Template:Eyring-log mle

Maximum Likelihood Estimation Method
The complete Eyring-lognormal log-likelihood function is composed of two summation portions:


 * $$\begin{align}

& \ln (L)= & \Lambda =\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}\phi \left( \frac{\ln \left( {{T}_{i}} \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right) \right] \\ & & \text{ }+\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\ln \left[ 1-\Phi \left( \frac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right) \right] \\ & & +\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:


 * $$z_{Li}^{\prime \prime }=\frac{\ln T_{Li}^{\prime \prime }+\ln {{V}_{i}}+A-\tfrac{B}}{\sigma _{T}^{\prime }}$$


 * $$z_{Ri}^{\prime \prime }=\frac{\ln T_{Ri}^{\prime \prime }+\ln {{V}_{i}}+A-\tfrac{B}}{\sigma _{T}^{\prime }}$$


 * and:

•	 $${{F}_{e}}$$ is the number of groups of exact times-to-failure data points. •	 $${{N}_{i}}$$ is the number of times-to-failure data points in the  $${{i}^{th}}$$  time-to-failure data group. •	 $${{\sigma }_}$$ is the standard deviation of the natural logarithm of the times-to-failure (unknown, the first of three parameters to be estimated). •	 $$A$$ is the Eyring parameter (unknown, the second of three parameters to be estimated). •	 $$C$$ is the second Eyring parameter (unknown, the third of three parameters to be estimated). •	 $${{V}_{i}}$$ is the stress level of the  $${{i}^{th}}$$  group. •	 $${{T}_{i}}$$ 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 running 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. The solution (parameter estimates) will be found by solving for $${{\widehat{\sigma }}_},$$   $$\widehat{A},$$   $$\widehat{B}$$  so that  $$\tfrac{\partial \Lambda }{\partial {{\sigma }_}}=0,$$   $$\tfrac{\partial \Lambda }{\partial A}=0$$  and  $$\tfrac{\partial \Lambda }{\partial B}=0$$ :


 * $$\begin{align}

& \frac{\partial \Lambda }{\partial A}= & -\frac{1}{\sigma _^{2}}\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}(\ln ({{T}_{i}})+\ln ({{V}_{i}})+A-\frac{B}) \\ & & -\frac{1}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)} \\ & & \overset{FI}{\mathop{+\underset{i=1}{\mathop{\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{\varphi (z_{Ri}^{\prime \prime })-\varphi (z_{Li}^{\prime \prime })}{\sigma _{T}^{\prime }(\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime }))} \end{align}$$


 * $$\begin{align}

& \frac{\partial \Lambda }{\partial B}= & \frac{1}{\sigma _^{2}}\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\frac{1}(\ln ({{T}_{i}})+\ln ({{V}_{i}})+A-\frac{B}) \\ & & +\frac{1}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{1}\frac{\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)} \\ & & \overset{FI}{\mathop{\underset{i=1}{\mathop{-\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{\varphi (z_{Ri}^{\prime \prime })-\varphi (z_{Li}^{\prime \prime })}{\sigma _{T}^{\prime }{{V}_{i}}(\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime }))} \\ & \frac{\partial \Lambda }{\partial {{\sigma }_}}= & \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\left( \frac{\sigma _^{3}}-\frac{1} \right) \\ & & +\frac{1}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)+\ln ({{V}_{i}})+A-\tfrac{B}} \right)} \\ & & \overset{FI}{\mathop{\underset{i=1}{\mathop{-\underset{}{\overset{}{\mathop \sum }}\,}}\,}}\,N_{i}^{\prime \prime }\frac{z_{Ri}^{\prime \prime }\varphi (z_{Ri}^{\prime \prime })-z_{Li}^{\prime \prime }\varphi (z_{Li}^{\prime \prime })}{\sigma _{T}^{\prime }(\Phi (z_{Ri}^{\prime \prime })-\Phi (z_{Li}^{\prime \prime }))} \end{align}$$


 * and:


 * $$\phi \left( x \right)=\frac{1}{\sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( x \right)}^{2}}}}$$


 * $$\Phi (x)=\frac{1}{\sqrt{2\pi }}\mathop{}_{-\infty }^{x}{{e}^{-\tfrac{2}}}dt$$