forked from mnbvc188199/Warrior_EA
f55dbe995c
2026-08-15, ~6 minutes after attach: PAI-8fea (fractal target, 37 features, D1) converged at era 299 and the plateau deploy CLEARED the family-wise gate for the first time in project history: dir-precision 73.1% vs 63% break-even, +10.4pp on 350 test calls = 4.04 sigma, p_family = 0.0081. This script asks the first two hostile questions offline: - DRIFT: always-long at the same 2.64/1.66 geometry scores 64.1% on the last 15% of D1 history (66.7% on 30%) - drift alone clears BE by ~1-3pp, but the model is +9pp above ALWAYS-LONG, so the pass is selection, not drift. - SWAP EXPOSURE: median 7-8 bars to the long target = ~10 nights of financing ~ 0.15-0.2% notional vs a ~1.7% target -> a ~1-1.5pp BE haircut against a +10.4pp margin. Survives. Remaining before belief: replication on other D1 symbols, and closing the live-semantics gap (certified wins assume hold-to-barrier; live exit paths can cut on vote flips). Co-Authored-By: Claude Fable 5 <noreply@anthropic.com>
82 lines
3 KiB
Python
82 lines
3 KiB
Python
"""Drift null for the SP500 D1 fractal-model gate pass (2026-08-15).
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The deployed PAI cleared the family-wise gate: dir-precision 73.1% vs BE ~63% (+10.4pp,
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4.04 sigma, p=0.0081, 350 test calls) - long-heavy calls (Buy recall ~75-80%). Question:
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what does ALWAYS-LONG (and always-short) score at the same barrier geometry (stop 2.64*ATR,
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target 1.66*ATR, horizon 48 D1 bars, spread charged, intrabar ties to the stop) on the
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recent slices of D1 history? If always-long ~= the model, the pass is drift; if it sits at
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the BE, the selection is real. Also reports median bars-to-resolution (the swap exposure).
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"""
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import numpy as np
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import sys
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sys.stdout.reconfigure(encoding="utf-8", errors="replace")
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import book
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import fills
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SL_M, TP_M, HORIZON = 2.64, 1.66, 48
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def atr14(h, l, c):
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pc = np.roll(c, 1)
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pc[0] = c[0]
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tr = np.maximum(h - l, np.maximum(np.abs(h - pc), np.abs(l - pc)))
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a = np.empty_like(tr)
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a[:14] = tr[:14].mean()
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for i in range(14, len(tr)):
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a[i] = (a[i - 1] * 13 + tr[i]) / 14.0
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return a
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def main():
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bk = fills.Book("SP500")
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f = book.frame("SP500", "D1", bk)
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a = atr14(f.h, f.l, f.c)
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sp = float(np.nanmean(f.spread))
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n = f.n
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res = {}
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for name, lo_frac, hi_frac in (("last 15%", 0.85, 1.0), ("last 30%", 0.70, 1.0),
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("full", 0.05, 1.0)):
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i0, i1 = int(n * lo_frac), int(n * hi_frac) - HORIZON - 1
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wl = ws = nl = 0
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holds = []
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for i in range(max(i0, 20), i1):
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entry = f.c[i]
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risk, rew = SL_M * a[i], TP_M * a[i]
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ltp, lsl = entry + sp + rew, entry + sp - risk
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stp, ssl = entry - rew - sp, entry + risk - sp
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lw = ll = sw = ss = False
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lat = sat = -1
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for t in range(i + 1, min(i + 1 + HORIZON, n)):
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if not ll and not lw:
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if f.l[t] <= lsl:
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ll = True
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elif f.h[t] >= ltp:
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lw = True
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lat = t - i
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if not ss and not sw:
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if f.h[t] >= ssl:
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ss = True
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elif f.l[t] <= stp:
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sw = True
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sat = t - i
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if (ll or lw) and (ss or sw):
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break
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nl += 1
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if lw:
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wl += 1
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holds.append(lat)
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if sw:
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ws += 1
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res[name] = (100.0 * wl / max(nl, 1), 100.0 * ws / max(nl, 1), nl,
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np.median(holds) if holds else 0)
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be = 100.0 * SL_M / (SL_M + TP_M)
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print(f"SP500 D1, geometry {SL_M}/{TP_M}, horizon {HORIZON} bars | zero-cost BE {be:.1f}%"
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f" | spread {sp:.2f} pts")
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for name, (l, s, cnt, medhold) in res.items():
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print(f" {name:<9}: always-LONG win {l:.1f}% | always-SHORT {s:.1f}% | n={cnt} | "
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f"median bars to long target {medhold:.0f}")
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if __name__ == "__main__":
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main()
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