Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

What is the most common definition of long tailed distribution for physicists? I am looking for definition and examples. Examples should have arguments why the distribution is or is not long tailed.

I know that there are a few definitions.

share|improve this question
1  
Interesting. All the times I have worked with "long tailed distributions" the term was used somewhat colloquially for distributions with a divergent mean. Ex: the hopping time distribution for a dispersive continuous time random walk (Scher-Montroll theory). I must confess that I've never bothered to look up the "official" definition. :S –  Michael Brown Apr 9 '13 at 13:03
    
Here's the tome we were all given on the topic: Metzler, R. (2000). The random walk’s guide to anomalous diffusion: a fractional dynamics approach. Physics Reports, 339(1), 1–77. doi:10.1016/S0370-1573(00)00070-3 –  Michael Brown Apr 9 '13 at 13:05
1  
Make a list? I suspect that the multiple scattering angular distribution (which has annoyingly non-Gaussian tails) does not meet the forma definition (for one thing it is not clear what would be meant be taking $\theta \to \infty$), but the tail of the distribution is dominated by rare hard scattering events while the peak is controlled by the action of many weak scattering events. –  dmckee Apr 9 '13 at 13:13
add comment

1 Answer

Ok, I'll start. Definition and list (filled with mistakes). Please, comment/add/edit/etc.

Definition

${lim}_{x\to\infty} P(x) = P(x+t),$ for $\forall\; t > 0$

$P(x)$ is finite i.e. $P(x) = \int p(x) dx = $ const.

Clarifying definitions

$p(x) :=$ "probability density function of measurable $x \in \Re_+$"

$P(x) :=$ "cumulative distribution function" = $\int_{0}^x p(x)\;dx$

$a > 0$

$0 < f(x) < const $

$g(x) \ne 0$

Long tailed

  • all exponential distributions: $p = f(x) exp(-ax)$

  • all power law distributions: $p = x^{-\alpha}$, where $\alpha < -1$

  • all "stretched exponentials": $p = f(x) exp(-ax^{g(x)})$

Non Long tailed

  • Uniform distribution: p(x) = const

  • All "increasing or equal" distributions: $|p(x)| \le |p(x+a)| $ for $\forall a$

  • power law distributions with $p(x) = x^\alpha$, where $-1 \le \alpha \le 0$ (integral not finite)

share|improve this answer
    
That increasing condition in your last line looks weird. Take $a \to \infty$, and you get $p(x) = 0$ for all $x$. –  user1504 Apr 15 '13 at 14:42
    
hmm, Not if $p(x) \to \infty$ when $x \to \infty$. The point was to say that if the distribution was increasing, like p(x) = x, it does not have a long tail. Perhaps a better formulation for this? –  Juha Apr 16 '13 at 11:13
    
Honestly, I wouldn't call a function $p$ a distribution if it isn't integrable. –  user1504 Apr 16 '13 at 13:39
    
@user1504: This is what I was also thinking... This also essentially rules out all power law distributions with exponent from 0 to -1. –  Juha Apr 23 '13 at 11:05
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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