Lognormal Log-Likelihood Functions and their Partials
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Lognormal Log-Likelihood Functions and their Partials
The general log-likelihood function (without the constant) for the lognormal distribution 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[ \frac{1}{{{\sigma }_{{{T}'}}}}\phi \left( \frac{\ln \left( {{T}_{i}} \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \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)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right) \right] \\ & \text{ }+\underset{i=1}{\overset{FI}{\mathop \sum }}\,N_{i}^{\prime \prime }\ln \left[ \Phi \left( \frac{\ln \left( T_{Ri}^{\prime \prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)-\Phi \left( \frac{\ln \left( T_{Li}^{\prime \prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \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{ {\mu }' }[/math] is the mean of the natural logarithms of the times-to-failure (unknown a priori, the first of two parameters to be found)
- • [math]\displaystyle{ {{\sigma }_{{{T}'}}} }[/math] is the standard deviation of the natural logarithms of the times-to-failure (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 solution will be found by solving for a pair of parameters [math]\displaystyle{ \left( {\mu }',{{\sigma }_{{{T}'}}} \right) }[/math] so that [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial {\mu }'}=0 }[/math] and [math]\displaystyle{ \tfrac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}=0 }[/math]:
- [math]\displaystyle{ \begin{align} \frac{\partial \Lambda }{\partial {\mu }'}= & \frac{1}{\sigma _{{{T}'}}^{2}}\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,{{N}_{i}}(\ln ({{T}_{i}})-{\mu }') \\ & +\frac{1}{{{\sigma }_{{{T}'}}}}\underset{i=1}{\overset{S}{\mathop \sum }}\,N_{i}^{\prime }\frac{\phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}{1-\Phi \left( \tfrac{\ln \left( T_{i}^{\prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)} \\ & \ \ -\underset{i=1}{\overset{FI}{\mathop \sum }}\,\frac{N_{i}^{\prime \prime }}{\sigma }\frac{\phi \left( \tfrac{\ln \left( T_{Ri}^{\prime \prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)-\phi \left( \tfrac{\ln \left( T_{Li}^{\prime \prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}{\Phi \left( \tfrac{\ln \left( T_{Ri}^{\prime \prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)-\Phi \left( \tfrac{\ln \left( T_{Li}^{\prime \prime } \right)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)} \end{align} }[/math]
[math]\displaystyle{ }[/math]
- where:
- [math]\displaystyle{ \phi \left( x \right)=\frac{1}{\sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( x \right)}^{2}}}} }[/math]
- and:
- [math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]