Double Weibull Distribution#
This is a signed form of the Weibull distribution. There is one shape parameter \(c>0\). Support is \(x\in\mathbb{R}\).
\begin{eqnarray*} f\left(x;c\right) & = & \frac{c}{2}\left|x\right|^{c-1}\exp\left(-\left|x\right|^{c}\right)\\
F\left(x;c\right) & = & \left\{
\begin{array}{ccc}
\frac{1}{2}\exp\left(-\left|x\right|^{c}\right) & & x\leq0\\
1-\frac{1}{2}\exp\left(-\left|x\right|^{c}\right) & & x>0
\end{array}
\right.\\
G\left(q;c\right) & = & \left\{
\begin{array}{ccc}
-\log^{1/c}\left(\frac{1}{2q}\right) & & q\leq\frac{1}{2}\\
\log^{1/c}\left(\frac{1}{2q-1}\right) & & q>\frac{1}{2}
\end{array}
\right.\end{eqnarray*}
\[\begin{split}\mu_{n}^{\prime}=\mu_{n}=\begin{cases}
\Gamma\left(1+\frac{n}{c}\right) & n\text{ even}\\
0 & n\text{ odd}
\end{cases}\end{split}\]
\begin{eqnarray*} m_{n}=\mu & = & 0\\
\mu_{2} & = & \Gamma\left(\frac{c+2}{c}\right)\\
\gamma_{1} & = & 0\\
\gamma_{2} & = & \frac{\Gamma\left(1+\frac{4}{c}\right)}{\Gamma^{2}\left(1+\frac{2}{c}\right)}\\
m_{d} & = & \text{NA bimodal}\end{eqnarray*}
Implementation: scipy.stats.dweibull