MLE for Weibull scale with known shape

Question

Time-to-failure of a wind-turbine bearing (in thousands of hours) is Weibull with known shape $k = 2$ and unknown scale $\lambda$ , density

$f(x \mid \lambda) = \frac{k}{\lambda}\left(\frac{x}{\lambda}\right)^{k-1} e^{-(x/\lambda)^k}, \qquad x > 0.$

A field study of $n = 100$ bearings has $\sum_i x_i^{\,2} = 40000$ .

Derive the maximum likelihood estimator of $\lambda$ and compute it.

Accepted Answer

The log of one density is \ln f = \ln k + (k-1)\ln x - k\ln\lambda - (x/\lambda)^k. Summing and keeping the \lambda-dependent terms, \ell(\lambda) = -\,nk\ln\lambda - \lambda^{-k}\sumi xi^{\,k} + 	ext{const}. Differentiate with respect to \lambda: \ell'(\lambda) = -\frac{nk}{\lambda} + k\,\lambda^{-k-1}\sumi xi^{\,k} = 0 \;\Longrightarrow\; n\lambda^{k} = \sumi xi^{\,k}. So \hat\lambda = \left(\frac{1}{n}\sumi xi^{\,k}ight)^{1/k}. With k = 2, \sumi xi^2 = 40000, n = 100: the mean of x^2 is 400, and \hat\lambda = \sqrt{400} = 20 	ext{ (thousand hours)}. Takeaway. With the Weibull shape kno…

MLE for Weibull scale with known shape

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