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Find structural breaks when you don't know the dates

If you already know the date, tsecon.chow_test is your answer. You usually do not — and searching for the break yourself invalidates the standard critical values, because you have implicitly run one test per candidate date. Two functions handle the search honestly: sup_f_test for a single break at an unknown date, and bai_perron when there may be several.

Step 1: is there a break at all?

import numpy as np, tsecon

rng = np.random.default_rng(4)
n = 300
mu = np.repeat([0.0, 2.0, -1.0], 100)          # true mean shifts after t=99 and t=199
y = mu + rng.standard_normal(n)
X = np.ones((n, 1))                            # intercept-only: a pure mean-shift scan

s = tsecon.sup_f_test(y, X)
print(f"sup-F {s['stat']:.2f}   p {s['p_value']:.4f}   most likely break at t={s['break_date']}")

sup-F 122.66   p 0.0000   most likely break at t=199

The Andrews (1993) sup-F statistic is the largest Chow statistic over all candidate dates in the trimmed interior; the p-value comes from Hansen's (1997) response surfaces and already accounts for the search. f_path holds the whole scan if you want to plot it.

Note what it reports: one break, at t=199. There are two. A single-break test finds the most prominent one, which is why a rejection is a reason to keep going, not to stop.

Step 2: how many, when, and how sure?

bp = tsecon.bai_perron(y, X, max_breaks=5)
print("breaks selected:", bp["n_breaks"], " dates:", np.asarray(bp["break_dates"]).tolist())
for k, d in enumerate(bp["break_dates"]):
    print(f"  break {k+1} at t={d}   95% CI [{bp['ci_lower_95'][k]}, {bp['ci_upper_95'][k]}]")
for k in range(bp["n_breaks"] + 1):
    print(f"  regime {k+1}: t={bp['regime_starts'][k]}..{bp['regime_ends'][k]}"
          f"   mean {bp['params'][k][0]:+.3f} (se {bp['bse'][k][0]:.3f})")

breaks selected: 2  dates: [99, 199]
  break 1 at t=99   95% CI [95, 103]
  break 2 at t=199   95% CI [196, 202]
  regime 1: t=0..99   mean -0.063 (se 0.102)
  regime 2: t=100..199   mean +2.159 (se 0.097)
  regime 3: t=200..299   mean -0.790 (se 0.102)

Both true dates recovered, with Bai (1997) confidence intervals a few observations wide, and per-regime OLS estimates that reproduce the true means (0, 2, -1). Report the break-date confidence intervals. A break date is an estimate like any other; quoting it as a fact is the most common way this literature is misread.

How the count was chosen

print("sequential supF(l+1|l):", np.round(bp["sup_f_seq"], 2))
print("5% critical values   :", np.round(bp["sup_f_crit"], 2))

sequential supF(l+1|l): [122.66 248.82   1.45   1.11   0.38]
5% critical values   : [ 8.58 10.13 11.14 11.83 12.25]

The sequential procedure tests l+1 breaks against l. The first two statistics blow past their critical values; the third (1.45 against 11.14) does not, so the procedure stops at two. break_dates_by_m carries the optimal partition for every m, so you can see what a three- or four-break model would have claimed before deciding.

Gotchas

  • x is the design whose coefficients all switch, and you supply the constant. np.ones((n, 1)) scans for mean shifts; add columns to let slopes break too. Every coefficient in x switches at every break — there is no partial-break option.
  • Autocorrelation inflates sup-F. The scan assumes serially uncorrelated errors under the null. On a persistent series, a rejection may be persistence, not a break.
  • An unremoved trend looks like a cascade of mean shifts. Detrend, or put the trend in x.
  • trim=0.15 leaves 15% of the sample at each end unsearchable — breaks near the endpoints cannot be found, by construction.
  • Breaks and unit roots are confusable in both directions: a broken-mean stationary series fools ADF (Perron 1989), and a random walk fools a break scan. Run the stationarity workflow alongside this one and reconcile the two.

See also