# Artificial Age Score (AAS): Proof-Theoretic Foundations and Axiomatic Uniqueness

Seyma Yaman Kayadibi, Seyma Yaman Kayadibi

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Seyma Yaman Kayadibi, Seyma Yaman Kayadibi. Artificial Age Score (AAS): Proof-Theoretic Foundations and Axiomatic Uniqueness. 2026. hal-05472679

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# Artificial Age Score (AAS): Proof-Theoretic Foundations and Axiomatic Uniqueness

Seyma Yaman Kayadibi

Institute for Sustainable Industries & Liveable Cities (ISILC)

Victoria University

Melbourne, VIC, Australia

Email: [seyma.yamankayadibi@live.vu.edu.au](mailto:seyma.yamankayadibi@live.vu.edu.au)

## Abstract

An axiomatic and fully proved foundation is provided for the Artificial Age Score (AAS), a penalty-based metric designed to quantify output-level aging-like degradation in observable recall accuracy under redundancy-aware channel weighting. The score is analyzed as a separable aggregation of channel contributions, enabling transparent decomposition across subsystems and consistent comparison across sessions via normalization. Core analytical properties are proved, including well-definedness on the admissible accuracy domain, convexity, and Jensen-type coarsening bounds that quantify the cost of heterogeneity, finite additivity over disjoint index sets, and positive homogeneity of degree one in the effective weights. Sensitivity ceilings are derived to control worst-case perturbation effects, and the small-noise regime is characterized to clarify limiting behavior as the regularization vanishes. Interactions between weights and redundancy gates are formalized, yielding principled targeting rules on both raw and normalized scales. Finally, an axiomatic representation and uniqueness theorem is proved, showing that under weak functional requirements, the AAS form is uniquely determined up to bit calibration. Complete proofs are provided for all stated results.

## 1. Introduction

Artificial intelligence systems may exhibit “aging-like” degradation not as a function of chronological time, but as a change in observable recall outputs across repeated interactions. To formalize and measure such effects in a model-agnostic manner at the output level, the Artificial Age Score (AAS) is studied as a multi-channel metric that aggregates accuracy and redundancy under a single penalty kernel (Kayadibi, 2025b). The central objective is to establish that AAS is mathematically well-posed and to show that, under an explicit axiom system, its functional form is not merely convenient but necessary up to calibration, consistent with axiomatic characterizations of information measures (Aczél, 1966; Khinchin, 1957; Rényi, 1961).

Under the Redundancy-as-Masking specification, the session score is
$$AAS_t = \sum_{i=1}^{m} w_i (1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}).$$

All quantities entering the score are defined at the level of observable recall outcomes and overlap statistics; no latent-state variables enter the construction (Kayadibi, 2025b).

A system with channels indexed by $i \in \{1, ..., m\}$ and sessions time steps indexed by $t \in \{1, ..., T\}$ is considered. Channel importances are encoded by weights $w_i \ge 0$ satisfying $\sum_{i=1}^{m} w_i = 1$. For each session t and channel i, an accuracy/recall score $x_{t,i} \in (0,1]$ and a redundancy level $R_{t,i} \in [0,1]$ are assumed. A regularization parameter $\varepsilon > 0$ is fixed so that boundary behavior is controlled, and a well-conditioned penalty is maintained for small accuracies.

The effective contribution weight is defined as

$$\alpha_{t,i} := w_i(1 - R_{t,i}) \ge 0,$$

so that redundancy is represented as a reduction in the impact of a channel's penalty while nonnegativity is preserved.

A log-based kernel is used to construct AAS. The penalty kernel is defined by

$$\phi_{\varepsilon}(x) := \log_2 \frac{1+\varepsilon}{x+\varepsilon} = -\log_2 \frac{x+\varepsilon}{1+\varepsilon}, \quad x \in (0,1], \varepsilon > 0,$$

and the per-channel contribution and total score are defined by

$$c_{t,i} := \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad AAS_t := \sum_{i=1}^{m} c_{t,i}.$$

The kernel $\phi_{\varepsilon}$ is decreasing and convex on $(0,1]$. Moreover, the boundary values satisfy

$$\phi_{\varepsilon}(1) = 0, \quad \phi_{\varepsilon}(0^{+}) = \lim_{x \to 0^{+}} \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{\varepsilon}.$$

Differentiability on $(0,1]$ is obtained, and derivatives are expressed explicitly as

$$\frac{d}{dx} \phi_{\varepsilon}(x) = - \frac{1}{(x+\varepsilon) \ln 2}, \quad x \in (0,1],$$

$$\frac{d^2}{dx^2} \phi_{\varepsilon}(x) = \frac{1}{(x+\varepsilon)^2 \ln 2}, \quad x \in (0,1].$$

Consequently, since $\alpha_{t,i} \ge 0$ and $\phi_{\varepsilon}(x) \ge 0$ for $x \in (0,1]$, the score is nonnegative and a uniform upper bound is admitted. Writing $\alpha_t := \sum_{i=1}^{m} \alpha_{t,i}$, and using; $\alpha_t \le \sum_{i=1}^{m} w_i = 1$, the bound

$$0 \le AAS_t \le \alpha_t \phi_{\varepsilon}(0^+) \le \phi_{\varepsilon}(0^+).$$
is obtained.

## 1.1 Problem definition

Current evaluations of repeated-interaction recall in AI systems lack a proof-grounded, redundancy-aware score that is (i) well-posed on an explicit admissible domain, (ii) comparable across sessions via normalization, and (iii) equipped with coarsening and stability bounds. This absence makes it difficult to quantify the penalty induced by heterogeneity across channels and to control worst-case perturbation effects under redundancy gating. The problem addressed here is to provide such a metric with complete analytical guarantees and an axiomatic uniqueness characterization up to calibration.

## 1.2 Contributions.
This paper provides a complete proof-theoretic foundation for AAS. First, the score is shown to be well-defined on the admissible domain and to satisfy separability, boundedness, differentiability, monotonicity, and normalization/scale invariance (4.1–4.6). Second, stability guarantees are established via global Lipschitz-type controls, boundary regularity, and explicit $\epsilon$ sensitivity and redundancy saturation results (5.1–5.4). Third, compositional invariances are proved, including refinement merge-split identities, Jensen coarsening inequalities with quantitative Jensen-gap refinements, and order preservation in homogeneous regimes (6.1–6.4, 7.1–7.6). Finally, an axiomatic representation and uniqueness theorem is proved, showing that under weak structural requirements, the AAS form is uniquely determined up to calibration (8).

## 2. Context and summary of contributions

Two principles are used to guide the construction and analysis. First, compositional consistency is imposed: when sequential independent components of performance within a channel are combined, the corresponding penalties are required to combine additively. This mechanism is the standard route by which logarithmic forms arise from multiplicative or compositional axioms in information theory and related functional-equation settings (Aczél, 1966; Cover & Thomas, 2006; Khinchin, 1957). Second, scale-independent comparability is imposed: fair comparisons across sessions and architectures are supported by using a common normalized scale. To that end, when $\alpha_t > 0$; a normalized score is defined as

$$ \overline{\text{AAS}}_t := \frac{\text{AAS}_t}{\alpha_t \phi_{\epsilon}(0^+)} \in [0,1], $$

so that trivial scaling effects due to differences in effective weights $\alpha_t$ are removed while the ordering induced by AAS$_t$ is preserved. When $\alpha_t = 0$, the denominator equals 0, and the ratio is undefined; in applications, the convention $\overline{\text{AAS}}_t := 0$ may be adopted, provided that this convention is stated explicitly.

These principles are aligned with the broader interpretation of logarithmic penalties as information cost surprisal in information theory (Shannon, 1948; Cover & Thomas, 2006). In particular, under classical product-type composition, the kernel $\phi(x) = -\log_2 x$ is recovered up to calibration; the present $\epsilon$-regularized form remains finite as $x \to 0^+$ while preserving exact additivity under a $\epsilon$-deformed composition rule, which is analytically convenient in bounded-performance settings.
The contributions are summarized as follows.

**(C1) Kernel theorems.** Well-definedness, boundedness, differentiability, monotonicity, and convexity-based lower bounds (Jensen-type) are established; Lipschitz-type upper bounds are derived on each interval $[\delta, 1]$ for any $\delta > 0$; and the regime $\epsilon \to 0^+$ is characterized.

**(C2) Stability sensitivity analysis.** Weight redundancy interactions are analyzed, and gradient ceilings are derived so that robustness and sensitivity control are supported.

**(C3) Axiomatic uniqueness.** Under axioms Ax1-Ax7, every admissible score is determined uniquely up to calibration and is required to take the form

$$AAS_t = \sum_{i=1}^{m} w_i (1 - R_{t,i}) \log_2 \frac{1 + \epsilon}{x_{t,i} + \epsilon},$$

(Kayadibi, 2025b) and, when the classical product axiom is selected, the limit

$$\lim_{\epsilon \to 0^+} \log_2 \frac{1 + \epsilon}{x + \epsilon} = -\log_2 x, \quad x \in (0,1],$$

yields the canonical kernel $-\log_2 x$. This representation style follows established axiomatic traditions for information measures (Aczél, 1966; Cover & Thomas, 2006; Khinchin, 1957; Rényi, 1961).

### 3. Notation and standing assumptions

Throughout, sessions are indexed by $t \in \{1, \dots, T\}$ and channels are indexed by $i \in \{1, \dots, m\}$. An accuracy/recall score $x_{t,i} \in (0,1]$ and a redundancy level $R_{t,i} \in [0,1]$ are assumed for each (t, i). Channel importances are encoded by weights $w_i \ge 0$ satisfying $\sum_{i=1}^m w_i = 1$. A fixed regularization parameter $\epsilon > 0$ is used, and the penalty kernel is defined by

$$\phi_{\epsilon}(x) := \log_2 \frac{1 + \epsilon}{x + \epsilon}, \quad x \in (0,1].$$

The effective weights are defined by

$$\alpha_{t,i} := w_i (1 - R_{t,i}) \ge 0, \quad \alpha_t := \sum_{i=1}^{m} \alpha_{t,i} \le 1.$$

Under these definitions, the AAS score is expressed as

$$AAS_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_{t,i}) = \sum_{i=1}^{m} w_i (1 - R_{t,i}) \phi_{\epsilon}(x_{t,i}),$$
(Kayadibi, 2025b) and it satisfies the bounds

$$0 \le AAS_t \le \alpha_t \phi_{\varepsilon}(0^+) \le \phi_{\varepsilon}(0^+), \quad \phi_{\varepsilon}(0^+) = \lim_{x \to 0^+} \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{\varepsilon}.$$

If $\alpha_t > 0$, then the normalized score $\overline{AAS}_t$ defined previously is contained in $[0,1]$ by construction.

**Definition 3.1 Effective weight**

For each session t and channel i, the effective weight is defined by

$$\alpha_{t,i} := w_i(1 - R_{t,i}) \ge 0, \quad \alpha_t := \sum_{i=1}^{m} \alpha_{t,i} = \sum_{i=1}^{m} w_i(1 - R_{t,i}).$$

This quantity encodes the attenuation of a channel's contribution induced by redundancy under a fixed importance weight $w_i$ (Kayadibi, 2025a).

**Lemma 3.1 Basic bounds and extreme cases**

For every t and i,

$$0 \le \alpha_{t,i} \le w_i$$

Moreover,

$$\alpha_{t,i} = 0 \Leftrightarrow (w_i = 0) \text{ or } (R_{t,i} = 1),$$

$$\alpha_{t,i} = w_i \Leftrightarrow (R_{t,i} = 0) \text{ or } (w_i = 0).$$

In particular, if $w_i > 0$, then $\alpha_{t,i} = w_i$ holds if and only if $R_{t,i} = 0$. Finally,

$$0 \le \alpha_t \le 1.$$

$$\text{and } \alpha_t = 1 \Leftrightarrow \sum_{i=1}^{m} w_i R_{t,i} = 0 \Leftrightarrow \forall i: w_i > 0 \Rightarrow R_{t,i} = 0.$$

**Proof.** Since $w_i \ge 0$ and $1 - R_{t,i} \in [0,1]$, it follows that

$$0 \le w_i(1 - R_{t,i}) \le w_i,$$

and hence $0 \le \alpha_{t,i} \le w_i$. The characterizations of $\alpha_{t,i} = 0$ and $\alpha_{t,i} = w_i$ follow from the equalities $w_i(1 - R_{t,i}) = 0$ and $w_i(1 - R_{t,i}) = w_i$, respectively. For the aggregate bound,
$$ \alpha_t = \sum_{i=1}^{m} w_i (1 - R_{t,i}) = \sum_{i=1}^{m} w_i - \sum_{i=1}^{m} w_i R_{t,i} \le \sum_{i=1}^{m} w_i = 1, $$

and equality holds if and only if $\sum_{i=1}^{m} w_i R_{t,i} = 0$. Since each term $w_i R_{t,i} \ge 0$, this is equivalent to $w_i R_{t,i} = 0$ for all i, i.e., $w_i > 0 \Rightarrow R_{t,i} = 0$. ■

### Lemma 3.2 Monotonicity and partial derivatives

The effective weight is linear in each argument. In particular,

$$ \frac{\partial \alpha_{t,i}}{\partial R_{t,i}} = -w_i \le 0, \quad \frac{\partial \alpha_{t,i}}{\partial w_i} = 1 - R_{t,i} \ge 0 $$

Thus, an increase in redundancy decreases $\alpha_{t,i}$ linearly, whereas an increase in $w_i$ increases $\alpha_{t,i}$ linearly. Since $\alpha_{t,i} = w_i(1 - R_{t,i}) = w_i - w_i R_{t,i}$ is affine in each argument (bi-affine) and has an indefinite Hessian, it is neither convex nor concave over the $(w_i, R_{t,i})$ plane. The mixed second derivative is given by

$$ \frac{\partial^2 \alpha_{t,i}}{\partial R_{t,i} \partial w_i} = -1, $$

which is consistent with the interaction term $-w_i R_{t,i}$ in $\alpha_{t,i}$.

**Proof.** The identities follow by differentiating $\alpha_{t,i} = w_i(1 - R_{t,i})$ with respect to $R_{t,i}$ and $w_i$. ■

### Lemma 3.3 Permutation invariance and split-merge invariance

The effective-weight construction is invariant under channel relabeling: if channels are permuted, then the family $\{\alpha_{t,i}\}_{i=1}^m$ is permuted accordingly and

$$ \alpha_t = \sum_{i=1}^{m} \alpha_{t,i} $$

is left unchanged.

Consistent refinement of the channelization is also supported. If a channel $(w_i, R)$ is split into two subchannels with the same redundancy R and weights $w_i^{(1)}, w_i^{(2)} \ge 0$ satisfying $w_i = w_i^{(1)} + w_i^{(2)}$, then

$$ w_i(1-R) = w_i^{(1)}(1-R) + w_i^{(2)}(1-R), $$

and the split preserves the effective weight. If redundancies may differ, preservation of $\alpha$ is equivalent to preservation of the weighted average redundancy:
$$w_i R = w_i^{(1)} R^{(1)} + w_i^{(2)} R^{(2)} \Leftrightarrow w_i(1-R) = w_i^{(1)}(1-R^{(1)}) + w_i^{(2)}(1-R^{(2)}).$$

Conversely, merging two channels $(w_1, R_1)$ and $(w_2, R_2)$ preserves $\alpha$ when

$$w := w_1 + w_2, \quad w_1 + w_2 > 0, \quad R := \frac{w_1 R_1 + w_2 R_2}{w_1 + w_2},$$

because then

$$w(1-R) = w_1(1-R_1) + w_2(1-R_2).$$

Boundary cases with $w_i = 0$ are trivial: $\alpha_{t,i} = 0$ and $R_{t,i}$ is unrestricted. In particular, if $w_1 = w_2 = 0$, then $\alpha = 0$ and $R$ need not be defined.

**Proof.** Each claim is obtained by direct substitution into $\alpha = w(1-R)$. ■

### 3.3 Connection to AAS

Given effective weights, the per-channel contribution and total score are defined by

$$c_{t,i} := \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \text{AAS}_t := \sum_{i=1}^{m} c_{t,i}, \quad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}.$$

Logarithmic penalty forms are consistent with classical entropy-style constructions in which additive costs arise from compositional rules (Khinchin, 1957). The score satisfies the bound

$$0 \le \text{AAS}_t \le \alpha_t \phi_{\varepsilon}(0^+), \quad \phi_{\varepsilon}(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon}.$$

**Proof.** Since $\phi_{\varepsilon}$ is decreasing on $(0,1]$ and $x_{t,i} \in (0,1]$, it follows that

$$\phi_{\varepsilon}(x_{t,i}) \le \phi_{\varepsilon}(0^{+}).$$

By Lemma 3.1, $\alpha_{t,i} \ge 0$, and hence

$$c_{t,i} = \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) \le \alpha_{t,i} \phi_{\varepsilon}(0^{+}).$$

Summation over $i$ yields

$$\text{AAS}_t = \sum_{i=1}^{m} c_{t,i} \le \left( \sum_{i=1}^{m} \alpha_{t,i} \right) \phi_{\varepsilon}(0^+) = \alpha_t \phi_{\varepsilon}(0^+).$$

Nonnegativity follows because $\alpha_{t,i} \ge 0$ and $\phi_{\varepsilon}(x) \ge 0$ for $x \in (0,1]$. ■
The equality conditions are immediate: AAS$_t = 0$ holds if and only if $x_{t,i} = 1$ for every channel with $\alpha_{t,i} > 0$, whereas the upper bound is approached as $x_{t,i} \to 0^+$ on all active channels.

## 3.4 The penalty kernel

### Definition 3.2 Penalty kernel

For $\epsilon > 0$, the penalty kernel is defined by

$$ \phi_{\epsilon}(x) := \log_{2} \frac{1 + \epsilon}{x + \epsilon}, \quad x \in (0,1]. $$

Equivalently,

$$ \phi_{\epsilon}(x) = -\log_{2} \frac{x + \epsilon}{1 + \epsilon}, \quad x \in (0,1]. $$

This kernel assigns a larger cost as $x$ decreases, aligning with the $-\log$ form used for information-like penalties and with axiomatic treatments of entropy (Cover & Thomas, 2006; Khinchin, 1957; Shannon, 1948).

### Lemma 3.4 Range and extreme values

For $x \in (0,1]$ and $\epsilon > 0$,

$$ 0 \le \phi_{\epsilon}(x) \le \phi_{\epsilon}(0^{+}), $$

$$ \phi_{\epsilon}(1) = 0, $$

$$ \phi_{\epsilon}(0^{+}) = \log_{2} \frac{1 + \epsilon}{\epsilon}. $$

**Proof.** If $x \le 1$, then $\frac{1+\epsilon}{x+\epsilon} \ge 1$, hence $\phi_{\epsilon}(x) \ge 0$. Since $\phi_{\epsilon}$ is decreasing on $(0,1]$, it follows that

$$ \phi_{\epsilon}(1) \le \phi_{\epsilon}(x) \le \lim_{x \to 0^{+}} \phi_{\epsilon}(x). $$

Direct evaluation yields $\phi_{\epsilon}(1) = \log_{2} \frac{1+\epsilon}{1+\epsilon} = 0$ and $\lim_{x \to 0^{+}} \phi_{\epsilon}(x) = \log_{2} \frac{1+\epsilon}{\epsilon}$. ■

### Lemma 3.5 Derivatives, monotonicity, and convexity

For $x \in (0,1]$ and $\epsilon > 0$, the kernel is infinitely differentiable on $(0,1]$. In particular,

$$ \frac{d}{dx} \phi_{\epsilon}(x) = - \frac{1}{(x + \epsilon) \ln 2} < 0, $$
$$ \frac{d^2}{dx^2} \phi_{\varepsilon}(x) = \frac{1}{(x + \varepsilon)^2 \ln 2} > 0. $$

Hence, $\phi_{\varepsilon}$ is strictly decreasing and strictly convex on $(0,1]$, and it is $C^{\infty}$ indeed real-analytic on $(0,1]$.

**Proof.** The derivatives are obtained by differentiating the closed form. Since $x + \varepsilon > 0$ and $\ln 2 > 0$, the stated signs follow. ■

A global Lipschitz bound on $(0,1]$ is obtained as a corollary. Since $x + \varepsilon \ge \varepsilon$ for all $x \in (0,1]$,

$$ \left| \frac{d}{dx} \phi_{\varepsilon}(x) \right| = \frac{1}{(x + \varepsilon) \ln 2} \le \frac{1}{\varepsilon \ln 2}. $$

Therefore, $\phi_{\varepsilon}$ is Lipschitz on $(0,1]$ with Lipschitz constant

$$ L = \frac{1}{\varepsilon \ln 2}. $$

### 3.5 Regularization and continuity properties of the kernel

#### Lemma 3.6 $\varepsilon$-sensitivity / regularization effect

For $x \in (0,1]$ and $\varepsilon > 0$, the kernel

$$ \phi_{\varepsilon}(x) = \log_{2} \frac{1 + \varepsilon}{x + \varepsilon}, $$

satisfies

$$ \frac{\partial \phi_{\varepsilon}(x)}{\partial \varepsilon} = \frac{x - 1}{(1 + \varepsilon)(x + \varepsilon) \ln 2} \le 0, $$

with equality if and only if $x=1$. Consequently, for any fixed $x \in (0,1)$, increasing $\varepsilon$ strictly decreases the penalty, reflecting a smoothing, regularization effect.

**Proof.** The representation

$$ \phi_{\varepsilon}(x) = \log_{2}(1 + \varepsilon) - \log_{2}(x + \varepsilon) $$

is used, and differentiation with respect to $\varepsilon$ yields the stated expression. Since $(1 + \varepsilon)(x + \varepsilon) \ln 2 > 0$, the sign is determined by $x - 1 \le 0$. ■

#### Lemma 3.7 Lipschitz continuity: local and global constants

Fix $\varepsilon > 0$. For any $\delta > 0$ and all $x_1, x_2 \in [\delta, 1]$,
$$|\phi_{\varepsilon}(x_1) - \phi_{\varepsilon}(x_2)| \le \frac{1}{(\delta + \varepsilon) \ln 2} |x_1 - x_2|.$$

Moreover, for all $x \in (0,1]$,

$$\left|\frac{d}{dx}\phi_{\varepsilon}(x)\right| = \frac{1}{(x + \varepsilon) \ln 2} \le \frac{1}{\varepsilon \ln 2},$$

so $\phi_{\varepsilon}$ is globally $\frac{1}{\varepsilon \ln 2}$ Lipschitz on $(0,1]$.

