Template:Example: Lognormal Distribution Likelihood Ratio Bound (Parameters)

Lognormal Distribution Likelihood Ratio Bound Example (Parameters)

Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assuming a lognormal distribution, the MLE parameter estimates are calculated to be $${{\widehat{\mu }}^{\prime }}=4.2926$$  and  $${{\widehat{\sigma }}_}=0.32361.$$  Calculate the two-sided 75% confidence bounds on these parameters using the likelihood ratio method.

Solution

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

$$\begin{align} L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_}), \\ = & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma }}_}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma }}_}} \right)}^{2}}}} \\ L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\ L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_})= & 1.115256\times {{10}^{-10}} \end{align}$$

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


 * $$L({\mu }',{{\sigma }_})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0$$

Since our specified confidence level, $$\delta $$, is 75%, we can calculate the value of the chi-squared statistic,  $$\chi _{0.75;1}^{2}=1.323303.$$  We can now substitute this information into the equation:


 * $$\begin{align}

& L({\mu }',{{\sigma }_})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ & L({\mu }',{{\sigma }_})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\ & L({\mu }',{{\sigma }_})-5.754703\times {{10}^{-11}}= & 0 \end{align}$$

It now remains to find the values of $${\mu }'$$  and  $${{\sigma }_}$$  which satisfy this equation. This is an iterative process that requires setting the value of $${{\sigma }_}$$  and finding the appropriate values of  $${\mu }'$$, and vice versa.

The following table gives the values of $${\mu }'$$  based on given values of  $${{\sigma }_}$$.

$$\begin{matrix} {{\sigma }_} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma }_} & \mu _{1}^{\prime } & \mu _{2}^{\prime } \\ 0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708 \\   0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701  \\   0.26 & 4.1909 & 4.3944 & 0.39 & 4.1170 & 4.4683  \\   0.27 & 4.1748 & 4.4105 & 0.40 & 4.1200 & 4.4653  \\   0.28 & 4.1618 & 4.4235 & 0.41 & 4.1244 & 4.4609  \\   0.29 & 4.1509 & 4.4344 & 0.42 & 4.1302 & 4.4551  \\   0.30 & 4.1419 & 4.4434 & 0.43 & 4.1377 & 4.4476  \\   0.31 & 4.1343 & 4.4510 & 0.44 & 4.1472 & 4.4381  \\   0.32 & 4.1281 & 4.4572 & 0.45 & 4.1591 & 4.4262  \\   0.33 & 4.1231 & 4.4622 & 0.46 & 4.1742 & 4.4111  \\   0.34 & 4.1193 & 4.4660 & 0.47 & 4.1939 & 4.3914  \\   0.35 & 4.1166 & 4.4687 & 0.48 & 4.2221 & 4.3632  \\   0.36 & 4.1150 & 4.4703 & {} & {} & {}  \\ \end{matrix}$$

These points 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  $${\mu }'$$  is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of $${{\sigma }_}$$  below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on $${{\sigma }_}$$, we can perform the same procedure as before, but finding the two values of  $$\sigma $$  that correspond with a given value of  $${\mu }'.$$  Using this method, we find that the 75% confidence limits on  $${{\sigma }_}$$  are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.