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.

Note regarding the Crow Bounds calculations: The equations that involve the use of the Chi-Squared distribution assume left-tail probability.

Individual (Non-Grouped) Data
This section presents the confidence bounds for the Crow-AMSAA model under developmental testing when the failure times are known. The confidence bounds for when the failure times are not known are presented in the Grouped Data section.

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 =\hat{\beta },\lambda =\hat{\lambda }}^{-1}=\left[ \begin{matrix} Var(\hat{\lambda }) & Cov(\hat{\beta },\hat{\lambda }) \\ Cov(\hat{\beta },\hat{\lambda }) & Var(\hat{\beta }) \\ \end{matrix} \right]\,\!$$

$$\Lambda \,\!$$ is the natural log-likelihood function:


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

And:


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


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


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

Crow Bounds
Failure Terminated Data

For the 2-sided $$(1-\alpha )\,\!$$ 100% 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}\,\!$$

Time Terminated Data

For the 2-sided $$(1-\alpha )\,\!$$ 100% 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 confidence bounds on $$\beta \,\!$$ are:


 * $$\begin{align}

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

Growth Rate
Since the growth rate, $$\alpha \,\!$$, is equal to $$1-\beta \,\!$$, the confidence bounds for both the Fisher Matrix and Crow methods are:


 * $$\alpha_L=1-\beta_U\,\!$$
 * $$\alpha_U=1-\beta_L\,\!$$

$${{\beta }_{L}}\,\!$$ and $${{\beta }_{U}}\,\!$$ are obtained using the methods described above in the confidence bounds on Beta section.

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 given above in the confidence bounds on Beta section.

Crow Bounds
Failure Terminated Data

For the 2-sided $$(1-\alpha )\,\!$$ 100% 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}\,\!$$

where:
 * $$N\,\!$$ = total number of failures.
 * $$T\,\!$$ = termination time.

Time Terminated Data

For the 2-sided $$(1-\alpha )\,\!$$ 100% 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}\,\!$$

where:
 * $$N\,\!$$ = total number of failures.
 * $$T\,\!$$ = termination time.

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 given above in the confidence bounds on Beta section. 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}{IFI}{{(t)}_{L}} \\ {{N}_{U}}(t)= & \frac{t}{IFI}{{(t)}_{U}} \end{align}\,\!$$

where $$IFI{{(t)}_{L}}\,\!$$ and $$IFI{{(t)}_{U}}\,\!$$ are calculated using the process for the confidence bounds on instantaneous failure intensity.

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 given above in the confidence bounds on Beta section. 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 bounds on the cumulative failure intensity $$(CFI)\,\!$$ are calculated by first estimating the number of failiures, $$N\,\!$$, such that:


 * $$N=\hat{\lambda }{{t}^}\,\!$$

Failure Terminated
 * $$\begin{align}

CFI{{(t)}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ \end{align}\,\!$$


 * $$\begin{align}

CFI{{(t)}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \end{align}\,\!$$

Time Terminated


 * $$\begin{align}

CFI{{(t)}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ CFI{{(t)}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \end{align}\,\!$$

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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on the cumulative MTBF $$(CMTBF)\,\!$$ are given by:


 * $$\begin{align}

& CMTBF_{L}=\frac{1}{CFI_{U}} \\ & CMTBF_{U}=\frac{1}{CFI_{L}} \end{align}\,\!$$

where $$CFI_L\,\!$$ and $$CFI_U\,\!$$ are calculated using the process for the confidence bounds on cumulative failure intensity.

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 given above in the confidence bounds on Beta section. 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

For failure terminated data and the 2-sided confidence bounds on instantaneous MTBF $$(IMTBF)\,\!$$, 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 $$G\left( \left. \frac{c} \right|n \right)=\frac{\alpha }{2}$$ and $$G\left( \left. \frac{c} \right|n \right)=1-\frac{\alpha }{2}$$ for the lower and upper bounds, respectively.

If using the biased parameters, $$\hat{\beta }\,\!$$ and $$\hat{\lambda }\,\!$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

{{IMTBF}_{L}}= & IMTBF\cdot {{p}_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot {{p}_{2}} \end{align}\,\!$$

where $$IMTBF=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}\,\!$$.

If using the unbiased parameters, $$\bar{\beta }\,\!$$ and $$\bar{\lambda }\,\!$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

{{IMTBF}_{L}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{1}} \\ {{IMTBF}_{U}}= & MTB{{F}_{i}}\cdot \left( \frac{N-2}{N} \right)\cdot {{p}_{2}} \end{align}\,\!$$

where $$IMTBF=\tfrac{1}{\bar{\lambda }\bar{\beta }{{t}^{\bar{\beta }-1}}}\,\!$$.

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}

{{IMTBF}_{L}}= & IMTBF\cdot {{\Pi }_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot {{\Pi }_{2}} \end{align}\,\!$$

where $$IMTBF=\tfrac{1}{\hat{\lambda }\hat{\beta }{{t}^{\hat{\beta }-1}}}\,\!$$.

