Normal Log-Likelihood Functions and their Partials

Normal Log-Likelihood Functions and their Partials
The complete normal likelihood function (without the constant) is composed of three summation portions:


 * $$\begin{align}

\ln (L)= & \Lambda =\underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\ln \left[ \frac{1}{\sigma }\phi \left( \frac{{{T}_{i}}-\mu }{\sigma } \right) \right] \\ & +\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{^{\prime }}\ln \left[ 1-\Phi \left( \frac{T_{i}^{^{\prime }}-\mu }{\sigma } \right) \right] \\ & \text{ }+\underset{i=1}{\overset{\mathop \sum }}\,N_{i}^{^{\prime \prime }}\ln \left[ \Phi \left( \frac{T_^{^{\prime \prime }}-\mu }{\sigma } \right)-\Phi \left( \frac{T_^{^{\prime \prime }}-\mu }{\sigma } \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
 * •	$$\mu $$ is the mean parameter (unknown a priori, the first of two parameters to be found)
 * •	$$\sigma $$ is the standard deviation 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
 * •	$${{F}_{i}}$$ 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 pair of parameters $$\left( {{\mu }_{0}},{{\sigma }_{0}} \right)$$ so that $$\tfrac{\partial \Lambda }{\partial \mu }=0$$ and $$\tfrac{\partial \Lambda }{\partial \sigma }=0.$$


 * $$\begin{align}

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


 * $$\begin{align}

\frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{\mathop \sum }}\,{{N}_{i}}\left( \frac-\frac{1}{\sigma } \right) \\ & +\frac{1}{\sigma }\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\left( \tfrac{T_{i}^{\prime }-\mu }{\sigma } \right)\phi \left( \tfrac{T_{i}^{\prime }-\mu }{\sigma } \right)}{1-\Phi \left( \tfrac{T_{i}^{\prime }-\mu }{\sigma } \right)} \\ & -\frac{1}{\sigma }\underset{i=1}{\overset{\mathop \sum }}\,N_{i}^{\prime \prime }\frac{\left( \tfrac{T_{Ri}^{\prime \prime }-\mu }{\sigma } \right)\phi \left( \tfrac{T_{Ri}^{\prime \prime }-\mu }{\sigma } \right)-\left( \tfrac{T_{Li}^{\prime \prime }-\mu }{\sigma } \right)\phi \left( \tfrac{T_{Li}^{\prime \prime }-\mu }{\sigma } \right)}{\Phi \left( \tfrac{T_{Ri}^{\prime \prime }-\mu }{\sigma } \right)-\Phi \left( \tfrac{T_{Li}^{\prime \prime }-\mu }{\sigma } \right)} \end{align}$$


 * where:


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


 * and:


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

Complete Data
Note that for the normal distribution, and in the case of complete data only (as was shown in Chapter 3), there exists a closed-form solution for both of the parameters or:


 * $$\widehat{\mu }=\widehat=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{T}_{i}}$$


 * and:


 * $$\begin{align}

\hat{\sigma }_{T}^{2}= & \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \\ {{{\hat{\sigma }}}_{T}}= & \sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}} \end{align}$$