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.

Fisher Matrix Bounds
The parameter $$\beta $$  must be positive, thus  $$\ln \beta $$  is treated as being normally distributed as well.


 * $$\frac{\ln \hat{\beta }-\ln \beta }{\sqrt{Var(\ln \hat{\beta }})}\ \tilde{\ }\ N(0,1)$$

The approximate confidence bounds are given as:


 * $$C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}}$$

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


 * $$\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]$$
 * $$\Lambda $$ is the natural log-likelihood function:


 * $$\Lambda =N\ln \lambda +N\ln \beta -\lambda {{T}^{\beta }}+(\beta -1)\underset{i=1}{\overset{N}{\mathop \sum }}\,\ln {{T}_{i}}$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{N}$$


 * and:


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}=-\frac{N}-\lambda {{T}^{\beta }}{{(\ln T)}^{2}}$$


 * also:


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }=-{{T}^{\beta }}\ln T$$

Crow Bounds
Time Terminated Data

For the 2-sided $$(1-\alpha )$$ 100-percent confidence interval on  $$\beta $$, calculate:


 * $$\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}$$

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


 * $$\begin{align}

& {{\beta }_{L}}= & {{D}_{L}}\cdot \hat{\beta } \\ & {{\beta }_{U}}= & {{D}_{U}}\cdot \hat{\beta } \end{align}$$

Failure Terminated Data For the 2-sided $$(1-\alpha )$$ 100-percent confidence interval on  $$\beta $$, calculate:


 * $$\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}$$

Thus the confidence bounds on $$\beta $$  are:


 * $$\begin{align}

& {{\beta }_{L}}= & {{D}_{L}}\cdot \hat{\beta } \\ & {{\beta }_{U}}= & {{D}_{U}}\cdot \hat{\beta } \end{align}$$

Fisher Matrix Bounds
The parameter $$\lambda $$  must be positive, thus  $$\ln \lambda $$  is treated as being normally distributed as well. These bounds are based on:
 * $$\frac{\ln \hat{\lambda }-\ln \lambda }{\sqrt{Var(\ln \hat{\lambda }})}\ \tilde{\ }\ N(0,1)$$

The approximate confidence bounds on $$\lambda $$  are given as:
 * $$C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}}$$


 * where:


 * $$\hat{\lambda }=\frac{n}$$

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

Crow Bounds
Time Terminated Data For the 2-sided $$(1-\alpha )$$ 100-percent confidence interval, the confidence bounds on  $$\lambda $$  are:


 * $$\begin{align}

& {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2{{T}^}} \\ & {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2{{T}^}} \end{align}$$

The fractiles can be found in the tables of the $${{\chi }^{2}}$$  distribution. Failure Terminated Data For the 2-sided $$(1-\alpha )$$ 100-percent confidence interval, the confidence bounds on  $$\lambda $$  are:


 * $$\begin{align}

& {{\lambda }_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2{{T}^}} \\ & {{\lambda }_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2{{T}^}} \end{align}$$

Bounds on Growth Rate
Since the growth rate is equal to $$1-\beta $$, the confidence bounds for both the Fisher Matrix and Crow methods are:
 * $$G\text{row}th Rate_L=1-\beta_U$$
 * $$G\text{row}th Rate_U=1-\beta_L$$

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

Fisher Matrix Bounds
The cumulative MTBF, $${{m}_{c}}(t)$$, must be positive, thus  $$\ln {{m}_{c}}(t)$$  is treated as being normally distributed as well.


 * $$\frac{\ln {{{\hat{m}}}_{c}}(t)-\ln {{m}_{c}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{c}}(t)})}\ \tilde{\ }\ N(0,1)$$

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


 * $$CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}}$$


 * where:


 * $${{\hat{m}}_{c}}(t)=\frac{1}{{t}^{1-\hat{\beta }}}$$


 * $$\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}$$ The variance calculation is the same as Eqn. (variance1) and:


 * $$\begin{align}

& \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{t}^{1-\hat{\beta }}} \end{align}$$

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


 * $$C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}$$


 * $$C{{(t)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}$$


 * Then:


 * $$\begin{align}

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

Fisher Matrix Bounds
The instantaneous MTBF, $${{m}_{i}}(t)$$, must be positive, thus  $$\ln {{m}_{i}}(t)$$  is treated as being normally distributed as well.


