Multifractal Width Lite

Problem

Estimate multifractal width as the gap between two scaling exponents. Let d1 be absolute one-step changes and d2 absolute two-step changes; compute H1 = log2(mean(d2) / mean(d1)) and H2 = log2(rms(d2) / rms(d1)); return abs(H1 - H2).

Return 0.0 when there are fewer than three points or a denominator is zero.

Hint

A monofractal signal gives nearly equal exponents, so the width is near zero.

Test cases

case 1

in: [[1,2,3,4,5]]

expected: 0

case 2

in: [[1,1,1,2,10]]

expected: 0.19192974728965728

case 3

in: [[1,2,4,8,16]]

expected: 0.11651983276797817

case 4

in: [[3,3,3,3]]

expected: 0

Starter codePython

import math
def multifractal_width_lite(series):
    # Your code here
    pass

4 test cases · run your code against them in the editor