Fisher Matrix Confidence Bounds

This section presents an overview of the theory on obtaining approximate confidence bounds on suspended (multiply censored) data. The methodology used is the so-called Fisher matrix bounds (FM), described in Nelson [30] and Lloyd and Lipow [24]. These bounds are employed in most other commercial statistical applications. In general, these bounds tend to be more optimistic than the non-parametric rank based bounds. This may be a concern, particularly when dealing with small sample sizes. Some statisticians feel that the Fisher matrix bounds are too optimistic when dealing with small sample sizes and prefer to use other techniques for calculating confidence bounds, such as the likelihood ratio bounds.

Approximate Estimates of the Mean and Variance of a Function
In utilizing FM bounds for functions, one must first determine the mean and variance of the function in question (i.e. reliability function, failure rate function, etc.). An example of the methodology and assumptions for an arbitrary function $$G$$ is presented next.

Single Parameter Case
For simplicity, consider a one-parameter distribution represented by a general function, $$G,$$ which is a function of one parameter estimator, say $$G(\widehat{\theta }).$$ For example, the mean of the exponential distribution is a function of the parameter $$\lambda $$: $$G(\lambda )=1/\lambda =\mu $$. Then, in general, the expected value of $$G\left( \widehat{\theta } \right)$$ can be found by:


 * $$E\left( G\left( \widehat{\theta } \right) \right)=G(\theta )+O\left( \frac{1}{n} \right)$$

where $$G(\theta )$$ is some function of $$\theta $$, such as the reliability function, and $$\theta $$ is the population parameter where $$E\left( \widehat{\theta } \right)=\theta $$ as $$n\to \infty $$. The term $$O\left( \tfrac{1}{n} \right)$$ is a function of $$n$$, the sample size, and tends to zero, as fast as $$\tfrac{1}{n},$$ as $$n\to \infty .$$ For example, in the case of $$\widehat{\theta }=1/\overline{x}$$ and $$G(x)=1/x$$, then $$E(G(\widehat{\theta }))=\overline{x}+O\left( \tfrac{1}{n} \right)$$ where $$O\left( \tfrac{1}{n} \right)=\tfrac{n}$$. Thus as $$n\to \infty $$, $$E(G(\widehat{\theta }))=\mu $$ where $$\mu $$ and $$\sigma $$ are the mean and standard deviation, respectively. Using the same one-parameter distribution, the variance of the function $$G\left( \widehat{\theta } \right)$$ can then be estimated by:


 * $$Var\left( G\left( \widehat{\theta } \right) \right)=\left( \frac{\partial G}{\partial \widehat{\theta }} \right)_{\widehat{\theta }=\theta }^{2}Var\left( \widehat{\theta } \right)+O\left( \frac{1} \right)$$

Two-Parameter Case
Consider a Weibull distribution with two parameters $$\beta $$ and $$\eta $$. For a given value of $$T$$, $$R(T)=G(\beta ,\eta )={{e}^{-{{\left( \tfrac{T}{\eta } \right)}^{\beta }}}}$$. Repeating the previous method for the case of a two-parameter distribution, it is generally true that for a function $$G$$, which is a function of two parameter estimators, say $$G\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right)$$, that:


 * $$E\left( G\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right) \right)=G\left( {{\theta }_{1}},{{\theta }_{2}} \right)+O\left( \frac{1}{n} \right)$$

and:


 * $$\begin{align}

Var( G( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}}))= &{(\frac{\partial G}{\partial {{\widehat{\theta }}_{1}}})^2}_{{\widehat{\theta_{1}}}={\theta_{1}}}Var(\widehat{\theta_{1}})+{(\frac{\partial G}{\partial {{\widehat{\theta }}_{2}}})^2}_{{\widehat{\theta_{2}}}={\theta_{1}}}Var(\widehat{\theta_{2}})\\

& +2{(\frac{\partial G}{\partial {{\widehat{\theta }}_{1}}})^2}_{{\widehat{\theta_{1}}}={\theta_{1}}}{(\frac{\partial G}{\partial {{\widehat{\theta }}_{2}}})^2}_{{\widehat{\theta_{2}}}={\theta_{1}}}Cov(\widehat{\theta_{1}},\widehat{\theta_{2}}) \\

& +O(\frac{1}{n^{\tfrac{3}{2}}}) \end{align}

$$

Note that the derivatives of Eqn. (var) are evaluated at $${{\widehat{\theta }}_{1}}={{\theta }_{1}}$$ and $${{\widehat{\theta }}_{2}}={{\theta }_{1}},$$ where E $$\left( {{\widehat{\theta }}_{1}} \right)\simeq {{\theta }_{1}}$$ and E $$\left( {{\widehat{\theta }}_{2}} \right)\simeq {{\theta }_{2}}.$$ Parameter Variance and Covariance Determination The determination of the variance and covariance of the parameters is accomplished via the use of the Fisher information matrix. For a two-parameter distribution, and using maximum likelihood estimates (MLE), the log-likelihood function for censored data is given by:


