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| ===Maximum Likelihood Estimators===
| | #REDIRECT [[Lloyd-Lipow#Maximum_Likelihood_Estimators]] |
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| For the <math>{{k}^{th}}</math> stage:
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| ::<math>{{L}_{k}}=const.\text{ }R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}}</math>
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| And assuming that the results are independent between stages:
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| ::<math>L=\underset{k=1}{\overset{N}{\mathop \prod }}\,R_{k}^{{{S}_{k}}}{{(1-{{R}_{k}})}^{{{n}_{k}}-{{S}_{k}}}}</math>
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| Then taking the natural log gives:
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| ::<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>
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| Differentiating with respect to <math>{{R}_{\infty }}</math> and <math>\alpha ,</math> yields:
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| ::<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>
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| ::<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>
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| Rearranging Eqns. (R1) and (alpha1) and setting equal to zero gives:
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| ::<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>
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| ::<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>
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| 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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