WeibullParetoDist

This package enables you to implement the Weibull distribution parameterised to include the Pareto-approximated parameter.

The following parameterisation of the Weibull distribution is implemented in Distribution.jl package, and see the source file here.
$f(x;\alpha, \theta) = \frac{\alpha}{\theta} \left( \frac{x}{\theta} \right)^{\alpha-1} e^{-(x/\theta)^\alpha}$

The Weibull distribution has a property to be power-law if $\alpha$ takes less than 1 value. Following the definition Ivan et al., 2018 , the power-law distribution is power-law if the complementary cumulative function (that is a survival function), $\bar{F(k)}$, follows $\bar{F}(k)=l(k)k^{-\alpha}$. Since the form of the complementary cumulative function for the Weibull distribution is $\bar{F(k)} = e^{-(x/\theta)^\alpha}$, which satisfies the definition of power-law.

However, when the Weibull distribution is fitted to the data which follows the power-law, the order of estimates of $\theta$ becomes $10{^-3}$ and $10^{-8}$. Also, the strong correlation between $\alpha$ and $\theta$ is observed, which hampers the stable estimation.

As Akira et al., 2022 proved in the supplementary material, let $\kappa$ be $\alpha/\theta^\alpha$, and replace $\kappa$ with $\alpha$. At the region of 0 in log-log plot of the distribution, $\log \bar{F}(k)=-\kappa/\alpha e^{\alpha \log x} $, the slope becomes $-\kappa$, which does not vary so much, and empirically the correlation between $\alpha$ and $\kappa$ becomes very week.

In this package, we implemented the above-parameterised version of the Weibull distribution.

α = 0.5
κ = 1.1
wb = WeibullPareto(α, κ)
x = 1:100
pdf.(wb, x)
cdf.(wb, x)
ccdf.(wb, x)
mean(wb)
var(wb)
sd(wb)

Also, truncated version is surpported.

α = 0.5
κ = 1.1
wb = WeibullPareto(α, κ)
l = 1
u = 10
wb_doubly = truncated(wb, l, u)
wb_left = truncated(wb, l, nothing)
wb_left = truncated(wb; lower=l)
wb_right = truncated(wb, nothing, u)
wb_right = truncated(wb; upper=u)

The relationship between $\kappa$ and $\theta$.
$\theta = (\frac{\alpha}{\kappa})^{1/\alpha}$ and $\kappa=\frac{\alpha}{\theta^\alpha}$.

$pdf(x) = \kappa x^{\alpha-1}e^{-x^\alpha \kappa/\alpha}$
$cdf(x) = 1-e^{-x^\alpha\kappa/\alpha}$

Let $l$ be the left truncation point.

$pdf(x) = \kappa x^{\alpha-1}e^{-(x^\alpha - l^\alpha)\kappa/\alpha} $
$cdf(x) = 1 - e^{-(x^\alpha - l^\alpha)\kappa/\alpha} $

Let $r$ be the right truncation point.

$pdf(x) = (\kappa x^{\alpha-1}e^{-x^\alpha\kappa/\alpha})/( {e^{-l^\alpha \kappa/\alpha} - e^{-r^\alpha \kappa/\alpha}}) $
$cdf(x) = (1 - e^{-x^\alpha\kappa/\alpha})/({e^{-l^\alpha \kappa/\alpha} - e^{-r^\alpha \kappa/\alpha}}) $