Geometric helper: ε-net inner product bound #
For a 1/2-net N of the unit ball and any nonzero vector x, there exists z ∈ N with ⟪z, x⟫ ≥ (1/2)‖x‖.
Proof: Let θ = x/‖x‖ be the unit direction of x. Since θ ∈ B₂, there exists z ∈ N with ‖θ - z‖ ≤ 1/2. Then: ‖x‖ = ⟪θ, x⟫ = ⟪z, x⟫ + ⟪θ-z, x⟫ ≤ ⟪z, x⟫ + (1/2)‖x‖ So ⟪z, x⟫ ≥ (1/2)‖x‖.
Event containment via ε-net reduction #
If some θ ∈ B₂ achieves ⟪θ, x⟫ > t (with t > 0), then some z ∈ N achieves ⟪z, x⟫ > t/2.
Proof: From ⟪θ, x⟫ > t and ‖θ‖ ≤ 1, we get ‖x‖ ≥ ⟪θ, x⟫ > t (Cauchy-Schwarz). By the geometric helper, ∃ z ∈ N with ⟪z, x⟫ ≥ (1/2)‖x‖ > t/2.
Main theorem #
Theorem 1.19 (tail bound version): For a sub-Gaussian random vector X ∈ ℝ^d, P(max_{θ ∈ B₂} θᵀX > t) ≤ 6^d · exp(-t²/(8σ²)).
Here "sub-Gaussian random vector" means every unit-ball projection is sub-Gaussian: for all a with ‖a‖ ≤ 1, the scalar ⟪a, X⟫ is sub-Gaussian with parameter σ².
Proof: By the ε-net argument with a 1/2-net from Lemma 1.18, the event {∃θ ∈ B₂, ⟪θ,X⟫ > t} is contained in {∃z ∈ N, ⟪z,X⟫ > t/2}. Applying the union bound and Lemma 1.3 to each net element gives P ≤ |N| · exp(-(t/2)²/(2σ²)) ≤ 6^d · exp(-t²/(8σ²)).
Expectation and high-probability bounds #
Theorem 1.19 (expectation form). E[max_{θ ∈ B₂} θᵀX] ≤ 4σ√d for a sub-Gaussian random vector X with variance proxy σ².
Since max_{θ ∈ B₂} θᵀX(ω) = ‖X(ω)‖ (the supremum of the inner product over the unit ball equals the Euclidean norm), the statement simplifies to E[‖X‖] ≤ 4√σ² · √d.
Proof outline (from Rigollet):
- Let N be a 1/2-net of B₂ with |N| ≤ 6^d.
- For every θ ∈ B₂, write θ = z + x with z ∈ N and ‖x‖ ≤ 1/2.
- max_{θ∈B₂} θᵀX ≤ max_{z∈N} zᵀX + (1/2) max_{θ∈B₂} θᵀX.
- Rearranging: max_{θ∈B₂} θᵀX ≤ 2 max_{z∈N} zᵀX.
- By Theorem 1.14 (expectation form): E[max_{z∈N} zᵀX] ≤ σ√(2 log|N|) ≤ σ√(2d log 6).
- Therefore E[‖X‖] ≤ 2σ√(2d log 6) ≤ 4σ√d since 2 log 6 < 4.
Theorem 1.19 (high probability form). With probability at least 1 - δ: ‖X‖ ≤ 4σ√d + 2σ√(2 log(1/δ)).
Since max_{θ ∈ B₂} θᵀX(ω) = ‖X(ω)‖, this gives a high-probability upper bound on ‖X‖.
Proof outline (from Rigollet):
- From the tail bound (theorem_1_19_tail_bound): P(‖X‖ > t) ≤ 6^d · exp(-t²/(8σ²)).
- Set 6^d · exp(-t²/(8σ²)) ≤ δ, which gives t² ≥ 8σ² (d log 6 + log(1/δ)).
- Taking t = √(8 log 6) · σ · √d + 2σ · √(2 log(1/δ)) suffices.
- Since √(8 log 6) < 4, we can use t = 4σ√d + 2σ√(2 log(1/δ)).