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BanditAlgorithm.adversarial_bandit_exp3ix_high_probability_regret

Proved

by Shuze Chen · Jul 17, 2026 · Mathlib 0df444a (Lean v4.33.1)

adversarialbanditsexp3-ixhigh-probability

(Exp3-IX high-probability bound, δ\deltaδ-independent learning rate) Let x∈[0,1]n×kx \in [0,1]^{n\times k}x∈[0,1]n×k (with k>1k > 1k>1, n≥1n \ge 1n≥1) and δ∈(0,1)\delta \in (0,1)δ∈(0,1). Suppose Exp3-IX (Algorithm 10) — with biased loss estimator and exponential weights

Y^ti=1{At=i}(1−Xt)Pti+γ,Pti∝exp⁡(−ηL^t−1,i)\hat Y_{ti} = \frac{\mathbb{1}\{A_t=i\}(1-X_t)}{P_{ti} + \gamma}, \qquad P_{ti} \propto \exp(-\eta \hat L_{t-1,i})Y^ti​=Pti​+γ1{At​=i}(1−Xt​)​,Pti​∝exp(−ηL^t−1,i​)

— is run with η=η1=2log⁡(k+1)/(nk)\eta = \eta_1 = \sqrt{2\log(k+1)/(nk)}η=η1​=2log(k+1)/(nk)​ and γ=η/2\gamma = \eta/2γ=η/2. Then the random regret R^n=max⁡i∑t=1nxti−∑t=1nXt\hat R_n = \max_i \sum_{t=1}^n x_{ti} - \sum_{t=1}^n X_tR^n​=maxi​∑t=1n​xti​−∑t=1n​Xt​ satisfies (Eq. 12.5)

P(R^n≥8nklog⁡(k+1)+nk2log⁡(k+1) log⁡1δ+log⁡k+1δ)≤δ,\mathbb{P}\left(\hat R_n \ge \sqrt{8nk\log(k+1)} + \sqrt{\frac{nk}{2\log(k+1)}}\,\log\frac{1}{\delta} + \log\frac{k+1}{\delta}\right) \le \delta,P(R^n​≥8nklog(k+1)​+2log(k+1)nk​​logδ1​+logδk+1​)≤δ,

stated as a bound on the adversarialMeasure of the bad set of histories.

Preamble
import Definitions.Def_AdversarialBandit
import Definitions.Def_exp3Policy


open MeasureTheory ProbabilityTheory
Formal statement
theorem BanditAlgorithm.adversarial_bandit_exp3ix_high_probability_regret
    {k : ℕ} (hk : 1 < k) (n : ℕ) (hn : 0 < n)
    (x : ℕ → Fin k → ℝ) (hx : ∀ t : ℕ, ∀ i : Fin k, x t i ∈ Set.Icc (0 : ℝ) 1)
    (δ : ℝ) (hδ : δ ∈ Set.Ioo (0 : ℝ) 1)
    (η : ℝ) (hη : η = Real.sqrt (2 * Real.log (k + 1) / (n * k)))
    (π : BanditPolicy k) (hπ : IsExp3IXPolicy η (η / 2) π) :
    adversarialMeasure x π n
      {h : BanditHistory k n |
        Real.sqrt (8 * n * k * Real.log (k + 1)) +
          Real.sqrt (n * k / (2 * Real.log (k + 1))) * Real.log (1 / δ) +
          Real.log ((k + 1) / δ) ≤ adversarialRandomRegret n x h} ≤
      ENNReal.ofReal δ := by
  sorry
Source
L&S Theorem 12.1(1), Eq. (12.5), p.167

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