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===Likelihood Ratio Confidence Bounds===
#REDIRECT [[The_Lognormal_Distribution#Likelihood_Ratio_Confidence_Bounds]]
 
====Bounds on Parameters====
As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for  <math>{{\theta }_{1}}</math>  and  <math>{{\theta }_{2}}</math>  that satisfy:
 
 
::<math>-2\cdot \text{ln}\left( \frac{L({{\theta }_{1}},{{\theta }_{2}})}{L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})} \right)=\chi _{\alpha ;1}^{2}</math>
 
This equation can be rewritten as:
 
 
::<math>L({{\theta }_{1}},{{\theta }_{2}})=L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}</math>
 
For complete data, the likelihood formula for the normal distribution is given by:
 
 
::<math>L({\mu }',{{\sigma }_{{{T}'}}})=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{\mu }',{{\sigma }_{{{T}'}}})=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\sigma }_{{{T}'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}</math>
 
where the  <math>{{x}_{i}}</math>  values represent the original time-to-failure data.  For a given value of  <math>\alpha </math> , values for  <math>{\mu }'</math>  and  <math>{{\sigma }_{{{T}'}}}</math>  can be found which represent the maximum and minimum values that satisfy Eqn. (lratio3). These represent the confidence bounds for the parameters at a confidence level  <math>\delta ,</math>  where  <math>\alpha =\delta </math>  for two-sided bounds and  <math>\alpha =2\delta -1</math>  for one-sided.
 
====Example 5====
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 }}_{{{T}'}}}=0.32361.</math>  Calculate the two-sided 75% confidence bounds on these parameters using the likelihood ratio method.
=====Solution to Example 5=====
The first step is to calculate the likelihood function for the parameter estimates:
 
<center><math>\begin{align}
  L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}}), \\
  = & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma }}_{{{T}'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma }}_{{{T}'}}}} \right)}^{2}}}} \\
  L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & \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 }}_{{{T}'}}})= & 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 Eqn. (lratio3) to the form:
 
::<math>L({\mu }',{{\sigma }_{{{T}'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})\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 }_{{{T}'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\
& L({\mu }',{{\sigma }_{{{T}'}}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\
& L({\mu }',{{\sigma }_{{{T}'}}})-5.754703\times {{10}^{-11}}= & 0 
\end{align}</math>
 
It now remains to find the values of  <math>{\mu }'</math>  and  <math>{{\sigma }_{{{T}'}}}</math>  which satisfy this equation. This is an iterative process that requires setting the value of  <math>{{\sigma }_{{{T}'}}}</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 }_{{{T}'}}}</math> .
 
 
<center><math>\begin{matrix}
  {{\sigma }_{{{T}'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma }_{{{T}'}}} & \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 }_{{{T}'}}}</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 }_{{{T}'}}}</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 }_{{{T}'}}}</math>  are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.
 
====Bounds on Time and Reliability====
In order to calculate the bounds on a time estimate for a given reliability, or on a reliability estimate for a given time, the likelihood function needs to be rewritten in terms of one parameter and time/reliability, so that the maximum and minimum values of the time can be observed as the parameter is varied. This can be accomplished by substituting a form of the normal reliability equation into the likelihood function. The normal reliability equation can be written as:
 
::<math>R=1-\Phi \left( \frac{\text{ln}(t)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)</math>
 
This can be rearranged to the form:
 
::<math>{\mu }'=\text{ln}(t)-{{\sigma }_{{{T}'}}}\cdot {{\Phi }^{-1}}(1-R)</math>
 
where  <math>{{\Phi }^{-1}}</math>  is the inverse standard normal. This equation can now be substituted into Eqn. (lognormlikelihood) to produce a likelihood equation in terms of  <math>{{\sigma }_{{{T}'}}},</math>  <math>t</math>  and  <math>R\ \ :</math> 
 
::<math>L({{\sigma }_{{{T}'}}},t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\sigma }_{{{T}'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-\left( \text{ln}(t)-{{\sigma }_{{{T}'}}}\cdot {{\Phi }^{-1}}(1-R) \right)}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}</math>
 
The unknown variable  <math>t/R</math>  depends on what type of bounds are being determined.  If one is trying to determine the bounds on time for a given reliability, then  <math>R</math>  is a known constant and  <math>t</math>  is the unknown variable. Conversely, if one is trying to determine the bounds on reliability for a given time, then  <math>t</math>  is a known constant and  <math>R</math>  is the unknown variable. Either way, Eqn. (lognormliketr) can be used to solve Eqn. (lratio3) for the values of interest.
 
