I've been struggling to unwind the following definition and understand it heuristically.
Definition 1 (Tagged Poisson point processes). Let $\Omega$ be a locally compact metric space. The tagged particle $z_{1}$ is an independent random variable with law $\mu / \mu(\Omega)$. The random variable $\tilde{z} \in \bigcup_{r=0}^{\infty} \Omega^{r}$ forms a Poisson point process with density $\mu \in M_{+}(\Omega)$ (non-negative Radon measures, i.e. positive elements of $\left.\left(C_{c}^{0}(\Omega)\right)^{*}\right)$ if $$ \operatorname{Prob}\left(\tilde{z} \in \Omega^{r}\right)=e^{-\mu(\Omega)} \frac{\mu(\Omega)^{r}}{r !}, \quad \operatorname{law}\left(\tilde{z}_{i}\right)=\mu / \mu(\Omega) $$ and $\tilde{z}_{1}, \ldots, \tilde{z}_{r}$ are independent. Now letting $N=r+1 \in\{1,2, \ldots\}$, realizations of the tagged Poisson point process (tppp) are obtained by letting $z=\left(z_{1}, \ldots, z_{N}\right)=\left(z_{1}, \tilde{z}\right)$, i.e. one obtains for symmetric $A \subset \bigcup_{N=1}^{\infty} \Omega^{N}$ that $$ \operatorname{Prob}_{\mathrm{tppp}}\left(\left(z_{1}, \ldots, z_{N}\right) \in A\right)=\frac{1}{\mu(\Omega) e^{\mu(\Omega)}} \sum_{N=1}^{\infty} \frac{1}{(N-1) !} \int_{A \cap \Omega^{N}} \mathrm{~d} \mu\left(z_{1}\right) \ldots \mathrm{d} \mu\left(z_{N}\right)$$
Where we mean that a set is symmetric if any point in $A$ is still in $A$ after any permutation of the coordinates
From what I understand we are interested in describing the distribution of the first particle generated by the PPP with density $\mu$ which is given by the first equation. However the bit that confuses me is what the tagged PPP is meant to be. It feels like they are trying to say that Prob$_{\mathrm{tppp}}$ gives probability that there exist N particles and that the tagged particle $z_1$ lies in $A$ but then looking at the formula for Prob$_{\mathrm{tppp}}$ this is clearly not the case.
My questions are:
- What is Prob$_{\mathrm{tppp}}$ trying to tell us?
- How do we get to the formula given in the definition?