Take the 2-minute tour ×

If we suppose an instrument goes up or down 1 tick per $\Delta t$ (binary model), its long term distribution will be normal, per the Central Limit Theorem.

However, suppose we model as follows:

  • The first tick is up or down with 50% probability.

  • Every future tick is: 60% likely to be in the same direction as the previous tick, 40% likely to be in the opposite direction.

The Central Limit Theorem doesn't apply here, because these are no longer independent random variables.

However, as this post shows us, the resulting distribution is still normal.

My question: can I construct a binary model that yields a non-normal distribution, ideally a "fat tailed" distribution?

share|improve this question
Why don't you accept any answers? –  strimp099 Nov 4 '11 at 20:25
You can. One simple way is to have the tick size vary over time. Equivalently, time between ticks could be time varying, then distribution measured in ordinary time will have fat tails. –  Tal Fishman Nov 4 '11 at 21:09
Tal: what's the precise statement? barrycarter: if you're asking, are there simple fat-tailed distrtibutions with a large basin of attraction, that appear in finance, then the answer is probably, no. –  LazyCat Nov 6 '11 at 2:37
you would have to violate the conditions of the Central Limit Theorem. For practical purposes this means having a non-finite variance to the binary jumps. –  shabbychef Nov 8 '11 at 6:48

1 Answer 1

Yes you can.

Begin by choosing your favorite stochastic differential equation with a fat-tailed terminal distribution, for example a local volatility model. Use the usual techniques to convert to a partial differential equation (PDE).

Construct an explicit finite difference scheme for solving the PDE, and make your number $M$ of grid points in time $\tau$ sufficiently large compared to the number of grid points $N$ in asset price. Specifically, you need $$ N\geq 2M. $$ Your choice of boundary conditions has no influence on the central node since the grid is so wide.

This explicit scheme is therefore trivially equivalent to a trinomial tree. Now, noting that one step of a trinomial tree can be constructed from two steps of a binomial tree, insert those binomial half-steps to obtain a $2M$ step binomial tree whose terminal distibution matches that of your SDE.

share|improve this answer

Your Answer


