Let $X_{1},\dots,X_{n}$ be a random sample from a Pareto distribution with pdf $f(x,\theta)=\frac{2\theta^{2}}{x^{3}}$, $x\geq\theta$; $0$ otherwise. Here, $\theta\ (>0)$ is a parameter.

Find the MLE $\hat{\theta}$ of $\theta$.

We have $$L(\theta|\mathbf{x})=\prod_{i=1}^{n}\frac{2\theta^{2}}{x_{i}^{3}}=\frac{2^{n}\theta^{2n}}{\prod_{i=1}^{n}x_{i}^{3}}$$ for $x_{i}\geq\theta\iff\min_{i}x_{i}\geq\theta$. Finding the log-likelihood yields $$\ell(\theta,\mathbf{x})=n\log 2+2n\log \theta-\sum\log x_{i}^{3}$$

This does not give me the correct MLE (at least I don't think it does).

I have a feeling that we need to use some sort of indicator function in the likelihood, but I am not sure what.

  • 1
    $\begingroup$ I think the MLE is $\hat{\theta}=x_{(1)}$, since the largest $\theta$ can be is the $\min x_{i}$. I might be wrong, but I am pretty sure the MLE will be some order statistic. $\endgroup$ – Kirk Fogg Jul 17 '13 at 22:29
  • $\begingroup$ I guess your issue is setting the derivative to zero. Remember, that works only for maxima that are in the interior of the domain. $\endgroup$ – S.B. Jul 17 '13 at 23:05

As you stated, you have to maximize the log likelihood over $(0,\min_i x_i]$. Since the $\log$ is a strictly increasing function, your answer is simply $\theta_\text{MLE}=\min_i x_i$.


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