**Proof.** By the Mean Value Theorem, for $x_1, x_2 \in [\delta, 1]$ there exists $\xi \in [\delta, 1]$ such that

$$\phi_{\varepsilon}(x_1) - \phi_{\varepsilon}(x_2) = \frac{d}{dx} \phi_{\varepsilon}(\xi) (x_1 - x_2).$$

Taking absolute values and using

$$\left|\frac{d}{dx}\phi_{\varepsilon}(\xi)\right| = \frac{1}{(\xi + \varepsilon) \ln 2} \le \frac{1}{(\delta + \varepsilon) \ln 2}$$

yields the local Lipschitz constant. The global bound follows from $x + \varepsilon \ge \varepsilon$ on $(0,1]$. ■

**Remark.** As $\varepsilon \to 0^+$, the global Lipschitz constant $\frac{1}{\varepsilon \ln 2}$ diverges, which is consistent with the re-emergence of the logarithmic singularity when regularization is removed. In addition, Lemma 3.6 implies the monotonicity

$$\varepsilon_2 > \varepsilon_1 \Rightarrow \phi_{\varepsilon_2}(x) \le \phi_{\varepsilon_1}(x), \quad x \in (0,1],$$

which is used in later comparison arguments.

### Result 3.8 Endpoint extension and continuity

Continuity at $x=1$ is satisfied and $\phi_{\varepsilon}(1) = 0$.

At $x=0$, the continuous right-endpoint extension is defined by

$$\phi_{\varepsilon}(0^{+}) := \lim_{x \to 0^{+}} \phi_{\varepsilon}(x) = \log_{2} \frac{1 + \varepsilon}{\varepsilon},$$

and $\phi_{\varepsilon}(0^+)$ is used as the canonical endpoint value in upper bounds for the AAS. ■

## 3.6 Per-channel contributions and the total score

### Definition 3.3 Per-channel contribution and total score
For each $t \in \{1, ..., T\}$ and $i \in \{1, ..., m\}$, the per-channel contribution and the total score are defined by

$$c_{t,i} := \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \text{AAS}_t := \sum_{i=1}^{m} c_{t,i}$$

where the effective weight is given by

$$\alpha_{t,i} = w_i(1 - R_{t,i}) \geq 0,$$

and the penalty kernel is defined by

$$\phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon}, \quad x \in (0,1]$$

(Definition 3.2; Lemmas 3.4–3.7). Logarithmic penalties of this type align with the information-theoretic self-information (surprisal) form and its additivity under independent composition (Shannon, 1948; Cover & Thomas, 2006; Khinchin, 1957). The per-channel contribution and total score definitions follow the original AAS construction (Kayadibi, 2025b).

**Lemma 3.9 Nonnegativity, equality conditions, and per-channel upper bound**

For every t and i, $c_{t,i} \geq 0$.

Moreover,

$$c_{t,i} = 0 \Leftrightarrow [x_{t,i} = 1] \text{ or } [R_{t,i} = 1] \text{ or } [w_i = 0],$$

and the per-channel upper bound

$$c_{t,i} \leq \alpha_{t,i} \phi_{\varepsilon}(0^{+}), \quad \phi_{\varepsilon}(0^{+}) = \log_2 \frac{1 + \varepsilon}{\varepsilon},$$

is satisfied.

**Proof.** Nonnegativity is obtained since $\alpha_{t,i} \geq 0$ and $\phi_{\varepsilon}(x) \geq 0$ for all $x \in (0,1]$. The equality characterization follows from the equivalence

$$c_{t,i} = 0 \Leftrightarrow [\alpha_{t,i} = 0] \text{ or } [\phi_{\varepsilon}(x_{t,i}) = 0],$$

together with $\phi_{\varepsilon}(1) = 0 \Leftrightarrow x_{t,i} = 1$ and

$$\alpha_{t,i} = 0 \Leftrightarrow [w_i = 0] \text{ or } [R_{t,i} = 1].$$

For the upper bound, $\phi_{\varepsilon}$ is decreasing on $(0,1]$ and $x_{t,i} \in (0,1]$ is assumed, hence
$$ \Phi_{\varepsilon}(x_{t,i}) \leq \Phi_{\varepsilon}(0^{+}). $$

Multiplication by $\alpha_{t,i} \geq 0$ yields $c_{t,i} \leq \alpha_{t,i} \Phi_{\varepsilon}(0^{+})$. ■

**Remark (tightness).** If $\alpha_{t,i} = 0$, then the inequality holds with equality on both sides. If $\alpha_{t,i} > 0$, the bound is approached only asymptotically as $x_{t,i} \to 0^{+}$; since $x_{t,i} \in (0,1]$, the supremum is not attained at any finite $x_{t,i}$.

Summation of Lemma 3.9 over i yields

$$ \mathrm{AAS}_{t} = \sum_{i=1}^{m} c_{t,i} \leq \left( \sum_{i=1}^{m} \alpha_{t,i} \right) \Phi_{\varepsilon}(0^{+}) = \alpha_{t} \Phi_{\varepsilon}(0^{+}), $$

which is consistent with the endpoint extension used in Result 3.8.

**Lemma 3.10 Additivity over subsets and permutation invariance**

For any subset $S \subseteq \{1, ..., m\}$, the partial score is defined by

$$ \mathrm{AAS}_{t}(S) := \sum_{i \in S} c_{t,i}. $$

If $S \cap T = \emptyset$, then

$$ \mathrm{AAS}_{t}(S \cup T) = \mathrm{AAS}_{t}(S) + \mathrm{AAS}_{t}(T). $$

In addition, reordering channels by any permutation leaves $\mathrm{AAS}_t$ unchanged.

**Proof.** Both statements follow from commutativity and associativity of finite sums. ■

**Remark.** For arbitrary $S, T \subseteq \{1, ..., m\}$,

$$ \mathrm{AAS}_{t}(S \cup T) = \mathrm{AAS}_{t}(S) + \mathrm{AAS}_{t}(T) - \mathrm{AAS}_{t}(S \cap T). $$

**Lemma 3.11 Continuity and differentiability in the interior**

Fix $\varepsilon > 0$. Viewed as a function of $\{(x_{t,i}, R_{t,i}, w_i)\}_{i=1}^m$, the mapping $\mathrm{AAS}_t$ is $C^1$ on the interior domain, e.g., $x_{t,i} \in (0,1)$, $R_{t,i} \in (0,1)$, $w_i > 0$, together with $\sum_{i=1}^m w_i = 1$. The componentwise partial derivatives are

$$ \frac{\partial \mathrm{AAS}_t}{\partial x_{t,i}} = \alpha_{t,i} \frac{d}{dx} \Phi_{\varepsilon}(x)|_{x=x_{t,i}} = - \frac{\alpha_{t,i}}{(x_{t,i} + \varepsilon) \ln 2}, $$
$$ \frac{\partial \text{AAS}_t}{\partial R_{t,i}} = -w_i \phi_{\varepsilon}(x_{t,i}), \qquad \frac{\partial \text{AAS}_t}{\partial w_i} = (1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}). $$

**Proof.** The chain rule is applied to

$$ \text{AAS}_t = \sum_{i=1}^{m} w_i (1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}), $$

together with the identity
$$ \frac{d}{dx} \phi_{\varepsilon}(x) = - \frac{1}{(x+\varepsilon) \ln 2}, \quad x \in (0,1]. \blacksquare $$

**Note (weight constraint).** Under the constraint $\sum_{i=1}^{m} w_i = 1$, derivatives with respect to $w = (w_1, ..., w_m)$ are interpreted as constrained tangent space gradients; equivalently, the constraint may be incorporated via a Lagrange multiplier. The expressions above are the unconstrained componentwise partial derivatives.

**Corollary 3.12 Basic bounds for the total score**

Let $\alpha_t = \sum_{i=1}^{m} \alpha_{t,i}$. Then

$$ 0 \le \text{AAS}_t = \sum_{i=1}^{m} c_{t,i} \le \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(0^+) = \alpha_t \phi_{\varepsilon}(0^+). $$

The lower bound is attained when either (i) $x_{t,i} = 1$ for all active channels $\alpha_{t,i} > 0$, or (ii) $\alpha_t = 0$, e.g., $R_{t,i} = 1$ for every channel with $w_i > 0$. The upper bound is a supremum, approached as $x_{t,i} \to 0^+$ on all active channels, but not attained for any finite $x_{t,i} \in (0,1]$.

**Proof.** The statement follows by summing the per-channel inequality $c_{t,i} \le \alpha_{t,i} \phi_{\varepsilon}(0^+)$ from Lemma 3.9 over i. ■

**3.7 Maximum score and normalization**

**Definition 3.4 Maximum score**

For each session t, the session-wise maximum (supremal) score is defined by

$$ \text{AAS}_t^{\max} := \alpha_t \phi_{\varepsilon}(0^+), \qquad \alpha_t := \sum_{i=1}^{m} \alpha_{t,i} = \sum_{i=1}^{m} w_i (1 - R_{t,i}). $$

Here $\phi_{\varepsilon}(0^+)$ denotes the right-endpoint value of the kernel, defined by

$$ \phi_{\varepsilon}(0^+) := \lim_{x \to 0^+} \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{\varepsilon}, \qquad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}. $$
These definitions are understood under the standing assumptions $\epsilon > 0$, $x_{t,i} \in (0,1]$, and

$$ \alpha_{t,i} = w_i(1 - R_{t,i}) \ge 0. \text{ (Kayadibi, 2025a).} $$

### Lemma 3.13 Upper bound and attainability in the limit

For every t,

$$ 0 \le \mathrm{AAS}_t \le \mathrm{AAS}_t^{\mathrm{max}} = \alpha_t \phi_{\epsilon}(0^+) $$

The lower equality $\mathrm{AAS}_t = 0$ holds either when all active channels achieve perfect accuracy $x_{t,i} = 1$ for each i with $\alpha_{t,i} > 0$ or when $\alpha_t = 0$, i.e., no channel is effectively active. Let

$$ I_t := \{i: \alpha_{t,i} > 0\} $$

denote the active-channel set. The upper equality is not attained on the domain $x_{t,i} \in (0,1]$; it is achieved only in the closure via the limit $x_{t,i} \downarrow 0$ for all $i \in I_t$. Consequently, $\mathrm{AAS}_t^{\mathrm{max}}$ is a supremum rather than a maximum.

**Proof.** Since $\phi_{\epsilon}$ is decreasing on $(0,1]$ and $x_{t,i} \in (0,1]$ is assumed, it follows that

$$ \phi_{\epsilon}(x_{t,i}) \le \phi_{\epsilon}(0^{+}). $$

With $c_{t,i} = \alpha_{t,i}\phi_{\epsilon}(x_{t,i})$ and $\alpha_{t,i} \ge 0$, it is obtained that

$$ c_{t,i} \le \alpha_{t,i} \phi_{\epsilon}(0^{+}). $$

Summation over i yields

$$ \mathrm{AAS}_t = \sum_{i=1}^{m} c_{t,i} \le \left( \sum_{i=1}^{m} \alpha_{t,i} \right) \phi_{\epsilon}(0^{+}) = \alpha_t \phi_{\epsilon}(0^{+}). $$

Attainability requires $x_{t,i} \to 0^+$ on all active channels, which lies outside the domain $x_{t,i} \in (0,1]$ but belongs to its closure. ■

### Corollary 3.14 Normalized score and scale invariance

When $\alpha_t > 0$, the normalized score is defined by $\overline{\mathrm{AAS}}_t := \frac{\mathrm{AAS}_t}{\mathrm{AAS}_t^{\mathrm{max}}} = \frac{\sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_{t,i})}{(\sum_{i=1}^{m} \alpha_{t,i}) \phi_{\epsilon}(0^+)} \in [0,1]$

If $\alpha_t = 0$, then both numerator and denominator are equal to 0; a convenient convention is to set
$\overline{\text{AAS}}_t := 0$ alternatively, it may be left undefined, and this choice should be stated explicitly. The normalization is scale-invariant in $\alpha$: if $\alpha_{t,i} \mapsto \lambda\alpha_{t,i}$ for any $\lambda > 0$, then both numerator and denominator scale by $\lambda$, and $\overline{\text{AAS}}_t$ is left unchanged.

**Proof.** The claims follow immediately from Definition 3.4 and Lemma 3.13. ■

**Lemma 3.15 Parameter dependencies**

The maximum score

$\text{AAS}_t^{\max} = \alpha_t \phi_{\varepsilon}(0^+)$ is linear and increasing in $\alpha_t$. Since $0 \le \alpha_t \le 1$,

$$0 \le \text{AAS}_t^{\max} \le \phi_{\varepsilon}(0^+).$$

Moreover, $\text{AAS}_t^{\max}$ is decreasing in $\varepsilon$. In particular,

$$\frac{\partial \text{AAS}_t^{\max}}{\partial \varepsilon} = \alpha_t \frac{\partial \phi_{\varepsilon}(0^+)}{\partial \varepsilon} = \alpha_t \left( - \frac{1}{\varepsilon(1 + \varepsilon) \ln 2} \right) \le 0,$$

with equality only in the trivial case $\alpha_t = 0$.

**Proof.** Linearity and monotonicity in $\alpha_t$ follow directly from Definition 3.4 together with $0 \le \alpha_t \le 1$. For the $\varepsilon$-dependence, the identity

$$\phi_{\varepsilon}(0^+) = \log_2 \frac{1 + \varepsilon}{\varepsilon} = \frac{1}{\ln 2} [\ln(1 + \varepsilon) - \ln \varepsilon]$$

is used. Differentiation with respect to $\varepsilon$ yields

$$\frac{\partial \phi_{\varepsilon}(0^+)}{\partial \varepsilon} = \frac{1}{\ln 2} \left( \frac{1}{1 + \varepsilon} - \frac{1}{\varepsilon} \right) = - \frac{1}{\varepsilon(1 + \varepsilon) \ln 2}.$$

Multiplication by $\alpha_t \ge 0$ gives the stated result. ■

**Contextual remark.** Logarithmic penalties and regularized variants formalize the principle that “cost” increases as accuracy or likelihood decreases, consistent with the classical information-theoretic interpretation of surprisal and additive information costs (Cover & Thomas, 2006; Shannon, 1948).

**Remark 3.1 Practical attainability**

Although $\text{AAS}_t^{\max}$ not attained because $x_{t,i} = 0$ is excluded by the standing assumption $x_{t,i} \in (0,1]$, it is arbitrarily approximable. Specifically, for any $\eta > 0$ and any session with $\alpha_t > 0$, there exists $\delta > 0$ such that
$$x_{t,i} \in (0, \delta] (\forall i \in I_t) \Rightarrow \mathrm{AAS}_t \ge \mathrm{AAS}_t^{\max} - \eta.$$

**Justification.** Right-continuity at 0 implies $\phi_{\epsilon}(x) \uparrow \phi_{\epsilon}(0^{+})$ as $x \to 0^{+}$. Hence, there exists $\delta > 0$ such that for all $x \in (0, \delta]$,

$$0 \le \phi_{\epsilon}(0^{+}) - \phi_{\epsilon}(x) \le \frac{\eta}{\alpha_t}.$$

Therefore,

$$\mathrm{AAS}_t^{\max} - \mathrm{AAS}_t = \sum_{i \in I_t} \alpha_{t,i} (\phi_{\epsilon}(0^{+}) - \phi_{\epsilon}(x_{t,i})) \le \alpha_t \cdot \frac{\eta}{\alpha_t} = \eta.$$

If $\alpha_t = 0$, then $\mathrm{AAS}_t^{\max} = 0 = \mathrm{AAS}_t$ holds trivially. ■

## Interpretation for AAS design

The quantity $\mathrm{AAS}_t^{\max} = \alpha_t \phi_{\epsilon}(0^{+})$ovides a session-specific ceiling that separates two effects that would otherwise be conflated: structural activation through $\alpha_t$ the effective weight remaining after redundancy attenuation and penalty geometry through $\phi_{\epsilon}$ the degree to which low accuracy is penalized. This separation is useful when architectures or sessions with different redundancy profiles are compared: normalization by $\mathrm{AAS}_t^{\max}$ yields $\overline{\mathrm{AAS}}_t \in [0,1]$, which is invariant to uniform rescaling of the effective weights and therefore supports cross-session comparability without changing within-session ordering. In addition, Lemma 3.15 shows that increasing $\epsilon$ reduces $\mathrm{AAS}_t^{\max}$ thereby decreasing sensitivity to very small accuracies. Accordingly, $\epsilon$ can be treated as an explicit robustness parameter that trades sharpness near $x \approx 0$ for smoother behavior.

# 4. Core properties

## 4.1 Theorem Well-definedness

**Theorem 4.1 Well-definedness.** Under the standing assumptions stated in 3.1–3.5, for each $t \in \{1, ..., T\}$,

$$0 \le \mathrm{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_{t,i}) \le \alpha_t \phi_{\epsilon}(0^{+}) < \infty.$$

Hence, $\mathrm{AAS}_t$ is well-defined finite and nonnegative. Moreover,

$$\mathrm{AAS}_t = 0 \Leftrightarrow \forall i: [x_{t,i} = 1] \text{ or } [\alpha_{t,i} = 0],$$

equivalently, for each channel either perfect accuracy is achieved $x_{t,i} = 1$ or the channel is inactive:

$$\alpha_{t,i} = w_i(1 - R_{t,i}) = 0 \Leftrightarrow [w_i = 0] \text{ or } [R_{t,i} = 1].$$
If there exists at least one channel with $\alpha_{t,i} > 0$ and $x_{t,i} < 1$, then $\text{AAS}_t > 0$.

**Proof.** For $\varepsilon > 0$ and $x \in (0,1]$,

$$ \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon} \ge 0, \quad \phi_{\varepsilon}(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon} < \infty. $$

Since $\alpha_{t,i} \ge 0$, each term $c_{t,i} = \alpha_{t,i} \phi_{\varepsilon}(x_{t,i})$ is nonnegative. Moreover, because $\phi_{\varepsilon}$ is decreasing on $(0,1]$,

$$ \phi_{\varepsilon}(x_{t,i}) \le \phi_{\varepsilon}(0^{+}), $$

and therefore

$$ c_{t,i} \le \alpha_{t,i} \phi_{\varepsilon}(0^{+}). $$

Summation over finitely many channels yields

$$ \text{AAS}_t = \sum_{i=1}^{m} c_{t,i} \le \left( \sum_{i=1}^{m} \alpha_{t,i} \right) \phi_{\varepsilon}(0^{+}) = \alpha_t \phi_{\varepsilon}(0^{+}), $$

which establishes finiteness and the stated upper bound. Since each $c_{t,i} \ge 0$, the equality $\text{AAS}_t = 0$ holds if and only if $c_{t,i} = 0$ for all i. Now

$$ c_{t,i} = 0 \Leftrightarrow [\alpha_{t,i} = 0] \text{ or } [\phi_{\varepsilon}(x_{t,i}) = 0], $$

and $\phi_{\varepsilon}(x) = 0$ holds only at $x=1$. Finally, if some $\alpha_{t,i} > 0$ and $x_{t,i} < 1$, then $\phi_{\varepsilon}(x_{t,i}) > 0$, hence $c_{t,i} > 0$ and therefore $\text{AAS}_t > 0$. ■

**Note.** The upper bound $\alpha_t \phi_{\varepsilon}(0^+)$ is approached only as $x_{t,i} \to 0^+$ on active channels. Since $x_{t,i} \in (0,1]$, the bound is a supremum and is not attained.

## 4.2 Decomposability and order invariance

To formalize decomposability, for any subset $S \subseteq \{1, ..., m\}$, the partial score is defined by

$$ \text{AAS}_t(S) := \sum_{i \in S} c_{t,i}, \quad c_{t,i} := \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}). $$

This definition aggregates the channel-wise contributions over the selected subcollection S.

**Theorem 4.2 Decomposability/additivity; permutation invariance**

Under the standing assumptions in 3.1–3.5 and a finite channel set $\{1, ..., m\}$, the following properties are satisfied.
(i) Additivity over disjoint subsets. If $S, T \subseteq \{1, ..., m\}$ are disjoint $S \cap T = \emptyset$, then

$$AAS_t(S \cup T) = AAS_t(S) + AAS_t(T).$$

(i') Additivity over partitions. If $G = \{G_1, ..., G_K\}$ is a partition of $\{1, ..., m\}$, then

$$AAS_t = \sum_{i=1}^{m} c_{t,i} = \sum_{G \in G} \sum_{i \in G} c_{t,i} = \sum_{G \in G} AAS_t(G).$$

(ii) Permutation invariance. For any permutation $\pi$ of $\{1, ..., m\}$,

$$\sum_{i=1}^{m} c_{t,i} = \sum_{i=1}^{m} c_{t,\pi(i)},$$

so reordering channels leaves $AAS_t$ unchanged. In particular, $AAS_t$ is a symmetric function of the multiset $\{c_{t,1}, ..., c_{t,m}\}$.

**Proof.** If $S \cap T = \emptyset$, then each index in $S \cup T$ appears exactly once in either S or T, and therefore

$$\sum_{i \in S \cup T} c_{t,i} = \sum_{i \in S} c_{t,i} + \sum_{i \in T} c_{t,i},$$

which yields (i). The partition form (i') is obtained by repeated application of (i), since the blocks $G_k$ are pairwise disjoint and their union is $\{1, ..., m\}$. For (ii), since $\pi$ is a bijection, the change of index $j = \pi(i)$ yields

$$\sum_{i=1}^{m} c_{t,\pi(i)} = \sum_{j=1}^{m} c_{t,j}.$$

**Remark non-disjoint case.** For arbitrary $S, T \subseteq \{1, ..., m\}$, the inclusion-exclusion identity holds:

$$AAS_t(S \cup T) = AAS_t(S) + AAS_t(T) - AAS_t(S \cap T).$$

**Interpretation.** Theorem 4.2 shows that AAS is order-independent and finitely additive across disjoint channel groups by construction. Such additive ‘information cost’ structures are also classical in information theory, where logarithmic penalties yield additivity under composition (Cover & Thomas, 2006; Shannon, 1948). The present additivity, however, is an algebraic consequence of the per-channel summation (Kayadibi, 2025a).

### Corollary 4.2.1 Group-sum representation

Let $g: \{1, ..., m\} \to \{1, ..., K\}$ be a grouping function and define group-level contributions by
$$AAS_t^{(k)} := \sum_{i: g(i)=k} c_{t,i}, \quad k \in \{1, \dots, K\}.$$

Then

$$AAS_t = \sum_{k=1}^{K} AAS_t^{(k)}.$$

**Proof.** The mapping g induces a partition $G_k := \{i: g(i) = k\}$ of $\{1, \dots, m\}$ empty groups contribute 0, and the claim follows from Theorem 4.2(i'). ■

### Corollary 4.2.2 Refinement merging identity under fixed x and R

Fix $x \in (0,1]$ and $R \in [0,1]$. If a channel is split into two subchannels that share the same $(x,R)$ and the effective weight is split as $\alpha = \alpha_1 + \alpha_2$, then the total contribution is preserved:

$$\alpha \phi_{\varepsilon}(x) = \alpha_1 \phi_{\varepsilon}(x) + \alpha_2 \phi_{\varepsilon}(x).$$

Conversely, merging two such subchannels by setting $\alpha = \alpha_1 + \alpha_2$ leaves the total unchanged. This is the simplest refinement-invariance case; more general refinement principles are developed later (see 6.1).

