Template:Lognormal distribution Likelihood ratio confidence bounds

Bounds on Parameters
As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for $${{\theta }_{1}}$$  and  $${{\theta }_{2}}$$  that satisfy:


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

This equation can be rewritten as:


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

For complete data, the likelihood formula for the normal distribution is given by:


 * $$L({\mu }',{{\sigma }_})=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{\mu }',{{\sigma }_})=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\sigma }_}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{\mu }'}{{{\sigma }_}} \right)}^{2}}}}$$

where the $${{x}_{i}}$$  values represent the original time-to-failure data. For a given value of $$\alpha $$, values for  $${\mu }'$$  and  $${{\sigma }_}$$  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 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 to Example 5
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.

Bounds on Time and Reliability
In order to calculate the bounds on a time estimate for a given reliability, or on a reliability estimate for a given time, the likelihood function needs to be rewritten in terms of one parameter and time/reliability, so that the maximum and minimum values of the time can be observed as the parameter is varied. This can be accomplished by substituting a form of the normal reliability equation into the likelihood function. The normal reliability equation can be written as:


 * $$R=1-\Phi \left( \frac{\text{ln}(t)-{\mu }'} \right)$$

This can be rearranged to the form:


 * $${\mu }'=\text{ln}(t)-{{\sigma }_}\cdot {{\Phi }^{-1}}(1-R)$$

where $${{\Phi }^{-1}}$$  is the inverse standard normal. This equation can now be substituted into Eqn. (lognormlikelihood) to produce a likelihood equation in terms of $${{\sigma }_},$$   $$t$$  and  $$R\ \ :$$


 * $$L({{\sigma }_},t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\sigma }_}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-\left( \text{ln}(t)-{{\sigma }_}\cdot {{\Phi }^{-1}}(1-R) \right)}{{{\sigma }_}} \right)}^{2}}}}$$

The unknown variable $$t/R$$  depends on what type of bounds are being determined. If one is trying to determine the bounds on time for a given reliability, then $$R$$  is a known constant and  $$t$$  is the unknown variable. Conversely, if one is trying to determine the bounds on reliability for a given time, then $$t$$  is a known constant and  $$R$$  is the unknown variable. Either way, Eqn. (lognormliketr) can be used to solve Eqn. (lratio3) for the values of interest.

Example 6
For the data given in Example 5, determine the two-sided 75% confidence bounds on the time estimate for a reliability of 80%. The ML estimate for the time at $$R(t)=80%$$  is 55.718.

Solution to Example 6
In this example, we are trying to determine the two-sided 75% confidence bounds on the time estimate of 55.718. This is accomplished by substituting $$R=0.80$$  and  $$\alpha =0.75$$  into Eqn. (lognormliketr), and varying $${{\sigma }_}$$  until the maximum and minimum values of  $$t$$  are found. The following table gives the values of $$t$$  based on given values of  $${{\sigma }_}$$.

$$\begin{matrix} {{\sigma }_} & {{t}_{1}} & {{t}_{2}} & {{\sigma }_} & {{t}_{1}} & {{t}_{2}} \\ 0.24 & 56.832 & 62.879 & 0.37 & 44.841 & 64.031 \\   0.25 & 54.660 & 64.287 & 0.38 & 44.494 & 63.454  \\   0.26 & 53.093 & 65.079 & 0.39 & 44.200 & 62.809  \\   0.27 & 51.811 & 65.576 & 0.40 & 43.963 & 62.093  \\   0.28 & 50.711 & 65.881 & 0.41 & 43.786 & 61.304  \\   0.29 & 49.743 & 66.041 & 0.42 & 43.674 & 60.436  \\   0.30 & 48.881 & 66.085 & 0.43 & 43.634 & 59.481  \\   0.31 & 48.106 & 66.028 & 0.44 & 43.681 & 58.426  \\   0.32 & 47.408 & 65.883 & 0.45 & 43.832 & 57.252  \\   0.33 & 46.777 & 65.657 & 0.46 & 44.124 & 55.924  \\   0.34 & 46.208 & 65.355 & 0.47 & 44.625 & 54.373  \\   0.35 & 45.697 & 64.983 & 0.48 & 45.517 & 52.418  \\   0.36 & 45.242 & 64.541 & {} & {} & {}  \\ \end{matrix}$$

This data set is represented graphically in the following contour plot:



As can be determined from the table, the lowest calculated value for $$t$$  is 43.634, while the highest is 66.085. These represent the two-sided 75% confidence limits on the time at which reliability is equal to 80%.

Example 7
For the data given in Example 5, determine the two-sided 75% confidence bounds on the reliability estimate for $$t=65$$. The ML estimate for the reliability at $$t=65$$  is 64.261%.

Solution to Example 7
In this example, we are trying to determine the two-sided 75% confidence bounds on the reliability estimate of 64.261%. This is accomplished by substituting $$t=65$$  and  $$\alpha =0.75$$  into Eqn. (lognormliketr), and varying $${{\sigma }_}$$  until the maximum and minimum values of  $$R$$  are found. The following table gives the values of $$R$$  based on given values of  $${{\sigma }_}$$.

$$\begin{matrix} {{\sigma }_} & {{R}_{1}} & {{R}_{2}} & {{\sigma }_} & {{R}_{1}} & {{R}_{2}} \\ 0.24 & 61.107% & 75.910% & 0.37 & 43.573% & 78.845% \\   0.25 & 55.906% & 78.742% & 0.38 & 43.807% & 78.180%  \\   0.26 & 55.528% & 80.131% & 0.39 & 44.147% & 77.448%  \\   0.27 & 50.067% & 80.903% & 0.40 & 44.593% & 76.646%  \\   0.28 & 48.206% & 81.319% & 0.41 & 45.146% & 75.767%  \\   0.29 & 46.779% & 81.499% & 0.42 & 45.813% & 74.802%  \\   0.30 & 45.685% & 81.508% & 0.43 & 46.604% & 73.737%  \\   0.31 & 44.857% & 81.387% & 0.44 & 47.538% & 72.551%  \\   0.32 & 44.250% & 81.159% & 0.45 & 48.645% & 71.212%  \\   0.33 & 43.827% & 80.842% & 0.46 & 49.980% & 69.661%  \\   0.34 & 43.565% & 80.446% & 0.47 & 51.652% & 67.789%  \\   0.35 & 43.444% & 79.979% & 0.48 & 53.956% & 65.299%  \\   0.36 & 43.450% & 79.444% & {} & {} & {}  \\ \end{matrix}$$

This data set is represented graphically in the following contour plot:



As can be determined from the table, the lowest calculated value for $$R$$  is 43.444%, while the highest is 81.508%. These represent the two-sided 75% confidence limits on the reliability at $$t=65$$.