Half-life of mean reversion

mediumQuantacceptance —~25 min

Implement half_life(spread) — the mean-reversion half-life of a spread series. Regress Δs_t = s_t - s_{t-1} on s_{t-1} with an intercept by ordinary least squares to get the slope b. If b < 0 the spread reverts and the half-life is -ln(2) / ln(1 + b); if b >= 0 it does not revert, and there is no half-life — return None.

Half-life tells a pairs trader how long a divergence should take to close. Return it rounded to 4 decimal places.

example 1

input

spread=[1.0, 0.6, 0.3, 0.2, 0.1, 0.05, 0.02]

output

1.3207

example 2

input

spread=[1.0, 1.1, 1.25, 1.4, 1.6, 1.85]

output

None

note

A diverging spread — no half-life.

constraints

4 ≤ len(spread) ≤ 5_000
spread is not constant

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


def half_life(spread: list):
    """Half-life of mean reversion in periods (4 dp), or None if it does not revert."""
    x = spread[:-1]
    y = [b - a for a, b in zip(spread[:-1], spread[1:])]
    n = len(x)
    mx = sum(x) / n
    my = sum(y) / n
    sxx = sum((xi - mx) ** 2 for xi in x)
    sxy = sum((xi - mx) * (yi - my) for xi, yi in zip(x, y))
    b = sxy / sxx
    # TODO: a spread that does not revert (b >= 0) has no half-life
    return round(-log(2) / log(1 + b), 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to half_life()

spread=[1.0, 0.6, 0.3, 0.2, 0.1, 0.05, 0.02]

expected

1.3207

the judge calls half_life() 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