 * $$\begin{align}

\ln [L]= & \Lambda =\underset{i=1}{\overset{R}{\mathop \sum }}\,\ln [f({{T}_{i}};{{\theta }_{1}},{{\theta }_{2}})] \\ & \text{ }+\underset{j=1}{\overset{M}{\mathop \sum }}\,\ln [1-F({{S}_{j}};{{\theta }_{1}},{{\theta }_{2}})] \\ & \text{ }+\underset{l=1}{\overset{P}{\mathop \sum }}\,\ln \left\{ F({{I}_};{{\theta }_{1}},{{\theta }_{2}})-F({{I}_};{{\theta }_{1}},{{\theta }_{2}}) \right\} \end{align}$$

In the equation above, the first summation is for complete data, the second summation is for right censored data, and the third summation is for interval or left censored data. For more information on these data types, see Chapter 4. Then the Fisher information matrix is given by:


 * $${{F}_{0}}=\left[ \begin{matrix}

{{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{1}^{2}} \right]}_{0}} & {} & {{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{1}}\partial {{\theta }_{2}}} \right]}_{0}} \\ {} & {} & {} \\   {{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{2}}\partial {{\theta }_{1}}} \right]}_{0}} & {} & {{E}_{0}}{{\left[ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{2}^{2}} \right]}_{0}}  \\ \end{matrix} \right]$$

The subscript $$0$$ indicates that the quantity is evaluated at $${{\theta }_{1}}={{\theta }_}$$ and $${{\theta }_{2}}={{\theta }_},$$ the true values of the parameters. So for a sample of $$N$$ units where $$R$$ units have failed, $$S$$ have been suspended, and $$P$$ have failed within a time interval, and $$N=R+M+P,$$ one could obtain the sample local information matrix by:


 * $$F={{\left[ \begin{matrix}

-\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{1}^{2}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{1}}\partial {{\theta }_{2}}} \\ {} & {} & {} \\   -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{2}}\partial {{\theta }_{1}}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{2}^{2}}  \\ \end{matrix} \right]}^{}}$$

Substituting in the values of the estimated parameters, in this case $${{\widehat{\theta }}_{1}}$$ and $${{\widehat{\theta }}_{2}}$$, and then inverting the matrix, one can then obtain the local estimate of the covariance matrix or:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( {{\widehat{\theta }}_{1}} \right) & {} & \widehat{Cov}\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right) \\ {} & {} & {} \\   \widehat{Cov}\left( {{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}} \right) & {} & \widehat{Var}\left( {{\widehat{\theta }}_{2}} \right)  \\ \end{matrix} \right]={{\left[ \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{1}^{2}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{1}}\partial {{\theta }_{2}}} \\ {} & {} & {} \\   -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\theta }_{2}}\partial {{\theta }_{1}}} & {} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \theta _{2}^{2}}  \\ \end{matrix} \right]}^{-1}}$$

Then the variance of a function ($$Var(G)$$) can be estimated using Eqn. (var). Values for the variance and covariance of the parameters are obtained from Eqn. (Fisher2). Once they have been obtained, the approximate confidence bounds on the function are given as:


 * $$C{{B}_{R}}=E(G)\pm {{z}_{\alpha }}\sqrt{Var(G)}$$

which is the estimated value plus or minus a certain number of standard deviations. We address finding $${{z}_{\alpha }}$$ next.

Approximate Confidence Intervals on the Parameters
In general, MLE estimates of the parameters are asymptotically normal, meaning for large sample sizes that a distribution of parameter estimates from the same population would be very close to the normal distribution. Thus if $$\widehat{\theta }$$ is the MLE estimator for $$\theta $$, in the case of a single parameter distribution, estimated from a large sample of $$n$$ units and if:


 * $$z\equiv \frac{\widehat{\theta }-\theta }{\sqrt{Var\left( \widehat{\theta } \right)}}$$

then using the normal distribution of $$z\ \ :$$


 * $$P\left( x\le z \right)\to \Phi \left( z \right)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z}{{e}^{-\tfrac{2}}}dt$$

for large $$n$$. We now place confidence bounds on $$\theta ,$$ at some confidence level $$\delta $$, bounded by the two end points $${{C}_{1}}$$ and $${{C}_{2}}$$ where:


 * $$P\left( {{C}_{1}}<\theta <{{C}_{2}} \right)=\delta $$

From Eqn. (e729):


 * $$P\left( -{{K}_{\tfrac{1-\delta }{2}}}<\frac{\widehat{\theta }-\theta }{\sqrt{Var\left( \widehat{\theta } \right)}}<{{K}_{\tfrac{1-\delta }{2}}} \right)\simeq \delta $$

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


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

Now by simplifying Eqn. (e731), one can obtain the approximate two-sided confidence bounds on the parameter $$\theta ,$$ at a confidence level $$\delta ,$$ or:


 * $$\left( \widehat{\theta }-{{K}_{\tfrac{1-\delta }{2}}}\cdot \sqrt{Var\left( \widehat{\theta } \right)}<\theta <\widehat{\theta }+{{K}_{\tfrac{1-\delta }{2}}}\cdot \sqrt{Var\left( \widehat{\theta } \right)} \right)$$

