Crow-AMSAA Confidence Bounds

In this appendix, we will present the two methods used in the RGA software to estimate the confidence bounds for the Crow-AMSAA (NHPP) model when applied to developmental testing data. The Fisher Matrix approach is based on the Fisher Information Matrix and is commonly employed in the reliability field. The Crow bounds were developed by Dr. Larry Crow.

Individual (Non-Grouped) Data

Bounds on Beta

Fisher Matrix Bounds

The parameter [math]\displaystyle{ \beta }[/math] must be positive, thus [math]\displaystyle{ \ln \beta }[/math] is treated as being normally distributed as well.

[math]\displaystyle{ \frac{\ln \hat{\beta }-\ln \beta }{\sqrt{Var(\ln \hat{\beta }})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds are given as:

[math]\displaystyle{ C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} }[/math]

[math]\displaystyle{ \alpha }[/math] in [math]\displaystyle{ {{z}_{\alpha }} }[/math] is different ( [math]\displaystyle{ \alpha /2 }[/math] , [math]\displaystyle{ \alpha }[/math] ) according to a 2-sided confidence interval or a 1-sided confidence interval, and variances can be calculated using the Fisher Matrix.

[math]\displaystyle{ \left[ \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta } \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta } & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}} \\ \end{matrix} \right]_{\beta =\widehat{\beta },\lambda =\widehat{\lambda }}^{-1}=\left[ \begin{matrix} Var(\widehat{\lambda }) & Cov(\widehat{\beta },\widehat{\lambda }) \\ Cov(\widehat{\beta },\widehat{\lambda }) & Var(\widehat{\beta }) \\ \end{matrix} \right] }[/math]

[math]\displaystyle{ \Lambda }[/math] is the natural log-likelihood function:

[math]\displaystyle{ \Lambda =N\ln \lambda +N\ln \beta -\lambda {{T}^{\beta }}+(\beta -1)\underset{i=1}{\overset{N}{\mathop \sum }}\,\ln {{T}_{i}} }[/math]

[math]\displaystyle{ \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{N}{{{\lambda }^{2}}} }[/math]

and:

[math]\displaystyle{ \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}=-\frac{N}{{{\beta }^{2}}}-\lambda {{T}^{\beta }}{{(\ln T)}^{2}} }[/math]

also:

[math]\displaystyle{ \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }=-{{T}^{\beta }}\ln T }[/math]

Crow Bounds

Time Terminated Data

For the 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval on [math]\displaystyle{ \beta }[/math] , calculate:

[math]\displaystyle{ \begin{align} & {{D}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2(N-1)} \\ & {{D}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2(N-1)} \end{align} }[/math]

The fractiles can be found in the tables of the [math]\displaystyle{ {{\chi }^{2}} }[/math] distribution. Thus the confidence bounds on [math]\displaystyle{ \beta }[/math] are:

[math]\displaystyle{ \begin{align} & {{\beta }_{L}}= & {{D}_{L}}\cdot \hat{\beta } \\ & {{\beta }_{U}}= & {{D}_{U}}\cdot \hat{\beta } \end{align} }[/math]

Failure Terminated Data
For the 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval on [math]\displaystyle{ \beta }[/math] , calculate:

[math]\displaystyle{ \begin{align} & {{D}_{L}}= & \frac{N\cdot \chi _{\tfrac{\alpha }{2},2(N-1)}^{2}}{2(N-1)(N-2)} \\ & {{D}_{U}}= & \frac{N\cdot \chi _{1-\tfrac{\alpha }{2},2(N-1)}^{2}}{2(N-1)(N-2)} \end{align} }[/math]

Thus the confidence bounds on [math]\displaystyle{ \beta }[/math] are:

[math]\displaystyle{ \begin{align} & {{\beta }_{L}}= & {{D}_{L}}\cdot \hat{\beta } \\ & {{\beta }_{U}}= & {{D}_{U}}\cdot \hat{\beta } \end{align} }[/math]

Bounds on Lambda

Fisher Matrix Bounds

The parameter [math]\displaystyle{ \lambda }[/math] must be positive, thus [math]\displaystyle{ \ln \lambda }[/math] is treated as being normally distributed as well. These bounds are based on:

[math]\displaystyle{ \frac{\ln \hat{\lambda }-\ln \lambda }{\sqrt{Var(\ln \hat{\lambda }})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on [math]\displaystyle{ \lambda }[/math] are given as:

[math]\displaystyle{ C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} }[/math]

where:

[math]\displaystyle{ \hat{\lambda }=\frac{n}{{{T}^{*\hat{\beta }}}} }[/math]

The variance calculation is the same as Eqn. (variance1).

