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===Likelihood Ratio Confidence Bounds===
#REDIRECT [[The Exponential Distribution]]
 
{{Bounds on Parameters.LRCB.FMB.ED}}
 
====Example 5====
 
Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming an exponential distribution, the MLE parameter estimate is calculated to be  <math>\hat{\lambda }=0.013514.</math>  Calculate the 85% two-sided 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:
 
::<math>\begin{align}
  L(\hat{\lambda })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\hat{\lambda })=\underset{i=1}{\overset{N}{\mathop \prod }}\,\hat{\lambda }\cdot {{e}^{-\hat{\lambda }\cdot {{x}_{i}}}} \\
  L(\hat{\lambda })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,0.013514\cdot {{e}^{-0.013514\cdot {{x}_{i}}}} \\
  L(\hat{\lambda })= & 3.03647\times {{10}^{-12}} 
\end{align}</math>
 
where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:
 
::<math>L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
 
Since our specified confidence level, <math>\delta </math>, is 85%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.85;1}^{2}=2.072251.</math> We can now substitute this information into the equation:
 
::<math>\begin{align}
  L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\
  L(\lambda )-3.03647\times {{10}^{-12}}\cdot {{e}^{\tfrac{-2.072251}{2}}}= & 0, \\
  L(\lambda )-1.07742\times {{10}^{-12}}= & 0. 
\end{align}</math>
 
It now remains to find the values of <math>\lambda </math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>\lambda </math> that will satisfy the equation.  These values represent the <math>\delta =85%</math> two-sided confidence limits of the parameter estimate <math>\hat{\lambda }</math>. For our problem, the confidence limits are:
 
::<math>{{\lambda }_{0.85}}=(0.006572,0.024172)</math>
 
 
====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 exponential reliability equation into the likelihood function. The exponential reliability equation can be written as:
 
::<math>R={{e}^{-\lambda \cdot t}}</math>
 
This can be rearranged to the form:
 
::<math>\lambda =\frac{-\text{ln}(R)}{t}</math>
 
This equation can now be substituted into Eqn. (explikelihood) to produce a likelihood equation in terms of <math>t</math> and <math>R\ \ :</math>
 
::<math>L(t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\left( \frac{-\text{ln}(R)}{t} \right)\cdot {{e}^{\left( \tfrac{\text{ln}(R)}{t} \right)\cdot {{x}_{i}}}}</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. (expliketr) can be used to solve Eqn. (lratio3) for the values of interest.
 
====Example 6====
For the data given in Example 5, determine the 85% two-sided confidence bounds on the time estimate for a reliability of 90%. The ML estimate for the time at <math>R(t)=90%</math> is <math>\hat{t}=7.797</math>.
 
=====Solution to Example 6=====
In this example, we are trying to determine the 85% two-sided confidence bounds on the time estimate of 7.797. This is accomplished by substituting <math>R=0.90</math> and <math>\alpha =0.85</math> into Eqn. (expliketr). It now remains to find the values of <math>t</math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>t</math> that will satisfy the equation. These values represent the <math>\delta =85%</math> two-sided confidence limits of the time estimate <math>\hat{t}</math>. For our problem, the confidence limits are:
 
::<math>{{\hat{t}}_{R=0.9}}=(4.359,16.033).</math>
 
 
====Example 7====
For the data given in Example 5, determine the 85% two-sided confidence bounds on the reliability estimate for a <math>t=50</math>. The ML estimate for the time at <math>t=50</math> is <math>\hat{R}=50.881%</math>.
 
=====Solution to Example 7=====
In this example, we are trying to determine the 85% two-sided confidence bounds on the reliability estimate of 50.881%. This is accomplished by substituting <math>t=50</math> and <math>\alpha =0.85</math> into Eqn. (expliketr). It now remains to find the values of <math>R</math> which satisfy this equation. Since there is only one parameter, there are only two values of <math>t</math> that will satisfy the equation. These values represent the <math>\delta =85%</math> two-sided confidence limits of the reliability estimate <math>\hat{R}</math>. For our problem, the confidence limits are:
 
::<math>{{\hat{R}}_{t=50}}=(29.861%,71.794%)</math>

Latest revision as of 08:21, 10 August 2012

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