I'm trying to set straight various $q$-deformations of the standard geometric distribution.

The geometric distribution on $\left\{ 0,1,\ldots \right\}$ is well-known, it has $$ \mu_1(X=j)=(1-p)p^j,\qquad j=0,1,2,\ldots. $$ Here $0<p<1$ is the parameter.

This distribution can be $q$-deformed in at least two ways.

(1) One way uses the Heine's $q$-binomial theorem and has $$ \mu_2(X=j)=p^j\frac{(a;q)_{j}^{}}{(q;q)_{j}^{}}\frac{(p;q)_{\infty}^{}}{(ap;q)_{\infty}^{}},\qquad j=0,1,2,\ldots. $$ where $0<q<1$ is the main $q$-deformation parameter, and $0<a<1$ is an additional parameter. Here a$(z;q)_{k}^{}=(1-z)(1-zq)\ldots(1-zq_{}^{k-1})$ is the $q$-Pochhammer symbol, and it makes sense for $k=\infty$, too. The distribution $\mu_2$ should not be viewed as a $q$-deformation of $\mu_1$ because if one sets $q=0$ then $\mu_2$ does not become $\mu_1$.

EDIT: But when $a=q$ and for any $q$ $\mu_2$ does not depend on $q$ and becomes $\mu_1$.

$\mu_2$ is the orthogonality weight of the little $q$-Jacobi polynomials.

(1a) For $a=0$ the above distribution simplifies to $$ \mu_3(X=j)=\frac{p^j}{(q;q)_{j}^{}}(p;q)_{\infty}^{},\qquad j=0,1,2,\ldots. $$ And this for $q=0$ turns into $\mu_1$.

$\mu_1$ is the orthogonality weight of the little $q$-Laguerre / Wall polynomials or maybe $q$-Charlier polynomials.

(2) Now take another (probably more naive) way to $q$-deform $\mu_1$, and define $$ \mu_4(j)\propto\frac{p^j}{1-q^j},\qquad j=1,2,\ldots. $$ The normalization constant in $\mu_4$ is not that nice, it can be expressed through the $q$-digamma function. The distribution $\mu_4$ can in fact be obtained from $\mu_2$ by taking $a\to1$ and conditioning $\mu_2$ on $\left\{ 1,2,\ldots \right\}$. Clearly, for $q=0$ the distribution $\mu_4$ also becomes the standard geometric distribution (but on $\left\{ 1,2,\ldots \right\}$).

After a brief skimming through Koekoek et al ("Hypergeometric Orthogonal Polynomials and Their q-Analogues") I could not find orthogonal polynomials with the weight $\mu_4$.

My main questions are:

Are there any more or less standard names for $\mu_2,\mu_3$, and $\mu_4$? And which of $\mu_3$ or $\mu_4$ should be more naturally called the $q$-geometric distribution?

EDIT: I would like to have distributions which become the original ones as $q\to0$ and not $q\to1$, this is related to how the $q$-Whittaker polynomials (same as $t=0$ Macdonald polynomials) turn into the classical Schur polynomials when $q=0$.

PS. I do not even start with taking analogous distributions on a finite lattice, where there are more numerous interesting $q$-deformations of the binomial distribution. For a rather general $q$-deformation of the binomial distribution e.g., see formula (1.3.1) in https://arxiv.org/abs/1401.3321 - this is the orthogonality weight of the $q$-Hahn orthogonal polynomials.


1 Answer 1


Kupershmidt argues for $$\mu_{K}([j]_q)\propto(p;q)_{j-1}\,q^j,\;\;[j]_q=\frac{q^j-1}{q-1},$$ as the "natural" $q$-geometric distribution, because it produces the $q$-analog of the Pascal (negative binomial) distribution. It has a simple normalization, $$\mu_{K}([j]_q)=\frac{p}{1-(p;q)_\infty}\,(p;q)_{j-1}\,q^j.$$ Since $(p;1)_n=(1-p)^n$ and $[j]_1=j$, for $q\rightarrow 1$ the classical geometric distribution $\mu(j)=p(1-p)^{j-1}$ is recovered (the $\mu_1$ from the OP with $p\mapsto 1-p$).

  • $\begingroup$ Thank you, I didn't know about this distribution! But somehow I want things to become the original geometric distributions when $q=0$ and to $q\to1$ - this is related to Macdonald symmetric polynomials and how they degenerate to the Schur ones $\endgroup$ Dec 19, 2016 at 13:52

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