For many basic random processes (like fractional Brownian motion) Hurst exponent "H" and fractal dimensions ( Hausdorf = Minkoswki typically) "D" are realated by simple formula D = 2-H. I want to clarify what are exact statemetns known about this relationships.

Consider some function y = f(t) (not a random process, but just function). Assume that fractal dimensions of its graph equals to "D".

Is it true that Hurst exponent is somehow correctly defined and equals to H=2-D ?

There possibly can be some subtleties defining the Hurst exponent. Let me try to give the following precise definition. Assume function f(t) is defined on the interval [0, 1]. Let us look on the values of function at points t=k/n, k=0...n. So we obtain discrete time series f_0, ...,f_n. And use the standard way of calculating the rescaled range $(R/S)_{n}$, and existence of the Hurst exponent means that $(R/S)_{n}$ ~ $C n^H$. I am not quite sure this definition is the correct one, but I think it can be suitable modified.

  • $\begingroup$ Do you want your function to be continuous? $\endgroup$
    – Goldstern
    Oct 19 '13 at 23:19
  • $\begingroup$ @Goldstern Yes, I mean continuous. $\endgroup$ Oct 20 '13 at 16:10

In principle, fractal dimension and Hurst exponent are independent of each other: fractal dimension is a local property, while the long-memory dependence characterized by the Hurst exponent is a global characteristic. For self-affine processes, the local properties are reflected in the global ones, resulting in the celebrated relationship $D+H=n+1$ between fractal dimension $D$ and Hurst exponent $H$ for a self-affine surface in $n$-dimensional space.

More generally, local and global behavior are decoupled. Stochastic models of the socalled Cauchy class separate fractal dimension and Hurst exponent and allow for any combination of the two parameters, without any linear relation.

Stochastic Models That Separate Fractal Dimension and Hurst Effect, Tilmann Gneiting and Martin Schlather (2001).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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