If using the unbiased parameters, $$\bar{\beta }\,\!$$ and $$\bar{\lambda }\,\!$$, then the upper and lower confidence bounds are:


 * $$\begin{align}

{{IMTBF}_{L}}= & IMTBF\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{1}} \\ {{IMTBF}_{U}}= & IMTBF\cdot \left( \frac{N-1}{N} \right)\cdot {{\Pi }_{2}} \end{align}\,\!$$

where $$IMTBF=\tfrac{1}{\bar{\lambda }\bar{\beta }{{t}^{\bar{\beta }-1}}}\,\!$$.

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 given above in the confidence bounds on Beta section. 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 2-sided confidence bounds on the instantaneous failure intensity $$(IFI)\,\!$$ are given by:


 * $$\begin{align}

{IFI_{L}}= & \frac{1}{{IMTBF}_{U}} \\ {IFI_{U}}= & \frac{1}{{IMTBF}_{L}} \end{align}\,\!$$

where $$IMTB{{F}_{L}}\,\!$$ and $$IMTB{{F}_{U}}\,\!$$ are calculated using the process presented for the confidence bounds on 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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on time given cumulative failure intensity $$(CFI)\,\!$$ are given by:


 * $$\hat{t}={{\left( \frac{CFI}{{\hat{\lambda }}} \right)}^{\tfrac{1}{\hat{\beta }-1}}}\,\!$$

Then estimate, the number of failures, $$N\,\!$$, such that:


 * $$N=\hat{\lambda }{{\hat{t}}^}\,\!$$

The lower and upper confidence bounds on time are then estimated using:


 * $$\begin{align}

{{t}_{L}}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot CFI} \\ {{t}_{U}}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot CFI} \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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on time given cumulative MTBF $$(CMTBF)\,\!$$ are estimated using the process for the confidence bounds on time given cumulative failure intensity where $$CFI=\frac{1}{CMTBF}\,\!$$.

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 given above in the confidence bounds on Beta section. 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
Failure Terminated Data

If the unbiased value $$\bar{\beta }\,\!$$ is used then:

$$IMTBF=IMTBF\cdot \frac{N-1}{N}\,\!$$

where:
 * $$IMTBF\,\!$$ = instantaneous MTBF.
 * $$N\,\!$$ = total number of failures.

Calculate the constants $$p_1\,\!$$ and $$p_2\,\!$$ using procedures described for the confidence bounds on instantaneous MTBF. The lower and upper confidence bounds on time are then given by:

So the lower an upper bounds on time are:


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


 * $${{\hat{T}}_{U}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{p}_{2}}} \right)}^{\tfrac{1}{1-\beta }}}$$

Time Terminated Data

If the unbiased value $$\bar{\beta }\,\!$$ is used then:

$$IMTBF=IMTBF\cdot \frac{N-2}{N}\,\!$$

where:
 * $$IMTBF\,\!$$ = instantaneous MTBF.
 * $$N\,\!$$ = total number of failures.

Calculate the constants $${{\Pi }_{1}}\,\!$$ and $${{\Pi }_{2}}\,\!$$ using procedures described for the confidence bounds on instantaneous MTBF. The lower and upper confidence bounds on time are then given by:


 * $${{\hat{T}}_{L}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{\Pi }_{1}}} \right)}^{\tfrac{1}{1-\beta }}}\,\!$$


 * $${{\hat{T}}_{U}}={{\left( \frac{\lambda \beta \cdot IMTBF}{{{\Pi }_{2}}} \right)}^{\tfrac{1}{1-\beta }}}\,\!$$

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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on time given instantaneous failure intensity $$(IFI)\,\!$$ are estimated using the process for the confidence bounds on time given instantaneous MTBF where $$IMTBF=\frac{1}{IFI}\,\!$$.

Grouped Data
This section presents the confidence bounds for the Crow-AMSAA model when using Grouped data.

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:


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


 * $$\hat{\beta }\,\!$$ 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 =\hat{\beta },\lambda =\hat{\lambda }}^{-1}=\left[ \begin{matrix} Var(\hat{\lambda }) & Cov(\hat{\beta },\hat{\lambda }) \\ Cov(\hat{\beta },\hat{\lambda }) & Var(\hat{\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
The 2-sided confidence bounds on $$\hat{\beta }\,\!$$ are given by first calculating:


 * $$P\left( i \right)=\frac;\text{ }i=1,2,...,K$$

where:


 * $$T_i\,\!$$ = interval end time for the $${{i}^{th}}\,\!$$ interval.
 * $$K\,\!$$ = number of intervals.
 * $$T_K\,\!$$ = end time for the last interval.

Next:


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

And:


 * $$c=\tfrac{1}{\sqrt{A}}\,\!$$

Then:


 * $$S=\tfrac{({{z}_{1-\alpha /2}})\cdot C}{\sqrt{N}}\,\!$$

where:


 * $${{z}_{1-\tfrac{\alpha }{2}}}\,\!$$ = inverse standard normal.
 * $$N\,\!$$ = number of failures.