The upper one-sided bounds are given by:


 * $$\theta <\widehat{\theta }+{{K}_{1-\delta }}\sqrt{Var(\widehat{\theta })}$$

while the lower one-sided bounds are given by:


 * $$\theta >\widehat{\theta }-{{K}_{1-\delta }}\sqrt{Var(\widehat{\theta })}$$

If $$\widehat{\theta }$$ must be positive, then $$\ln \widehat{\theta }$$ is treated as normally distributed. The two-sided approximate confidence bounds on the parameter $$\theta $$, at confidence level $$\delta $$, then become:


 * $$\begin{align}

& {{\theta }_{U}}= & \widehat{\theta }\cdot {{e}^{\tfrac{{{K}_{\tfrac{1-\delta }{2}}}\sqrt{Var\left( \widehat{\theta } \right)}}{\widehat{\theta }}}}\text{ (Two-sided upper)} \\ & {{\theta }_{L}}= & \frac{\widehat{\theta }}\text{    (Two-sided lower)} \end{align}$$

The one-sided approximate confidence bounds on the parameter $$\theta $$, at confidence level $$\delta ,$$ can be found from:


 * $$\begin{align}

& {{\theta }_{U}}= & \widehat{\theta }\cdot {{e}^{\tfrac{{{K}_{1-\delta }}\sqrt{Var\left( \widehat{\theta } \right)}}{\widehat{\theta }}}}\text{ (One-sided upper)} \\ & {{\theta }_{L}}= & \frac{\widehat{\theta }}\text{    (One-sided lower)} \end{align}$$

The same procedure can be extended for the case of a two or more parameter distribution. Lloyd and Lipow [24] further elaborate on this procedure.

Confidence Bounds on Time (Type 1)
Type 1 confidence bounds are confidence bounds around time for a given reliability. For example, when using the one-parameter exponential distribution, the corresponding time for a given exponential percentile (i.e. y-ordinate or unreliability, $$Q=1-R)$$ is determined by solving the unreliability function for the time, $$T$$, or:


 * $$\begin{align}\widehat{T}(Q)= &-\frac{1}{\widehat{\lambda }}

\ln (1-Q)= & -\frac{1}{\widehat{\lambda }}\ln (R) \end{align}$$

Bounds on time (Type 1) return the confidence bounds around this time value by determining the confidence intervals around $$\widehat{\lambda }$$ and substituting these values into Eqn. (cb). The bounds on $$\widehat{\lambda }$$ were determined using Eqns. (cblmu) and (cblml), with its variance obtained from Eqn. (Fisher2). Note that the procedure is slightly more complicated for distributions with more than one parameter.

Confidence Bounds on Reliability (Type 2)
Type 2 confidence bounds are confidence bounds around reliability. For example, when using the two-parameter exponential distribution, the reliability function is:


 * $$\widehat{R}(T)={{e}^{-\widehat{\lambda }\cdot T}}$$

Reliability bounds (Type 2) return the confidence bounds by determining the confidence intervals around $$\widehat{\lambda }$$ and substituting these values into Eqn. (cbr). The bounds on $$\widehat{\lambda }$$ were determined using Eqns. (cblmu) and (cblml), with its variance obtained from Eqn. (Fisher2). Once again, the procedure is more complicated for distributions with more than one parameter.

Beta Binomial Confidence Bounds
Another less mathematically intensive method of calculating confidence bounds involves a procedure similar to that used in calculating median ranks (see Chapter 4). This is a non-parametric approach to confidence interval calculations that involves the use of rank tables and is commonly known as beta-binomial bounds (BB). By non-parametric, we mean that no underlying distribution is assumed. (Parametric implies that an underlying distribution, with parameters, is assumed.) In other words, this method can be used for any distribution, without having to make adjustments in the underlying equations based on the assumed distribution. Recall from the discussion on the median ranks that we used the binomial equation to compute the ranks at the 50% confidence level (or median ranks) by solving the cumulative binomial distribution for $$Z$$ (rank for the $${{j}^{th}}$$ failure):


 * $$P=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}

N \\ k \\ \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}$$

where $$N$$ is the sample size and $$j$$ is the order number. The median rank was obtained by solving the following equation for $$Z$$:


 * $$0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}

N \\ k \\ \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}$$

The same methodology can then be repeated by changing $$P$$ for $$0.50$$ $$(50%)$$ to our desired confidence level. For $$P=90%$$  one would formulate the equation as


 * $$0.90=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}

N \\ k \\ \end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}$$

Keep in mind that one must be careful to select the appropriate values for $$P$$ based on the type of confidence bounds desired. For example, if two-sided 80% confidence bounds are to be calculated, one must solve the equation twice (once with $$P=0.1$$ and once with $$P=0.9$$) in order to place the bounds around 80% of the population. Using this methodology, the appropriate ranks are obtained and plotted based on the desired confidence level. These points are then joined by a smooth curve to obtain the corresponding confidence bound. This non-parametric methodology is only used by Weibull++ when plotting bounds on the mixed Weibull distribution. Full details on this methodology can be found in Kececioglu [20]. These binomial equations can again be transformed using the beta and F distributions, thus the name beta binomial confidence bounds.

