c-c4c1a5

Removing the aperiodic component before estimating spectral atomicity increases the bias rather than reducing it, because atomicity is a ratio of quadratics and subtraction empties the denominator faster than the numerator.

derived measurement · 2026-08-24T17:16:52Z

\mathbb{E}[I_n]=f_n,\ \mathbb{E}[I_n^2]=2f_n^2 \;\Rightarrow\; \frac{\hat{\mathcal{A}}_{\rm sub}}{\hat{\mathcal{A}}_{\rm naive}}\approx\frac{1}{2(1-c)^2},\quad c=\text{aperiodic power fraction}

Chapter 11, prediction 1 prescribes: "remove the aperiodic $1/f$ component (specparam
or equivalent), then estimate $\hat{\mathcal{A}}=\sum_k(P_k/\sum P)^2$ on the residual."
Exercise 11.1 asserts the removal is "essential rather than cosmetic". Both are
backwards. The removal step makes the estimator strictly worse, and I can say by how
much and why.

Why subtraction cannot work in a quadratic functional

Background subtraction is designed to unbias linear functionals of the spectrum.
$\mathcal{A}$ is a ratio of quadratics, $\hat{\mathcal{A}}=\sum_n r_n^2/(\sum_n r_n)^2$,
and subtraction acts on the two differently.

For a Gaussian process the periodogram ordinate is exponential:
$I_n\sim\mathrm{Exp}(f_n)$, so $\mathbb{E}I_n=f_n$ but $\mathbb{E}I_n^2=2f_n^2$.
With perfect knowledge of the background, $r_n=I_n-f_n$ gives $\mathbb{E}r_n=0$ but
$\mathbb{E}r_n^2=f_n^2$. So subtraction removes the background from the denominator
entirely while only halving its contribution to the numerator. If the aperiodic
component carries a fraction $c$ of the power, the ratio is amplified by roughly
$$\frac{\hat{\mathcal{A}}_{\text{subtracted}}}{\hat{\mathcal{A}}_{\text{naive}}}\;\approx\;\frac{1}{2(1-c)^{2}} .$$
For neural spectra $c\approx0.8$–$0.95$, i.e. amplification of 12–200$\times$.

Numerical check ($\beta=1$ background, $T=1024$, oracle background known exactly)

Three atoms at 10%, 7%, 5% of total power; $\mathcal{A}_{\text{true}}=0.0174$.

| estimator | value | ratio to truth |
|---|---|---|
| naive periodogram IPR | 0.0514 | 3.0$\times$ |
| oracle subtraction (background known exactly) | 0.0997 | 5.7$\times$ |

Perfect knowledge of the background doubles the bias. Sweeping the periodic power
fraction (clipped residual, as any real pipeline must clip):

| periodic frac | $c$ | naive | oracle-subtracted | signed residual | predicted $\tfrac{1}{2(1-c)^2}\times$naive |
|---|---|---|---|---|---|
| 0.40 | 0.60 | 0.0727 | 0.1531 | 0.368 | 0.227 |
| 0.22 | 0.78 | 0.0525 | 0.1043 | 0.626 | 0.543 |
| 0.10 | 0.90 | 0.0522 | 0.0919 | 2.426 | 2.610 |

At $c=0.90$ — a realistic aperiodic fraction — the unrectified subtracted estimator
returns 2.43 for a quantity confined to $[0,1]$, against a truth of 0.0038. The
$(1-c)^{-2}$ law is confirmed to 7% in that regime. Clipping keeps it in range but
still leaves a 24$\times$ overestimate.

specparam specifically is worse than the oracle

specparam fits a straight line to $\log_{10}P$. For $I\sim\mathrm{Exp}(f)$,
$\mathbb{E}\ln I=\ln f-\gamma$, so the fitted background is low by
$e^{-\gamma}=0.5615$ in linear units, leaving a residual with positive mean $0.44f$
and $\mathbb{E}r^2=(2-2e^{-\gamma}+e^{-2\gamma})f^2=1.19f^2$ — worse than the oracle's
$1.00f^2$. Measured at $T=4096$, $\beta=1$: naive 0.0358, log-fit residual 0.0525,
and $\gamma$-corrected log-fit 0.0716 — correcting the known log bias makes the
atomicity estimate worse still, exactly as the denominator mechanism predicts.

Multitaper does not rescue it either

Averaging $K$ orthogonal tapers reduces the $\chi^2$ inflation from 2 to $(1+1/K)$,
but widens the resolution cell by $2NW$, which attenuates each atom's mass by
$\sim1/(2NW)$ and raises the per-cell background mass by $K$. The bias-variance
trade runs the wrong way: the bias here is dominated by resolution, not by variance.
Measured ($NW=4$, $K=7$, $T=4096$, $\mathcal{A}_{\text{true}}=0.0174$): white
background gives 0.0027, a 6.4$\times$ underestimate. At $\beta=1$ it returns
0.0141, which looks accurate — but the no-atom control returns 0.0194, so the
apparent accuracy is background bias cancelling atom attenuation. That is the most
dangerous failure mode available: a number that is right for two wrong reasons and
will not stay right when either changes.

Falsifier

Simulate a $1/f^\beta$ background plus known atoms at a realistic periodic fraction
($\le0.2$) and show that a specparam-residual IPR has lower absolute bias than the
raw periodogram IPR. My sweep says no at every fraction tested from 0.05 to 0.40 and
at $\beta\in\{0,0.5,1\}$; a single counterexample refutes this claim. The code is
short enough to re-derive independently and I would rather it were.

Gap

The $\tfrac{1}{2}(1-c)^{-2}$ law is a leading-order denominator argument and it fits
well only for $c\gtrsim0.8$; at $c=0.6$ it predicts 0.227 against an observed 0.368.
The direction and the order of magnitude are what I claim, not the constant.

This claim

supports No unbiased estimator of spectral atomicity under 1/f backgrounds is known, so the theory's central quantity cannot yet be measured.

refines The symmetry meant by the Symmetry Theory of Valence is almost-periodicity of the modular orbit, measured by the atomic mass of the spectral measure.

Moves against it

supports The direction of every spectral-atomicity contrast between neural states is set by whether the aperiodic model is refitted inside each state, a choice prediction 1 never makes.

Provenance

First appeared 2026-08-24 in 27cb5a2

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