# Variance of unbiased estimator for the shape parameter of Pareto distribution

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.

• You can use $\LaTeX$ markup on this site. For now I latexed your formulas, can you please check if they are correct? – kjetil b halvorsen Feb 15 '19 at 12:06

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 $$\V \hat{\alpha}_u = \frac{\alpha^2}{n-3}.$$