ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

A finite two-expert Exp4 instance

Example

Consider one round with two actions, two experts, reward vector $x_{1} = (1, 0)$ , and deterministic advice rows $E (1)_{1} = (1, 0)$ and $E (1)_{2} = (0, 1)$ . For any learning rate $η > 0$ , Exp4 starts with equal expert weights and samples each action with probability $1 / 2$ . The expected learner reward is $1 / 2$ , the best fixed expert reward is $1$ , and the expected regret is $1 / 2$ , independently of $η$ . For the theorem-bound clause, use the tuned learning rate $η = \sqrt{\log 2}$ ; the general Exp4 theorem then gives $R_{1} \leq \sqrt{4 \log 2}$ .

Example references

Contextual bandit protocol: the finite expert advice and expected-regret conventions used in the calculation.

Exp4 initialization and action mixture: the uniform initial expert weights and mixture action distribution used here.

Expected-regret theorem: the stated general bound specialized to this one-round instance.

Facts & Assumptions

Given: One round, actions $[2]$ , experts $[2]$ , reward vector $(1, 0)$ , advice rows $(1, 0)$ and $(0, 1)$ , and Exp4 with $η = \sqrt{\log 2}$ for the theorem-bound clause.

[F1]

Contextual bandit protocol with finite expert advice: an expert's round reward is the dot product of its advice row with the reward vector.

[F2]

Exp4 with unbiased estimated rewards: $Q_{1, m} = 1 / M$ and $P_{1, i} = \sum_{m} Q_{1, m} E (1)_{m, i}$ .

[F3]

Exp4 expected regret against the best fixed expert: for $n = 1$ , $k = 2$ , and $M = 2$ , the expected regret is at most $\sqrt{4 \log 2}$ .

Verification

1.1F2algebra

The uniform initial weights give $Q_{1, 1} = Q_{1, 2} = 1 / 2$ . Hence $P_{1, 1} = 1 / 2$ and $P_{1, 2} = 1 / 2$ . The realized reward equals one with probability $1 / 2$ and zero with probability $1 / 2$ , so its expectation is $1 / 2$ .

2.1F1step 1.1algebra

The two experts' expected rewards are $⟨ (1, 0), (1, 0) ⟩ = 1$ and $⟨ (0, 1), (1, 0) ⟩ = 0$ . Thus the best fixed expert reward is one and the expected regret is $1 - 1 / 2 = 1 / 2$ .

3.1F3algebra∎

Substituting $n = 1$ , $k = 2$ , and $M = 2$ into [F3] gives the stated bound $\sqrt{4 \log 2}$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Tor Lattimore and Csaba Szepesvári, Bandit Algorithms (standard reference, not scraped)