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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.

Adversarial bandits and importance-weighted feedback

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

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

Second-order exponential upper bound on the nonpositive axis

Statement

For every real z≤0,

exp⁡(z)≤1+z+z22.

Facts & Assumptions

Given: A real number z≤0 and the real exponential function.

[F2]

The exponential function is smooth and (exp⁡)′=exp⁡: the real exponential is C∞, with derivative (exp⁡)′=exp⁡.

[F3]

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

[F4]

Mean value theorem on a closed real interval: a function continuous on [a,b] and differentiable on (a,b) has f(b)−f(a)=f′(c)(b−a) for some c∈(a,b).

Proof

1.1givenF1F2F3F4construct

If z=0, the claim is equality. For z<0, fix any a<b≤0 and set g(t)=1+t−exp⁡(t). By [F2], g′(t)=1−exp⁡(t)≥0 for t≤0, because [F3] and [F1] give exp⁡(t)≤exp⁡(0)=1. Applying [F4] to g on [a,b] shows g(b)−g(a)=g′(c)(b−a)≥0, so g is nondecreasing on (−∞,0].

2.1givenF1F2F4step 1.1algebra∎

For the remaining case z<0, step 1.1 shows g(t)≤0 for every t≤0. Set f(t)=1+t+t2/2−exp⁡(t); then f′(t)=g(t). Applying [F4] on [z,0] gives f(0)−f(z)=g(c)(−z)≤0. Since f(0)=0, f(z)≥0, which is the required inequality.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Bandit importance-weighted estimator

Definition

Let k,n≥1 be integers, let [k]={1,…,k}, and let

Δk={p∈[0,1]k:∑i=1kpi=1}.

On a probability space (Ω,F,P), let (Ft)t=0n be the discrete filtration generated by the complete histories through round t, with F0 trivial; a complete pre-action history records past reward vectors, actions, and observed rewards. A reward vector is either fixed in advance or a deterministic function of the complete pre-action history. Thus, at a finite horizon with finitely many arms, there are finitely many possible histories. Before action At is drawn, the current reward vector xt=(xt,1,…,xt,k)∈[0,1]k and sampling distribution Pt=(Pt,1,…,Pt,k)∈Δk are Ft−1-measurable, and Pt,i>0 for every i. The finite-valued measurable action At∈[k] has conditional law P(At=i∣Ft−1)=Pt,i. The learner observes only Xt=xt,At and sets

Yt=1−Xt,yt,i=1−xt,i,Y^t,i=1{At=i}YtPt,i,X^t,i=1−Y^t,i.

The quantity Y^t,i is the importance-weighted loss estimate for arm i, and X^t,i is its corresponding reward estimate. A fixed reward sequence is the oblivious case; allowing xt to depend on the history available before At is drawn gives the non-anticipating past-adaptive case. In either case the current reward vector is chosen before the current action. For finite horizons and finite action histories generated by such a fixed or deterministic history-dependent sequence, all these estimates are finite and integrable.

Definition references

Online decision protocol: the learner selects its round-t action using prior history before round-t feedback.

Adversary and feedback models: bandit feedback reveals only the played reward, and non-anticipating rewards are fixed before the current action.

Random elements and real random variables: a measurable finite-valued action is a random element of [k].

Filtration and filtered probability space: the round-history sigma-algebras are increasing and record information available before each action.

Conditional expectation given a sigma algebra: conditional expectation is characterized by equality of integrals on every event in the conditioning sigma-algebra.

Expectation of a nonnegative or integrable random variable: expectations of the finite integrable estimates are Lebesgue integrals on the probability space.

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

Conditional moments of the bandit importance-weighted estimator

Statement

In the bandit importance-weighted estimator definition, fix a round t and arm i, and write p=Pt,i>0 and y=yt,i. Then

E[Y^t,i∣Ft−1]=yt,i,E[Y^t,i2∣Ft−1]=yt,i2Pt,i.

With conditional variance defined by Var⁡(Z∣Ft−1)=E[(Z−E[Z∣Ft−1])2∣Ft−1], one also has

Var⁡(Y^t,i∣Ft−1)=yt,i2(1−Pt,i)Pt,i.

Statement references

Bandit importance-weighted estimator: the arm probabilities, observed loss, and importance-weighted loss estimate used here.

Facts & Assumptions

Given: A finite arm set [k], a pre-action history at round t, a reward vector xt∈[0,1]k, and a distribution Pt∈Δk with Pt,i>0.

