Template:MLE lloyd-l

Maximum Likelihood Estimators

For the [math]\displaystyle{ {{k}^{th}} }[/math] stage:

[math]\displaystyle{ {{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]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ {{R}_{\infty }} }[/math] and [math]\displaystyle{ \alpha , }[/math] yields:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \widehat{\alpha } }[/math] and [math]\displaystyle{ {{\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.