1
$\begingroup$

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!

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $x\in(0,1)$,$$\log(x)<0$$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.