Template:Example: Weibull MLE: Difference between revisions

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'''Maximum Likelihood Estimation Example'''
#REDIRECT [[The Weibull Distribution]]
 
Repeat [[Weibull Example 1 Data|Example 1]] using maximum likelihood estimation.
 
 
'''Solution'''
 
In this case, we have non-grouped data with no suspensions or intervals, i.e. complete data. The equations for the partial derivatives of the log-likelihood function are derived in [[Appendix: Distribution Log-Likelihood Equations|Appendix]] and given next:
::<math> \frac{\partial \Lambda }{\partial \beta }=\frac{6}{\beta } +\sum_{i=1}^{6}\ln \left( \frac{T_{i}}{\eta }\right) -\sum_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }\ln \left( \frac{T_{i}}{\eta }\right) =0
</math>
 
:and:
 
::<math> \frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{ \beta }{\eta }\sum\limits_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }=0 </math>
 
Solving the above equations simultaneously we get:
 
::<math> \hat{\beta }=1.933,</math> <math>\hat{\eta }=73.526 </math>
 
<br>
The variance/covariance matrix is found to be,
 
::<math> \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.4211 & \hat{Cov}( \hat{\beta },\hat{\eta })=3.272  \\
 
\hat{Cov}(\hat{\beta },\hat{\eta })=3.272 & \hat{Var} \left( \hat{\eta }\right) =266.646 \end{array} \right] </math>
 
The results and the associated graph using Weibull++ (MLE) are shown next.
 
[[Image:Weibull Distribution Example 5 Plot.png|thumb|center|400px| ]]
 
You can view the variance/covariance matrix directly by clicking the ''Analysis Summary'' at the '''Main''' Panel.
 
[[Image: Weibull Distribution Example 5 Variance Matrix.png|thumb|center|400px| ]]
 
<br> Note that the decimal accuracy displayed and used is based on your individual User Setup.

Latest revision as of 09:04, 8 August 2012

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