Template:Bounds on Parameters.LRCB.FMB.ED

Bounds on Parameters
For one-parameter distributions such as the exponential, the likelihood confidence bounds are calculated by finding values for $$\theta $$ that satisfy:


 * $$-2\cdot \text{ln}\left( \frac{L(\theta )}{L(\hat{\theta })} \right)=\chi _{\alpha ;1}^{2}$$

This equation can be rewritten as:


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

For complete data, the likelihood function for the exponential distribution is given by:


 * $$L(\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\lambda \cdot {{e}^{-\lambda \cdot {{x}_{i}}}}$$

where the $${{x}_{i}}$$ values represent the original time-to-failure data. For a given value of $$\alpha $$, values for $$\lambda $$ 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 $$\delta ,$$ where $$\alpha =\delta $$ for two-sided bounds and $$\alpha =2\delta -1$$ for one-sided.

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 $$\hat{\lambda }=0.013514.$$  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:


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

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 $$R(t)=90%$$ is $$\hat{t}=7.797$$.

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 $$R=0.90$$ and $$\alpha =0.85$$ into Eqn. (expliketr). It now remains to find the values of $$t$$ which satisfy this equation. Since there is only one parameter, there are only two values of $$t$$ that will satisfy the equation. These values represent the $$\delta =85%$$ two-sided confidence limits of the time estimate $$\hat{t}$$. For our problem, the confidence limits are:


 * $${{\hat{t}}_{R=0.9}}=(4.359,16.033).$$

Example 7
For the data given in Example 5, determine the 85% two-sided confidence bounds on the reliability estimate for a $$t=50$$. The ML estimate for the time at $$t=50$$ is $$\hat{R}=50.881%$$.

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 $$t=50$$ and $$\alpha =0.85$$ into Eqn. (expliketr). It now remains to find the values of $$R$$ which satisfy this equation. Since there is only one parameter, there are only two values of $$t$$ that will satisfy the equation. These values represent the $$\delta =85%$$ two-sided confidence limits of the reliability estimate $$\hat{R}$$. For our problem, the confidence limits are:


 * $${{\hat{R}}_{t=50}}=(29.861%,71.794%)$$