When working with a stochastic process based on brownian motion, the increments have normal (gaussian) distribution.

However, it seems that a Laplace distribution, with density:

$$f(t) = \frac{\lambda}{2} e^{-\lambda |t|} \qquad (t \in \mathbb R)$$

would fit much more returns of EUR/USD for example than a normal distribution. (Especially, it has fatter tails than normal distribution, as required).

Here in blue is the density of returns, based on 10 years of historical data of 5-minutes chart of EUR/USD. In green, the density of a Laplace distribution.

enter image description here


Are there some financial models, in which the stochastic process used is:

$$d \, X_t = ... + c \, d \, W_t$$

where $d\, W_t$ has a Laplace distribution instead of a normal distribution?

  • $\begingroup$ Errors in observations are usually either normal or Laplace (Source: Wilmott: FAQ in Quant Finance) $\endgroup$
    – vonjd
    Commented Feb 14, 2016 at 19:26
  • $\begingroup$ @vonjd do you have a link for this topic? $\endgroup$
    – Basj
    Commented Feb 14, 2016 at 20:05
  • $\begingroup$ wilmottwiki.com/wiki/index.php?title=Laplace_distribution $\endgroup$
    – vonjd
    Commented Feb 15, 2016 at 8:27

2 Answers 2


It is very natural to think that why assumption of Normal distribution is made for stochastic process $W_t$ when other more appropriate and valid distribution is available specially for modelling stock price. Before writing answer to your question explicitly, first look at definition of Wiener process:

Wiener Process: A Wiener process $W_t$, relative to family of information set {$\mathscr{F}_t$}, is a stochastic process such that:

  1. $W_t$ is square integrable martingale with $W_0=0$ and $$\mathbb{E}\big[(W_t-W_s)^2 \big]=t-s, \quad s\leq t$$
  2. The path of $W_t$ is continuous over $t$.

The following definition lead to following characteristics of Wiener process:

  • $W_t$ has independent increments because it is martingale.

  • $W_t$ has zero mean and and mean of every increment equals zero.

  • $W_t$ has variance $t$

  • $W_t$ has continuous paths.

Note that, in the above definition nothing is said about the distribution of increments. Normal distribution follows from the assumption stated in the definition. This is known as famous Levy Theorem. If assumptions under Wiener process are satisfied then Levy theorem prove that Wiener increments, $W_t - W_s$, are normally distributed with mean zero and variance $|t-s|$.

In short, for a stochastic process having continuous path and independent increments (natural properties of stock price), normality is not an assumption but gets derived from the basics assumption of Wiener process. And this is the reason no other assumption about the distribution of $dW_t$is made in literature because normality is not an assumption at all.

  • $\begingroup$ Thanks for this point of view. But two remarks : in all documents I have found, gaussian is an assumption (i.e. part of the definition of "Wiener process"), example : here and here. Do you have a reference showing that "normal distribution" is not in the usual definition but follows from other assumptions? $\endgroup$
    – Basj
    Commented Feb 14, 2016 at 19:59
  • $\begingroup$ Second thing @Neeraj : the real historical data for 10 years of EUR/USD shows really how close to Laplace are the increments. Wouldn't make it sense to study a stochastic process for which $d\, W_t \sim Laplace(...)$ ? $\endgroup$
    – Basj
    Commented Feb 14, 2016 at 20:02
  • $\begingroup$ @Basj it is very misleading that various books and website mention Gaussian distribution as assumption of Wiener process. But Neftci in his book "An introduction to mathematics of financial derivative" has explicitly mentioned about this point. I did not find pdf of the book on any site, but if you have physical copy you may read chapter 8 from 3rd edition. $\endgroup$
    – Neeraj
    Commented Feb 14, 2016 at 20:34
  • $\begingroup$ Futher, I have already mentioned in my answer that assumption of Wiener process are natural properties of stock price which are difficult to relax. $\endgroup$
    – Neeraj
    Commented Feb 14, 2016 at 20:48
  • 1
    $\begingroup$ @Basj Your plot of historical data assume each day data point is independent and identically distributed. If sample is large, former assumption can be relaxed but not the later. If each day data point is not identically distributed then your distributional plot of historical data is meaningless. $\endgroup$
    – Neeraj
    Commented Feb 14, 2016 at 21:25

If you're willing to drop the requirement to have continuous paths, or rather, if you're willing to relax it, it is possible to have a bigger class of stochastic processes called Lévy processes. The requirement for it to work is that the probability distribution of your variable is infinitely divisible. The easiest way to formulate this is in terms of the characteristic function


If for any positive integer $n$ the characteristic function $\phi_X(u)$ is the $n$th power of a characteristic function, then we have the infinite divisibility property. For such distributions, it is possible to construct Lévy processes, i.e. a process for which $X_0=0$ and the increments $X_{s+t}-X_s$ have $\phi_X(u)^t$ as their characteristic function.

The Laplace distribution possesses the infinite divisibility property, its characteristic function is


and there are indeed random variables with as characteristic functions powers of this. Those variables possess a variance gamma or generalized Laplace distribution and the Laplace distribution is of course a special case of those distributions.

The associated process is known as a variance gamma process and it was introduced in financial mathematics by Madan, Seneta, Carr and Chang in the 90's.

There is also a jump-diffusion model called the Kou model which has jump sizes which are Laplace distributed.


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