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I'm interested in getting the error bounds of the unbiased estimator of the shape parameter ($\alpha$) using maximum likelihood method of Pareto distribution.

The unbiased estimator is known to be $$\frac{n-2}{\sum_i \log(x_i/\min_j(x_j))}.$$ In order to find the error bounds, I would like to get the variance of the above estimator where $n$ is supposed to be the size of the sample.

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    $\begingroup$ You can use $\LaTeX$ markup on this site. For now I latexed your formulas, can you please check if they are correct? $\endgroup$ – kjetil b halvorsen Feb 15 '19 at 12:06
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I will write the standard Pareto distribution with density $$ f(x;\alpha,x_m)=\frac{\alpha x_m^\alpha}{x^{\alpha+1}}\cdot I(x > x_m), $$ for some $\alpha>0, x_m>0$. Then the loglikelihood function can be written $$ \ell(\alpha,x_m)=n\log\alpha + n\alpha\log x_m - (\alpha+1) \sum_i \log x_i $$ (for $x_m<\min_i x_i$, otherwise $-\infty$.)

So the maximum likelihood estimators can be found to be $$ \begin{align} \hat{x}_m &= \min_i x_i \\ \hat{\alpha}&= \frac{n}{\sum_i \log(x_i/\hat{x}_m)} \end{align} $$ but these are not unbiased!

In Barry C. Arnold: Pareto Distributions Second Edition the following results are given, which can be used then to find an unbiased estimator and its variance (for $\alpha$, as asked). First, the exact distributions of the ML estimators can be found, $\hat{x}_m \sim \mathcal{P}(x_m,n \alpha)$ and with $Y_i=\log X_i$ we have $Y_i - \log x_m \sim \Gamma(1,\alpha^{-1})$ (that is, exponential) and we have the following densities $$ \begin{align} f_{\hat{x}_m}(u)&=n\alpha x_m^{n\alpha} u^{-(n\alpha+1)},\quad u>x_m \\ f_\hat{\alpha}(v)&= \frac{(\alpha n)^{n-1}}{\Gamma(n-1)v^n}e^{-(n\alpha/v)},\quad v>0 (\text{Inverse gamma}) \end{align} $$ Then we can find expectation and variances $$ \DeclareMathOperator{\E}{\mathbb{E}}\DeclareMathOperator{\V}{\mathbb{V}} \begin{align} \E \hat{x}_m &= x_m (1-\frac1{n\alpha})^{-1} \\ \V \hat{x}_m&= \alpha n (n-2)^{-1} \\ \E \hat{\alpha} &=\alpha n (n-2)^{-1} \\ \V \hat{\alpha}&= \alpha^2 n^2 (n-2)^{-2} (n-3)^{-1} \end{align} $$ Using this we can modify the ML estimator to find the unbiased estimator $$ \hat{\alpha}_u=\frac{n-2}{\sum_i \log(x_i/\hat{x}_m)} $$ with variance $$ \V \hat{\alpha}_u = \frac{\alpha^2}{n-3}.$$

Addendum: The referenced book states, but does not prove (p 229) that the family of distributions of the maximum likelihood estimators is complete. A consequence of this is that the unbiased estimator above is UMVUE, that is, it has lowest variance among all unbiased estimators. Two papers are referred, paper1 and paper2. I did not find ungated links.

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