Crow Bounds

Time Terminated Data
For the 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval, the confidence bounds on [math]\displaystyle{ \lambda }[/math] are:

[math]\displaystyle{ \begin{align} & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2{{T}^{{\hat{\beta }}}}} \\ & {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2{{T}^{{\hat{\beta }}}}} \end{align} }[/math]

The fractiles can be found in the tables of the [math]\displaystyle{ {{\chi }^{2}} }[/math] distribution.

Failure Terminated Data
For the 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval, the confidence bounds on [math]\displaystyle{ \lambda }[/math] are:

[math]\displaystyle{ \begin{align} & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2{{T}^{{\hat{\beta }}}}} \\ & {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2{{T}^{{\hat{\beta }}}}} \end{align} }[/math]

Bounds on Growth Rate

Since the growth rate is equal to [math]\displaystyle{ 1-\beta }[/math] , the confidence bounds for both the Fisher Matrix and Crow methods are:

[math]\displaystyle{ G\text{row}th Rate_L=1-\beta_U }[/math]

[math]\displaystyle{ G\text{row}th Rate_U=1-\beta_L }[/math]

For the Fisher Matrix confidence bounds, [math]\displaystyle{ {{\beta }_{L}} }[/math] and [math]\displaystyle{ {{\beta }_{U}} }[/math] are obtained from Eqn. (amsaac1). For the Crow bounds, [math]\displaystyle{ {{\beta }_{L}} }[/math] and [math]\displaystyle{ {{\beta }_{U}} }[/math] are obtained from Eqns. (amsaac2) and (amsaac22) depending on whether the analysis is for time terminated data or failure terminated data.

Bounds on Cumulative MTBF

Fisher Matrix Bounds

The cumulative MTBF, [math]\displaystyle{ {{m}_{c}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{m}_{c}}(t) }[/math] is treated as being normally distributed as well.

[math]\displaystyle{ \frac{\ln {{{\hat{m}}}_{c}}(t)-\ln {{m}_{c}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{c}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the cumulative MTBF are then estimated from:

[math]\displaystyle{ CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} }[/math]

where:

[math]\displaystyle{ {{\hat{m}}_{c}}(t)=\frac{1}{{\hat{\lambda }}}{{t}^{1-\hat{\beta }}} }[/math]

[math]\displaystyle{ \begin{align} & Var({{{\hat{m}}}_{c}}(t))= & {{\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda })\, \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{{\hat{\lambda }}}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}}{{t}^{1-\hat{\beta }}} \end{align} }[/math]

Crow Bounds

To calculate the Crow confidence bounds on cumulative MTBF, first calculate the Crow cumulative failure intensity confidence bounds:

[math]\displaystyle{ C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} }[/math]

[math]\displaystyle{ C{{(t)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} }[/math]

Then:

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\ & {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} \end{align} }[/math]

Bounds on Instantaneous MTBF

Fisher Matrix Bounds

The instantaneous MTBF, [math]\displaystyle{ {{m}_{i}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{m}_{i}}(t) }[/math] is treated as being normally distributed as well.

[math]\displaystyle{ \frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the instantaneous MTBF are then estimated from:

[math]\displaystyle{ CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} }[/math]

where:

[math]\displaystyle{ {{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}} }[/math]

[math]\displaystyle{ \begin{align} & Var({{{\hat{m}}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda }). \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\hat{\lambda }{{{\hat{\beta }}}^{2}}}{{t}^{1-\hat{\beta }}}-\frac{1}{\hat{\lambda }\hat{\beta }}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}\hat{\beta }}{{t}^{1-\hat{\beta }}} \end{align} }[/math]

Crow Bounds

Failure Terminated Data
Consider the following equation:

