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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.

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    $\begingroup$ Looks right. You can also think about extreme cases: what should $\hat{\theta}$ do when all the samples are very small or very large? It should be very small or very large, respectively. And it has this behavior. The important thing that is not 100% obvious from the actual formula is that the denominator is also negative, because the $x_i$ will be in $(0,1)$ with probability 1. $\endgroup$ – Ian Sep 3 '19 at 1:32
  • $\begingroup$ Thanks.... could i have a different question? If i have a function and the first derivative is n. And the second derivative is 0. Is it MLE? $\endgroup$ – user516076 Sep 3 '19 at 1:44
  • $\begingroup$ math.stackexchange.com/q/2339322/321264 $\endgroup$ – StubbornAtom Apr 28 at 12:11
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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:

0.423195, 0.904185, 0.768371, 0.709028, 0.82171, 0.98423, 0.833026, 0.967757, 0.831291, 0.942482, 0.804875, 0.936858, 0.703554, 0.924037, 0.910317, 0.889744, 0.964979, 0.937217, 0.959913, 0.901902, 0.941537, 0.847308, 0.987343, 0.824175, 0.766928, 0.649205, 0.996147, 0.870595, 0.923796, 0.818284, 0.779518, 0.914382, 0.996359, 0.904565, 0.821205, 0.936537, 0.926851, 0.907287, 0.916814, 0.754686, 0.422209, 0.977708, 0.885583, 0.913659, 0.72229, 0.932748, 0.786656, 0.9677, 0.6759, 0.939519, 0.939619, 0.976682, 0.901609, 0.877578, 0.957474, 0.659596, 0.812173, 0.77898, 0.93857, 0.846139, 0.970211, 0.991469, 0.941998, 0.99228, 0.941368, 0.919001, 0.877389, 0.964053, 0.62892, 0.974418, 0.836646, 0.747614, 0.994126, 0.784671, 0.886119, 0.957653, 0.723802, 0.763465, 0.849859, 0.972564, 0.818141, 0.864318, 0.955251, 0.994557, 0.999883, 0.84261, 0.792689, 0.976037, 0.378906, 0.854057, 0.881663, 0.962214, 0.882605, 0.633129, 0.979769, 0.984534, 0.972508, 0.934276, 0.970884, 0.687226

What is your MLE for this sample? Can you calculate the variance of this estimator; i.e., what is $\operatorname{Var}[\hat \theta]$?

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