Geometric distributions

geometric_distribution q_deformation

Definition. $$X \sim Geom(p)$$ if

$f_X(k) = p (1 - p)^k, \qquad k \ge 0$

Then

$\prob(X \ge n) = p \sum_{k \ge n} (1 - p)^k = (1 - p)^n$

Claim. For $$\theta > 0$$, if $$X_n \sim Geom(\theta / n)$$ then $$X_n / n \to Exp(\theta)$$ the exponential distribution with parameter $$\theta$$. Write this as $$Geom(\theta / n) / n \to Exp(\theta)$$.

Proof.

$\prob(X_n / n \ge t) = (1 - \theta / n)^{n t} \to e^{- \theta t}.$

$$\square$$

$$q$$-deformation

Let $$q, \alpha \in [0, 1]$$. With parameter $$\alpha$$, the $$q$$-geometric distribution $$q$$Geom$$(\alpha)$$ has pmf (see e.g. [{matveev-petrov15}])

$f(k) = {\alpha^k \over (q)_k} (\alpha)_\infty$

when $$q = 0$$ it turns into $$Geom(1 - \alpha)$$.

$$qt$$-deformation

Let $$q, t, \alpha \in [0, 1]$$. The $$qt$$-geometric distribution $$qt$$Geom$$(\alpha)$$ has pmf

$f(k) = \alpha^k {(t)_k \over (q)_k} {(\alpha)_\infty \over (\alpha t)_\infty}$

It is a probability distribution due to Cauchy's $$q$$-binomial series formula, (3) in q-Pochhammers.

• When $$t = 0$$, $$qt$$Geom = $$q$$Geom.
• When $$t = q$$, $$qt$$Geom$$(\alpha) =$$Geom$$(1 - \alpha)$$.

References

• [matveev-petrov15] q-randomized Robinson-Schensted-Knuth correspondences and random polymers, , 2016.