Template:Aae mle

Maximum Likelihood Estimation Method
The log-likelihood function for the exponential distribution is as shown next:


 * $$\begin{align}

& \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 } \\ & & \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}$$ where:


 * $$R_{Li}^{\prime \prime }={{e}^{-\lambda T_{Li}^{\prime \prime }}}$$


 * $$R_{Ri}^{\prime \prime }={{e}^{-\lambda T_{Ri}^{\prime \prime }}}$$

and: •	 $${{F}_{e}}$$ is the number of groups of exact 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).

•	 $${{T}_{i}}$$ is the exact failure time of the  $${{i}^{th}}$$  group.

•	 $$S$$ 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.

•	 $$T_{Ri}^{\prime \prime }$$ is the ending of the i $$^{th}$$  interval. Substituting the Arrhenius-exponential model into the log-likelihood function yields:


 * $$\begin{align}

& \Lambda = & \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}{C\cdot {{e}^{\tfrac{B}}}}{{e}^{-\tfrac{1}{C\cdot {{e}^{\tfrac{B}}}}{{T}_{i}}}} \right] \\ & & -\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{1}{C\cdot {{e}^{\tfrac{B}}}}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}$$

where:


 * $$R_{Li}^{\prime \prime }={{e}^{-\tfrac{T_{Li}^{\prime \prime }}{C{{e}^{\tfrac{B}}}}}}$$


 * $$R_{Ri}^{\prime \prime }={{e}^{-\tfrac{T_{Ri}^{\prime \prime }}{C{{e}^{\tfrac{B}}}}}}$$

The solution (parameter estimates) will be found by solving for the parameters $$\widehat{B},$$   $$\widehat{C}$$  so that  $$\tfrac{\partial \Lambda }{\partial B}=0$$  and  $$\tfrac{\partial \Lambda }{\partial C}=0$$, where:


 * $$\begin{align}

& \frac{\partial \Lambda }{\partial B}= & \frac{1}{C}\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\left( \frac-\frac{C} \right)+\frac{1}{C}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{T_{i}^{\prime }} \\ & & \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}}}} \end{align}$$


 * $$\begin{align}

& \frac{\partial \Lambda }{\partial C}= & \frac{1}{C}\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\left( \frac{C{{e}^{\tfrac{B}}}}-1 \right)+\frac{1}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{T_{i}^{\prime }} \\ & & \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}}}} \end{align}$$