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BanditAlgorithm.bandit_ucb_regret_bound

Proved

by Shuze Chen · Jul 17, 2026 · Mathlib c5ea003 (Lean v4.30.0)

banditsregretucb

(UCB regret bound, GOAL) Consider UCB (L&S Algorithm 3) on a stochastic kkk-armed 1-subgaussian bandit: play each arm once, then play

At=arg⁡max⁡i(μ^i(t−1)+2log⁡(1/δ)Ti(t−1)).A_t = \arg\max_i \left( \hat\mu_i(t-1) + \sqrt{\frac{2\log(1/\delta)}{T_i(t-1)}} \right).At​=argimax​(μ^​i​(t−1)+Ti​(t−1)2log(1/δ)​​).

For any horizon nnn, if δ=1/n2\delta = 1/n^2δ=1/n2 then

Rn≤3∑i=1kΔi+∑i:Δi>016log⁡nΔi.R_n \le 3\sum_{i=1}^k \Delta_i + \sum_{i : \Delta_i > 0} \frac{16\log n}{\Delta_i}.Rn​≤3i=1∑k​Δi​+i:Δi​>0∑​Δi​16logn​.
Preamble
import Definitions.Def_banditRegret
import Definitions.Def_ucbPolicy


open MeasureTheory ProbabilityTheory
Formal statement
theorem BanditAlgorithm.bandit_ucb_regret_bound {k : ℕ} (hk : 0 < k) {ν : StochasticBandit k}
    (hν : IsSubgaussianBandit 1 ν) {n : ℕ} (hn : 0 < n)
    {π : BanditPolicy k} (hπ : IsUCBPolicy (1 / (n : ℝ) ^ 2) π) :
    banditRegret ν π n ≤
      3 * ∑ i, banditGap ν i +
        ∑ i ∈ Finset.univ.filter (fun i ↦ 0 < banditGap ν i),
          16 * Real.log n / banditGap ν i := by
  sorry
Source
L&S Theorem 7.1, p.105
Read-back

What the Lean code literally says, in plain math · claude-fable-5

Notation (all unfolded from the imported definitions). Fix kkk arms with distributions P0,…,Pk−1P_0, \dots, P_{k-1}P0​,…,Pk−1​ (probability measures on R\mathbb{R}R), means μi=∫x dPi\mu_i = \int x \, dP_iμi​=∫xdPi​, optimal mean μ∗=sup⁡iμi\mu^* = \sup_i \mu_iμ∗=supi​μi​, gaps Δi=μ∗−μi\Delta_i = \mu^* - \mu_iΔi​=μ∗−μi​. For a length-ttt history hhh of (arm, reward) pairs, Ti(h)T_i(h)Ti​(h) is the pull count of arm iii and μ^i(h)\hat\mu_i(h)μ^​i​(h) its empirical mean (=0= 0=0 when Ti(h)=0T_i(h) = 0Ti​(h)=0). The index with parameter δ\deltaδ is Uiδ(h)=μ^i(h)+2log⁡(1/δ)/Ti(h)U_i^{\delta}(h) = \hat\mu_i(h) + \sqrt{2\log(1/\delta)/T_i(h)}Uiδ​(h)=μ^​i​(h)+2log(1/δ)/Ti​(h)​, with the junk conventions 1/0=01/0 = 01/0=0, log⁡x=0\log x = 0logx=0 for x≤0x \le 0x≤0, x=0\sqrt{x} = 0x​=0 for x<0x < 0x<0 (so an unpulled arm has index 000). Pν,πn\mathbb{P}^n_{\nu,\pi}Pν,πn​ is the history measure generated by π\piπ interacting with ν\nuν (Dirac at the empty history, then iteratively compose with the arm-then-reward step kernel and push forward along append), and Rn(ν,π)=n μ∗−∫∑t<nXt dPν,πnR_n(\nu,\pi) = n\,\mu^* - \int \sum_{t<n} X_t \, d\mathbb{P}^n_{\nu,\pi}Rn​(ν,π)=nμ∗−∫∑t<n​Xt​dPν,πn​ is the regret.

Claim. Suppose:

  • k>0k > 0k>0;
  • ν\nuν is 111-subgaussian: every arm is integrable and for every arm iii and every t∈Rt \in \mathbb{R}t∈R, ∫et(x−μi)dPi≤et2/2\int e^{t(x-\mu_i)} dP_i \le e^{t^2/2}∫et(x−μi​)dPi​≤et2/2;
  • n>0n > 0n>0;
  • π\piπ is a UCB policy at confidence parameter δ=1/n2\delta = 1/n^2δ=1/n2: at every round and every history there is an arm aaa with π(⋅∣h)=δa\pi(\cdot \mid h) = \delta_aπ(⋅∣h)=δa​ (deterministic), such that aaa is unpulled whenever some arm is unpulled, and otherwise aaa maximizes Uj1/n2(h)U_j^{1/n^2}(h)Uj1/n2​(h) over all arms jjj. (For n≥1n \ge 1n≥1 and a pulled arm, Ui1/n2(h)=μ^i(h)+4log⁡n/Ti(h)U_i^{1/n^2}(h) = \hat\mu_i(h) + \sqrt{4 \log n / T_i(h)}Ui1/n2​(h)=μ^​i​(h)+4logn/Ti​(h)​, since log⁡(n2)=2log⁡n\log(n^2) = 2\log nlog(n2)=2logn.)

Then

Rn(ν,π)  ≤  3∑i=0k−1Δi  +  ∑i : Δi>016log⁡nΔi.R_n(\nu, \pi) \;\le\; 3 \sum_{i=0}^{k-1} \Delta_i \;+\; \sum_{i \,:\, \Delta_i > 0} \frac{16 \log n}{\Delta_i}.Rn​(ν,π)≤3i=0∑k−1​Δi​+i:Δi​>0∑​Δi​16logn​.

Edge cases:

  • The second sum runs only over arms with strictly positive gap, so its divisions are never by zero; arms with Δi=0\Delta_i = 0Δi​=0 (in particular any optimal arm) contribute only to the first sum, where their term is 000.
  • log⁡\loglog is the natural logarithm; for n=1n = 1n=1 one has δ=1\delta = 1δ=1, every index bonus is 000 (the index degenerates to the empirical mean), log⁡n=0\log n = 0logn=0, and the bound reads R1≤3∑iΔiR_1 \le 3\sum_i \Delta_iR1​≤3∑i​Δi​.
  • The UCB predicate leaves tie-breaking and the choice among unpulled arms free; the bound is asserted for every policy satisfying the predicate.
Human review
  • Endorsed by Community (Bot) · Jul 17, 2026

  • Endorsed by Shuze Chen · Jul 17, 2026

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

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