Template:Gll exponential: Difference between revisions
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(Created page with '====GLL Exponential==== <br> The GLL-exponential model can be derived by setting <math>m=L(\underline{X})</math> in Eqn. (GLL1), yielding the following GLL-exponential <math>p…') |
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<math>f(t,\underline{X})={{e}^{-\left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}{{e}^{-\left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)\cdot t}}</math> | |||
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The total number of unknowns to solve for in this model is <math>n+1</math> (i.e. <math>{{a}_{0}},{{a}_{1}},...{{a}_{n}}).</math> | The total number of unknowns to solve for in this model is <math>n+1</math> (i.e. <math>{{a}_{0}},{{a}_{1}},...{{a}_{n}}).</math> | ||
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Revision as of 18:01, 14 February 2012
GLL Exponential
The GLL-exponential model can be derived by setting [math]\displaystyle{ m=L(\underline{X}) }[/math] in Eqn. (GLL1), yielding the following GLL-exponential [math]\displaystyle{ pdf }[/math] :
[math]\displaystyle{ f(t,\underline{X})={{e}^{-\left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)}}{{e}^{-\left( {{\alpha }_{0}}+\underset{j=1}{\overset{n}{\mathop{\sum }}}\,{{\alpha }_{j}}{{X}_{j}} \right)\cdot t}} }[/math]
The total number of unknowns to solve for in this model is [math]\displaystyle{ n+1 }[/math] (i.e. [math]\displaystyle{ {{a}_{0}},{{a}_{1}},...{{a}_{n}}). }[/math]