Template:Aw mle: Difference between revisions
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====MLE Parameter Estimation==== | ====MLE Parameter Estimation==== | ||
The parameters of the 2-parameter Weibull distribution can also be estimated using | 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> | <br> | ||
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::<math>R_{Ri}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}}}</math> | ::<math>R_{Ri}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}}}</math> | ||
<br> | <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 <math>{{i}^{th}}</math> group of data intervals. | ||
: | :* <math>T_{Li}^{\prime \prime }</math> is the beginning of the <math>{{i}^{th}}</math> interval. | ||
: | :* <math>T_{Ri}^{\prime \prime }</math> is the ending of the <math>{{i}^{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.) | |||
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 | |||
<br> | <br> | ||
::<math>\begin{align} | ::<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 \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 }} | &\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> | \end{align}</math>. | ||
=====Example 4===== | =====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. | 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> | <br> | ||
'''Solution''' | '''Solution''' | ||
In this case we have non-grouped data with no suspensions, therefore the above equations become: | In this case we have non-grouped data with no suspensions, therefore the above equations become: | ||
<br> | <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> | ::<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> | ||
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<br> | <br> | ||
Solving the above equations simultaneously we get: | Solving the above equations simultaneously we get: | ||
<br> | <br> | ||
::<math>\begin{matrix} | ::<math>\begin{matrix} |
Revision as of 00:10, 7 March 2012
MLE Parameter Estimation
The parameters of the 2-parameter Weibull distribution can also be estimated using maximum likelihood estimation (MLE). This log-likelihood function is composed of :
- [math]\displaystyle{ \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]
- where:
- [math]\displaystyle{ R_{Li}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}}} }[/math]
- [math]\displaystyle{ R_{Ri}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}}} }[/math]
- [math]\displaystyle{ {{F}_{e}} }[/math] is the number of groups of times-to-failure data points.
- [math]\displaystyle{ {{N}_{i}} }[/math] is the number of times-to-failure in the [math]\displaystyle{ {{i}^{th}} }[/math] time-to-failure data group.
- [math]\displaystyle{ \beta }[/math] is the Weibull shape parameter (unknown a priori, the first of two parameters to be found).
- [math]\displaystyle{ \eta }[/math] is the Weibull scale parameter (unknown a priori, the second of two parameters to be found).
- [math]\displaystyle{ {{T}_{i}} }[/math] is the time of the [math]\displaystyle{ {{i}^{th}} }[/math] group of time-to-failure data.
- [math]\displaystyle{ S }[/math] is the number of groups of suspension data points.
- [math]\displaystyle{ N_{i}^{\prime } }[/math] is the number of suspensions in [math]\displaystyle{ {{i}^{th}} }[/math] group of suspension data points.
- [math]\displaystyle{ T_{i}^{\prime } }[/math] is the time of the [math]\displaystyle{ {{i}^{th}} }[/math] suspension data group.
- [math]\displaystyle{ FI }[/math] is the number of interval data groups.
- [math]\displaystyle{ N_{i}^{\prime \prime } }[/math] is the number of intervals in the [math]\displaystyle{ {{i}^{th}} }[/math] group of data intervals.
- [math]\displaystyle{ T_{Li}^{\prime \prime } }[/math] is the beginning of the [math]\displaystyle{ {{i}^{th}} }[/math] interval.
- [math]\displaystyle{ T_{Ri}^{\prime \prime } }[/math] is the ending of the [math]\displaystyle{ {{i}^{th}} }[/math] interval.
The solution is found by solving for a pair of parameters [math]\displaystyle{ \left( \widehat{\beta },\widehat{\eta } \right) }[/math] so that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial \beta }=0 }[/math] and [math]\displaystyle{ \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.)
- [math]\displaystyle{ \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].
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.
Solution
In this case we have non-grouped data with no suspensions, therefore the above equations become:
- [math]\displaystyle{ \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]
- and:
- [math]\displaystyle{ \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]
Solving the above equations simultaneously we get:
- [math]\displaystyle{ \begin{matrix} \widehat{\beta }=1.933 \\ \widehat{\eta }=73.526 \\ \end{matrix} }[/math]