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Let $X$ a random variable with density function $f(x)=\theta x^{\theta -1}\mathbb I_{(0,1)}(x)$, with $\theta>0$ unknown. I would like to compute the maximum likelihood estimator of $\theta$.

My idea is the following. I write the likelihood function: $$G(x_1, \cdots, x_n)=\theta^n\prod_{i=1}^nx_i \mathbb I_{(0,1)}(x_i). $$ My problem is how to deal with the indicator function. Without it I would consider the $\log G$ and I would compute its derivative to see where it is equal to $0$. Doing this I find $$\hat \theta=-n\sum_{i=1}^n\log x_i.$$

Is this correct? How can I deal with the indicator function?

@edit The maximum likelihood estimator I found, that is $\hat \theta=-n\sum_{i=1}^n\log x_i$ is not a sufficient statistics for $\theta$. Could someone telling me how I could find a sufficient statistics for $\theta$?

Thank you

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Recall $$ \Bbb{I}_{(0,1)}(x_i) = \begin{cases} 1 ,& x_i \in (0,1) \\ 0 ,& \text{otherwise} \end{cases} \text{.} $$

Then \begin{align*} \prod_{i=1}^n x_i \Bbb{I}_{(0,1)}(x_i) &= \prod_{i=1}^n x_i \prod_{i=1}^n \Bbb{I}_{(0,1)}(x_i) \\ &= \begin{cases} \prod_{i=1}^n x_i ,& \text{ all the $x_i \in (0,1)$} \\ 0,& \text{otherwise} \end{cases} \end{align*}

The upshot is you get your $\hat{\theta}$ conditional on all the $x_i \in (0,1)$.

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  • $\begingroup$ Thank you very much! I was confused. Now s clear. I edited the question adding another part. It would be great if you could help me also in this second part. Thank you again! $\endgroup$ – m91c Aug 31 '20 at 21:53
  • $\begingroup$ @user268193 : Why do you say your $\hat{\theta}$ is not sufficient? It has no dependence on $\theta$ (which is the only condition I recall for detecting a sufficient statistic). $\endgroup$ – Eric Towers Aug 31 '20 at 22:02
  • $\begingroup$ Thank you for the answer. For what I know a statistic $\hat \theta$ is sufficient if there exist two functions $F$ and $H$ such that the likelihood function $G(x_1, \cdots, x_n, \theta)=F(\hat\theta, \theta)H(x_1, \cdots,x_n)$. In this case I don't see how to construct the functions $F$ and $H$. $\endgroup$ – m91c Aug 31 '20 at 22:09
  • $\begingroup$ @user268193 : Relative to the Fisher-Neyman factorization, $G(\vec{x}) = h(\vec{x})g(\theta,T(\vec{x}))$, the product of the indicator functions lands in the $h(\vec{x})$ and the rest lands in $g(\theta, T(\vec{x}))$. You've solved stationarity of $g$ for $\theta$ in terms of $T(x)$. It appears to me you have a sufficient statistic. (Maybe it's not minimally sufficient?) $\endgroup$ – Eric Towers Aug 31 '20 at 22:11
  • $\begingroup$ Also the product of the $x_i$ lands in $h$. Am I wrong? It should be $h(x)=\prod_i x_i^{\theta-1}\mathbb I_{(0,1)}(x_i)$ and $g(\theta, T(x))=\theta^n$? $\endgroup$ – m91c Aug 31 '20 at 22:14

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