Template:Maximum likelihood estimators camsaa-pe: Difference between revisions

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(Created page with ' ===Maximum Likelihood Estimators=== The probability density function ( <math>pdf</math> ) of the <math>{{i}^{th}}</math> event given that the <math>{{(i-1)}^{th}}</math> eve…')
 
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Now differentiate Eqn. (amsaa4) with respect to  <math>\beta </math> :  
Now differentiate with respect to  <math>\beta </math> :  


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Revision as of 12:23, 23 August 2012

Maximum Likelihood Estimators

The probability density function ( [math]\displaystyle{ pdf }[/math] ) of the [math]\displaystyle{ {{i}^{th}} }[/math] event given that the [math]\displaystyle{ {{(i-1)}^{th}} }[/math] event occurred at [math]\displaystyle{ {{T}_{i-1}} }[/math] is:


[math]\displaystyle{ f({{T}_{i}}|{{T}_{i-1}})=\frac{\beta }{\eta }{{\left( \frac{{{T}_{i}}}{\eta } \right)}^{\beta -1}}\cdot {{e}^{-\tfrac{1}{{{\eta }^{\beta }}}\left( T_{i}^{\beta }-T_{i-1}^{\beta } \right)}} }[/math]

The likelihood function is:


[math]\displaystyle{ L={{\lambda }^{n}}{{\beta }^{n}}{{e}^{-\lambda {{T}^{*\beta }}}}\underset{i=1}{\overset{n}{\mathop \prod }}\,T_{i}^{\beta -1} }[/math]

where [math]\displaystyle{ {{T}^{*}} }[/math] is the termination time and is given by:


[math]\displaystyle{ {{T}^{*}}=\left\{ \begin{matrix} {{T}_{n}}\text{ if the test is failure terminated} \\ T\gt {{T}_{n}}\text{ if the test is time terminated} \\ \end{matrix} \right\} }[/math]

Taking the natural log on both sides:


[math]\displaystyle{ \Lambda =n\ln \lambda +n\ln \beta -\lambda {{T}^{*\beta }}+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}} }[/math]


And differentiating with respect to [math]\displaystyle{ \lambda }[/math] yields:


[math]\displaystyle{ \frac{\partial \Lambda }{\partial \lambda }=\frac{n}{\lambda }-{{T}^{*\beta }} }[/math]


Set equal to zero and solve for [math]\displaystyle{ \lambda }[/math] :


[math]\displaystyle{ \widehat{\lambda }=\frac{n}{{{T}^{*\beta }}} }[/math]


Now differentiate with respect to [math]\displaystyle{ \beta }[/math] :


[math]\displaystyle{ \frac{\partial \Lambda }{\partial \beta }=\frac{n}{\beta }-\lambda {{T}^{*\beta }}\ln {{T}^{*}}+\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}} }[/math]


Set equal to zero and solve for [math]\displaystyle{ \beta }[/math] :


[math]\displaystyle{ \widehat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}} }[/math]