**Proof.** Linearity in $\alpha$ yields

$$\alpha_1 \phi_{\varepsilon}(x) + \alpha_2 \phi_{\varepsilon}(x) = (\alpha_1 + \alpha_2) \phi_{\varepsilon}(x) = \alpha \phi_{\varepsilon}(x). \blacksquare$$

**Remark generalizations.** If $x$ is fixed but redundancies differ, writing $\alpha = w(1-R)$ and

$\alpha_j = w_j(1 - R_j)$ shows that preservation holds under the condition

$$w(1-R) = w_1(1-R_1) + w_2(1-R_2),$$

equivalently,

$$R = \frac{w_1 R_1 + w_2 R_2}{w_1 + w_2}.$$

If accuracies differ, equality generally fails unless additional conditions hold e.g., $x_1 = x_2$.

## 4.3 Boundedness, endpoint behavior, and normalization

### Theorem 4.3 (Boundedness, endpoint behavior, and normalization).

Under the standing assumptions in 3.1–3.5 with $\varepsilon > 0$, for each session $t \in \{1, \dots, T\}$,
$$0 \leq \mathrm{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \Phi_{\varepsilon}(x_{t,i}) \leq \alpha_t \Phi_{\varepsilon}(0^+) = : \mathrm{AAS}_t^{\max}, \text{ where}$$

$$\alpha_t := \sum_{i=1}^{m} \alpha_{t,i}, \quad \Phi_{\varepsilon}(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon} < \infty.$$

Moreover,

$$\mathrm{AAS}_t = 0 \Leftrightarrow \forall i: [x_{t,i} = 1] \text{ or } [\alpha_{t,i} = 0].$$

Let

$$I_t := \{i: \alpha_{t,i} > 0\}$$

denote the active-channel set. Then $\mathrm{AAS}_t^{\max}$ not attained on the domain $x_{t,i} \in (0,1]$; it is approached only as a supremum in the limit $x_{t,i} \downarrow 0$ for all $i \in I_t$.

**Proof.** For $\varepsilon > 0$ and $x \in (0,1]$,

$$\Phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon} \geq 0, \quad \Phi_{\varepsilon}(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon} < \infty,$$

and $\alpha_{t,i} \geq 0$ holds by definition. Hence, each term $c_{t,i} := \alpha_{t,i} \Phi_{\varepsilon}(x_{t,i})$ is finite and nonnegative, and the finite sum $\mathrm{AAS}_t = \sum_{i=1}^{m} c_{t,i}$ is finite and nonnegative.

For the upper bound, $\Phi_{\varepsilon}$ is decreasing on $(0,1]$, so

$$\Phi_{\varepsilon}(x_{t,i}) \leq \Phi_{\varepsilon}(0^{+}).$$

Multiplication by $\alpha_{t,i} \geq 0$ and summation yield

$$\mathrm{AAS}_t \leq \left( \sum_{i=1}^{m} \alpha_{t,i} \right) \Phi_{\varepsilon}(0^+) = \alpha_t \Phi_{\varepsilon}(0^+) = \mathrm{AAS}_t^{\max}.$$

To characterize $\mathrm{AAS}_t = 0$, it is noted that $c_{t,i} \geq 0$ for all $i$, hence $\mathrm{AAS}_t = 0$ holds if and only if $c_{t,i} = 0$ for every $i$. Now,

$$c_{t,i} = 0 \Leftrightarrow [\alpha_{t,i} = 0] \text{ or } [\Phi_{\varepsilon}(x_{t,i}) = 0],$$

and $\Phi_{\varepsilon}(x) = 0$ holds only at $x=1$, which yields the stated equivalence.
Finally, if $\alpha_{t,i} > 0$ and $x_{t,i} \in (0,1]$, then $\phi_{\varepsilon}(x_{t,i}) < \phi_{\varepsilon}(0^{+})$, so equality cannot occur within the domain. Since $\phi_{\varepsilon}(x) \uparrow \phi_{\varepsilon}(0^{+})$ as $x \to 0^{+}$, choosing $x_{t,i} \downarrow 0$ for all $i \in I_t$ implies $\text{AAS}_t \to \text{AAS}_t^{\text{max}}$. ■

**Interpretation.** The bound follows directly from $\alpha_{t,i} \ge 0$ and the fact that $\phi_{\varepsilon}$ is strictly decreasing in $x$ (so the penalty increases as accuracy decreases). This is consistent with classical information-theoretic and axiomatic treatments in which logarithmic forms quantify surprisal/uncertainty and are compatible with additive information-like cost constructions under suitable composition principles (Cover & Thomas, 2006; Khinchin, 1957; Shannon, 1948).

## Endpoint behavior

### Proposition 4.3.1 Behavior as $x \to 1$

If $x_{t,i} \to 1$ for all $i$, then

$$ \text{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) \to 0. $$

In particular, if $x_{t,i} = 1$ for some channel, then $c_{t,i} = 0$ for that channel.

**Proof.** The kernel $\phi_{\varepsilon}$ is continuous on $(0,1]$ and satisfies $\phi_{\varepsilon}(1) = 0$. For each fixed $\alpha_{t,i} \ge 0$, it follows that $\alpha_{t,i}\phi_{\varepsilon}(x_{t,i}) \to 0$ as $x_{t,i} \to 1$. Passage of the limit through a finite sum yields the claim. ■

### Proposition 4.3.2 Behavior as $x \to 0^{+}$ on active channels

Let $I_t = \{i: \alpha_{t,i} > 0\}$. If $x_{t,i} \downarrow 0$ for all $i \in I_t$, then

$$ \text{AAS}_t = \sum_{i \in I_t} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) \to \sum_{i \in I_t} \alpha_{t,i} \phi_{\varepsilon}(0^{+}) = \alpha_t \phi_{\varepsilon}(0^{+}) = \text{AAS}_t^{\text{max}}. $$

Inactive channels $\alpha_{t,i} = 0$ do not affect this limit.

**Proof.** Right-continuity at 0 gives $\phi_{\varepsilon}(x_{t,i}) \to \phi_{\varepsilon}(0^{+})$ as $x_{t,i} \downarrow 0$. Multiplication by $\alpha_{t,i}$ and passage of the limit through a finite sum over $I_t$ yield the result. ■

**Note.** Since $x_{t,i} = 0$ is excluded by the domain $(0,1]$, $\text{AAS}_t^{\text{max}}$ a supremum cannot be attained exactly.

## Normalized score and scale invariance

When $\alpha_t > 0$, the normalized score is defined by

$$ \overline{\text{AAS}}_t := \frac{\text{AAS}_t}{\text{AAS}_t^{\text{max}}} = \frac{\sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i})}{\left(\sum_{i=1}^{m} \alpha_{t,i}\right) \phi_{\varepsilon}(0^{+})} \in [0,1] $$
If $\alpha_t = 0$, then both numerator and denominator are equal to 0; a common convention is to set $\overline{\text{AAS}_t} := 0$ or to leave it undefined, provided that the chosen convention is stated explicitly. The normalization is invariant under common scaling of effective weights: if $\alpha_{t,i} \mapsto \lambda\alpha_{t,i}$ for $\lambda > 0$, then $\overline{\text{AAS}_t}$ is unchanged because the numerator and denominator scale by the same factor.

**Practical note.** This scale invariance holds at the level of $\alpha_{t,i}$. Rescaling the original weights $w_i$ is generally not meaningful under the constraint $\sum_{i=1}^m w_i = 1$; accordingly, normalization is applied to $\alpha$ or directly to $\text{AAS}_t$ rather than to $w$.

## Parameter dependence summary

The session ceiling $\text{AAS}_t^{\max} = \alpha_t \phi_{\varepsilon}(0^+)$ increases linearly with $\alpha_t \in [0,1]$. It decreases with $\varepsilon$, since

$$ \phi_{\varepsilon}(0^{+}) = \log_2 \frac{1 + \varepsilon}{\varepsilon} $$

is strictly decreasing in $\varepsilon$, which reflects the smoothing effect of regularization.

## 4.4 Differentiability and local sensitivity

### Theorem 4.4 (Differentiability)

Assume the standing conditions in 3.1–3.5 and fix $\varepsilon > 0$. Let

$$ \text{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \alpha_{t,i} = w_i(1 - R_{t,i}), \quad \phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon}. $$

On the interior domain

$$ x_{t,i} \in (0,1), \quad R_{t,i} \in (0,1), \quad w_i > 0, \quad \sum_{i=1}^{m} w_i = 1, $$

the mapping $x_{t,i}, R_{t,i}, w_i \mapsto \text{AAS}_t$ is $C^1$, and the componentwise partial derivatives are

$$ \frac{\partial \text{AAS}_t}{\partial x_{t,i}} = \alpha_{t,i} \frac{d}{dx} \phi_{\varepsilon}(x) |_{x=x_{t,i}} = - \frac{\alpha_{t,i}}{(x_{t,i} + \varepsilon) \ln 2} \quad (4.4a) $$

$$ \frac{\partial \text{AAS}_t}{\partial R_{t,i}} = -w_i \phi_{\varepsilon}(x_{t,i}) \quad (4.4b) $$

$$ \frac{\partial \text{AAS}_t}{\partial w_i} = (1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}) \quad (4.4c) $$
In particular, if $\alpha_{t,i} > 0$, then the derivative in (4.4a) is strictly negative; the derivative in (4.4b) is nonpositive; and the derivative in (4.4c) is nonnegative.

**Proof.** For $\varepsilon > 0$, the kernel $\phi_{\varepsilon}$ is $C^{\infty}$ on $(0,1)$, and extends continuously indeed smoothly to $x=1$.

$$ \frac{d}{dx} \phi_{\varepsilon}(x) = - \frac{1}{(x + \varepsilon) \ln 2}, \quad \frac{d^2}{dx^2} \phi_{\varepsilon}(x) = \frac{1}{(x + \varepsilon)^2 \ln 2}, \quad x \in (0,1]. $$

The effective weight $\alpha_{t,i} = w_i(1 - R_{t,i})$ is bilinear in $(w_i, R_{t,i})$ and is therefore smooth on the interior domain $w_i > 0, R_{t,i} \in (0,1)$. Each term

$$ c_{t,i}(x_{t,i}, R_{t,i}, w_i) = w_i(1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}) $$

is thus a product of smooth functions, and the finite sum $\text{AAS}_t = \sum_{i=1}^m c_{t,i}$ is $C^1$. Differentiation of $c_{t,i}$ with respect to $x_{t,i}, R_{t,i}$, and $w_i$ yields (4.4a)–(4.4c). No cross-dependence appears because $x_{t,i}$ enters only through the i-th term. Continuity of the partial derivatives follows since $x_{t,i} + \varepsilon > 0$ on the interior.

## Immediate monotonicity consequences

With all other variables held fixed:

i. From (4.4a), $\text{AAS}_t$ decreases as $x_{t,i}$ increases (higher accuracy reduces the penalty).

ii. From (4.4b), $\text{AAS}_t$ decreases as $R_{t,i}$ increases (higher redundancy lowers the effective weight $\alpha_{t,i}$, reducing the channel's contribution).

iii. From (4.4c), $\text{AAS}_t$ increases as $w_i$ increases (a larger importance weight increases the channel's contribution).

These signs are consistent with interpreting $\phi_{\varepsilon}$ as an information-like logarithmic penalty (surprisal-like), in the same spirit as $-\log(\cdot)$ terms in classical information theory (Cover & Thomas, 2006; Shannon, 1948).

## Useful bounds for sensitivity control

From (4.4a),

$$ \left| \frac{\partial \text{AAS}_t}{\partial x_{t,i}} \right| = \frac{\alpha_{t,i}}{(x_{t,i} + \varepsilon) \ln 2} \le \frac{\alpha_{t,i}}{\varepsilon \ln 2'} $$

which provides a uniform ceiling on local sensitivity with respect to $x_{t,i}$ when $\varepsilon > 0$. The dependence on $1/\varepsilon$ matches the earlier observation that decreasing regularization increases sharpness near $x \approx 0$.

## Note on the weight-sum constraint

$$ \sum_{i=1}^{m} w_i = 1 $$
The derivative in (4.4c) is the unconstrained componentwise partial derivative. Under the affine constraint $\sum_{i=1}^{m} w_i = 1$, feasible perturbations $\delta w = (\delta w_1, ..., \delta w_m)$ satisfy $\sum_{i=1}^{m} \delta w_i = 0$. The directional derivative along such feasible directions is

$$D_w AAS_t[\delta w] = \sum_{i=1}^{m} \frac{\partial AAS_t}{\partial w_i} \delta w_i = \sum_{i=1}^{m} (1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}) \delta w_i.$$

Let

$$g_i := (1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}), \quad g := (g_1, ..., g_m).$$

Under the Euclidean inner product, the orthogonal projection of $g$ onto the tangent space

$$\{v \in \mathbb{R}^m : \sum_{i=1}^{m} v_i = 0\}$$

is given componentwise by

$$\hat{g}_l = g_i - \bar{g}, \quad \bar{g} := \frac{1}{m} \sum_{j=1}^{m} g_j,$$

so that $\sum_{i=1}^{m} \hat{g}_l = 0$. This projected vector $\hat{g}$ is the appropriate constrained gradient for gradient-based updates on the simplex (Boyd & Vandenberghe, 2004; Nocedal & Wright, 2006).

## 4.5 Monotonicity revised to reflect the weight-simplex constraint

Let

$$AAS_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \alpha_{t,i} = w_i(1 - R_{t,i}) \ge 0, \quad \phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon},$$

with $x_{t,i} \in (0,1], R_{t,i} \in [0,1], w_i \ge 0$, and $\sum_{i=1}^{m} w_i = 1$. (Kayadibi, 2025a, 2025b).

### Theorem 4.5 (Monotonicity)

Under the standing assumptions in 3.1–3.5 and $\varepsilon > 0$, the score is coordinate-wise monotone in accuracy and redundancy, and it is monotone with respect to feasible weight transfers on the simplex, all other variables held fixed.

(i) **Monotonicity in accuracy** x. For each i, $AAS_t$ is nonincreasing in $x_{t,i}$, and it is strictly decreasing when $\alpha_{t,i} > 0$ and $x_{t,i} \in (0,1]$. In particular,
$$ \frac{\partial \text{AAS}_t}{\partial x_{t,i}} = \alpha_{t,i} \frac{d}{dx} \phi_{\varepsilon}(x)|_{x=x_{t,i}} = - \frac{\alpha_{t,i}}{(x_{t,i} + \varepsilon) \ln 2} \le 0, \quad (4.5a) $$

with strict inequality when $\alpha_{t,i} > 0$.

(ii) Monotonicity in redundancy R. For each i, AAS$_t$ is nonincreasing in $R_{t,i}$, and it is strictly decreasing when $w_i > 0$ and $x_{t,i} < 1$. In particular,

$$ \frac{\partial \text{AAS}_t}{\partial R_{t,i}} = -w_i \phi_{\varepsilon}(x_{t,i}) \le 0, \quad (4.5b) $$

with equality only if $w_i = 0$ or $x_{t,i} = 1$.

(iii) Monotonicity under feasible weight transfers on $\sum_{i=1}^m w_i = 1$.

On the simplex $\sum_{i=1}^m w_i = 1$, monotonicity with respect to weights is interpreted through feasible transfers of mass between coordinates. Let $i \ne j$ and consider a transfer $j \to i$:

$$ w_i \mapsto w_i + \delta, \quad w_j \mapsto w_j - \delta, \quad \delta > 0, $$

with all other $w_k$ held fixed and chosen so that feasibility $w_i + \delta \ge 0$, $w_j - \delta \ge 0$ is preserved. Holding $x_{t,i}$ and $R_{t,i}$ fixed, the map $w \mapsto \text{AAS}_t$ is affine, and the induced change satisfies

$$ \Delta \text{AAS}_t = \delta[(1 - R_{t,i}) \phi_{\varepsilon}(x_{t,i}) - (1 - R_{t,j}) \phi_{\varepsilon}(x_{t,j})]. \quad (4.5c) $$

Consequently, transferring weight toward a channel with larger $(1 - R_{t,i})\phi_{\varepsilon}(x_{t,i})$ increases AAS$_t$, whereas transferring weight in the opposite direction decreases AAS$_t$. This provides the simplex-constrained analogue of the unconstrained partial derivative $\partial \text{AAS}_t / \partial w_i = (1 - R_{t,i})\phi_{\varepsilon}(x_{t,i}) \ge 0$.

**Proof.** The kernel satisfies

$$ \frac{d}{dx} \phi_{\varepsilon}(x) = - \frac{1}{(x + \varepsilon) \ln 2} < 0, \quad x \in (0,1], $$

and $\phi_{\varepsilon}(x) \ge 0$ holds for all $x \in (0,1]$. For (i), differentiation of the i-th term yields (4.5a), and strictness follows when $\alpha_{t,i} > 0$. For (ii), since

$$ \frac{\partial \alpha_{t,i}}{\partial R_{t,i}} = -w_i, $$

it follows that $\frac{\partial \text{AAS}_t}{\partial R_{t,i}} = -w_i \phi_{\varepsilon}(x_{t,i}) \le 0$, giving (4.5b), with equality only if
$w_i = 0$ or $\phi_{\varepsilon}(x_{t,i}) = 0 \Leftrightarrow x_{t,i} = 1$. For (iii), under fixed $x_{t,i}$ and $R_{t,i}$,

$$ \mathrm{AAS}_t = \sum_{k=1}^{m} w_k (1 - R_{t,k}) \phi_{\varepsilon}(x_{t,k}) $$

is affine in $w$, and substituting the transfer $j \to i$ yields (4.5c) by direct algebra. ■

**Interpretation.** The monotone signs match the intended design of the score: increasing accuracy decreases the penalty through the strictly decreasing kernel; increasing redundancy decreases the penalty by reducing the effective weight; and, on the weight simplex, shifting mass toward channels with larger $(1 - R)\phi_{\varepsilon}(x)$ increases the total score.

**Proposition 4.5.1 Aggregate improvement under componentwise accuracy gains**

If accuracies improve componentwise according to

$$ x_{t,i} \mapsto x_{t,i} + \Delta_i, \quad \Delta_i \ge 0, \quad x_{t,i} + \Delta_i \in (0,1], $$

then

$$ \mathrm{AAS}_t^{\mathrm{new}} = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i} + \Delta_i) \le \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) = \mathrm{AAS}_t^{\mathrm{old}}. $$

The inequality is strict if there exists at least one channel with $\alpha_{t,i} > 0$ and $\Delta_i > 0$.

**Proof.** Since $\phi_{\varepsilon}$ is decreasing on $(0,1]$, $\phi_{\varepsilon}(x_{t,i} + \Delta_i) \le \phi_{\varepsilon}(x_{t,i})$ holds for each $i$. Multiplication by

$\alpha_{t,i} \ge 0$ and summation yield the result. ■

**4.6 Theorem Normalized score and scale invariance**

**4.6.1 Definition Normalized score**

For a fixed session $t$, the raw score $\mathrm{AAS}_t$ depends jointly on the penalty kernel and on the effective weights. To support comparability across sessions and architectures, normalization is performed by the session-wise supremal penalty under the same effective mass $\alpha_t$, noting that $x_{t,i} = 0$ is excluded by the standing domain $x_{t,i} \in (0,1]$ and hence this supremum is attainable only in the limit $x_{t,i} \downarrow 0$ on active channels.

Where $\alpha_t > 0$, define

$$ \widetilde{\mathrm{AAS}}_t := \frac{\mathrm{AAS}_t}{\mathrm{AAS}_t^{\mathrm{max}}} = \frac{\sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i})}{\alpha_t \phi_{\varepsilon}(0^+)} , \quad \alpha_t := \sum_{i=1}^{m} \alpha_{t,i} \cdot (4.6a) $$
If $\alpha_t = 0$, then $\overline{\text{AAS}}_t^{\max} = 0$ and the numerator is also 0, so the ratio is undefined (0/0). In applications, the convention $\overline{\text{AAS}}_t := 0$ may be adopted the chosen convention should be stated explicitly. Under

$\alpha_t > 0$, the normalized score is dimensionless and lies in [0,1], thereby enabling cross-session comparability.

### 4.6.2 Theorem 4.6

Under the standing assumptions in 3.1–3.5 and $\varepsilon > 0$, the following properties hold.

#### (i) Bounds.

$$0 \le \overline{\text{AAS}}_t \le 1. \quad (4.6b)$$

#### (ii) Scale invariance in $\alpha$.

For any $\lambda > 0$, define a common rescaling $\alpha'_{t,i} := \lambda \alpha_{t,i}$ for all i. Then

$$\overline{\text{AAS}}'_t = \frac{\sum_{i=1}^m \alpha'_{t,i} \phi_{\varepsilon}(x_{t,i})}{\alpha'_t \phi_{\varepsilon}(0^+)} = \frac{\lambda \sum_{i=1}^m \alpha_{t,i} \phi_{\varepsilon}(x_{t,i})}{\lambda \alpha_t \phi_{\varepsilon}(0^+)} = \overline{\text{AAS}}_t. \quad (4.6c)$$

Thus, $\overline{\text{AAS}}_t$ is 0-homogeneous in $\{\alpha_{t,i}\}$, i.e., invariant under common scaling.

#### Proof of Theorem 4.6

Because the denominator $\alpha_t \phi_{\varepsilon}(0^+)$ vanishes when $\alpha_t = 0$, the degenerate “no active channels” case is treated separately from the regular case $\alpha_t > 0$.

### 4.6.3 Preparation

$$\text{The kernel} \qquad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}$$

is decreasing, convex, and continuous on (0,1], with endpoint values

$$\phi_{\varepsilon}(1) = 0, \qquad \phi_{\varepsilon}(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon} < \infty.$$

These properties are obtained from standard calculus facts for $\log(\cdot)$ and from the convexity of $-\log(\cdot)$ on $(0, \infty)$ (Boyd & Vandenberghe, 2004; Rudin, 1976). The effective weights satisfy

$$\alpha_{t,i} = w_i(1 - R_{t,i}) \ge 0, \qquad \alpha_t := \sum_{i=1}^m \alpha_{t,i} = \sum_{i=1}^m w_i(1 - R_{t,i}).$$
Recalling the standing weight normalization $\sum_{i=1}^{m} w_i = 1$, it follows that $\alpha_t \in [0,1]$. By boundedness (Theorem 4.3),

$$0 \le AAS_t \le \alpha_t \phi_{\varepsilon}(0^+) = :AAS_t^{\max}. \quad (4.6d)$$

$$4.6.4 \text{ Bounds } 0 \le \overline{AAS}_t \le 1$$

**Case A: $\alpha_t = 0$.**

If $\alpha_t = 0$, then necessarily $\alpha_{t,i} = 0$ for all i, hence each channel contribution is

$$c_{t,i} = \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) = 0,$$

and therefore

$$AAS_t = 0, \quad AAS_t^{\max} = 0.$$

Thus, $\overline{AAS}_t = AAS_t/AAS_t^{\max}$ is undefined (0/0). If the convention $\overline{AAS}_t := 0$ is adopted, then (4.6b) holds trivially.

