Lecture 8: Markings and Stationairy

Let $K: E \times \mathcal{G} \rightarrow [0,1]$.
That is, for all $x \in E$, $K(x, \cdot)$ is a probability measure on $(G, \mathcal{G})$,
and for all $B \in \mathcal{G}$, the mapping $x \mapsto K(x, B)$ is measurable.

Definition. Let $\Phi = \sum_{k=1}^{N} \delta_{X_k}$ be a point process on $E$.
A $K$-marking of $\Phi$ is the point process $\tilde{\Phi} = \sum_{k=1}^{N} \delta_{(X_k, Z_k)}$ on $E \times G$,
where, conditioned on $\Phi$, $\{Z_k\}_{k=1}^{N}$ is a collection of independent random elements of $G$ such that
$\mathbb{P}[Z_k \in \cdot \mid \Phi] = K(X_k, \cdot)$.

If $K(x, \cdot) = \mathbb{Q}$ for all $x \in E$, then $\tilde{\Phi}$ is called an independent $\mathbb{Q}$-marking or an i.i.d. marked point process.

Proposition. Let $\tilde{\Phi}$ be a $K$-marking of a point process $\Phi$ with intensity measure $\Lambda$. Then:

[(i)] The intensity measure $\tilde{\Lambda}$ of $\tilde{\Phi}$ is given by
$\tilde{\Lambda}(B) = \int_{E} \int_{G} \mathbf{1}_{B}(x, y) \, K(x, dy) \, \Lambda(dx)$,
for $B \in \mathcal{B}(E) \otimes \mathcal{G}$.
[(ii)] For all measurable $f: E \times G \rightarrow \mathbb{R}_{+}$,
$L_{\tilde{\Phi}}(f) = L_{\Phi}(\tilde{f})$,
where $\tilde{f}(x) = -\log \left[ \int_{G} e^{-f(x, z)} K(x, dz) \right]$.

Theorem (Marking Theorem). Let $\tilde{\Phi}$ be a $K$-marking of a Poisson process on $E$ with intensity measure $\Lambda$.
Then, $\tilde{\Phi}$ is a Poisson process on $E \times G$ with intensity measure
$\tilde{\Lambda}(dx \, dz) = \Lambda(dx) \, K(x, dz)$ (This is the same as $\tilde{\Lambda}$ from (i) of the previous proposition.).

Proof. Use the Laplace functional characterization of Poisson processes and the previous proposition (ii).

$\square$

Let $E = \mathbb{R}^d$ and $(G, \mathcal{G}) = (\mathcal{M}(\mathbb{R}^d), \mathcal{M}(\mathcal{B}(\mathbb{R}^d)))$,
the measurable space of counting measures on $\mathbb{R}^d$.

Let $K: \mathbb{R}^d \times \mathcal{M}(\mathbb{R}^d) \rightarrow [0,1]$ be a probability kernel from $\mathbb{R}^d$ to $\mathcal{M}(\mathbb{R}^d)$.
That is, for all $x \in \mathbb{R}^d$, $K(x, \cdot)$ is a probability measure on $\mathcal{M}(\mathbb{R}^d)$, i.e., the distribution of a point process.

Then, let $\tilde{\Phi} = \sum_{k=1}^{N} \delta_{(X_k, \Psi_k)}$ be a $K$-marking of a Poisson process $\Phi = \sum_{k=1}^{N} \delta_{X_k}$ on $\mathbb{R}^d$.
Here $\{\Psi_k\}_{k=1}^{N}$ are point processes on $\mathbb{R}^d$ that, conditioned on $\Phi$, are independent, and
$\mathbb{P}[\Psi_k \in \cdot \mid \Phi] = K(X_k, \cdot)$.

Then, define $\hat{\Phi} = \sum_{k=1}^{N} \sum_{y \in \Psi_k} \delta_{X_k + y}$.
Note that $\hat{\Phi}(B) = \sum_{k=1}^{N} \Psi_k(B - X_k)$ for $B \in \mathcal{B}(\mathbb{R}^d)$.
Hence, $\hat{\Phi}$ is a Poisson cluster process.

Example (Special Case: Matérn Cluster Process). Let $\{\Psi_k\}_{k=1}^{N}$ be i.i.d. homogeneous Poisson processes in $B_d(0, R) = \{x \in \mathbb{R}^d : \|x\| \le R\}$.
Then each cluster $\Psi_k$ is centered at $X_k$, and
$B_d(X_k, R)$ denotes the cluster region associated with the parent point $X_k$.

Stationarity of Point Processes on $

Define shift maps $\{t_x\}_{x \in \mathbb{R}^d}$ where $t_x : \mathcal{M}(\mathbb{R}^d) \rightarrow \mathcal{M}(\mathbb{R}^d)$
is defined by $t_x \gamma(B) = \gamma(B + x)$.

Note that $t_x \delta_x(B) = \delta_{x}(B+x) = \delta_0(B)$, thus $t_x \delta_x = \delta_0$.

Definition. A point process $\Phi$ on $\mathbb{R}^d$ is stationary if
$t_x \Phi \overset{d}{=} \Phi$ for all $x \in \mathbb{R}^d$.

Proposition. Let $\Phi$ be a stationary point process on $\mathbb{R}^d$ with intensity measure $\Lambda$ such that
$\lambda := \Lambda([0,1]^d) < \infty$.
Then $\Lambda(dx) = \lambda \, dx$, and $\lambda$ is the intensity of $\Phi$.

Proof. Stationarity of $\Phi$ implies that for all $x \in \mathbb{R}^d$,
$\Lambda(B + x) = \mathbb{E}[\Phi(B + x)] = \mathbb{E}[\Phi(B)] = \Lambda(B)$.
That is, $\Lambda$ is a translation-invariant measure on $\mathbb{R}^d$ such that
$\Lambda([0,1]^d) = \lambda < \infty$.
By uniqueness of the Lebesgue measure, $\Lambda(dx) = \lambda \, dx$.

$\square$

Proposition. A Poisson process $\Phi$ on $\mathbb{R}^d$ is stationary
if and only if $\Lambda(dx) = \lambda \, dx$ for some $\lambda \in \mathbb{R}_{+}$.