[math]\displaystyle{ G(\mu |n)=\mathop{}_{0}^{\infty }\frac{{{e}^{-x}}{{x}^{n-2}}}{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx }[/math]

Find the values [math]\displaystyle{ {{p}_{1}} }[/math] and [math]\displaystyle{ {{p}_{2}} }[/math] by finding the solution [math]\displaystyle{ c }[/math] to [math]\displaystyle{ G({{n}^{2}}/c|n)=\xi }[/math] for [math]\displaystyle{ \xi =\tfrac{\alpha }{2} }[/math] and [math]\displaystyle{ \xi =1-\tfrac{\alpha }{2} }[/math] , respectively. If using the biased parameters, [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] , then the upper and lower confidence bounds are:

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} \end{align} }[/math]

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}} }[/math] . If using the unbiased parameters, [math]\displaystyle{ \bar{\beta } }[/math] and [math]\displaystyle{ \bar{\lambda } }[/math] , then the upper and lower confidence bounds are:

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} \end{align} }[/math]

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}} }[/math] .

Time Terminated Data
Consider the following equation where [math]\displaystyle{ {{I}_{1}}(.) }[/math] is the modified Bessel function of order one:

[math]\displaystyle{ H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{x}^{2j-1}}}{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)} }[/math]

Find the values [math]\displaystyle{ {{\Pi }_{1}} }[/math] and [math]\displaystyle{ {{\Pi }_{2}} }[/math] by finding the solution [math]\displaystyle{ x }[/math] to [math]\displaystyle{ H(x|k)=\tfrac{\alpha }{2} }[/math] and [math]\displaystyle{ H(x|k)=1-\tfrac{\alpha }{2} }[/math] in the cases corresponding to the lower and upper bounds, respectively. Calculate [math]\displaystyle{ \Pi =\tfrac{4{{n}^{2}}}{{{x}^{2}}} }[/math] for each case. If using the biased parameters, [math]\displaystyle{ \hat{\beta } }[/math] and [math]\displaystyle{ \hat{\lambda } }[/math] , then the upper and lower confidence bounds are:

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \end{align} }[/math]

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}} }[/math] . If using the unbiased parameters, [math]\displaystyle{ \bar{\beta } }[/math] and [math]\displaystyle{ \bar{\lambda } }[/math] , then the upper and lower confidence bounds are:

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} \end{align} }[/math]

where [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}} }[/math] .

Bounds on Cumulative Failure Intensity

Fisher Matrix Bounds

The cumulative failure intensity, [math]\displaystyle{ {{\lambda }_{c}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{c}}(t) }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln {{{\hat{\lambda }}}_{c}}(t)-\ln {{\lambda }_{c}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{c}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the cumulative failure intensity are then estimated from:

[math]\displaystyle{ CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{c}}(t))}/{{{\hat{\lambda }}}_{c}}(t)}} }[/math]

where:

[math]\displaystyle{ {{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}} }[/math]

and:

[math]\displaystyle{ \begin{align} & Var({{{\hat{\lambda }}}_{c}}(t))= & {{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}\ln t \\ & \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda }= & {{t}^{\hat{\beta }-1}} \end{align} }[/math]

Crow Bounds

The Crow cumulative failure intensity confidence bounds are given as:

[math]\displaystyle{ \begin{align} & C{{(t)}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ & C{{(t)}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \end{align} }[/math]

Bounds on Instantaneous Failure Intensity

Fisher Matrix Bounds

The instantaneous failure intensity, [math]\displaystyle{ {{\lambda }_{i}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{i}}(t) }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln {{{\hat{\lambda }}}_{i}}(t)-\ln {{\lambda }_{i}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{i}}(t)})}\text{ }\tilde{\ }\text{ }N(0,1) }[/math]

The approximate confidence bounds on the instantaneous failure intensity are then estimated from:

[math]\displaystyle{ CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}} }[/math]

where

[math]\displaystyle{ {{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}} }[/math]

[math]\displaystyle{ \begin{align} & Var({{{\hat{\lambda }}}_{i}}(t))= & {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}\ln t \\ & \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }= & \hat{\beta }{{t}^{\hat{\beta }-1}} \end{align} }[/math]

