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====MLE Parameter Estimation====
#REDIRECT [[Distributions_Used_in_Accelerated_Testing]]
The parameters of the 2-parameter Weibull distribution can also be estimated using Maximum Likelihood Estimation (MLE). This log-likelihood function is composed of :
 
<br>
::<math>\begin{align}
  & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{\beta }{\eta }{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta -1}}{{e}^{-{{\left( \tfrac{{{T}_{i}}}{\eta } \right)}^{\beta }}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }{{\left( \frac{T_{i}^{\prime }}{\eta } \right)}^{\beta }}\overset{FI}{\mathop{+\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }]
\end{align}</math>
<br>
:where:
<br>
::<math>R_{Li}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}}}</math>
<br>
::<math>R_{Ri}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}}}</math>
<br>
:• <math>{{F}_{e}}</math>  is the number of groups of times-to-failure data points.
:• <math>{{N}_{i}}</math>  is the number of times-to-failure in the  <math>{{i}^{th}}</math>  time-to-failure data group.
:• <math>\beta </math>  is the Weibull shape parameter (unknown a priori, the first of two parameters to be found).
:• <math>\eta </math>  is the Weibull scale parameter (unknown a priori, the second of two parameters to be found).
:• <math>{{T}_{i}}</math>  is the time of the  <math>{{i}^{th}}</math>  group of time-to-failure data.
:• <math>S</math>  is the number of groups of suspension data points.
:• <math>N_{i}^{\prime }</math>  is the number of suspensions in  <math>{{i}^{th}}</math>  group of suspension data points.
:• <math>T_{i}^{\prime }</math>  is the time of the  <math>{{i}^{th}}</math>  suspension data group.
:• <math>FI</math>  is the number of interval data groups.
:• <math>N_{i}^{\prime \prime }</math>  is the number of intervals in the i <math>^{th}</math>  group of data intervals.
:• <math>T_{Li}^{\prime \prime }</math>  is the beginning of the i <math>^{th}</math>  interval.
:• <math>T_{Ri}^{\prime \prime }</math>  is the ending of the i <math>^{th}</math>  interval.
<br>
<br>
The solution is found by solving for a pair of parameters  <math>\left( \widehat{\beta },\widehat{\eta } \right)</math>  so that  <math>\tfrac{\partial \Lambda }{\partial \beta }=0</math>  and  <math>\tfrac{\partial \Lambda }{\partial \eta }=0.</math>  (Other methods can also be used, such as direct maximization of the likelihood function, without having to compute the derivatives.)
<br>
::<math>\begin{align}
&\frac{\partial \Lambda }{\partial \beta }= \frac{1}{\beta }\underset{i=1}{\overset{{{F}_{e}}}{\mathop{\sum }}}\,{{N}_{i}}+\underset{i=1}{\overset{{{F}_{e}}}{\mathop{\sum }}}\,{{N}_{i}}\ln \left( \frac{{{T}_{i}}}{\eta } \right) -\underset{i=1}{\overset{{{F}_{e}}}{\mathop{\sum }}}\,{{N}_{i}}{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta }}\ln \left( \frac{{{T}_{i}}}{\eta } \right)-\underset{i=1}{\overset{S}{\mathop{\sum }}}\,N_{i}^{\prime }{{\left( \frac{T_{i}^{\prime }}{\eta } \right)}^{\beta }}\ln \left( \frac{T_{i}^{\prime }}{\eta } \right)+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{-{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}\ln (\tfrac{T_{Li}^{\prime \prime }}{\eta })R_{Li}^{\prime \prime }+{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}\ln (\tfrac{T_{Ri}^{\prime \prime }}{\eta })R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }} \\
&\frac{\partial \Lambda }{\partial \eta }= \frac{-\beta }{\eta }\underset{i=1}{\overset{{{F}_{e}}}{\mathop{\sum }}}\,{{N}_{i}}+\frac{\beta }{\eta }\underset{i=1}{\overset{{{F}_{e}}}{\mathop{\sum }}}\,{{N}_{i}}{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta }}+\frac{\beta }{\eta }\underset{i=1}{\overset{S}{\mathop{\sum }}}\,N_{i}^{\prime }{{\left( \frac{T_{i}^{\prime }}{\eta } \right)}^{\beta }}+\overset{FI}{\mathop{\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{\beta }{\eta }\frac{{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}R_{Li}^{\prime \prime }-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}R_{Ri}^{\prime \prime }}{R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }}. 
\end{align}</math>
<br>
 
=====Example 4=====
Using the same data as in the probability plotting example (Example 3), and assuming a 2-parameter Weibull distribution, estimate the parameter using the MLE method.
<br>
'''Solution'''
<br>
In this case we have non-grouped data with no suspensions, therefore the above equations become:
<br>
::<math>\frac{\partial \Lambda }{\partial \beta }=\frac{6}{\beta }+\underset{i=1}{\overset{6}{\mathop{\sum }}}\,\ln \left( \frac{{{T}_{i}}}{\eta } \right)-\underset{i=1}{\overset{6}{\mathop{\sum }}}\,{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta }}\ln \left( \frac{{{T}_{i}}}{\eta } \right)=0</math>
<br>
:and:
<br>
::<math>\frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{\beta }{\eta }\underset{i=1}{\overset{6}{\mathop \sum }}\,{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta }}=0</math>
<br>
Solving the above equations simultaneously we get:
<br>
::<math>\begin{matrix}
  \widehat{\beta }=1.933  \\
  \widehat{\eta }=73.526  \\
\end{matrix}</math>

Latest revision as of 10:17, 9 August 2012

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