**Case B: $\alpha_t > 0$.**

From (4.6d),

$$0 \le AAS_t \le \alpha_t \phi_{\varepsilon}(0^+).$$

Since $\alpha_t > 0$ and $\phi_{\varepsilon}(0^+) > 0$, division by $\alpha_t \phi_{\varepsilon}(0^+)$ yields

$$0 \le \frac{AAS_t}{\alpha_t \phi_{\varepsilon}(0^+)} \le 1,$$

which is exactly

$$0 \le \overline{AAS}_t \le 1. \quad \text{This proves (4.6b).} \quad \blacksquare$$

### Equality conditions

$\overline{AAS}_t = 0$ occurs if and only if $AAS_t = 0$, i.e., all active channels satisfy $x_{t,i} = 1$ or $\alpha_t = 0$.

$\overline{AAS}_t = 1$ is approached only as a supremum, in the limit $x_{t,i} \downarrow 0$ on all active channels; it is not attained on the domain $x_{t,i} \in (0,1]$.

### Alternative representation weighted-average form

Assume $\alpha_t > 0$. Define
$$ \beta_{t,i} := \frac{\alpha_{t,i}}{\alpha_t}, \quad \left( \beta_{t,i} \ge 0, \sum_{i=1}^{m} \beta_{t,i} = 1 \right), $$

and

$$ \rho_{\epsilon}(x) := \frac{\phi_{\epsilon}(x)}{\phi_{\epsilon}(0^{+})} \in [0,1]. $$

Then

$$ \overline{\text{AAS}}_t = \frac{\sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_{t,i})}{\alpha_t \phi_{\epsilon}(0^{+})} = \sum_{i=1}^{m} \beta_{t,i} \rho_{\epsilon}(x_{t,i}). $$

Thus, $\overline{\text{AAS}}_t$ is expressed as a weighted average of values in $[0,1]$, and the bounds in (4.6b) follow immediately. Scale invariance is preserved because $\beta_{t,i}$ is unchanged under $\alpha_{t,i} \mapsto \lambda\alpha_{t,i}$.

### Corollary 4.6.1 Comparability and a normalized “order” metric

#### Definition normalized “order”

Define

$$ O_t := 1 - \overline{\text{AAS}}_t = 1 - \frac{\text{AAS}_t}{\alpha_t \phi_{\epsilon}(0^+)} \in [0,1]. \quad (4.6e) $$

### Properties

**Range.** Since $\overline{\text{AAS}}_t \in [0,1]$, it follows that $O_t \in [0,1]$.

#### Endpoints (regular case $\alpha_t > 0$)

$O_t = 1$ holds if and only if $\text{AAS}_t = 0$ equivalently, $x_{t,i} = 1$ for all i with $\alpha_{t,i} > 0$.

$O_t \to 0$ as $\text{AAS}_t \to \alpha_t \phi_{\epsilon}(0^+)$, i.e., as $x_{t,i} \downarrow 0^+$ on all active channels.

**Scale invariance.** Under $\alpha_{t,i} \mapsto \lambda\alpha_{t,i}$ with $\lambda > 0$, $\overline{\text{AAS}}_t$ is invariant; hence $O_t$ is invariant as well.

**Permutation invariance and decomposability.** Since $O_t$ is an affine transform of $\overline{\text{AAS}}_t$, it inherits invariance under channel permutations and the decompositions established in 4.2.

**Interpretation.** $O_t$ is a normalized “order/goodness” metric: larger $O_t$ indicates a smaller normalized penalty and hence a more ordered state.
**Note on parameterization under $\alpha_{t,i} = w_i(1 - R_{t,i})$**

Scale invariance refers to common multiplicative scaling at the $\alpha$-level. Directly scaling the original weights $w_i$ by a common factor is generally incompatible with $\sum_{i=1}^m w_i = 1$. Where admissible, i.e., provided $\lambda(1 - R_{t,i}) \in [0,1]$ for all i, a common scaling of the redundancy complements

$$ (1 - R_{t,i}) \mapsto \lambda(1 - R_{t,i}), \quad \lambda > 0, $$

induces $\alpha'_{t,i} = \lambda\alpha_{t,i}$, and therefore leaves $\overline{\text{AAS}}_t$ and hence $O_t$ unchanged by (4.6c).

**Addendum continuity and monotonicity of the normalized score**

Assuming $\alpha_t > 0$, the map $\overline{\text{AAS}}_t$ is $C^1$ on the interior region and preserves the same sign structure implied by 4.4–4.5:

1. As $x_{t,i}$ increases, $\overline{\text{AAS}}_t$ decreases, and $O_t$ increases.

2. As $R_{t,i}$ increases, $\text{AAS}_t$ decreases; however, $\overline{\text{AAS}}_t$ (and hence $O_t$) need not be monotone in $R_{t,i}$ because the normalizing factor $\alpha_t$ also changes.

3. Under $\sum_{i=1}^m w_i = 1$, feasible transfers $w_i \mapsto w_i + \delta$, $w_j \mapsto w_j - \delta$ give

$$ \Delta \overline{\text{AAS}}_t = \frac{\delta}{\alpha_t \phi_{\epsilon}(0^+)} [(1 - R_{t,i})\phi_{\epsilon}(x_{t,i}) - (1 - R_{t,j})\phi_{\epsilon}(x_{t,j})], \text{ so the sign is governed by the same bracket term as in (4.5c).} $$

**5. Stability and Sensitivity**

The normalization established in Theorem 4.6 yields a dimensionless, scale-invariant quantity that supports cross-session comparability. For practical use, stability is additionally required: small perturbations in the observed accuracy vector should not induce disproportionate changes in the score. Sensitivity is therefore quantified by proving global Lipschitz continuity of $\text{AAS}_t$ in $x_t$ with respect to standard norms on $(0,1]^m$ (Kreyszig, 1978; Rudin, 1976).

**Theorem 5.1 Global Lipschitz continuity in two norms**

**5.1.1 Statement**

Let $m \in \mathbb{N}$ be fixed and let $t$ denote a fixed session. For $x = (x_1, ..., x_m) \in (0,1]^m$, define

$$ \text{AAS}_t(x) := \sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_i), \quad \phi_{\epsilon}(x) := \log_2 \frac{1+\epsilon}{x+\epsilon}, \quad \epsilon > 0, \quad \alpha_{t,i} \ge 0. \quad (5.1a) $$

Let $\alpha_t := \sum_{i=1}^m \alpha_{t,i}$. Under the standing parameterization $\alpha_{t,i} = w_i(1 - R_{t,i})$ with $w_i \ge 0$, $R_{t,i} \in [0,1]$, and the normalization $\sum_{i=1}^m w_i = 1$, it is ensured that $\alpha_t \in [0,1]$, and hence $\alpha_t \le 1$.

Then for all $x, y \in (0,1]^m$, the following bounds hold.
### 5.1.2 $l_\infty$-Lipschitz bound

$$|\text{AAS}_t(x) - \text{AAS}_t(y)| \le \frac{\alpha_t}{\varepsilon \ln 2} |x - y|_\infty. \quad (5.1b)$$

### 5.1.3 $l_1$-Lipschitz bound

$$|\text{AAS}_t(x) - \text{AAS}_t(y)| \le \frac{\max_{1 \le i \le m} \alpha_{t,i}}{\varepsilon \ln 2} |x - y|_1. \quad (5.1c)$$

The proof is obtained by combining (i) a uniform bound on the derivative of $\phi_\varepsilon$ over $(0,1]$, (ii) a line-segment integral representation, and (iii) the dual-norm Hölder inequality (Kreyszig, 1978; Rudin, 1976).

## Proof of Theorem 5.1

### 5.1.4 Step 1 Derivative bounds

For each coordinate $i \in \{1, ..., m\}$, the partial derivative of $\text{AAS}_t$ with respect to $x_i$ is computed on the interior domain $(0,1)^m$ as

$$\frac{\partial}{\partial x_i} \text{AAS}_t(x) = \alpha_{t,i} \frac{d}{dx} \phi_\varepsilon(x) |_{x=x_i}. \quad (5.1d)$$

Since

$$\phi_\varepsilon(x) = \log_2(1 + \varepsilon) - \log_2(x + \varepsilon),$$

it follows that

$$\frac{d}{dx} \phi_{\varepsilon}(x) = - \frac{1}{(x + \varepsilon) \ln 2}, \quad x > 0, \quad (5.1e)$$

and therefore

$$\frac{\partial}{\partial x_i} \text{AAS}_t(x) = - \frac{\alpha_{t,i}}{(x_i + \varepsilon) \ln 2}. \quad (5.1f)$$

Because $x_i \in (0,1]$ implies $x_i + \varepsilon \ge \varepsilon$, the uniform bound

$$\left| \frac{\partial}{\partial x_i} \text{AAS}_t(x) \right| = \frac{\alpha_{t,i}}{(x_i + \varepsilon) \ln 2} \le \frac{\alpha_{t,i}}{\varepsilon \ln 2} \quad (5.1g)$$

is obtained for all $x \in (0,1]^m$ (Rudin, 1976).

Consequently, the following x-independent gradient bounds are obtained:
$$|\nabla \text{AAS}_t(x)|_1 = \sum_{i=1}^{m} \left| \frac{\partial}{\partial x_i} \text{AAS}_t(x) \right| \le \frac{1}{\varepsilon \ln 2} \sum_{i=1}^{m} \alpha_{t,i} = \frac{\alpha_t}{\varepsilon \ln 2}, \quad (5.1h)$$

and

$$|\nabla \text{AAS}_t(x)|_{\infty} = \max_{1 \le i \le m} \left| \frac{\partial}{\partial x_i} \text{AAS}_t(x) \right| \le \frac{1}{\varepsilon \ln 2} \max_{1 \le i \le m} \alpha_{t,i}. \quad (5.1i)$$

Finally, the domain $(0,1]^m$ is convex, so the line segment connecting any two points $x, y \in (0,1]^m$ remains within $(0,1]^m$ (Rudin, 1976).

### 5.1.5 Step 2 Line segment integral identity and dual norms

Let $x, y \in (0,1]^m$ be arbitrary and define the segment parameterization

$$z(s) := x + s(y - x), \quad s \in [0,1]. \quad (5.1j)$$

By the fundamental theorem of calculus applied along the $C^1$ path $z(\cdot)$, the line-integral identity

$$\text{AAS}_t(y) - \text{AAS}_t(x) = \int_0^1 \langle \nabla \text{AAS}_t(z(s)), y - x \rangle \, ds. \quad (5.1k)$$

is obtained. (Rudin, 1976).

Moreover, for any norm $|\cdot|$ with dual norm $|\cdot|_*$, Hölder's inequality yields

$$|\langle u, v \rangle| \le |u|_* |v|, \quad (5.1l)$$

which is applied below (Kreyszig, 1978).

### 5.1.6 $l_\infty$-Lipschitz bound

The norm $|\cdot| = |\cdot|_\infty$ is selected, whose dual norm is $|\cdot|_* = |\cdot|_1$. (Kreyszig, 1978). From (5.1k)–(5.1l),

$$|\text{AAS}_t(y) - \text{AAS}_t(x)| \le \int_0^1 |\nabla \text{AAS}_t(z(s))|_1 \, ds \cdot |y - x|_\infty. \quad (5.1m)$$

Using the uniform bound (5.1h),

$$\sup_{s \in [0,1]} |\nabla \text{AAS}_t(z(s))|_1 \le \frac{\alpha_t}{\varepsilon \ln 2}, \quad (5.1n)$$

and therefore
$$|\text{AAS}_t(y) - \text{AAS}_t(x)| \le \frac{\alpha_t}{\varepsilon \ln 2} |x - y|_{\infty},$$

which proves (5.1b). ■

### 5.1.7 $l_1$-Lipschitz bound

The norm $| \cdot | = | \cdot |_1$ is selected, whose dual norm is $| \cdot |_* = | \cdot |_{\infty}$ (Kreyszig, 1978). From (5.1k)–(5.1l),

$$|\text{AAS}_t(y) - \text{AAS}_t(x)| \le \int_0^1 |\nabla \text{AAS}_t(z(s))|_{\infty} \, ds \cdot |y - x|_1. \quad (5.1o)$$

Using the uniform bound (5.1i),

$$\sup_{s \in [0,1]} |\nabla \text{AAS}_t(z(s))|_{\infty} \le \frac{\max_{t,i} \alpha_{t,i}}{\varepsilon \ln 2}, \quad (5.1p)$$

and hence

$$|\text{AAS}_t(y) - \text{AAS}_t(x)| \le \frac{\max_{t,i} \alpha_{t,i}}{\varepsilon \ln 2} |x - y|_1,$$

which proves (5.1c). ■

### 5.1.8 Notes.
If $\alpha_t = 0$, then $\alpha_{t,i} = 0$ for all $i$ and hence $\text{AAS}_t(\cdot) \equiv 0$ on $(0,1]^m$, therefore, the Lipschitz bounds are trivial.

## 5.2 Theorem Boundary Continuities

Fix a session t. For notational simplicity, suppress the index t and write

$$\alpha_i := \alpha_{t,i} = w_i(1 - R_{t,i}) \ge 0, \quad \alpha := \sum_{i=1}^{m} \alpha_i, \quad \text{AAS}(x) := \text{AAS}_t(x) = \sum_{i=1}^{m} \alpha_i \phi_{\varepsilon}(x_i), \\ \varepsilon > 0. \quad (5.2-1)$$

The kernel is:

$$\phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon}, \quad x \in (0,1]. \quad (5.2-2)$$

Since $\phi_{\varepsilon}$ admits a finite right limit as $x \to 0^+$, AAS admits a natural continuous extension from $(0,1]^m$ to the closed cube $[0,1]^m$.

### Continuous extension
Extend the kernel at 0 by its right limit:

$$ \hat{\phi}_{\varepsilon}(x) = \phi_{\varepsilon}(x) (0 < x \le 1), \quad \hat{\phi}_{\varepsilon}(0) = \phi_{\varepsilon}(0^{+}) = \log_{2} \frac{1+\varepsilon}{\varepsilon}. \quad (5.2-3) $$

and define the extended score on $[0,1]^m$ by

$$ \widehat{\text{AAS}}(x) := \sum_{i=1}^{m} \alpha_i \hat{\phi}_{\varepsilon}(x_i), \quad x \in [0,1]^m. \quad (5.2-4) $$

**Theorem 5.2 Boundary continuities**

(a) The extended kernel $\hat{\phi}_{\varepsilon}$ is continuous on $[0,1]$ and satisfies $\hat{\phi}_{\varepsilon}(1) = 0$. (5.2a)

(b) The extended score $\widehat{\text{AAS}}$ is continuous on $[0,1]^m$.

(c) For each coordinate $i$, the one-dimensional boundary limits satisfy

$$ \lim_{x_i \to 1^-} \alpha_i \phi_{\varepsilon}(x_i) = \alpha_i \hat{\phi}_{\varepsilon}(1) = 0, (5.2b) $$

$$ \lim_{x_i \to 0^+} \alpha_i \phi_{\varepsilon}(x_i) = \alpha_i \hat{\phi}_{\varepsilon}(0) = \alpha_i \log_2 \frac{1+\varepsilon}{\varepsilon}. (5.2c) $$

More generally, for any sequence $x^{(k)} \in (0,1]^m$ with $x^{(k)} \to x^* \in [0,1]^m$,

$$ \lim_{k \to \infty} \text{AAS}(x^{(k)}) = \widehat{\text{AAS}}(x^*). \quad (5.2d) $$

**Proof.**

(a) The function $\phi_{\varepsilon}$ is continuous on $(0,1]$ and satisfies $\phi_{\varepsilon}(1) = 0$. Moreover,

$$ \lim_{x \to 0^+} \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{\varepsilon} < \infty. \quad (5.2-5) $$

Defining $\hat{\phi}_{\varepsilon}(0)$ by this right limit yields continuity at 0, hence $\hat{\phi}_{\varepsilon}$ is continuous on $[0,1]$ and $\hat{\phi}_{\varepsilon}(1) = 0$ (Rudin, 1976).

(b) For each $i$, the map $x \mapsto \hat{\phi}_{\varepsilon}(x_i)$ is continuous on $[0,1]^m$. Multiplication by the scalar $\alpha_i \ge 0$ and finite summation preserve continuity, hence $\widehat{\text{AAS}}$ is continuous on $[0,1]^m$ (Rudin, 1976).

(c) The limits in (5.2b)– (5.2c) follow from (a) and linearity of limits. For (5.2d), continuity in (b) gives $\widehat{\text{AAS}}(x^{(k)}) \to \widehat{\text{AAS}}(x^*)$ whenever $x^{(k)} \to x^*$. Since $\widehat{\text{AAS}} = \text{AAS}$ on $(0,1]^m$, identity (5.2d) follows. ■

**Remarks and manuscript links**
1. **Vanishing at perfect accuracy.** As $x_i \to 1^-$, $\phi_\epsilon(x_i) \to 0$, hence $\alpha_i \phi_\epsilon(x_i) \to 0$, consistent with the endpoint behavior discussed in 4.3.1.

2. **Maximal boundary limit.** As $x_i \to 0^+$, $\alpha_i \phi_\epsilon(x_i) \to \alpha_i \phi_\epsilon(0^+)$; over active channels this yields $\alpha_i \phi_\epsilon(0^+) = \text{AAS}_t^{\text{max}}$ (see 4.3.2).

3. **Closure-based arguments.** Since a continuous extension to $[0,1]^m$ is available, “supremum on the boundary” arguments may be stated on the closure, which supports limit-attainability reasoning.

## Quantitative refinement-Lipschitz control on the closure

A stronger, quantitative form of continuity is obtained via Lipschitz bounds.

### (i) Lipschitz continuity of $\widehat{\phi}_\epsilon$ on $[0,1]$

For $u, v \in (0,1]$, the derivative exists and is given by

$$ \frac{d}{dx} \phi_{\epsilon}(x) = - \frac{1}{(x + \epsilon) \ln 2}, \quad x \in (0,1], (5.2-6) $$

so $|\phi'_\epsilon(x)| \le \frac{1}{\epsilon \ln 2}$ for all $x \in (0,1]$. By the Mean Value Theorem,

$$ |\phi_{\epsilon}(u) - \phi_{\epsilon}(v)| \le \frac{1}{\epsilon \ln 2} |u - v|, \quad u, v \in (0,1]. (5.2-7) $$

If $v=0$ and $u \in (0,1]$, then by the extension definition,

$$ |\widehat{\phi}_{\epsilon}(u) - \widehat{\phi}_{\epsilon}(0)| = \phi_{\epsilon}(0^{+}) - \phi_{\epsilon}(u) = \log_{2} \frac{u + \epsilon}{\epsilon} = \frac{1}{\ln 2} \ln \left(1 + \frac{u}{\epsilon}\right) \le \frac{u}{\epsilon \ln 2}, (5.2-8) $$

using $\ln(1+z) \le z$ for $z \ge 0$ (Rudin, 1976). The same bound holds when $u = 0$ and $v \in (0,1]$. Hence,

$$ |\widehat{\phi}_{\epsilon}(u) - \widehat{\phi}_{\epsilon}(v)| \le \frac{1}{\epsilon \ln 2} |u - v|, \quad u, v \in [0,1]. (5.2e) $$

Therefore $\widehat{\phi}_\epsilon$ is Lipschitz on $[0,1]$, hence uniformly continuous (Rudin, 1976).

### (ii) Lipschitz continuity of $\widehat{\text{AAS}}$ on $[0, 1]^m$

For $x, y \in [0,1]^m$, applying (5.2e) coordinatewise yields

$$ \begin{aligned} |\widehat{\text{AAS}}(x) - \widehat{\text{AAS}}(y)| & \le \sum_{i=1}^{m} \alpha_i |\widehat{\phi}_{\epsilon}(x_i) - \widehat{\phi}_{\epsilon}(y_i)| \\ & \le \frac{1}{\epsilon \ln 2} \sum_{i=1}^{m} \alpha_i |x_i - y_i| \\ & = \frac{1}{\epsilon \ln 2} |\alpha \odot (x - y)|_1, (5.2f) \end{aligned} $$
where ⊙ denotes coordinatewise multiplication and α := (α₁, ..., αₘ). Since αᵢ ≤ α for every i,

$$ \sum_{i=1}^{m} \alpha_i |x_i - y_i| \le \sum_{i=1}^{m} \alpha |x_i - y_i| = \alpha |x - y|_1. $$

Therefore,

$$ |\widehat{\text{AAS}}(x) - \widehat{\text{AAS}}(y)| \le \frac{\alpha}{\varepsilon \ln 2} |x - y|_1. \quad (5.2g) $$

Thus, AAS is Lipschitz on [0,1]ᵀ, hence uniformly continuous, and the convergence statement (5.2d) follows (Rudin, 1976).

### Mixed boundary regimes, multiple coordinates

Let S₀, S₁ ⊆ {1, ..., m} be disjoint, with xᵢ → 0⁺ for i ∈ S₀, xᵢ → 1⁻ for i ∈ S₁, and remaining coordinates fixed at x̄ⱼ ∈ (0,1]. Then by (5.2b)–(5.2c),

a) for i ∈ S₀: αᵢφᵢ(xᵢ) → αᵢφᵢ(0⁺);

b) for i ∈ S₁: αᵢφᵢ(xᵢ) → 0;

c) otherwise: αⱼφᵢ(xⱼ) → αⱼφᵢ(x̄ⱼ).

Summation yields exactly AAS evaluated at the corresponding boundary point, confirming continuity in all mixed cases.

### Corollary 5.2.1 Limit attainability of the upper and lower bounds

(i) **Upper bound attainable in the limit.** Let I := {i: αᵢ > 0}. If xᵢ ↓ 0 for all i ∈ I, then

$$ \text{AAS}(x) \to \sum_{i \in I} \alpha_i \phi_{\varepsilon}(0^+) = \alpha \phi_{\varepsilon}(0^+) = \text{AAS}_{\text{I}}^{\text{max}}, \quad (5.2h) $$

consistent with the maximal-limit characterization in 4.3.2.

