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Lecture 17: Set Processes

Introduction

Recall from Last Lecture:

\end{itemize}

Example (Boolean Model). Let $ = {n} {X_n} $ be a stationary Poisson process on \(\mathbb{R}^d\) with intensity \(\lambda>0\).

Fix \(r>0\) and define the union of balls $ Z_:= _{n} B(X_n,r). $ Then \(Z_\Phi\) is a RACS.

Volume fraction:

\[\begin{align*} p &:= \mathbb{P}(0\in Z_\Phi) = \mathbb{P}(\Phi(B(0,r))>0) = 1 - e^{-\lambda |B(0,r)|}. \end{align*}\]

Also, \[ \int_{\mathbb{R}^d} \mathbb{P}(x\in Z_\Phi)\, dx = \mathbb{E}\left[\int_{\mathbb{R}^d} \mathbf{1}_{\{x\in Z_\Phi\}}\, dx\right] = \mathbb{E}\big[ \mathrm{vol}(Z_\Phi)\big]. \]

Locally Finite Measures on $

Proposition. A measure \(\mu\) on \((\mathcal{F}', \mathcal{B}(\mathcal{F}'))\) is locally finite if and only if \[ \mu(\mathcal{F}_C) < \infty \qquad \forall\, C \in \mathcal{C}. \]

Proof. (1) Show that \(\mathcal{F}_C\) is compact in \(\mathcal{F}'\) for all \(C \in \mathcal{C}\).

Recall that \(\mathcal{F}\) is a compact space.

We observe that \(\mathcal{F}_C\) is closed in \(\mathcal{F}\), since $ _C = ^{,C}, $ i.e., it is the complement of an open set.
Since \(\mathcal{F}\) is compact, any closed subset is compact, hence \(\mathcal{F}_C\) is compact.

Note that \(\varnothing \in \mathcal{F}_C\) since \(\varnothing \cap C = \varnothing\).
Thus \(\mathcal{F}_C\) is compact in \(\mathcal{F}'\).

(2) For any compact subset \(\mathcal{H} \subset \mathcal{F}'\), show that there exists a compact \(C \in \mathcal{C}\) such that \(\mathcal{H} \subset \mathcal{F}_C\).

First, note that the family \[ \{ \mathcal{F}^{C} : C \in \mathcal{C} \} \] forms a basis of neighborhoods of \(\varnothing\) in \(\mathcal{F}\).
That is, for any open neighborhood \(\mathcal{U}\) of \(\varnothing\) (an open set containing \(\varnothing\)), there exists \(C \in \mathcal{C}\) such that $ ^{,C} . $

Next observe that if \(\mathcal{H}\) is compact in \(\mathcal{F}'\), then it is also compact in \(\mathcal{F}\).
Indeed, let \(\{ \mathcal{U}_i : i \in I \}\) be an open cover of \(\mathcal{H}\) by open sets of \(\mathcal{T}'\).
Let \(\mathcal{U}'_i = \mathcal{U}_i \setminus \{\varnothing\}\). Then \(\{ \mathcal{U}'_i : i \in I \}\) is also a cover of \(\mathcal{H}\) in the \(\mathcal{T}'\)–topology.
Since \(\mathcal{H}\) is compact, there exists a finite subcover \(\{ \mathcal{U}'_{i}\}_{i \in I_0}\) of \(\mathcal{H}\), implying that \(\{ \mathcal{U}_{i}\}_{i \in I_0}\) is a finite subcover in the \(\mathcal{T}\)–topology.
Thus \(\mathcal{H}\) is compact in \(\mathcal{F}\) for \(\mathcal{T}\) topology.

Since \(\mathcal{T}\) is Hausdorff, \(\mathcal{H}\) is closed, meaning that its complement \(\mathcal{F} \setminus \mathcal{H}\) is open.
Since \(\varnothing \in \mathcal{F}' \setminus \mathcal{H}\), by the first observation, there exists \(C \in \mathcal{C}\) such that $ ^{,C} . $ Taking complements gives $ _C. $

\(\square\)

Point Processes on $

Definition (Set Process). A point process \(\Phi\) on \(\mathcal{F}'\) is called a set process.

One may write a set process as \[ \Phi = \sum_{n=1}^{\Phi(\mathcal{F})} \delta_{F_n}, \] where each \(F_n\) is a RACS.

The union \[ Z_\Phi := \bigcup_{F\in\mathrm{supp}(\Phi)} F \] is called the coverage model.

Theorem. \(Z_\Phi\) is a RACS.

Proof. We need to show:

Let \((\Omega, \mathcal{A},\mathbb{P})\) be a probability space.

  1. Fix \(\omega \in \Omega\). Let \(\{x_j\}_{j\in\mathbb{N}}\) be a sequence in \(Z_\Phi(\omega)\) converging to \(x\).
    Then $ C = {x,x_1,x_2,} $ is compact in \(\mathbb{R}^d\).
    Since \(\Phi(\omega)\) is locally finite and \(\mathcal{F}_C\) is compact,
    we have \(\Phi(\omega)(\mathcal{F}_C) < \infty\), and so only a finite number of closed sets $ F_{i_1}(), , F_{i_k}() $ in \(\mathrm{supp}(\Phi(\omega))\) hit \(C\). Thus, $ { x_j } {j=1}^k F{i_j}() := F $ is a closed set. This implies \(x \in F \subset Z_{\Phi}(\omega)\) and \(Z_{\Phi}(\omega) \text{ is closed}.\)

We have shown that \(Z_{\Phi}(\omega)\) is closed for all \(\omega \in \Omega\) such that
\(\Phi(\omega)\) is locally finite.
Since \(\Phi\) is locally finite almost surely, it follows that \(Z_{\Phi}\) is closed almost surely.

  1. Let \(C \in \mathcal{C}\). Then \[ Z_{\Phi} \in \mathcal{F}^C \quad\Longleftrightarrow\quad \Phi(\mathcal{F}_C) = 0. \]

Since \(\{ \Phi(\mathcal{F}_C) = 0 \}\) is measurable, so is $ { Z_{} ^C }. $

We are done, since \(\{ \mathcal{F}_C : C \in \mathcal{C} \}\) generates \(\mathcal{B}(\mathcal{F})\).

\(\square\)

The capacity functional of \(Z_\Phi\) is

\[\begin{equation*} T_{Z_\Phi}(C) = \mathbb{P}(Z_\Phi \cap C \neq \varnothing) = \mathbb{P}(Z_\Phi \in \mathcal{F}_C) = \mathbb{P}(\Phi(\mathcal{F}_C) > 0) = 1 - \mathbb{P}\big( \Phi(\mathcal{F}_C)=0 \big). \end{equation*}\]

Stationarity

Definition. A set process \(\Phi = \sum_{n\in\mathbb{N}} \delta_{F_n}\) is stationary if \[ \Phi \stackrel{d}{=} t_x \Phi := \sum_{n\in\mathbb{N}} \delta_{F_n - x}, \qquad \forall\, x \in \mathbb{R}^d. \]

Note: \(\Lambda\) is a measure on \(\mathcal{F}'\) and is translation invariant if \[ \Lambda(A) = \Lambda\big( \{ F + x : F \in A \} \big), \qquad \forall\, A \in \mathcal{B}(\mathcal{F}'),\ \forall\, x \in \mathbb{R}^d. \]

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