Example: Gumbel Distribution: Difference between revisions

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Verify using Monte Carlo simulation that if  <math>{{t}_{i}}</math>  follows a Weibull distribution with  <math>\beta </math>  and  <math>\eta </math> , then the  <math>Ln({{t}_{i}})</math>  follows a Gumbel distribution with  <math>\mu =\ln (\eta )</math>  and  <math>\sigma =1/\beta.</math>
 
Let us assume that  <math>{{t}_{i}}</math>  follows a Weibull distribution with  <math>\beta =0.5</math>  and  <math>\eta =10000.</math>  The Monte Carlo simulation tool in Weibull++ can be used to generate a set of random numbers that follow a Weibull distribution with the specified parameters.
 
[[Image:montecarlo4eva.png|thumb|center|400px| ]]
 
Other simulation settings are:
 
[[Image:Gumbel Distribution Example 1 Simulation Setting.png|thumb|center|400px| ]]
 
After obtaining the random time values  <math>{{t}_{i}}</math> , insert a new data sheet into the folio. In this sheet enter the  <math>Ln({{t}_{i}})</math>  values using the LN function and referring to the cells in the sheet that contains the  <math>{{t}_{i}}</math>  values. Delete any negative values, if there are any (since Weibull++ expects all time values to be positive). Calculate the parameters of the Gumbel distribution that fits the  <math>Ln({{t}_{i}})</math>  values.
 
Using maximum likelihood as the analysis method, the estimated parameters are:
 
::<math>\begin{align}
  & \hat{\mu }= & 9.3816 \\
& \hat{\sigma }= & 1.9717 
\end{align}</math>
 
 
Since  <math>\ln (\eta )=</math>  9.2103 ( <math>\simeq 9.3816</math> ) and  <math>1/\beta =2</math>  <math>(\simeq 1.9717),</math>  then this simulation verifies that  <math>Ln({{t}_{i}})</math>  follows a Gumbel distribution with  <math>\mu =\ln (\eta )</math>  and  <math>\delta =1/\beta .</math>
 
Note: This example illustrates a property of the Gumbel distribution; it is not meant to be a formal proof.

Latest revision as of 06:42, 15 August 2012

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