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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Adversarial robustness and robust risk — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Gaussian robust risk across dimensions and radii

Example

Assume AC. For an integer d≥2, set σ=d1/4/32 and ϵ=1/4. For the fixed-parameter Gaussian model with ∥θ∥2=d, the class-weighted sample-mean classifier has ℓ∞ robust risk at most 1/100 with training-sample probability at least 1−2e−d/(8(σ2+1)) whenever the integer sample size satisfies n≥4d. In the separate Gaussian-prior experiment Θ∼Nd(0,Id), every Borel learning rule F:(Rd×{−1,+1})n×Rd→{−1,+1} has expected ℓ∞ robust risk at least (1−1/d)/2 whenever n≤d/(131072log⁡d). The first assertion concerns a fixed normalized parameter and an upper bound for one classifier; the second averages over a Gaussian prior and lower-bounds every such Borel learning rule. Expected robust risk in the prior experiment uses outer expectation over the parameter and training sample.

Facts & Assumptions

Given: The Example's Gaussian parameter, radius, and dimension.

[A1]

Assume AC as required by both Gaussian robust-risk bounds. The Axiom of Choice.

[F1]

If σ≤d1/4/32, d≥1, and 1/(4d1/4)≤ϵ≤1/4, then n≥64ϵ2d suffices for the class-weighted mean classifier to have robust risk at most 1/100 with failure probability at most 2e−d/(8(σ2+1)). Robust sample upper bound in the Gaussian model.

[F2]

If d≥2 and n≤ϵ2σ2/(8log⁡d), every Borel learning rule under prior Θ∼Nd(0,Id) has expected robust risk at least (1−1/d)/2. Robust sample lower bound in the Gaussian model.

Proof

1.1givenA1F1algebra

The chosen variance satisfies σ=d1/4/32, and d≥2 makes 1/(4d1/4)≤1/4=ϵ. The upper-bound condition becomes n≥64(1/4)2d=4d. Apply [F1] to obtain the first assertion.

2.1givenA1F2algebra∎

For the lower bound, ϵ2σ2/(8log⁡d)=(1/16)(d/1024)/(8log⁡d)=d/(131072log⁡d). Thus the stated lower sample-size condition is exactly the premise of [F2], which gives the second assertion.

References

Robust sample upper bound in the Gaussian model: supplies the exact one-percent sufficient sample size.

Robust sample lower bound in the Gaussian model: supplies the learner-independent Gaussian-prior necessary sample size.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Thresholding removes Bernoulli sign perturbations

Example

Let d≥1, θ∈{−1,+1}d, 0<τ≤1/2, 0≤ϵ<1, and let (X,Y)∼Pθ,τ be the binary-feature Bernoulli model. Let T(x)j=+1 for xj≥0 and T(x)j=−1 for xj<0. For every classifier g:{−1,+1}d→{−1,+1}, the classifier g∘T has exactly the same ℓ∞ robust risk at radius ϵ as g has natural risk: RPθ,τ,Δϵrob(g∘T)=P(g(X)≠Y). For w∈Rd, define fw(x)=+1 when ⟨w,x⟩≥0 and fw(x)=−1 otherwise. In particular, if 0<β<1 and τ≥(log⁡(1/β)/(2d))1/4, then the one-sample classifier fY0X0∘T has robust risk at most β with training-sample probability at least 1−e−τ2d/2.

Facts & Assumptions

Given: The hypotheses in the Example.

[F1]

The Bernoulli model is supported on {−1,+1}d×{−1,+1}. Binary-feature Bernoulli model.

[F2]

T maps a nonnegative coordinate to +1 and a negative coordinate to −1. Coordinatewise threshold preprocessing.

[F3]

Robust zero-one loss is one exactly when some admissible perturbation changes the label; robust risk is the expectation of this loss. Measurable threat correspondences and robust risk.

