# Prove that exists, and is $\lim_{M\to\infty}\frac{M F'(M)}{1 - F(M)} < \infty$ iff $F(z)$ is a power-law

Standard Definitions Define regularly varying with index $a \in \mathbb{R}$ as $$\lim_{z\to\infty} \frac{L(t x)}{L(x)} = t^{-a}, \forall t>0$$ A slowly varying function is nested, where $a=0$. i.e. Use a standard definition of slowly-varying as $$\lim_{z\to\infty} \frac{L(t x)}{L(x)} = 1, \forall t>0$$ Power Law Define a distribution with a differentiable CDF $F(z)$, counter-CDF $1 - F(z)$ is a power-law with tail index $\alpha > 0$ if $1-F(z)$ is regularly varying with index $\alpha > 0$

(This should be equivalent to other definitions based on decomposition to a slowly varying function)

Proposition: Assume $F(z)$ is a infinitely differentiable CDF defined on $[0,\infty)$

Then, if the following limit is defined, then $\lim_{M\to\infty}\frac{M F'(M)}{1 - F(M)} < \infty$ iff $F(z)$ is a power-law with index $\alpha > 0$

(NOTE: $\impliedby$ direction is certainly true. We are not 100% sure about the other direction. Is there a chance this isn't actually true for some other class of distributions not power-law (e.g. "heavy tailed" but not power-law, or whatever class you want to check). You can check specifics on that limit using L'Hopitals rule to find it seems to hold for anything you can think of such as the Lognormal, etc.)

One more point: The counter-cdf for the Cauchy is slowly-varying. The counter-cdf for normal, log-normal, etc. is not regularly varying (or slowly-varying)

Start of Proof: See http://evm.ivic.gob.ve/LibroSoulier.pdf page 6, corollary 1.9 and http://www.eurandom.tue.nl/reports/1999/013-report.pdf page 11

This says that $l(x)$ is regularly varying with index $\alpha$ if and only if $\lim_{x\to\infty}\frac{x l'(x)}{l(x)} = \alpha$.

If so, can this help prove both directions by taking $l(z) = 1 - F(z)$, the counter-cdf (note that for a power-law distribution, the counter-cdf is regularly varying). This is just a start because showing that it is slowly varying is only part of the problem as we need to show divergence of the limit?

• Isn't this just a modification of your previous question at stats.stackexchange.com/questions/117391/…? If so, we would appreciate you simply editing that question (and deleting this one) so that the comment thread remains intact and only one main thread is devoted to the issue.
– whuber
Oct 1, 2014 at 16:45
• Yes, I just posted a comment to that effect. I think they are different questions though, as I don't use the previous question in my attempted proof here (i.e., not using proof by negation which requires the definition of thin-tailed). Up to you if you want the other one deleted. Oct 1, 2014 at 16:47
• If you are abandoning the other one, then consider closing or deleting it. As far as this question goes, it seems it needs more work. For instance, let $X$ be a mixture of Normal variables with means $n$, SDs $1/n^{2+\alpha}$, and proportions $6/(\pi n)^2$ for $n=1,2,3,\ldots$. Let $L$ be its CDF. Although this function is slowly-varying (like any CDF is), which makes your $F$ a power law, the left hand side of your proposition has no limit (not even a finite lim sup or lim inf). The point is that your power law definition provides no effective control over the derivative of $F$.
– whuber
Oct 1, 2014 at 17:17
• Concerning your link to Soulier, please see his Remark 1.8 (p. 3): it points out the conditions you are omitting (about being in the Zygmund class).
– whuber
Oct 1, 2014 at 18:46
• @whuber Thanks for your comments. For your first one, I believe you may be thinking about the CDF rather than the counter-cdf. Note that my definition of the power-law has the index of regular variation strictly positive. In the case of normal, lognormal, etc., you can show it is slowly varying and $alpha = 0$ Oct 1, 2014 at 21:30