Template:Normal Distribution likelihood ratio confidence bounds

Bounds on Parameters
As covered in Chapter Confidence Bounds, 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 the above likelihood ratio equation. 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:

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 $$t=30$$.