Lambda\}=(\int h\,d\Lambda)^2+\int h^2\,d\Lambda\). Subtracting \((\mathbb{E}X)^2\) proves the variance formula. \(\square\)
For two response functions \(h_x\) and \(h_y\) satisfying the finite second-moment conditions, the same calculation gives \(\operatorname{Cov}(X(x),X(y))=\int_E h_x(u)h_y(u)\,\Lambda(du)\). Indeed, apply the variance formula to \(h_x+h_y\) and subtract the two individual variances; expanding \((h_x+h_y)^2\) leaves twice the integral of the product, which equals twice the covariance by the definition of variance.
For a homogeneous Poisson process in \(\mathbb{R}^d\), these results take the integral forms \(\mathbb{E}X(x)=\lambda\int h_x(u)\,du\), \(\operatorname{Var}(X(x))=\lambda\int h_x(u)^2\,du\), and \(\mathbb{E}e^{-sX(x)}=\exp\{-\lambda\int(1-e^{-s h_x(u)})\,du\}\). The mean and variance require their respective integrals to be finite. The Laplace transform formula remains meaningful under the weaker condition that its exponent integral is finite, and still holds with transform zero if that integral diverges.
Worked Calculations
Worked Example: A Finite-Region Shot Noise
On the real line, let \(N\) be a homogeneous Poisson random measure with intensity \(\lambda=\tfrac32\) per unit length. At a fixed observation location, suppose each point in the interval \(A=[0,2]\) contributes \(2\), and points outside \(A\) contribute zero. Find the mean, variance, and Laplace transform of \(X=2N(A)\).
The length of \(A\) is \(2\), so \(\Lambda(A)=\tfrac32\cdot2=3\). Thus \(N(A)\) is Poisson with mean \(3\). The response is \(h(u)=2\mathbf{1}_{[0,2]}(u)\), so the integrals give \(\mathbb{E}X=2\cdot3=6\) and \(\operatorname{Var}(X)=2^2\cdot3=12\). For \(s\ge0\), the Laplace transform is \(\exp\{-3(1-e^{-2s})\}\).
Check: Directly, \(X=2N(A)\). A Poisson variable of mean \(3\) has variance \(3\), so scaling by \(2\) gives variance \(2^2\cdot3=12\), and scaling its mean gives \(2\cdot3=6\). Its probability generating calculation gives \(\mathbb{E}e^{-sX}=\exp\{3(e^{-2s}-1)\}\), which equals the displayed transform.
Worked Example: Probability of Coverage at a Location
In the plane, let the points form a homogeneous Poisson process with intensity \(\lambda=\tfrac1\pi\) per unit area. Each point covers every location within distance \(R=2\) of itself. At the origin, define \(X\) to be the number of points that cover it. Find the probability that the origin is covered by at least one point.
A point covers the origin exactly when it lies in the disk of radius \(2\) centered at the origin. The disk has area \(\pi R^2=\pi\cdot2^2=4\pi\). Therefore its point count is Poisson with mean \(\lambda\cdot4\pi=(1/\pi)\cdot4\pi=4\). Coverage means \(X\ge1\), so the probability is one minus the probability of no points: \(\mathbb{P}(X\ge1)=1-\mathbb{P}(X=0)=1-e^{-4}\).
Check: The response is \(h(u)=\mathbf{1}_{\{|u|\le2\}}\), so \(X=\int h\,dN=N(\{|u|\le2\})\). The Laplace transform is \(\exp\{-4(1-e^{-s})\}\), the transform of a Poisson variable with mean \(4\). Its probability of zero is \(e^{-4}\), confirming the coverage calculation. The boundary circle does not affect the answer because it has area zero.
Worked Example: Interference with Inverse-Fourth-Power Path Loss
Let planar transmitters form a homogeneous Poisson process of intensity \(\lambda=\tfrac2\pi\). Each has an independent exponential mark \(H\) with mean \(1\), so \(\mathbb{P}(H>h)=e^{-h}\) for \(h\ge0\). At the origin, a point at \(u\ne0\) contributes \(H|u|^{-4}\). Find the Laplace transform of total interference \(I\), and determine whether its mean is finite.
