Weibull scale MLE with shape three

Question

Crack-growth times are Weibull with known shape $k = 3$ and unknown scale $\lambda$ , density $f(x \mid \lambda) = \dfrac{k}{\lambda}\left(\dfrac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k}$ for $x > 0$ . A sample of $n = 10$ specimens has $\sum_i x_i^{\,3} = 270$ .

What is the maximum likelihood estimate of $\lambda$ ?

Accepted Answer

The scale MLE is \hat\lambda = \left(\frac{1}{n}\sumi xi^{\,k}\right)^{1/k}. With k = 3 the average cube is 270/10 = 27, so \hat\lambda = 27^{1/3} = 3. Takeaway. The exponent in the root always matches the shape: with k = 3 you take a cube root of the mean cube.

Weibull scale MLE with shape three

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