Template:Bounds on beta camsaa-cb

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}$$