[F1]

Bandit importance-weighted estimator: given the pre-action history, P(At=i∣Ft−1)=Pt,i, Yt=yt,At, and Y^t,i=1{At=i}Yt/Pt,i.

[F2]

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

[F3]

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

[F4]

Filtration and filtered probability space: a filtration consists of nested sub-sigma-algebras of the ambient sigma-algebra. In Bandit importance-weighted estimator, Ft−1 is specifically generated by the complete histories before the round-t action.

Proof

1.1F1F2F3F4

Fix a positive-probability atom H of the finite pre-action history. On H, yt,i and Pt,i are fixed. By [F1], P(H∩{At=i})=Pt,iP(H); the estimator is yt,i/Pt,i on that event and zero on the rest of H. Hence ∫HY^t,i dP=yt,iP(H). The function with value yt,i on each history atom is measurable and integrable, and finite additivity gives the same integral identity on every event in Ft−1. By [F2], it is a conditional-expectation version, proving E[Y^t,i∣Ft−1]=yt,i.

1.2F1F2F3algebra

On the same atom H, Y^t,i 2 is yt,i2/Pt,i2 on H∩{At=i} and zero otherwise. Thus ∫HY^t,i 2 dP=(yt,i2/Pt,i2)Pt,iP(H)=(yt,i2/Pt,i)P(H). Extending over the finite atoms and applying [F2] proves E[Y^t,i 2∣Ft−1]=yt,i2/Pt,i.

2.1F1F2F4step 1.1algebra∎

At a fixed history, the centered estimator has value yt,i(1−Pt,i)/Pt,i with probability Pt,i and value −yt,i with probability 1−Pt,i, by step 1.1. Its conditional variance is therefore Pt,i[yt,i(1−Pt,i)/Pt,i]2+(1−Pt,i)yt,i2=yt,i2(1−Pt,i)/Pt,i.

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

Exp3 with importance-weighted loss feedback

Definition

Let k>1 and n≥1 be integers, and let η>0. Use the finite-arm bandit protocol with rewards in [0,1] and conditional sampling as specified in the bandit importance-weighted estimator definition. Initialize S^0,i=0 for each i∈[k]. For rounds t=1,…,n, perform the following operations:

  1. Set Pt,i=exp⁡(ηS^t−1,i)∑j=1kexp⁡(ηS^t−1,j),i∈[k].
  2. Draw a measurable finite-valued action At with conditional probabilities P(At=i∣Ft−1)=Pt,i, and observe only Xt=xt,At.
  3. Form the loss estimates Y^t,i=1{At=i}(1−Xt)/Pt,i and reward estimates X^t,i=1−Y^t,i for every i∈[k].
  4. Update S^t,i=S^t−1,i+X^t,i for every i∈[k].

The output is (Pt,At,Xt,X^t,Y^t)t=1n and S^n. These operations use exact real arithmetic. The procedure assumes a probability space carrying the stated measurable draws; it makes no probability-space construction claim. Normalizing the weights exp⁡(ηS^t−1,i) is the exponential-weights step used by Hedge.

Definition references

Bandit importance-weighted estimator: the finite-arm reward protocol and loss/reward estimate definitions used at each round.

Hedge: initialize equal weights, normalize exponentials, draw an action, and update after feedback.

The real exponential function and the number e by a power series: the real exponential used in the unnormalized weights.

The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x): every exponential weight is positive.

Random elements and real random variables: each sampled arm is a measurable finite-valued random element.

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

Exp3 expected regret under bandit feedback

Statement

Let n≥1 and k>1 be integers. Let x1,…,xn∈[0,1]k be a reward sequence fixed before the learner's random choices, and run Exp3 with learning rate

η=2log⁡knk.

Here log⁡ is the natural logarithm. For the well-definedness, finite-horizon completion, and two weight identities below, one may instead take any learning rate η>0 and any reward sequence admitted by the algorithm: rewards in [0,1]k fixed in advance or deterministic functions of the complete pre-action history. These procedure properties do not require the tuned learning rate. Assume the run is on a probability space carrying the measurable, finite-valued actions with the conditional probabilities specified by Exp3; no probability-space construction is asserted. Every normalizing denominator is finite and positive, all estimates and cumulative estimates are finite, and the procedure completes its n rounds. For every t∈{0,…,n} and i∈[k],

S^t,i=t−∑s=1tY^s,i,

where the sum is empty when t=0. Consequently, for every t∈{1,…,n},

Pt,i=exp⁡(−η∑s<tY^s,i)∑j=1kexp⁡(−η∑s<tY^s,j),i∈[k].

