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High within-class variance raises contrastive loss despite perfect mean separation

Example

Let $C = {a, b}$ be uniform and let each class contain two equally likely inputs indexed by $η \in {- 1, + 1}$ . In $R^{2}$ , define $f (a, η) = (1, η r)$ and $f (b, η) = (- 1, η r)$ for $r \geq 1$ . Each class mean is respectively $(1, 0)$ and $(- 1, 0)$ , so the mean classifier separates both classes with hinge loss zero. Yet for one independent marginal negative and hinge contrastive loss, the total risk is $L_{u n} (f) = \frac{1}{4} + \frac{1}{2} r^{2}$ , which grows without bound as the within-class variance $r^{2}$ increases. The norm bound holds with $R = \sqrt{1 + r^{2}}$ for each fixed $r$ .

Example references

Latent-class contrastive representation learning: defines the latent-class triplet law, mean classifier, covariance and contrastive risk.

Convexity, Lipschitz bounds, and oscillation of the source contrastive losses: defines the hinge loss.

Saunshi et al., Section 5.1 and Figure 1(b): gives a two-class example where high within-class variation raises the unsupervised loss despite a good mean classifier.

Facts & Assumptions

Given: The two-class distribution and representation in the Example, one independent marginal negative, and hinge loss.

[F1]

The class means are $μ_{a} = (1, 0)$ and $μ_{b} = (- 1, 0)$ ; the mean classifier for the ordered task $(a, b)$ has weight $(2, 0)$ . Latent-class contrastive representation learning

[F2]

The class-mean hinge risk uses loss $(1 - Y w^{⊤} f (X))_{+}$ , and the contrastive hinge loss is $(1 + q_{f})_{+}$ . Latent-class contrastive representation learning, Convexity, Lipschitz bounds, and oscillation of the source contrastive losses

[F3]

Given the positive class, the negative class equals it with probability $1 / 2$ ; conditional on collision, the score's random part is $r^{2} η_{0} (η_{-} - η_{+})$ , whose values $2 r^{2}, - 2 r^{2}, 0$ have probabilities $1 / 4, 1 / 4, 1 / 2$ . Conditional on distinct classes, the score is $- 2 + r^{2} η_{0} (η_{-} - η_{+})$ with the same three probabilities. Latent-class contrastive representation learning

Proof

Given: The finite class-conditional laws in the Example.

1.1F1F2givenalgebra

For the task $(a, b)$ , the mean weight is $(2, 0)$ . On class $a$ with label $+ 1$ , $Y w^{⊤} f (a, η) = 2$ for either $η$ ; on class $b$ with label $- 1$ , it is also $2$ . The hinge loss is therefore zero for both labels. Swapping the ordered task gives the same result, so the average mean-classifier hinge risk is zero.

1.2F2F3givenalgebra

Conditional on a class collision, [F3] gives score values $2 r^{2}, - 2 r^{2}, 0$ with probabilities $1 / 4, 1 / 4, 1 / 2$ . Since $r \geq 1$ , their hinge losses are $1 + 2 r^{2}, 0, 1$ , respectively. Thus $L_{u n}^{=} (f) = \frac{1}{4} (1 + 2 r^{2}) + \frac{1}{2} = \frac{3}{4} + \frac{1}{2} r^{2}$ .

1.3F2F3givenalgebra

Conditional on distinct classes, the score values are $- 2 + 2 r^{2}, - 2 - 2 r^{2}, - 2$ with probabilities $1 / 4, 1 / 4, 1 / 2$ . For $r \geq 1$ , their hinge losses are $2 r^{2} - 1, 0, 0$ . Hence $L_{u n}^{\neq} (f) = \frac{1}{4} (2 r^{2} - 1) = \frac{1}{2} r^{2} - \frac{1}{4}$ .

2.1F3step 1.2step 1.3step 1.1givenalgebra∎

Collision and distinct-class events each have probability $1 / 2$ . Combining steps 1.2 and 1.3 yields $L_{u n} (f) = \frac{1}{2} (\frac{3}{4} + \frac{1}{2} r^{2}) + \frac{1}{2} (\frac{1}{2} r^{2} - \frac{1}{4}) = \frac{1}{4} + \frac{1}{2} r^{2}$ . This tends to infinity with $r$ , while step 1.1 keeps the mean-classifier hinge risk zero.

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Sources

Nikunj Saunshi et al., A Theoretical Analysis of Contrastive Unsupervised Representation Learning, Section 5.1 and Figure 1(b) (standard reference, not scraped)