Introduction
A third method for calculating confidence bounds is the likelihood ratio bounds (LRB) method. Conceptually, this method is a great deal simpler than that of the Fisher matrix, although that does not mean that the results are of any less value. In fact, the LRB method is often preferred over the FM method in situations where there are smaller sample sizes. Likelihood ratio confidence bounds are based on the equation:


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

where: If $$\delta $$ is the confidence level, then $$\alpha =\delta $$ for two-sided bounds and $$\alpha =(2\delta -1)$$ for one-sided. Recall from Chapter 3 that if $$x$$ is a continuous random variable with $$pdf$$:
 * 1) $$L(\theta )$$ is the likelihood function for the unknown parameter vector $$\theta $$
 * 2) $$L(\widehat{\theta })$$ is the likelihood function calculated at the estimated vector $$\widehat{\theta }$$
 * 3) $$\chi _{\alpha ;k}^{2}$$ is the chi-squared statistic with probability $$\alpha $$ and $$k$$ degrees of freedom, where $$k$$ is the number of quantities jointly estimated
 * $$f(x;{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})$$,

where $${{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}}$$ are $$k$$ unknown constant parameters that need to be estimated, one can conduct an experiment and obtain $$R$$ independent observations, $${{x}_{1}},$$ $${{x}_{2}},$$$$...,{{x}_{R}}$$, which correspond in the case of life data analysis to failure times. The likelihood function is given by:


 * $$L({{x}_{1}},{{x}_{2}},...,{{x}_{R}}|{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})=L=\underset{i=1}{\overset{R}{\mathop \prod }}\,f({{x}_{i}};{{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}})$$


 * $$i=1,2,...,R$$

The maximum likelihood estimators (MLE) of $${{\theta }_{1}},{{\theta }_{2}},...,{{\theta }_{k}},$$ are obtained by maximizing $$L.$$ These are represented by the $$L(\widehat{\theta })$$ term in the denominator of the ratio in Eqn. (lratio1). Since the values of the data points are known, and the values of the parameter estimates $$\widehat{\theta }$$ have been calculated using MLE methods, the only unknown term in Eqn. (lratio1) is the $$L(\theta )$$ term in the numerator of the ratio. It remains to find the values of the unknown parameter vector $$\theta $$ that satisfy Eqn. (lratio1). For distributions that have two parameters, the values of these two parameters can be varied in order to satisfy Eqn. (lratio1). The values of the parameters that satisfy this equation will change based on the desired confidence level $$\delta ;$$ but at a given value of $$\delta $$ there is only a certain region of values for $${{\theta }_{1}}$$ and $${{\theta }_{2}}$$ for which Eqn. (lratio1) holds true. This region can be represented graphically as a contour plot, an example of which is given in the following graphic.

The region of the contour plot essentially represents a cross-section of the likelihood function surface that satisfies the conditions of Eqn. (lratio1).

Note on Contour Plots in Weibull++
Contour plots can be used for comparing data sets. Consider two data sets, e.g. old and new design where the engineer would like to determine if the two designs are significantly different and at what confidence. By plotting the contour plots of each data set in a multiple plot (the same distribution must be fitted to each data set), one can determine the confidence at which the two sets are significantly different. If, for example, there is no overlap (i.e. the two plots do not intersect) between the two 90% contours, then the two data sets are significantly different with a 90% confidence. If there is an overlap between the two 95% contours, then the two designs are NOT significantly different at the 95% confidence level. An example of non-intersecting contours is shown next. Chapter 12 discusses comparing data sets.

Confidence Bounds on the Parameters
The bounds on the parameters are calculated by finding the extreme values of the contour plot on each axis for a given confidence level. Since each axis represents the possible values of a given parameter, the boundaries of the contour plot represent the extreme values of the parameters 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:


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

The task now becomes to find the values of the parameters $${{\theta }_{1}}$$ and $${{\theta }_{2}}$$ so that the equality in Eqn. (lratio3) is satisfied. Unfortunately, there is no closed-form solution, thus these values must be arrived at numerically. One method of doing this is to hold one parameter constant and iterate on the other until an acceptable solution is reached. This can prove to be rather tricky, since there will be two solutions for one parameter if the other is held constant. In situations such as these, it is best to begin the iterative calculations with values close to those of the MLE values, so as to ensure that one is not attempting to perform calculations outside of the region of the contour plot where no solution exists.

Example 1
Five units were put on a reliability test and experienced failures at 10, 20, 30, 40, and 50 hours. Assuming a Weibull distribution, the MLE parameter estimates are calculated to be $$\widehat{\beta }=2.2938$$ and $$\widehat{\eta }=33.9428.$$ Calculate the 90% two-sided confidence bounds on these parameters using the likelihood ratio method.

