# Laplace functional of a point process

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!

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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 $a$, $$E(\exp(-aX))=\sum_{n\ge0}\mathrm{e}^{-an}\mathrm{e}^{-\lambda}\lambda^n/n!=\exp(-(1-\mathrm{e}^{-a})\lambda).$$
@Didier Piau, the statement "$N(f)$ is an integer valued nonnegative random variable and is a deterministic nonnegative number" is a contradiction, since it implies that random variable is deterministic. I may be missing something so I would be grateful if you could clarify. – mpiktas Apr 20 '11 at 14:36
But nonnegative holds only for nonnegative functions $f$, hence I will cancel it. – Did Apr 20 '11 at 14:40
@Didier Piau, ah, sorry, this is a formatting bug. The formula for $\Psi_N(f)$ jumped to the beginning of the page, so I read the sentence with out it. – mpiktas Apr 20 '11 at 14:42