For the following regret guarantees, use the tuned learning rate displayed above. Write Xt=xt,At and define expected regret against the best fixed arm by

Rn=max⁡i∈[k]∑t=1nxt,i−E[∑t=1nXt].

Then

Rn≤2nklog⁡k.

The same bound holds when each xt is a deterministic function of the complete history available before At is drawn, provided xt∈[0,1]k. In that past-adaptive case define

Rnpast=max⁡i∈[k]E[∑t=1n(xt,i−Xt)],

with the maximum outside the expectation; then Rnpast≤2nklog⁡k. This is an expected fixed-arm regret bound, not a high-probability bound, an expected in-hindsight maximum bound for random reward vectors, or a policy-regret bound for a reward rule that responds to the current action.

Statement references

Exp3 with importance-weighted loss feedback: the sampling distribution, loss estimate, reward estimate, and cumulative update.

Bandit importance-weighted estimator: the bandit feedback and importance-weighted estimate conventions.

Adversary and feedback models: oblivious and non-anticipating timing, distinguished from current-action-reactive timing.

Static, dynamic, interval, and strongly adaptive regret: fixed-comparator and expected-regret conventions.

The natural logarithm as the inverse of the exponential function: the natural logarithm is the inverse of the exponential on positive reals.

Facts & Assumptions

Given: A finite horizon n≥1, k>1 arms, reward vectors in [0,1]k, Exp3 with any η>0 for its procedure properties and with the displayed tuned rate for the regret guarantees, and its probability space of learner draws. In the optional past-adaptive case each reward vector is selected from the pre-action history.

[A1]

Adversary and feedback models: an oblivious adversary fixes the entire reward sequence before learner randomization; a non-anticipating adversary selects the current vector from past history before the current action; a reactive adversary may select it after seeing that action.

[F1]

Exp3 with importance-weighted loss feedback: Pt,i=exp⁡(ηS^t−1,i)/∑jexp⁡(ηS^t−1,j).

[F2]

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

[F3]

Conditional moments of the bandit importance-weighted estimator: E[Y^t,i∣Ft−1]=yt,i and E[Y^t,i2∣Ft−1]=yt,i2/Pt,i.

[F4]

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

[F6]

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

[F7]

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

[F8]

The exponential function is strictly increasing: the real exponential is strictly increasing.

[F9]

The natural logarithm as the inverse of the exponential function: for q>0, exp⁡(log⁡q)=q.

[F11]

Online decision protocol: the learner chooses from prior history before current-round feedback.

[F12]

Filtration and filtered probability space defines nested information sigma-algebras; Exp3 with importance-weighted loss feedback specifies the complete pre-action history recorded by Ft−1.

[F13]

Conditional expectation given a sigma algebra: conditional expectation versions have the same integrals as the original variable over every event in the conditioning sigma-algebra.

[F14]

Linearity, monotonicity, and the modulus bound for expectation: expectation is linear on integrable random variables and preserves almost-sure order.

[F15]

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

[F16]

Static, dynamic, interval, and strongly adaptive regret: expected regret is expectation over the specified randomness when the regret variable is measurable and integrable.

[F17]

The real exponential function and the number e by a power series: exp⁡ is a real-valued function on R and exp⁡(0)=1.

[F18]

Exp3 with importance-weighted loss feedback: Exp3 initializes S^0,i=0.

[F19]

Exp3 with importance-weighted loss feedback: Exp3 updates S^t,i=S^t−1,i+X^t,i.

[F21]

Exp3 with importance-weighted loss feedback: the procedure uses exact real arithmetic and assumes a probability space carrying the stated measurable finite-valued actions with their conditional probabilities.

Proof

1.1givenF1F7F9F10F17F18F19F20F21induction

First fix any positive finite real learning rate η and any reward sequence admitted by the algorithm. Initially S^0,i=0 is finite for every i by [F18]. Suppose all S^t−1,i are finite. Then each exp⁡(ηS^t−1,i) is a positive finite real by [F7] and the real-valued exponential definition [F17]. Its finite sum over [k] is positive and finite, so [F1] gives Pt,i>0 and ∑iPt,i=1. The assumed measurable action draw exists by [F21]; its observed reward is in [0,1]. Thus each Y^t,i=1{At=i}(1−Xt)/Pt,i and X^t,i=1−Y^t,i is finite, and [F19] makes every S^t,i finite. Induction proves these facts through round n. The prescribed loop has the finite index set {1,…,n} and each round consists of the finite operations just checked together with an assumed draw, so it completes exactly n rounds. This argument uses only η>0 and the algorithm's reward range, for either fixed or deterministic past-adaptive rewards. For the rate in the regret clauses, k>1, [F10] and [F20] give log⁡k>log⁡1=0, and [F9] makes log⁡k real; hence that tuned η is also a positive finite real and the same argument applies to it.

