1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65
|
.. _continuous-genpareto:
Generalized Pareto Distribution
===============================
There is one shape parameter :math:`c\neq0`. The support is :math:`x\geq0` if :math:`c>0`,
and :math:`0\leq x<\frac{1}{\left|c\right|}` if :math:`c` is negative.
.. math::
:nowrap:
\begin{eqnarray*} f\left(x;c\right) & = & \left(1+cx\right)^{-1-\frac{1}{c}}\\
F\left(x;c\right) & = & 1-\frac{1}{\left(1+cx\right)^{1/c}}\\
G\left(q;c\right) & = & \frac{1}{c}\left[\left(\frac{1}{1-q}\right)^{c}-1\right]\end{eqnarray*}
.. Comment Is it \gamma\left(-\frac{1}{c},-\frac{t}{c}\right) or \Gamma\left(-\frac{1}{c},-\frac{t}{c}\right) below?
.. math::
M\left(t\right) = \left\{
\begin{array}{cc}
\left(-\frac{t}{c}\right)^{\frac{1}{c}}
e^{-\frac{t}{c}}
\left[
\Gamma\left(1-\frac{1}{c}\right)
+ \left(\gamma\left(-\frac{1}{c},-\frac{t}{c}\right) / \Gamma\left(\frac{1}{-c}\right)\right)
- \pi\csc\left(\frac{\pi}{c}\right)/\Gamma\left(\frac{1}{c}\right)
\right] & c>0\\
\left(
\frac{\left|c\right|}{t}\right)^{1/\left|c\right|}
\Gamma\left(\frac{1}{\left|c\right|}, \frac{t}{\left|c\right|}\right)
\frac{1}{\Gamma\left(\frac{1}{|c|}\right)}
& c<0
\end{array}
\right.
.. math::
\mu_{n}^{\prime}=\frac{\left(-1\right)^{n}}{c^{n}}\sum_{k=0}^{n}\binom{n}{k}\frac{\left(-1\right)^{k}}{1-ck}\quad \text{ if }cn<1
.. math::
:nowrap:
\begin{eqnarray*} \mu_{1}^{\prime} & = & \frac{1}{1-c}\quad c<1\\
\mu_{2}^{\prime} & = & \frac{2}{\left(1-2c\right)\left(1-c\right)}\quad c<\frac{1}{2}\\
\mu_{3}^{\prime} & = & \frac{6}{\left(1-c\right)\left(1-2c\right)\left(1-3c\right)}\quad c<\frac{1}{3}\\
\mu_{4}^{\prime} & = & \frac{24}{\left(1-c\right)\left(1-2c\right)\left(1-3c\right)\left(1-4c\right)}\quad c<\frac{1}{4}\end{eqnarray*}
Thus,
.. math::
:nowrap:
\begin{eqnarray*} \mu & = & \mu_{1}^{\prime}\\
\mu_{2} & = & \mu_{2}^{\prime}-\mu^{2}\\
\gamma_{1} & = & \frac{\mu_{3}^{\prime}-3\mu\mu_{2}-\mu^{3}}{\mu_{2}^{3/2}}\\
\gamma_{2} & = & \frac{\mu_{4}^{\prime}-4\mu\mu_{3}-6\mu^{2}\mu_{2}-\mu^{4}}{\mu_{2}^{2}}-3\end{eqnarray*}
.. math::
h\left[X\right]=1+c\quad c>0
Implementation: `scipy.stats.genpareto`
|