Crow Bounds

The Crow instantaneous failure intensity confidence bounds are given as:

[math]\displaystyle{ \begin{align} & {{\lambda }_{i}}{{(t)}_{L}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ & {{\lambda }_{i}}{{(t)}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \end{align} }[/math]

Bounds on Time Given Cumulative Failure Intensity

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & \frac{-{{\left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\ln \left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}{{{(1-\beta )}^{2}}} \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate:

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}} }[/math]

Step 2: Estimate the number of failures:

[math]\displaystyle{ N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}} }[/math]

Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for [math]\displaystyle{ {{t}_{l}} }[/math] and [math]\displaystyle{ {{t}_{u}} }[/math] in the following equations:

[math]\displaystyle{ \begin{align} & {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\ & {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} \end{align} }[/math]

Bounds on Time Given Cumulative MTBF

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot \,{{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot \text{ }{{m}_{c}})}{{{(1-\beta )}^{2}}} \\ & \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \text{ }\cdot \text{ }{{m}_{c}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ {{\lambda }_{c}}(T)=\tfrac{1}{MTB{{F}_{c}}} }[/math] .

Step 2: Use the equations from 5.2.8.2 to calculate the bounds on time given the cumulative failure intensity.

Bounds on Time Given Instantaneous MTBF

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \hat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & {{\left( \lambda \beta \cdot MTB{{F}_{i}} \right)}^{1/(1-\beta )}}\left[ \frac{1}{{{(1-\beta )}^{2}}}\ln (\lambda \beta \cdot MTB{{F}_{i}})+\frac{1}{\beta (1-\beta )} \right] \\ & \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate the confidence bounds on the instantaneous MTBF as presented in Section 5.5.2.

Step 2: Calculate the bounds on time as follows.

Failure Terminated Data

[math]\displaystyle{ \hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}} }[/math]

So the lower an upper bounds on time are:

[math]\displaystyle{ {{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ {{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}} }[/math]

Time Terminated Data

[math]\displaystyle{ \hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}} }[/math]

So the lower and upper bounds on time are:

[math]\displaystyle{ {{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ {{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{2}}})}^{1/(1-\beta )}} }[/math]

Bounds on Time Given Instantaneous Failure Intensity

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}\left[ -\frac{\ln (\tfrac{{{\lambda }_{i}}(T)}{\lambda \beta })}{{{(\beta -1)}^{2}}}+\frac{1}{\beta (1-\beta )} \right] \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)} }[/math] .

Step 2: Use the equations from 5.2.10.2 to calculate the bounds on time given the instantaneous failure intensity.

Bounds on Cumulative Number of Failures

Fisher Matrix Bounds

The cumulative number of failures, [math]\displaystyle{ N(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln N(t) }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{N}(t)-\ln N(t)}{\sqrt{Var(\ln \hat{N}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

[math]\displaystyle{ N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}} }[/math]

where:

[math]\displaystyle{ \hat{N}(t)=\hat{\lambda }{{t}^{{\hat{\beta }}}} }[/math]

[math]\displaystyle{ \begin{align} & Var(\hat{N}(t))= & {{\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variance1) and:

[math]\displaystyle{ \begin{align} & \frac{\partial \hat{N}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{{\hat{\beta }}}}\ln t \\ & \frac{\partial \hat{N}(t)}{\partial \lambda }= & {{t}^{{\hat{\beta }}}} \end{align} }[/math]

Crow Bounds

The Crow cumulative number of failure confidence bounds are:

[math]\displaystyle{ \begin{align} & {{N}_{L}}(T)= & \frac{T}{{\hat{\beta }}}{{\lambda }_{i}}{{(T)}_{L}} \\ & {{N}_{U}}(T)= & \frac{T}{{\hat{\beta }}}{{\lambda }_{i}}{{(T)}_{U}} \end{align} }[/math]

where [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{L}} }[/math] and [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{U}} }[/math] can be obtained from Eqn. (amsaac14).

Grouped Data

Bounds on Beta

Fisher Matrix Bounds

The parameter [math]\displaystyle{ \beta }[/math] must be positive, thus [math]\displaystyle{ \ln \beta }[/math] is treated as being normally distributed as well.

