Template:Bounds on beta camsaa-gd

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: