What processes could generate Laplace-distributed (double exponential) data or parameters?

Lots of distributions have "origin myths", or examples of physical processes that they describe well:

• You can get normally distributed data from sums of uncorrelated errors via the Central Limit Theorem
• You can get binomially distributed data from independent coin flips, or Poisson-distributed variables from a limit of that process
• You can get exponentially distributed data from waiting times under a constant decay rate.

And so on.

But what about the Laplace distribution? It's useful for L1 regularization and LAD regression, but it's hard for me to think of a situation where one should actually expect to see it in nature. Diffusion would be Gaussian, and all the examples I can think of with exponential distributions (e.g. waiting times) involve non-negative values.

• – StubbornAtom Jan 4 '18 at 19:07

• If $X_1$ and $X_2$ are IID exponential distributions, $X_1 - X_2$ has a Laplace distribution.

• If $X_1, X_2, X_3, X_4$ are IID standard normal distributions, $X_1X_4 - X_2X_3$ has a standard Laplace distribution. So, the determinant of a random $2\times 2$ matrix with IID standard normal entries $\begin{pmatrix}X_1 & X_2 \\\ X_3 & X_4 \end{pmatrix}$ has a Laplace distribution.

• If $X_1, X_2$ are IID uniform on $[0,1]$, then $\log \frac{X_1}{X_2}$ has a standard Laplace distribution.

• +1 It may be worth noticing that all three examples are really the same: #2 can be rewritten as $((X_1+X_4)^2 + (X_2+X_3)^2 - [(X_1-X_4)^2 + (X_2-X_3)^2])/4$, a scaled difference of two scaled $\chi^2(2)$ (Exponential) distributions, and #3 is the difference of two Exponential distributions because the $\log(X_i)$ are Exponential. – whuber Sep 24 '12 at 20:59
• @whuber: Thanks for that explanation for why the determinant was the same as the others! I was surprised to see it, since I would have guessed that the density of the determinant would vary smoothly, as it does everywhere except $0$. – Douglas Zare Sep 24 '12 at 21:03
• So I'm trying to think of a "story" that would fit any of the examples on wikipedia. Say I'm playing pinball with my equally lousy brother. Each game we play one ball each. Roughly any given moment there is an equal chance that I (or he) will lose a ball and the score is basically a linear function of for how long I play. Then my score (and his) could be modelled by an exponential distribution and the difference between me and my brother's score each round will be Laplace distributed. Sort of works? – Rasmus Bååth Jan 13 '14 at 22:16

Define a compound geometric distribution as the sum of $$N_p$$ iid random variables $$X_N = \sum_i^{N_p} X_i$$, where $$N_p$$ is distributed like a geometric distribution with parameter $$p$$. Assume that the iid random variables $$X_i$$ have finite mean $$\mu$$ and variance $$v$$.

It was shown by Gnedenko that in the limit $$p\to 0$$, the compound geometric distribution approaches a Laplace distribution.

$$Y:= \lim_{p\to 0} \sqrt{p} (X_N - N_p\mu) = Laplace(0,\sqrt{\frac{v}{2}})$$

The density of the $$Laplace(a,b)$$ is $$\phi(x) = \frac{1}{2b} \exp\left( - \frac{|x-a|}{2b}\right)$$

B.V Gnedenko, Limit theorems for Sums of random number of positive independent random variables, Proc. 6th Berkeley Syposium Math. Stat. Probabil. 2, 537-549, 1970.