Exponential Log-Likelihood Functions and their Partials

The One-Parameter Exponential
This log-likelihood function is composed of three summation portions:


 * $$\ln (L)=\Lambda =\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \lambda {{e}^{-\lambda {{T}_{i}}}} \right]-\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda T_{i}^{\prime }+\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\ln \left[ {{e}^{-\lambda T_{Li}^{\prime \prime }}}-{{e}^{-\lambda T_{Ri}^{\prime \prime }}} \right]$$


 * where:
 * •	$${{F}_{e}}$$ is the number of groups of times-to-failure data points
 * •	$${{N}_{i}}$$ is the number of times-to-failure in the $${{i}^{th}}$$ time-to-failure data group
 * •	$$\lambda $$ is the failure rate parameter (unknown a priori, the only parameter to be found)
 * •	$${{T}_{i}}$$ is the time of the $${{i}^{th}}$$ group of time-to-failure data
 * •	$$S$$ is the number of groups of suspension data points
 * •	$$N_{i}^{\prime }$$ is the number of suspensions in the $${{i}^{th}}$$ group of suspension data points
 * •	$$T_{i}^{\prime }$$ is the time of the $${{i}^{th}}$$ suspension data group
 * •	$$FI$$ is the number of interval data groups
 * •	$$N_{i}^{\prime \prime }$$ is the number of intervals in the $${{i}^{th}}$$ group of data intervals
 * •	$$T_{Li}^{\prime \prime }$$ is the beginning of the $${{i}^{th}}$$ interval
 * •	and $$T_{Ri}^{\prime \prime }$$ is the ending of the $${{i}^{th}}$$ interval

The solution will be found by solving for a parameter $$\widehat{\lambda }$$ so that $$\tfrac{\partial \Lambda }{\partial \lambda }=0.$$ Note that for $$FI=0$$ there exists a closed form solution.


 * $$\begin{align}

& \frac{\partial \Lambda }{\partial \lambda }= & \underset{i=1}{\overset{\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}$$

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:


 * $$\begin{align}

& \ln (L)= & \Lambda =\underset{i=1}{\overset{\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}$$


 * where,
 * •	$${{F}_{e}}$$ is the number of groups of times-to-failure data points
 * •	$${{N}_{i}}$$ is the number of times-to-failure in the $${{i}^{th}}$$ time-to-failure data group
 * •	$$\lambda $$ is the failure rate parameter (unknown a priori, the first of two parameters to be found)
 * •	$$\gamma $$ is the location parameter (unknown a priori, the second of two parameters to be found)
 * •	$${{T}_{i}}$$ is the time of the $${{i}^{th}}$$ group of time-to-failure data
 * •	$$S$$ is the number of groups of suspension data points
 * •	$$N_{i}^{\prime }$$ is the number of suspensions in the $${{i}^{th}}$$ group of suspension data points
 * •	$$T_{i}^{\prime }$$ is the time of the $${{i}^{th}}$$ suspension data group
 * •	$$FI$$ is the number of interval data groups
 * •	$$N_{i}^{\prime \prime }$$ is the number of intervals in the $${{i}^{th}}$$ group of data intervals
 * •	$$T_{Li}^{\prime \prime }$$ is the beginning of the $${{i}^{th}}$$ interval
 * •	and $$T_{Ri}^{\prime \prime }$$ is the ending of the $${{i}^{th}}$$ interval

The two-parameter solution will be found by solving for a pair of parameters ($$\widehat{\lambda },\widehat{\gamma }),$$ such that $$\tfrac{\partial \Lambda }{\partial \lambda }=0,\tfrac{\partial \Lambda }{\partial \gamma }=0.$$ For the one-parameter case, solve for $$\tfrac{\partial \Lambda }{\partial \lambda }=0.$$


 * $$\begin{align}

\frac{\partial \Lambda }{\partial \lambda }= & \underset{i=1}{\overset{\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}$$


 * and:


 * $$\frac{\partial \Lambda }{\partial \gamma }=\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\lambda +\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\lambda +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\lambda $$

Examination of Eqn. (expll1) will reveal that:


 * $$\frac{\partial \Lambda }{\partial \gamma }=\left( \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ \ +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime } \right)\lambda \equiv 0$$


 * or Eqn. (expll2) will be equal to zero only if either:


 * $$\lambda =0$$


 * or:


 * $$\left( \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ \ +\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime } \right)=0$$

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 $$\tfrac{\partial \Lambda }{\partial \lambda }=0,\tfrac{\partial \Lambda }{\partial \gamma }=0.$$ It can be shown that the best solution for $$\gamma ,$$ satisfying the constraint that $$\gamma \le {{T}_{1}}$$ is $$\gamma ={{T}_{1}}.$$ 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 $$\lambda $$ such that $$\tfrac{\partial \Lambda }{\partial \lambda }=0.$$

Using this methodology, a maximum can be achieved along the $$\lambda $$-axis, and a local maximum along the $$\gamma $$-axis at $$\gamma ={{T}_{1}}$$, constrained by the fact that $$\gamma \le {{T}_{1}}$$. The 3D Plot utility in Weibull++ illustrates this behavior of the log-likelihood function, as shown next:

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