1.2F1F6F18F19inductionalgebra

For t=0, the claimed identity S^t,i=t−∑s=1tY^s,i holds by [F18] and the empty-sum convention. If it holds at t−1, then [F19] and X^t,i=1−Y^t,i give

S^t,i=S^t−1,i+X^t,i=t−∑s=1tY^s,i.

Substituting this identity into [F1] and using [F6] factors out the common positive term exp⁡(η(t−1)) from numerator and denominator, giving the stated loss-weight formula for Pt,i.

2.1F1F7step 1.1

For each t, set Wt=∑j=1kexp⁡(ηS^t,j) and Vt=∑j=1kPt,jY^t,j2. By [F1], [F7], and step 1.1, Wt is positive and finite, and Pt is a probability vector.

2.2A1F2F3F11F12F13F14F19step 1.1

By [A1] and [F11], each reward vector is fixed at the pre-action history. At each such history, [F3] and X^t,i=1−Y^t,i give the conditional mean E[X^t,i∣Ft−1]=xt,i by the finite conditional sum. Averaging these sums over histories and using [F14] and [F19] gives E[S^n,i]=E[∑txt,i].

2.3F2step 1.1algebra

On every round and every action outcome, ∑jPt,jX^t,j=1−∑jPt,jY^t,j=1−Yt=Xt, because only the sampled arm contributes to the middle sum.

3.1givenF2F3F12F14F15step 2.1algebra

Fix a pre-action history at round t. If At=a, only coordinate a contributes to Vt, with value yt,a2/Pt,a. Its conditional expectation is therefore ∑a=1kPt,a(yt,a2/Pt,a)=∑a=1kyt,a2≤k, since 0≤yt,a≤1. Averaging over the finite histories and summing the n rounds gives E[∑tVt]≤nk.

3.2F1F2F4F5step 2.1algebra

By [F1]–[F2], Wt/Wt−1=∑j=1kPt,jexp⁡(ηX^t,j)=exp⁡(η)∑j=1kPt,jexp⁡(−ηY^t,j). Each Y^t,j≥0, so [F4] with z=−ηY^t,j gives Wt/Wt−1≤exp⁡(η)(1−η∑jPt,jY^t,j+η2Vt/2). Put ut=−η∑jPt,jY^t,j+η2Vt/2. By [F5] and ∑jPt,jX^t,j=1−∑jPt,jY^t,j, this is at most exp⁡(η+ut)=exp⁡(η∑jPt,jX^t,j+η2Vt/2).

4.1F6F8F9F17F18F19step 3.2algebra

Multiplying step 3.2 over t=1,…,n and using [F6] gives Wn/W0≤exp⁡(η∑t=1n∑jPt,jX^t,j+η2∑t=1nVt/2). Here W0=k by [F17]–[F18], and Wn≥exp⁡(ηS^n,i) for every i by [F19]. Since k=exp⁡(log⁡k) by [F9], strict increase [F8] and η>0 yield S^n,i−∑t=1n∑jPt,jX^t,j≤log⁡k/η+(η/2)∑t=1nVt for every arm i.

5.1F14F15F16step 4.1step 3.1step 2.2step 2.3algebra

For the oblivious case set Rn,i=∑txt,i−E[∑tXt]; for the past-adaptive case set Rn,i=E[∑t(xt,i−Xt)]. By steps 2.2–2.3, both equal E[S^n,i−∑t∑jPt,jX^t,j]. Taking expectations in step 4.1, using [F14]–[F16], and applying step 3.1 gives Rn,i≤log⁡k/η+ηnk/2 for every i.

6.1givenstep 1.1step 5.1algebra∎

Using η>0 from step 1.1, substitution gives log⁡k/η=ηnk/2=nklog⁡k/2. The per-arm bound in step 5.1 therefore equals 2nklog⁡k; taking the finite maximum over i proves both regret bounds in the Statement.

4 · Examples, counterexamples and false statements

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