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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.

Contextual expert bandits

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Contextual bandit protocol with finite expert advice

Definition

Fix integers n≥1, k>1, and M>1. Write [r]={1,…,r} for a positive integer r, and let

Δk={p∈[0,1]k:∑i=1kpi=1},E=ΔkM.

An oblivious contextual bandit instance consists of reward vectors xt∈[0,1]k and advice matrices E(t)∈E, for t∈[n], all fixed before the learner's private randomization. The matrix E(t) is revealed before the round-t action; its row E(t)m=(E(t)m,1,…,E(t)m,k) is expert m's probability distribution over the k actions. The expected reward of expert m on round t is ⟨E(t)m,xt⟩=∑i=1kE(t)m,ixt,i.

At round t, the learner's pre-action history consists of its earlier actions and observed rewards. Given that history and the revealed advice, it selects a distribution Pt∈Δk, draws a measurable finite-valued action At∈[k] with conditional probabilities P(At=i∣Ft−1)=Pt,i, and observes only Xt=xt,At. Here Ft−1 is the sigma-algebra generated by the pre-action history. For every fixed expert sequence and reward sequence, the expected regret against the best fixed expert is

Rn=E[max⁡m∈[M]∑t=1n⟨E(t)m,xt⟩−∑t=1nXt],

where expectation is over the learner's action draws. The regret random variable is bounded, so this expectation is finite. This protocol makes no stochastic generalization claim and imposes no shared-structure condition on the experts; it does not cover advice or rewards selected in response to the learner's current action.

Definition references

Online decision timing: the learner chooses each action from its pre-action history before observing current-round feedback.

Oblivious adversary and bandit feedback: an oblivious sequence is fixed before learner randomization, and bandit feedback reveals only the played reward.

Expected-regret convention: expected regret is the expectation of the specified measurable, integrable regret random variable.

Measurable random actions: a finite action draw is a measurable random element of the discrete action set.

Pre-action filtration: the sigma-algebras generated by successive histories form a filtration.

Expectation: the expectation of an integrable real random variable is its Lebesgue integral on the underlying probability space.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Exp4 with unbiased estimated rewards

Definition

Given integers n≥1, k>1, M>1, a learning rate η>0, and an instance from the contextual bandit protocol with finite expert advice, initialize Q1,m=1/M for m∈[M]. For rounds t=1,…,n, perform the following operations:

  1. Receive the advice matrix E(t), whose mth row is E(t)m∈Δk.

  2. Form the action distribution

Pt,i=∑m=1MQt,mE(t)m,i,i∈[k].
  1. Draw a measurable action At∈[k] with conditional probabilities P(At=i∣Ft−1)=Pt,i and observe only Xt=xt,At.

  2. For each i∈[k], form

X^t,i=1−1{At=i}(1−Xt)Pt,i.

When Pt,i=0, the action At=i is impossible; in this case the quotient term is defined to be zero on every possible outcome.

  1. For each expert m∈[M], form its estimated reward
X~t,m=∑i=1kE(t)m,iX^t,i

and update the expert distribution by

Qt+1,m=Qt,mexp⁡(ηX~t,m)∑r=1MQt,rexp⁡(ηX~t,r).

The output is (Qt,Pt,At,Xt,X^t,X~t)t=1n and Qn+1. All arithmetic is over exact real numbers, and the procedure assumes a probability space carrying the stated finite-valued measurable draws. This is the γ=0 Exp4 update; no regret guarantee for bounded-loss Hedge is part of the algorithm.

Definition references

Contextual bandit protocol: the fixed oblivious reward/advice instance, action set, observed reward, and pre-action history used here.

Normalized exponential weights: the full-information algorithm uses normalized exponential weights; this is notation only, not a regret result for the present unbounded estimates.

Importance-weighted loss estimate: on coordinates with positive sampling probability, the quotient term is the standard estimate of loss 1−Xt; the zero-probability convention is handled locally here.