[math]\displaystyle{ \frac{\ln \hat{\beta }-\ln \beta }{\sqrt{Var(\ln \hat{\beta }})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds are given as:

[math]\displaystyle{ C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} }[/math]

[math]\displaystyle{ \widehat{\beta } }[/math] can be obtained by [math]\displaystyle{ \underset{i=1}{\overset{K}{\mathop{\sum }}}\,{{n}_{i}}\left( \tfrac{T_{i}^{{\hat{\beta }}}\ln {{T}_{i}}-T_{i-1}^{{\hat{\beta }}}\ln \,{{T}_{i-1}}}{T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}}}-\ln {{T}_{k}} \right)=0 }[/math] .

All variance can be calculated using the Fisher Matrix:

[math]\displaystyle{ \left[ \begin{matrix} -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta } \\ -\tfrac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta } & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}} \\ \end{matrix} \right]_{\beta =\widehat{\beta },\lambda =\widehat{\lambda }}^{-1}=\left[ \begin{matrix} Var(\widehat{\lambda }) & Cov(\widehat{\beta },\widehat{\lambda }) \\ Cov(\widehat{\beta },\widehat{\lambda }) & Var(\widehat{\beta }) \\ \end{matrix} \right] }[/math]

[math]\displaystyle{ \Lambda }[/math] is the natural log-likelihood function where ln [math]\displaystyle{ ^{2}T={{\left( \ln T \right)}^{2}} }[/math] and:

[math]\displaystyle{ \Lambda =\underset{i=1}{\overset{k}{\mathop \sum }}\,\left[ {{n}_{i}}\ln (\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })-(\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })-\ln {{n}_{i}}! \right] }[/math]

[math]\displaystyle{ \begin{align} & \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}= & -\frac{n}{{{\lambda }^{2}}} \\ & \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}= & \underset{i=1}{\overset{k}{\mathop \sum }}\,\left[ \begin{matrix} {{n}_{i}}\left( \tfrac{(T_{i}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i}}-T_{i-1}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i-1}})(T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}})-{{\left( T_{i}^{{\hat{\beta }}}\ln {{T}_{i}}-T_{i-1}^{{\hat{\beta }}}\ln {{T}_{i-1}} \right)}^{2}}}{{{(T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}})}^{2}}} \right) \\ -\left( \lambda T_{i}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i}}-\lambda T_{i-1}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i-1}} \right) \\ \end{matrix} \right] \\ & \frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }= & -T_{K}^{\beta }\ln {{T}_{k}} \end{align} }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K }[/math] .

Step 2: Calculate:

[math]\displaystyle{ A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{{{[P{{(i)}^{{\hat{\beta }}}}\ln P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{\widehat{\beta }}}\ln P{{(i-1)}^{{\hat{\beta }}}}]}^{2}}}{[P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{{\hat{\beta }}}}]} }[/math]

Step 3: Calculate [math]\displaystyle{ c=\tfrac{1}{\sqrt{A}} }[/math] and [math]\displaystyle{ S=\tfrac{({{z}_{1-\alpha /2}})\cdot C}{\sqrt{N}} }[/math] . Thus an approximate 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval on [math]\displaystyle{ \widehat{\beta } }[/math] is:

Bounds on Lambda

Fisher Matrix Bounds

The parameter [math]\displaystyle{ \lambda }[/math] must be positive, thus [math]\displaystyle{ \ln \lambda }[/math] is treated as being normally distributed as well. These bounds are based on:

[math]\displaystyle{ \frac{\ln \hat{\lambda }-\ln \lambda }{\sqrt{Var(\ln \hat{\lambda }})}\ \tilde{\ }\ \lt math\gt \hat{\beta }(1\pm S) }[/math]

[math]\displaystyle{ N(0,1) }[/math]

The approximate confidence bounds on [math]\displaystyle{ \lambda }[/math] are given as:

[math]\displaystyle{ C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}} }[/math]

where:

[math]\displaystyle{ \hat{\lambda }=\frac{n}{T_{k}^{{\hat{\beta }}}} }[/math]

The variance calculation is the same as Eqn. (variances).

