Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer 1

up vote 3 down vote accepted

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 $a$, $$ E(\exp(-aX))=\sum_{n\ge0}\mathrm{e}^{-an}\mathrm{e}^{-\lambda}\lambda^n/n!=\exp(-(1-\mathrm{e}^{-a})\lambda). $$

share|improve this answer
    
@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
    
@mpiktas The statement is not in my post. –  Did Apr 20 '11 at 14:39
    
But nonnegative holds only for nonnegative functions $f$, hence I will cancel it. –  Did Apr 20 '11 at 14:40
    
@Didier: Thanks! I was wondering why N(f) is an integer valued nonnegative random variable? What is the codomain of f? –  Tim Apr 20 '11 at 14:42
    
@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
show 7 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.