Last time: We defined the Palm version \(\Phi^0\) of a stationary point process \(\Phi\) on \(\mathbb{R}^d\) with distribution \(\mathbb{P}_\Phi^0\) as described in the Refined Campbell’s Theorem.
Theorem (Refined Campbell’s Theorem). Let \(\Phi\) be a stationary point process on \(\mathbb{R}^{d}\) with intensity \(\lambda \in (0,\infty)\).
Then there is a unique probability measure \(P_{\Phi}^{0}\) on \(\mathcal{M}(\mathbb{R}^{d})\) such that, for all measurable
\(f : \mathbb{R}^{d} \times \mathcal{M}(\mathbb{R}^{d}) \rightarrow \mathbb{R}_{+}\), \[
\mathbb{E}\bigg[ \int_{\mathbb{R}^{d}} f(x, t_x \Phi)\, \Phi(dx) \bigg]
=
\lambda \int_{\mathbb{R}^{d}} \int_{\mathcal{M}(\mathbb{R}^{d})} f(x,\nu)\, P_{\Phi}^{0}(d\nu)\, dx.
\]
We also saw Slivnyak’s Theorem, which states that if \(\Phi\) is a stationary Poisson process on \(\mathbb{R}^{d}\), then \[ \mathbb{P}_{\Phi}^{0}(A) = \mathbb{P}\big(\Phi + \delta_{0} \in A\big), \] or equivalently, \[ \mathbb{P}_{\Phi}^{0,-}(A) := \mathbb{P}_{\Phi}^{0}(\Phi^{0} - \delta_{0} \in A) = \mathbb{P}(\Phi \in A). \]
Example (Matérn I Hard-Core Process). Let \(\Phi = \sum_{k \in \mathbb{N}} \delta_{X_k}\) be a stationary Poisson process on \(\mathbb{R}^{d}\) with intensity \(\lambda\).
Define
$$
\widetilde{\Phi}
=
\sum_{k \in \mathbb{N}}
\delta_{X_k}\,
\mathbf{1}_{\{\Phi(B_{d}(X_k,r)) = 1\}}.
$$
**Question:** What is the intensity measure of $\widetilde{\Phi}$?
For $B \in \mathcal{B}(\mathbb{R}^d)$,
$$
\mathbb{E}[\widetilde{\Phi}(B)]
= \mathbb{E}\!\left[
\sum_{k\in\mathbb{N}}
\mathbf{1}_{\{X_k \in B\}}
\mathbf{1}_{\{\Phi(B_d(X_k,r)) = 1\}}
\right].
$$
$$
= \mathbb{E}\!\left[
\int_{\mathbb{R}^d}
\mathbf{1}_{\{x \in B\}}
\mathbf{1}_{\{\Phi(B_d(x,r)) = 1\}}
\, \Phi(dx)
\right].
$$
Note that
$$
\mathbf{1}_{\{\Phi(B_d(x,r)) = 1\}}
= \mathbf{1}_{\{(t_x \Phi)(B_d(0,r)) = 1\}}.
$$
*Refined Campbell's theorem* gives
$$
= \lambda
\int_{\mathbb{R}^d}
\int_{\mathcal{M}(\mathbb{R}^d)}
\mathbf{1}_{\{x \in B\}}
\mathbf{1}_{\{\nu(B_d(0,r)) = 1\}}
\, \mathbb{P}^0_{\Phi}(d\nu)
\, dx.
$$
$$
= \lambda |B|
\mathbb{P}^0_{\Phi}\!\left(\Phi(B_d(0,r)) = 1\right).
$$
By Slivnyak's theorem,
$$
\mathbb{P}^0_{\Phi}\!\left(\Phi(B_d(0,r)) = 1\right)
= \mathbb{P}\!\left(\Phi(B_d(0,r)) = 0\right).
$$
$$
= e^{-\lambda |B_d(0,r)|}.
$$
Hence,
$$
\mathbb{E}[\widetilde{\Phi}(B)]
= \lambda |B| e^{-\lambda |B_d(0,r)|}.
