Palm formula for Poisson processes

Question

I'm reading some lecture notes and the following gets stated without proof

If $N \sim Poisson(\lambda)$ and $f$ and $G$ are functions of $x \in S$ and of $(x, N)$ respectively, then $$ \mathbb E \int f(x) G(x, N) N(dx) = \int \mathbb E[G(x, \delta_x + N)] f(x) \lambda(x) $$

I think this has something to do with Campbell's theorem, but I cannot find any proof of this. How can one show this? Is it valid for other point processes with intesity functions?

Answer 1

Note first, by absorbing $f$ into $G$ by setting $G'(x,N)=f(x)G(x,N)$, it suffices to assume $f \equiv 1$. Then the resulting equation

$$ \mathbb{E} \biggl[ \int_S G(x, N) \, N(\mathrm{d}x) \biggr] = \int_S \mathbb{E}[G(x, N+\delta_x)] \, \lambda(\mathrm{d}x) \tag{*} $$

is also called the Mecke equation. It is well-known that a random point process $N$ is Poisson if and only if this equation holds for any non-negative $G$. Its proof may be found in the book Lectures on the Poisson Process.

We first gain insight on this result from:

Observation. Let $N$ be a $\mathbb{N}_0$-valued random variable. Then $N$ has a Poisson distribution with parameter $\lambda$ if and only if $$ \mathbb{E}[N G(N)] = \lambda \mathbb{E}[G(N+1)] \tag{1} $$ for any non-negative function $G : \mathbb{N}_0 \to [0, \infty)$.

Its proof is easy, so I will skip this. But the general case is not so different from this one. Indeed, let $N$ be a Poisson point process. Assume first that $\lambda$ is finite and $G$ takes the form

$$ G(x, N) = \mathbf{1}_{B_1}(x) \prod_{i=1}^{n} \mathbf{1}_{\{N(B_i) = k_i\}} \tag{2} $$

for disjoint measurable sets $B_1,\dots,B_n$ and $k_1,\dots,k_n \in \mathbb{N}_0$. Then by the independence of $N(B_i)$'s,

\begin{align*} \mathbb{E} \biggl[ \int_S G(x, N) \, N(\mathrm{d}x) \biggr] &= \mathbb{E} \biggl[ \int_{B_1} \prod_{i=1}^{n} \mathbf{1}_{\{N(B_i) = k_i\}} \, N(\mathrm{d}x) \biggr] \\ &= \mathbb{E} \biggl[ k_1 \prod_{i=1}^{n} \mathbf{1}_{\{N(B_i) = k_i\}} \biggr] \\ &= k_1 \prod_{i=1}^{n} \mathbb{P}(N(B_i) = k_i). \tag{3} \end{align*}

On the other hand,

\begin{align*} &\int_S \mathbb{E}[G(x, N+\delta_x)] \, \lambda(\mathrm{d}x) \\ &= \int_{B_1} \mathbb{E}\biggl[ \prod_{i=1}^{n} \mathbf{1}_{\{N(B_i)+\delta_x(B_i) = k_i\}} \biggr] \, \lambda(\mathrm{d}x) \\ &= \int_{B_1} \prod_{i=1}^{n} \mathbb{P}(N(B_i)+\delta_x(B_i) = k_i) \, \lambda(\mathrm{d}x)\\ &= \int_{B_1} \mathbb{P}(N(B_1)+1=k_1) \prod_{i=2}^{n} \mathbb{P}(N(B_i) = k_i) \, \lambda(\mathrm{d}x)\\ &= \lambda(B_1)\mathbb{P}(N(B_1)+1=k_1) \prod_{i=2}^{n} \mathbb{P}(N(B_i) = k_i). \tag{4} \end{align*}

Now by noting that $N(B_1)\sim\text{Poisson}(\lambda(B_1))$, $\text{(3)}$ and $\text{(4})$ are equal by $\text{(1)}$. Therefore $\text{(*)}$ holds when $G$ takes the form $\text{(2)}$. The general case follows from a routine argument by approximating $G$ by the sum of functions of the form $\text{(2)}$ and writing $\lambda$ as an increasing limit of finite measures.

Answer 2

See Mathew Penrose's chapter on random geometric graphs in Random Graphs, Geometry and Asymptotic Structure (LMS Student Text) for a discussion and proof of the formula.