Multifractals

The Method of Moments for Planar Data

Suppose the data are a collection of points (x1, y1), ..., (xN, yN) lying in some square region of the plane.
Subdivide the square into smaller squares of side length s1, then into still smaller squares of side length s2, and so on, down to some fairly tiny size sT.
   
Count the number of points that lie in each square of side length s1. Call these n(1,s1), n(2,s1), n(3,s1), ... .
Count the number of points that lie in each square of side length s2. Call these n(1,s2), n(2,s2), n(3,s2), ... .
Repeat the same counting for all the square sizes down to sT.
 
Now compute the moments for a range of q-values. That is, let q range from, say, -20 to 20 in steps of 0.25. For each q value compute the qth moment
M(s1,q) = (n(1,s1)/N)q + (n(2,s1)/N)q + ...
where the sum omits any term with n(i,s1) = 0. (Think of a negative power of 0 and you'll see why.)
Now for each q-value compute the moments M(s2,q), M(s3,q), ... M(sT,q).
 
We hope to discover a power law relation between M(si,q) and (1/si)τ(q), so for each q find the best-fitting line through the points
(log(1/s1), log(M(r1,q))), (log(1/s2), log(M(r2,q))), ... (log(1/sT), log(M(rT,q)))
The slope of this line is τ(q).
 
Finally, to find the f(α) curve, start with α small, say around 0.1. Find the minimum value of
τ(q) + α⋅q
using all the q-values. (Why does this give f(α)?)
Now we have a fairly delicate point. What if we have started with an α-value below the minimum, or continued to an α-value above the maximum? For the moment we adopt an imperfect solution:
If this minimum of τ(q) + α⋅q is negative, ignore that α-value.
Otherwise, the minumum of τ(q) + α⋅q is the f(α) for that α.

Return to Multifractals.