Template:Example: MLE for Exponential Distribution: Difference between revisions

Latest revision as of 08:32, 10 August 2012

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+#REDIRECT [[The Exponential Distribution]]
-'''MLE for Exponential Distribution'''
-Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters using the MLE method.
-'''Solution to Example 4'''
-In this example we have complete data only. The partial derivative of the log-likelihood function, <math>\Lambda ,</math> is given by:
-::<math>\frac{\partial \Lambda }{\partial \lambda }=\underset{i=1}{\overset{{{F}_{e}}}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}}-\gamma  \right) \right]=\underset{i=1}{\overset{14}{\mathop \sum }}\,\left[ \frac{1}{\lambda }-\left( {{T}_{i}}-\gamma  \right) \right]=0</math>
-Complete descriptions of the partial derivatives can be found in [[Appendix: Distribution Log-Likelihood Equations|Appendix]]. Recall that when using the MLE method for the exponential distribution, the value of <math>\gamma </math> is equal to that of the first failure time. The first failure occurred at 5 hours, thus <math>\gamma =5</math> hours<math>.</math> Substituting the values for <math>T</math> and <math>\gamma </math> we get:
-::<math>\frac{14}{\hat{\lambda }}=560</math>
-or:
-::<math>\hat{\lambda }=0.025\text{ failures/hour}.</math>
-Using Weibull++:
-[[Image:weibullfolio1.png|thumb|center|400px|]]
-The probability plot is:
-[[Image:weibullfolioplot1.png|thumb|center|400px|]]