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

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 $$\beta> 0$$.

1. Write $$\alpha$$ in terms of $$\beta$$ to obtain $$f(x, \beta)$$
2. Find its likelihood function and draw its graph
3. 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 $1=\int_{\beta}^{\infty} \frac{\alpha}{x^2}dx$ ? Jun 12, 2020 at 12:54
• Jun 12, 2020 at 14:34

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

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

• So obvious now. I accepted the answer, thank you. Jun 12, 2020 at 13:04
• I finished the exercise :) Jun 12, 2020 at 13:06