Template:Gll weibull: Difference between revisions

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(Created page with '====GLL Weibull==== <br> The GLL-Weibull model can be derived by setting <math>\eta =L(\underline{X})</math> in Eqn. (GLL1), yielding the following GLL-Weibull <math>pdf</math…')
 
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====GLL Weibull====
====GLL Weibull====
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The GLL-Weibull model can be derived by setting  <math>\eta =L(\underline{X})</math>  in Eqn. (GLL1), yielding the following GLL-Weibull  <math>pdf</math> :  
The GLL-Weibull model can be derived by setting  <math>\eta =L(\underline{X})</math>  in Eqn. (GLL1), yielding the following GLL-Weibull  <math>pdf</math> :  
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::<math>f(t,\underline{X})=\beta \cdot {{t}^{\beta -1}}{{e}^{-\beta \left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}{{e}^{-{{t}^{\beta }}{{e}^{-\beta \left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}}}</math>
<math>f(t,\underline{X})=\beta \cdot {{t}^{\beta -1}}{{e}^{-\beta \left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}{{e}^{-{{t}^{\beta }}{{e}^{-\beta \left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}}}</math>
 
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The total number of unknowns to solve for in this model is  <math>n+2</math>  (i.e.  <math>\beta ,{{a}_{0}},{{a}_{1}},...{{a}_{n}}).</math>  
The total number of unknowns to solve for in this model is  <math>n+2</math>  (i.e.  <math>\beta ,{{a}_{0}},{{a}_{1}},...{{a}_{n}}).</math>  
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Revision as of 22:53, 10 February 2012

GLL Weibull

The GLL-Weibull model can be derived by setting [math]\displaystyle{ \eta =L(\underline{X}) }[/math] in Eqn. (GLL1), yielding the following GLL-Weibull [math]\displaystyle{ pdf }[/math] :
[math]\displaystyle{ f(t,\underline{X})=\beta \cdot {{t}^{\beta -1}}{{e}^{-\beta \left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}{{e}^{-{{t}^{\beta }}{{e}^{-\beta \left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}}} }[/math]
The total number of unknowns to solve for in this model is [math]\displaystyle{ n+2 }[/math] (i.e. [math]\displaystyle{ \beta ,{{a}_{0}},{{a}_{1}},...{{a}_{n}}). }[/math]