Measurable finite actions: each action draw is a measurable finite-valued random element.

Real exponential: the update uses the real exponential function.

Exponential positivity: exponential weights are positive for every finite real input.

Exp4 well-definedness ↗: the separate result proves finite estimates, simplex mixtures, positive expert weights and finite-horizon completion, including zero-probability actions.

LemmaStatement: AI-adaptedProof: AI-generatedjudge pass (gpt-6-luna)Open item page →

Exp4 updates are well-defined, including zero-probability actions

Statement

For the exact-real Exp4 procedure, take any integers n≥1, k>1, M>1, any finite real learning rate η>0, and every oblivious reward/advice instance and supplied measurable draws in its input scope. At every round t∈[n], the expert vector Qt is a probability vector with strictly positive entries, the action mixture Pt belongs to Δk, and Pt,i=0 implies E(t)m,i=0 for every expert m. Every action and expert reward estimate is finite; the corresponding estimated losses 1−X^t,i and 1−X~t,m are nonnegative. The normalizing denominator is finite and strictly positive, so Qt+1 again has strictly positive entries summing to one. The procedure completes its n prescribed rounds and returns all specified outputs.

The zero-quotient convention is retained exactly. These conclusions do not require positive sampling probability for every action, a uniform exploration mixture, a bound on the magnitude of an importance-weighted estimate, or a restriction on the learning rate beyond η>0.

Facts & Assumptions

Given: The full finite-horizon Exp4 inputs in the Statement, using exact real arithmetic and the probability space carrying the algorithm's prescribed finite-valued measurable action draws.

[F1]

The input has n≥1, k>1, M>1, fixed rewards xt∈[0,1]k, and advice rows with E(t)m,i≥0 and ∑iE(t)m,i=1. The observed reward is Xt=xt,At∈[0,1]; action draws are measurable, and the pre-action history consists of earlier actions and observed rewards. Contextual bandit protocol with finite expert advice

[F2]

The algorithm initializes Q1,m=1/M, sets Pt,i=∑mQt,mE(t)m,i, and uses X^t,i=1−L^t,i with L^t,i=1{At=i}(1−Xt)/Pt,i when Pt,i>0, and L^t,i=0 when Pt,i=0. It sets X~t,m=∑iE(t)m,iX^t,i and normalizes Qt,mexp⁡(ηX~t,m) to obtain Qt+1,m. Exp4 with unbiased estimated rewards

[F4]

Inverses and products of positive real numbers are positive, and positive multiplication preserves inequalities. Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication

[F5]

Induction propagates a base assertion preserved at each successor through the finite sequence of rounds. The principle of mathematical induction

Proof

1.1givenF2F4basealgebra

At the first round, M>1 makes 1/M a finite strictly positive real by [F4]. The M entries sum to one, so Q1 is a strictly positive probability vector.

1.2F1F2F4IHalgebra

Assume Qt is a strictly positive probability vector at a round t≤n. Every Pt,i is a finite nonnegative sum. Reordering its finite double sum gives ∑iPt,i=∑mQt,m∑iE(t)m,i=∑mQt,m=1. Hence each Pt,i∈[0,1] and Pt∈Δk. If Pt,i=0, every term Qt,mE(t)m,i in this nonnegative sum is zero, and strict positivity of Qt,m forces E(t)m,i=0 for every m. Thus unsupported actions carry no expert advice.

2.1F1F2F4step 1.2algebra

Fix the round's action. Since 0≤Xt≤1, a supported coordinate has finite nonnegative L^t,i by [F4], while an unsupported coordinate has the defined value zero. Thus every X^t,i=1−L^t,i is finite, with no division by zero. The finite sum defining X~t,m is also finite. As each advice row sums to one, L~t,m:=1−X~t,m=∑iE(t)m,iL^t,i≥0. The reasoning uses the zero convention even on null outcomes, so no pointwise undefined quotient is hidden behind a probability-zero claim.

