Template:Exponential Distribution Example: Likelihood Ratio Bound for lambda

Example 5: Likelihood Ratio Bound for $$\lambda $$
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 $$\hat{\lambda }=0.013514.$$  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:


 * $$\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}$$

where $${{x}_{i}}$$ are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:


 * $$L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0$$

Since our specified confidence level, $$\delta $$, is 85%, we can calculate the value of the chi-squared statistic, $$\chi _{0.85;1}^{2}=2.072251.$$ We can now substitute this information into the equation:


 * $$\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}$$

It now remains to find the values of $$\lambda $$ which satisfy this equation. Since there is only one parameter, there are only two values of $$\lambda $$ that will satisfy the equation. These values represent the $$\delta =85%$$ two-sided confidence limits of the parameter estimate $$\hat{\lambda }$$. For our problem, the confidence limits are:


 * $${{\lambda }_{0.85}}=(0.006572,0.024172)$$