Finding MLE estimator for given density $f(x, \alpha, \beta)$

Question

I'm having trouble with the following example problem of MLE:

Let $X = (X_1, ..., X_n)$ be a trial from i.i.d r.v. with density: $$ g(x) = \frac{\alpha}{x^2}\mathbb{1}_{[\beta, \infty)}(x) $$ where $\beta> 0$.

Write $\alpha$ in terms of $\beta$ to obtain $f(x, \beta)$

Find its likelihood function and draw its graph

Using above result get MLE estimator of $\beta$

Could anyone give me a hint on the first task? I'm banging my head against the wall but can't see how $\alpha$ may be written only in terms of $\beta$.

I derived $L$ as $$ L(\textbf{x}, \alpha, \beta) = \frac{\alpha^n}{\prod_\limits{i=1}^n x_i^2}\mathbb{1}_{[\beta, \infty)(X(1))} $$ to maybe find some clues there but without any meaningful effect.

Have you tried $1=\int_{\beta}^{\infty} \frac{\alpha}{x^2}dx$ ? — Philipp123, Jun 12 '20 at 12:54
https://math.stackexchange.com/q/2659520/321264, https://math.stackexchange.com/q/2949033/321264 — StubbornAtom, Jun 12 '20 at 14:34

tommik · Accepted Answer · 2020-06-12T13:06:18.580

1

It is enough to set

$$\int_{\beta}^{+\infty}\frac{\alpha}{x^2}dx=1$$

to find $\alpha=\beta$

Point 3) observing that the domain depends on the parameter, the likellihood is

$$L(\beta) \propto \beta^n\mathbb{1}_{(0;x_{(1)}]}(\beta)$$

it is self evident that the likelihood is strictly increasing so the MLE estimator is

$\hat{\beta}=x_{(1)}=min(x)$

edited Jun 12 '20 at 13:06

answered Jun 12 '20 at 13:00

tommik

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So obvious now. I accepted the answer, thank you. – blahblah Jun 12 '20 at 13:04
I finished the exercise :) – tommik Jun 12 '20 at 13:06

Finding MLE estimator for given density $f(x, \alpha, \beta)$

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