When recognition earns its cost, and when it fails

Recognition earns its value through warranted selectivity. Blind neutralization can match raw accuracy, so a useful comparison must include the cost of unnecessary change and the actual cost of inquiry. The noiseless and noisy calculations make that comparison exact. Model misspecification, selected task distributions and relapse then expose specific failure conditions rather than treating every revision as improvement.

13 When recognition earns its cost, and when it fails

13.1 A costed selective target

Because blind neutralization already matches raw accuracy, the stronger comparison includes an actual cost of unnecessary revision. The operational boundary must be explicit. One retain-or-revise installation is chosen before an $H$-task block. Initial accessible advice is independent of $b$. During the block, the authority-write port is closed until the next audit; action follows 2-4, with $k=0$. The comparator has no free compensator, additional policy rewrite or source-sensitive audit. Its class consists of the two installed retain/revise rules and their source-blind random mixtures. Task observations do not silently open an amendment route.

This is a finite audit grammar, not a claim that humans cannot revise online or that these are all possible algorithms. An alternative controller with a free $C$-to-action bypass could match the desired behavior. An online learner with access to $R\oplus C$ on an $SD=1$ case and a free write route could discover $b$ during action. Either expansion changes the comparison. The present target is costed selectivity within the declared grammar.

Let $p=\mathbb P(b=1)$, let $c\gt 0$ be a revision charge, and let $\kappa$ be the total diagnostic, retention and assessment charge, all in declared common reward or opportunity-cost units. These charges are not identities for heat or gate count. In the scored block take $D=K=1$, $P=J$, and put $B_H=H(q-1/2)\gt c$. Relative to retaining, revision gains $B_H-c$ in class $b=1$ and loses $c$ in class $b=0$. Every scored task has the stated law conditional on its mechanism class. Additivity of expected reward gives $B_H=H(q-1/2)$ without requiring independence among the $H$ task outcomes. These class conditions do not follow from pooled fairness: with fair $b$ and $S=b$, an independent fair target and cue of quality $q$, the tag is marginally fair and independent of $(Y,C)$, yet the old class-one accuracy is $1-q$ rather than $1/2$.

Proposition 13.1 (Selective value of an acquired self-mechanism model). Perfect diagnosis followed by the selective rule has expected gain $p(B_H-c)-\kappa$ over always retaining. The best blind rule in the declared class has gain $\max(0,pB_H-c)$. Their difference is

$$\Delta=\min\{p(B_H-c),(1-p)c\}-\kappa.\tag{5}$$

Proof. Selective revision is charged only in class $b=1$, giving the first formula. Blind randomization is a convex combination of retain gain zero and revise gain $pB_H-c$, so its maximum is an endpoint. If $pB_H\geq c$, subtracting yields $(1-p)c-\kappa$; otherwise it yields $p(B_H-c)-\kappa$. The same case split identifies the smaller term in 5. ◻

This is a worked application of elementary value-of-information mathematics. At $q=3/4$, $H=4$, $p=1/2$, $c=1/4$ and $\kappa=1/16$, one has $B_H=1$. Selective net gain is $5/16$, best blind gain is $1/4$, and $\Delta=1/16$. The acquired account earns a role because it guides when to pay for change. The retained rule in class $b=0$ is a positive outcome of inquiry, not failed liberation.

The connection to target-relative completion is now exact. With $B_H\gt c\gt 0$, the uniquely preferred installed rule in each known class is $\lambda^*(0)=1$, $\lambda^*(1)=0$. This target varies across the initially constant-readout fibre. An exact descriptor sufficient for that target must distinguish the two classes; $\hat b$ supplies a sufficient binary completion. The target was independently defined by task law, action grammar and cost. It is not imposed solely to make information necessary. If $c=0$, blind neutralization matches accuracy and the selective-information advantage disappears. The proof does not establish that every successful controller must know its source, or that model-based learning dominates every model-free alternative.

13.2 Noisy acquisition

Suppose each of the two diagnostic returns is independently flipped with probability $\eta$, conditional on the fixed mechanism and probe settings. Their parity is wrong exactly when one return is flipped, with probability $\delta=2\eta(1-\eta)$. For an odd number $n$ of independent paired contrasts, majority error is

$$\alpha_n=\sum_{j=(n+1)/2}^{n}\binom nj\delta^j(1-\delta)^{n-j}.$$

Independence is substantive; marginal error rates alone do not justify either formula. Every additional contrast must be included in the actual cost $\kappa_n$, memory, time and error budget. With full-support noise, finite acquisition generally gives risk rather than certainty.

Let $D_H$ be the raw block benefit of revision relative to retention on the same fresh task inputs, before the installation charge. Under the declared class-conditional task family, $\mathbb E[D_H\mid b=1]=B_H$, while $D_H=0$ pointwise for $b=0$. Acquisition error can select a different subensemble of later tasks. Define the selected benefit

$$B_H^+=\mathbb E[D_H\mid b=1,\hat b=1]$$

when that event has positive probability.

Proposition 13.2 (Diagnostic errors and selected benefit). Let $\alpha_0=\mathbb P(\hat b=1\mid b=0)$ and $\alpha_1=\mathbb P(\hat b=0\mid b=1)$. The same rule, revising exactly when $\hat b=1$, has expected gain

$$p(1-\alpha_1)(B_H^+-c)-(1-p)\alpha_0c-\kappa .$$

A zero-probability selected event contributes zero. If the selected mean satisfies $B_H^+=B_H$, this reduces to $p(1-\alpha_1)(B_H-c)-(1-p)\alpha_0c-\kappa$. For symmetric error $\alpha$ under that mean condition, its advantage over the best blind rule is

$$\Delta_\alpha=\Delta-\alpha\{p(B_H-c)+(1-p)c\},$$

where $\Delta$ uses the same actual cost $\kappa$.

