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The Hurst exponent is related to the fractal dimension by noticing that the fractal dimension $D$ is equal to $2-H$, where $d$ is the intrinsic dimension and $H$ is the Hurst exponent, for 1-D fractional Brownian motion.

I can generate an fBm series in Mathematica as follows:

tlow = 1;
thigh = 1000;
tinc = 1;
hurst = 0.4;    
dataz=RandomFunction[FractionalBrownianMotionProcess[hurst], {tlow, thigh, tinc}, 1];

However, using the method described here to compute the fractal dimension, I am unable to get close to recovering the Hurst exponent used to generate the fBm.

For example:

stdev[{x__}] := StandardDeviation[{x}]
stdev[{x_}] := 1
dataf = RotateRight[Mean[R /@ #/stdev /@ #] & /@ (Partition[dataz, 2^#] & /@ Range[10])]
Fit[Log[dataf], {1, x}, x]

Is this a numerical error or are there better ways to compute Hurst exponents and fractal dimension in Mathematica?

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4  
What is R[] ? –  belisarius Aug 31 at 4:08
    
I edited your code for readability. Please undo the edit if you disagree –  belisarius Aug 31 at 4:29
    
I think he uses R as Rescaled Range (The rescaled range is a statistical measure of the variability of a time series...) Rescaled Range. Also see: Hust Exponent. I think I've got a notebook on this I'll try to dig it out of my archive. –  Jagra Aug 31 at 5:17
    
You have some programs in Mathematica to compute fractal dimensions, Hurst coefficient and multi fractals for time series and images in the book Fractal Geography by Andre Dauphine (2013) –  Dauphine Aug 31 at 7:14

1 Answer 1

up vote 11 down vote accepted

Mathematica's estimation routines are able to recover the Hurst exponent from the sample:

BlockRandom[SeedRandom["mathematica.SE/58539"];
  tlow = 1; thigh = 1000; tinc = 1; hurst = 0.4; 
  dataz = RandomFunction[FractionalBrownianMotionProcess[hurst], {tlow, thigh, tinc}, 1]];

FindProcessParameters[dataz, FractionalBrownianMotionProcess[h]]

{h -> 0.397063}

so the issue must be with your code.

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