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.

References