(ii) **Lower bound.** If xᵢ ↑ 1 for all i, then φᵢ(1) = 0 implies

$$ \text{AAS}(x) = \sum_{i=1}^{m} \alpha_i \phi_{\varepsilon}(x_i) \to 0. \quad (5.2i) $$

### Remark 5.2.1 Uniform continuity and numerical robustness

By (5.2g), AAS is Lipschitz on [0,1]ᵀ, hence uniformly continuous; the Lipschitz constant depends only on α and ε. Consequently, if measurement perturbations satisfy |Δx|₁ small, then the score variation is
linearly controlled by $|\Delta x|_1$, ensuring predictable stability even near boundary regions. (Rudin, 1976; Kreyszig, 1978).

## 5.3 Theorem $\varepsilon$-sensitivity / regularization

### Abbreviation and setup

Fix a session t. For $x = (x_1, ..., x_m) \in (0,1]^m$ and $\varepsilon > 0$, define the $\varepsilon$-parametrized score

$$ \text{AAS}_t(\varepsilon; x) := \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_i), \quad \alpha_{t,i} \ge 0, (5.3-1) $$

with

$$ \phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon} = \frac{1}{\ln 2} [\ln(1 + \varepsilon) - \ln(x + \varepsilon)]. \quad (5.3-2) $$

Let $\alpha_t := \sum_{i=1}^{m} \alpha_{t,i}$.

### Theorem 5.3 Monotonicity, convexity, global band, mixed derivatives, and extreme regimes

Fix $x \in (0,1]^m$ and assume $\alpha_{t,i} \ge 0$. Then:

#### (a) First derivative regularization effect

$$ \frac{\partial \text{AAS}_t}{\partial \varepsilon}(\varepsilon; x) = \frac{1}{\ln 2} \sum_{i=1}^{m} \alpha_{t,i} \left( \frac{1}{1+\varepsilon} - \frac{1}{x_i + \varepsilon} \right) = \frac{1}{\ln 2} \sum_{i=1}^{m} \alpha_{t,i} \frac{x_i - 1}{(1+\varepsilon)(x_i + \varepsilon)} \le 0. \quad (5.3a) $$

Equality holds if and only if $x_i = 1$ for every active coordinate $\alpha_{t,i} > 0$.

#### (b) Second derivative convexity in $\varepsilon$

$$ \frac{\partial^2 \text{AAS}_t}{\partial \varepsilon^2}(\varepsilon; x) = \frac{1}{\ln 2} \sum_{i=1}^{m} \alpha_{t,i} \left( \frac{1}{(x_i + \varepsilon)^2} - \frac{1}{(1 + \varepsilon)^2} \right) \ge 0. \quad (5.3b) $$

Equality holds if and only if $x_i = 1$ for every active coordinate.

#### (c) Global band Lipschitz-type bound on $\varepsilon$-bands

Fix $\varepsilon_0 > 0$. If $\varepsilon, \varepsilon' \ge \varepsilon_0$, then

$$ |\text{AAS}_t(\varepsilon; x) - \text{AAS}_t(\varepsilon'; x)| \le \frac{\alpha_t}{\varepsilon_0(1 + \varepsilon_0) \ln 2} |\varepsilon - \varepsilon'|. \quad (5.3c) $$
(d) Mixed derivative and damping of x-sensitivity as ε increases

For each i,

$$ \frac{\partial^2 \text{AAS}_t}{\partial x_i \partial \varepsilon}(\varepsilon; x) = \alpha_{t,i} \frac{\partial}{\partial \varepsilon} (\phi'_{\varepsilon}(x_i)) = \frac{\alpha_{t,i}}{(x_i + \varepsilon)^2 \ln 2} \ge 0, (5.3d) $$

with strict inequality if and only if $\alpha_{t,i} > 0$. Moreover,

$$ \partial_{x_i} \text{AAS}_t(\varepsilon; x) = - \frac{\alpha_{t,i}}{(x_i + \varepsilon) \ln 2} \le 0, $$

and for active channels $\alpha_{t,i} > 0$,

$$ \frac{\partial}{\partial \varepsilon} |\partial_{x_i} \text{AAS}_t(\varepsilon; x)| = - \frac{\alpha_{t,i}}{(x_i + \varepsilon)^2 \ln 2} < 0, $$

so the magnitude of the $x_i$ sensitivity decreases as $\varepsilon$ increases.

(e) Extreme regimes and normalized asymptotics

$$ \lim_{\varepsilon \to 0^+} \text{AAS}_t(\varepsilon; x) = \sum_{i=1}^{m} \alpha_{t,i} \log_2 \frac{1}{x_i}, \quad \lim_{\varepsilon \to \infty} \text{AAS}_t(\varepsilon; x) = 0. (5.3e) $$

Define

$$ \text{AAS}_t^{\max}(\varepsilon) := \alpha_t \phi_{\varepsilon}(0^+) = \alpha_t \log_2 \frac{1 + \varepsilon}{\varepsilon} $$

Then

$$ \lim_{\varepsilon \to 0^+} \text{AAS}_t^{\max}(\varepsilon) = +\infty, \quad \lim_{\varepsilon \to \infty} \text{AAS}_t^{\max}(\varepsilon) = 0. (5.3f) $$

If $\alpha_t > 0$ and

$$ \overline{\text{AAS}}_t(\varepsilon; x) := \text{AAS}_t(\varepsilon; x) / \text{AAS}_t^{\max}(\varepsilon), $$

Then

$$ \text{as } \varepsilon \to 0^+, \overline{\text{AAS}}_t(\varepsilon; x) \to 0; $$

$$ \text{as } \varepsilon \to \infty, \overline{\text{AAS}}_t(\varepsilon; x) \to \frac{\sum_{i=1}^{m} \alpha_{t,i}(1-x_i)}{\alpha_t} = 1 - \frac{\sum_{i=1}^{m} \alpha_{t,i}x_i}{\alpha_t}. (5.3g) $$

**Proof**
(a) Differentiating (5.3-2) gives

$$ \partial_{\varepsilon} \phi_{\varepsilon}(x) = \frac{1}{\ln 2} \left( \frac{1}{1+\varepsilon} - \frac{1}{x+\varepsilon} \right) = \frac{1}{\ln 2} \frac{x-1}{(1+\varepsilon)(x+\varepsilon)}. $$

Summing with weights $\alpha_{t,i} \ge 0$ yields (5.3a). Since $x_i \le 1$, each summand is nonpositive; equality requires $x_i = 1$ on all active coordinates.

(b) A second differentiation gives

$$ \frac{\partial^2}{\partial \varepsilon^2} \phi_{\varepsilon}(x) = \frac{1}{\ln 2} \left( \frac{1}{(x+\varepsilon)^2} - \frac{1}{(1+\varepsilon)^2} \right) $$

Since $x_i + \varepsilon \le 1 + \varepsilon$, the bracket is nonnegative; summation yields (5.3b).

(c) By the Mean Value Theorem in $\varepsilon$, for $\varepsilon, \varepsilon' \ge \varepsilon_0$ there exists $\tilde{\varepsilon}$ between them such that

$$ \text{AAS}_t(\varepsilon; x) - \text{AAS}_t(\varepsilon'; x) = (\partial_{\varepsilon}\text{AAS}_t)(\tilde{\varepsilon}; x) (\varepsilon - \varepsilon'). $$

Using (5.3a) and $x_i \in (0,1]$,

$$ \left| \frac{1}{1+\tilde{\varepsilon}} - \frac{1}{x_i + \tilde{\varepsilon}} \right| = \frac{1}{x_i + \tilde{\varepsilon}} - \frac{1}{1+\tilde{\varepsilon}} \le \frac{1}{\tilde{\varepsilon}(1+\tilde{\varepsilon})} \le \frac{1}{\varepsilon_0(1+\varepsilon_0)}. $$

Therefore,

$$ |\partial_{\varepsilon} \text{AAS}_t(\tilde{\varepsilon}; x)| \le \frac{1}{\ln 2} \sum_{i=1}^{m} \alpha_{t,i} \frac{1}{\varepsilon_0(1+\varepsilon_0)} = \frac{\alpha_t}{\varepsilon_0(1+\varepsilon_0) \ln 2}, $$

which implies (5.3c). (Rudin, 1976).

(d) Since $\phi'_{\varepsilon}(x) = -1/((x+\varepsilon) \ln 2)$, differentiating in $\varepsilon$ gives

$$ \partial_{\varepsilon} \phi'_{\varepsilon}(x) = \frac{1}{(x+\varepsilon)^2 \ln 2}. $$

Multiplying by $\alpha_{t,i}$ yields (5.3d).

(e) The limits in (5.3e)–(5.3f) follow directly from $\phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}$:

$$ \varepsilon \to 0^+ : \quad \frac{1+\varepsilon}{x+\varepsilon} \to \frac{1}{x}, \qquad \varepsilon \to \infty : \quad \frac{1+\varepsilon}{x+\varepsilon} \to 1 $$

For (5.3g), use $\ln(1+u) = u+o(u)$ as $u \to 0$:
$$ \phi_{\varepsilon}(x) = \frac{1}{\ln 2} [\ln(1 + 1/\varepsilon) - \ln(1 + x/\varepsilon)] = \frac{1}{\ln 2} \left( \frac{1-x}{\varepsilon} + o(1/\varepsilon) \right), $$

$$ \phi_{\varepsilon}(0^{+}) = \frac{1}{\ln 2} \ln(1 + 1/\varepsilon) = \frac{1}{\ln 2} \left( \frac{1}{\varepsilon} + o(1/\varepsilon) \right), $$

and divide numerator and denominator to obtain (5.3g). (Rudin, 1976). ■

## 5.4 Theorem Redundancy saturation

### Setup and affine form in redundancy

Fix a session t. With $x_i := x_{t,i} \in (0,1]$ and $R_i := R_{t,i} \in [0,1]$, define

$$ \mathrm{AAS}_t(R; x) = \sum_{i=1}^{m} w_i (1 - R_i) \phi_{\varepsilon}(x_i), \quad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}, \quad \varepsilon > 0. \quad (5.4-1) $$

Introduce the nonnegative coefficients

$$ b_i := w_i \phi_{\varepsilon}(x_i) \ge 0, \quad C(x) := \sum_{i=1}^{m} b_i < \infty. \quad (5.4-2) $$

Then, for fixed x, the score is affine in R:

$$ \mathrm{AAS}_t(R; x) = C(x) - \sum_{i=1}^{m} b_i R_i. \quad (5.4-3) $$

In particular, as $R_i \uparrow 1$, the term $b_i R_i \to b_i$ and the $i$-th channel contribution is driven to 0.

### Theorem 5.4 (a–g): Saturation laws in R

Under (5.4-1)–(5.4-2), the following hold.

#### (a) Single-channel saturation

Fix $R_{-i}$ and let $R_i \to 1$. Then

$$ \lim_{R_i \to 1} \mathrm{AAS}_t(R; x) = \sum_{j \neq i} w_j (1 - R_j) \phi_{\varepsilon}(x_j). \quad (5.4a) $$

#### (b) Multi-channel saturation
Fix $S \subseteq \{1, ..., m\}$ and hold $R_j$ fixed for $j \notin S$. If $R_i \to 1$ for all $i \in S$, then

$$ \lim_{R_i \to 1, i \in S} \mathrm{AAS}_t(R; x) = \sum_{j \notin S} w_j (1 - R_j) \phi_{\epsilon}(x_j). \quad (5.4b) $$

In particular, if $S = \{1, ..., m\}$, then $\lim_{R \to 1} \mathrm{AAS}_t(R; x) = 0$.

### (c) Monotone linear decrease

For each $i$ on the interior $R_i \in (0,1)$,

$$ \frac{\partial \mathrm{AAS}_t}{\partial R_i} = -w_i \phi_{\epsilon}(x_i) = -b_i \le 0. \quad (5.4c) $$

The inequality is strict whenever $b_i > 0$ equivalently $w_i > 0$ and $x_i < 1$. At the boundary $R_i = 1$, further increases are infeasible, so the monotonicity is interpreted along feasible directions in $[0,1]^m$.

### (d) Global Lipschitz bounds on $[0, 1]^m$

For all $R, R' \in [0,1]^m$, using the affine form

$$ \mathrm{AAS}_t(R; x) = C(x) - \sum_{i=1}^{m} b_i R_i, $$

it follows that

$$ \mathrm{AAS}_t(R; x) - \mathrm{AAS}_t(R'; x) = \left( C(x) - \sum_{i=1}^{m} b_i R_i \right) - \left( C(x) - \sum_{i=1}^{m} b_i R'_i \right) = \sum_{i=1}^{m} b_i (R'_i - R_i). $$

Hence,

$$ |\mathrm{AAS}_t(R; x) - \mathrm{AAS}_t(R'; x)| = \left| \sum_{i=1}^{m} b_i (R'_i - R_i) \right| = \left| \sum_{i=1}^{m} b_i (R_i - R'_i) \right| \le \sum_{i=1}^{m} b_i |R_i - R'_i|. \quad (5.4d) $$

### $l_\infty$-Lipschitz bound

Since $|R_i - R'_i| \le |R - R'|_\infty$ for every $i$, it follows that

$$ \sum_{i=1}^{m} b_i |R_i - R'_i| \le \sum_{i=1}^{m} b_i |R - R'|_\infty = \left( \sum_{i=1}^{m} b_i \right) |R - R'|_\infty = C(x) |R - R'|_\infty, \quad (5.4d1) $$
which yields the $l_\infty$-Lipschitz bound.

**l₁-Lipschitz bound.**

Also, since $b_i \le \max_k b_k$ for all i,

$$ \sum_{i=1}^{m} b_i |R_i - R'_i| \le \sum_{i=1}^{m} (\max_k b_k) |R_i - R'_i| = (\max_k b_k) \sum_{i=1}^{m} |R_i - R'_i| \\ = (\max_i b_i) |R - R'|_1, \quad (5.4d2) $$

which yields the l₁-Lipschitz bound (Boyd & Vandenberghe, 2004; Kreyszig, 1978).

**(e) Order invariance of sequential saturations**

For any $i \neq j$,

$$ \mathrm{AAS}_t(R_i = 1, R_j = 1; x) = \mathrm{AAS}_t(R_j = 1, R_i = 1; x) = \sum_{k \notin \{i,j\}} w_k (1 - R_k) \phi_{\varepsilon}(x_k). \quad (5.4e) $$

**(f) Clamp update: linear decrease until saturation**

Fix $S \subseteq \{1, ..., m\}$ and $\delta > 0$. Define

$$ R_i^{\text{new}} = \min\{1, R_i + \delta\} \quad (i \in S), \quad R_j^{\text{new}} = R_j \quad (j \notin S). $$

Then

$$ \mathrm{AAS}_t(R; x) - \mathrm{AAS}_t(R^{\text{new}}; x) = \sum_{i \in S} b_i \min\{\delta, 1 - R_i\}. \quad (5.4f) $$

**(g) Saturation rate via constant gradient ceilings.**

Since $\nabla_R \mathrm{AAS}_t(R; x) = -(b_1, ..., b_m)$ is constant,

$$ |\nabla_R \mathrm{AAS}_t|_\infty = \max_i b_i, \quad |\nabla_R \mathrm{AAS}_t|_1 = \sum_{i=1}^{m} b_i = C(x). \quad (5.4g) $$

Thus, the largest instantaneous decrease per unit increase in a single redundancy coordinate is achieved on any channel attaining $\max_i b_i = w_i \phi_\varepsilon(x_i)$.

**Proof.**

From (5.4-1),
$$AAS_t(R; x) = \sum_{i=1}^{m} w_i \phi_{\varepsilon}(x_i) - \sum_{i=1}^{m} w_i \phi_{\varepsilon}(x_i) R_i = C(x) - \sum_{i=1}^{m} b_i R_i,$$

which is (5.4-0).

Parts (a)–(b) follow by taking coordinatewise limits in (5.4-0).

Part (c) follows by differentiating (5.4-0) on the interior and restricting to feasible directions on $[0,1]^m$ at the boundary.

Part (d) follows from $AAS_t(R; x) - AAS_t(R'; x) = \sum_i b_i(R_i - R'_i)$ and standard norm inequalities.

Part (e) is immediate from separability in (5.4-0).

Part (f) uses $R_i^{\text{new}} - R_i = \min\{\delta, 1 - R_i\}$ and linearity in each $R_i$.

Part (g) holds because the gradient in R is constant. ■

## Corollaries 5.4 Design consequences of redundancy saturation

Recall

$$AAS_t(R; x) = C(x) - \sum_{i=1}^{m} b_i R_i, \quad b_i := w_i \phi_{\varepsilon}(x_i) \ge 0, \quad C(x) := \sum_{i=1}^{m} b_i,$$

with $x_i \in (0,1], R_i \in [0,1], \varepsilon > 0$, and $\phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}$.

### Corollary 5.4.1 Dominance ordering

If $b_i \ge b_j$, then for any feasible increment $\Delta \in [0, \min\{1 - R_i, 1 - R_j\}]$,

$$AAS_t(R; x) - AAS_t(R + \Delta e_i; x) \ge AAS_t(R; x) - AAS_t(R + \Delta e_j; x).$$

**Proof.** By (5.4-0),

$$AAS_t(R; x) - AAS_t(R + \Delta e_i; x) = b_i \Delta, \quad AAS_t(R; x) - AAS_t(R + \Delta e_j; x) = b_j \Delta.$$

Since $\Delta \ge 0$ and $b_i \ge b_j$, the claim follows. ■

### Corollary 5.4.2 Optimal allocation under a redundancy budget

Consider
$$ \max_{\Delta R \in R^m} \sum_{i=1}^{m} b_i \Delta R_i \quad \text{subject to} \quad 0 \le \Delta R_i \le 1 - R_i, \quad \sum_{i=1}^{m} \Delta R_i \le B. $$

where $B \ge 0$ is a redundancy budget. An optimal solution allocates increments in nonincreasing order of $b_i$, saturating the largest-$b_i$ coordinates first and if needed assigning the remaining fractional budget to the next coordinate.

**Justification.** This is a linear program, so an optimizer occurs at an extreme point; an exchange argument shows that reallocating any positive mass from a smaller coefficient to a larger one cannot decrease the objective. (Boyd & Vandenberghe, 2004). ■

### Corollary 5.4.3 Effect on the normalized score

Let

$$ \alpha_t(R) := \sum_{i=1}^{m} w_i(1 - R_i), \quad \text{AAS}_t^{\max}(R) := \alpha_t(R) \phi_{\varepsilon}(0^{+}), \\ \widetilde{\text{AAS}}_t(R; x) := \frac{\text{AAS}_t(R; x)}{\text{AAS}_t^{\max}(R)} \quad (\alpha_t(R) > 0). $$

Although $\text{AAS}_t(R; x)$ decreases coordinatewise in $R, \widetilde{\text{AAS}}_t(R; x)$ need not be monotone in $R$ because the denominator depends on $\alpha_t(R)$, which also decreases. For $\alpha_t(R) > 0$,

$$ \widetilde{\text{AAS}}_t(R; x) = \sum_{i=1}^{m} \beta_i(R) \rho_{\varepsilon}(x_i), \quad \beta_i(R) := \frac{w_i(1 - R_i)}{\alpha_t(R)}, \quad \sum_{i=1}^{m} \beta_i(R) = 1, \quad \rho_{\varepsilon}(x) := \frac{\phi_{\varepsilon}(x)}{\phi_{\varepsilon}(0^{+})} \in [0,1]. $$

Hence changing $R$ changes the mixture weights $\beta_i(R)$, so $\widetilde{\text{AAS}}_t$ may increase or decrease depending on which channels are attenuated. A special case recovers the scale-invariance principle in 4.6: if

$(1 - R_i) \mapsto \lambda(1 - R_i)$ for all $i$ with feasibility maintained, then $\beta_i(R)$ is unchanged and $\widetilde{\text{AAS}}_t$ is unchanged. ■

### Corollary 5.4.4 Gating interpretation

If $R_i \uparrow 1$, then $\alpha_{t,i} = w_i(1 - R_i) \downarrow 0$ and $w_i(1 - R_i)\phi_{\varepsilon}(x_i) \to 0$; redundancy saturation therefore gates channel $i$ off in the aggregation. ■

## 6. Refinement and coarsening principles

This section formalizes refinement invariance merge-split identities for AAS under fixed accuracy, and then characterizes the corresponding coarsening inequalities that arise when accuracies differ across
subchannels. The results clarify when channelization changes are score-neutral and when they necessarily alter the total penalty due to the convex geometry of the kernel $\phi_{\epsilon}$.

## 6.1 Theorem Refinement invariance; merge split identity under fixed accuracy

**Setup refinement under fixed x. Fix t. Recall**

$$ \mathrm{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_{t,i}), \quad \phi_{\epsilon}(x) = \log_2 \frac{1+\epsilon}{x+\epsilon}, \quad \epsilon > 0, $$

with effective weights $\alpha_{t,i} \ge 0$. Choose a channel $i^*$ with parameters $(\alpha, x) = (\alpha_{t,i^*}, x_{t,i^*})$. For an integer $k \ge 2$, replace $i^*$ by subchannels $\{i_1^*, \dots, i_k^*\}$ such that

$$ x_{t,i_j^*} = x, \quad \alpha_{t,i_j^*} = \alpha^{(j)} \ge 0, \quad \sum_{j=1}^{k} \alpha^{(j)} = \alpha. \quad (6.1a) $$

All other channels remain unchanged. This refinement constraint is imposed at the $\alpha$-level; any factorization $\alpha^{(j)} = w^{(j)}(1-R^{(j)})$ consistent with (6.1a) is admissible.

