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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

user268193
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1 Answers1

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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 $x_i \in (0,1)$.

Eric Towers
  • 70,953