Example: Gumbel Distribution

Verify using Monte Carlo simulation that if $${{t}_{i}}$$  follows a Weibull distribution with  $$\beta $$  and  $$\eta $$, then the  $$Ln({{t}_{i}})$$  follows a Gumbel distribution with  $$\mu =\ln (\eta )$$  and  $$\sigma =1/\beta.$$

Let us assume that $${{t}_{i}}$$  follows a Weibull distribution with  $$\beta =0.5$$  and  $$\eta =10000.$$  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.



Other simulation settings are:



After obtaining the random time values $${{t}_{i}}$$, insert a new data sheet into the folio. In this sheet enter the $$Ln({{t}_{i}})$$  values using the LN function and referring to the cells in the sheet that contains the  $${{t}_{i}}$$  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 $$Ln({{t}_{i}})$$  values.

Using maximum likelihood as the analysis method, the estimated parameters are:


 * $$\begin{align}

& \hat{\mu }= & 9.3816 \\ & \hat{\sigma }= & 1.9717 \end{align}$$

Since $$\ln (\eta )=$$  9.2103 ( $$\simeq 9.3816$$ ) and  $$1/\beta =2$$   $$(\simeq 1.9717),$$  then this simulation verifies that  $$Ln({{t}_{i}})$$  follows a Gumbel distribution with  $$\mu =\ln (\eta )$$  and  $$\delta =1/\beta .$$

Note: This example illustrates a property of the Gumbel distribution; it is not meant to be a formal proof.