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| | | #REDIRECT [[Crow-AMSAA - NHPP]] |
| ===Maximum Likelihood Estimators===
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| The probability density function ( <math>pdf</math> ) of the <math>{{i}^{th}}</math> event given that the <math>{{(i-1)}^{th}}</math> event occurred at <math>{{T}_{i-1}}</math> is:
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| <br>
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| ::<math>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>
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| The likelihood function is:
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| <br>
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| ::<math>L={{\lambda }^{n}}{{\beta }^{n}}{{e}^{-\lambda {{T}^{*\beta }}}}\underset{i=1}{\overset{n}{\mathop \prod }}\,T_{i}^{\beta -1}</math>
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| where <math>{{T}^{*}}</math> is the termination time and is given by:
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| <br>
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| ::<math>{{T}^{*}}=\left\{ \begin{matrix}
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| {{T}_{n}}\text{ if the test is failure terminated} \\
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| T>{{T}_{n}}\text{ if the test is time terminated} \\
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| \end{matrix} \right\}</math>
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| Taking the natural log on both sides:
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| <br>
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| ::<math>\Lambda =n\ln \lambda +n\ln \beta -\lambda {{T}^{*\beta }}+(\beta -1)\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}</math>
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| <br>
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| And differentiating with respect to <math>\lambda </math> yields:
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| <br>
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| ::<math>\frac{\partial \Lambda }{\partial \lambda }=\frac{n}{\lambda }-{{T}^{*\beta }}</math>
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| <br>
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| Set equal to zero and solve for <math>\lambda </math> :
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| <br>
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| ::<math>\widehat{\lambda }=\frac{n}{{{T}^{*\beta }}}</math>
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| <br>
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| Now differentiate with respect to <math>\beta </math> :
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| <br>
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| ::<math>\frac{\partial \Lambda }{\partial \beta }=\frac{n}{\beta }-\lambda {{T}^{*\beta }}\ln {{T}^{*}}+\underset{i=1}{\overset{n}{\mathop \sum }}\,\ln {{T}_{i}}</math>
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| <br>
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| Set equal to zero and solve for <math>\beta </math> :
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| <br>
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| ::<math>\widehat{\beta }=\frac{n}{n\ln {{T}^{*}}-\underset{i=1}{\overset{n}{\mathop{\sum }}}\,\ln {{T}_{i}}}</math>
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