Solution to Example 1
The first step is to calculate the likelihood function for the parameter estimates:


 * $$\begin{align}

L(\widehat{\beta },\widehat{\eta })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\widehat{\beta },\widehat{\eta })=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\widehat{\beta }}{\widehat{\eta }}\cdot {{\left( \frac{{{x}_{i}}}{\widehat{\eta }} \right)}^{\widehat{\beta }-1}}\cdot {{e}^{-{{\left( \tfrac{{{x}_{i}}}{\widehat{\eta }} \right)}^{\widehat{\beta }}}}} \\ \\  L(\widehat{\beta },\widehat{\eta })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{2.2938}{33.9428}\cdot {{\left( \frac{{{x}_{i}}}{33.9428} \right)}^{1.2938}}\cdot {{e}^{-{{\left( \tfrac{{{x}_{i}}}{33.9428} \right)}^{2.2938}}}} \\ \\  L(\widehat{\beta },\widehat{\eta })= & 1.714714\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:


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

Since our specified confidence level, $$\delta $$, is 90%, we can calculate the value of the chi-squared statistic, $$\chi _{0.9;1}^{2}=2.705543.$$ We then substitute this information into the equation:


 * $$\begin{align}

L(\beta ,\eta )-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\ L(\beta ,\eta )-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ \\  L(\beta ,\eta )-4.432926\cdot {{10}^{-10}}= & 0 \end{align}$$

The next step is to find the set of values of $$\beta $$  and  $$\eta $$  that satisfy this equation, or find the values of $$\beta $$ and $$\eta $$ such that $$L(\beta ,\eta )=4.432926\cdot {{10}^{-10}}.$$

The solution is an iterative process that requires setting the value of $$\beta $$ and finding the appropriate values of $$\eta $$, and vice versa. The following table gives values of $$\beta $$ based on given values of $$\eta $$.

These data 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 $$\beta $$ is 1.142, while the highest is 3.950. These represent the two-sided 90% confidence limits on this parameter. Since solutions for the equation do not exist for values of $$\eta $$ below 23 or above 50, these can be considered the 90% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on $$\eta $$, we can perform the same procedure as before, but finding the two values of $$\eta $$ that correspond with a given value of $$\beta .$$ Using this method, we find that the 90% confidence limits on $$\eta $$ are 22.474 and 49.967, which are close to the initial estimates of 23 and 50. Note that the points where $$\beta $$ are maximized and minimized do not necessarily correspond with the points where $$\eta $$ are maximized and minimized. This is due to the fact that the contour plot is not symmetrical, so that the parameters will have their extremes at different points.

Confidence Bounds on Time (Type 1)
The manner in which the bounds on the time estimate for a given reliability are calculated is much the same as the manner in which the bounds on the parameters are calculated. The difference lies in the form of the likelihood functions that comprise the likelihood ratio. In the preceding section we used the standard form of the likelihood function, which was in terms of the parameters $${{\theta }_{1}}$$ and $${{\theta }_{2}}$$. In order to calculate the bounds on a time estimate, the likelihood function needs to be rewritten in terms of one parameter and time, so that the maximum and minimum values of the time can be observed as the parameter is varied. This process is best illustrated with an example.

Example 2
For the data given in Example 1, determine the 90% two-sided confidence bounds on the time estimate for a reliability of 50%. The ML estimate for the time at which $$R(t)=50%$$ is 28.930.

Solution to Example 2
In this example, we are trying to determine the 90% two-sided confidence bounds on the time estimate of 28.930. As was mentioned, we need to rewrite Eqn. (lrbexample) so that it is in terms of $$t$$ and $$\beta .$$ This is accomplished by using a form of the Weibull reliability equation, $$R={{e}^{-{{\left( \tfrac{t}{\eta } \right)}^{\beta }}}}.$$ This can be rearranged in terms of $$\eta $$, with $$R$$ being considered a known variable or:


 * $$\eta =\frac{t}$$

This can then be substituted into the $$\eta $$ term in Eqn. (lrbexample) to form a likelihood equation in terms of $$t$$ and $$\beta $$ or:


 * $$\begin{align}

& L(\beta ,t)= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\beta ,t,R) \\ & &   \end{align}$$


 * $$=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\beta }{\left( \tfrac{t} \right)}\cdot {{\left( \frac{{{x}_{i}}}{\left( \tfrac{t} \right)} \right)}^{\beta -1}}\cdot \text{exp}\left[ -{{\left( \frac{{{x}_{i}}}{\left( \tfrac{t} \right)} \right)}^{\beta }} \right]$$

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


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

Since our specified confidence level, $$\delta $$, is 90%, we can calculate the value of the chi-squared statistic, $$\chi _{0.9;1}^{2}=2.705543.$$ We can now substitute this information into the equation:


 * $$\begin{align}

L(\beta ,t)-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\  L(\beta ,t)-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ & \\   L(\beta ,t)-4.432926\cdot {{10}^{-10}}= & 0 \end{align}$$

Note that the likelihood value for $$L(\widehat{\beta },\widehat{\eta })$$ is the same as it was for Example 1. This is because we are dealing with the same data and parameter estimates or, in other words, the maximum value of the likelihood function did not change. It now remains to find the values of $$\beta $$ and $$t$$ which satisfy this equation. This is an iterative process that requires setting the value of $$\beta $$ and finding the appropriate values of $$t$$. The following table gives the values of $$t$$ based on given values of $$\beta $$.