Crow Bounds

Time Terminated Data
For the 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval, the confidence bounds on [math]\displaystyle{ \lambda }[/math] are:

[math]\displaystyle{ \begin{align} & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot T_{k}^{\beta }} \\ & {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot T_{k}^{\beta }} \end{align} }[/math]

Failure Terminated Data
For the 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval, the confidence bounds on [math]\displaystyle{ \lambda }[/math] are:

[math]\displaystyle{ \begin{align} & {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot T_{k}^{\beta }} \\ & {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot T_{k}^{\beta }} \end{align} }[/math]

Bounds on Growth Rate

Fisher Matrix Bounds

Since the growth rate is equal to [math]\displaystyle{ 1-\beta }[/math] , the confidence bounds are calculated from:

[math]\displaystyle{ \begin{align} & G\operatorname{row}th\text{ }Rat{{e}_{L}}= & 1-{{\beta }_{U}} \\ & G\operatorname{row}th\text{ }Rat{{e}_{U}}= & 1-{{\beta }_{L}} \end{align} }[/math]

For the Fisher Matrix confidence bounds, [math]\displaystyle{ {{\beta }_{L}} }[/math] and [math]\displaystyle{ {{\beta }_{U}} }[/math] are obtained from Eqn. (Gcbb). For the Crow bounds, [math]\displaystyle{ {{\beta }_{L}} }[/math] and [math]\displaystyle{ {{\beta }_{U}} }[/math] are obtained from Eqn. (gcbb).

Bounds on Cumulative MTBF

Fisher Matrix Bounds

The cumulative MTBF, [math]\displaystyle{ {{m}_{c}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{m}_{c}}(t) }[/math] is treated as being normally distributed as well.

[math]\displaystyle{ \frac{\ln {{{\hat{m}}}_{c}}(t)-\ln {{m}_{c}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{c}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the cumulative MTBF are then estimated from:

[math]\displaystyle{ CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}} }[/math]

where:

[math]\displaystyle{ {{\hat{m}}_{c}}(t)=\frac{1}{{\hat{\lambda }}}{{t}^{1-\hat{\beta }}} }[/math]

[math]\displaystyle{ \begin{align} & Var({{{\hat{m}}}_{c}}(t))= & {{\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{c}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda })\, \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{{\hat{\lambda }}}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}}{{t}^{1-\hat{\beta }}} \end{align} }[/math]

Crow Bounds

Calculate the Crow cumulative failure intensity confidence bounds:

[math]\displaystyle{ C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} }[/math]

[math]\displaystyle{ C{{(t)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} }[/math]

Then:

[math]\displaystyle{ \begin{align} & {{[MTB{{F}_{c}}]}_{L}}= & \frac{1}{C{{(t)}_{U}}} \\ & {{[MTB{{F}_{c}}]}_{U}}= & \frac{1}{C{{(t)}_{L}}} \end{align} }[/math]

Bounds on Instantaneous MTBF

Fisher Matrix Bounds

The instantaneous MTBF, [math]\displaystyle{ {{m}_{i}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{m}_{i}}(t) }[/math] is approximately treated as being normally distributed as well.

[math]\displaystyle{ \frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the instantaneous MTBF are then estimated from:

[math]\displaystyle{ CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}} }[/math]

where:

[math]\displaystyle{ {{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}} }[/math]

[math]\displaystyle{ \begin{align} & Var({{{\hat{m}}}_{i}}(t))= & {{\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{m}_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{m}_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{m}_{i}}(t)}{\partial \beta }= & -\frac{1}{\hat{\lambda }{{{\hat{\beta }}}^{2}}}{{t}^{1-\hat{\beta }}}-\frac{1}{\hat{\lambda }\hat{\beta }}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{i}}(t)}{\partial \lambda }= & -\frac{1}{{{{\hat{\lambda }}}^{2}}\hat{\beta }}{{t}^{1-\hat{\beta }}} \end{align} }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K }[/math] .

