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Ti(n)≤κiT_i(n) \le \kappa_iTi​(n)≤κi​ when 2Δ~<Δi2\tilde\Delta < \Delta_i2Δ~<Δi​

Disproved
BanditAlgorithm.moss_pull_count_le_kappa_on_large_gap

by tianyipeng · Jul 29, 2026 · Mathlib c5ea003 (Lean v4.30.0)

banditsmachine-learning

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Disproved status carries no information about that problem. Do not import this node or use it as a dependency.

Under MOSS, on the event that an arm's suboptimality gap exceeds twice the optimal arm's index shortfall, the arm's pull count is dominated by its MOSS index count:

2Δ~<Δi  ⟹  Ti(n)≤κi,2\tilde\Delta < \Delta_i \;\Longrightarrow\; T_i(n) \le \kappa_i,2Δ~<Δi​⟹Ti​(n)≤κi​,

almost surely under the canonical bandit measure. Here Δ~\tilde\DeltaΔ~ is the amount by which the optimal arm's MOSS index ever drops below its true mean, and i∗i^{*}i∗ is an optimal arm.

This is the justification given on printed p. 126 of Lattimore--Szepesvári: "for arms iii with Δi>2Δ~\Delta_i > 2\tilde\DeltaΔi​>2Δ~, the index of the optimal arm is always larger than μi+Δi/2\mu_i + \Delta_i/2μi​+Δi​/2, so κi\kappa_iκi​ is an upper bound on Ti(n)T_i(n)Ti​(n)." Indeed, whenever MOSS plays arm iii its index is maximal, hence at least the optimal arm's index, which on this event exceeds μi+Δi/2\mu_i + \Delta_i/2μi​+Δi​/2 — so that pull is counted by κi\kappa_iκi​. The hypothesis is essential: without it the domination genuinely fails.

Why this node was retired

The posted statement is

theorem BanditAlgorithm.moss_pull_count_le_kappa_on_large_gap
    {k : ℕ} (hk : 0 < k)
    {ν : BanditAlgorithm.StochasticBandit k}
    {n : ℕ} {π : BanditAlgorithm.BanditPolicy k}
    (hπ : BanditAlgorithm.IsMOSSPolicy n π)
    (iStar : Fin k)
    (hiStar : BanditAlgorithm.banditArmMean ν iStar =
      BanditAlgorithm.banditOptimalMean ν)
    (i : Fin k) :
    ∀ᵐ h ∂(BanditAlgorithm.banditMeasure ν π n),
      2 * BanditAlgorithm.mossOptimalShortfall ν iStar h <
          BanditAlgorithm.banditGap ν i →
        (BanditAlgorithm.armPullCount i h : ℝ) ≤
          (BanditAlgorithm.mossKappa ν i h : ℝ) := by sorry

The positive-sample-size κ can be zero although initialization already pulled the arm once; the posted finite-history stream additionally distorts unavailable sample means.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

Restore the full potential reward-stream definition and the source counting convention, explicitly accounting for the initialization pull (e.g. a 1+κ bound when κ counts only positive sample sizes below threshold). Retain the random large-gap event from the source and re-derive the exact comparison.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.

Preamble
import Definitions.Def_mossKappa
import Definitions.Def_banditRegret

open MeasureTheory ProbabilityTheory
Formal statement
theorem BanditAlgorithm.moss_pull_count_le_kappa_on_large_gap
    {k : ℕ} (hk : 0 < k)
    {ν : BanditAlgorithm.StochasticBandit k}
    {n : ℕ} {π : BanditAlgorithm.BanditPolicy k}
    (hπ : BanditAlgorithm.IsMOSSPolicy n π)
    (iStar : Fin k)
    (hiStar : BanditAlgorithm.banditArmMean ν iStar =
      BanditAlgorithm.banditOptimalMean ν)
    (i : Fin k) :
    ∀ᵐ h ∂(BanditAlgorithm.banditMeasure ν π n),
      2 * BanditAlgorithm.mossOptimalShortfall ν iStar h <
          BanditAlgorithm.banditGap ν i →
        (BanditAlgorithm.armPullCount i h : ℝ) ≤
          (BanditAlgorithm.mossKappa ν i h : ℝ) := by sorry
Source
Lattimore and Szepesvari, Bandit Algorithms (CUP 2020), https://tor-lattimore.com/downloads/book/book.pdf, proof of Theorem 9.1, printed p. 126 / PDF p. 135: 'for arms i with Delta_i > 2 Delta, the index of the optimal arm is always larger than mu_i + Delta_i/2, so kappa_i is an upper bound on T_i(n)'.

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