Linear-regression channel

mediumQuantacceptance —~20 min

Implement linreg_channel(values, k) — a linear-regression channel over the whole series: fit y = a + b·t for t = 0..n-1 by ordinary least squares, take the population standard deviation of the RESIDUALS y - (a + b·t), and return the channel at the last bar: [slope, fitted_last - k·std, fitted_last + k·std].

It is the channel the workbench's regression bot draws and the /pro chart lets you drag. Return the three values rounded to 4 decimal places.

example 1

input

values=[100.0, 102.0, 98.0, 98.0, 102.0, 100.0], k=2.0

output

[0, 96.734, 103.266]

example 2

input

values=[100.0, 101.0, 103.0, 102.0, 105.0, 106.0], k=2.0

output

[1.1714, 104.391, 107.1328]

constraints

3 ≤ len(values) ≤ 5_000
0.5 ≤ k ≤ 4

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 pstdev


def linreg_channel(values: list, k: float) -> list:
    """[slope, lower, upper] at the last bar, rounded to 4 dp."""
    n = len(values)
    t = list(range(n))
    mt = sum(t) / n
    my = sum(values) / n
    stt = sum((ti - mt) ** 2 for ti in t)
    sty = sum((ti - mt) * (yi - my) for ti, yi in zip(t, values))
    b = sty / stt
    a = my - b * mt
    # TODO: the channel width is the scatter around the FITTED line, not around the mean
    s = pstdev(values)
    last = a + b * (n - 1)
    return [round(b, 4), round(last - k * s, 4), round(last + k * s, 4)]

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to linreg_channel()

values=[100.0, 102.0, 98.0, 98.0, 102.0, 100.0], k=2.0

expected

[0, 96.734, 103.266]

the judge calls linreg_channel() 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