Step 2: Calculate:

[math]\displaystyle{ A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{{{\left[ P{{(i)}^{{\hat{\beta }}}}\ln P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{\widehat{\beta }}}\ln P{{(i-1)}^{{\hat{\beta }}}} \right]}^{2}}}{\left[ P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{{\hat{\beta }}}} \right]} }[/math]

Step 3: Calculate [math]\displaystyle{ D=\sqrt{\tfrac{1}{A}+1} }[/math] and [math]\displaystyle{ W=\tfrac{({{z}_{1-\alpha /2}})\cdot D}{\sqrt{N}} }[/math] . Thus an approximate 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval on [math]\displaystyle{ {{\hat{m}}_{i}}(t) }[/math] is:

[math]\displaystyle{ MTB{{F}_{i}}={{\widehat{m}}_{i}}(1\pm W) }[/math]

Bounds on Cumulative Failure Intensity

Fisher Matrix Bounds

The cumulative failure intensity, [math]\displaystyle{ {{\lambda }_{c}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{c}}(t) }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln {{{\hat{\lambda }}}_{c}}(t)-\ln {{\lambda }_{c}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{c}}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

The approximate confidence bounds on the cumulative failure intensity are then estimated from:

[math]\displaystyle{ CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{c}}(t))}/{{{\hat{\lambda }}}_{c}}(t)}} }[/math]

where:

[math]\displaystyle{ {{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}} }[/math]

and:

[math]\displaystyle{ \begin{align} & Var({{{\hat{\lambda }}}_{c}}(t))= & {{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{c}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}\ln t \\ & \frac{\partial {{\lambda }_{c}}(t)}{\partial \lambda }= & {{t}^{\hat{\beta }-1}} \end{align} }[/math]

Crow Bounds

The Crow cumulative failure intensity confidence bounds are given as:

[math]\displaystyle{ \begin{align} & C{{(t)}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ & C{{(t)}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \end{align} }[/math]

Bounds on Instantaneous Failure Intensity

Fisher Matrix Bounds

The instantaneous failure intensity, [math]\displaystyle{ {{\lambda }_{i}}(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln {{\lambda }_{i}}(t) }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln {{{\hat{\lambda }}}_{i}}(t)-\ln {{\lambda }_{i}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{i}}(t)})}\tilde{\ }N(0,1) }[/math]

The approximate confidence bounds on the instantaneous failure intensity are then estimated from:

[math]\displaystyle{ CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}} }[/math]

where [math]\displaystyle{ {{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}} }[/math] and:

[math]\displaystyle{ \begin{align} & Var({{{\hat{\lambda }}}_{i}}(t))= & {{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta } \right)\left( \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \begin{align} & \frac{\partial {{\lambda }_{i}}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{\hat{\beta }-1}}+\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}\ln t \\ & \frac{\partial {{\lambda }_{i}}(t)}{\partial \lambda }= & \hat{\beta }{{t}^{\hat{\beta }-1}} \end{align} }[/math]

Crow Bounds

The Crow instantaneous failure intensity confidence bounds are given as:

[math]\displaystyle{ \begin{align} & {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{U}}} \\ & {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1}{{{[MTB{{F}_{i}}]}_{L}}} \end{align} }[/math]

Bounds on Time Given Cumulative MTBF

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot \,{{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot \text{ }{{m}_{c}})}{{{(1-\beta )}^{2}}} \\ & \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ {{\lambda }_{c}}(T)=\tfrac{1}{MTB{{F}_{c}}} }[/math] .

Step 2: Use equations in 5.4.10.1 to calculate the bounds on time given the cumulative failure intensity.

Bounds on Time Given Instantaneous MTBF

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \hat{T}={{(\lambda \beta \cdot {{m}_{i}}(T))}^{1/(1-\beta )}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & {{\left( \lambda \beta \cdot \text{ }{{m}_{i}}(T) \right)}^{1/(1-\beta )}}\left[ \frac{1}{{{(1-\beta )}^{2}}}\ln (\lambda \beta \cdot {{m}_{i}}(T))+\frac{1}{\beta (1-\beta )} \right] \\ & \frac{\partial T}{\partial \lambda }= & \frac{{{(\lambda \beta \cdot \text{ }{{m}_{i}}(T))}^{1/(1-\beta )}}}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate the confidence bounds on the instantaneous MTBF:

[math]\displaystyle{ MTB{{F}_{i}}={{\widehat{m}}_{i}}(1\pm W) }[/math]

Step 2: Use equations in 5.4.5.2 to calculate the time given the instantaneous MTBF.