====Example 6====
For the data given in Example 5, determine the two-sided 75% confidence bounds on the time estimate for a reliability of 80%.  The ML estimate for the time at  <math>R(t)=80%</math>  is 55.718.
=====Solution to Example 6=====
In this example, we are trying to determine the two-sided 75% confidence bounds on the time estimate of 55.718. This is accomplished by substituting  <math>R=0.80</math>  and  <math>\alpha =0.75</math>  into Eqn. (lognormliketr), and varying  <math>{{\sigma }_{{{T}'}}}</math>  until the maximum and minimum values of  <math>t</math>  are found. The following table gives the values of  <math>t</math>  based on given values of  <math>{{\sigma }_{{{T}'}}}</math> .
 
 
<center><math>\begin{matrix}
  {{\sigma }_{{{T}'}}} & {{t}_{1}} & {{t}_{2}} & {{\sigma }_{{{T}'}}} & {{t}_{1}} & {{t}_{2}}  \\
  0.24 & 56.832 & 62.879 & 0.37 & 44.841 & 64.031  \\
  0.25 & 54.660 & 64.287 & 0.38 & 44.494 & 63.454  \\
  0.26 & 53.093 & 65.079 & 0.39 & 44.200 & 62.809  \\
  0.27 & 51.811 & 65.576 & 0.40 & 43.963 & 62.093  \\
  0.28 & 50.711 & 65.881 & 0.41 & 43.786 & 61.304  \\
  0.29 & 49.743 & 66.041 & 0.42 & 43.674 & 60.436  \\
  0.30 & 48.881 & 66.085 & 0.43 & 43.634 & 59.481  \\
  0.31 & 48.106 & 66.028 & 0.44 & 43.681 & 58.426  \\
  0.32 & 47.408 & 65.883 & 0.45 & 43.832 & 57.252  \\
  0.33 & 46.777 & 65.657 & 0.46 & 44.124 & 55.924  \\
  0.34 & 46.208 & 65.355 & 0.47 & 44.625 & 54.373  \\
  0.35 & 45.697 & 64.983 & 0.48 & 45.517 & 52.418  \\
  0.36 & 45.242 & 64.541 & {} & {} & {}  \\
\end{matrix}</math></center>
 
 
This data set is represented graphically in the following contour plot:
 
[[Image:ldachp9ex6.gif|thumb|center|400px| ]]
 
As can be determined from the table, the lowest calculated value for  <math>t</math>  is 43.634, while the highest is 66.085. These represent the two-sided 75% confidence limits on the time at which reliability is equal to 80%.
 
====Example 7====
For the data given in Example 5, determine the two-sided 75% confidence bounds on the reliability estimate for  <math>t=65</math> .  The ML estimate for the reliability at  <math>t=65</math>  is 64.261%.
=====Solution to Example 7=====
In this example, we are trying to determine the two-sided 75% confidence bounds on the reliability estimate of 64.261%. This is accomplished by substituting  <math>t=65</math>  and  <math>\alpha =0.75</math>  into Eqn. (lognormliketr), and varying  <math>{{\sigma }_{{{T}'}}}</math>  until the maximum and minimum values of  <math>R</math>  are found. The following table gives the values of  <math>R</math>  based on given values of  <math>{{\sigma }_{{{T}'}}}</math> .
 
 
<center><math>\begin{matrix}
  {{\sigma }_{{{T}'}}} & {{R}_{1}} & {{R}_{2}} & {{\sigma }_{{{T}'}}} & {{R}_{1}} & {{R}_{2}}  \\
  0.24 & 61.107% & 75.910% & 0.37 & 43.573% & 78.845%  \\
  0.25 & 55.906% & 78.742% & 0.38 & 43.807% & 78.180%  \\
  0.26 & 55.528% & 80.131% & 0.39 & 44.147% & 77.448%  \\
  0.27 & 50.067% & 80.903% & 0.40 & 44.593% & 76.646%  \\
  0.28 & 48.206% & 81.319% & 0.41 & 45.146% & 75.767%  \\
  0.29 & 46.779% & 81.499% & 0.42 & 45.813% & 74.802%  \\
  0.30 & 45.685% & 81.508% & 0.43 & 46.604% & 73.737%  \\
  0.31 & 44.857% & 81.387% & 0.44 & 47.538% & 72.551%  \\
  0.32 & 44.250% & 81.159% & 0.45 & 48.645% & 71.212%  \\
  0.33 & 43.827% & 80.842% & 0.46 & 49.980% & 69.661%  \\
  0.34 & 43.565% & 80.446% & 0.47 & 51.652% & 67.789%  \\
  0.35 & 43.444% & 79.979% & 0.48 & 53.956% & 65.299%  \\
  0.36 & 43.450% & 79.444% & {} & {} & {}  \\
\end{matrix}</math></center>
 
 
This data set is represented graphically in the following contour plot:
 
[[Image:ldachp9ex7.gif|thumb|center|400px| ]]
 
As can be determined from the table, the lowest calculated value for  <math>R</math>  is 43.444%, while the highest is 81.508%. These represent the two-sided 75% confidence limits on the reliability at  <math>t=65</math> .

Latest revision as of 06:06, 13 August 2012

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