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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: Eugene Lytvynov

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)...

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Published in: Infinite Dimensional Analysis, Quantum Probability and Related Topics
ISSN: 0219-0257 1793-6306
Published: 2018
Online Access: Check full text

URI: https://cronfa.swan.ac.uk/Record/cronfa43227
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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: College of Science
Issue: 3
Start Page: 1850020-1
End Page: 1850020-25