 * $$\frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1)$$

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


 * $$CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}}$$


 * where:


 * $${{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}$$


 * $$\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}$$

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


 * $$\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}$$

Crow Bounds
Failure Terminated Data Consider the following equation:


 * $$G(\mu |n)=\mathop{}_{0}^{\infty }\frac{(n-2)!}\underset{i=0}{\overset{n-1}{\mathop \sum }}\,\frac{1}{i!}{{\left( \frac{\mu }{x} \right)}^{i}}\exp (-\frac{\mu }{x})\,dx$$

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


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{p}_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{p}_{2}} \end{align}$$

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


 * $$\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}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$. Time Terminated Data Consider the following equation where $${{I}_{1}}(.)$$  is the modified Bessel function of order one:


 * $$H(x|k)=\underset{j=1}{\overset{k}{\mathop \sum }}\,\frac{{{2}^{2j-1}}(j-1)!j!{{I}_{1}}(x)}$$

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


 * $$\begin{align}

& {{[MTB{{F}_{i}}]}_{L}}= & MTB{{F}_{i}}\cdot {{\Pi }_{1}} \\ & {{[MTB{{F}_{i}}]}_{U}}= & MTB{{F}_{i}}\cdot {{\Pi }_{2}} \end{align}$$

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


 * $$\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}$$

where $$MTB{{F}_{i}}=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}$$.

Fisher Matrix Bounds
The cumulative failure intensity, $${{\lambda }_{c}}(t)$$, must be positive, thus  $$\ln {{\lambda }_{c}}(t)$$  is treated as being normally distributed.


 * $$\frac{\ln {{{\hat{\lambda }}}_{c}}(t)-\ln {{\lambda }_{c}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{c}}(t)})}\ \tilde{\ }\ N(0,1)$$

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


 * $$CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{c}}(t))}/{{{\hat{\lambda }}}_{c}}(t)}}$$


 * where:


 * $${{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}}$$


 * and:


 * $$\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}$$

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


 * $$\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}$$

Crow Bounds
The Crow cumulative failure intensity confidence bounds are given as:


 * $$\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}$$

Fisher Matrix Bounds
The instantaneous failure intensity, $${{\lambda }_{i}}(t)$$, must be positive, thus  $$\ln {{\lambda }_{i}}(t)$$  is treated as being normally distributed.


 * $$\frac{\ln {{{\hat{\lambda }}}_{i}}(t)-\ln {{\lambda }_{i}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{i}}(t)})}\text{ }\tilde{\ }\text{ }N(0,1)$$

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


 * $$CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}}$$


 * where


 * $${{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}$$


 * $$\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}$$

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


 * $$\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}$$

Crow Bounds
The Crow instantaneous failure intensity confidence bounds are given as:


 * $$\begin{align}

& {{\lambda }_{i}}{{(t)}_{L}}= & \frac{1} \\ & {{\lambda }_{i}}{{(t)}_{U}}= & \frac{1} \end{align}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:
 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$ The variance calculation is the same as Eqn. (variance1) and:
 * $$\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)} \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate:


 * $$\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}$$


 * Step 2: Estimate the number of failures:


 * $$N(\hat{T})=\hat{\lambda }{{\hat{T}}^}$$


 * Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for $${{t}_{l}}$$  and  $${{t}_{u}}$$  in the following equations:


 * $$\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}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.
 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:
 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$
 * where:
 * $$\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}$$ The variance calculation is the same as Eqn. (variance1) and:
 * $$\hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}$$


 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot \,{{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot \text{ }{{m}_{c}})} \\ & \frac{\partial T}{\partial \lambda }= & \frac{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate $${{\lambda }_{c}}(T)=\tfrac{1}{MTB{{F}_{c}}}$$.
 * Step 2: Use the equations from 5.2.8.2 to calculate the bounds on time given the cumulative failure intensity.

