Template:Example: Lognormal Distribution Likelihood Ratio Bound (Parameters): Difference between revisions

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'''Lognormal Distribution Likelihood Ratio Bound Example (Parameters)'''
#REDIRECT [[The Lognormal Distribution]]
 
Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assuming a lognormal distribution, the MLE parameter estimates are calculated to be  <math>{{\widehat{\mu }}^{\prime }}=4.2926</math>  and  <math>{{\widehat{\sigma'}}}=0.32361.</math>  Calculate the two-sided 75% confidence bounds on these parameters using the likelihood ratio method.
 
'''Solution'''
 
The first step is to calculate the likelihood function for the parameter estimates:
 
<center><math>\begin{align}
  L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}}), \\
  = & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma' }}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma' }}}} \right)}^{2}}}} \\
  L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\
  L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & 1.115256\times {{10}^{-10}} 
\end{align}</math></center>
 
where  <math>{{x}_{i}}</math>  are the original time-to-failure data points. We can now rearrange the likelihod ratio equation to the form:
 
::<math>L({\mu }',{{\sigma' }})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
 
Since our specified confidence level,  <math>\delta </math> , is 75%, we can calculate the value of the chi-squared statistic,  <math>\chi _{0.75;1}^{2}=1.323303.</math>  We can now substitute this information into the equation:
 
::<math>\begin{align}
  & L({\mu }',{{\sigma' }})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\
& L({\mu }',{{\sigma'}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\
& L({\mu }',{{\sigma'}})-5.754703\times {{10}^{-11}}= & 0 
\end{align}</math>
 
It now remains to find the values of  <math>{\mu }'</math>  and  <math>{{\sigma'}}</math>  which satisfy this equation. This is an iterative process that requires setting the value of  <math>{{\sigma'}}</math>  and finding the appropriate values of  <math>{\mu }'</math> , and vice versa.
 
The following table gives the values of  <math>{\mu }'</math>  based on given values of  <math>{{\sigma'}}</math> .
 
 
<center><math>\begin{matrix}
  {{\sigma' }} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma' }} & \mu _{1}^{\prime } & \mu _{2}^{\prime }  \\
  0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708  \\
  0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701  \\
  0.26 & 4.1909 & 4.3944 & 0.39 & 4.1170 & 4.4683  \\
  0.27 & 4.1748 & 4.4105 & 0.40 & 4.1200 & 4.4653  \\
  0.28 & 4.1618 & 4.4235 & 0.41 & 4.1244 & 4.4609  \\
  0.29 & 4.1509 & 4.4344 & 0.42 & 4.1302 & 4.4551  \\
  0.30 & 4.1419 & 4.4434 & 0.43 & 4.1377 & 4.4476  \\
  0.31 & 4.1343 & 4.4510 & 0.44 & 4.1472 & 4.4381  \\
  0.32 & 4.1281 & 4.4572 & 0.45 & 4.1591 & 4.4262  \\
  0.33 & 4.1231 & 4.4622 & 0.46 & 4.1742 & 4.4111  \\
  0.34 & 4.1193 & 4.4660 & 0.47 & 4.1939 & 4.3914  \\
  0.35 & 4.1166 & 4.4687 & 0.48 & 4.2221 & 4.3632  \\
  0.36 & 4.1150 & 4.4703 & {} & {} & {}  \\
\end{matrix}</math></center>
 
These points are represented graphically in the following contour plot:
 
[[Image:ldachp9ex5.gif|thumb|center|400px| ]]  
 
(Note that this plot is generated with degrees of freedom  <math>k=1</math> , as we are only determining bounds on one parameter. The contour plots generated in Weibull++ are done with degrees of freedom  <math>k=2</math> , for use in comparing both parameters simultaneously.) As can be determined from the table the lowest calculated value for  <math>{\mu }'</math>  is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of  <math>{{\sigma'
}}</math>  below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on  <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of  <math>\sigma </math>  that correspond with a given value of  <math>{\mu }'.</math>  Using this method, we find that the 75% confidence limits on  <math>{{\sigma'}}</math>  are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.

Latest revision as of 03:36, 13 August 2012

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