Template:Bounds on beta rsa

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


 * $$\frac{\ln (\widehat{\beta })-\ln (\beta )}{\sqrt{Var\left[ \ln (\widehat{\beta }) \right]}}\ \tilde{\ }\ N(0,1)$$


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


 * $$\widehat{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}{\widehat{\lambda }\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\left[ (T_{q}^{\widehat{\beta }}\ln ({{T}_{q}})-S_{q}^{\widehat{\beta }}\ln ({{S}_{q}}) \right]-\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{\mathop{\sum }}}\,\ln ({{X}_{i}}{{}_{q}})}$$

All variance can be calculated using the Fisher Information Matrix. $$\Lambda $$ is the natural log-likelihood function.


 * $$\Lambda =\underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ {{N}_{q}}(\ln (\lambda )+\ln (\beta ))-\lambda (T_{q}^{\beta }-S_{q}^{\beta })+(\beta -1)\underset{i=1}{\overset{\mathop \sum }}\,\ln ({{x}_{iq}}) \right]$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial \lambda \partial \beta }=-\underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ T_{q}^{\beta }\ln ({{T}_{q}})-S_{q}^{\beta }\ln ({{S}_{q}}) \right]$$


 * $$\frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}=-\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{N}_{q}}}-\lambda \underset{q=1}{\overset{K}{\mathop \sum }}\,\left[ T_{q}^{\beta }{{(\ln ({{T}_{q}}))}^{2}}-S_{q}^{\beta }{{(\ln ({{S}_{q}}))}^{2}} \right]$$

Crow Bounds
Calculate the conditional maximum likelihood estimate of $$\tilde{\beta }$$ :


 * $$\tilde{\beta }=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{M}{\mathop{\sum }}}\,\ln \left( \tfrac \right)}$$

The Crow 2-sided $$(1-a)$$ 100-percent confidence bounds on  $$\beta $$  are:


 * $$\begin{align}

& {{\beta }_{L}}= & \tilde{\beta }\frac{\chi _{\tfrac{\alpha }{2},2M}^{2}}{2M} \\ & {{\beta }_{U}}= & \tilde{\beta }\frac{\chi _{1-\tfrac{\alpha }{2},2M}^{2}}{2M} \end{align}$$