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$

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


 * $$\hat{T}={{(\lambda \beta \cdot MTB{{F}_{i}})}^{1/(1-\beta )}}$$


 * $$\begin{align}

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

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

 * $$\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{c})}^{1/(1-\beta )}}$$

So the lower an upper bounds on time are:


 * $${{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{1}}})}^{1/(1-\beta )}}$$


 * $${{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{c}_{2}}})}^{1/(1-\beta )}}$$

Time Terminated Data


 * $$\hat{T}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{\Pi })}^{1/(1-\beta )}}$$

So the lower and upper bounds on time are:


 * $${{\hat{T}}_{L}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{1}}})}^{1/(1-\beta )}}$$


 * $${{\hat{T}}_{U}}={{(\frac{\lambda \beta \cdot MTB{{F}_{i}}}{{{\Pi }_{2}}})}^{1/(1-\beta )}}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$

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


 * $$\hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}$$


 * $$\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 })}+\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}$$

Crow Bounds

 * Step 1: Calculate $$MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)}$$.
 * Step 2: Use the equations from 5.2.10.2 to calculate the bounds on time given the instantaneous failure intensity.

Fisher Matrix Bounds
The cumulative number of failures, $$N(t)$$, must be positive, thus  $$\ln N(t)$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{N}(t)-\ln N(t)}{\sqrt{Var(\ln \hat{N}(t)})}\ \tilde{\ }\ N(0,1)$$


 * $$N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}}$$


 * where:


 * $$\hat{N}(t)=\hat{\lambda }{{t}^}$$


 * $$\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}$$

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


 * $$\begin{align}

& \frac{\partial \hat{N}(t)}{\partial \beta }= & \hat{\lambda }{{t}^}\ln t \\ & \frac{\partial \hat{N}(t)}{\partial \lambda }= & {{t}^} \end{align}$$

Crow Bounds
The Crow cumulative number of failure confidence bounds are:


 * $$\begin{align}

& {{N}_{L}}(T)= & \frac{T}{{\lambda }_{i}}{{(T)}_{L}} \\ & {{N}_{U}}(T)= & \frac{T}{{\lambda }_{i}}{{(T)}_{U}} \end{align}$$

where $${{\lambda }_{i}}{{(T)}_{L}}$$  and  $${{\lambda }_{i}}{{(T)}_{U}}$$  can be obtained from Eqn. (amsaac14).

Fisher Matrix Bounds
The parameter $$\beta $$  must be positive, thus  $$\ln \beta $$  is treated as being normally distributed as well.


 * $$\frac{\ln \hat{\beta }-\ln \beta }{\sqrt{Var(\ln \hat{\beta }})}\ \tilde{\ }\ N(0,1)$$

The approximate confidence bounds are given as:


 * $$C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}}$$


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

All variance can be calculated using the Fisher Matrix:


 * $$\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]$$

$$\Lambda $$ is the natural log-likelihood function where ln $$^{2}T={{\left( \ln T \right)}^{2}}$$  and:


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


 * $$\begin{align}

& \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}= & -\frac{n} \\ & \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}= & \underset{i=1}{\overset{k}{\mathop \sum }}\,\left[ \begin{matrix} {{n}_{i}}\left( \tfrac{(T_{i}^{{\ln }^{2}}{{T}_{i}}-T_{i-1}^{{\ln }^{2}}{{T}_{i-1}})(T_{i}^-T_{i-1}^)-{{\left( T_{i}^\ln {{T}_{i}}-T_{i-1}^\ln {{T}_{i-1}} \right)}^{2}}} \right) \\ -\left( \lambda T_{i}^{{\ln }^{2}}{{T}_{i}}-\lambda T_{i-1}^{{\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}$$

Crow Bounds

 * Step 1: Calculate $$P(i)=\tfrac,\,\,i=1,2,\ldots ,K$$.
 * Step 2: Calculate:


 * $$A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{[P{{(i)}^}-P{{(i-1)}^}]}$$


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

Fisher Matrix Bounds
The parameter $$\lambda $$  must be positive, thus  $$\ln \lambda $$  is treated as being normally distributed as well. These bounds are based on:


 * \frac{\ln \hat{\lambda }-\ln \lambda }{\sqrt{Var(\ln \hat{\lambda }})}\ \tilde{\ }\

$$\hat{\beta }(1\pm S)$$
 * $$N(0,1)$$

The approximate confidence bounds on $$\lambda $$  are given as:


 * $$C{{B}_{\lambda }}=\hat{\lambda }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\lambda })}/\hat{\lambda }}}$$


 * where:


 * $$\hat{\lambda }=\frac{n}{T_{k}^}$$

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

Crow Bounds
Time Terminated Data For the 2-sided $$(1-\alpha )$$ 100-percent confidence interval, the confidence bounds on  $$\lambda $$  are:


 * $$\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}$$

Failure Terminated Data For the 2-sided $$(1-\alpha )$$ 100-percent confidence interval, the confidence bounds on  $$\lambda $$  are:


 * $$\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}$$

Fisher Matrix Bounds
Since the growth rate is equal to $$1-\beta $$, the confidence bounds are calculated from:


 * $$\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}$$

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

Fisher Matrix Bounds
The cumulative MTBF, $${{m}_{c}}(t)$$, must be positive, thus  $$\ln {{m}_{c}}(t)$$  is treated as being normally distributed as well.


