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===Likelihood Ratio Confidence Bounds===
#REDIRECT [[The_Normal_Distribution#Likelihood_Ratio_Confidence_Bounds]]
====Bounds on Parameters====
As covered in Chapter [[Confidence Bounds]], 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 )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{t}_{i}};\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{t}_{i}}-\mu }{\sigma } \right)}^{2}}}}</math>
 
where the  <math>{{t}_{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 </math>  can be found which represent the maximum and minimum values that satisfy the above likelihood ratio equation. 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:'''
{{Example: Normal Distribution Likelihood Ratio Bound (Parameters)}}
 
====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{t-\mu }{\sigma } \right)</math>
 
This can be rearranged to the form:
 
::<math>\mu =t-\sigma \cdot {{\Phi }^{-1}}(1-R)</math>
 
where  <math>{{\Phi }^{-1}}</math>  is the inverse standard normal. This equation can now be substituted into Eqn. (normlikelihood), to produce a likelihood equation in terms of  <math>\sigma ,</math>  <math>t</math>  and  <math>R\ \ :</math> 
 
::<math>L(\sigma ,t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{t}_{i}}-\left[ t-\sigma \cdot {{\Phi }^{-1}}(1-R) \right]}{\sigma } \right)}^{2}}}}</math>
 
The unknown parameter  <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 parameter. 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 parameter. Either way, Eqn. (normliketr) 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 80% confidence bounds on the time estimate for a reliability of 40%.  The ML estimate for the time at  <math>R(t)=40%</math>  is 31.637.
=====Solution to Example 6=====
In this example, we are trying to determine the two-sided 80% confidence bounds on the time estimate of 31.637. This is accomplished by substituting  <math>R=0.40</math>  and  <math>\alpha =0.8</math>  into Eqn. (normliketr), and varying  <math>\sigma </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 </math> .
 
[[Image:tabletbasedonsigma.gif|thumb|center|400px| ]]
<math></math>
 
This data set is represented graphically in the following contour plot:
 
[[Image:ovalplot.gif|thumb|center|400px| ]]
 
 
As can be determined from the table, the lowest calculated value for  <math>t</math>  is 25.046, while the highest is 39.250. These represent the 80% confidence limits on the time at which reliability is equal to 40%.
 
====Example 7====
For the data given in Example 5, determine the two-sided 80% confidence bounds on the reliability estimate for  <math>t=30</math> .  The ML estimate for the reliability at  <math>t=30</math>  is 45.739%.
 
=====Solution to Example 7=====
In this example, we are trying to determine the two-sided 80% confidence bounds on the reliability estimate of 45.739%. This is accomplished by substituting  <math>t=30</math>  and  <math>\alpha =0.8</math>  into Eqn. (normliketr), and varying  <math>\sigma </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 </math> .
 
[[Image:tablerbasedonsigma.gif|thumb|center|400px| ]]
 
This data set is represented graphically in the following contour plot:
 
[[Image:crazyoplot.gif|thumb|center|400px| ]]
 
 
As can be determined from the table, the lowest calculated value for  <math>R</math>  is 24.776%, while the highest is 68.000%. These represent the 80% two-sided confidence limits on the reliability at  <math>t=30</math> .

Latest revision as of 04:27, 13 August 2012

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