Regime split by VIX median

easyMLacceptance —~15 min

Implement regime(vix, window) — for every day t >= window, flag 1 if the VIX close is STRICTLY above the median of the previous window closes (vix[t - window : t], excluding today), else 0. Return the list of flags, one per day from t = window onward.

The regime model in ai quants routes each day to a high-vol or low-vol specialist on exactly this switch. Use statistics.median.

example 1

input

vix=[15.0, 16.0, 14.0, 17.0, 13.0, 18.0], window=3

output

[1, 0, 1]

example 2

input

vix=[15.0, 16.0, 14.0, 15.0, 13.0, 18.0], window=3

output

[0, 0, 1]

note

Day 3 equals its trailing median exactly — that is a 0.

constraints

window + 1 ≤ len(vix) ≤ 10_000
1 ≤ window ≤ 120

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 statistics import median


def regime(vix: list, window: int) -> list:
    """1 = high-vol day (above the trailing median), 0 = low-vol."""
    out = []
    for t in range(window, len(vix)):
        m = median(vix[t - window:t])
        # TODO: a tie is not a high-vol day — strictly above
        out.append(1 if vix[t] >= m else 0)
    return out

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python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to regime()

vix=[15.0, 16.0, 14.0, 17.0, 13.0, 18.0], window=3

expected

[1, 0, 1]

the judge calls regime() 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