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Uniform generalization bound for the one-negative contrastive loss

Statement

Assume the Axiom of Choice. In the one-negative latent-class model, fix R>0, a source loss ℓ∈{ℓh,ℓlog}, a sample size M≥1, and a sample S of M IID triplets. Assume the loss risks and all sample, independent-copy, and sign suprema below are measurable. Let Bℓ,R=ℓ(2R2)−ℓ(−2R2) and define

ΓM(S,δ):=218L0R RS(F)M+3Bℓ,Rln⁡(4/δ)2M,0<δ<1.

With probability at least 1−δ over S,

∀f∈F,∣Lun(f)−L^un(f)∣≤ΓM(S,δ).

Here RS(F) is the unnormalized vector Rademacher complexity of the three embeddings in each triplet.

Statement references

Latent-class contrastive representation learning: fixes the one-negative triplet loss, IID sample, representation bound and unnormalized vector Rademacher complexity.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: supplies L0 and the positive oscillation Bℓ,R of either source loss on the score range.

A finite-dimensional vector contraction bound for one-sided Rademacher averages: bounds the one-sided vector Rademacher average, with constant 3.

Rademacher uniform generalization bounds: supplies the two-sided high-probability deviation bound for measurable [0,1]-valued classes under AC.

Saunshi et al., Supplementary Appendix A.3: applies vector contraction and bounded-loss uniform generalization to the contrastive class.

Mohri et al., Theorem 3.3 and complete proof, printed pp. 30–33: proves the bounded-loss Rademacher uniform-deviation step.

Maurer, Theorem 3, Corollary 4 and complete proof: gives a sharper vector-contraction constant; the proof here instead uses the local 3 lemma.

Facts & Assumptions

Given: AC, the one-negative latent-class model with R>0, either source loss, IID triplets, M≥1, 0<δ<1, and the measurability assumptions in the Statement.

[A1]

AC is the set-theoretic choice axiom used by the uniform-generalization supplier through its bounded-differences theorem. The Axiom of Choice, Rademacher uniform generalization bounds

[F1]

The triplets are IID; the population and empirical risks are the expectation and sample average of ℓ(qf). Latent-class contrastive representation learning, Identical distribution and IID families, Independent random elements have product joint law

[F2]

The score obeys ∣qf(x,x+,x−)∣≤2R2, the loss is nondecreasing and L0-Lipschitz, and its range on that interval is Bℓ,R>0. Latent-class contrastive representation learning, Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

[F3]

The two-sided empirical Rademacher deviation theorem for a measurable class G of functions into [0,1] gives, with probability at least 1−δ, sup⁡g∈G∣Pg−PMg∣≤2R^S(G)+3ln⁡(4/δ)/(2M). It assumes AC. Rademacher uniform generalization bounds

[F4]

For a nonempty bounded trace set and coordinatewise functions with common Euclidean Lipschitz constant L, one-sided vector contraction bounds the scalar sign supremum by 3L times the independent coordinate-sign supremum. A finite-dimensional vector contraction bound for one-sided Rademacher averages

[F5]

For vectors u,v, ∣u⊤v∣≤∥u∥2∥v∥2; integrable expectations are linear. Cauchy–Schwarz: ∣⟨u,v⟩∣≤∥u∥∥v∥, with equality exactly for linearly dependent vectors, The Lebesgue integral is linear on L1(μ)

Proof

Given: The hypotheses in the Statement and Facts & Assumptions.

1.1A1F1F2F5given

Set a=ℓ(−2R2) and B=Bℓ,R. For each triplet z=(x,x+,x−), define gf(z)=(ℓ(qf(z))−a)/B. By [F2], gf takes values in [0,1]. Since the shift a is common to every f, averaging over the signs shows that the empirical one-sided Rademacher complexity of {gf:f∈F} equals B−1 times that of {ℓ(qf):f∈F}: the extra term is −(a/B)∑jεj, whose sign expectation is zero.

1.2F2F5givenalgebra

For a fixed triplet, write v=(u0,u+,u−)∈(Rd)3 and h(v)=ℓ(u0⊤(u−−u+)) on K=(B‾R)3, where B‾R={u:∥u∥2≤R}. For v,v′∈K, expansion of the score difference and [F5] give ∣h(v)−h(v′)∣≤L0R(2∥u0−u0′∥2+∥u−−u−′∥2+∥u+−u+′∥2)≤6L0R∥v−v′∥2. The last inequality is Cauchy–Schwarz for the three nonnegative coordinate distances. This Lipschitz function extends to all of (Rd)3 with the same constant by h~(v)=inf⁡w∈K{h(w)+6L0R∥v−w∥2}: on K the Lipschitz inequality gives equality with h, and the triangle inequality gives the global Lipschitz bound.

2.1F1F4step 1.1step 1.2givenalgebra

For the observed triplets, let V={(f(xj),f(xj+),f(xj−))j=1M:f∈F}, a nonempty bounded trace set. Apply [F4] to the extensions in step 1.2 with L=6L0R. The independent coordinate signs are exactly those defining RS(F), so R^S({gf:f∈F})≤36L0R RS(F)BM=18L0R RS(F)BM. The normalization 1/M comes from the empirical complexity convention; RS itself is unnormalized.

3.1A1F1F2F3step 1.1step 2.1givenalgebra∎

Apply [F3] to the normalized class in step 1.1. With probability at least 1−δ, its uniform deviation is at most twice the complexity in step 2.1 plus 3ln⁡(4/δ)/(2M). Multiplying by B converts that deviation to the unnormalized losses and yields exactly ΓM(S,δ) for every f∈F.

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