Exponential Confidence Bounds

Confidence Bounds
In this section, we present the methods used in the application to estimate the different types of confidence bounds for exponentially distributed data. The complete derivations were presented in detail (for a general function) in the chapter for Confidence Bounds. At this time we should point out that exact confidence bounds for the exponential distribution have been derived, and exist in a closed form, utilizing the $${{\chi }^{2}}\,\!$$ distribution. These are described in detail in Kececioglu [20], and are covered in the section in the test design chapter. For most exponential data analyses, Weibull++ will use the approximate confidence bounds, provided from the Fisher information matrix or the likelihood ratio, in order to stay consistent with all of the other available distributions in the application. The $${{\chi }^{2}}\,\!$$ confidence bounds for the exponential distribution are discussed in more detail in the test design chapter.

Bounds on the Parameters
For the failure rate $$\hat{\lambda }\,\!$$ the upper ($${{\lambda }_{U}}\,\!$$) and lower ($${{\lambda }_{L}}\,\!$$) bounds are estimated by Nelson [30]:


 * $$\begin{align}

& {{\lambda }_{U}}= & \hat{\lambda }\cdot {{e}^{\left[ \tfrac{{{K}_{\alpha }}\sqrt{Var(\hat{\lambda })}}{\hat{\lambda }} \right]}} \\ & &  \\  & {{\lambda }_{L}}= & \frac{\hat{\lambda }} \end{align}\,\!$$

where $${{K}_{\alpha }}\,\!$$ is defined by:


 * $$\alpha =\frac{1}{\sqrt{2\pi }}\int_^{\infty }{{e}^{-\tfrac{2}}}dt=1-\Phi ({{K}_{\alpha }})\,\!$$

If $$\delta \,\!$$ is the confidence level, then $$\alpha =\tfrac{1-\delta }{2}\,\!$$ for the two-sided bounds, and $$\alpha =1-\delta \,\!$$ for the one-sided bounds.

The variance of $$\hat{\lambda },\,\!$$ $$Var(\hat{\lambda }),\,\!$$ is estimated from the Fisher matrix, as follows:


 * $$Var(\hat{\lambda })={{\left( -\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} \right)}^{-1}}\,\!$$

where $$\Lambda \,\!$$ is the log-likelihood function of the exponential distribution, described in Appendix D.

Note that, for fixed $$\lambda \,\!$$, the log-likelihood function is increasing in $$\gamma$$. This means that the MLE solution for $$\gamma$$ cannot be found by setting $$\tfrac{\partial \Lambda}{\partial \gamma}$$ to zero. (Since $$\Lambda$$ is increasing in $$\gamma$$, $$\tfrac{\partial \Lambda}{\partial \gamma}$$ can never be zero.) Instead, the MLE solution for $$\gamma$$ is simply its largest possible value allowed by the sample; namely, $$\gamma = t_1$$, the first failure time. The MLE solution for $$\lambda$$ is found in the usual fashion by setting $$\tfrac{\partial \Lambda}{\partial \lambda}$$ to zero and solving. Weibull++ treats $$\gamma$$ as a constant when computing bounds; i.e., $$Var(\hat{\gamma}) = 0$$. (See the discussion in Appendix D for more information.)

Bounds on Reliability
The reliability of the two-parameter exponential distribution is:


 * $$\hat{R}(t;\hat{\lambda })={{e}^{-\hat{\lambda }(t-\hat{\gamma })}}\,\!$$

The corresponding confidence bounds are estimated from:


 * $$\begin{align}

& {{R}_{L}}= & {{e}^{-{{\lambda }_{U}}(t-\hat{\gamma })}} \\ & {{R}_{U}}= & {{e}^{-{{\lambda }_{L}}(t-\hat{\gamma })}} \end{align}\,\!$$

These equations hold true for the 1-parameter exponential distribution, with $$\gamma =0\,\!$$.

