Exponential Log-Likelihood Functions and their Partials
Exponential Log-Likelihood Functions and their Partials
The One-Parameter Exponential
This log-likelihood function is composed of three summation portions:
- [math]\displaystyle{ \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]\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{ \lambda }[/math] is the failure rate parameter (unknown a priori, the only parameter 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 the [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
- • and [math]\displaystyle{ T_{Ri}^{\prime \prime } }[/math] is the ending of the [math]\displaystyle{ {{i}^{th}} }[/math] interval
The solution will be found by solving for a parameter [math]\displaystyle{ \widehat{\lambda } }[/math] so that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial \lambda }=0. }[/math] Note that for [math]\displaystyle{ FI=0 }[/math] there exists a closed form solution.
- [math]\displaystyle{ \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]\displaystyle{ \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]\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{ \lambda }[/math] is the failure rate parameter (unknown a priori, the first of two parameters to be found)
- • [math]\displaystyle{ \gamma }[/math] is the location 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 the [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
- • and [math]\displaystyle{ T_{Ri}^{\prime \prime } }[/math] is the ending of the [math]\displaystyle{ {{i}^{th}} }[/math] interval
The two-parameter solution will be found by solving for a pair of parameters ([math]\displaystyle{ \widehat{\lambda },\widehat{\gamma }), }[/math] such that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial \lambda }=0,\tfrac{\partial \Lambda }{\partial \gamma }=0. }[/math] For the one-parameter case, solve for [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial \lambda }=0. }[/math]
- [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \lambda =0 }[/math]
- or:
- [math]\displaystyle{ \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]\displaystyle{ \tfrac{\partial \Lambda }{\partial \lambda }=0,\tfrac{\partial \Lambda }{\partial \gamma }=0. }[/math] It can be shown that the best solution for [math]\displaystyle{ \gamma , }[/math] satisfying the constraint that [math]\displaystyle{ \gamma \le {{T}_{1}} }[/math] is [math]\displaystyle{ \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]\displaystyle{ \lambda }[/math] such that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial \lambda }=0. }[/math]
Using this methodology, a maximum can be achieved along the [math]\displaystyle{ \lambda }[/math]-axis, and a local maximum along the [math]\displaystyle{ \gamma }[/math]-axis at [math]\displaystyle{ \gamma ={{T}_{1}} }[/math], constrained by the fact that [math]\displaystyle{ \gamma \le {{T}_{1}} }[/math]. The 3D Plot utility in Weibull++ illustrates this behavior of the log-likelihood function, as shown next:
[math]\displaystyle{ }[/math]