Weibull scale MLE reduces to the mean at shape one

Question

Renewal times are Weibull with known shape $k = 1$ and unknown scale $\lambda$ , density $f(x \mid \lambda) = \dfrac{k}{\lambda}\left(\dfrac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k}$ for $x > 0$ . Four times are observed: 3, 5, 7, 9.

What is the maximum likelihood estimate of $\lambda$ , and what familiar model does $k = 1$ give?

Accepted Answer

The scale MLE is \hat\lambda = \left(\frac{1}{n}\sumi xi^{\,k}\right)^{1/k}. At k = 1 the k-th power and k-th root are trivial, so this collapses to the sample mean: \hat\lambda = \frac{1}{n}\sumi xi = \frac{3 + 5 + 7 + 9}{4} = \frac{24}{4} = 6. At k = 1 the Weibull density is \frac{1}{\lambda}e^{-x/\lambda}, exactly an exponential with mean \lambda, so its MLE being the sample mean is no surprise. Takeaway. Weibull with shape one is the exponential, and the general scale MLE correctly reduces to the plain sample mean.

Weibull scale MLE reduces to the mean at shape one

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