# Maximum Likelihood Estimator of : $f(x) = \theta x^{-2}, \; \; 0< \theta \leq x < \infty$

Exercise :

Find a maximum likelihood estimator of $\theta$ for : $f(x) = \theta x^{-2}, \; \; 0< \theta \leq x < \infty$.

Attempt :

$$L(x;\theta) = \prod_{i=1}^n \theta x^{-2} \mathbb{I}_{[\theta, + \infty)}(x_i) = \theta^n \mathbb{I}_{[\theta, + \infty)}(\min x_i)$$

How should one proceed from now on to find a MLE ?

I think it should be such as :

$$\begin{cases} \theta \; \text{sufficiently large} \\ \min x_i \geq \theta \end{cases} \implies \hat{\theta} = \min x_i$$

Is my approach correct ?

• Are you sure $f(x)$ has a maximum? – qwr May 27 '18 at 6:09
• @qwr Yes, it does. Look at the support of the sample $X$. – Rebellos May 27 '18 at 6:10
• – StubbornAtom Sep 17 at 15:19

$$L(x;\theta) = \prod_{i=1}^n \theta x_{\color{red}i}^{-2} \mathbb{I}_{[\theta, + \infty)}(x_i)$$
If $\theta > x_i$ for any of the $x_i$, then the likelihood dropped to $0$.
Hence we need $\theta \le x_i, \forall i \in \{ 1, \ldots, n\}$ .
Also, note that $\theta_1^n \le \theta_2^n$ if and only if $\theta_1 \le \theta_2$. hence we want $\theta$ to be as big as possible but it needs to be upper bounded by the minimal of $x_i$.
Hence, you are right that $\hat{\theta}=\min_{i \in \{1, \ldots, n\}} x_i$.