I need to calculate the ML estimator of the following probability function and I have doubts about the solution.

$$f(x; \theta) = \theta x^{\theta -1} $$

$$L(\theta) = \theta^n \left(\prod_{i=1}^n x_i\right)^{\theta-1} = $$ $$l(\theta) =\log(L)= n\log(\theta) + (\theta -1) \log \left(\prod_{i=1}^n xi\right) = $$ $$n\log(\theta) + (\theta -1) \sum_{i=1}^n \log(xi)$$

$$l'(\theta) = \cfrac{\partial l}{\partial \theta} = \cfrac{n}{\theta} + \sum_{i=1}^n\log(x_i)$$

$$l'(\theta) = 0 \Leftrightarrow \hat{\theta} = -\cfrac{n}{\sum_{i=1}^n \log(x_i)}$$

Is my result correct? How can it be negative if $\theta > 0$ and $x \in \{0,1\}$?

Does $f(x; \theta) = \theta x^{\theta -1} $ have a specific name? It looks like a Weibull, but it does not have the exponential component.

Many thanks in advance!


1 Answer 1


The description is missing the important part that the support is $(0,1)$, i.e. $$f_\theta(x) = \theta x^{\theta-1}\mathbb I_{(0,1)}(x)$$ This is a special case of the Beta distribution, namely the Beta$(\theta,1)$ distribution. And, since $x\in(0,1)$,$$\log(x)<0$$

  • 1
    $\begingroup$ Right, I was not paying attention to the support. Thank you! $\endgroup$ Oct 28, 2020 at 13:51

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