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Quoted from Wikipedia

The Laplace functional $Ψ_N(f)$ of a point process $N$ is a map from the set of all positive valued functions $f$ on the state space of $N$, to $[0,\infty)$ defined as follows:

$ Ψ_N(f) = E[\exp( - N(f))]$

They play a similar role as the characteristic functions for random variable.

I was wondering what $N(f)$ means? How to understand it as the image of $f$ under the mapping $N$?

Is $ Ψ_N(f)$ a measure on $[0,\infty)$?

Thanks and regards!

1 Answers 1

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A point process can be viewed as a locally finite random subset $\mathcal{N}$ of the ambient space or as a random locally finite point measure $N$, the translation from the subset presentation to the measure presentation being that $N(B)=\#(\mathcal{N}\cap B)$ for every measurable subset $B$ and $ N(f)=\sum_{x\in\mathcal{N}}f(x), $ for every measurable function $f$. Hence $N(f)$ is a random variable (nonnegative if $f$ is nonnegative, integer valued if $f$ is integer valued) and $\Psi_N(f)=E(\exp(-N(f)))$ is a deterministic nonnegative number.

The so-called intensity measure $\mu$ of the Poisson process $\mathcal{N}$ is the deterministic measure defined by $\mu(B)=E(N(B))=E(\#(\mathcal{N}\cap B))$, hence $\mu(f)=E(N(f))$.

The characteristic functional $\Psi_N$ is such that $ \Psi_N(f)=\displaystyle \exp\left(-\int(1-\mathrm{e}^{-f(x)})\mathrm{d}\mu(x)\right). $ This formula is a generalization and a consequence of two simple facts: first, for every measurable subset $B$, $N(B)$ is Poisson distributed with parameter $\mu(B)$ and, second, for every Poisson random variable $X$ of mean $\lambda$ and every real number $a$, $ E(\exp(-aX))=\sum_{n\ge0}\mathrm{e}^{-an}\mathrm{e}^{-\lambda}\lambda^n/n!=\exp(-(1-\mathrm{e}^{-a})\lambda). $

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    @Tim You could even begin with computing $\Psi_N(f)$ for $f=\mathbf{1}_B$, then for $f=a\mathbf{1}_B$, then for the simple function $f$ written in my last comment.2011-04-21