Template:Bounds on beta camsaa-gd: Difference between revisions
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(Created page with '===Bounds on <math>\beta </math>=== ====Fisher Matrix Bounds==== The parameter <math>\beta </math> must be positive, thus <math>\ln \beta </math> is treated as being normall…') |
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===Bounds on <math>\beta </math>=== | ====Bounds on <math>\beta </math>==== | ||
====Fisher Matrix Bounds==== | =====Fisher Matrix Bounds===== | ||
The parameter <math>\beta </math> must be positive, thus <math>\ln \beta </math> is treated as being normally distributed as well. | The parameter <math>\beta </math> must be positive, thus <math>\ln \beta </math> is treated as being normally distributed as well. | ||
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\end{align}</math> | \end{align}</math> | ||
====Crow Bounds==== | =====Crow Bounds===== | ||
:Step 1: Calculate <math>P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K</math> . | :Step 1: Calculate <math>P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K</math> . | ||
:Step 2: Calculate: | :Step 2: Calculate: |
Revision as of 22:58, 23 August 2012
Bounds on [math]\displaystyle{ \beta }[/math]
Fisher Matrix Bounds
The parameter [math]\displaystyle{ \beta }[/math] must be positive, thus [math]\displaystyle{ \ln \beta }[/math] is treated as being normally distributed as well.
- [math]\displaystyle{ \frac{\ln \hat{\beta }-\ln \beta }{\sqrt{Var(\ln \hat{\beta }})}\ \tilde{\ }\ N(0,1) }[/math]
The approximate confidence bounds are given as:
- [math]\displaystyle{ C{{B}_{\beta }}=\hat{\beta }{{e}^{\pm {{z}_{\alpha }}\sqrt{Var(\hat{\beta })}/\hat{\beta }}} }[/math]
- [math]\displaystyle{ \widehat{\beta } }[/math] can be obtained by [math]\displaystyle{ \underset{i=1}{\overset{K}{\mathop{\sum }}}\,{{n}_{i}}\left( \tfrac{T_{i}^{{\hat{\beta }}}\ln {{T}_{i}}-T_{i-1}^{{\hat{\beta }}}\ln \,{{T}_{i-1}}}{T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}}}-\ln {{T}_{k}} \right)=0 }[/math] .
All variance can be calculated using the Fisher Matrix:
- [math]\displaystyle{ \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] }[/math]
[math]\displaystyle{ \Lambda }[/math] is the natural log-likelihood function where ln [math]\displaystyle{ ^{2}T={{\left( \ln T \right)}^{2}} }[/math] and:
- [math]\displaystyle{ \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] }[/math]
- [math]\displaystyle{ \begin{align} & \frac{{{\partial }^{2}}\Lambda }{\partial {{\lambda }^{2}}}= & -\frac{n}{{{\lambda }^{2}}} \\ & \frac{{{\partial }^{2}}\Lambda }{\partial {{\beta }^{2}}}= & \underset{i=1}{\overset{k}{\mathop \sum }}\,\left[ \begin{matrix} {{n}_{i}}\left( \tfrac{(T_{i}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i}}-T_{i-1}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i-1}})(T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}})-{{\left( T_{i}^{{\hat{\beta }}}\ln {{T}_{i}}-T_{i-1}^{{\hat{\beta }}}\ln {{T}_{i-1}} \right)}^{2}}}{{{(T_{i}^{{\hat{\beta }}}-T_{i-1}^{{\hat{\beta }}})}^{2}}} \right) \\ -\left( \lambda T_{i}^{{\hat{\beta }}}{{\ln }^{2}}{{T}_{i}}-\lambda T_{i-1}^{{\hat{\beta }}}{{\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} }[/math]
Crow Bounds
- Step 1: Calculate [math]\displaystyle{ P(i)=\tfrac{{{T}_{i}}}{{{T}_{K}}},\,\,i=1,2,\ldots ,K }[/math] .
- Step 2: Calculate:
- [math]\displaystyle{ A=\underset{i=1}{\overset{K}{\mathop \sum }}\,\frac{{{[P{{(i)}^{{\hat{\beta }}}}\ln P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{\widehat{\beta }}}\ln P{{(i-1)}^{{\hat{\beta }}}}]}^{2}}}{[P{{(i)}^{{\hat{\beta }}}}-P{{(i-1)}^{{\hat{\beta }}}}]} }[/math]
- Step 3: Calculate [math]\displaystyle{ c=\tfrac{1}{\sqrt{A}} }[/math] and [math]\displaystyle{ S=\tfrac{({{z}_{1-\alpha /2}})\cdot C}{\sqrt{N}} }[/math] . Thus an approximate 2-sided [math]\displaystyle{ (1-\alpha ) }[/math] 100-percent confidence interval on [math]\displaystyle{ \widehat{\beta } }[/math] is: