4
$\begingroup$

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%?

Update:

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?

$\endgroup$
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.