### Approximation of a free Poisson process by systems of freely independent particles / Marek Bożejko, José Luís da Silva, Tobias Kuna, Eugene Lytvynov

Infinite Dimensional Analysis, Quantum Probability and Related Topics, Volume: 21, Issue: 3, Pages: 1850020-1 - 1850020-25

Swansea University Author:

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DOI (Published version): 10.1142/S0219025718500200

Abstract

Let $\sigma$ be a non-atomic, infinite Radon measure on $\mathbb R^d$, for example, $d\sigma(x)=z\,dx$ where $z>0$. We consider a system of freely independent particles $x_1,\dots,x_N$ in a bounded set $\Lambda\subset\mathbb R^d$, where each particle $x_i$ has distribution $\frac1{\sigma(\Lambda)... Full description Published in: Infinite Dimensional Analysis, Quantum Probability and Related Topics 0219-0257 1793-6306 2018 https://cronfa.swan.ac.uk/Record/cronfa43227 No Tags, Be the first to tag this record! Abstract: Let$\sigma$be a non-atomic, infinite Radon measure on$\mathbb R^d$, for example,$d\sigma(x)=z\,dx$where$z>0$. We consider a system of freely independent particles$x_1,\dots,x_N$in a bounded set$\Lambda\subset\mathbb R^d$, where each particle$x_i$has distribution$\frac1{\sigma(\Lambda)}\,\sigma$on$\Lambda$and the number of particles,$N$, is random and has Poisson distribution with parameter$\sigma(\Lambda)$. If the particles were classically independent rather than freely independent, this particle system would be the restriction to$\Lambda$of the Poisson point process on$\mathbb R^d$with intensity measure$\sigma$. In the case of free independence, this particle system is not the restriction of the free Poisson process on$\mathbb R^d$with intensity measure$\sigma$. Nevertheless, we prove that this is true in an approximative sense: if bounded sets$\Lambda^{(n)}$($n\in\mathbb N$) are such that$\Lambda^{(1)}\subset\Lambda^{(2)}\subset\Lambda^{(3)}\subset\dotsm$and$\bigcup_{n=1}^\infty \Lambda^{(n)}=\mathbb R^d$, then the corresponding particle system in$\Lambda^{(n)}$converges (as$n\to\infty$) to the free Poisson process on$\mathbb R^d$with intensity measure$\sigma$. We also prove the following$N/V$-limit: Let$N^{(n)}$be a determinstic sequence of natural numbers such that$\lim_{n\to\infty}N^{(n)}/\sigma(\Lambda^{(n)})=1$. Then the system of$N^{(n)}$freely independent particles in$\Lambda^{(n)}$converges (as$n\to\infty\$) to the free Poisson process. We finally extend these results to the case of a free L\'evy white noise (in particular, a free L\'evy process) without free Gaussian part. College of Science 3 1850020-1 1850020-25