Hurst exponent by R/S

hardQuantacceptance —~40 min

Implement hurst_rs(returns) — the rescaled-range estimate of the Hurst exponent on a single window: subtract the mean from the series, cumulate the deviations, take R = max - min of the cumulative series and S the population standard deviation of the original series, and return H = log(R / S) / log(n).

H near 0.5 is a random walk, above it trends persist, below it they revert — the regime call the workbench's Hurst bot makes. Return H rounded to 4 decimal places.

example 1

input

returns=[0.01, -0.01, 0.02, -0.02, 0.015, -0.015, 0.005, -0.005]

output

0.1822

example 2

input

returns=[0.01, 0.02, 0.015, 0.03, 0.005, 0.01, 0.02, 0.025]

output

0.4168

note

A trending series: every return is positive, so the mean is not zero and it matters.

constraints

8 ≤ len(returns) ≤ 5_000
S > 0

teacher · 2 hints · the first is free

↗ read the theory

the chapter this problem is cut from

▶ open in the workbench

run the same math on a real tape

●solution.py· utf-8 · spaces:4

from math import log
from statistics import pstdev


def hurst_rs(returns: list) -> float:
    """Rescaled-range Hurst exponent on one window, rounded to 4 dp."""
    n = len(returns)
    cum = 0.0
    walk = []
    for r in returns:
        # TODO: subtract the series mean before cumulating
        cum += r
        walk.append(cum)
    R = max(walk) - min(walk)
    S = pstdev(returns)
    return round(log(R / S) / log(n), 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to hurst_rs()

returns=[0.01, -0.01, 0.02, -0.02, 0.015, -0.015, 0.005, -0.005]

expected

0.1822

the judge calls hurst_rs() with exactly these inputs · outputs are compared to 6 decimal places

● standing bypython 3.13 in your browser · pyodide 0.28.3 · judged on the server to 6 dp⌘↵ run · ⌘⇧↵ submit

paper only · educational · not advice