The 2-sided confidence bounds on $$\beta\,\!$$ are then $$\hat{\beta }\left( 1\pm S \right)\,\!$$.

Growth Rate (Grouped)
Since the growth rate, $$\alpha \,\!$$, is equal to $$1-\beta \,\!$$, the confidence bounds for both the Fisher Matrix and Crow methods are:


 * $$\alpha_L=1-\beta_U\,\!$$
 * $$\alpha_U=1-\beta_L\,\!$$

$${{\beta }_{L}}\,\!$$ and $${{\beta }_{U}}\,\!$$ are obtained using the methods described above in the confidence bounds on Beta section.

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:


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

The variance calculation is the same as given above in the confidence bounds on Beta section.

Crow Bounds
Failure Terminated Data

For failure terminated data, the 2-sided $$(1-\alpha )\,\!$$ 100% 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}\,\!$$

where:
 * $$N\,\!$$ = total number of failures.
 * $$T_K\,\!$$ = end time of last interval.

Time Terminated Data

For time terminated data, the 2-sided $$(1-\alpha )\,\!$$ 100% 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}\,\!$$

where:
 * $$N\,\!$$ = total number of failures.
 * $$T_K\,\!$$ = end time of last interval.

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 given above in the confidence bounds on Beta section. 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 2-sided confidence bounds on teh cumulative number of failures are given by:


 * $$N{{(t)}_{L}}=\frac{t}CF{{I}_{L}}\,\!$$


 * $$N{{(t)}_{U}}=\frac{t}CF{{I}_{U}}\,\!$$

where $$IFI_L\,\!$$ and $$IFI_U\,\!$$ are calculated based on the procedures for the confidence bounds on the instantaneous failure intensity.

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 given above in the confidence bounds on Beta section. 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 2-sided confidence bounds on the cumulative failure intensity $$(CFI\,\!)$$ are given below. Let:


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

Then:


 * $$\begin{align}

CFI_{L}= & \frac{\chi _{\tfrac{\alpha }{2},2N}^{2}}{2\cdot t} \\ CFI_{U}= & \frac{\chi _{1-\tfrac{\alpha }{2},2N+2}^{2}}{2\cdot t} \end{align}\,\!$$

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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on cumulative MTBF $$(CMTBF)\,\!$$ are given by:


 * $$CMTB{{F}_{L}}=\frac{1}{CF{{I}_{U}}}\,\!$$


 * $$CMTB{{F}_{U}}=\frac{1}{CF{{I}_{L}}}\,\!$$

where $$CFI_{L}\,\!$$ and $$CFI_{U}\,\!$$ are calculating using the process for the confidence bounds on the cumulative failure intensity.

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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on instantaneous MTBF $$(IMTBF)\,\!$$ are given by first calculating:


 * $$P\left( i \right)=\frac;\text{ }i=1,2,...,K$$

where:


 * $$T_i\,\!$$ = interval end time for the $${{i}^{th}}\,\!$$ interval.
 * $$K\,\!$$ = number of intervals.
 * $$T_K\,\!$$ = end time for the last interval.

Calculate:


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

Next:


 * $$D=\sqrt{\frac{1}{A}+1}$$

And:


 * $$W=\frac{\left( {{z}_{1-\tfrac{\alpha }{2}}} \right)\cdot D}{\sqrt{N}}$$

where:


 * $${{z}_{1-\tfrac{\alpha }{2}}}\,\!$$ = inverse standard normal.
 * $$N\,\!$$ = number of failures.

The 2-sided confidence bounds on instantaneous MTBF are then $$IMTBF\left( 1\pm W \right)\,\!$$.

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 given above in the confidence bounds on Beta section. 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 2-sided confidence bounds on the instantaneous failure intensity $$(IFI)\,\!$$ are given by:

$$\begin{align} IF{{I}_{U}}= & \frac{1}{IMTB{{F}_{L}}} \\ IF{{I}_{L}}= & \frac{1}{IMTB{{F}_{U}}} \end{align}\,\!$$ where $$IMTB{{F}_{L}}\,\!$$and $$IMTB{{F}_{U}}\,\!$$ are calculated using the process for the confidence bounds on 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 given above in the confidence bounds on Beta section. 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 given above in the confidence bounds on Beta section. 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 the Bounds on Time Given Cumulative Failure Intensity section to calculate the bounds.

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 given above in the confidence bounds on Beta section. 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:


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

Step 2: Use equations in the confidence bounds on Time Given Instantaneous MTBF section 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 given above in the confidence bounds on Beta section. 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
The 2-sided confidence bounds on time given instantaneous failure intensity $$(IFI)\,\!$$ are estimated using the process for the confidence bounds on time given instantaneous MTBF where $$IMTBF=\frac{1}{IFI}\,\!$$.