Let $\underline{Y}= (Y_1,..., Y_n)$ be an i.i.d. random sample from a Weibull distribution, with probability density function given by $f(x; \lambda)= \frac{k}{\lambda}(\frac{y}{\lambda})^{k-1}exp$ {$-(\frac{y}{\lambda})^k$} where k > $0$ is a known shape parameter, and λ is an unknown scale parameter taking values in $\mathbb{R^+}$.

Consider the parametrisation $\theta= \lambda^k$

Derive the likelihood function $L(\theta; \underline{Y})$ and thus the Maximum likelihood estimator $\hat{\theta}(\underline{Y})$ for $\theta.$ Show that the MLE is unbiased.

What I know so far

take the sum of the pdf up to n to find the likelihood function. take the log and differentiate and then set to $0$ and solve for the MLE. If the expectation is 0 then the estimator is unbiased. I know the method but I am unsure of how to actually put it into practice. Any help would be greatly appreciated.


1 Answer 1


First rewrite the density with the new parametrization


Calculate the likelihood

$$L(\theta)\propto \theta^{-n}e^{-\frac{\Sigma_i y_i^k}{\theta}}$$

proceeding in the calculation you find that the score function (derivative of the log likelihood with respect to $\theta$) is

$$l^*=-\frac{n}{\theta}+\frac{1}{\theta^2}\Sigma_i y_i^k$$

And thus

$$T=\hat{\theta}_{ML}=\frac{\Sigma_i y_i^k}{n}$$

To show that $\mathbb{E}[T]=\theta$ let's rewrite the score function in the following way


Now simply remembering that (First Bartlett Identity)


you get


that is also


To calculate its variance, using II Bartlett Identity, that is


This identity leads to


that is



Alternative method to calculate expectation and variance of T

Simply transforming


you get that $W\sim Exp\Big(\frac{1}{\theta}\Big)$ thus

$$T\sim Gamma\Big(n;\frac{n}{\theta}\Big)$$

thus immediately you get



  • $\begingroup$ Please don't use \frac in exponents or limits of integrals. It looks bad and confusing, and it rarely appears in professional mathematics typesetting. $\endgroup$ Commented Dec 10, 2020 at 10:34
  • $\begingroup$ im unsure of what the MLE is... $\endgroup$
    – user848358
    Commented Dec 10, 2020 at 10:44
  • $\begingroup$ @PadChipper : I wrote all the passages. You have the score function $l^*$. Set it =0 and solve in $\theta$ $\endgroup$
    – tommik
    Commented Dec 10, 2020 at 10:50
  • $\begingroup$ Great thank you so much! How would I show that the variance of the MLE is $\theta^2$/n? $\endgroup$
    – user848358
    Commented Dec 10, 2020 at 11:03
  • 1
    $\begingroup$ @PadChipper: any other questions? It looks like I did all for you...see my edits $\endgroup$
    – tommik
    Commented Dec 10, 2020 at 11:21

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