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 [math]\displaystyle{ \theta }[/math] that satisfy:

[math]\displaystyle{ -2\cdot \text{ln}\left( \frac{L(\theta )}{L(\hat{\theta })} \right)=\chi _{\alpha ;1}^{2} }[/math]

This equation can be rewritten as:

[math]\displaystyle{ L(\theta )=L(\hat{\theta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}} }[/math]

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

[math]\displaystyle{ 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}}}} }[/math]

where the [math]\displaystyle{ {{x}_{i}} }[/math] values represent the original time-to-failure data. For a given value of [math]\displaystyle{ \alpha }[/math], values for [math]\displaystyle{ \lambda }[/math] 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 [math]\displaystyle{ \delta , }[/math] where [math]\displaystyle{ \alpha =\delta }[/math] for two-sided bounds and [math]\displaystyle{ \alpha =2\delta -1 }[/math] 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 [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ {{x}_{i}} }[/math] are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:

[math]\displaystyle{ L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0 }[/math]

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

[math]\displaystyle{ \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]\displaystyle{ \lambda }[/math] which satisfy this equation. Since there is only one parameter, there are only two values of [math]\displaystyle{ \lambda }[/math] that will satisfy the equation. These values represent the [math]\displaystyle{ \delta =85% }[/math] two-sided confidence limits of the parameter estimate [math]\displaystyle{ \hat{\lambda } }[/math]. For our problem, the confidence limits are:

[math]\displaystyle{ {{\lambda }_{0.85}}=(0.006572,0.024172) }[/math]