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Let $X_1, ..., X_n$ be a random sample from a continuous distribution with pdf $$f_{\theta,\kappa}(x) = \frac{\kappa\theta^\kappa}{x^{\kappa+1}}, x\geq \theta, \theta > 0, \kappa > 0.$$

How to find the MLE of the pair $(\theta,\kappa)$? I tried to first define the likelihood function: $$f_{\theta,\kappa}(x_1, ..., x_n)= (\kappa\theta^\kappa)^n\Pi_{i=1}^{n}\frac{1}{x_i^{\kappa+1}}$$

Then take the log of the likelihood function:

$$l(\theta,\kappa) = n log(\kappa) + nklog(\theta)+(-\kappa-1)\Sigma_{i=1}^{n}log(X_i)$$

Fix $\kappa$, and set the derivative of $l(\theta, \kappa)$ with respect to $\theta$ to zero, this results in:

$$\frac{n\kappa}{\theta} = 0$$

This doesn't seem right, how can I find the MLEs?

Jack
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    The trouble is that you forgot the indicator function. Consider $$m=\inf\limits_{1\leqslant i\leqslant n}x_i\qquad p=\prod\limits_{i=1}^nx_i,$$ then the log-likelihood is actually $L=n\log\kappa+n\kappa\log\theta-(\kappa+1)\log p$ when $\theta<m$, and is $-\infty$ otherwise. From there one can check (using the fact that $\theta<m$ implies $p>\theta^n$) that the MLE of $(\theta,\kappa)$ is $(\hat\theta,\hat\kappa)$ with $$\hat\theta=m\qquad\hat\kappa=\frac{n}{\log p-n\log m}$$ – Did Jun 19 '15 at 11:13
  • @Did, what is this $p$ here? – Indominus Jun 19 '15 at 11:22
  • @Indominus Somebody should read more carefully others' comments. – Did Jun 19 '15 at 11:22
  • @Did, sorry. I now see why the indicator matters now. – Indominus Jun 19 '15 at 11:30

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