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Exercise 10.3: the error of hₜ with respect to the updated distribution D⁽ᵗ⁺¹⁾ is exactly 1/2

Proved
UnderstandingML.adaboost_reweighted_error

by naimengye · Sep 24, 2026 · Mathlib 0df444a (Lean v4.33.1)

adaboostboosting

Exercise 10.3. The AdaBoost weighting mechanism forces the weak learner to focus on the problematic examples: the error of hth_tht​ with respect to the distribution D(t+1)D^{(t+1)}D(t+1) is exactly 1/21/21/2, that is, for every t∈[T]t \in [T]t∈[T], ∑i=1mDi(t+1)1[yi≠ht(xi)]=1/2\sum_{i=1}^m D^{(t+1)}_i \mathbb{1}[y_i \ne h_t(x_i)] = 1/2∑i=1m​Di(t+1)​1[yi​=ht​(xi​)]=1/2.

Formally: for a nonempty sample and ϵt∈(0,1)\epsilon_t \in (0,1)ϵt​∈(0,1), the weighted error of hth_tht​ under adaDist S h (t+1) equals 1/21/21/2.

Preamble
import Definitions.Def_UnderstandingML_Boosting

open MeasureTheory
Formal statement
namespace UnderstandingML

/-- **Exercise 10.3** (p. 143). The error of `hₜ` with respect to the updated distribution
`D⁽ᵗ⁺¹⁾` is exactly `1/2`: `∑ᵢ Dᵢ⁽ᵗ⁺¹⁾ 𝟙[yᵢ ≠ hₜ(xᵢ)] = 1/2`. Stated for a nonempty sample and
`εₜ ∈ (0, 1)`, where the weight `wₜ` is defined. -/
theorem adaboost_reweighted_error {X : Type*} {m : ℕ} (hm : 0 < m) (S : Fin m → X × Bool)
    (h : ℕ → X → Bool) (t : ℕ) (hε : 0 < adaError S h t ∧ adaError S h t < 1) :
    weightedError (adaDist S h (t + 1)) S (h t) = 1 / 2 := by sorry

end UnderstandingML
Source
Shalev-Shwartz and Ben-David, Understanding Machine Learning: From Theory to Algorithms, Cambridge University Press 2014, doi:10.1017/CBO9781107298019, §10.7 Exercise 10.3 pp. 142-143
Human review
  • Endorsed by Shuze Chen · Sep 25, 2026

    Confirmed by the moderator at approval.

  • Endorsed by naimengye · Sep 25, 2026

    Confirmed by the mission captain (proposal self-audit).

  • Endorsed by mikedeng1 · Oct 7, 2026

    Confirmed by the mission captain (proposal self-audit).

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