3.1F2F3F4step 2.1IHalgebra

By step 2.1, ηX~t,m is a finite real number for every expert. By [F3], every weight wm=Qt,mexp⁡(ηX~t,m) is finite and strictly positive. Since M is finite and nonzero, Zt=∑mwm is finite and strictly positive. Therefore every Qt+1,m=wm/Zt is finite and strictly positive by [F4], and their sum is one. This is the induction step.

4.1givenF1F2F3F5step 1.1step 1.2step 2.1step 3.1discharge-induction∎

By [F5], all stated invariants and finite estimates hold through round n, including Qn+1. For fixed inputs, Qt and Pt are functions of the finite past action sequence, and all other round records are functions of that sequence and the current action. The finite measurable action-history partitions therefore make the states and records measurable with finite ranges; in particular Pt is pre-action measurable. There is at least one supported arm at each round because ∑iPt,i=1, and the algorithm's probability space supplies its measurable draw; no sampler construction is being assumed beyond the input contract. Each round performs only finite sums and the prescribed finite evaluations and draw. After exactly n rounds the loop ends and returns (Qt,Pt,At,Xt,X^t,X~t)t=1n and Qn+1. This proves well-definedness and finite-horizon completion without any extra exploration or learning-rate hypothesis.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Exp4 expert reward estimates are conditionally unbiased

Statement

In Exp4, at every round t∈[n] and for every expert m∈[M], the expert's estimated reward satisfies

E[X~t,m∣Ft−1]=⟨E(t)m,xt⟩.

The assertion includes rounds where some actions have zero sampling probability; no componentwise unbiasedness claim is made for an unsupported action.

Statement references

Exp4 estimates and zero-probability convention: the estimated action rewards and expert rewards are defined in the algorithm.

Facts & Assumptions

Given: A fixed oblivious instance with n≥1, k>1, M>1, the Exp4 algorithm, and a round t.

[F1]

Exp4 with unbiased estimated rewards: Pt,i=∑r=1MQt,rE(t)r,i, every Qt,r>0, X^t,i=1−Y^t,i, and the quotient in Y^t,i is set to zero when Pt,i=0.

[F2]

Contextual bandit protocol with finite expert advice: xt and E(t) are fixed before the learner's action, each advice row is a probability vector, and Ft−1 records the pre-action history.

[F3]

Exp4 with unbiased estimated rewards: P(At=i∣Ft−1)=Pt,i, Xt=xt,At, and for Pt,i>0, Y^t,i=1{At=i}(1−Xt)/Pt,i.

[F4]

Expectation of a nonnegative or integrable random variable: expectation is the Lebesgue integral; a measurable finite-range real random variable on a probability space is integrable.

[F5]

Conditional expectation given a sigma algebra: conditional expectation is characterized by its integrals on pre-action history events; with a finite action set, its value on each positive-probability history is the finite conditional sum over actions.

Proof

1.1F1F2F3F4algebra

For the fixed instance, there are finitely many action histories. The weights and finite estimates are functions of those histories and the current action, so they are measurable with finite ranges and hence integrable. Fix a positive-probability pre-action history; on it, Qt and E(t) are fixed. If Pt,i=0, then 0=∑rQt,rE(t)r,i is a sum of nonnegative terms and every Qt,r>0; hence E(t)r,i=0 for every expert r. Such a coordinate contributes nothing to any expert reward estimate.

2.1F1F2F3F4F5step 1.1algebra∎

For Pt,i>0 on this history, the estimate is (1−xt,i)/Pt,i when At=i and zero otherwise. Thus its conditional mean, computed directly from [F3], is Pt,i(1−xt,i)/Pt,i=1−xt,i, regardless of other arms' probabilities. For fixed expert m, its row sums to one, so X~t,m=1−∑iE(t)m,iY^t,i. Distributing the finite conditional sum and omitting the zero coefficients from step 1.1 gives E[X~t,m∣Ft−1]=1−∑i:Pt,i>0E(t)m,i(1−xt,i)=∑i=1kE(t)m,ixt,i. These values on the finitely many positive-probability history atoms satisfy the integral characterization [F5] on every history event; values on null histories do not affect the conditional-expectation identity.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Exp4 weighted estimated loss has conditional second moment at most k

Statement

At every round t in Exp4, define the estimated action loss by L^t,i=1−X^t,i and the estimated loss of expert m by L~t,m=1−X~t,m. Then

E[∑m=1MQt,mL~t,m2 | Ft−1]≤k.

