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:• <math>{{F}_{e}}</math>  is the number of groups of times-to-failure data points.
:• <math>{{F}_{e}}</math>  is the number of groups of times-to-failure data points.
:• .. is the number of times-to-failure in the  <math>{{i}^{th}}</math>  time-to-failure data group.
:• <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>\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>\eta </math>  is the Weibull scale parameter (unknown a priori, the second of two parameters to be found).

Revision as of 00:36, 7 February 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 i [math]\displaystyle{ ^{th} }[/math] group of data intervals.
[math]\displaystyle{ T_{Li}^{\prime \prime } }[/math] is the beginning of the i [math]\displaystyle{ ^{th} }[/math] interval.
[math]\displaystyle{ T_{Ri}^{\prime \prime } }[/math] is the ending of the i [math]\displaystyle{ ^{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, thus Eqns. (mle2w1) and (mle2w2) 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]