I have a random variable $x \in (0,\infty)$ with distribution $P(x)$ falling off slowly $P(x) \sim 1/x^3$ for large $x$. So the expectation value $\bar{x}$ is finite but the second moment $\bar{x^2}$ is divergent.

If I remember correctly this class of distributions also falls into the Gaussian universality class, i.e. the distribution of the mean $\bar{x} = \frac{1}{N} \sum_{i=1}^{N} x_i $ will be Gaussian for large $N$ where of course all $x_i$ are distributed according to $P(x)$. If the tail of $P(x)$ is $1/x^{\alpha+1}$ then (again if I remember correctly) for $1 < \alpha < 2$ we have the $\alpha$-stable Levy distributions for the central limit distribution, but for $\alpha = 2$ it is already Gaussian. Obviously for $\alpha > 2$ it will definitely be Gaussian since both first and second moments exist in this case.

Now what I'm not sure about is whether if I have the $x_i$ "measurements" and estimate $\bar{x}$ by

$$\bar{x} = \frac{1}{N} \sum_{i=1}^N x_i$$

then can I use the usual $\sigma_N / \sqrt{N}$ formula for the "error" where

$$\sigma_N^2 = \frac{1}{N-1} \sum_{i=1}^N \left( x_i - \bar{x} \right)^2$$

i.e. will it be true that if $N$ is large enough then the estimated $\bar{x}$ will be within $\sigma_N / \sqrt{N}$ of the true mean with probability 68%?

On the one hand since for large $N$ the central limit theorem will give me a Gaussian where $1-\sigma$ deviation is certainly 68% so I'd think the above is correct but on the other hand $\sigma_N$ is divergent for all finite $N$. But I guess $\sigma_N$ is only divergent as $\sim \log(N)$ so $\sigma_N / \sqrt{N}$ still goes to zero as it should. But I'm still kinda confused.

If the above doesn't make any sense, how would you determine the interval around the estimated mean $\bar{x}$ such that the true mean is within that with probability 68%?


Is it known what the first correction is to the $\sigma_N / \sqrt{N} \sim \sqrt{\log(N)/N}$ form? Is it only logarithmically small like

$$\sqrt{\log(N)/N} \left( 1 + const / \log(N) + \ldots \right)$$

or faster?


1 Answer 1


Indeed: For $P(x)\propto 1/x^3$ the estimated $\bar{x}$ will deviate from the true mean by an amount that decays with increasing $N$ as $(\log N/N)^{1/2}$, so only slightly less rapidly than the $1/\sqrt N$ decay expected from the central limit theorem. See chapter 2, Elementary introduction to the theory of stable laws, in Chance and Stability: Stable Distributions and their Applications.


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