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Consider the simple cluster process: $$\sum_n \xi_n \epsilon_{X_n}$$ where $\{X_n\}$ are Poisson points independent of the iid non-negative integer sequence $\{\xi_n\}$. How do I find the Laplace functional? I am a bit confused reading about it online.

This is what I have so far. Let $\psi_N(f)$ be the Laplace functional and suppose $\mu$ is the mean measure of the Poisson random measure defined by $N$. \begin{align*} N(A) &= \sum_n \xi_n \epsilon_{X_n}(A) \\ \psi_N(f) &= E[e^{-\sum_n \xi_n \epsilon_{X_n(A_n)}}] \hbox{ where $A_n$ are disjoint} \\ &= E[\prod_{n} e^{-\xi_n N(A_n)}]\\ &= \prod_{n} E[e^{-\xi_n N(A_n)}]\\ &= \prod_{n} \exp(-\int (1-e^{-\xi_n 1_{A_n}(x)}) \mu(dx)) \end{align*}

bilbo
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1 Answers1

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Let us denote $g$ the probability mass function of $\xi_1$ and $\mu(A)$ the mean measure of the Poisson process on $A$. Then $N(A)$ has the following Laplace functional at point $\lambda$:

$E[e^{-\lambda N(A)}]$ = $E[E[e^{-\lambda \sum_{n=1}^\infty \xi_n\epsilon_{X_n}} | \sum_{n=1}^\infty\epsilon_{X_n} = k]] = \sum_{k=0}^\infty \big[\frac{\mu(A)^ke^{-\mu(A)}}{k!}\big(\sum_{i=0}^\infty e^{-\lambda i} g(i)\big)^k \big]$

$=\exp[-\mu(A)(1-\sum_{i=0}^\infty e^{-\lambda i}g(i))] = \exp[-\mu(A)\sum_{i=0}^\infty (1- e^{-\lambda i})g(i)]$,

where the second equality follows because ${\xi_n}$ and ${X_n}$ are independent and ${\xi_n}$ are iid.

TOMILO87
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