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.

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
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}\,\!$$

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}\,\!$$

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
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}\,\!$$

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}\,\!$$

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 $$IFI{{(t)}_{U}}\,\!$$ can be obtained from 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:


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

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}\,\!$$

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}\,\!$$

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
For the 2-sided confidence bounds on the cumulative MTBF, first calculate the confidence bounds on the cumulative failure intensity. Then:


 * $$\begin{align}

& CMTBF{{(t)}_{L}}=\frac{1}{CFI{{(t)}_{U}}} \\ & CMTBF{{(t)}_{U}}=\frac{1}{CFI{{(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 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, 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}

{{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}

{{[MTB{{F}_{i}}]}_{L}}= & IMTBF\cdot {{\Pi }_{1}} \\ {{[MTB{{F}_{i}}]}_{U}}= & IMTBF\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 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 failure intensity are given by:


 * $$\begin{align}

{IFI}{{(t)}_{L}}= & \frac{1} \\ {IFI}{{(t)}_{U}}= & \frac{1} \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 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 the equations from the confidence bounds on Time Given Instantaneous 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:


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


 * $$\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 the confidence bounds on Beta section. 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 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
Step 1: Calculate $$MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)}\,\!$$. Step 2: Use the equations from the confidence bounds on Time Given Instantaneous MTBF section to calculate the bounds on time given the instantaneous failure intensity.

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
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% confidence interval on $$\hat{\beta }\,\!$$ 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:


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

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 (Grouped) section.

Crow Bounds
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\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% 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}\,\!$$

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

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 (Grouped) 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
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 given above in the confidence bounds on Beta (Grouped) 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
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% confidence interval on $${{\hat{m}}_{i}}(t)\,\!$$ is:


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

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 given above in the confidence bounds on Beta (Grouped) 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 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 given above in the confidence bounds on Beta (Grouped) 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 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 given above in the confidence bounds on Beta (Grouped) 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 (Grouped) 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 (Grouped) 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 (Grouped) 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
Step 1: Calculate $$MTB{{F}_{i}}=\tfrac{1}{{{\lambda }_{i}}(T)}\,\!$$. Step 2: Follow the same process as in the confidence bounds on Time Given Instantaneous MTBF section 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 given above in the confidence bounds on Beta (Grouped) 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 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 the equations given above for Crow instantaneous failure intensity confidence bounds with grouped data.