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The question is just like the title. But...$\alpha$-stable distribution (for $\alpha\in (1,2)$) does not have the second moment, so the sample mean doesn't have variance well defined. Then for such a case, it seems sample mean is clearly not efficient?

Is it meaningful to talk about efficiency for the heavy-tailed distribution like $\alpha$-stable distributions? If yes, and if the sample mean is not an efficient estimator of mean, then what it is? (I even don't know how Fisher information is derived for $\alpha$-stable distribution.

Could anyone share an idea or some reference in terms of this topic?

kjetil b halvorsen
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    For distributions without finite variance, the sample mean does not converge to the population mean for large $n$. So that is not a reasonable estimator at all. Finding an optimal estimator is still an active research topic. See for example this recent paper https://arxiv.org/pdf/2011.12433.pdf . This is not an easy problem. Good luck :) – Simon Jun 16 '21 at 23:09
  • https://stats.stackexchange.com/questions/47310/weak-law-of-large-numbers-redundant – Christoph Hanck Jun 17 '21 at 16:51

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