C3.8T · STATISTICAL PN.2 AUDIT

Accuracy limits when branches are distinguishable

The C3.8T audit confirms the candidate bound for worst-branch expected metric risk. Explicit assumptions, a proof, sharp examples and counterexamples delimit the result and its physical interpretation.

Ivan Borisovich Kurpishev · KURPISHEV LOGIC 2 · 6 September 2026

$\mathcal R_R\mathcal R_D\geq\frac{\|K\|\Delta_D}{4}(1-\tau)^2$

Mathematical audit: passed under A1–A6
Canonical patch: not applied
Physical validation: not performed

PN.2 formula workbench

16 PN.2 formulas, lower-bound calculations, a c-separation check and direct links to verified proof pages.

Open the formula workbench

The precise result

The formula multiplies two maxima of expected error over the branches. It is neither an expected product of errors nor necessarily the product of risks in a common worst branch. Randomisation is allowed; independent errors are not required.

$\mathcal R_X=\max_{b\in\{L,R\}}\mathbb E_b\,d_X(\widehat X,X_b),\quad\tau=\operatorname{TV}(P_L,P_R)$

Assumptions A1–A6

A1 · Complete observations

One fixed protocol, one measurable space and two laws P_L and P_R. Y includes every available feature and repeated measurement.

A2 · Targets and metric

Two fixed targets for each coordinate, measurable metric loss and a finite separation Δ.

A3 · Common decision rule

One estimator or Markov kernel for both branches. Any additional information about the branch must be included in the data.

A4 · Finite expected losses

Risk means max_b E_b of metric error, not variance, a realised error or an arbitrarily weighted prior average.

A5 · PN.2 coordinates

S and D use absolute loss; R uses norm loss. K = r_L − r_R is specified in one normed space.

A6 · One experiment

The coordinates use the same pair of branches and statistical experiment. The c-form additionally requires Δ_S ≥ c‖K‖.

Proof using the common submeasure

Let μ = P_L + P_R and let p, q be the densities relative to μ. The common submeasure ν has density min(p,q), is dominated by both laws and has mass 1 − τ.

$d\nu=\min(p,q)\,d\mu,\qquad\nu(\mathcal Y)=1-\tau$

For every action, the sum of the two metric errors is at least the target separation. Integrating against the same decision kernel preserves this inequality.

$\int[d_X(a,x_L)+d_X(a,x_R)]Q_X(da\mid y)\geq\Delta_X$

Nonnegative losses and integration over ν yield a risk sum of at least Δ(1 − τ). Its maximum is at least half its sum, proving T1.

$\begin{aligned}r_{X,L}+r_{X,R}&\geq\int(L_L+L_R)\,d\nu\\&\geq\Delta_X(1-\tau)\end{aligned}$

Apply T1 separately to S, D and R and multiply the nonnegative lower bounds. This gives T2-S, T2-R and, with c-separation, T2-c.

$\mathcal R_X\geq\frac{\Delta_X}{2}(1-\tau)$

$\mathcal R_S\mathcal R_D\geq\frac{\Delta_S\Delta_D}{4}(1-\tau)^2$

$\mathcal R_S\mathcal R_D\geq\frac{c\|K\|\Delta_D}{4}(1-\tau)^2$

Endpoints τ = 0 and τ = 1

τ = 0

The observation laws coincide: the constants 1/2 and 1/4 are recovered. An admissible midpoint estimator in a normed space attains equality. A midpoint need not exist in an arbitrary metric space.

τ = 1

The laws are mutually singular and the lower bound is zero. If both target actions are allowed, an error-free rule exists. An arbitrary estimator need not be accurate.

A sharp model: the erasure channel

With probability τ the observation reveals the branch; otherwise it yields a common symbol ⊥. At that symbol the estimator returns the midpoint. This model attains equality and explains the squared overlap factor.

Branch distinguishability ττ = 0.50

R risk with ‖K‖ = 2: 0.5000
D risk with Δ_D = 3: 0.7500
Product of risks: 0.3750

These values describe the illustrative channel in the chosen units. They are not measurements of physical PN.2.

Discrete and continuous examples

D1 · Erasure

Y ∈ {L, ⊥, R}; P_L = (τ, 1 − τ, 0), P_R = (0, 1 − τ, τ). Each risk is Δ(1 − τ)/2, so 1/4 is sharp for every τ < 1.

D2 · Binary channel

P_L(1) = q, P_R(1) = 1 − q, 0 ≤ q ≤ 1/2. Then τ = 1 − 2q, and the label-based estimator has risk qΔ.

D3 · Asymmetry

P_L = (1, 0), P_R = (1/2, 1/2), targets 0 and 1. At τ = 1/2 the bound is 1/4 but the minimax risk is 1/3: TV does not determine every experiment’s exact risk.

C1 · Uniform shift

Unif[0,1] and Unif[t,1+t], 0 ≤ t ≤ 1. Use the midpoint on the overlap: τ = t and each risk is Δ(1 − t)/2.

C2 · Normal laws

N(−a,σ²) and N(a,σ²), a ≥ 0, σ > 0. Here τ = 2Φ(a/σ) − 1; the sign rule has risk ΔΦ(−a/σ) and attains the bound.

For n independent normal observations, τ_n = 2Φ(√n a/σ) − 1. Single-observation distinguishability cannot be retained when collecting more data. For general independent identically distributed samples, 1 − τ_n ≥ (1 − τ)^n.

Seven invalid strengthenings

CE1 · Every branch

With identical laws, targets 0 and 1 and the constant estimator 0, r_L = 0. The lower bound concerns the maximum.

CE2 · A common worst branch

Targets (0,0)/(1,1) and the estimate (0,1) yield risks (0,1)/(1,0). The maximum of products is 0; the product of maxima is 1.

CE3 · Expected product

A random choice between (0,1)/(1,0) has mean error 1/2 in each coordinate, while the product of errors is always zero.

CE4 · First power

At τ = 1/2 with unit separations, erasure gives 1/16. The proposed bound (1 − τ)/4 = 1/8 fails.

CE5 · Standard deviations

A constant estimator has zero variance despite different targets. Squared loss requires a different formula and units.

CE6 · Arbitrary prior weights

With identical laws and prior probability 0.9 for the left target, the estimate 0 has Bayes risk 0.1, below 1/2.

CE7 · Selection and free abstention

Discarding ⊥ can make conditional risk zero. Coverage and the new conditional experiment must be accounted for; free ABSTAIN does not satisfy metric loss.

Method attribution and evidence boundaries

The general two-point bound is classical Le Cam methodology. The PN.2 formulation supplies S, 𝔇, R, the structural separation K and application conditions. No novelty claim is made for the general statistical method.

Risks have the units of their target separations; τ is dimensionless. An instrument needs actual P_L/P_R, calibration, losses and a systematic-error budget. This audit establishes no identification of ‖K‖ with ℏ.

A separate canonical decision

A conditional statistical appendix is recommended, containing A1–A6, T1/T2, counterexamples and attribution to classical decision theory. Inclusion requires a separate author decision under C3.8T. The original candidate is preserved; independent review has not been performed.

Physical PO and F statuses are unchanged. truth_layer_promotion=0 · EXT_RUN=NOT_RUN · NO_LOSS_FREEZE=BLOCKED.

Complete proof and sources

The complete Russian paper has 8 pages: assumptions, T1/T2 proofs, sharp examples, seven counterexamples, units and bibliography. The 27 computational checks reproduce examples; they do not replace the proof.

Full audit · PDF (RU)TeX source (RU)Reproduction package · ZIP Evidence anchor ledger