Proof of Theorem 11:
ProvedStabGen.Hypothesis.second_moment_looalgorithmic-stabilitylearning-theoryp2o-batch-pfp1ap2o-gran-per-chapterp2o-plan-paperp2o-v1
Let be a probability distribution on , let the cost satisfy , and let be a symmetric deterministic learning algorithm whose losses are measurable. Suppose that, at sample size , has hypothesis stability . Then, with ,
Chebyshev's inequality applied to this second-moment bound gives the leave-one-out half of Theorem 11.
Formalization Note The bound is as printed. The paper says it follows "by (10)"; it follows from Lemma 9, eq. (9), together with Definition 3, whereas (10) would give a constant .
Preamble
import Mathlib import Definitions.Def_FoundationsML_Stability_Loss import Definitions.Def_FoundationsML_Stability_EmpiricalError import Definitions.Def_FoundationsML_Stability_GeneralizationError import Definitions.Def_StabGen_Hypothesis_Setting import Definitions.Def_StabGen_Hypothesis_Stability open MeasureTheory
Formal statement
namespace StabGen.Hypothesis
open FoundationsML.Stability
theorem second_moment_loo {X Y Y' : Type*} [MeasurableSpace X] [MeasurableSpace Y]
(D : Measure (X × Y)) [IsProbabilityMeasure D] (L : Y' → Y → ℝ) (M : ℝ)
(hL : ∀ (y' : Y') (y : Y), 0 ≤ L y' y ∧ L y' y ≤ M)
(A : LearningAlgorithm X Y Y')
(hA : ∀ n : ℕ, Measurable (fun p : (Fin n → X × Y) × (X × Y) =>
Loss L (A (trainingSet p.1)) p.2))
(m : ℕ) (hm : 2 ≤ m) (β₁ : ℝ) (h₁ : HypothesisStable D L A m β₁) :
∫ S : Fin m → X × Y,
(GeneralizationError D L (A (trainingSet S)) - looError L A S) ^ 2
∂(Measure.pi fun _ : Fin m => D)
≤ M ^ 2 / (2 * (m : ℝ)) + 3 * M * β₁ := by sorry
end StabGen.Hypothesis
Source
Bousquet & Elisseeff, Stability and Generalization, JMLR 2 (2002), p. 506, proof of Theorem 11 ("Also, we have by (10)")
Human review
Confirmed by the mission captain (proposal self-audit).
Confirmed by the moderator at approval.