Return to Answer

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, $$\mu_{K}([j]_q)=\frac{p}{1-(p;q)_\infty}\,(p;q)_{j-1}\,q^j.$$

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 $p\mapsto 1-p$).

Kupershmidt argues for $$\mu_{K}([j]_q)\propto(pq)^j,\;\;[j]_q=\frac{q^j-1}{q-1},$$ as the "natural" $q$-geometric distribution, because it produces the $q$-analog of the Pascal (negative binomial) distribution. It has a simple normalization, $$\mu_{K}([j]_q)=\frac{1-p}{1-(p;q)_\infty}\,(pq)^{j-1}.$$

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, $$\mu_{K}([j]_q)=\frac{p}{1-(p;q)_\infty}\,(p;q)_{j-1}\,q^j.$$

Kupershmidt argues for $$\mu_{K}([j]_q)\propto(pq)^j,\;\;[j]_q=\frac{q^j-1}{q-1},$$ as the "natural" $q$-geometric distribution, because it produces the $q$-analog of the Pascal (negative binomial) distribution. It has a simple normalization,

Kupershmidt argues for $$\mu_{K}([j]_q)\propto(pq)^j,\;\;[j]_q=\frac{q^j-1}{q-1},$$ as the "natural" $q$-geometric distribution, because it produces the $q$-analog of the Pascal (negative binomial) distribution. It has a simple normalization, $$\mu_{K}([j]_q)=\frac{1-p}{1-(p;q)_\infty}\,(pq)^{j-1}.$$