# What is a long-tailed distribution for physicists?

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.

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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
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

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)

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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