 * $$\frac{\ln {{{\hat{m}}}_{c}}(t)-\ln {{m}_{c}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{c}}(t)})}\ \tilde{\ }\ N(0,1)$$

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


 * $$CB={{\hat{m}}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{c}}(t))}/{{{\hat{m}}}_{c}}(t)}}$$


 * where:


 * $${{\hat{m}}_{c}}(t)=\frac{1}{{t}^{1-\hat{\beta }}}$$


 * $$\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}$$

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


 * $$\begin{align}

& \frac{\partial {{m}_{c}}(t)}{\partial \beta }= & -\frac{1}{{t}^{1-\hat{\beta }}}\ln t \\ & \frac{\partial {{m}_{c}}(t)}{\partial \lambda }= & -\frac{1}{{t}^{1-\hat{\beta }}} \end{align}$$

Crow Bounds
Calculate the Crow cumulative failure intensity confidence bounds:


 * $$C{{(t)}_{L}}=\frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t}$$


 * $$C{{(t)}_{U}}=\frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t}$$


 * Then:


 * $$\begin{align}

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

Fisher Matrix Bounds
The instantaneous MTBF, $${{m}_{i}}(t)$$, must be positive, thus  $$\ln {{m}_{i}}(t)$$  is approximately treated as being normally distributed as well.


 * $$\frac{\ln {{{\hat{m}}}_{i}}(t)-\ln {{m}_{i}}(t)}{\sqrt{Var(\ln {{{\hat{m}}}_{i}}(t)})}\ \tilde{\ }\ N(0,1)$$

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


 * $$CB={{\hat{m}}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{m}}}_{i}}(t))}/{{{\hat{m}}}_{i}}(t)}}$$


 * where:


 * $${{\hat{m}}_{i}}(t)=\frac{1}{\lambda \beta {{t}^{\beta -1}}}$$


 * $$\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}$$

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


 * $$\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}$$

Crow Bounds

 * Step 1: Calculate $$P(i)=\tfrac,\,\,i=1,2,\ldots ,K$$.
 * Step 2: Calculate:


 * $$A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{\left[ P{{(i)}^}-P{{(i-1)}^} \right]}$$


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


 * $$MTB{{F}_{i}}={{\widehat{m}}_{i}}(1\pm W)$$

Fisher Matrix Bounds
The cumulative failure intensity, $${{\lambda }_{c}}(t)$$, must be positive, thus  $$\ln {{\lambda }_{c}}(t)$$  is treated as being normally distributed.


 * $$\frac{\ln {{{\hat{\lambda }}}_{c}}(t)-\ln {{\lambda }_{c}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{c}}(t)})}\ \tilde{\ }\ N(0,1)$$

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


 * $$CB={{\hat{\lambda }}_{c}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{c}}(t))}/{{{\hat{\lambda }}}_{c}}(t)}}$$


 * where:


 * $${{\hat{\lambda }}_{c}}(t)=\hat{\lambda }{{t}^{\hat{\beta }-1}}$$


 * and:


 * $$\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}$$

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


 * $$\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}$$

Crow Bounds
The Crow cumulative failure intensity confidence bounds are given as:


 * $$\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}$$

Fisher Matrix Bounds
The instantaneous failure intensity, $${{\lambda }_{i}}(t)$$, must be positive, thus  $$\ln {{\lambda }_{i}}(t)$$  is treated as being normally distributed.