These points are represented graphically in the following contour plot: As can be determined from the table, the lowest calculated value for $$t$$ is 17.389, while the highest is 41.714. These represent the 90% two-sided confidence limits on the time at which reliability is equal to 50%.

Confidence Bounds on Reliability (Type 2)
The likelihood ratio bounds on a reliability estimate for a given time value are calculated in the same manner as were the bounds on time. The only difference is that the likelihood function must now be considered in terms of $$\beta $$ and $$R$$. The likelihood function is once again altered in the same way as before, only now $$R$$ is considered to be a parameter instead of $$t$$, since the value of $$t$$ must be specified in advance. Once again, this process is best illustrated with an example.

Example 3
For the data given in Example 1, determine the 90% two-sided confidence bounds on the reliability estimate for $$t=45$$. The ML estimate for the reliability at $$t=45$$ is 14.816%.

Solution to Example 3
In this example, we are trying to determine the 90% two-sided confidence bounds on the reliability estimate of 14.816%. As was mentioned, we need to rewrite Eqn. (lrbexample) so that it is in terms of $$R$$ and $$\beta .$$ This is again accomplished by substituting the Weibull reliability equation into the $$\eta $$ term in Eqn. (lrbexample) to form a likelihood equation in terms of $$R$$ and $$\beta $$:


 * $$\begin{align}

& L(\beta ,R)= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\beta ,t,R) \\ & &   \end{align}$$


 * $$=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{\beta }{\left( \tfrac{t} \right)}\cdot {{\left( \frac{{{x}_{i}}}{\left( \tfrac{t} \right)} \right)}^{\beta -1}}\cdot \text{exp}\left[ -{{\left( \frac{{{x}_{i}}}{\left( \tfrac{t}{{{(-\text{ln}(R))}^{\tfrac{1}{\beta }}}} \right)} \right)}^{\beta }} \right]$$

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


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

Since our specified confidence level, $$\delta $$, is 90%, we can calculate the value of the chi-squared statistic, $$\chi _{0.9;1}^{2}=2.705543.$$ We can now substitute this information into the equation:


 * $$\begin{align}

L(\beta ,R)-L(\widehat{\beta },\widehat{\eta })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\ \\  L(\beta ,R)-1.714714\times {{10}^{-9}}\cdot {{e}^{\tfrac{-2.705543}{2}}}= & 0 \\ \\  L(\beta ,R)-4.432926\cdot {{10}^{-10}}= & 0 \end{align}$$

It now remains to find the values of $$\beta $$ and $$R$$ that satisfy this equation. This is an iterative process that requires setting the value of $$\beta $$ and finding the appropriate values of $$R$$. The following table gives the values of $$R$$ based on given values of $$\beta $$.

These points are represented graphically in the following contour plot:

As can be determined from the table, the lowest calculated value for $$R$$ is 2.38%, while the highest is 44.26%. These represent the 90% two-sided confidence limits on the reliability at $$t=45$$.

Bayesian Confidence Bounds
A fourth method of estimating confidence bounds is based on the Bayes theorem. This type of confidence bounds relies on a different school of thought in statistical analysis, where prior information is combined with sample data in order to make inferences on model parameters and their functions. An introduction to Bayesian methods is given in Chapter 3. Bayesian confidence bounds are derived from Bayes rule, which states that:


 * $$f(\theta |Data)=\frac{L(Data|\theta )\varphi (\theta )}{\underset{\varsigma }{\int{\mathop{}_{}^{}}}\,L(Data|\theta )\varphi (\theta )d\theta }$$

where: In other words, the prior knowledge is provided in the form of the prior $$pdf$$ of the parameters, which in turn is combined with the sample data in order to obtain the posterior $$pdf.$$ Different forms of prior information exist, such as past data, expert opinion or non-informative (refer to Chapter 3). It can be seen from Eqn. (BayesRule) that we are now dealing with distributions of parameters rather than single value parameters. For example, consider a one-parameter distribution with a positive parameter $${{\theta }_{1}}$$. Given a set of sample data, and a prior distribution for $${{\theta }_{1}},$$  $$\varphi ({{\theta }_{1}}),$$ Eqn. (BayesRule) can be written as:
 * 1) $$f(\theta |Data)$$ is the $$posterior$$ $$pdf$$ of $$\theta $$
 * 2) $$\theta $$ is the parameter vector of the chosen distribution (i.e. Weibull, lognormal, etc.)
 * 3) $$L(\bullet )$$ is the likelihood function
 * 4) $$\varphi (\theta )$$ is the $$prior$$ $$pdf$$ of the parameter vector $$\theta $$
 * 5) $$\varsigma $$ is the range of $$\theta $$.


