Template:Normal 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:


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

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 12, 24, 28, 34, and 46 hours. Assuming a normal distribution, the MLE parameter estimates are calculated to be $$\widehat{\mu }=28.8$$  and  $$\widehat{\sigma }=11.2143.$$  Calculate the two-sided 80% 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 },\widehat{\sigma })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\widehat{\mu },\widehat{\sigma })=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{\widehat{\sigma }\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\widehat{\mu }}{\widehat{\sigma }} \right)}^{2}}}} \\ L(\widehat{\mu },\widehat{\sigma })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{11.2143\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-28.8}{11.2143} \right)}^{2}}}} \\ L(\widehat{\mu },\widehat{\sigma })= & 4.676897\times {{10}^{-9}} \end{align}$$

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


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

Since our specified confidence level, $$\delta $$, is 80%, we can calculate the value of the chi-squared statistic,  $$\chi _{0.8;1}^{2}=1.642374.$$  We can now substitute this information into the equation:


 * $$\begin{align}

L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\ \\  L(\mu ,\sigma )-4.676897\times {{10}^{-9}}\cdot {{e}^{\tfrac{-1.642374}{2}}}= & 0, \\ \\  L(\mu ,\sigma )-2.057410\times {{10}^{-9}}= & 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 $$\mu $$  and finding the appropriate values of  $$\sigma $$, and vice versa.

The following table gives the values of $$\sigma $$  based on given values of  $$\mu $$.



$$$$

$$\begin{matrix} \text{ }\!\!\mu\!\!\text{ } & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{1}}} & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{2}}} & \text{ }\!\!\mu\!\!\text{ } & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{1}}} & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{2}}} \\ \text{22}\text{.0} & \text{12}\text{.045} & \text{14}\text{.354} & \text{29}\text{.0} & \text{7.849}& \text{19.909}   \\ \text{22}\text{.5} & \text{11}\text{.004} & \text{15}\text{.310} & \text{29}\text{.5} & \text{7}\text{.876} & \text{17}\text{.889} \\ \text{23}\text{.0} & \text{10}\text{.341} & \text{15}\text{.894} & \text{30}\text{.0} & \text{7}\text{.935} & \text{17}\text{.844} \\ \text{23}\text{.5} & \text{9}\text{.832} & \text{16}\text{.328} & \text{30}\text{.5} & \text{8}\text{.025} & \text{17}\text{.776} \\ \text{24}\text{.0} & \text{9}\text{.418} & \text{16}\text{.673} & \text{31}\text{.0} & \text{8}\text{.147} & \text{17}\text{.683} \\ \text{24}\text{.5} & \text{9}\text{.074} & \text{16}\text{.954} & \text{31}\text{.5} & \text{8}\text{.304} & \text{17}\text{.562} \\ \text{25}\text{.0} & \text{8}\text{.784} & \text{17}\text{.186} & \text{32}\text{.0} & \text{8}\text{.498} & \text{17}\text{.411} \\ \text{25}\text{.5} & \text{8}\text{.542} & \text{17}\text{.377} & \text{32}\text{.5} & \text{8}\text{.732} & \text{17}\text{.227} \\ \text{26}\text{.0} & \text{8}\text{.340} & \text{17}\text{.534} & \text{33}\text{.0} & \text{9}\text{.012} & \text{17}\text{.004} \\ \text{26}\text{.5} & \text{8}\text{.176} & \text{17}\text{.661} & \text{33}\text{.5} & \text{9}\text{.344} & \text{16}\text{.734} \\ \text{27}\text{.0} & \text{8}\text{.047} & \text{17}\text{.760} & \text{34}\text{.0} & \text{9}\text{.742} & \text{16}\text{.403} \\ \text{27}\text{.5} & \text{7}\text{.950} & \text{17}\text{.833} & \text{34}\text{.5} & \text{10}\text{.229} & \text{15}\text{.990} \\ \text{28}\text{.0} & \text{7}\text{.885} & \text{17}\text{.882} & \text{35}\text{.0} & \text{10}\text{.854} & \text{15}\text{.444} \\ \text{28}\text{.5} & \text{7}\text{.852} & \text{17}\text{.907} & \text{35}\text{.5} & \text{11}\text{.772} & \text{14}\text{.609} \\ \end{matrix}$$

This data set is 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  $$\sigma $$  is 7.849, while the highest is 17.909. These represent the two-sided 80% confidence limits on this parameter. Since solutions for the equation do not exist for values of $$\mu $$  below 22 or above 35.5, these can be considered the two-sided 80% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on $$\mu $$, we can perform the same procedure as before, but finding the two values of  $$\mu $$  that correspond with a given value of  $$\sigma .$$  Using this method, we find that the two-sided 80% confidence limits on  $$\mu $$  are 21.807 and 35.793, which are close to the initial estimates of 22 and 35.5.

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:


 * $$R=1-\Phi \left( \frac{t-\mu }{\sigma } \right)$$

This can be rearranged to the form:


 * $$\mu =t-\sigma \cdot {{\Phi }^{-1}}(1-R)$$

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


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

The unknown parameter $$t/R$$  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 parameter. 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 parameter. Either way, Eqn. (normliketr) 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 80% confidence bounds on the time estimate for a reliability of 40%. The ML estimate for the time at $$R(t)=40%$$  is 31.637.

Solution to Example 6
In this example, we are trying to determine the two-sided 80% confidence bounds on the time estimate of 31.637. This is accomplished by substituting $$R=0.40$$  and  $$\alpha =0.8$$  into Eqn. (normliketr), 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  $$\sigma $$.

$$$$

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 25.046, while the highest is 39.250. These represent the 80% confidence limits on the time at which reliability is equal to 40%.

Example 7
For the data given in Example 5, determine the two-sided 80% confidence bounds on the reliability estimate for $$t=30$$. The ML estimate for the reliability at $$t=30$$  is 45.739%.

Solution to Example 7
In this example, we are trying to determine the two-sided 80% confidence bounds on the reliability estimate of 45.739%. This is accomplished by substituting $$t=30$$  and  $$\alpha =0.8$$  into Eqn. (normliketr), 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  $$\sigma $$.



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 24.776%, while the highest is 68.000%. These represent the 80% two-sided confidence limits on the reliability at $$t=30$$.