Recall our main example of a Binomial point process.
For \(n \in \mathbb{N}\) and \(X_1, \dots, X_n\) i.i.d. random elements of CSMS \((E, \mathcal{B})\) with distribution \(\mathbb{Q}\), letd \[
\Phi = \sum_{i=1}^n \delta_{X_i}.
\]
For \(B \in \mathcal{B}\), \(\Phi(B)\) has a binomial distribution: \[ \mathbb{P}[\Phi(B) = k] = \binom{n}{k} \mathbb{Q}(B)^k (1 - \mathbb{Q}(B))^{n-k}. \]
A binomial distribution \(\mathrm{Binomial}(n, p_n)\) converges to a Poisson distribution \(\mathrm{Poisson}(\gamma)\) if \(n p_n \to \gamma \in (0, \infty)\).
Thus, letting \(n \to \infty\) and \(\mathbb{Q}(B) \to 0\) in an appropriate way leads us to a Poisson process.
Let \(\Lambda\) be a locally finite measure on \(E\), and let \(\{C_j\}_{j \in \mathbb{N}}\) be a sequence of compact sets such that: \(C_j \uparrow E\), \(\Lambda(C_j) < \infty\), and \(\Lambda(C_j) \to \infty\) as \(j \to \infty\).
Define a sequence of point processes \[ \Phi_j = \sum_{i=1}^{n_j} \delta_{X_i^{(j)}}, \] where \[ X_i^{(j)} \sim \frac{\Lambda(\cdot \cap C_j)}{\Lambda(C_j)}, \quad i = 1, \dots, n_j \] are i.i.d. random variables.
Assume that \[ \frac{n_j}{\Lambda(C_j)} \to 1 \quad \text{as } j \to \infty. \]
Then, for \(B \in \mathcal{B}\): \[ \mathbb{P}\left[ \Phi_j(B) = k \right] = \binom{n_j}{k} \left( \frac{\Lambda(B \cap C_j)}{\Lambda(C_j)} \right)^k \left[ 1 - \frac{\Lambda(B \cap C_j)}{\Lambda(C_j)} \right]^{n_j - k}. \]
Taking the limit \(j \to \infty\), we obtain: \[ \mathbb{P}\left[ \Phi(B) = k \right] = \frac{\Lambda(B)^k}{k!} e^{-\Lambda(B)}. \]
Thus, the limiting process \(\Phi\) is a Poisson point process with mean measure \(\Lambda\).
Definition. Let \(\Lambda\) be a locally finite measure on the CSMS space \((E, \mathcal{B})\).
A Poisson point process \(\Phi\) on \(E\) with intensity measure \(\Lambda\) is a point process such that:
That is, \[ \mathbb{P}\left[ \Phi(B_1) = n_1, \dots, \Phi(B_k) = n_k \right] = \prod_{i=1}^k e^{-\Lambda(B_i)} \frac{[\Lambda(B_i)]^{n_i}}{n_i!}. \]
The use of the intensity measure is consistent since \[ \mathbb{E}\big[\Phi(B)\big] = \Lambda(B), \quad \forall B \in \mathcal{B}. \]
Some further Observations:
Proposition. Let \(N\) be a Poisson random variable with parameter \(\gamma > 0\).
Let \(X_1, X_2, \dots\) be i.i.d. random elements in \(E\) with distribution \(\mathbb{Q}\), independent of \(N\).
Then \[
\Phi = \sum_{i=1}^{N} \delta_{X_i}
\] is a Poisson point process with intensity measure \(\gamma \mathbb{Q}\).
Proof. It suffices to compute the joint distribution of \(\Phi(B_1), \dots, \Phi(B_k)\) for pairwise disjoint sets \(B_1, \dots, B_k \in \mathcal{B}\) such that \[ \bigcup_{i=1}^k B_i = E. \] If not, we can simply add the complement of the union as an additional set.
Let \(n_1, \dots, n_k \in \mathbb{N}_0\) and define \(n = n_1 + \cdots + n_k\). Then,
\[\begin{align*} &~\mathbb{P}\big[\Phi(B_1) = n_1, \dots, \Phi(B_k) = n_k \big]\\ &= \sum_{m=0}^\infty \mathbb{P}\big[\Phi(B_1) = n_1, \dots, \Phi(B_k) = n_k \,\big|\, N = m \big] \mathbb{P}[N = m] \\ &= \mathbb{P}\big[\Phi(B_1) = n_1, \dots, \Phi(B_k) = n_k \,\big|\, N = n \big] \mathbb{P}[N = n]\\ &= \mathbb{P}\left[\sum_{j=1}^n 1_{\{X_j \in B_1\}} = n_1, \dots, \sum_{j=1}^n 1_{\{X_j \in B_k\}} = n_k \,\middle|\, N = n\right] \mathbb{P}[N=n]. \end{align*}\]
The conditional distribution of \((\Phi(B_1),\dots,\Phi(B_k))\) given \(N=n\) is multinomial: \[ \mathbb{P}\left[\Phi(B_1)=n_1, \dots, \Phi(B_k)=n_k \,\middle|\, N=n\right] = \frac{n!}{n_1!\cdots n_k!} \mathbb{Q}(B_1)^{n_1}\cdots \mathbb{Q}(B_k)^{n_k}. \]
Multiplying by \(\mathbb{P}[N=n] = \dfrac{\gamma^n e^{-\gamma}}{n!}\) gives: \[ \mathbb{P}\big[\Phi(B_1)=n_1,\dots,\Phi(B_k)=n_k\big] = \prod_{i=1}^k \frac{(\gamma \mathbb{Q}(B_i))^{n_i}}{n_i!} e^{-\gamma \mathbb{Q}(B_i)}. \]
Here we have used that \(\sum_{i=1}^k \mathbb{Q}(B_i) = 1\).
\(\square\)
Proposition. Let \(\Phi\) be a Poisson process on \(E\) with intensity measure \(\Lambda\). Then, for all measurable functions \(f: E \to \mathbb{R}_+\), \[ L_\Phi(f) := \mathbb{E}\left[ e^{-\int_E f\, d\Phi} \right] = \exp\left( - \int_E (1 - e^{-f})\, d\Lambda \right). \]
Note: the Laplace transform of a Poisson random variable \(X\) with parameter \(\gamma\) is \[ \mathbb{E}\left[ e^{-tX} \right] = e^{-\gamma (1 - e^{-t})}, \quad t > 0. \]
Proof. Consider first a simple function \[ f = \sum_{i=1}^k a_i 1_{B_i} \] for pairwise disjoint sets \(B_1, \dots, B_k \in \mathcal{B}\) and constants \(a_i \in (0,\infty)\). Then,
\[\begin{align*} \mathbb{E}\left[ e^{-\int_E f\, d\Phi} \right] &= \mathbb{E}\left[ e^{ - \sum_{i=1}^k a_i \Phi(B_i)} \right] \\ &= \prod_{i=1}^k \mathbb{E}\left[ e^{- a_i \Phi(B_i)} \right] \\ &= \prod_{i=1}^k e^{ - \Lambda(B_i) (1 - e^{-a_i}) } \\ &= \exp\left( - \sum_{i=1}^k \Lambda(B_i)(1 - e^{-a_i}) \right) \\ &= \exp\left( - \int_E (1 - e^{-f})\, d\Lambda \right). \end{align*}\]
For a general measurable function \(f\), we approximate \(f\) by an increasing sequence of simple functions and use the monotone convergence theorem to obtain the result.
\(\square\)