Proposition. Let \(\Phi_1, \Phi_2, \dots\) be a sequence of independent Poisson processes on \(E\) with intensity measures \(\Lambda_1, \Lambda_2, \dots\). Then their superposition \[ \Phi = \sum_{i=1}^\infty \Phi_i \] is itself a Poisson process with intensity measure \[ \Lambda = \sum_{i=1}^\infty \Lambda_i \] provided that \(\Lambda\) is locally finite.
Proof. If \(\Lambda = \sum_{i=1}^\infty \Lambda_i\) is locally finite, then \(\Phi\) is a point process (see Proposition 2.2.1 in [baccelli2024random]). It remains to check that \(\Phi\) is a Poisson process.
We do this by computing its Laplace functional:
\[\begin{align*} L_\Phi(f) &= \mathbb{E}\left[ e^{ - \int_E f\, d\Phi } \right] = \mathbb{E}\left[ e^{ - \int_E f\, d\left( \sum_{i=1}^\infty \Phi_i \right) } \right] \\ &= \mathbb{E}\left[ \prod_{i=1}^\infty e^{ - \int_E f\, d\Phi_i } \right] = \prod_{i=1}^\infty \mathbb{E}\left[ e^{ - \int_E f\, d\Phi_i } \right] \\ &= \prod_{i=1}^\infty \exp\left( - \int_E (1 - e^{-f})\, d\Lambda_i \right) \\ &= \exp\left( - \sum_{i=1}^\infty \int_E (1 - e^{-f})\, d\Lambda_i \right) \\ &= \exp\left( - \int_E (1 - e^{-f})\, d\Lambda \right). \end{align*}\]
Thus, \(L_\Phi(f)\) has the Laplace functional of a Poisson process with intensity measure \(\Lambda\), and the result follows.
\(\square\)
Theorem. Let \(\Lambda\) be a locally finite measure on a locally compact CSMS \(E\). Then, there exists a Poisson process on \(E\) with intensity measure \(\Lambda\).
Proof. Suppose first that \(0 < \Lambda(E) < \infty\).
Let \(X_1, X_2, \dots\) be i.i.d. random elements of \(E\) with distribution $ . $ That is, $ [X_i B] = B . $
Also, let \(N\) be a Poisson random variable with parameter \(\Lambda(E)\).
By a proposition from the previous lecture, $ = {i=1}^N {X_i} $ is a Poisson process with intensity measure $ (E) = (). $
Now, assume \(\Lambda(E) = \infty\).
Fact: Due to the local compactness assumption on \(E\), there exist pairwise disjoint relatively compact (closure is compact) Borel sets \(A_1, A_2, \dots\) such that \[ \bigcup_{i \in \mathbb{N}} A_i = E. \] (see Lemma 1.1.4 in [baccelli2024random])
Since \(\Lambda\) is locally finite, we have \(\Lambda(A_i) < \infty\) for all \(i \in \mathbb{N}\).
Note: If \(\Lambda(A_i) = 0\), we can either omit \(A_i\) or combine it with another \(A_j\) so that \(\Lambda(A_j) > 0\).
Let \(\{\Phi_i\}_{i \in \mathbb{N}}\) be independent point processes such that for each \(i \in \mathbb{N}\), \(\Phi_i\) is a Poisson process with intensity measure \[ \Lambda_i(\cdot) = \Lambda(\cdot \cap A_i). \]
By the superposition property, \[ \Phi = \sum_{i \in \mathbb{N}} \Phi_i \] is a Poisson process on \(E\) with intensity measure \(\Lambda\). Indeed, \[ \sum_i \Lambda_i(B) = \sum_i \Lambda(B \cap A_i) = \Lambda(B). \]
\(\square\)
Definition. Let $ = {k=1}^{N} {X_k} $ be a point process on \(E\) with \(N = \Phi(E)\) a random variable taking values in \(\mathbb{N} \cup \{\infty\}\), let \(p: E \to [0,1]\) be a measurable function, and let \(U_1, U_2, \dots\) be a sequence of i.i.d. \(\mathrm{Uniform}([0,1])\) random variables, independent of \(\Phi\).
Then the \(p\)-thinning of \(\Phi\) is the point process \[ \widetilde{\Phi} = \sum_{k=1}^{N} \mathbf{1}_{\{U_k \leq p(X_k)\}} \, \delta_{X_k}. \]
That is, \(\widetilde{\Phi}\) is obtained from \(\Phi\) by deleting each point \(X_k\) with probability \(1 - p(X_k)\) and retaining it with probability \(p(X_k)\), independently of the other points.