\qed
$$Proof. Let \(\Phi\) be a stationary point process on \(\mathbb{R}^d\) with intensity \(\lambda\). For each \(A \in \mathcal{M}(\mathbb{R}^d)\), define a measure \(\mu_A\) on \(\mathbb{R}^d\) by \[ \mu_A(B) := \mathbb{E}\!\left[ \int_{\mathbb{R}^d} \mathbf{1}_B(x)\, \mathbf{1}_A(t_x \Phi)\, \Phi(dx) \right], \qquad B \in \mathcal{B}(\mathbb{R}^d). \]
We claim that $\mu_A$ is a translation-invariant measure.
Indeed, for all $y \in \mathbb{R}^d$,
$$
\mu_A(B+y)
= \mathbb{E}\!\left[
\int_{\mathbb{R}^d}
\mathbf{1}_B(x-y)\,
\mathbf{1}_A(t_x \Phi)\,
\Phi(dx)
\right].
$$
If $\Phi = \sum_k \delta_{X_k}$, then
$$
= \mathbb{E}\!\left[
\sum_k
\mathbf{1}_B(X_k - y)\,
\mathbf{1}_A(t_{X_k}\Phi)
\right].
$$
Let $Y_k := X_k - y$. Then
$$
= \mathbb{E}\!\left[
\sum_k
\mathbf{1}_B(Y_k)\,
\mathbf{1}_A(t_{Y_k + y}\Phi)
\right]
= \mathbb{E}\!\left[
\sum_k
\mathbf{1}_B(Y_k)\,
\mathbf{1}_A(t_{Y_k} t_y \Phi)
\right].
$$
Hence,
$$
= \mathbb{E}\!\left[
\int_{\mathbb{R}^d}
\mathbf{1}_B(x)\,
\mathbf{1}_A(t_x t_y \Phi)\,
t_y \Phi(dx)
\right].
$$
By stationarity of $\Phi$, $t_y \Phi \overset{d}{=} \Phi$, and therefore
$$
= \mathbb{E}\!\left[
\int_{\mathbb{R}^d}
\mathbf{1}_B(x)\,
\mathbf{1}_A(t_x \Phi)\,
\Phi(dx)
\right]
= \mu_A(B).
$$
That is, $\mu_A$ is a translation-invariant measure on $\mathbb{R}^d$.
By uniqueness of Lebesgue measure, there exists $\lambda_A \in [0,\infty)$ such that
$$
\mu_A(B) = \lambda_A |B|,
\qquad \forall B \in \mathcal{B}(\mathbb{R}^d).
$$
We can take $B = [0,1]^d$, so
$$
\lambda_A = \mu_A([0,1]^d).
$$
Also,
$$
\lambda_{\mathcal{M}(\mathbb{R}^d)}
= \mathbb{E}\!\left[\Phi([0,1]^d)\right]
= \lambda,
$$
so we can define a probability measure on $\mathcal{M}(\mathbb{R}^d)$ by
$$
\mathbb{P}^0_\Phi(A)
:= \frac{\lambda_A}{\lambda},
\qquad A \in \mathcal{M}(\mathbb{R}^d).
$$
We then have
$$
\mathbb{E}\!\left[
\int_{\mathbb{R}^d}
\mathbf{1}_B(x)\,
\mathbf{1}_A(t_x \Phi)\,
\Phi(dx)
\right]
= \lambda\, \mathbb{P}^0_\Phi(A)\, |B|,
$$
which is the claim of the theorem for
$f(x,\nu) = \mathbf{1}_{\{x \in B\}} \mathbf{1}_{\{\nu \in A\}}$,
and can be built up to non-negative measurable functions in the usual way.\(\square\)
We also have the ergodic interpretation of the Palm distribution. Let \[ B^d := B_d(0,1) \subset \mathbb{R}^d . \]
Then \[ \mathbb{P}_{\Phi}^{0}(A) = \lim_{r \to \infty} \frac{\displaystyle \int_{r B^d} \mathbf{1}_A\!\left(t_x \Phi\right)\,\Phi(dx)} {\displaystyle \Phi\!\left(r B^d\right)} \quad \text{a.s.} \]
for an ergodic stationary point process on \(\mathbb{R}^d\).
Ergodicity is a topic we won’t go into further in this class, but see Section 8.4 in [last2018lectures] or Chapter 8 in [baccelli2024random] for definitions and additional theory.