Finding the solution to this MLE problem

Question

Suppose $X_1,X_2,\ldots,X_n$ be a random sample of distribution with probability density function

$$f(x\mid\theta) = \theta x^{\theta-1},\quad 0\lt x \lt 1,\quad 0\lt \theta \lt \infty$$

how can i find MLE of parameter $θ$?

My working is like this:

$$L(\theta\mid x) = \prod f(x\mid\theta) = \prod \theta x^{\theta-1} = \theta^n \prod x ^{\theta-1}$$

Then I got stuck...

"$\theta x\theta^{-1}$" doesn't make a lot of sense. Can you explain what you intended here? — Michael Hardy, Sep 03 '13 at 19:41

score 1 · Accepted Answer · answered Sep 03 '13 at 19:51

1

$$ \prod_{i=1}^n x_i^{\theta-1} = \left( \prod_{i=1}^n x_i \right)^{\theta-1}, $$ so $$ L(\theta) = \theta^n P^{\theta-1}, $$ where $P$ is the product. Therefore $$ \ell(\theta)=\log L(\theta) = n\log\theta + (\theta-1)\log P, $$ so $\ell\,'(\theta) = \dfrac n\theta+\log P$, etc. Can you take it from there?

answered Sep 03 '13 at 19:51

Michael Hardy

1

can explain why the $\theta-1 $ was taken out? – lakshmen Sep 03 '13 at 20:04
Because $\log\left(P^{\theta-1}\right) = (\theta-1)\log P$, and differentiation is simple when it's in that form. – Michael Hardy Sep 03 '13 at 20:06
i mean the earlier one.. at the top... – lakshmen Sep 03 '13 at 20:14
Just so that the number of times the expression $\theta-1$ appears would be only one. And for some other purposes, if not for the present problem, it is of interest to know that this function depends on the data only through the product of the $x$ values. – Michael Hardy Sep 03 '13 at 20:28
do u know how to show that it is consistent estimator for this as well? – lakshmen Sep 08 '13 at 12:02

Finding the solution to this MLE problem

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