Editing Weibull distribution (section)

==Parameter estimation==
===Ordinary least square using Weibull plot===
[[File:Weibull qq.svg|thumb|right|Weibull plot]]
The fit of a Weibull distribution to data can be visually assessed using a Weibull plot.<ref>{{cite web|url=http://www.itl.nist.gov/div898/handbook/eda/section3/weibplot.htm|title=1.3.3.30. Weibull Plot|website=www.itl.nist.gov}}</ref> The Weibull plot is a plot of the [[empirical cumulative distribution function]] <math>\widehat F(x)</math> of data on special axes in a type of [[Q–Q plot]]. The axes are <math>\ln(-\ln(1-\widehat F(x)))</math> versus <math>\ln(x)</math>. The reason for this change of variables is the cumulative distribution function can be linearized:
:<math>\begin{align}
F(x) &= 1-e^{-(x/\lambda)^k}\\[4pt]
-\ln(1-F(x)) &= (x/\lambda)^k\\[4pt]
\underbrace{\ln(-\ln(1-F(x)))}_{\textrm{'y'}} &= \underbrace{k\ln x}_{\textrm{'mx'}} - \underbrace{k\ln \lambda}_{\textrm{'c'}}
\end{align}
</math>
which can be seen to be in the standard form of a straight line. Therefore, if the data came from a Weibull distribution then a straight line is expected on a Weibull plot.

There are various approaches to obtaining the empirical distribution function from data. One method is to obtain the vertical coordinate for each point using
:<math>\widehat F = \frac{i-0.3}{n+0.4}</math>,
where <math>i</math> is the rank of the data point and <math>n</math> is the number of data points.<ref>Wayne Nelson (2004) ''Applied Life Data Analysis''. Wiley-Blackwell {{ISBN|0-471-64462-5}}</ref><ref>{{Cite journal |last=Barnett |first=V. |date=1975 |title=Probability Plotting Methods and Order Statistics |url=https://www.jstor.org/stable/2346708 |journal=Journal of the Royal Statistical Society. Series C (Applied Statistics) |volume=24 |issue=1 |pages=95–108 |doi=10.2307/2346708 |jstor=2346708 |issn=0035-9254}}</ref> Another common estimator<ref>{{Cite ISO standard | csnumber = 69875 | title = ISO 20501:2019 – Fine ceramics (advanced ceramics, advanced technical ceramics) – Weibull statistics for strength data}}</ref> is 
:<math>\widehat F = \frac{i-0.5}{n}</math>.

Linear regression can also be used to numerically assess goodness of fit and estimate the parameters of the Weibull distribution. The gradient informs one directly about the shape parameter <math>k</math> and the scale parameter <math>\lambda</math> can also be inferred.

===Method of moments===
The [[coefficient of variation]] of Weibull distribution depends only on the shape parameter:<ref name="Cohen1965">{{cite journal | url=https://www.stat.cmu.edu/technometrics/59-69/VOL-07-04/v0704579.pdf |title=Maximum Likelihood Estimation in the Weibull Distribution Based on Complete and on Censored Samples |first=A. Clifford |last=Cohen | journal=Technometrics |issue=4 |volume=7 |date=Nov 1965 | pages=579–588|doi=10.1080/00401706.1965.10490300 }}</ref> 

:<math>CV^2 = \frac{\sigma^2}{\mu^2} 
 = \frac{\Gamma\left(1+\frac{2}{k}\right) - \left(\Gamma\left(1+\frac{1}{k}\right)\right)^2}{\left(\Gamma\left(1+\frac{1}{k}\right)\right)^2}.</math>

Equating the sample quantities <math>s^2/\bar{x}^2</math> to <math>\sigma^2/\mu^2</math>, the moment estimate of the shape parameter <math>k</math> can be read off either from a look up table or a graph of <math>CV^2</math> versus <math>k</math>. A more accurate estimate of <math>\hat{k}</math> can be found using a root finding algorithm to solve

:<math>\frac{\Gamma\left(1+\frac{2}{k}\right) - \left(\Gamma\left(1+\frac{1}{k}\right)\right)^2}{\left(\Gamma\left(1+\frac{1}{k}\right)\right)^2} = \frac{s^2}{\bar{x}^2}.</math>

The moment estimate of the scale parameter can then be found using the first moment equation as 

:<math>\hat{\lambda} = \frac{\bar{x}}{\Gamma\left(1 + \frac{1}{\hat{k}}\right)}.</math>

===Maximum likelihood===
The [[Maximum likelihood estimation|maximum likelihood estimator]] for the <math>\lambda</math> parameter given <math>k</math> is<ref name="Cohen1965"/>

:<math>\widehat \lambda = \left(\frac{1}{n} \sum_{i=1}^n x_i^k \right)^\frac{1}{k} </math>

The maximum likelihood estimator for <math>k</math> is the solution for ''k'' of the following equation<ref name="Sornette, D. 2004">{{cite book
 | author = Sornette, D.
 | year = 2004
 | title = Critical Phenomena in Natural Science: Chaos, Fractals, Self-organization, and Disorder}}.</ref>
:<math>
  0 =  \frac{\sum_{i=1}^n x_i^k \ln x_i }{\sum_{i=1}^n x_i^k }
                  - \frac{1}{k} - \frac{1}{n} \sum_{i=1}^n \ln x_i
</math>

This equation defines <math>\widehat k</math> only implicitly, one must generally solve for <math>k</math> by numerical means.

When <math>x_1 > x_2 > \cdots > x_N</math> are the <math>N</math> largest observed samples from a dataset of more than <math>N</math> samples, then the maximum likelihood estimator for the <math>\lambda</math> parameter given <math>k</math> is<ref name="Sornette, D. 2004"/>

:<math>\widehat \lambda^k = \frac{1}{N} \sum_{i=1}^N (x_i^k - x_N^k)</math>

Also given that condition, the maximum likelihood estimator for <math>k</math> is{{citation needed|date=December 2017}}
:<math>
  0 = \frac{\sum_{i=1}^N (x_i^k \ln x_i -  x_N^k \ln x_N)}
                       {\sum_{i=1}^N (x_i^k - x_N^k)}
                  - \frac{1}{N} \sum_{i=1}^N \ln x_i
</math>

Again, this being an implicit function, one must generally solve for <math>k</math> by numerical means.