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Lisa Hacker: Redirected page to Distributions Used in Accelerated Testing

2012-08-09T10:16:47Z

Redirected page to Distributions Used in Accelerated Testing

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	~~{{aw mle}}~~		#REDIRECT [[Distributions_Used_in_Accelerated_Testing]]

Nicolette Young: /* MLE Parameter Estimation */

2012-01-12T18:46:50Z

MLE Parameter Estimation

← Older revision		Revision as of 18:46, 12 January 2012
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	~~==MLE Parameter Estimation==~~		{{aw mle}}
	~~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.~~
	~~:• .. 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, thus Eqns. (mle2w1) and (mle2w2) 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>~~

Nicolette Young: Created page with '==MLE Parameter Estimation== The parameters of the 2-parameter Weibull distribution can also be estimated using Maximum Likelihood Estimation (MLE). This log-likelihood function …'

2012-01-12T18:15:33Z

Created page with '==MLE Parameter Estimation== The parameters of the 2-parameter Weibull distribution can also be estimated using Maximum Likelihood Estimation (MLE). This log-likelihood function …'

New page

==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>\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>R_{Li}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Li}^{\prime \prime }}{\eta })}^{\beta }}}}</math>
 
::<math>R_{Ri}^{\prime \prime }={{e}^{-{{(\tfrac{T_{Ri}^{\prime \prime }}{\eta })}^{\beta }}}}</math>
 
:• <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>\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.
 
 
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.)
 
::<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>
 

====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>\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>\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>\begin{matrix}
\widehat{\beta }=1.933 \\
\widehat{\eta }=73.526 \\
\end{matrix}</math>