Let $X_i$ be Weibull distributed, $X_i\sim W_{2,\theta}$. The Distribution function is given by $$F(t) = 1 - \exp \Big\{-\frac{t^2}{\theta}\Big\}, t>0$$

How can one calculate the Maximum-Likelihood estimator of this sample $X_1, \ldots, X_n$ for $\tau = \frac{1}{\theta}$? Furthermore, is this estimator an UMVU-estimator?

  • $\begingroup$ I suppose you mean $F(t) = 1-\exp\{-t^2/\theta\}$? $\endgroup$
    – Therkel
    Jan 7, 2017 at 14:56
  • $\begingroup$ Have you found the density of the distribution, and then the likelihood of the observations? $\endgroup$
    – Henry
    Jan 7, 2017 at 23:32
  • $\begingroup$ The denisty of $X_i$ should be equal to $f(x) = \frac{2}{\theta} x \exp \Big\{ - \frac{x^2}{\theta} \Big\}$. $\endgroup$
    – ducks17
    Jan 8, 2017 at 15:39

1 Answer 1


Is this way correct:

The density is given by $f(x \mid \theta) = \frac{2}{\theta} x \exp \{-\frac{x^2}{\theta}\}$. Then, $L(x_1, x_2, \ldots, x_n \mid \theta) = f(x_1 \mid \theta) \cdots f(x_n \mid \theta) = ... = \big( \frac{2}{\theta} \big)^n \exp \{- \frac{u}{\theta}\}(x_1 \cdots x_n) = g(u,\theta)h(x_1,\ldots, x_n)$

with $u = \sum_{i=1}^n x_i^2$. Furthermore: $\ln(L) = n \ln(2)-n\ln(\theta) - \frac{u}{\theta}+ \ln(x_1 \cdots x_n)$

$\frac{d \ln(L)}{d \theta} = -\frac{n}{\theta} + \frac{u}{\theta^2} = 0$

$\Rightarrow \hat{\theta} = \frac{u}{n} ? \frac{1}{n} \sum_{i=1}^n x_i^2$

Finally, since $E[\hat{\theta}]=\theta$, it follows by factorization theorem and Lehmann-Scheffe, that this is an UMVE.

  • $\begingroup$ I believe you meant $E[\hat \theta] = \theta$. And by factorization theorem maybe you mean exponential-family factorization, not Neymann-Fisher Factorization theorem, since you need to show $\frac{1}{n} \sum_{i=1}^n X_i^2$ is both sufficient & complete. But Lehmann-Scheffe indeed shows it's an UMVUE $\endgroup$
    – user365239
    Jan 8, 2017 at 21:25
  • $\begingroup$ Yeah, thank you, that was a typo... $\endgroup$
    – ducks17
    Jan 9, 2017 at 15:18

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