View source for Template:Exponential Distribution Example: Likelihood Ratio Bound for lambda

=====Example 5: Likelihood Ratio Bound on <math>\lambda </math>=====

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>