Bounds on Time Given Cumulative Failure Intensity

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & \frac{-{{\left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\ln \left( \tfrac{{{\lambda }_{c}}(T)}{\lambda } \right)}{{{(1-\beta )}^{2}}} \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate:

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}} }[/math]

Step 2: Estimate the number of failures:

[math]\displaystyle{ N(\hat{T})=\hat{\lambda }{{\hat{T}}^{{\hat{\beta }}}} }[/math]

Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for [math]\displaystyle{ {{t}_{l}} }[/math] and [math]\displaystyle{ {{t}_{u}} }[/math] in the following equations:

[math]\displaystyle{ \begin{align} & {{t}_{l}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot {{\lambda }_{c}}(T)} \\ & {{t}_{u}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot {{\lambda }_{c}}(T)} \end{align} }[/math]

Bounds on Time Given Instantaneous Failure Intensity

Fisher Matrix Bounds

The time, [math]\displaystyle{ T }[/math] , must be positive, thus [math]\displaystyle{ \ln T }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1) }[/math]

Confidence bounds on the time are given by:

[math]\displaystyle{ CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}} }[/math]

where:

[math]\displaystyle{ \begin{align} & Var(\hat{T})= & {{\left( \frac{\partial T}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial T}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial T}{\partial \beta } \right)\left( \frac{\partial T}{\partial \lambda } \right)cov(\hat{\beta },\,\,\,\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}} }[/math]

[math]\displaystyle{ \begin{align} & \frac{\partial T}{\partial \beta }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}\left[ -\frac{\ln (\tfrac{{{\lambda }_{i}}(T)}{\lambda \beta })}{{{(\beta -1)}^{2}}}+\frac{1}{\beta (1-\beta )} \right] \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align} }[/math]

Crow Bounds

Step 1: Calculate [math]\displaystyle{ MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)} }[/math] .

Step 2: Follow the same process as in 5.4.9.2 to calculate the bounds on time given the instantaneous failure intensity.

Bounds on Cumulative Number of Failures

Fisher Matrix Bounds

The cumulative number of failures, [math]\displaystyle{ N(t) }[/math] , must be positive, thus [math]\displaystyle{ \ln N(t) }[/math] is treated as being normally distributed.

[math]\displaystyle{ \frac{\ln \hat{N}(t)-\ln N(t)}{\sqrt{Var(\ln \hat{N}(t)})}\ \tilde{\ }\ N(0,1) }[/math]

[math]\displaystyle{ N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}} }[/math]

where:

[math]\displaystyle{ \hat{N}(t)=\hat{\lambda }{{t}^{{\hat{\beta }}}} }[/math]

[math]\displaystyle{ \begin{align} & Var(\hat{N}(t))= & {{\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)}^{2}}Var(\hat{\beta })+{{\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)}^{2}}Var(\hat{\lambda }) \\ & & +2\left( \frac{\partial \hat{N}(t)}{\partial \beta } \right)\left( \frac{\partial \hat{N}(t)}{\partial \lambda } \right)cov(\hat{\beta },\hat{\lambda }) \end{align} }[/math]

The variance calculation is the same as Eqn. (variances) and:

[math]\displaystyle{ \begin{align} & \frac{\partial \hat{N}(t)}{\partial \beta }= & \hat{\lambda }{{t}^{{\hat{\beta }}}}\ln t \\ & \frac{\partial \hat{N}(t)}{\partial \lambda }= & {{t}^{{\hat{\beta }}}} \end{align} }[/math]

Crow Bounds

The Crow confidence bounds on cumulative number of failures are:

[math]\displaystyle{ \begin{align} & {{N}_{L}}(T)= & \frac{T}{{\hat{\beta }}}{{\lambda }_{i}}{{(T)}_{L}} \\ & {{N}_{U}}(T)= & \frac{T}{{\hat{\beta }}}{{\lambda }_{i}}{{(T)}_{U}} \end{align} }[/math]

where [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{L}} }[/math] and [math]\displaystyle{ {{\lambda }_{i}}{{(T)}_{U}} }[/math] can be obtained from Eqn. (dsaf).