Template:Example: Likelihood Ratio Bounds on Parameters

Likelihood Ratio Bounds on Parameters

Five units were put on a reliability test and experienced failures at 10, 20, 30, 40 and 50 hours. Assuming a Weibull distribution, the MLE parameter estimates are calculated to be $$\widehat{\beta }=2.2938\,\!$$ and $$\widehat{\eta }=33.9428.\,\!$$ Calculate the 90% two-sided confidence bounds on these parameters using the likelihood ratio method.

Solution

The first step is to calculate the likelihood function for the parameter estimates:


 * $$\begin{align}

L(\widehat{\beta },\widehat{\eta })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\widehat{\beta },\widehat{\eta })=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\widehat{\beta }}{\widehat{\eta }}\cdot {{\left( \frac{{{x}_{i}}}{\widehat{\eta }} \right)}^{\widehat{\beta }-1}}\cdot {{e}^{-{{\left( \tfrac{{{x}_{i}}}{\widehat{\eta }} \right)}^{\widehat{\beta }}}}} \\ \\  L(\widehat{\beta },\widehat{\eta })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{2.2938}{33.9428}\cdot {{\left( \frac{{{x}_{i}}}{33.9428} \right)}^{1.2938}}\cdot {{e}^{-{{\left( \tfrac{{{x}_{i}}}{33.9428} \right)}^{2.2938}}}} \\ \\  L(\widehat{\beta },\widehat{\eta })= & 1.714714\times {{10}^{-9}} \end{align}\,\!$$

where $${{x}_{i}}\,\!$$ are the original time-to-failure data points. We can now rearrange the likelihood ratio equation to the form:


 * $$L(\beta ,\eta )-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0\,\!$$

Since our specified confidence level, $$\delta\,\!$$, is 90%, we can calculate the value of the chi-squared statistic, $$\chi _{0.9;1}^{2}=2.705543.\,\!$$ We then substitute this information into the equation:


 * $$\begin{align}

L(\beta ,\eta )-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\ L(\beta ,\eta )-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ \\  L(\beta ,\eta )-4.432926\cdot {{10}^{-10}}= & 0 \end{align}\,\!$$

The next step is to find the set of values of $$\beta\,\!$$ and $$\eta\,\!$$ that satisfy this equation, or find the values of $$\beta\,\!$$ and $$\eta\,\!$$ such that $$L(\beta ,\eta )=4.432926\cdot {{10}^{-10}}.\,\!$$

The solution is an iterative process that requires setting the value of $$\beta\,\!$$ and finding the appropriate values of $$\eta\,\!$$, and vice versa. The following table gives values of $$\beta\,\!$$ based on given values of $$\eta\,\!$$.



These data are represented graphically in the following contour plot:



(Note that this plot is generated with degrees of freedom $$k = 1\,\!$$, as we are only determining bounds on one parameter. The contour plots generated in Weibull++ are done with degrees of freedom $$k = 2\,\!$$, for use in comparing both parameters simultaneously.) As can be determined from the table, the lowest calculated value for $$\beta\,\!$$ is 1.142, while the highest is 3.950. These represent the two-sided 90% confidence limits on this parameter. Since solutions for the equation do not exist for values of $$\eta\,\!$$ below 23 or above 50, these can be considered the 90% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on $$\eta\,\!$$, we can perform the same procedure as before, but finding the two values of $$\eta\,\!$$ that correspond with a given value of $$\beta\,\!$$ Using this method, we find that the 90% confidence limits on $$\eta\,\!$$ are 22.474 and 49.967, which are close to the initial estimates of 23 and 50.

Note that the points where $$\beta\,\!$$ are maximized and minimized do not necessarily correspond with the points where $$\eta\,\!$$ are maximized and minimized. This is due to the fact that the contour plot is not symmetrical, so that the parameters will have their extremes at different points.