User:Camiloortizas/Sandbox

Gaussian q-distribution

The Gaussian $$q$$-distribution introduced by Diaz and Teruel is a q-analogue of the Gaussian or Normal distribution.

Let $$q$$ be a real number in the interval [0,1). The Gaussian $$q$$-density is the function

$$s_q:\mathbb{R} \rightarrow \mathbb{R}$$

given by


 * $$s_q = \begin{cases} 0 & \mbox{if } x < -\nu  \\ \frac{1}{c(q)}E_{q^2}^{\frac{-q^2x^2}{[2]_q}}  & \mbox{if } -\nu \leq x \leq \nu \\ 0 & \mbox{if } x >\nu. \end{cases} $$

where


 * $$\nu = \nu(q) = \frac{1}{\sqrt{1-q}} $$

$$c(q)=2(1-q)^{1/2}\sum_{m=0}^{\infty} \frac{(-1)^m q^{(m)(m+1)}}{(1-q^{2m+1})(1-q^2)_{q^2}^m}$$. The $$q$$-analogue $$ [t]_q $$ of the real number $$ t $$ is given by

$$ [t]_q=\frac{q^t-1}{q-1} $$.

The $$q$$-analogue of the exponential function $$ E_q^{x} $$ is given by

$$ E_q^{x}=\sum_{j=0}^{\infty}q^{j(j-1)/2}\frac{x^{j}}{[j]!}$$

where the $$q$$-analogue of the factorial $$ [n]_q!$$ is given by

$$ [n]_q!=[n]_q[n-1]_q...[2]_q$$

for an integer $$ n>2 $$ and $$ [1]_q!=[0]_q!=1$$

The cumulative Gaussian $$q$$-distribution



$$ G_q: \mathbb{R} \rightarrow \mathbb{R}$$

is given by

$$\begin{cases} 0 & \mbox{if }  x < -\nu \\ \frac{1}{c(q)}\int_{-\nu}^{x} E_{q^{2}}^{\frac{-q^{2}t^{2}}{[2]}}d_qt& \mbox{if }  -\nu \leq x \leq \nu  \\ 1 & \mbox{if }  x>\nu \end{cases}$$ where the integration symbol denotes the Jackson integral.

Explicitly the function $$G_q$$ is given by

$$G_q(x)= \begin{cases} 0 & \mbox{if} x < -\nu, \\ {\displaystyle \frac{1}{2} + \frac{1-q}{c(q)} \sum_{n=0}^{\infty}\frac{q^{n(n+1)}(q-1)^n  }{(1-q^{2n+1})(1-q^2)_{q^2}^{n}}x^{2n+1}} &  \mbox{if } -\nu \leq x \leq \nu \\  1 &  \mbox{if}\ x > \nu \end{cases}$$

where $$(a+b)_q^n=\prod_{i=0}^{n-1}(a+q^ib)$$

The moment (mathematics) of the Gaussian $$q$$-distribution are given by

$$\frac{1}{c(q)}\int_{-\nu}^{\nu} E_{q^{2}}^{\frac{-q^{2}x^{2}}{[2]}}x^{2n}d_qx=[2n-1]!!$$

$$\frac{1}{c(q)}\int_{-\nu}^{\nu} E_{q^{2}}^{\frac{-q^{2}x^{2}}{[2]}}x^{2n+1}d_qx=0$$

Where the symbol $$[2n-1]!!$$ is the q-analogue of the double factorial given by

$$ [2n-1][2n-3]\cdots[1]= [2n-1]!!$$