Bounds on Time
The bounds around time for a given exponential percentile, or reliability value, are estimated by first solving the reliability equation with respect to time, or reliable life:


 * $$\hat{t}=-\frac{1}\cdot \ln (R)+\hat{\gamma }\,\!$$

The corresponding confidence bounds are estimated from:


 * $$\begin{align}

& {{t}_{U}}= & -\frac{1}\cdot \ln (R)+\hat{\gamma } \\ & {{t}_{L}}= & -\frac{1}\cdot \ln (R)+\hat{\gamma } \end{align}\,\!$$

The same equations apply for the one-parameter exponential with $$\gamma =0.\,\!$$

Bounds on Parameters
For one-parameter distributions such as the exponential, the likelihood confidence bounds are calculated by finding values for $$\theta \,\!$$ that satisfy:


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

This equation can be rewritten as:


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

For complete data, the likelihood function for the exponential distribution is given by:


 * $$L(\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{t}_{i}};\lambda )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\lambda \cdot {{e}^{-\lambda \cdot {{t}_{i}}}}\,\!$$

where the $${{t}_{i}}\,\!$$ values represent the original time-to-failure data. For a given value of $$\alpha \,\!$$, values for $$\lambda \,\!$$ 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: LR Bounds for Lambda
Five units are put on a reliability test and experience failures at 20, 40, 60, 100, and 150 hours. Assuming an exponential distribution, the MLE parameter estimate is calculated to be $$\hat{\lambda }=0.013514\,\!$$. Calculate the 85% 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(\hat{\lambda })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\hat{\lambda })=\underset{i=1}{\overset{N}{\mathop \prod }}\,\hat{\lambda }\cdot {{e}^{-\hat{\lambda }\cdot {{x}_{i}}}} \\ L(\hat{\lambda })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,0.013514\cdot {{e}^{-0.013514\cdot {{x}_{i}}}} \\ L(\hat{\lambda })= & 3.03647\times {{10}^{-12}} \end{align}\,\!$$

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


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

Since our specified confidence level, $$\delta \,\!$$, is 85%, we can calculate the value of the chi-squared statistic, $$\chi _{0.85;1}^{2}=2.072251.\,\!$$ We can now substitute this information into the equation:


 * $$\begin{align}

L(\lambda )-L(\hat{\lambda })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\ L(\lambda )-3.03647\times {{10}^{-12}}\cdot {{e}^{\tfrac{-2.072251}{2}}}= & 0, \\ L(\lambda )-1.07742\times {{10}^{-12}}= & 0. \end{align}\,\!$$

It now remains to find the values of $$\lambda \,\!$$ which satisfy this equation. Since there is only one parameter, there are only two values of $$\lambda \,\!$$ that will satisfy the equation. These values represent the $$\delta =85%\,\!$$ two-sided confidence limits of the parameter estimate $$\hat{\lambda }\,\!$$. For our problem, the confidence limits are:


 * $$\begin{align}

{{\lambda }_{0.85}}=(0.006572,0.024172) \end{align}\,\!$$

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 exponential reliability equation into the likelihood function. The exponential reliability equation can be written as:


 * $$R={{e}^{-\lambda \cdot t}}\,\!$$

This can be rearranged to the form:


 * $$\lambda =\frac{-\text{ln}(R)}{t}\,\!$$

This equation can now be substituted into the likelihood ratio equation to produce a likelihood equation in terms of $$t\,\!$$ and $$R:\,\!$$


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

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 the equation for the mean and the Bayes's rule equation for single parametera 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, the likelihood ratio function can be solved for the values of interest.

Example: LR Bounds on Time
For the data given above for the LR Bounds on Lambda example (five failures at 20, 40, 60, 100 and 150 hours), determine the 85% two-sided confidence bounds on the time estimate for a reliability of 90%. The ML estimate for the time at $$R(t)=90%\,\!$$ is $$\hat{t}=7.797\,\!$$. Solution

In this example, we are trying to determine the 85% two-sided confidence bounds on the time estimate of 7.797. This is accomplished by substituting $$R=0.90\,\!$$ and $$\alpha =0.85\,\!$$ into the likelihood ratio bound equation. It now remains to find the values of $$t\,\!$$ which satisfy this equation. Since there is only one parameter, there are only two values of $$t\,\!$$ that will satisfy the equation. These values represent the $$\delta =85%\,\!$$ two-sided confidence limits of the time estimate $$\hat{t}\,\!$$. For our problem, the confidence limits are:


 * $${{\hat{t}}_{R=0.9}}=(4.359,16.033)\,\!$$

Example: LR Bounds on Reliability
Again using the data given above for the LR Bounds on Lambda example (five failures at 20, 40, 60, 100 and 150 hours), determine the 85% two-sided confidence bounds on the reliability estimate for a $$t=50\,\!$$. The ML estimate for the time at $$t=50\,\!$$ is $$\hat{R}=50.881%\,\!$$.

