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Wikipedia: http://en.wikipedia.org/wiki/Weibull_distribution gives a nice description on what the shape parameter (they call it k) means in the Weibull distribution, but I can't find anywhere what the scale parameter (denoted as λ on Wikipedia) means or how to estimate it. Obviously for estimation one could employ a MLE method, but I'm thinking more in the lines of real world examples. For instance if a bug has an average lifespan of a year and we know λ (scale parameter), can I just find k by solving for it in the equation for the mean or would that produce bias? Thanks for the help.

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Actually there are many estimators of the scale parameter of the Weibull distribution. If you'd like to you can turn to "Continuous univariate distributions. Vol. 1" by N.L. Johnson, S. Kotz and N. Balakrishnan. But they are not that nice as one can think of looking at the pdf.
For instance the moment estimator, based on the sample of size $n$: $$\hat{\lambda}_{MM}=\exp{\left(\frac{1}{n}\sum_{i=1}^n\log(X_i)+\gamma\frac{\sqrt{6}}{\pi}\sqrt{\frac{1}{n-1}\sum_{i=1}^n(\log(X_i)-\overline{\log(X)})^2}\right)},$$ where $\gamma$ - is Euler's constant.
The estimator is asymptotically unbiased, with variance: $$\mathrm{Var}(\hat{\lambda}_{MM})=1.2\frac{k^{-2}}{n}+k^{-2}O(n^{-\frac{3}{2}})$$ $\hat{\lambda}_{MM}$ has the asymptotic efficiency of $95$%.
Maximum likelihood estimator of $\lambda$ (when $k$ is unknown) is given by the statistics: $$\hat{\lambda}_{ML}=\left(\frac{1}{n}\sum_{i=1}^n X_i^{\hat{k}_{ML}}\right)^{\frac{1}{\hat{k}_{ML}}}$$ And $\hat{k}_{ML}$ is the solution of the following equation: $$\hat{k}_{ML}=\left(\left(\sum_{i=1}^n X_i^{\hat{k}_{ML}}\log(X_i)\right)\left(\sum_{i=1}^n X_i^{\hat{k}_{ML}}\right)^{-1}\!\!\!\!-\frac{1}{n}\sum_{i=1}^n\log(X_i)\right)^{-1}$$ Other estimators are even more peculiar but much harder in understanding ;).

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  • $\begingroup$ nice thanks for the source too $\endgroup$ Jul 16, 2013 at 18:18

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