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The hinge margin comparison does not extend to logistic contrastive loss

Example

Consider two equally likely deterministic classes $a, b$ and the scalar representation $f (a) = 2$ , $f (b) = 4$ , so $R = 4$ , $τ = 1 / 2$ , and the classwise centered representation has sub-Gaussian proxy $σ^{2} = 0$ . For the distinct-class tasks, the mean classifier has logistic risk

$L_{s u p}^{μ} (f) = \frac{1}{2} [\log_{2} (1 + e^{4}) + \log_{2} (1 + e^{- 8})] > \frac{5}{2},$

while its source hinge margin risk at $γ = 1$ is exactly $L_{1, s u p}^{μ} (f) = 5 / 2$ . The logistic contrastive risk conditioned on distinct classes has the same value as the displayed logistic mean-classifier risk. Consequently, for every $0 < ϵ \leq 1$ smaller than the positive gap, the source hinge margin comparison $L_{u n}^{\neq} (f) \leq γ L_{γ, s u p}^{μ} (f) + ϵ$ fails for logistic contrastive loss. In particular, the hinge-margin corollary cannot be transferred unchanged to logistic training.

For the optional transfer comparison, assume the Axiom of Choice, fix $M \geq 1$ and $0 < δ < 1$ , impose the corollary's remaining model, risk and supremum measurability, and measurable exact-empirical-minimizer hypotheses, and use the singleton $F = {f}$ . The proposed extension is

$L_{s u p}^{μ} (\hat{f}) \leq γ L_{γ, s u p}^{μ} (f) + ϵ + \frac{\sqrt{2} L_{0} τ s (f) + 2 Γ_{M}^{l o g} (S, δ)}{1 - τ},$

where $L_{s u p}^{μ} (\hat{f})$ uses logistic task loss, $L_{γ, s u p}^{μ} (f)$ uses the scaled hinge margin loss, $γ = 1$ here, and

$Γ_{M}^{l o g} (S, δ) : = \frac{2 \sqrt{18} L_{0} R R_{S} (F)}{M} + 3 B_{ℓ_{l o g}, R} \sqrt{\frac{\ln (4 / δ)}{2 M}}$

uses the logistic loss oscillation. The example tests this claimed mean-classifier bound with logistic contrastive training.

Example references

Latent-class contrastive representation learning: defines the latent-class risks, scores and centered classwise representations.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: defines the base-2 logistic loss and hinge margin loss.

Class-conditional sub-Gaussian concentration yields a contrastive margin bound: states the one-negative margin inequality for hinge contrastive loss.

Sub-Gaussian contrastive transfer with a mean-classifier margin loss: states the source margin transfer corollary with its hinge-only scope.

The Axiom of Choice: supplies the choice assumption for the optional transfer comparison.

Saunshi et al., Lemma 5.1 and Corollary 5.1.1: states the margin comparison in the hinge-loss section.

Facts & Assumptions

Given: The deterministic two-class model and representation in the Example; base-2 logistic contrastive loss; hinge loss for the scaled mean-classifier margin; and $0 < ϵ \leq 1$ . For the optional transfer comparison, also assume AC, $M \geq 1$ , $0 < δ < 1$ , the stated risk and supremum measurability hypotheses, and a measurable exact empirical selector for $F = {f}$ .

[A1]

The optional transfer comparison assumes the Axiom of Choice required by the transfer corollary's uniform-generalization route. The Axiom of Choice, Sub-Gaussian contrastive transfer with a mean-classifier margin loss

[F1]

The ordered distinct class pairs $(a, b)$ and $(b, a)$ have equal conditional probability, the means equal their deterministic embeddings, and the positive-pair and negative sampling rules are those in the latent-class model. Latent-class contrastive representation learning

[F2]

The contrastive score is $q_{f} = f (x)^{⊤} (f (x^{-}) - f (x^{+}))$ , logistic loss is $\log_{2} (1 + e^{q})$ , and the scaled hinge mean-classifier loss is $(1 - Y w_{μ}^{⊤} f (x) / γ)_{+}$ . Latent-class contrastive representation learning, Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

[F3]

The source sub-Gaussian margin parameter equals $γ = 1$ when $σ = 0$ , and the source margin corollary is hinge-only. Class-conditional sub-Gaussian concentration yields a contrastive margin bound, Sub-Gaussian contrastive transfer with a mean-classifier margin loss

Proof

Given: The finite deterministic model in the Example.