## Theorem 6.1 Refinement invariance

Under (6.1a),

$$ \alpha \phi_{\epsilon}(x) = \sum_{j=1}^{k} \alpha^{(j)} \phi_{\epsilon}(x). \quad (6.1b) $$

Equivalently, splitting a channel $(\alpha, x)$ into finitely many subchannels $\{(\alpha^{(j)}, x)\}_{j=1}^k$ with $\sum_j \alpha^{(j)} = \alpha$ leaves the total score unchanged:

$$ \mathrm{AAS}_t^{\mathrm{after}} = \mathrm{AAS}_t^{\mathrm{before}}. \quad (6.1c) $$

**Proof.** Since $x_{t,i_j^*} = x$ for all $j$, the factor $\phi_{\epsilon}(x)$ is common on the refined block, so

$$ \sum_{j=1}^{k} \alpha^{(j)} \phi_{\epsilon}(x) = \left( \sum_{j=1}^{k} \alpha^{(j)} \right) \phi_{\epsilon}(x) = \alpha \phi_{\epsilon}(x), $$

which proves (6.1b). Writing the full sum before and after refinement gives

$$ \mathrm{AAS}_t^{\mathrm{before}} = \sum_{i \neq i^*} \alpha_{t,i} \phi_{\epsilon}(x_{t,i}) + \alpha \phi_{\epsilon}(x), \quad \mathrm{AAS}_t^{\mathrm{after}} = \sum_{i \neq i^*} \alpha_{t,i} \phi_{\epsilon}(x_{t,i}) + \sum_{j=1}^{k} \alpha^{(j)} \phi_{\epsilon}(x), $$
and substituting (6.1b) yields (6.1c). Boundary cases are immediate: if $\alpha = 0$ then all $\alpha^{(j)} = 0$; if $x=1$ then $\phi_{\varepsilon}(1) = 0$. ■

**Remarks.**

1. **Block contribution form.** If $x$ is constant on a block, then

$$ \text{Contribution(block)} = \left( \sum_{i \in \text{block}} \alpha_{t,i} \right) \phi_{\varepsilon}(x), $$

hence refinement preserves the contribution by preserving both the $\alpha$-mass and the common $x$.

2. **Measure-theoretic representation.** Let $\mu = \sum_i \alpha_{t,i} \delta_{x_{t,i}}$ on $(0,1]$. Then

$$ \text{AAS}_{t} = \int \phi_{\varepsilon}(x) \, d\mu(x), $$

and replacing $\alpha \delta_x$ by $\sum_j \alpha^{(j)} \delta_x$ leaves $\mu$ unchanged at that atom.

3. **Merging inverse operation.** If $\{(\alpha^{(j)}, x)\}_{j=1}^k$ share the same $x$, merging them into $(\sum_j \alpha^{(j)}, x)$ preserves the score; this is the inverse of refinement.

4. **Why fixed $x$ matters.** If accuracies differ across parts $\{x^{(j)}\}$ and $\sum_j \alpha^{(j)} > 0$, convexity of $\phi_{\varepsilon}$ yields the Jensen-type inequality

$$ \sum_{j=1}^{k} \alpha^{(j)} \phi_{\varepsilon}(x^{(j)}) \geq \left( \sum_{j=1}^{k} \alpha^{(j)} \right) \phi_{\varepsilon} \left( \frac{\sum_{j=1}^{k} \alpha^{(j)} x^{(j)}}{\sum_{j=1}^{k} \alpha^{(j)}} \right) $$

with equality if and only if all $x^{(j)}$ are equal (Boyd & Vandenberghe, 2004). Thus, Theorem 6.1 identifies the equality regime of the coarsening inequality formalized in Theorem 6.2.

## 6.2 Theorem Coarsening and Jensen inequality

### Setup and notation

Fix a session t. Recall

$$ \text{AAS}_{t} = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \alpha_{t,i} \ge 0, \quad \alpha_{t} := \sum_{i=1}^{m} \alpha_{t,i} > 0, $$

with kernel
$$ \phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon}, \quad x \in (0,1], \quad \varepsilon > 0. $$

Since

$$ \phi_{\varepsilon}''(x) = \frac{1}{(x + \varepsilon)^2 \ln 2} > 0 \quad (x \in (0,1]), $$

$\phi_{\varepsilon}$ is strictly convex on $(0,1]$. Define normalized weights

$$ p_i := \frac{\alpha_{t,i}}{\alpha_t} \left( \sum_{i=1}^{m} p_i = 1 \right), $$

and the $\alpha$-weighted mean accuracy

$$ \bar{x} := \sum_{i=1}^{m} p_i x_{t,i} \in (0,1]. $$

**Theorem 6.2 Jensen coarsening inequality**

$$ \text{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) \geq \alpha_t \phi_{\varepsilon}(\bar{x}). \quad (6.2a) $$

Equality holds if and only if $x_{t,i} = \bar{x}$ for every active coordinate $\alpha_{t,i} > 0$. Equivalently, replacing any collection of channels by a single representative channel at the $\alpha$-weighted mean accuracy weakly decreases the total penalty, with strict decrease unless the block has constant accuracy, the equality regime of Theorem 6.1.

**Proof**

By Jensen's inequality for the strictly convex $\phi_{\varepsilon}$ with weights $\{p_i\}$,

$$ \sum_{i=1}^{m} p_i \phi_{\varepsilon}(x_{t,i}) \geq \phi_{\varepsilon} \left( \sum_{i=1}^{m} p_i x_{t,i} \right) = \phi_{\varepsilon}(\bar{x}). $$

Multiplying by $\alpha_t$ yields (6.2a). Strict convexity implies equality iff $x_{t,i}$ is constant on the support of $\{p_i\}$, i.e., on $\{i: \alpha_{t,i} > 0\}$. ■

**Quantitative refinement: Jensen gap lower bound**

Assume $x_{min} \leq x_{t,i} \leq x_{max}$ define the $\alpha$-weighted variance
$$Var_p(x) := \sum_{i=1}^{m} p_i (x_{t,i} - \bar{x})^2.$$

A second-order Taylor expansion of $\phi_{\epsilon}$ around $\bar{x}$ gives, for each i,

$$\phi_{\epsilon}(x_{t,i}) = \phi_{\epsilon}(\bar{x}) + \phi'_{\epsilon}(\bar{x})(x_{t,i} - \bar{x}) + \frac{1}{2} \phi''_{\epsilon}(\xi_i)(x_{t,i} - \bar{x})^2.$$

for some $\xi_i$ between $x_{t,i}$ and $\bar{x}$. Averaging with weights $p_i$ cancels the linear term, hence

$$\sum_{i=1}^{m} p_i \phi_{\epsilon}(x_{t,i}) - \phi_{\epsilon}(\bar{x}) = \frac{1}{2} \sum_{i=1}^{m} p_i \phi_{\epsilon}''(\xi_i) (x_{t,i} - \bar{x})^2 \geq \frac{1}{2} \left( \min_{x \in [x_{min}, x_{max}]} \phi_{\epsilon}''(x) \right) Var_p(x). \text{ Since} \\ \phi_{\epsilon}''(x) = 1/((x + \epsilon)^2 \ln 2) \text{ decreases in } x, \text{ the minimum over } [x_{min}, x_{max}]$$

occurs at $x_{max}$. Therefore,

$$\text{AAS}_t - \alpha_t \phi_{\epsilon}(\bar{x}) \geq \frac{\alpha_t}{2(x_{max}+\epsilon)^2 \ln 2} Var_p(x) \geq \frac{\alpha_t}{2(1+\epsilon)^2 \ln 2} Var_p(x). \quad (6.2)$$

The bound is zero iff $Var_p(x) = 0$, i.e., iff all active accuracies coincide, matching the equality case in (6.2a). ■

## Two-level grouped coarsening

Let $G = \{G_1, ..., G_K\}$ be a partition of $\{1, ..., m\}$. Define

$$\alpha_{t,G} := \sum_{i \in G} \alpha_{t,i}, \quad \bar{x}_G := \frac{1}{\alpha_{t,G}} \sum_{i \in G} \alpha_{t,i} x_{t,i} \quad (\alpha_{t,G} > 0).$$

Applying Jensen within each group and summing yields

$$\sum_{i=1}^{m} \alpha_{t,i} \phi_{\epsilon}(x_{t,i}) \geq \sum_{G \in G} \alpha_{t,G} \phi_{\epsilon}(\bar{x}_G). \quad (6.2c)$$

Applying Jensen again across groups with weights $\alpha_{t,G}/\alpha_t$ gives

$$\sum_{G \in G} \alpha_{t,G} \phi_{\epsilon}(\bar{x}_G) \geq \alpha_t \phi_{\epsilon} \left( \sum_{G \in G} \frac{\alpha_{t,G}}{\alpha_t} \bar{x}_G \right) = \alpha_t \phi_{\epsilon}(\bar{x}). \quad (6.2d)$$

Hence
$$AAS_t \geq \sum_{G \in \mathcal{G}} \alpha_{t,G} \phi_{\varepsilon}(\overline{x_G}) \geq \alpha_t \phi_{\varepsilon}(\bar{x}). \quad (6.2e)$$

Equalities require constant accuracy within each active group and coincident group means, equivalently constant accuracy on the full active set.

## 6.3 Theorem Order Preservation-Homogeneous Case

**Setting and homogeneity.** Fix a session t. Recall

$$AAS_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}, \varepsilon > 0,$$

with $x_{t,i} \in (0,1]$ and $\alpha_{t,i} \geq 0$. Let

$$I_t := \{i \in \{1, \dots, m\} : \alpha_{t,i} > 0\}$$

be the active set. Assume homogeneity on active channels: there exists $\alpha > 0$ such that

$$\alpha_{t,i} = \alpha \text{ for all } i \in I_t, \quad \alpha_{t,i} = 0 \text{ for } i \notin I_t. \quad (6.3a)$$

**Theorem 6.3 Pairwise order preservation, Top K preservation.**

Under (6.3a), define the active contribution

$$c_{t,i} := \alpha \phi_{\varepsilon}(x_{t,i}) \quad (i \in I_t).$$

Then:

1. **Pairwise order active channels.** For any $i, j \in I_t$,

$$x_{t,i} < x_{t,j} \Leftrightarrow c_{t,i} > c_{t,j}. \quad (6.3b)$$

If $x_{t,i} = x_{t,j}$, then $c_{t,i} = c_{t,j}$ ties are preserved.

2. **Top K preservation.** For any $K \in \{1, \dots, |I_t|\}$,

$$\text{TopK}(c_{t,\cdot}) = \text{BottomK}(x_{t,\cdot}) \quad \text{on } I_t. \quad (6.3c)$$

**Proof.**

**Step 1: Order reversal.** On $(0,1]$,
$$ \phi'_{\varepsilon}(x) = - \frac{1}{(x + \varepsilon) \ln 2} < 0, $$

so $\phi_{\varepsilon}$ is strictly decreasing. Since $\alpha > 0$, the map $x \mapsto c = \alpha\phi_{\varepsilon}(x)$ is strictly decreasing, hence (6.3b) follows immediately.

**Step 2: two-sided band via MVT.** For $i, j \in I_t$, the Mean Value Theorem gives a $\xi$ between $x_{t,i}$ and $x_{t,j}$ such that

$$ c_{t,i} - c_{t,j} = \alpha (\phi_{\varepsilon}(x_{t,i}) - \phi_{\varepsilon}(x_{t,j})) = \alpha \phi'_{\varepsilon}(\xi) (x_{t,i} - x_{t,j}). $$

(Rudin, 1976). Thus

$$ |c_{t,i} - c_{t,j}| = \frac{\alpha}{(\xi + \varepsilon) \ln 2} |x_{t,i} - x_{t,j}|. $$

Let $x_{min} := \min\{x_{t,i}, x_{t,j}\}$ and $x_{max} := \max\{x_{t,i}, x_{t,j}\}$. Since $x \mapsto 1/(x + \varepsilon)$ is decreasing,

$$ \frac{1}{x_{max} + \varepsilon} \le \frac{1}{\xi + \varepsilon} \le \frac{1}{x_{min} + \varepsilon'} $$

Which yields

$$ \frac{\alpha}{(x_{max} + \varepsilon) \ln 2} |x_{t,i} - x_{t,j}| \le |c_{t,i} - c_{t,j}| \le \frac{\alpha}{(x_{min} + \varepsilon) \ln 2} |x_{t,i} - x_{t,j}|. (6.3d) $$

**Step 3: Top K.** Because $x \mapsto \alpha\phi_{\varepsilon}(x)$ is strictly decreasing, it reverses the order on $I_t$; therefore the largest K values of $\{c_{t,i}\}_{i \in I_t}$ correspond exactly to the smallest K values of $\{x_{t,i}\}_{i \in I_t}$, proving (6.3c). ■

## 6.4 Theorem Additivity over disjoint index sets

**Setup.** Fix t. Write

$$ \text{AAS}_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad c_{t,i} := \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}) \ge 0, $$

$$ \text{where } \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon} \text{ and } \alpha_{t,i} \ge 0. $$

For any $S \subseteq \{1, ..., m\}$, define the restricted score

$$ \text{AAS}_t(S) := \sum_{i \in S} c_{t,i}. $$
**Theorem 6.4 Finite additivity and inclusion-exclusion**

If $S \cap T = \emptyset$, then

$$ AAS_t(S \cup T) = AAS_t(S) + AAS_t(T) \quad (6.4a) $$

More generally, for any finite pairwise-disjoint family $\mathcal{G} = \{G_1, \dots, G_K\}$,

$$ AAS_t\left(\bigcup_{k=1}^{K} G_k\right) = \sum_{k=1}^{K} AAS_t(G_k) \quad (6.4b) $$

For arbitrary $S, T$ possibly overlapping,

$$ AAS_t(S \cup T) = AAS_t(S) + AAS_t(T) - AAS_t(S \cap T). \quad (6.4c) $$

Thus, disjoint buckets aggregate directly, while overlaps require subtracting the intersection contribution (Feller, 1968).

**Proof.** Let $\mathbf{1}_S(i)$ be the indicator of $i \in S$. Then

$$ AAS_t(S) = \sum_{i=1}^{m} \mathbf{1}_S(i) c_{t,i}. $$

If $S \cap T = \emptyset$, then $\mathbf{1}_{S \cup T}(i) = \mathbf{1}_S(i) + \mathbf{1}_T(i)$, so summing against $c_{t,i}$ gives (6.4a).

If $G_1, \dots, G_K$ are pairwise disjoint, then $\mathbf{1}_{\cup_k G_k}(i) = \sum_{k=1}^K \mathbf{1}_{G_k}(i)$; exchanging finite sums yields (6.4b).

For general $S, T$, the identity $\mathbf{1}_{S \cup T} = \mathbf{1}_S + \mathbf{1}_T - \mathbf{1}_{S \cap T}$ implies (6.4c) after summation. $\blacksquare$

**7.0 Preliminaries: curvature and smoothness tools**

This section records curvature and smoothness facts for the AAS objective that will be used repeatedly in the refinement/coarsening arguments and in later stability results. In particular, we establish: (i) convexity of $AAS_t$, (ii) restricted strong convexity on the active support $I_t = \{i: \alpha_{t,i} > 0\}$, and (iii) Lipschitz continuity of the gradient on the interior domain. These properties underpin the Jensen-type bounds in 6.2 and provide quantitative stability constants.

**7.1 Lemma Convexity, restricted strong convexity on the active support, and smoothness**

**Setup.** Fix a session $t$ and write $x_i := x_{t,i}$. Define

$$ AAS_t(x) := \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_i), \quad x \in (0,1]^m, \varepsilon > 0, \alpha_{t,i} \ge 0, $$
with kernel

$$ \phi_{\varepsilon}(x) := \log_2 \frac{1 + \varepsilon}{x + \varepsilon}, \quad x \in (0,1]. $$

**Lemma 7.1.** Under the setup above:

**(a) Kernel derivatives, strict convexity, smoothness**

For $x \in (0,1)$,

$$ \phi'_{\varepsilon}(x) = -\frac{1}{(x + \varepsilon) \ln 2}, \quad \phi''_{\varepsilon}(x) = \frac{1}{(x + \varepsilon)^2 \ln 2} > 0. \quad (7.1a) $$

Hence $\phi_{\varepsilon}$ is strictly convex on $(0,1]$, is $C^{\infty}$ on $(0,1)$, and satisfies the continuous boundary value

$$ \phi_{\varepsilon}(1) = 0. $$

**(b) Convexity of AAS$_t$**

Since AAS$_t$ is a nonnegative weighted sum of convex coordinate functions $x_i \mapsto \phi_{\varepsilon}(x_i)$,

$$ \text{AAS}_t(x) = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_i) (7.1b) $$

is convex on $(0,1]^m$ (Boyd & Vandenberghe, 2004). Moreover, it is strictly convex along any direction that changes at least one active coordinate i with $\alpha_{t,i} > 0$.

**(c) Restricted strong convexity on the active support**

Let

$I_t := \{i: \alpha_{t,i} > 0\}$, $\alpha_{min} := \min_{i \in I_t} \alpha_{t,i}$ ($I_t \neq \emptyset$). en AAS$_t$ is $\mu$-strongly convex on the active subspace, with

$$ \mu := \frac{\alpha_{min}}{(1 + \varepsilon)^2 \ln 2}. \quad (7.1c) $$

equivalently, for all $x, y \in (0,1)^m$,

$$ \text{AAS}_t(y) \ge \text{AAS}_t(x) + \langle \nabla \text{AAS}_t(x), y - x \rangle + \frac{\mu}{2} |(y - x)_{I_t}|_2^2. \quad (7.1d) $$

Coordinates with $\alpha_{t,i} = 0$ contribute neither curvature nor gradient signal. If an a priori bound $x_i \le b < 1$ holds on active coordinates, $\mu$ improves to $\alpha_{min}/((b + \varepsilon)^2 \ln 2)$
(Nesterov, 2004). The inequality extends to $x, y \in (0,1]^m$ by continuity when some coordinates approach 1.

### (d) Hessian form and L-smoothness on the interior

On $(0,1)^m$, AAS$_t \in C^2$ and has diagonal Hessian

$$ \nabla^2 \mathrm{AAS}_t(x) = \mathrm{diag} \left( \frac{\alpha_{t,i}}{(x_i + \varepsilon)^2 \ln 2} \right). (7.1e) $$

Thus, the Hessian is positive definite on the active block. Moreover,

$$ |\nabla^2 \mathrm{AAS}_t(x)|_2 = \max_{1 \le i \le m} \frac{\alpha_{t,i}}{(x_i + \varepsilon)^2 \ln 2} \le \frac{\alpha_{max}}{\varepsilon^2 \ln 2}, \quad \alpha_{max} := \max_i \alpha_{t,i}, (7.1f) $$

so $\nabla \mathrm{AAS}_t$ is L-Lipschitz in $l_2$ on $(0,1)^m$ with $L := \alpha_{max}/(\varepsilon^2 \ln 2)$

(Nesterov, 2004). This bound remains valid as coordinates approach 1, since $x_i + \varepsilon \ge \varepsilon$.

**Proof.** Differentiate $\phi_\varepsilon$ to obtain (7.1a) and strict convexity (Rudin, 1976). Convexity of AAS$_t$ follows from closure of convexity under nonnegative sums. Separability yields (7.1e). The lower Hessian bound on active coordinates, using $x_i \le 1$, gives $\mu$ and implies (7.1d) (Nesterov, 2004). The uniform upper bound in (7.1f) gives L-smoothness. ■

### Corollary 7.1.1 Bregman gap bounds

Define

$$ D_{\mathrm{AAS}_t}(y, x) := \mathrm{AAS}_t(y) - \mathrm{AAS}_t(x) - \langle \nabla \mathrm{AAS}_t(x), y - x \rangle. $$

This is the Bregman divergence induced by the convex generator AAS$_t$ (Bregman, 1967).

For all $x, y \in (0,1)^m$,

$$ \frac{\mu}{2} |(y - x)|_2^2 \le D_{\mathrm{AAS}_t}(y, x) \le \frac{L}{2} |y - x|^2, \quad (7.1.1) $$

where the lower bound acts only on the active support (Boyd & Vandenberghe, 2004; Nesterov, 2004). The bounds extend to $(0,1]^m$ by continuity at coordinates equal to 1.

### Corollary 7.1.2 Jensen lower bound basis for 6.2

Assume $\alpha_t := \sum_{i=1}^m \alpha_{t,i} > 0$ and set $p_i := \alpha_{t,i}/\alpha_t$ so $\sum_i p_i = 1$. Then, by Jensen's inequality,

$$ \mathrm{AAS}_t = \alpha_t \sum_{i=1}^m p_i \phi_\varepsilon(x_i) \ge \alpha_t \phi_\varepsilon \left( \sum_{i=1}^m p_i x_i \right), (7.1.2) $$

as used in 6.2 (Boyd & Vandenberghe, 2004).
### Corollary 7.1.3 Unique minimizer on active coordinates

Let $I_t = \{i: \alpha_{t,i} > 0\}$. Since $\phi_{\epsilon}$ is strictly decreasing on $(0,1]$ and $\phi_{\epsilon}(1) = 0$, the minimizer is uniquely determined on active coordinates:

$$x_i^* = 1 \quad \text{for all } i \in I_t, (7.1.3a)$$

while indices with $\alpha_{t,i} = 0$ are free. Equivalently,

$$\arg\min_{x \in (0,1]^m} \text{AAS}_t(x) = \{x \in (0,1]^m: x_i = 1 \ \forall i \in I_t\}, (7.1.3b)$$

and

$$\min_{x \in (0,1]^m} \text{AAS}_t(x) = 0. (7.1.3c)$$

### Corollary 7.1.4 Coordinate-wise strong convexity; active-subspace constant

For each $i \in I_t$, let $f_i(x_i) := \alpha_{t,i} \phi_{\epsilon}(x_i)$. On $(0,1)$,

$$f_i''(x_i) = \frac{\alpha_{t,i}}{(x_i + \epsilon)^2 \ln 2} \geq \frac{\alpha_{t,i}}{(1 + \epsilon)^2 \ln 2}$$

so $f_i$ is $\mu_i$ strongly convex with

$$\mu_i = \frac{\alpha_{t,i}}{(1 + \epsilon)^2 \ln 2}. (7.1.4a)$$

Hence the restricted constant in Lemma 7.1(c) is

$$\mu = \min_{i \in I_t} \mu_i = \frac{\alpha_{min}}{(1 + \epsilon)^2 \ln 2}, \quad \alpha_{min} := \min_{i \in I_t} \alpha_{t,i}. (7.1.4b)$$

(Nesterov, 2004).

### Corollary 7.1.5 Step-size bound from smoothness interior descent

On $(0,1)^m$, $\nabla \text{AAS}_t$ is L-Lipschitz in $l_2$ with $L \leq \alpha_{max}/(\epsilon^2 \ln 2)$ (Lemma 7.1(d)). For the gradient step

$$x^{k+1} = x^k - \eta \nabla \text{AAS}_t(x^k),$$

assuming iterates remain in $(0,1)^m$, the descent lemma yields for any $\eta \in (0,1/L]$,

$$\text{AAS}_t(x^{k+1}) \leq \text{AAS}_t(x^k) - \frac{\eta}{2} |\nabla \text{AAS}_t(x^k)|_2^2, (7.1.5a)$$

and in particular for $\eta = 1/L$,
$$ \mathrm{AAS}_t(x^{k+1}) \le \mathrm{AAS}_t(x^k) - \frac{1}{2L} |\nabla \mathrm{AAS}_t(x^k)|_2^2. (7.1.5b) $$

(Nesterov, 2004).

