Free Poisson distribution

In the mathematics of free probability theory, the free Poisson distribution is a counterpart of the Poisson distribution in conventional probability theory.

Definition

The free Poisson distribution with jump size $\alpha$ and rate $\lambda$ arises in free probability theory as the limit of repeated free convolution

${((1-\frac{\lambda }{N}){\delta }_0+\frac{\lambda }{N}{\delta }_{\alpha })}^{⊞N}$

as N → ∞.

In other words, let $X_N$ be random variables so that $X_N$ has value $\alpha$ with probability $\frac{\lambda }{N}$ and value 0 with the remaining probability. Assume also that the family $X_1,X_2,\dots$ are freely independent. Then the limit as $N\to \mathrm{\infty }$ of the law of $X_1+\cdots +X_N$ is given by the Free Poisson law with parameters $\lambda ,\alpha$.

This definition is analogous to one of the ways in which the classical Poisson distribution is obtained from a (classical) Poisson process.

The measure associated to the free Poisson law is given by

$\mu =\{\begin{array}{ll}(1-\lambda ){\delta }_0+\nu ,& \text{if }0\le \lambda \le 1\\ \nu ,& \text{if }\lambda >1,\end{array}$

where

$\nu =\frac{1}{2\pi \alpha t}\sqrt{4\lambda {\alpha }^2-(t-\alpha (1+\lambda ){)}^2}dt$

and has support $[\alpha (1-\sqrt{\lambda }{)}^2,\alpha (1+\sqrt{\lambda }{)}^2]$.

This law also arises in random matrix theory as the Marchenko–Pastur law. Its free cumulants are equal to ${\kappa }_n=\lambda {\alpha }^n$.

Some transforms of this law

We give values of some important transforms of the free Poisson law; the computation can be found in e.g. in the book Lectures on the Combinatorics of Free Probability by A. Nica and R. Speicher

The R-transform of the free Poisson law is given by

$R(z)=\frac{\lambda \alpha }{1-\alpha z}.$

The Cauchy transform (which is the negative of the Stieltjes transformation) is given by

$G(z)=\frac{z+\alpha -\lambda \alpha -\sqrt{(z-\alpha (1+\lambda ){)}^2-4\lambda {\alpha }^2}}{2\alpha z}$

The S-transform is given by

$S(z)=\frac{1}{z+\lambda }$

in the case that $\alpha =1$.