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====Maximum Likelihood Estimation Method====
#REDIRECT [[Arrhenius_Relationship#Parameter_Estimation]]
<br>
The log-likelihood function for the exponential distribution is as shown next:
 
<br>
::<math>\begin{align}
  & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime } \
& \overset{FI}{\mathop{\underset{i=1}{\mathop{+\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\ln [\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime })] 
\end{align}</math>
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where:
 
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::<math>R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}</math>
 
 
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::<math>R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}</math>
 
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and:
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• <math>{{F}_{e}}</math>  is the number of groups of exact 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>\lambda </math>  is the failure rate parameter (unknown).
 
• <math>{{T}_{i}}</math>  is the exact failure time of the  <math>{{i}^{th}}</math>  group.
 
• <math>S</math>  is the number of groups of suspension data points.
 
• <math>N_{i}^{\prime }</math>  is the number of suspensions in the  <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>
Substituting the Arrhenius-exponential model into the log-likelihood function yields:
 
<br>
::<math>\begin{align}
  & \Lambda = & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}{C\cdot {{e}^{\tfrac{B}{{{V}_{i}}}}}}{{e}^{-\tfrac{1}{C\cdot {{e}^{\tfrac{B}{{{V}_{i}}}}}}{{T}_{i}}}} \right] \\
&  & -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{1}{C\cdot {{e}^{\tfrac{B}{{{V}_{i}}}}}}T_{i}^{\prime }+\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 }}{C{{e}^{\tfrac{B}{{{V}_{i}}}}}}}}</math>
 
<br>
::<math>R_{Ri}^{\prime \prime }={{e}^{-\tfrac{T_{Ri}^{\prime \prime }}{C{{e}^{\tfrac{B}{{{V}_{i}}}}}}}}</math>
 
 
<br>
The solution (parameter estimates) will be found by solving for the parameters  <math>\widehat{B},</math>  <math>\widehat{C}</math>  so that  <math>\tfrac{\partial \Lambda }{\partial B}=0</math>  and  <math>\tfrac{\partial \Lambda }{\partial C}=0</math> , where:
 
 
<br>
::<math>\begin{align}
  & \frac{\partial \Lambda }{\partial B}= & \frac{1}{C}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{{{T}_{i}}}{{{V}_{i}}{{e}^{\tfrac{B}{{{V}_{i}}}}}}-\frac{C}{{{V}_{i}}} \right)+\frac{1}{C}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{T_{i}^{\prime }}{{{V}_{i}}{{e}^{\tfrac{B}{{{V}_{i}}}}}} \\
&  & \overset{FI}{\mathop{\underset{i=1}{\mathop{+\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{(R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime })C{{V}_{i}}{{e}^{\tfrac{B}{{{V}_{i}}}}}} 
\end{align}</math>
 
 
<br>
::<math>\begin{align}
  & \frac{\partial \Lambda }{\partial C}= & \frac{1}{C}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{{{T}_{i}}}{C{{e}^{\tfrac{B}{{{V}_{i}}}}}}-1 \right)+\frac{1}{{{C}^{2}}}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{T_{i}^{\prime }}{{{e}^{\tfrac{B}{{{V}_{i}}}}}} \\
&  & \overset{FI}{\mathop{\underset{i=1}{\mathop{+\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{T_{Li}^{\prime \prime }R_{Li}^{\prime \prime }-T_{Ri}^{\prime \prime }R_{Ri}^{\prime \prime }}{(R_{Li}^{\prime \prime }-R_{Ri}^{\prime \prime }){{C}^{2}}{{e}^{\tfrac{B}{{{V}_{i}}}}}} 
\end{align}</math>
 
<br>

Latest revision as of 04:05, 16 August 2012

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