Use the marked space with intensity measure \(\tfrac2\pi\,du\,e^{-m}dm\), for \(m\ge0\), and response \(h(u,m)=m|u|^{-4}\). At a fixed radius \(r>0\), integrating over the mark gives \(\int_0^\infty(1-e^{-s m r^{-4}})e^{-m}dm=1-1/(1+s r^{-4})=s/(r^4+s)\). Polar coordinates therefore give the Laplace exponent
where \(v=r^2\), so \(r\,dr=dv/2\). In the last integral, setting \(v=\sqrt{s}\,w\) gives \(\int_0^\infty s/(v^2+s)\,dv=\sqrt{s}\int_0^\infty 1/(1+w^2)\,dw=\pi\sqrt{s}/2\). Thus \(\mathbb{E}e^{-sI}=e^{-\pi\sqrt{s}}\) for \(s>0\), while at \(s=0\) the transform is \(1\).
The mean is infinite: the mean response from marks is \(\mathbb{E}[H]r^{-4}=r^{-4}\), and its spatial integral includes \(2\pi\lambda\int_0^1 r^{-4}r\,dr=2\pi\lambda\int_0^1r^{-3}dr=\infty\). Hence the finite-mean condition fails, but the Laplace transform still exists. As a check on the transform calculation, the exponent integral is finite for every \(s>0\): its radial integrand is \(4sr/(r^4+s)\), which is bounded by \(4r\) near zero up to a constant depending on \(s\), and by \(4s/r^3\) for \(r\ge1\). Both bounds are integrable on their respective ranges.
Interpretation and a Common Mistake
The coverage example is a special case of a general void-probability calculation: if response is one on a region \(A\) and zero elsewhere, then \(X=N(A)\), and the probability of at least one contributing point is \(1-e^{-\Lambda(A)}\). More general shot-noise coverage events, such as \(\mathbb{P}(X(x)\ge c)\) for a nonzero threshold \(c\), require the distribution of the sum, not just its mean. The Laplace transform is often useful for that distributional analysis.
Infinite mean does not by itself imply that the Laplace transform is unavailable. In the interference example, points arbitrarily near the origin make the mean response integral diverge, while the bounded quantity \(1-e^{-s h}\) yields a finite exponent. This distinction is important for singular path-loss models. In applications, a minimum distance, bounded path-loss function, or exclusion region may make the mean finite, but such a modification changes the model and must be stated.
Common mistake. For the finite-region example, an incorrect variance calculation is \(\operatorname{Var}(X)=(\mathbb{E}X)^2=6^2=36\). This confuses the square of the mean with the variance. The correct calculation uses the Poisson count: \(X=2N(A)\), \(\operatorname{Var}(N(A))=3\), and therefore \(\operatorname{Var}(X)=2^2\cdot3=12\). Substitution into the second-moment identity confirms it: \(\mathbb{E}X^2=12+6^2=48\), so \(\mathbb{E}X^2-(\mathbb{E}X)^2=48-36=12\).
Check Your Understanding
For each question, state any integrability condition you use and check the result against the defining Poisson count or integral.
- For a Poisson random measure with intensity \(\lambda\,du\), take \(h(u)=a\mathbf{1}_A(u)\), where \(a>0\) and \(A\) has finite volume. Find the mean, variance, and Laplace transform in terms of \(\lambda\), \(a\), and the volume of \(A\).
- Why does a finite value of \(\int h\,d\Lambda\) imply that the nonnegative shot noise is finite almost surely?
- For the disk-coverage model, how does the coverage probability change if the criterion is at least two covering points rather than at least one?
- In the planar interference example, which part of the radial integral causes the mean to diverge? Why can the Laplace exponent nevertheless be finite?
- For two observation locations with response functions \(h_x\) and \(h_y\), what integral gives their covariance, and what finiteness condition makes that formula applicable?