Solution

In this example, we are trying to determine the 85% two-sided confidence bounds on the reliability estimate of 50.881%. This is accomplished by substituting $$t=50\,\!$$ and $$\alpha =0.85\,\!$$ into the likelihood ratio bound equation. It now remains to find the values of $$R\,\!$$ which satisfy this equation. Since there is only one parameter, there are only two values of $$R\,\!$$ that will satisfy the equation. These values represent the $$\delta =85%\,\!$$ two-sided confidence limits of the reliability estimate $$\hat{R}\,\!$$. For our problem, the confidence limits are:


 * $${{\hat{R}}_{t=50}}=(29.861\%,71.794\%)\,\!$$.

Bounds on Parameters
From Confidence Bounds, we know that the posterior distribution of $$\lambda \,\!$$ can be written as:


 * $$f(\lambda |Data)=\frac{L(Data|\lambda )\varphi (\lambda )}{\int_{0}^{\infty }L(Data|\lambda )\varphi (\lambda )d\lambda }\,\!$$

where $$\varphi (\lambda )=\tfrac{1}{\lambda }\,\!$$, is the non-informative prior of $$\lambda \,\!$$.

With the above prior distribution, $$f(\lambda |Data)\,\!$$ can be rewritten as:


 * $$f(\lambda |Data)=\frac{L(Data|\lambda )\tfrac{1}{\lambda }}{\int_{0}^{\infty }L(Data|\lambda )\tfrac{1}{\lambda }d\lambda }\,\!$$

The one-sided upper bound of $$\lambda \,\!$$ is:


 * $$CL=P(\lambda \le {{\lambda }_{U}})=\int_{0}^f(\lambda |Data)d\lambda \,\!$$

The one-sided lower bound of $$\lambda \,\!$$ is:


 * $$1-CL=P(\lambda \le {{\lambda }_{L}})=\int_{0}^f(\lambda |Data)d\lambda \,\!$$

The two-sided bounds of $$\lambda \,\!$$ are:


 * $$CL=P({{\lambda }_{L}}\le \lambda \le {{\lambda }_{U}})=\int_^f(\lambda |Data)d\lambda \,\!$$

Bounds on Time (Type 1)
The reliable life equation is:


 * $$t=\frac{-\ln R}{\lambda }\,\!$$

For the one-sided upper bound on time we have:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(t\le {{T}_{U}})=\underset{}{\overset{}{\mathop{\Pr }}}\,(\frac{-\ln R}{\lambda }\le {{T}_{U}})\,\!$$

The above equation can be rewritten in terms of $$\lambda \,\!$$ as:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(\frac{-\ln R}\le \lambda )\,\!$$

From the above posterior distribuiton equation, we have:


 * $$CL=\frac{\int_{\tfrac{-\ln R}}^{\infty }L(Data|\lambda )\tfrac{1}{\lambda }d\lambda }{\int_{0}^{\infty }L(Data|\lambda )\tfrac{1}{\lambda }d\lambda }\,\!$$

The above equation is solved w.r.t. $${{t}_{U}}.\,\!$$ The same method is applied for one-sided lower and two-sided bounds on time.

Bounds on Reliability (Type 2)
The one-sided upper bound on reliability is given by:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(R\le {{R}_{U}})=\underset{}{\overset{}{\mathop{\Pr }}}\,(\exp (-\lambda t)\le {{R}_{U}})\,\!$$

The above equaation can be rewritten in terms of $$\lambda \,\!$$ as:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(\frac{-\ln {{R}_{U}}}{t}\le \lambda )\,\!$$

From the equation for posterior distribution we have:


 * $$CL=\frac{\int_{\tfrac{-\ln {{R}_{U}}}{t}}^{\infty }L(Data|\lambda )\tfrac{1}{\lambda }d\lambda }{\int_{0}^{\infty }L(Data|\lambda )\tfrac{1}{\lambda }d\lambda }\,\!$$

The above equation is solved w.r.t. $${{R}_{U}}.\,\!$$ The same method can be used to calculate one-sided lower and two sided bounds on reliability.