 * $$\frac{\ln {{{\hat{\lambda }}}_{i}}(t)-\ln {{\lambda }_{i}}(t)}{\sqrt{Var(\ln {{{\hat{\lambda }}}_{i}}(t)})}\tilde{\ }N(0,1)$$

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


 * $$CB={{\hat{\lambda }}_{i}}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var({{{\hat{\lambda }}}_{i}}(t))}/{{{\hat{\lambda }}}_{i}}(t)}}$$

where $${{\lambda }_{i}}(t)=\lambda \beta {{t}^{\beta -1}}$$  and:


 * $$\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}$$

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


 * $$\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}$$

Crow Bounds
The Crow instantaneous failure intensity confidence bounds are given as:


 * $$\begin{align}

& {{[{{\lambda }_{i}}(t)]}_{L}}= & \frac{1} \\ & {{[{{\lambda }_{i}}(t)]}_{U}}= & \frac{1} \end{align}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$

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


 * $$\hat{T}={{(\lambda \cdot {{m}_{c}})}^{1/(1-\beta )}}$$


 * $$\begin{align}

& \frac{\partial T}{\partial \beta }= & \frac{{{(\lambda \cdot \,{{m}_{c}})}^{1/(1-\beta )}}\ln (\lambda \cdot \text{ }{{m}_{c}})} \\ & \frac{\partial T}{\partial \lambda }= & \frac{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate $${{\lambda }_{c}}(T)=\tfrac{1}{MTB{{F}_{c}}}$$.
 * Step 2: Use equations in 5.4.10.1 to calculate the bounds on time given the cumulative failure intensity.

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$

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


 * $$\hat{T}={{(\lambda \beta \cdot {{m}_{i}}(T))}^{1/(1-\beta )}}$$


 * $$\begin{align}

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

Crow Bounds

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


 * $$MTB{{F}_{i}}={{\widehat{m}}_{i}}(1\pm W)$$


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

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$

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


 * $$\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)} \\ & \frac{\partial T}{\partial \lambda }= & {{\left( \frac{{{\lambda }_{c}}(T)}{\lambda } \right)}^{1/(\beta -1)}}\frac{1}{\lambda (1-\beta )} \end{align}$$

Crow Bounds

 * Step 1: Calculate:


 * $$\hat{T}={{\left( \frac{{{\lambda }_{c}}(T)}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\beta -1}}}$$


 * Step 2: Estimate the number of failures:


 * $$N(\hat{T})=\hat{\lambda }{{\hat{T}}^}$$


 * Step 3: Obtain the confidence bounds on time given the cumulative failure intensity by solving for $${{t}_{l}}$$  and  $${{t}_{u}}$$  in the following equations:


 * $$\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}$$

Fisher Matrix Bounds
The time, $$T$$, must be positive, thus  $$\ln T$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{T}-\ln T}{\sqrt{Var(\ln \hat{T}})}\ \tilde{\ }\ N(0,1)$$

Confidence bounds on the time are given by:


 * $$CB=\hat{T}{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{T})}/\hat{T}}}$$


 * where:


 * $$\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}$$

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


 * $$\hat{T}={{\left( \frac{{{\lambda }_{i}}(T)}{\lambda \beta } \right)}^{1/(\beta -1)}}$$


 * $$\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 })}+\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}$$

Crow Bounds

 * Step 1: Calculate $$MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)}$$.
 * Step 2: Follow the same process as in 5.4.9.2 to calculate the bounds on time given the instantaneous failure intensity.

Fisher Matrix Bounds
The cumulative number of failures, $$N(t)$$, must be positive, thus  $$\ln N(t)$$  is treated as being normally distributed.


 * $$\frac{\ln \hat{N}(t)-\ln N(t)}{\sqrt{Var(\ln \hat{N}(t)})}\ \tilde{\ }\ N(0,1)$$


 * $$N(t)=\hat{N}(t){{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{N}(t))}/\hat{N}(t)}}$$


 * where:


 * $$\hat{N}(t)=\hat{\lambda }{{t}^}$$


 * $$\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}$$

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


 * $$\begin{align}

& \frac{\partial \hat{N}(t)}{\partial \beta }= & \hat{\lambda }{{t}^}\ln t \\ & \frac{\partial \hat{N}(t)}{\partial \lambda }= & {{t}^} \end{align}$$

Crow Bounds
The Crow confidence bounds on cumulative number of failures are:


 * $$\begin{align}

& {{N}_{L}}(T)= & \frac{T}{{\lambda }_{i}}{{(T)}_{L}} \\ & {{N}_{U}}(T)= & \frac{T}{{\lambda }_{i}}{{(T)}_{U}} \end{align}$$

where $${{\lambda }_{i}}{{(T)}_{L}}$$  and  $${{\lambda }_{i}}{{(T)}_{U}}$$  can be obtained from Eqn. (dsaf).