## 7.2 Lemma Jensen lower bound, normalized form, and Jensen-gap representations

**Setup.** Fix a session t and write $x_i := x_{t,i}$. Define

$$ \mathrm{AAS}_t(x) := \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_i), \quad \alpha_{t,i} \ge 0, \quad \alpha_t := \sum_{i=1}^{m} \alpha_{t,i} > 0, $$

where

$$ \phi_{\varepsilon}(x) := \log_2 \frac{1 + \varepsilon}{x + \varepsilon}, \quad x \in (0,1], \varepsilon > 0. $$

Let $p_i := \alpha_{t,i}/\alpha_t$ so that $\sum_{i=1}^m p_i = 1$ and the $\alpha$-weighted mean

$$ \bar{x} := \sum_{i=1}^{m} p_i x_i \in (0,1]. $$

**Lemma 7.2.**

### 1. Jensen lower bound.

$$ \mathrm{AAS}_t = \alpha_t \sum_{i=1}^{m} p_i \phi_{\varepsilon}(x_i) \ge \alpha_t \phi_{\varepsilon}(\bar{x}) \quad (7.2a) $$

Equality holds iff $x_i = \bar{x}$ for all indices with $\alpha_{t,i} > 0$.

### 2. Normalized bound and induced order constraint. Define

$$ \overline{\mathrm{AAS}}_t := \frac{\mathrm{AAS}_t}{\alpha_t \phi_{\varepsilon}(0^+)}, \quad \phi_{\varepsilon}(0^+) = \log_2 \frac{1 + \varepsilon}{\varepsilon}. $$

Then

$$ \overline{\mathrm{AAS}}_t \ge \frac{\phi_{\varepsilon}(\bar{x})}{\phi_{\varepsilon}(0^+)} \in [0,1]. \quad (7.2b) $$

Consequently, for $O_t := 1 - \overline{\mathrm{AAS}}_t,$
$$O_t \le 1 - \frac{\phi_{\varepsilon}(\bar{x})}{\phi_{\varepsilon}(0^+)} \quad (7.2c)$$

**Proof.** By Lemma 7.1, $\phi_{\varepsilon}$ is strictly convex on $(0,1]$. Jensen's inequality with weights $\{p_i\}$ yields

$$\sum_{i=1}^{m} p_i \phi_{\varepsilon}(x_i) \ge \phi_{\varepsilon} \left( \sum_{i=1}^{m} p_i x_i \right) = \phi_{\varepsilon}(\bar{x})$$

(Hardy, Littlewood, & Pólya, 1952). Multiplying by $\alpha_t$ gives (7.2a). Strict convexity gives the stated equality condition. Dividing (7.2a) by $\alpha_t \phi_{\varepsilon}(0^+)$ yields (7.2b), and (7.2c) follows from $O_t = 1 - \overline{AAS}_t$.

## Jensen gap and equivalent representations

Define the Jensen gap

$$Gap := \sum_{i=1}^{m} p_i \phi_{\varepsilon}(x_i) - \phi_{\varepsilon}(\bar{x}) \ge 0.$$

### (i) Taylor/variance-weighted form.

For each $i$, there exists $\xi_i$ between $x_i$ and $\bar{x}$ such that

$$\phi_{\varepsilon}(x_i) = \phi_{\varepsilon}(\bar{x}) + \phi'_{\varepsilon}(\bar{x})(x_i - \bar{x}) + \frac{1}{2} \phi''_{\varepsilon}(\xi_i)(x_i - \bar{x})^2.$$

Averaging with weights $p_i$ cancels the linear term since $\sum_i p_i(x_i - \bar{x}) = 0$, hence

$$Gap = \frac{1}{2} \sum_{i=1}^{m} p_i \phi''_{\varepsilon}(\xi_i) (x_i - \bar{x})^2.$$

Let $x_{max} := \max\{x_i: p_i > 0\}$ and $Var_p(x) := \sum_{i=1}^{m} p_i(x_i - \bar{x})^2$. Since

$$\phi''_{\varepsilon}(x) = \frac{1}{(x + \varepsilon)^2 \ln 2} \quad \text{is decreasing in } x,$$

and $\xi_i \le x_{max}$, it follows that

$$Gap \ge \frac{1}{2(x_{max} + \varepsilon)^2 \ln 2} Var_p(x). \quad (7.2d)$$

### ii) Integral/curvature representations.
Using the first-order remainder identity,

$$ \Phi_{\varepsilon}(x_i) - \Phi_{\varepsilon}(\bar{x}) - \Phi'_{\varepsilon}(\bar{x})(x_i - \bar{x}) = \int_{\bar{x}}^{x_i} [\Phi'_{\varepsilon}(u) - \Phi'_{\varepsilon}(\bar{x})] \, du, $$

and averaging gives the exact representation

$$ \text{Gap} = \sum_{i=1}^{m} p_i \int_{\bar{x}}^{x_i} [\Phi'_{\varepsilon}(u) - \Phi'_{\varepsilon}(\bar{x})] \, du. \quad (7.2e') $$

(Rudin, 1976). Integrating once more yields

$$ \text{Gap} = \int_{0}^{1} (1-s) \sum_{i=1}^{m} p_i (x_i - \bar{x})^2 \Phi_{\varepsilon}''(\bar{x} + s(x_i - \bar{x})) ds \geq 0. \quad (7.2e'') $$

(iii) Normalized gap. Define

$$ \widetilde{Gap} := \overline{\text{AAS}}_t - \frac{\Phi_{\varepsilon}(\bar{x})}{\Phi_{\varepsilon}(0^+)} = \frac{\text{Gap}}{\Phi_{\varepsilon}(0^+)} \geq 0, $$

and combining with (7.2d) gives

$$ \widetilde{Gap} \geq \frac{1}{2 \Phi_{\varepsilon}(0^{+}) (x_{max} + \varepsilon)^2 \ln 2} Var_p(x). \quad (7.2f) $$

## 7.3 Theorem Positive homogeneity in $\alpha$; scale shape separation

### Setup

Fix t. Define

$$ \text{AAS}_t(\alpha; x) := \sum_{i=1}^{m} \alpha_{t,i} \Phi_{\varepsilon}(x_{t,i}), \quad \alpha_{t,i} = w_i(1 - R_{t,i}) \geq 0, $$

where

$$ \Phi_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon}, \quad x \in (0,1], \varepsilon > 0. $$

Write $\alpha_t := \sum_{i=1}^{m} \alpha_{t,i}$.

### Theorem 7.3 Positive homogeneity of degree 1
(i) Homogeneity in α-space. For every $\lambda \ge 0$,

$$ \mathrm{AAS}_t(\lambda\alpha; x) = \lambda \mathrm{AAS}_t(\alpha; x). \quad (7.3a) $$

(ii) Implication under the (w,R) parameterization.

Any operation that multiplies all $\alpha_{t,i}$ by the same $\lambda$ multiplies the score by $\lambda$. In particular, whenever feasibility is preserved,

$$ w_i \mapsto \lambda w_i \quad (\text{R fixed}), \quad \text{or} \quad (1-R_{t,i}) \mapsto \lambda(1-R_{t,i}) \quad (\text{w fixed}), $$

implies $\alpha_{t,i} \mapsto \lambda\alpha_{t,i}$ and therefore

$$ \mathrm{AAS}_t \mapsto \lambda \mathrm{AAS}_t. \quad (7.3b) $$

For redundancy scaling, feasibility requires

$$ R'_{t,i} = 1 - \lambda(1 - R_{t,i}) \in [0,1] \quad \Leftrightarrow \quad 0 \le \lambda \le \min_{i: 1-R_{t,i}>0} \frac{1}{1-R_{t,i}}. \quad (7.3b') $$

(iii) Scale invariance of the normalized score.

Define

$$ \overline{\mathrm{AAS}}_t(\alpha; x) := \frac{\mathrm{AAS}_t(\alpha; x)}{\alpha_t \phi_\varepsilon(0^+)}; \quad \phi_\varepsilon(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon}. $$

Then for every $\lambda > 0$,

$$ \overline{\mathrm{AAS}}_t(\lambda\alpha; x) = \overline{\mathrm{AAS}}_t(\alpha; x) \quad (7.3c) $$

Proof

Step 1: linearity $\Rightarrow$ degree-1 homogeneity.

$$ \mathrm{AAS}_t(\lambda\alpha; x) = \sum_{i=1}^{m} (\lambda\alpha_{t,i}) \phi_\varepsilon(x_{t,i}) = \lambda \sum_{i=1}^{m} \alpha_{t,i} \phi_\varepsilon(x_{t,i}) = \lambda \mathrm{AAS}_t(\alpha; x), $$

which gives (7.3a). The case $\lambda = 0$ yields $\mathrm{AAS}_t(0; x) = 0$.

Step 2: (w, R) interpretation.
Either scaling $w_i$ with R fixed, or scaling $(1 - R_{t,i})$ with w fixed, multiplies $\alpha_{t,i} = w_i(1 - R_{t,i})$ by $\lambda$, hence (7.3b). Condition (7.3b') is exactly $R'_{t,i} \in [0,1]$.

**Step 3: normalized invariance.**

Since $\alpha'_{t} = \sum_{i} \lambda \alpha_{t,i} = \lambda \alpha_{t}$,

$$ \overline{\text{AAS}}_{t}(\lambda\alpha; x) = \frac{\lambda \text{AAS}_{t}(\alpha; x)}{\lambda \alpha_{t} \phi_{\varepsilon}(0^{+})} = \overline{\text{AAS}}_{t}(\alpha; x), $$

proving (7.3c). ■

**Euler identity, elasticities, and contribution shares**

Because $\text{AAS}_t$ is differentiable and positively homogeneous of degree 1 in $\alpha$, Euler's theorem gives

$$ \sum_{i=1}^{m} \alpha_{t,i} \frac{\partial \text{AAS}_t}{\partial \alpha_{t,i}} = \text{AAS}_t. \quad (7.3d) $$

Moreover,

$$ \frac{\partial \text{AAS}_t}{\partial \alpha_{t,i}} = \phi_{\varepsilon}(x_{t,i}), $$

so (7.3d) is identical to $\text{AAS}_t = \sum_i \alpha_{t,i} \phi_{\varepsilon}(x_{t,i})$.

**(1) Elasticity equals contribution share**

Define

$$ \varepsilon_i^\alpha := \frac{\partial \text{AAS}_t}{\partial \alpha_{t,i}} \frac{\alpha_{t,i}}{\text{AAS}_t} = \frac{\alpha_{t,i} \phi_{\varepsilon}(x_{t,i})}{\text{AAS}_t} = \frac{c_{t,i}}{\text{AAS}_t} =: p_{t,i}. \quad (7.3e) $$

$p_{t,i}$ is the contribution share of channel i in $\text{AAS}_t$, $\sum_{i=1}^{m} p_{t,i} = 1$ ($\text{AAS}_t > 0$).

Hence $\varepsilon_i^\alpha$ equals the normalized contribution share, and $\sum_i \varepsilon_i^\alpha = 1$ follows from (7.3d).

**(2) Boundary regime**

If $\text{AAS}_t = 0$ e.g., $x_{t,i} = 1$ on the active set, or $\alpha_{t,i} \equiv 0$, then $p_{t,i}$ and $\varepsilon_i^\alpha$ are undefined division by zero; however, (7.3a) and (7.3d) remain valid identities.
## Scale versus shape on the normalized metric

Equation (7.3a) isolates a pure scale effect: multiplying $\alpha$ by $\lambda$ multiplies AAS$_t$ by $\lambda$. By (7.3c), $\overline{\text{AAS}}_t$ depends only on the ratios $\alpha_{t,i}/\alpha_t$, enabling scale-free comparisons across sessions/systems.

## Notes implementation and modeling nuances

1. Simplex constraint on $w$. If $\sum_i w_i = 1$ is imposed, $w \mapsto \lambda w$ generally violates the simplex. This does not affect (7.3c), but constrained perturbations of $w$ should respect the simplex (Boyd & Vandenberghe, 2004).

2. Redundancy feasibility / saturation. Scaling $(1 - R_{t,i})$ is admissible only while $R'_{t,i} \in [0,1]$; beyond feasibility, saturation effects arise (see 5.4).

3. No homogeneity in $\varepsilon$. The dependence on $\varepsilon$ enters through $\phi_{\varepsilon}$ and is not homogeneous (see 5.3).

## Corollaries

### Corollary 7.3.1 Homogeneity of the maximum.

Since $\text{AAS}_t^{\max} = \alpha_t \phi_{\varepsilon}(0^+)$ it is also degree-1 homogeneous in $\alpha$, consistent with (7.3c).

### Corollary 7.3.2 Total elasticity equals one.

For $\text{AAS}_t > 0$, the total elasticity in $\alpha$-space satisfies

$$ \sum_{i=1}^{m} \varepsilon_i^{\alpha} = 1, \text{ and each component equals the normalized contribution share, } \varepsilon_i^{\alpha} = p_{t,i}. $$

This follows from Euler's identity together with $\partial \text{AAS}_t / \partial \alpha_{t,i} = \phi_{\varepsilon}(x_{t,i})$ and the definition of $\varepsilon_i^{\alpha}$.

### Corollary 7.3.3 Greedy allocation under a scale budget.

Because $\text{AAS}_t(\alpha; x)$ is linear in $\alpha$, under $\sum_i \Delta\alpha_{t,i} \le B$ with $\Delta\alpha_{t,i} \ge 0$, the largest marginal increase is obtained by allocating $\Delta\alpha$ to channels with the largest

$$ \frac{\partial \text{AAS}_t}{\partial \alpha_{t,i}} = \phi_{\varepsilon}(x_{t,i}), $$

i.e., to those with the smallest accuracies $x_{t,i}$.

## 7.4 Theorem $\varepsilon \to 0^+$ regime

### Setup

Fix t. Let $x_i := x_{t,i} \in (0,1]$, $\alpha_i := \alpha_{t,i} \ge 0$, and $\alpha_t := \sum_{i=1}^{m} \alpha_i$. Define
$$ \mathrm{AAS}_t(\varepsilon; x) := \sum_{i=1}^{m} \alpha_i \phi_{\varepsilon}(x_i), \quad \phi_{\varepsilon}(x) := \log_2 \frac{1+\varepsilon}{x+\varepsilon}, \quad \varepsilon > 0, $$

and the $\varepsilon = 0$ limit functional

$$ \mathrm{AAS}_t^{(0)}(x) := \sum_{i=1}^{m} \alpha_i \log_2 \frac{1}{x_i}. $$

**Theorem 7.4 Pointwise limits, rates, and normalized behavior**

(a) Pointwise limit fixed x. For each fixed $x_i \in (0,1]$,

$$ \lim_{\varepsilon \to 0^+} \phi_{\varepsilon}(x_i) = \log_2 \frac{1}{x_i}, \quad \lim_{\varepsilon \to 0^+} \mathrm{AAS}_t^{(0)}(\varepsilon; x) = \mathrm{AAS}_t^{(0)}(x) \quad (7.4a) $$

(b) Asymptotic expansion scalar, fixed x. For fixed $x \in (0,1]$,

$$ \phi_{\varepsilon}(x) = \log_2 \frac{1}{x} + \frac{\varepsilon}{\ln 2} \left(1 - \frac{1}{x}\right) + \frac{\varepsilon^2}{2 \ln 2} \left(\frac{1}{x^2} - 1\right) + O(\varepsilon^3), (7.4b) $$

and the remainder is uniform on $x \in [\delta, 1]$ for every $\delta > 0$.

(c) Rate for the total score uniform away from zero. If $x_i \ge \delta > 0$ for all i, then

$$ \mathrm{AAS}_t(\varepsilon; x) = \mathrm{AAS}_t^{(0)}(x) + \frac{\varepsilon}{\ln 2} \sum_{i=1}^{m} \alpha_i \left(1 - \frac{1}{x_i}\right) + \frac{\varepsilon^2}{2 \ln 2} \sum_{i=1}^{m} \alpha_i \left(\frac{1}{x_i^2} - 1\right) + O(\varepsilon^3), \quad (7.4c) $$

hence $|\mathrm{AAS}_t(\varepsilon; x) - \mathrm{AAS}_t^{(0)}(x)| = O(\varepsilon)$ under $x_i \ge \delta > 0$.

(d) Growth of the maximum score. The boundary value satisfies

$$ \phi_{\varepsilon}(0^{+}) = \log_2 \frac{1+\varepsilon}{\varepsilon} = \log_2 \frac{1}{\varepsilon} + \frac{\varepsilon}{\ln 2} - \frac{\varepsilon^2}{2 \ln 2} + O(\varepsilon^3) \quad (7.4d) $$

Consequently, if $\alpha_t > 0$ and $\mathrm{AAS}_t^{\max}(\varepsilon) := \alpha_t \phi_{\varepsilon}(0^+)$, then

$$ \lim_{\varepsilon \to 0^+} \mathrm{AAS}_t^{\max}(\varepsilon) = +\infty. $$

(e) Normalized score for fixed x. If $\alpha_t > 0$ and

$$ \overline{\mathrm{AAS}}_t(\varepsilon; x) := \frac{\mathrm{AAS}_t(\varepsilon; x)}{\alpha_t \phi_{\varepsilon}(0^+)} $$
then

$$ \overline{\text{AAS}}_t(\varepsilon; x) \to 0 \quad (\varepsilon \to 0^+) \quad (7.4e) $$

**(f) Scaled x-regimes single-channel law.** Suppose a channel obeys

$$ x(\varepsilon) \sim c \varepsilon^{\gamma} \quad (c \in (0, \infty), \gamma \ge 0), $$

and define the normalized ratio

$$ r(\varepsilon) := \frac{\phi_{\varepsilon}(x(\varepsilon))}{\phi_{\varepsilon}(0^{+})}, \quad \phi_{\varepsilon}(0^{+}) = \log_2 \frac{1 + \varepsilon}{\varepsilon}. $$

Then

$$ r(\varepsilon) \to \rho \quad (\varepsilon \to 0^{+}), \quad \rho = \begin{cases} 0 & (\gamma = 0), \\ \gamma & (0 < \gamma < 1), \\ 1 & (\gamma \ge 1). \end{cases} \quad (7.4f) $$

For $\gamma = 1$, $x(\varepsilon) + \varepsilon \sim (c + 1)\varepsilon$ so $\phi_{\varepsilon}(x(\varepsilon)) = \phi_{\varepsilon}(0^{+}) - \log_2(c + 1) + o(1)$, hence $r(\varepsilon) \to 1$; for

$\gamma > 1$, $x(\varepsilon) = o(\varepsilon)$ so $x(\varepsilon) + \varepsilon \sim \varepsilon$ and again $r(\varepsilon) \to 1$.

**(g) Normalized total score under mixed scaling.** If $\alpha_t > 0$ and each $x_i = x_i(\varepsilon)$ satisfies the regime in (f) with limit $\rho_i$, then

$$ \overline{\text{AAS}}_t(\varepsilon; x(\varepsilon)) = \sum_{i=1}^{m} \frac{\alpha_i}{\alpha_t} r_i(\varepsilon) \to \sum_{i=1}^{m} \frac{\alpha_i}{\alpha_t} \rho_i. \quad (7.4g) $$

**Proof.**

Write

$$ \phi_{\varepsilon}(x) = \frac{1}{\ln 2} [\ln(1 + \varepsilon) - \ln(x + \varepsilon)]. $$

For fixed $x > 0$, $\ln(1 + \varepsilon) \to 0$ and $\ln(x + \varepsilon) \to \ln x$, giving $\phi_{\varepsilon}(x) \to -\ln x / \ln 2 = \log_2(1/x)$; summation yields (7.4a).

On $x \in [\delta, 1]$, Taylor expansions of $\ln(1 + \varepsilon)$ and $\ln(x + \varepsilon) = \ln x + \ln(1 + \varepsilon/x)$ give (7.4b) with uniform remainder, hence (7.4c).

Also $\phi_{\varepsilon}(0^{+}) = \log_2((1 + \varepsilon)/\varepsilon) = \log_2(1/\varepsilon) + \log_2(1 + \varepsilon)$, so (7.4d) follows; thus
$\alpha_t \phi_\varepsilon(0^+) \to \infty$ when $\alpha_t > 0$. Combining with the finite limit in (7.4a) proves (7.4e).

For (f), compare $x(\varepsilon)$ to $\varepsilon$: if $\gamma = 0$ then $\phi_\varepsilon(x(\varepsilon)) = O(1)$ while $\phi_\varepsilon(0^+) \to \infty$;

if $0 < \gamma < 1$ then $x(\varepsilon) \gg \varepsilon$ giving $\phi_\varepsilon(x(\varepsilon)) = \gamma \log_2(1/\varepsilon) + O(1)$; if $\gamma \ge 1$ then $x(\varepsilon) \lesssim \varepsilon$ so $\phi_\varepsilon(x(\varepsilon)) = \phi_\varepsilon(0^+) + O(1)$. Therefore (7.4f) holds.

Finally, (g) follows by componentwise convergence $r_i(\varepsilon) \to \rho_i$ and finiteness of $m$. ■

**Remark consistency with monotonicity in $\varepsilon$**

$$ \partial_\varepsilon \phi_\varepsilon(x) = \frac{1}{\ln 2} \left( \frac{1}{1+\varepsilon} - \frac{1}{x+\varepsilon} \right) \le 0, $$

so decreasing $\varepsilon$ increases the penalty, consistent with 5.3.

## 7.5 Theorem Weight Redundancy Interaction

### Setup

Fix t. Let

$$ \text{AAS}_t(w, R; x) = \sum_{i=1}^{m} w_i (1 - R_i) \phi_\varepsilon(x_i), \quad \phi_\varepsilon(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}, \quad \varepsilon > 0, $$

with $w_i \ge 0$, $\sum_{i=1}^m w_i = 1$, $R_i \in [0,1]$, and $x_i \in (0,1]$. (Kayadibi, 2025b)

Define

$$ \alpha_i := w_i(1 - R_i), \quad \alpha_t := \sum_{i=1}^{m} \alpha_i, \quad \phi_i := \phi_\varepsilon(x_i), \quad b_i := (1 - R_i)\phi_i, $$

$$ \text{and } \phi_0 := \phi_\varepsilon(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon}. $$

### Theorem 7.5 (a–g): Core interaction laws

#### (a) Affinity in $R$ and monotonicity fixed $w, x$.

For fixed $w, x$,

$$ \text{AAS}_t(R) = \sum_{i=1}^{m} w_i (1 - R_i) \phi_i = \left( \sum_{i=1}^{m} w_i \phi_i \right) - \sum_{i=1}^{m} w_i \phi_i R_i, $$
hence, for each i,

$$ \frac{\partial \text{AAS}_t}{\partial R_i} = -w_i \phi_i \le 0. $$

The decrease is strict whenever $w_i > 0$ and $x_i < 1$ equivalently $\phi_i > 0$.

