Richard House: Redirected page to Lloyd-Lipow#Maximum Likelihood Estimators

2012-08-27T02:31:38Z

Redirected page to Lloyd-Lipow#Maximum Likelihood Estimators

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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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Nicolette Young: Created page with '===Maximum Likelihood Estimators===
For the ${{k}^{th}}$ stage: :: ${{L}_{k}}=const.\text{ }R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}}$
2012-01-09T22:06:21Z

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…'

New page
===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>

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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Richard House: Redirected page to Lloyd-Lipow#Maximum Likelihood Estimators