This bound includes actions with zero sampling probability under the zero-quotient convention.

Statement references

Exp4 expert weights and estimated losses: the action distribution, arm reward estimates, and expert weights define the random variables in the statement.

Facts & Assumptions

Given: A contextual bandit instance, a round t, and the Exp4 algorithm with finite action and expert sets.

[F1]

Exp4 with unbiased estimated rewards: Pt,i=∑mQt,mE(t)m,i, Qt,m>0, L~t,m=∑iE(t)m,iL^t,i, and the quotient for L^t,i is zero when Pt,i=0.

[F2]

Contextual bandit protocol with finite expert advice: each advice row is a probability vector, 0≤xt,i≤1, and the learner's action has conditional probability Pt,i.

[F3]

Exp4 with unbiased estimated rewards: Xt=xt,At, and when Pt,i>0, L^t,i=1{At=i}(1−Xt)/Pt,i.

[F4]

Conditional expectation given a sigma algebra: a conditional-expectation version is a measurable integrable function with the same integrals as the original variable on every event in the conditioning sigma-algebra.

[F5]

Expectation of a nonnegative or integrable random variable: expectation is the Lebesgue integral; a measurable finite-range real random variable on a probability space is integrable.

Proof

1.1F1F2F3F5algebra

For the fixed instance, weights and finite estimates are functions of finitely many action histories; they and their squares are measurable with finite ranges, hence integrable. Fix a positive-probability pre-action history. If Pt,i=0, then 0=∑mQt,mE(t)m,i with all terms nonnegative and every Qt,m>0, so E(t)m,i=0 for every expert m. Thus these coordinates contribute zero. For each realized action, at most one arm loss estimate L^t,i is nonzero, and hence L~t,m2=∑i:Pt,i>0E(t)m,i2L^t,i2.

2.1F2F3F4step 1.1algebra

On this history, for Pt,i>0, the square L^t,i2 equals (1−xt,i)2/Pt,i2 when At=i and zero otherwise. By [F2], its conditional mean is therefore Pt,i(1−xt,i)2/Pt,i2=(1−xt,i)2/Pt,i. These finite conditional sums on the history atoms satisfy [F4]; null histories do not affect the identities. Applying this calculation to step 1.1 gives E[L~t,m2∣Ft−1]=∑i:Pt,i>0E(t)m,i2(1−xt,i)2/Pt,i≤∑i:Pt,i>0E(t)m,i(1−xt,i)/Pt,i, because 0≤E(t)m,i(1−xt,i)≤1, whose square is at most itself.

3.1F1F2step 2.1algebra∎

Multiply step 2.1 by Qt,m and sum over experts. For every supported i, ∑mQt,mE(t)m,i=Pt,i, so the result is at most ∑i:Pt,i>0(1−xt,i)≤k. This proves the conditional bound.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Exp4 exponential-weights potential inequality

Statement

For any Exp4 run with learning rate η>0 and any fixed expert m∗∈[M], define L~t,m=1−X~t,m. The following pathwise inequality holds for every realized action sequence:

∑t=1nX~t,m∗−∑t=1n∑m=1MQt,mX~t,m≤log⁡Mη+η2∑t=1n∑m=1MQt,mL~t,m2.

Statement references

Exp4 update: the expert weights are updated by normalized exponentials of the estimated rewards.

Nonpositive-axis exponential inequality: the potential proof applies this inequality to the negative estimated losses.

Facts & Assumptions

Given: Integers n≥1, M>1, a learning rate η>0, and an Exp4 run.