Proof. In class one the rule revises with probability $1-\alpha_1$ and its conditional raw benefit on precisely those blocks is $B_H^+$. The charge is paid on the same event. In class zero revision changes no task output and incurs only $c$, with probability $\alpha_0$. Average and subtract $\kappa$. The specialization and symmetric formula follow by substitution and subtraction of the perfect-diagnosis and blind gains. ◻

The selected-mean condition is weaker than independence of all fresh-task variables from every diagnostic record. A sufficient operational condition is that the declared fresh task law remains valid conditional on $b$ and the diagnostic outcome $\hat b$. Independent probe flips alone establish the diagnostic error rate, not this task-stability condition.

For the preceding ledger, the selected-mean condition $B_H^+=B_H$, and a single paired contrast with $\eta=1/16$, $\alpha=\delta=15/128$. The noisy protocol’s gain is $65/256$, best blind gain is $64/256$, and the remaining advantage is $1/256$. At that fixed ledger, positivity requires $\alpha\lt 1/8$. The small margin makes both the informational value and its vulnerability inspectable. It is not a physical claim that two real diagnostics cost exactly $1/16$ reward units.

For a concrete failure, retain $q=3/4$, $H=4$, $p=1/2$, $c=1/4$, $\kappa=1/16$ and two independent probe flips of rate $\eta=1/16$. Let $E$ be their parity error, so $\mathbb P(E=1)=15/128$. Take $E$ independent of $(Y,C)$ within each mechanism class. Choose fresh $S$ with $\mathbb P(S=1\mid E=1)=1$ and $\mathbb P(S=1\mid E=0)=49/113$, independently of the cue and target given $E$. The marginal tag is still fair and independent of $(Y,C)$ in each class, but correct diagnosis in class one selects $B_H^+=98/113$. Direct joint enumeration gives gain $25/128$ against blind gain $1/4$, hence advantage $-7/128$ rather than $1/256$. The original positive noisy example remains valid when its fresh task family is stable under the stated conditioning.

13.3 A concrete model error and a repair

A self-model can fit its diagnostic settings and still be wrong about fresh conduct. Consider the expanded generator

$$R=C\oplus bSD\oplus hCSD,\qquad b,h\in\{0,1\}.$$

The two $C=0$ probes still return $b$, so the restricted model fits them perfectly. If $h=1$, it fails at $C=S=D=1$. This is not merely a generic warning about fallibility: it is a finite counterexample to treating training fit as sufficient mechanism knowledge.

Four noiseless probes, using $(C,S)\in\{0,1\}^2$ with $D=1$, identify the expanded family. If $d_c=R(c,0,1)\oplus R(c,1,1)$, then $d_0=b$, $d_1=b\oplus h$, so $\hat h=d_0\oplus d_1$. Substitution proves agreement on all fresh triples within that family. The expanded family changes the performance law, however. For $h=1$, $D=K=1$ and the stated fair task, the old uncompensated rule has accuracy $q/2+1/4$: when $S=0$ it uses $C$ with accuracy $q$, and when $S=1$ it outputs the constant $b$ with accuracy $1/2$. The original selective-value formula cannot simply be reused. A broader diagnostic and a revised assessment are separate obligations.

This example parallels the learning corpus’s distinction between candidate capacity, data selection and validated deployment (Rodgers 2026l). A claim to understand a mechanism needs evidence for its family and relevant contexts. Better data can repair a specific failure without providing a universal guarantee of complete self-knowledge.

13.4 Retention, application and relapse

A retained explanation and its practical effect can come apart in both directions. The model $\hat b$ may remain correct while an old authority relation is reinstated. Conversely, $\lambda=0$ may remain installed after the explanatory model is lost. Reports alone cannot determine which occurred.

For a simple maintenance construction with $b=1$, let an available audit reset $\lambda$ to zero with probability $\gamma$ on each occasion. A subsequent independent relapse reinstates $\lambda=1$ with probability $\rho$. Mechanism and task law remain fixed, and each fresh scored task has that declared law conditional on the current selector $\lambda$; independent reset/relapse events and fresh task primitives are one sufficient way to ensure this. If $r_t=\mathbb P(\lambda_t=1)$ after these events, then

$$r_{t+1}=\rho+(1-\gamma)(1-\rho)r_t.$$

When $\rho+\gamma(1-\rho)\gt 0$, the unique stationary probability and active accuracy are

$$r_* =\frac{\rho}{\rho+\gamma(1-\rho)},\qquad \mathrm{Acc}_*=q-r_*(q-\tfrac12).$$

Solving the affine fixed-point equation proves the formula; its contraction factor is below one under the stated condition. At $q=3/4$, $\rho=1/10$, $\gamma=1/2$, stationary accuracy is $31/44$. If $\gamma=0$ and $\rho\gt 0$, it returns to $1/2$. If both rates vanish, the initial state persists and there is no unique stationary distribution. At the stated nonzero rates, keep $(Y,C)$ independent of the current selector and generate $S$, independently of $(Y,C)$ given that selector, with $\mathbb P(S=1\mid\lambda=1)=1$ and $\mathbb P(S=1\mid\lambda=0)=7/18$ preserves marginal tag fairness but gives accuracy $29/44$ at the same $r_*$. It violates the selector-conditional task law. The selector dynamics and their task-quality consequence therefore have separate carrying premises.

This is a transparent Markov model, not an empirical law of human relapse. It demonstrates why authority revision can differ from continuous downstream suppression without implying permanent immunity. Corrupted records, an unavailable write route, a changed generator and an inappropriate criterion produce further failures. More revision is not always better; a correct retained commitment may be precisely what supports competent action.