Exponential Log-Likelihood Functions and their Partials: Difference between revisions

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(Created page with '=== Exponential Log-Likelihood Functions and their Partials=== ==== The One-Parameter Exponential==== This log-likelihood function is composed of three summation portions: ::<m…')
 
 
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===  Exponential Log-Likelihood Functions and their Partials===
#REDIRECT [[Appendix:_Log-Likelihood_Equations]]
==== The One-Parameter Exponential====
This log-likelihood function is composed of three summation portions:
 
::<math>\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 }+\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\ln \left[ {{e}^{-\lambda T_{Li}^{\prime \prime }}}-{{e}^{-\lambda T_{Ri}^{\prime \prime }}} \right]</math>
 
:where:
::• <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>\lambda </math> is the failure rate parameter (unknown a priori, the only parameter 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 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 <math>{{i}^{th}}</math> group of data intervals
::• <math>T_{Li}^{\prime \prime }</math> is the beginning of the <math>{{i}^{th}}</math> interval
::• and <math>T_{Ri}^{\prime \prime }</math> is the ending of the  <math>{{i}^{th}}</math> interval
 
The solution will be found by solving for a parameter <math>\widehat{\lambda }</math> so that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0.</math> Note that for <math>FI=0</math> there exists a closed form solution.
 
::<math>\begin{align}
  & \frac{\partial \Lambda }{\partial \lambda }= & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left( \frac{1}{\lambda }-{{T}_{i}} \right)-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }T_{i}^{\prime } \\
&  & -\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\left[ \frac{T_{Li}^{\prime \prime }{{e}^{-\lambda T_{Li}^{\prime \prime }}}-T_{Ri}^{\prime \prime }{{e}^{-\lambda T_{Ri}^{\prime \prime }}}}{{{e}^{-\lambda T_{Li}^{\prime \prime }}}-{{e}^{-\lambda T_{Ri}^{\prime \prime }}}} \right] 
\end{align}</math>
 
====  The Two-Parameter Exponential====
This log-likelihood function for the two-parameter exponential distribution is very similar to that of the one-parameter distribution and is composed of three summation portions:
 
 
::<math>\begin{align}
  & \ln (L)= & \Lambda =\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda \left( {{T}_{i}}-\gamma  \right)}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda \left( T_{i}^{\prime }-\gamma  \right) \\
&  & \ \ +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\ln \left[ {{e}^{-\lambda \left( T_{Li}^{\prime \prime }-\gamma  \right)}}-{{e}^{-\lambda \left( T_{Ri}^{\prime \prime }-\gamma  \right)}} \right], 
\end{align}</math>
 
:where,
::• <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>\lambda </math> is the failure rate parameter (unknown a priori, the first of two parameters to be found)
::• <math>\gamma </math> is the location 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 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 <math>{{i}^{th}}</math> group of data intervals
::• <math>T_{Li}^{\prime \prime }</math> is the beginning of the <math>{{i}^{th}}</math> interval
::• and <math>T_{Ri}^{\prime \prime }</math> is the ending of the <math>{{i}^{th}}</math> interval
 
 
The two-parameter solution will be found by solving for a pair of parameters (<math>\widehat{\lambda },\widehat{\gamma }),</math> such that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0,\tfrac{\partial \Lambda }{\partial \gamma }=0.</math> For the one-parameter case, solve for <math>\tfrac{\partial \Lambda }{\partial \lambda }=0.</math>
 
::<math>\begin{align}
  \frac{\partial \Lambda }{\partial \lambda }= & \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\left[ \frac{1}{\lambda }-\left( {{T}_{i}}-\gamma  \right) \right] \\
  & -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\left( T_{i}^{\prime }-\gamma  \right) \\
  & -\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\left[ \frac{\left( T_{Li}^{\prime \prime }-\gamma  \right){{e}^{-\lambda \left( T_{Li}^{\prime \prime }-{{\gamma }_{0}} \right)}}-\left( T_{Ri}^{\prime \prime }-\gamma  \right){{e}^{-\lambda \left( T_{Ri}^{\prime \prime }-\gamma  \right)}}}{{{e}^{-\lambda \left( T_{Li}^{\prime \prime }-\gamma  \right)}}-{{e}^{-\lambda \left( T_{Ri}^{\prime \prime }-\gamma  \right)}}} \right]
\end{align}</math>
 
:and:
 
::<math>\frac{\partial \Lambda }{\partial \gamma }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}\lambda +\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\lambda </math>
 
Examination of Eqn. (expll1) will reveal that:
 
::<math>\frac{\partial \Lambda }{\partial \gamma }=\left( \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ \ +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime } \right)\lambda \equiv 0</math>
 
:or Eqn. (expll2) will be equal to zero only if either:
 
::<math>\lambda =0</math>
 
:or:
 
::<math>\left( \underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ \ +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime } \right)=0</math>
 
This is an unwelcome fact, alluded to earlier in the chapter, that essentially indicates that there is no realistic solution for the two-parameter MLE for exponential. The above equations indicate that there is no non-trivial MLE solution that satisfies both <math>\tfrac{\partial \Lambda }{\partial \lambda }=0,\tfrac{\partial \Lambda }{\partial \gamma }=0.</math>
It can be shown that the best solution for <math>\gamma ,</math> satisfying the constraint that <math>\gamma \le {{T}_{1}}</math> is <math>\gamma ={{T}_{1}}.</math> To then solve for the two-parameter exponential distribution via MLE, one can set  equal to the first time-to-failure, and then find a <math>\lambda </math> such that <math>\tfrac{\partial \Lambda }{\partial \lambda }=0.</math>
 
Using this methodology, a maximum can be achieved along the <math>\lambda </math>-axis, and a local maximum along the <math>\gamma </math>-axis at <math>\gamma ={{T}_{1}}</math>, constrained by the fact that <math>\gamma \le {{T}_{1}}</math>. The 3D Plot utility in Weibull++ illustrates this behavior of the log-likelihood function, as shown next:
 
<math></math>

Latest revision as of 20:08, 25 June 2015

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