 * $$f({{\theta }_{1}}|Data)=\frac{L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})}{\int_{0}^{\infty }L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})d{{\theta }_{1}}}$$

In other words, we now have the distribution of $${{\theta }_{1}}$$ and we can now make statistical inferences on this parameter, such as calculating probabilities. Specifically, the probability that $${{\theta }_{1}}$$ is less than or equal to a value $$x,$$ $$P({{\theta }_{1}}\le x)$$ can be obtained by integrating Eqn. (BayesEX), or:


 * $$P({{\theta }_{1}}\le x)=\int_{0}^{x}f({{\theta }_{1}}|Data)d{{\theta }_{1}}$$

Eqn. (IntBayes) essentially calculates a confidence bound on the parameter, where $$P({{\theta }_{1}}\le x)$$ is the confidence level and $$x$$ is the confidence bound. Substituting Eqn. (BayesEX) into Eqn. (IntBayes) yields:


 * $$CL=\frac{\int_{0}^{x}L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})d{{\theta }_{1}}}{\int_{0}^{\infty }L(Data|{{\theta }_{1}})\varphi ({{\theta }_{1}})d{{\theta }_{1}}}$$

The only question at this point is what do we use as a prior distribution of $${{\theta }_{1}}.$$. For the confidence bounds calculation application, non-informative prior distributions are utilized. Non-informative prior distributions are distributions that have no population basis and play a minimal role in the posterior distribution. The idea behind the use of non-informative prior distributions is to make inferences that are not affected by external information, or when external information is not available. In the general case of calculating confidence bounds using Bayesian methods, the method should be independent of external information and it should only rely on the current data. Therefore, non-informative priors are used. Specifically, the uniform distribution is used as a prior distribution for the different parameters of the selected fitted distribution. For example, if the Weibull distribution is fitted to the data, the prior distributions for beta and eta are assumed to be uniform. Eqn. (BayesCLEX) can be generalized for any distribution having a vector of parameters $$\theta ,$$ yielding the general equation for calculating Bayesian confidence bounds:


 * $$CL=\frac{\underset{\xi }{\int{\mathop{}_{}^{}}}\,L(Data|\theta )\varphi (\theta )d\theta }{\underset{\varsigma }{\int{\mathop{}_{}^{}}}\,L(Data|\theta )\varphi (\theta )d\theta }$$

where: If $$T$$ is given, then from Eqn. (BayesCL) and $$\Psi $$ and for a given $$CL,$$ the bounds on $$R$$ are calculated. If $$R$$ is given, then from Eqn. (BayesCL) and $$\Psi $$ and for a given $$CL,$$ the bounds on $$T$$ are calculated.
 * 1) $$CL$$ is confidence level
 * 2) $$\theta $$ is the parameter vector
 * 3) $$L(\bullet )$$ is the likelihood function
 * 4) $$\varphi (\theta )$$ is the prior $$pdf$$ of the parameter vector $$\theta $$
 * 5) $$\varsigma $$ is the range of $$\theta $$
 * 6) $$\xi $$ is the range in which $$\theta $$ changes from $$\Psi (T,R)$$ till $${\theta }'s$$ maximum value or from $${\theta }'s$$ minimum value till $$\Psi (T,R)$$
 * 7) $$\Psi (T,R)$$ is function such that if $$T$$ is given then the bounds are calculated for $$R$$ and if $$R$$ is given, then he bounds are calculated for $$T$$.

Confidence Bounds on Time (Type 1)
For a given failure time distribution and a given reliability $$R$$, $$T(R)$$ is a function of $$R$$ and the distribution parameters. To illustrate the procedure for obtaining confidence bounds, the two-parameter Weibull distribution is used as an example. Bounds, for the case of other distributions, can be obtained in similar fashion. For the two-parameter Weibull distribution:


 * $$T(R)=\eta \exp (\frac{\ln (-\ln R)}{\beta })$$

For a given reliability, the Bayesian one-sided upper bound estimate for $$T(R)$$ is:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(T\le {{T}_{U}})=\int_{0}^{{{T}_{U}}(R)}f(T|Data,R)dT$$

where $$f(T|Data,R)$$ is the posterior distribution of Time $$T.$$ Using Eqn. (T bayes), we have the following:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(T\le {{T}_{U}})=\underset{}{\overset{}{\mathop{\Pr }}}\,(\eta \exp (\frac{\ln (-\ln R)}{\beta })\le {{T}_{U}})$$

Eqn. (cl) can be rewritten in terms of $$\eta $$ as:


 * $$CL=\underset{}{\overset{}{\mathop{\Pr }}}\,(\eta \le {{T}_{U}}\exp (-\frac{\ln (-\ln R)}{\beta }))$$

From Eqns. (IntBayes), (BayesCLEX) and (BayesCL), by assuming the priors of $$\beta $$ and $$\eta $$ are independent, we then obtain the following relationship:


 * $$CL=\frac{\int_{0}^{\infty }\int_{0}^{{{T}_{U}}\exp (-\frac{\ln (-\ln R)}{\beta })}L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }$$

Eqn. (cl2) can be solved for $${{T}_{U}}(R)$$, where: The same method can be used to get the one-sided lower bound of $$T(R)$$ from:
 * 1) $$CL$$ is confidence level,
 * 2) $$\varphi (\beta )$$ is the prior $$pdf$$ of the parameter $$\beta $$. For non-informative prior distribution, $$\varphi (\beta )=\tfrac{1}{\beta }.$$
 * 3) $$\varphi (\eta )$$ is the prior $$pdf$$ of the parameter $$\eta .$$. For non-informative prior distribution, $$\varphi (\eta )=\tfrac{1}{\eta }.$$
 * 4) $$L(\bullet )$$ is the likelihood function.


