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#589's option-2 attempt (comment on #589, commit 8659c066) failed for a reason nobody has ever tried to fix: the angular amplitude A4 is barely resolved, not because the offset is large. Across the ten channel×size cells of the #205 block, |A4_z| ranges from 0.06 to 2.18, and the required sample count to bound the Smith contamination scales as 1/A4_z².
The spread between channels is enormous and entirely accidental:
Sp beats M by 4.5× and Dp by ~9,000×, and no design decision produced that — the five channels are just what obs() happens to return. Every large production this repository has commissioned (#205, #583, #581) picked its observable before this figure of merit existed.
The question
Over the archived threshold-rank histograms, and at zero new sampling cost:
which readout of a threshold-rank histogram maximizes |A4| / se(A4) at fixed sample count?
Concretely, the family to scan is at least:
the five existing obs() channels (M, S, D, Sp, Dp) — already measured, they are the baseline;
higher tail derivatives (R''), which the same histogram supports;
linear combinations of the above — the leverage-optimal combination is a generalized-eigenvalue problem against the same delete-one jackknife covariance, so it can be solved rather than searched.
Pre-registration is mandatory here and is the hard part
This is an optimization over observables scored against the same finite sample, so it will find leverage that is partly noise. GOVERNANCE §2D/§2E apply directly:
The scan may only report se(A4) and the amplification relative to M. It must never report delta, rho, a z-score, or any contamination bound for a channel it selected — that would be scoring a number the selection already saw.
The selected channel has to be declared prospectively and then bought with new samples before it may carry any verdict.
The honest deliverable is a design recommendation with a stated selection-optimism penalty, not a measurement.
A useful control: split the batches, select on one half, and report the amplification on the held-out half. If the amplification does not transport, the answer is "the spread is noise" and this issue closes negative — which is itself worth knowing, because it would mean Sp's 4.5× advantage over M is not real either.
What it changes if it works
It multiplies into every open design decision at once:
Why this exists
#589's option-2 attempt (comment on #589, commit
8659c066) failed for a reason nobody has ever tried to fix: the angular amplitudeA4is barely resolved, not because the offset is large. Across the ten channel×size cells of the #205 block,|A4_z|ranges from 0.06 to 2.18, and the required sample count to bound the Smith contamination scales as1/A4_z².The spread between channels is enormous and entirely accidental:
| channel | ×samples to reach the
|A8/A4| < 0.20band ||---|---|
|
Sp@ N=325 | 2.8 ||
Sp@ N=425 | 6.5 ||
S@ N=425 | 11.3 ||
M@ N=325 | 12.5 ||
Dp@ N=325 | 25,601 ||
Dp@ N=425 | 1,476,190 |SpbeatsMby 4.5× andDpby ~9,000×, and no design decision produced that — the five channels are just whatobs()happens to return. Every large production this repository has commissioned (#205, #583, #581) picked its observable before this figure of merit existed.The question
Over the archived threshold-rank histograms, and at zero new sampling cost:
which readout of a threshold-rank histogram maximizes
|A4| / se(A4)at fixed sample count?Concretely, the family to scan is at least:
obs()channels (M,S,D,Sp,Dp) — already measured, they are the baseline;p, not only the frozenp_ref = 0.59274605079.p_refwas frozen for [P2 control/analysis] Same-parent conjugate norm-5 coalescence tomography #205's null; there is no reason it maximizes angular leverage, and the derivative channels in particular will peak somewhere;R''), which the same histogram supports;Pre-registration is mandatory here and is the hard part
This is an optimization over observables scored against the same finite sample, so it will find leverage that is partly noise. GOVERNANCE §2D/§2E apply directly:
se(A4)and the amplification relative toM. It must never reportdelta,rho, a z-score, or any contamination bound for a channel it selected — that would be scoring a number the selection already saw.Sp's 4.5× advantage overMis not real either.What it changes if it works
It multiplies into every open design decision at once:
Blocking relations
results/server-20260829/P205-norm5-conjugate-coalescence/raw/).