### (b) Linearity in $w$ on the simplex fixed $R, x$.

For fixed $R, x$,

$$ \text{AAS}_t(w) = \sum_{i=1}^{m} w_i b_i, \quad b_i = (1 - R_i)\phi_i. $$

Thus $\text{AAS}_t$ is linear over $\{w \ge 0: \sum_i w_i = 1\}$. Consequently, maxima/minima are attained at simplex vertices: any maximizer can be chosen $w = e_{i^*}$ with $i^* \in \arg\max_i b_i$, and any minimizer $w = e_{j^*}$ with $j^* \in \arg\min_i b_i$; non-unique arg sets yield the corresponding convex-face of optimizers (Bertsimas & Tsitsiklis, 1997).

### (c) Negative cross-effect within channel substitutability.

$$ \frac{\partial \text{AAS}_t}{\partial w_i} = (1 - R_i)\phi_i, \quad \frac{\partial^2 \text{AAS}_t}{\partial R_i \partial w_i} = \frac{\partial}{\partial R_i}((1 - R_i)\phi_i) = -\phi_i < 0. $$

Hence, increasing $R_i$ reduces the marginal effect of $w_i$, and increasing $w_i$ amplifies sensitivity to $R_i$. If $x_i = 1$ so $\phi_i = 0$, the cross-effect vanishes.

### (d) $\alpha$-identification and equivalence classes fixed $x$.

Since

$$ \text{AAS}_t(w, R; x) = \sum_{i=1}^{m} \alpha_i \phi_i, \quad \alpha_i = w_i(1 - R_i), $$

the dependence on $(w, R)$ is mediated only through $\alpha = w \odot (1 - R)$. In particular,

$$ \tilde{\alpha} = \alpha \Rightarrow \text{AAS}_t(\tilde{w}, \tilde{R}; x) = \text{AAS}_t(w, R; x). $$

The converse need not hold, since distinct $\alpha$-vectors can yield the same inner product $\sum_i \alpha_i \phi_i$. Under the simplex constraint $\sum_i w_i = 1$, only factorizations $(\tilde{w}, \tilde{R})$ preserving feasibility and the simplex are admissible. Along a level set $\alpha_i \equiv \text{const}$,
$$d\alpha_i = (1 - R_i) dw_i - w_i dR_i = 0 \quad \Rightarrow \quad \frac{dw_i}{w_i} = \frac{dR_i}{1 - R_i},$$

interpreted for $w_i > 0$ and $R_i < 1$.

### (e) Joint perturbations: exact increment and envelope bound.

For perturbations $(\Delta w, \Delta R)$,

$$\Delta AAS_t = \sum_{i=1}^{m} \phi_i ((1 - R_i)\Delta w_i - w_i \Delta R_i - \Delta w_i \Delta R_i).$$

Using $0 \le \phi_i \le \phi_0$, $0 \le 1 - R_i \le 1$, and triangle inequality,

$$|\Delta AAS_t| \le \phi_0 \left( \sum_i |(1 - R_i)\Delta w_i| + \sum_i w_i |\Delta R_i| + \sum_i |\Delta w_i \Delta R_i| \right).$$

Moreover,

$$\sum_i |(1 - R_i)\Delta w_i| \le |\Delta w|_1, \quad \sum_i w_i |\Delta R_i| \le |w|_\infty |\Delta R|_1, \quad |\Delta w \odot \Delta R|_1 \le |\Delta w|_1 |\Delta R|_\infty,$$

So

$$|\Delta AAS_t| \le \phi_0 (|\Delta w|_1 + |w|_\infty |\Delta R|_1 + |\Delta w|_1 |\Delta R|_\infty).$$

### (f) Normalized derivative and the “signature rule” fixed w, x.

Let

$$\overline{AAS_t} := \frac{AAS_t}{\alpha_t \phi_0}, \quad \alpha_t = \sum_{i=1}^{m} w_i (1 - R_i).$$

Write $A := \sum_i w_i (1 - R_i) \phi_i = AAS_t$ and $B := \sum_i w_i (1 - R_i) = \alpha_t$, so

$\overline{AAS_t} = A/(B\phi_0)$, and define $\overline{\phi} := A/B = AAS_t/\alpha_t$. Then

$$\frac{\partial A}{\partial R_i} = -w_i \phi_i, \quad \frac{\partial B}{\partial R_i} = -w_i,$$

and the quotient rule gives
$$ \frac{\partial \overline{\text{AAS}_t}}{\partial R_i} = - \frac{w_i}{\alpha_t \phi_0} (\phi_i - \bar{\phi}). $$

Hence

$$ \operatorname{sign}\left(\frac{\partial \overline{\text{AAS}_t}}{\partial R_i}\right)=\left\{\begin{array}{ll}-, & \phi_i>\bar{\phi}, \\ 0, & \phi_i=\bar{\phi}, \\ +, & \phi_i<\bar{\phi}.\end{array}\right. $$

Contrast: $\partial_{R_i} \text{AAS}_t = -w_i \phi_i \le 0$ on the raw scale.

### (g) Greedy allocation under small redundancy budgets first order.

Assume $\sum_i \Delta R_i \le B$ with $0 \le \Delta R_i \le 1 - R_i$.

1. Raw score: $\Delta \text{AAS}_t \approx -\sum_i w_i \phi_i \Delta R_i$, so the fastest decrease allocates to the largest $w_i \phi_i$.

2. Normalized score: $\Delta \overline{\text{AAS}_t} \approx -\frac{1}{\alpha_t \phi_0} \sum_i w_i (\phi_i - \bar{\phi}) \Delta R_i$, so the fastest decrease allocates to the largest positive $w_i (\phi_i - \bar{\phi})$; indices with $\phi_i < \bar{\phi}$ can increase $\overline{\text{AAS}_t}$.

### Proof.

(a)–(d) follow by expansion and differentiation; (b) uses the extreme-point principle for linear objectives on the simplex (Bertsimas & Tsitsiklis, 1997). (e) follows from the exact increment identity plus triangle and Hölder/l₁ - l_∞ duality. (f) is the quotient rule applied to $A/(B\phi_0)$. (g) is the first-order linear allocation rule under budget and cap constraints. ■

## Theorem 7.6 Gradient-norm ceiling: component and vector norms

### Setup

Fix t. Let $x_i := x_{t,i} \in (0,1]$ and $\alpha_i := \alpha_{t,i} \ge 0$. Define

$$ \text{AAS}_t(x) = \sum_{i=1}^{m} \alpha_i \phi_{\varepsilon}(x_i), \quad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}, \quad \varepsilon > 0. $$

Write $\alpha_t := \sum_{i=1}^m \alpha_i$ and $\phi_0 := \phi_\varepsilon(0^+) = \log_2 \frac{1+\varepsilon}{\varepsilon}$.

(i) Componentwise ceiling. For each i,

$$ \frac{\partial \text{AAS}_t}{\partial x_i} = \alpha_i \phi'_\varepsilon(x_i) = - \frac{\alpha_i}{(x_i + \varepsilon) \ln 2}. $$
Since $x_i + \epsilon \ge \epsilon$,

$$|\partial_{x_i} \mathrm{AAS}_t(x)| \le \frac{\alpha_i}{\epsilon \ln 2} \quad (7.6a)$$

and the bound is asymptotically tight as $x_i \to 0^+$ for any active coordinate with $\alpha_i > 0$.

#### (ii) $l_p$ ceilings for the full gradient.

$$\nabla_x \mathrm{AAS}_t(x) = \left( -\frac{\alpha_1}{(x_1 + \epsilon) \ln 2}, \dots, -\frac{\alpha_m}{(x_m + \epsilon) \ln 2} \right).$$

For any $p \in [1, \infty]$,

$$|\nabla_x \mathrm{AAS}_t(x)|_p \le \frac{1}{\epsilon \ln 2} |\alpha|_p. \quad (7.6b)$$

in particular,

$$|\nabla_x \mathrm{AAS}_t(x)|_\infty \le \frac{|\alpha|_\infty}{\epsilon \ln 2}, \quad |\nabla_x \mathrm{AAS}_t(x)|_1 \le \frac{\alpha_t}{\epsilon \ln 2}, \quad |\nabla_x \mathrm{AAS}_t(x)|_2 \le \frac{|\alpha|_2}{\epsilon \ln 2}. \quad (7.6c)$$

These are sharp in the limit along sequences where the relevant active coordinates satisfy $x_i \to 0^+$.

#### (iii) Ceilings for the normalized score. Assume $\alpha_t > 0$ and define

$$\overline{\mathrm{AAS}}_t(x) := \frac{\mathrm{AAS}_t(x)}{\alpha_t \phi_0}.$$

Since $\alpha_t \phi_0$ is constant in $x$,

$$\nabla_x \overline{\mathrm{AAS}}_t(x) = \frac{1}{\alpha_t \phi_0} \nabla_x \mathrm{AAS}_t(x),$$

hence for any $p \in [1, \infty]$,

$$|\nabla_x \overline{\mathrm{AAS}}_t(x)|_p \le \frac{1}{\epsilon \ln 2} \frac{|\alpha|_p}{\alpha_t \phi_0}. \quad (7.6d)$$

and in particular

$$|\nabla_x \overline{\mathrm{AAS}}_t(x)|_1 \le \frac{1}{\epsilon \ln 2 \phi_0}, \quad |\nabla_x \overline{\mathrm{AAS}}_t(x)|_\infty \le \frac{|\alpha|_\infty}{\alpha_t} \cdot \frac{1}{\epsilon \ln 2 \phi_0}. \quad (7.6e)$$

#### (iv) Noise penalty inequality global sensitivity bound.
Let $\delta x \in R^m$ be such that the segment $\{x + \theta \delta x: \theta \in [0,1]\} \subseteq (0,1]^m$. Then the fundamental theorem of calculus along $x + \theta \delta x$ gives

$$AAS_t(x + \delta x) - AAS_t(x) = \int_0^1 \langle \nabla_x AAS_t(x + \theta \delta x), \delta x \rangle d\theta.$$

Applying Hölder with $1/p + 1/q = 1$ and $p = \infty \Rightarrow q = 1$,

$$|\Delta AAS_t| \le \int_0^1 |\nabla_x AAS_t(x + \theta \delta x)|_p |\delta x|_q d\theta \le \left( \sup_{z \in (0,1]^m} |\nabla_x AAS_t(z)|_p \right) |\delta x|_q.$$

Using (7.6b),

$$|\Delta AAS_t| \le \frac{|\alpha|_p}{\varepsilon \ln 2} |\delta x|_q, \quad 1/p + 1/q = 1. \quad (7.6f)$$

and, if $\alpha_t > 0$,

$$|\Delta \overline{AAS}_t| \le \frac{|\alpha|_p}{\alpha_t \phi_0} \cdot \frac{1}{\varepsilon \ln 2} |\delta x|_q (7.6g)$$

### Proof

Differentiate $\phi_{\varepsilon}(x) = (\ln(1 + \varepsilon) - \ln(x + \varepsilon))/\ln 2$ to obtain

$$\phi'_{\varepsilon}(x) = -(x + \varepsilon)^{-1}(\ln 2)^{-1}, \text{ yielding (i) and (7.6a).}$$

Since $(x_i + \varepsilon)^{-1} \le \varepsilon^{-1}$, componentwise bounds aggregate to (7.6b)– (7.6c).

Because $\alpha_t \phi_0$ does not depend on $x$, (7.6d)– (7.6e) follow by scaling.

For (iv), use the line-segment integral identity above and apply Hölder; substituting (7.6b) gives (7.6f), and dividing by $\alpha_t \phi_0$ gives (7.6g). ■

## 8. Axiomatic correctness: representation and uniqueness

Having established the regularity properties in 4–7, I now show that the AAS form is not merely admissible but axiomatically forced: under a weak set of structural requirements, every valid score must coincide with the AAS representation, up to a calibration constant.

### 8.1 Setup and axioms Ax1–Ax7

Fix a measurement level $\varepsilon > 0$. Consider any finite family of channels described by pairs $(\alpha_i, x_i)$ with $x_i \in (0,1]$ and $\alpha_i \ge 0$. Let the regularized score functional be
$$S_{\varepsilon}(\{(\alpha_i, x_i)\}_{i=1}^m) \in [0, \infty).$$

Assume:

(Ax1) Normalization and symmetry. $S_{\varepsilon}(\emptyset) = 0$, and $S_{\varepsilon}$ is invariant under index permutations.

(Ax2) Decomposition and additivity. $S_{\varepsilon}$ is additive over disjoint subfamilies. Moreover, refinement at constant accuracy is score-neutral: if $\alpha = \alpha_1 + \alpha_2$ and both parts share the same x, then splitting/merging does not change the score.

(Ax3) α-linearity. For a single channel, scaling the weight scales the

score: for all $\lambda \ge 0$,

$$S_{\varepsilon}(\{(\lambda\alpha, x)\}) = \lambda S_{\varepsilon}(\{(\alpha, x)\}).$$

(Ax4) Channel-wise regularity for the unregularized kernel. There exists a single-channel kernel $g: (0,1] \to [0, \infty)$ such that $g$ is continuous, strictly decreasing, convex, satisfies $g(1) = 0$, and diverges as $x \downarrow 0$.

(Ax5) Multiplicative composition unregularized. For sequential independent subtasks with total accuracy $x_{\text{total}} = x^{(1)}x^{(2)}$,

$$g(x^{(1)}x^{(2)}) = g(x^{(1)}) + g(x^{(2)})$$

(Aczél, 1966).

(Ax6) Regularization via a noise floor measurement map. With $\varepsilon > 0$, define

$$u(x) := \frac{x + \varepsilon}{1 + \varepsilon} \in [\frac{\varepsilon}{1 + \varepsilon}, 1], \quad g_{\varepsilon}(x) := g(u(x)).$$

Thus $g_{\varepsilon}$ inherits continuity/monotonicity/convexity from $g$ and remains finite at $x \downarrow 0$ because

$u(x) \ge \varepsilon/(1 + \varepsilon)$. In general, $g_{\varepsilon}(xy) = g_{\varepsilon}(x) + g_{\varepsilon}(y)$ need not hold in the original x-space; Ax5 is imposed on $g$ and transported to $\varepsilon > 0$ through $u$.

(Ax7) Bit calibration. $g(1/2) - g(1) = 1$ base-2 normalization.

**Compatibility notes.** The properties established in 4–7 are consistent with (Ax1)–(Ax7) and are invoked by cross-reference rather than reproved.

## 8.2 Compliance: the AAS satisfies (Ax1)–(Ax7)
Define

$$AAS_t = \sum_{i=1}^{m} \alpha_{t,i} \phi_{\varepsilon}(x_{t,i}), \quad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon}.$$

Then:

i. **Ax1** holds by $S(\emptyset) = 0$ and permutation invariance (4.1–4.2).

ii. **Ax2** holds by refinement equality (6.1) and finite additivity over disjoint index sets (6.4).

iii. **Ax3** holds by degree-1 positive homogeneity in $\alpha$ (7.3).

iv. **Ax4** holds for the unregularized kernel $\phi_0(x) = \log_2(1/x)$, which is continuous, strictly decreasing, strictly convex, satisfies $\phi_0(1) = 0$, and diverges as $x \downarrow 0$.

v. **Ax5** holds for $\phi_0$ since $\phi_0(xy) = \phi_0(x) + \phi_0(y)$.

vi. **Ax6** lifts $\phi_0$ to $\phi_{\varepsilon}$ via

$$\phi_{\varepsilon}(x) = \phi_0\left(\frac{x+\varepsilon}{1+\varepsilon}\right).$$

Composition is therefore preserved in the measured $u$-space:

$$\phi_0(u_1 u_2) = \phi_0(u_1) + \phi_0(u_2), \quad u_k = \frac{x^{(k)} + \varepsilon}{1 + \varepsilon}.$$

In general, $\phi_{\varepsilon}(xy) = \phi_{\varepsilon}(x) + \phi_{\varepsilon}(y)$ need not hold; the correct statement is transported through $u$.

vii. **Ax7** holds because $\phi_0(1/2) - \phi_0(1) = \log_2 2 = 1$.

## 8.3 Representation and uniqueness

### Theorem 8.1 Representation and uniqueness.

If $S_{\varepsilon}$ satisfies (Ax1)–(Ax7), then necessarily

$$S_{\varepsilon}(\{(\alpha_i, x_i)\}_{i=1}^m) = \sum_{i=1}^{m} \alpha_i \phi_{\varepsilon}(x_i), \quad \phi_{\varepsilon}(x) = \log_2 \frac{1+\varepsilon}{x+\varepsilon},$$

and the representation is unique Ax7 fixes the scale.

### Proof.

#### Step 1: separability and $\alpha$ factorization.

By (Ax3), define the single-channel kernel
$$g_{\varepsilon}(x) := S_{\varepsilon}({(1, x)}), \quad x \in (0,1].$$

Then for any $\alpha \ge 0$,

$$S_{\varepsilon}({(\alpha, x)}) = S_{\varepsilon}({(\alpha \cdot 1, x)}) = \alpha S_{\varepsilon}({(1, x)}) = \alpha g_{\varepsilon}(x).$$

By additivity over disjoint subfamilies (Ax2), any finite family decomposes into singletons, hence

$$S_{\varepsilon}({(\alpha_i, x_i)}_{i=1}^m) = \sum_{i=1}^m S_{\varepsilon}({(\alpha_i, x_i)}) = \sum_{i=1}^m \alpha_i g_{\varepsilon}(x_i). \quad (8.3a)$$

**Step 2: unregularized composition ⇒ logarithmic form.**

Ax5 is imposed on $g$, the unregularized kernel in Ax4: $g(xy) = g(x) + g(y)$ on $(0,1]$. Define

$f(t) := g(e^{-t})$ for $t \ge 0$. Then $f(t + s) = f(t) + f(s)$. By continuity (Ax4), $f(t) = kt$, hence

$$g(x) = k \ln \frac{1}{x}. \quad (8.3b)$$

This is the continuous Cauchy equation result (Aczél, 1966; Kuczma, 2009).

**Step 3: $\varepsilon$ lift through the measurement map.**

By Ax6,

$$g_{\varepsilon}(x) = g\left(\frac{x + \varepsilon}{1 + \varepsilon}\right) = k\ln\frac{1 + \varepsilon}{x + \varepsilon}. \quad (8.3c)$$

**Step 4: bit calibration fixes the constant.**

Ax7 and $g(1) = 0$ give

$$1 = g(1/2) - g(1) = k \ln 2 \Rightarrow k = \frac{1}{\ln 2}.$$

Therefore,

$$g(x) = \log_2 \frac{1}{x}, \quad g_{\varepsilon}(x) = \log_2 \frac{1 + \varepsilon}{x + \varepsilon} = \phi_{\varepsilon}(x).$$

Substituting into (8.3a) yields the stated representation; uniqueness follows because Steps 2–4 determine both the functional form and its scale. ■

## 8.4 Minimality of the axioms
Each axiom plays a distinct forcing role:

I. Without **Ax5**, many continuous strictly decreasing convex kernels satisfying Ax4 are possible that are not logarithmic, e.g., $g(x) = x^{-p} - 1, p > 0$, hence the logarithmic law is not forced; via Ax6, these induce non-logarithmic regularized families

$$g_{\epsilon}(x) = g\left(\frac{x + \epsilon}{1 + \epsilon}\right).$$

II. Without **Ax4** (notably continuity), Cauchy-type equations admit pathological solutions, so the logarithmic form cannot be guaranteed (Kuczma, 2009).

III. Without **Ax7**, the kernel is determined only up to a multiplicative constant $c>0$:

$$g_{\epsilon}(x) = c \log_2 \frac{1 + \epsilon}{x + \epsilon}.$$

IV. Without **Ax2** or **Ax3**, separability need not hold and interaction terms can enter, so $\sum_i \alpha_i g_{\epsilon}(x_i)$ is not forced.

Hence, (Ax1)–(Ax7) are collectively weak yet sufficient: Ax5 is imposed at $\epsilon = 0$ and transported to $\epsilon > 0$ via Ax6.

## 8.5 AAS correctness and uniqueness

### Theorem 8.2.

Let $S_{\epsilon}$ map any finite multiset $\{(\alpha_i, x_i)\} \subset R_{\ge 0} \times (0,1]$ to $[0, \infty)$ and satisfy (Ax1)–(Ax7) for fixed $\epsilon > 0$. Then

$$S_{\epsilon}(\{(\alpha_i, x_i)\}_{i=1}^{m}) = \sum_{i=1}^{m} \alpha_i \log_2 \frac{1 + \epsilon}{x_i + \epsilon}$$

and this representation is unique (Ax7 fixes the scale).

**Proof.** This is an immediate corollary of Theorem 8.1 Representation and uniqueness, since the axioms are identical and Ax7 fixes the calibration constant. ■

## 8.6 Closure

Two conclusions follow:
1. **Compliance (4–7).** The AAS construction satisfies the required decomposition, convexity, sensitivity, and limiting-regime properties.

2. **Necessity (8).** Under (Ax1)–(Ax7), every admissible score is forced to take the AAS form
   $$ \sum_i \alpha_i \phi_\epsilon(x_i). $$

## 9. Conclusion.

This paper introduced a mathematically well-founded scoring rule for artificial memory performance, the Artificial Age Score (AAS). The score combines two intuitive ingredients: (i) how accurate each channel is, and (ii) how much of that channel is effectively counted once redundancy is taken into account. The design ensures that lower accuracy leads to a larger cost, while redundancy reduces the effective contribution of a channel rather than artificially inflating the total score. Building on the original AAS proposal (Kayadibi, 2025b), the main contribution of this work is to show that AAS is not only a reasonable design choice, but also uniquely characterized by a small set of natural structural principles. In particular, if a score is required to be symmetric across channels, additive across disjoint collections, proportional to channel weight, well-behaved as accuracy varies, and consistent with how independent accuracies should combine, then the score must coincide with the AAS representation up to a fixed calibration set by a one-bit normalization. This establishes an axiomatic necessity result: under the stated requirements, no alternative functional form can satisfy the same constraints.

In addition, the mathematical soundness of AAS was established through the main results and supporting lemmas presented throughout the paper. Collectively, these results show that the score is well-defined and decomposable, admits quantitative bounds that support stability and comparability, and behaves monotonically in the expected direction as accuracy improves, while redundancy acts by down-weighting effective contribution rather than increasing the score. Together, these properties justify AAS as a stable, interpretable, and principled metric for evaluating multi-channel artificial memory behavior. Practically, the framework supports meaningful comparisons across sessions and systems on a common normalized scale, separates the cost attributable to heterogeneity in accuracies, and provides principled guidance on where redundancy reduction is most impactful. Overall, the results offer a defensible mathematical foundation for evaluating artificial memory behavior in multi-channel settings, with clear interpretability and a unique characterization under the proposed axioms.

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