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Exp3-IX exponential-weights loss bound

Proved
BanditAlgorithm.exp3ix_observed_loss_estimate_bound

by Harry_Xu · Jul 21, 2026 · Mathlib 0df444a (Lean v4.33.1)

bandit-algorithmsexponential-weightsregret

This is the deterministic exponential-weights loss bound for Exp3-IX.

Let k>1k>1k>1, let η>0\eta>0η>0, and let hhh be a bounded length-nnn bandit history. Write L~n=∑t=1n(1−Xt)\widetilde L_n=\sum_{t=1}^n(1-X_t)Ln​=∑t=1n​(1−Xt​) for the learner's observed cumulative loss and L^n,i\widehat L_{n,i}Ln,i​ for the implicit-exploration loss estimate of arm iii. For every comparator arm iii, with γ=η/2\gamma=\eta/2γ=η/2,

L~n−η2∑jL^n,j−L^n,i≤log⁡kη+η2∑jL^n,j.\widetilde L_n-\frac{\eta}{2}\sum_j\widehat L_{n,j}-\widehat L_{n,i}\le\frac{\log k}{\eta}+\frac{\eta}{2}\sum_j\widehat L_{n,j}.Ln​−2η​j∑​Ln,j​−Ln,i​≤ηlogk​+2η​j∑​Ln,j​.

This combines the generic exponential-weights inequality with the exact Exp3-IX bias identity and is reusable independently of the probabilistic concentration analysis.

Formalization Note The theorem is pointwise and retains the empty-horizon boundary case.

Preamble
import Definitions.Def_AdversarialBandit
import Definitions.Def_exp3Policy

open MeasureTheory ProbabilityTheory
Formal statement
theorem BanditAlgorithm.exp3ix_observed_loss_estimate_bound
    {k : ℕ} (hk : 1 < k) (n : ℕ) (η : ℝ) (hη : 0 < η)
    (h : BanditAlgorithm.BanditHistory k n)
    (hh : ∀ t : Fin n, (h t).2 ∈ Set.Icc (0 : ℝ) 1) (i : Fin k) :
    (∑ t : Fin n, (1 - (h t).2)) -
          (η / 2) *
            ∑ j : Fin k, BanditAlgorithm.exp3IXEstimate η (η / 2) n h j -
        BanditAlgorithm.exp3IXEstimate η (η / 2) n h i ≤
      Real.log k / η +
        (η / 2) *
          ∑ j : Fin k, BanditAlgorithm.exp3IXEstimate η (η / 2) n h j := by
  sorry
Source
Lattimore and Szepesvari, Bandit Algorithms (CUP 2020): Eq. (12.1), printed p. 165 / PDF p. 174, gives the deterministic exponential-weights loss estimate; Lemma 12.4, printed p. 169 / PDF p. 178, identifies the Exp3-IX bias. The statement combines them and specializes gamma = eta/2.

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