Alphabeta Math
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.

Exp3 and full-information Hedge update differently

Example

Take one round with two arms, reward vector (1,0), learning rate η>0, and uniform initial probabilities. Run full-information Hedge with ϵ=η on losses (0,1); its next distribution is (exp⁡(η)/(exp⁡(η)+1),1/(exp⁡(η)+1)). After the first round, Exp3 has next distribution (1/2,1/2) if it sampled arm 1 and (exp⁡(η)/(exp⁡(η)+exp⁡(−η)),exp⁡(−η)/(exp⁡(η)+exp⁡(−η))) if it sampled arm 2. Both outcomes differ from the full-information distribution.

Example references

Exp3 with importance-weighted loss feedback: the estimated rewards and exponential probability update.

Hedge: full-information losses, equal initial weights, and the exponential loss update.

Bandit importance-weighted estimator: the observed reward and loss-based estimator formulas.

Facts & Assumptions

Given: One round, two arms, reward vector (1,0), Exp3 learning rate η>0, full-information Hedge learning rate ϵ=η, uniform initial weights, and the two update rules.

[F1]

Exp3 with importance-weighted loss feedback: Exp3 samples proportionally to exp⁡(ηS^t−1,i) and updates S^t,i=S^t−1,i+X^t,i.

[F2]

Hedge: Hedge initializes both weights at one, samples from their normalized values, then updates Wt+1,i=Wt,iexp⁡(−ηℓt,i) after observing the entire loss vector.

[F3]

Bandit importance-weighted estimator: Y^t,i=1{At=i}(1−Xt)/Pt,i and X^t,i=1−Y^t,i.

[F5]

The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x): exp⁡(z)>0 and exp⁡(−z)=1/exp⁡(z) for real z.

[F6]

The exponential function is strictly increasing: exp⁡ is strictly increasing on R.

Verification

1.1givenF1F3

Since S^0,1=S^0,2=0, Exp3 starts with probabilities (1/2,1/2). If A1=1, then Y1=0 and X^1,1=X^1,2=1, so the next Exp3 distribution is uniform. If A1=2, then Y1=1, P1,2=1/2, and (Y^1,1,Y^1,2)=(0,2); hence X^1=(1,−1) and the next distribution is (exp⁡(η)/(exp⁡(η)+exp⁡(−η)),exp⁡(−η)/(exp⁡(η)+exp⁡(−η))).

1.2givenF2F5algebra

Full-information Hedge uses ϵ=η and sees losses (0,1), so its next weights are (1,exp⁡(−η)) and its normalized distribution is (exp⁡(η)/(exp⁡(η)+1),1/(exp⁡(η)+1)).

2.1F4F6step 1.1step 1.2algebra∎

By [F4] and [F6], exp⁡(η)>1, so Hedge's first probability exceeds 1/2 and differs from Exp3's first probability after A1=1. Also exp⁡(2η)>exp⁡(η), so the odds ratio after A1=2 is exp⁡(2η) for Exp3 and exp⁡(η) for Hedge; these distributions differ as well.

Depends on

Used by

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Dependency tree · two levels

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Sources