 * $$CL=\frac{\int_{0}^{\infty }\int_{{{T}_{L}}\exp (\frac{-\ln (-\ln R)}{\beta })}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }$$

Eqn. (cl5) can be solved to get $${{T}_{L}}(R)$$. The Bayesian two-sided bounds estimate for $$T(R)$$ is:


 * $$CL=\int_{{{T}_{L}}(R)}^{{{T}_{U}}(R)}f(T|Data,R)dT$$

which is equivalent to:


 * $$(1+CL)/2=\int_{0}^{{{T}_{U}}(R)}f(T|Data,R)dT$$

and:


 * $$(1-CL)/2=\int_{0}^{{{T}_{L}}(R)}f(T|Data,R)dT$$

Using the same method for the one-sided bounds, $${{T}_{U}}(R)$$  and  $${{T}_{L}}(R)$$  can be solved.

Confidence Bounds on Reliability (Type 2)
For a given failure time distribution and a given time $$T$$, $$R(T)$$ is a function of $$T$$ and the distribution parameters. To illustrate the procedure for obtaining confidence bounds, the two-parameter Weibull distribution is used as an example. Bounds, for the case of other distributions, can be obtained in similar fashion. For example, for two parameter Weibull distribution:


 * $$R=\exp (-{{(\frac{T}{\eta })}^{\beta }})$$

The Bayesian one-sided upper bound estimate for $$R(T)$$ is:


 * $$CL=\int_{0}^{{{R}_{U}}(T)}f(R|Data,T)dR$$

Similar with the bounds on Time, the following is obtained:


 * $$CL=\frac{\int_{0}^{\infty }\int_{0}^{T\exp (-\frac{\ln (-\ln {{R}_{U}})}{\beta })}L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }$$

Eqn. (cl3) can be solved to get $${{R}_{U}}(T)$$.

The Bayesian one-sided lower bound estimate for R(T) is:


 * $$1-CL=\int_{0}^{{{R}_{L}}(T)}f(R|Data,T)dR$$

Using the posterior distribution, the following is obtained:


 * $$CL=\frac{\int_{0}^{\infty }\int_{T\exp (-\frac{\ln (-\ln {{R}_{L}})}{\beta })}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }{\int_{0}^{\infty }\int_{0}^{\infty }L(\beta ,\eta )\varphi (\beta )\varphi (\eta )d\eta d\beta }$$

Eqn. (cl4) can be solved to get $${{R}_{L}}(T)$$. The Bayesian two-sided bounds estimate for $$R(T)$$ is:


 * $$CL=\int_{{{R}_{L}}(T)}^{{{R}_{U}}(T)}f(R|Data,T)dR$$

which is equivalent to:


 * $$\int_{0}^{{{R}_{U}}(T)}f(R|Data,T)dR=(1+CL)/2$$

and


 * $$\int_{0}^{{{R}_{L}}(T)}f(R|Data,T)dR=(1-CL)/2$$

Using the same method for one-sided bounds, $${{R}_{U}}(T)$$ and $${{R}_{L}}(T)$$ can be solved.

Simulation Based Bounds
The SimuMatic tool in Weibull++ can be used to perform a large number of reliability analyses on data sets that have been created using Monte Carlo simulation. This utility can assist the analyst to a) better understand life data analysis concepts, b) experiment with the influences of sample sizes and censoring schemes on analysis methods, c) construct simulation-based confidence intervals, d) better understand the concepts behind confidence intervals and e) design reliability tests. This section describes how to use simulation for estimating confidence bounds. SimuMatic generates confidence bounds and assists in visualizing and understanding them. In addition, it allows one to determine the adequacy of certain parameter estimation methods (such as rank regression on X, rank regression on Y and maximum likelihood estimation) and to visualize the effects of different data censoring schemes on the confidence bounds.

Example 4
The purpose of this example is to determine the best parameter estimation method for a sample of ten units following a Weibull distribution with $$\beta =2$$ and $$\eta =100$$ and with complete time-to-failure data for each unit (i.e. no censoring). The number of generated data sets is set to 10,000. The SimuMatic inputs are shown next.

The parameters are estimated using RRX, RRY and MLE. The plotted results generated by SimuMatic are shown next.

Using RRX:

Using RRY:

Using MLE:

The results clearly demonstrate that the median RRX estimate provides the least deviation from the truth for this sample size and data type. However, the MLE outputs are grouped more closely together, as evidenced by the bounds. The previous figures also show the simulation-based bounds, as well as the expected variation due to sampling error. This experiment can be repeated in SimuMatic using multiple censoring schemes (including Type I and Type II right censoring as well as random censoring) with various distributions. Multiple experiments can be performed with this utility to evaluate assumptions about the appropriate parameter estimation method to use for data sets.