[F1]

Exp4 with unbiased estimated rewards: Q1,m=1/M, Qt+1,m is proportional to Qt,mexp⁡(ηX~t,m), and L~t,m=1−X~t,m≥0 is finite.

[F2]

Second-order exponential upper bound on the nonpositive axis: exp⁡(z)≤1+z+z2/2 for every real z≤0.

[F3]

The exponential addition formula exp⁡(x+y)=exp⁡(x)exp⁡(y): exp⁡(a+b)=exp⁡(a)exp⁡(b) for all real a,b.

[F4]

The natural logarithm as the inverse of the exponential function: log⁡ is defined on positive reals as the inverse of exp⁡.

[F5]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: log⁡ is strictly increasing, log⁡1=0, and log⁡(u/v)=log⁡u−log⁡v for u,v>0.

[F6]

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t: log⁡ is differentiable on (0,∞) with log⁡′(u)=1/u.

[F7]

Mean value theorem on a closed real interval: for a continuous function on a closed interval differentiable in its interior, some interior point realizes its secant slope as the derivative.

Proof

1.1F4F5F6F7caseschoosealgebra

First, for every s>0, log⁡s≤s−1. If s=1 this is equality by [F5]. If s>1, apply [F7] to log⁡ on [1,s]; by [F6] there is c∈(1,s) with log⁡s=(s−1)/c≤s−1. If 0<s<1, apply [F7] on [s,1] to obtain c∈(s,1) with −log⁡s=(1−s)/c≥1−s, again giving log⁡s≤s−1. All intervals lie in (0,∞), where [F6] supplies continuity and differentiability.

2.1F1F2F3F5F8step 1.1algebra

Define W1,m=1, Wt+1,m=Wt,mexp⁡(ηX~t,m), and Zt=∑mWt,m. The update gives Qt,m=Wt,m/Zt. Since X~t,m=1−L~t,m, [F3] and [F2] give Zt+1/Zt=∑mQt,mexp⁡(ηX~t,m)=exp⁡(η)∑mQt,mexp⁡(−ηL~t,m)≤exp⁡(η)St, where St=∑mQt,m(1−ηL~t,m+η2L~t,m2/2)>0. Positivity follows termwise from 1−u+u2/2=((u−1)2+1)/2>0 and Qt,m>0. Applying [F5] and step 1.1 yields log⁡(Zt+1/Zt)≤η+log⁡St≤η+St−1=η∑mQt,mX~t,m+(η2/2)∑mQt,mL~t,m2.

3.1F1F5step 2.1algebra∎

Summing step 2.1 over t and using [F5] telescopes the left side to log⁡(Zn+1/Z1). Since Z1=M and Zn+1=∑mexp⁡(η∑t=1nX~t,m), the fixed expert term gives log⁡(Zn+1/Z1)≥η∑t=1nX~t,m∗−log⁡M. Combining this lower bound with the summed upper bound and dividing by η>0 proves the stated inequality.

TheoremStatement: Literature-sourcedProof: AI-adaptedOpen item page →

Exp4 expected regret against the best fixed expert

Statement

Let n≥1, k>1, and M>1 be integers. For each t∈[n], let xt∈[0,1]k and let E(t) be an M×k matrix with nonnegative entries whose every row sums to one. Assume the reward vectors and advice matrices are deterministic and fixed before learner randomization, each advice matrix is revealed before its action, and the learner observes only the reward of its sampled action. Run Exp4 with

η=2log⁡Mnk,

where log⁡ is the natural logarithm and the square root is the nonnegative real root. Define expected regret against the best fixed expert by

Rn=E[max⁡m∈[M]∑t=1n∑i=1kE(t)m,ixt,i−∑t=1nXt],[M]={1,…,M}.

Then

Rn≤2nklog⁡M.

Statement references

Oblivious contextual bandit protocol and expected regret: the fixed reward/advice scope, action feedback, and reward-regret convention are defined there.

