Determine the maximum likelihood estimator $\hat \theta$ from $\theta$. If $X_i\, , \quad i=1,\, 2,\, \ldots ,\, n$
Given function :
$$f(x)=\begin{cases} \theta x^{\theta -1}\,\quad ,\,0\lt x \lt 1\, ; \, 0\le \theta \le \infty\\~~~~ 0\qquad,\text{ for other }x\end{cases}$$
My answer :
$\begin{aligned} L(\theta\, ; \, x_i)&=\theta x_1^{\theta -1}\cdot \theta x_2^{\theta -1}\cdots \theta x_n^{\theta -1}\\ &= \displaystyle\prod_{i=1}^n \theta x_i^{\theta -1}\\ \ln{L(\theta\, ; \, x_i)}&=\ln{\displaystyle\prod_{i=1}^n \theta x_i^{\theta -1}}\\ &= \displaystyle\sum_{i=1}^n \ln{\theta x_i^{\theta -1}}\\ &= \displaystyle\sum_{i=1}^n \ln{\theta}+ \displaystyle\sum_{i=1}^n \theta \ln{x_i}- \displaystyle\sum_{i=1}^n \ln{x_i}\\ &=n\cdot \ln{\theta}+ \displaystyle\sum_{i=1}^n \theta \ln{x_i}- \displaystyle\sum_{i=1}^n \ln{x_i}\\ \dfrac{\partial \ln{L(\theta\, ; \, x_i)}}{\partial \theta}&= \dfrac{n}{\theta}+\displaystyle\sum_{i=1}^n \ln{x_i}=0\\ \hat \theta &= \dfrac{-n}{\displaystyle\sum_{i=1}^n \ln{x_i}} \end{aligned}$
I doubt my answer. Please gimme some corrections if anything is wrong. And tell me which one. Thanks.
Why do you doubt your answer? It is for the most part correct.
Recall here that the support of $X$ is the open interval $(0,1)$. Therefore, $\log x_i \in (-\infty, 0)$, hence $\hat \theta \in (0,\infty)$. It may be instructive to use a computer to generate some realizations of $X$ and calculate the MLE from the sample. I generated $n = 100$ such realizations:
What is your MLE for this sample? Can you calculate the variance of this estimator; i.e., what is $\operatorname{Var}[\hat \theta]$?