# Maximum likelihood estimator. How can I deal with the indicator function?

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

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 $$x_i \in (0,1)$$.
• @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). – Eric Towers Aug 31 '20 at 22:02
• 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 $H$. – m91c Aug 31 '20 at 22:09
• @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?) – Eric Towers Aug 31 '20 at 22:11
• 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 $g(\theta, T(x))=\theta^n$? – m91c Aug 31 '20 at 22:14