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Approximation of a free Poisson process by systems of freely independent particles
Infinite Dimensional Analysis, Quantum Probability and Related Topics, Volume: 21, Issue: 3, Pages: 1850020-1 - 1850020-25
Swansea University Author:
Eugene Lytvynov
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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)...
| Published in: | Infinite Dimensional Analysis, Quantum Probability and Related Topics |
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| ISSN: | 0219-0257 1793-6306 |
| Published: |
2018
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| Online Access: |
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| 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. |
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| College: |
Faculty of Science and Engineering |
| Issue: |
3 |
| Start Page: |
1850020-1 |
| End Page: |
1850020-25 |