1.1F1F2givenalgebra

The class means are $μ_{a} = 2$ and $μ_{b} = 4$ . For task $(a, b)$ , the mean weight is $w_{μ} = μ_{a} - μ_{b} = - 2$ . The positive-label example $a$ has logistic argument $- w_{μ} f (a) = 4$ , while the negative-label example $b$ has argument $w_{μ} f (b) = - 8$ . Swapping the classes gives the same two losses in reverse order, proving the displayed mean-classifier logistic risk.

1.2F2F3givenalgebra

For the mean classifier, the two hinge-margin arguments at $γ = 1$ are $4$ and $- 8$ , so the losses are $5$ and $0$ and their average is $5 / 2$ . The sub-Gaussian proxy is zero because each class is deterministic, so [F3] gives $γ = 1$ for every permitted $ϵ$ .

2.1F1F2step 1.1givenalgebra

For distinct classes, the contrastive score is $q_{f} = 2 (4 - 2) = 4$ for $(a, b)$ and $q_{f} = 4 (2 - 4) = - 8$ for $(b, a)$ . Therefore $L_{u n}^{\neq} (f) = \frac{1}{2} [\log_{2} (1 + e^{4}) + \log_{2} (1 + e^{- 8})]$ , equal to the value in step 1.1.

3.1F2step 2.1step 1.2givenalgebra

The exponential series has positive terms and $e^{4} > 1 + 4 + 4^{2} / 2! + 4^{3} / 3! + 4^{4} / 4! = 103 / 3 > 31$ . Hence $1 + e^{4} > 32$ , so $\log_{2} (1 + e^{4}) > 5$ ; also $\log_{2} (1 + e^{- 8}) > 0$ . Thus the value in step 2.1 exceeds $5 / 2$ by a positive gap $Δ$ . Choose $0 < ϵ < \min (Δ / 4, 1)$ . Then $L_{u n}^{\neq} (f) > L_{1, s u p}^{μ} (f) + ϵ$ , contradicting the proposed logistic extension.

4.1A1F1F2F3step 1.2givenalgebra∎

Under [A1] and the transfer hypotheses in the Statement, take $F = {f}$ and the logistic objective. The only element is an exact empirical minimizer, its vector Rademacher complexity is zero because the expectation of its finite signed sum is zero, and $s (f) = 0$ because both class covariances vanish. Therefore $Γ_{M}^{l o g} (S, δ) = 3 B_{ℓ_{l o g}, 4} \sqrt{\ln (4 / δ) / (2 M)} \to 0$ as $M \to \infty$ for the fixed $0 < δ < 1$ . The mean-classifier logistic risk $L_{s u p}^{μ} (f)$ exceeds $5 / 2$ by $Δ$ , while the proposed logistic-training extension would upper-bound that risk by $5 / 2 + ϵ + 4 Γ_{M}^{l o g} (S, δ)$ (since $1 - τ = 1 / 2$ ). Taking $0 < ϵ < \min (Δ / 4, 1)$ and then $M$ large enough that $4 Γ_{M}^{l o g} < Δ / 4$ makes that right side less than $5 / 2 + Δ / 2$ , contradicting $L_{s u p}^{μ} (f) = 5 / 2 + Δ$ .

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Sources

Nikunj Saunshi et al., A Theoretical Analysis of Contrastive Unsupervised Representation Learning, Lemma 5.1 and Corollary 5.1.1 (standard reference, not scraped)