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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Gap-free expected regret of UCB

Statement

Let n,k be integers with n≥max⁡{k,2}. In an n-round k-armed stochastic bandit, suppose all reward-stack entries are 1-sub-Gaussian. Run the upper-confidence-bound algorithm with δ=n−2. Its expected reward regret satisfies

Rn≤8nklog⁡n+3∑i=1kΔi,

where log⁡ is the natural logarithm. This is an expected-regret bound, not a high-probability guarantee.

Statement references

Finite stochastic bandit model, gaps, and regret: the finite reward-stack model, gaps, and expected reward regret.

Centered 1-sub-Gaussian reward: the exact mean-centred MGF condition for each reward entry.

Finite-horizon upper-confidence-bound bandit algorithm: the UCB index and finite-horizon schedule.

A sharper expected suboptimal-arm pull count for the unchanged UCB rule: the local once-per-prefix argument gives the sharper expected-pull estimate used in the finite Cauchy–Schwarz proof.

Finite-arm regret decomposes into expected action counts: expected regret is the gap-weighted sum of expected pull counts under the fixed conditional-mean law.

Tor Lattimore and Csaba Szepesvári, Bandit Algorithms: Theorem 7.2 and proof, Chapter 7 §7.1, PDF p. 117 / printed p. 108.

Facts & Assumptions

Given: The unchanged finite reward-stack bandit, horizon n≥max⁡{k,2}, centered 1-sub-Gaussian entries, and UCB with δ=n−2 in the Statement.

[F1]

The gaps are nonnegative and exactly one arm is pulled each round, so ∑iTi(n)=n. Finite stochastic bandit model, gaps, and regret

[F2]

Reward entries satisfy the exact centered MGF condition, and the algorithm uses the stated confidence schedule. Centered 1-sub-Gaussian reward Finite-horizon upper-confidence-bound bandit algorithm

[F3]

The reward-stack model supplies the eventwise conditional-mean hypothesis of the finite-arm regret decomposition. The independent reward-stack model has the prescribed conditional arm laws

[F4]

Under that hypothesis, expected reward regret equals ∑iΔiE[Ti(n)]. Finite-arm regret decomposes into expected action counts

[F5]

For every suboptimal arm, the unchanged UCB rule satisfies E[Ti(n)]≤3+8log⁡(n)/Δi2. A sharper expected suboptimal-arm pull count for the unchanged UCB rule

[F6]

For finite real lists, (∑iaibi)2≤(∑iai2)(∑ibi2), with the corresponding nonnegative square-root inequality. The Cauchy-Schwarz inequality for finite sums

[F8]

Expectation is linear for finite sums of integrable random variables. Linearity, monotonicity, and the modulus bound for expectation

Proof

1.1givenF1F3F4F8

By [F3], the actual reward table meets the conditional-mean premise of [F4], hence Rn=∑iΔiE[Ti(n)]. All counts are bounded, and [F1] gives ∑iE[Ti(n)]=n.

2.1F1F2F5F7step 1.1constructalgebra

Let I={i:Δi>0} and, for i∈I, put qi=(E[Ti(n)]−3)+. By [F5], Δi2qi≤8L. Also qi≤E[Ti(n)], so ∑i∈Iqi≤n by step 1.1. Therefore Rn≤3∑iΔi+∑i∈IΔiqi. Zero-gap arms contribute zero, and no reciprocal of a zero gap occurs.

3.1F6F7step 2.1casesalgebra∎

Apply [F6] to the finite nonnegative lists ai=Δiqi and bi=qi for i∈I. Step 2.1 yields ∑i∈IΔiqi≤(∑i∈IΔi2qi)(∑i∈Iqi)≤8∣I∣Ln≤8nklog⁡n. If I is empty both sums and the regret are zero; otherwise the same finite inequality applies. Substitution in step 2.1 proves the unchanged Statement.

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