[F4]

Under the stated τ,β condition, the one-sample thresholded Bernoulli classifier has robust risk at most β with the stated training-sample probability. Architecture-dependent robustness in the Bernoulli model.

Proof

1.1givenF1F2algebra

Fix x∈{−1,+1}d and any δ with ∥δ∥∞≤ϵ<1. If xj=+1, then xj+δj≥1−ϵ>0, so [F2] gives T(x+δ)j=+1=xj. If xj=−1, then xj+δj≤−1+ϵ<0, so T(x+δ)j=−1=xj. Hence T(x+δ)=x for every admissible perturbation.

2.1step 1.1F3

For each support point (x,y), every allowed perturbation therefore makes g(T(x+δ))=g(x). The robust zero-one loss of g∘T equals 1{g(x)≠y} pointwise. Taking expectation under Pθ,τ proves the displayed equality.

3.1step 2.1F4∎

Under the final τ,β condition, [F4] gives the one-sample risk bound for fY0X0∘T with the stated probability. The pointwise identity in step 2.1 explains why every radius ϵ<1 has the same risk.

References

Binary-feature Bernoulli model: the support is exactly the sign hypercube with a binary label.

Architecture-dependent robustness in the Bernoulli model: supplies the one-sample high-probability error guarantee.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedOpen item page →

Natural and robust risk in the Gaussian model

Example

Assume AC and set d=1014, σ=d1/4/32, β=1/100, and ϵ=1/4. In the fixed-parameter experiment with ∥θ∥2=d, one class-weighted training sample yields natural test error at most 1/100 with probability at least 1−2e−d/(8(σ2+1)). In the distinct Gaussian-prior experiment Θ∼Nd(0,Id), at the same n=1 and ϵ, every Borel learning rule F:(Rd×{−1,+1})n×Rd→{−1,+1} has expected robust risk at least (1−1/d)/2. The first assertion fixes a normalized parameter; the second averages over the stated Gaussian prior. Expected robust risk in the prior experiment uses outer expectation over the parameter and training sample.

Facts & Assumptions

Given: The Example's parameters and the two separate Gaussian experiments in its Statement.

[A1]

Assume AC as required by the Gaussian model and both risk bounds. The Axiom of Choice.

[F1]

For 0<β<1, if σ≤d1/4/(5log⁡(1/β)), then the one-sample class-weighted classifier has natural test error at most β with probability at least 1−2e−d/(8(σ2+1)). One-sample natural error in the Gaussian model.

[F2]

If d≥2 and n≤ϵ2σ2/(8log⁡d), every Borel learning rule under prior Θ∼Nd(0,Id) has expected robust risk at least (1−1/d)/2. Robust sample lower bound in the Gaussian model.

[F3]

The natural logarithm is increasing, obeys log⁡(2k)=klog⁡2, and satisfies log⁡b−log⁡a=∫abdx/x for 0<a≤b. Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t.

Proof

1.1givenA1F1F3algebra

Since 100<256=28 and log⁡2=∫12dx/x<1, [F3] gives log⁡100<8. Hence 5log⁡100<58<32, so σ=d1/4/32≤d1/4/(5log⁡100). Apply [F1] with β=1/100 to get the fixed-parameter natural-risk assertion.

2.1givenA1F2F3algebra∎

We have 210=1024>103, hence 247=27(210)4>128⋅1012>1014=d. Since log⁡2<1, [F3] gives log⁡d<log⁡(247)=47log⁡2<47. Therefore ϵ2σ28log⁡d=d131072log⁡d>107131072⋅47>1=n, where 131072⋅47=6,160,384<10,000,000. Since d≥2, [F2] applies and proves the Gaussian-prior robust-risk lower bound.

References

One-sample natural error in the Gaussian model: supplies the fixed-parameter one-sample natural-error guarantee.

Robust sample lower bound in the Gaussian model: supplies the Gaussian-prior robust-risk lower bound.

Sources