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 ?


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

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 $\hat{\theta}=\min_{i \in \{1, \ldots, n\}} x_i$.

  • $\begingroup$ Hi and thanks for the clarification regarding the correctness of my answer. If one wanted to proceed by finding an estimator with the method of moments, how should he proceed ? $\endgroup$ – Rebellos May 27 '18 at 6:25
  • $\begingroup$ start by finding the first moment. $\endgroup$ – Siong Thye Goh May 27 '18 at 7:09

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