Template:MLE lloyd-l: Difference between revisions

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(Created page with '===Maximum Likelihood Estimators=== <br> For the <math>{{k}^{th}}</math> stage: ::<math>{{L}_{k}}=const.\text{ }R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}}</math…')
 
 
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===Maximum Likelihood Estimators===
#REDIRECT [[Lloyd-Lipow#Maximum_Likelihood_Estimators]]
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For the  <math>{{k}^{th}}</math>  stage:
 
::<math>{{L}_{k}}=const.\text{ }R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}}</math>
 
And assuming that the results are independent between stages:
 
::<math>L=\underset{k=1}{\overset{N}{\mathop \prod }}\,R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}}</math>
 
Then taking the natural log gives:
 
::<math>\Lambda =\underset{k=1}{\overset{N}{\mathop \sum }}\,{{S}_{k}}\ln \left( {{R}_{\infty }}-\frac{\alpha }{k} \right)+\underset{k=1}{\overset{N}{\mathop \sum }}\,({{n}_{k}}-{{S}_{k}})\ln \left( 1-{{R}_{\infty }}+\frac{\alpha }{k} \right)</math>
 
Differentiating with respect to  <math>{{R}_{\infty }}</math>  and  <math>\alpha ,</math>  yields:
 
::<math>\frac{\partial \Lambda }{\partial {{R}_{\infty }}}=\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{S}_{k}}}{{{R}_{\infty }}-\tfrac{\alpha }{k}}-\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{{{n}_{k}}-{{S}_{k}}}{1-{{R}_{\infty }}+\tfrac{\alpha }{k}}</math>
 
 
::<math>\frac{\partial \Lambda }{\partial \alpha }=-\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{{{S}_{k}}}{k}}{{{R}_{\infty }}-\tfrac{\alpha }{k}}+\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{{{n}_{k}}-{{S}_{k}}}{k}}{1-{{R}_{\infty }}+\tfrac{\alpha }{k}}</math>
 
Rearranging Eqns. (R1) and (alpha1) and setting equal to zero gives:
 
::<math>\frac{\partial \Lambda }{\partial {{R}_{\infty }}}=\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{{{S}_{k}}}{{{n}_{k}}}-\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)}{\tfrac{1}{{{n}_{k}}}\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)\left( 1-{{R}_{\infty }}+\tfrac{\alpha }{k} \right)}=0</math>
 
 
::<math>\frac{\partial \Lambda }{\partial \alpha }=-\underset{k=1}{\overset{N}{\mathop \sum }}\,\frac{\tfrac{1}{k}\tfrac{{{S}_{k}}}{{{n}_{k}}}-\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)\tfrac{1}{k}}{\tfrac{1}{{{n}_{k}}}\left( {{R}_{\infty }}-\tfrac{\alpha }{k} \right)\left( 1-{{R}_{\infty }}+\tfrac{\alpha }{k} \right)}=0</math>
 
Eqns. (R2) and (alpha2) can be solved simultaneously for  <math>\widehat{\alpha }</math>  and  <math>{{\hat{R}}_{\infty }}</math> . It should be noted that a closed form solution does not exist for either of the parameters; thus they must be estimated numerically.
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Latest revision as of 02:31, 27 August 2012

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