Exp4 algorithm: the action mixture, importance-weighted estimates, and expert update used in the theorem are defined there.

Conditional unbiasedness of expert reward estimates: supported action estimates recover each fixed expert's expected reward, including the zero-support boundary.

Weighted estimated-loss second moment: the expected quadratic term in the potential bound is at most k per round.

Exp4 potential inequality: every fixed expert's estimated-reward advantage is bounded by the log-prior term and estimated-loss second moments.

Facts & Assumptions

Given: The finite oblivious contextual bandit instance, Exp4 with the stated learning rate, and expected regret against its best fixed expert.

[A1]

The reward vectors xt and advice matrices E(t) are fixed before learner randomization, as stated in the Statement; hence a best expert can be chosen deterministically.

[F1]

Contextual bandit protocol with finite expert advice: advice rows are probability vectors, the instance is oblivious, the learner observes Xt=xt,At, and Rn is the expectation of the best-fixed-expert reward minus the learner's cumulative reward.

[F2]

Exp4 with unbiased estimated rewards: Pt,i=∑mQt,mE(t)m,i, Qt is determined before the current draw, and all estimates are finite under the zero-quotient convention.

[F3]

Exp4 expert reward estimates are conditionally unbiased: E[X~t,m∣Ft−1]=∑iE(t)m,ixt,i for every expert, including zero-probability actions.

[F4]

Exp4 weighted estimated loss has conditional second moment at most k: E[∑mQt,mL~t,m2∣Ft−1]≤k.

[F5]

Exp4 exponential-weights potential inequality: for every fixed m∗, the pathwise estimated-reward gap is at most log⁡(M)/η+(η/2)∑t,mQt,mL~t,m2.

[F6]

Static, dynamic, interval, and strongly adaptive regret: expected regret is the expectation of the specified measurable, integrable regret random variable.

[F7]

Conditional expectation given a sigma algebra: a conditional expectation is characterized by its integrals on events in the pre-action sigma-algebra.

[F8]

Expectation of a nonnegative or integrable random variable: expectation of an integrable random variable is its Lebesgue integral.

[F9]

Linearity, monotonicity, and the modulus bound for expectation: finite sums of integrable random variables have the corresponding sum of expectations.

[F10]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: log⁡ is strictly increasing, log⁡1=0, and log⁡(ab)=log⁡a+log⁡b for positive a,b.

[F11]

Nonnegative real square roots exist uniquely: each nonnegative real number has a unique nonnegative square root.

Proof

1.1A1F1F2F3F6F7F8F9algebrachoose

Since [M] is finite and the advice and rewards are fixed, choose a maximizing expert m∗; it is nonrandom. At any positive-probability pre-action history, [F2] makes Qt and Pt fixed. By [F3], the conditional mean of the comparator estimate is its true expert reward. Finite summation and Pt,i=∑mQt,mE(t)m,i give E[∑mQt,mX~t,m∣Ft−1]=∑mQt,m∑iE(t)m,ixt,i=∑iPt,ixt,i=E[Xt∣Ft−1]. All histories and estimates are finite, so summing these conditional identities over histories and rounds gives Rn=E[∑tX~t,m∗−∑t,mQt,mX~t,m]. This uses finite conditional sums, not a general tower-property theorem.

2.1F4F5F7F8F9step 1.1algebra

Apply [F5] pathwise to m∗. Taking expectations, using [F9], and summing the conditional bound [F4] over the finite pre-action histories gives Rn≤log⁡M/η+(η/2)∑t=1nE[∑mQt,mL~t,m2]≤log⁡M/η+ηnk/2.

3.1F10F11step 2.1algebra∎

Since M>1, [F10] gives log⁡M>0. Thus 2log⁡M/(nk)>0, and [F11] supplies the positive learning rate in the Statement. Substituting it into step 2.1 makes both terms equal to nklog⁡M/2, whose sum is 2nklog⁡M.

4 · Examples, counterexamples and false statements

None yet.

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