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Stochastic finite-armed bandits — Examples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: AI-generatedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A two-arm deterministic gap and explore-then-commit

Example

Let m≥1 be an integer, let n≥2m+1, and fix 0<Δ<1. In a two-arm n-round bandit, let every reward-stack entry of arm 1 equal 1 and every reward-stack entry of arm 2 equal 1−Δ. Under explore-then-commit with exploration count m, the algorithm commits to arm 1 and has expected reward regret exactly mΔ. The expected-regret theorem gives the upper bound mΔ+(n−2m)Δe−mΔ2/4 for this instance.

Example references

Finite stochastic bandit model, gaps, and regret: arm means, gaps, and expected reward regret.

Centered 1-sub-Gaussian reward: the mean-centred MGF condition checked for deterministic rewards.

The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x): the exponential factor in the regret bound is positive.

Explore-then-commit for a finite stochastic bandit: the fixed round-robin exploration and empirical-mean commitment rule.

Expected regret of explore-then-commit: the expected-regret bound compared with the exact regret.

Facts & Assumptions

Given: The deterministic reward table and explore-then-commit parameters in the example.

[F1]

Arm means, gaps, and expected reward regret are defined by the stochastic-bandit model (Finite stochastic bandit model, gaps, and regret).

[F2]

Each reward is 1-sub-Gaussian exactly when its mean-centred MGF satisfies the stated bound (Centered 1-sub-Gaussian reward).

[F3]

Explore-then-commit samples each arm m times, commits to a least-index empirical-mean maximizer, then repeats it (Explore-then-commit for a finite stochastic bandit).

[F4]

Under the finite-mean 1-sub-Gaussian assumptions, the theorem bounds expected reward regret by the stated ETC expression (Expected regret of explore-then-commit).

[F5]

The real exponential is positive at every real input (The exponential is positive and satisfies exp⁡(−x)=1/exp⁡(x)).

Verification

1.1givenF1F2algebra

The arm laws are point masses at 1 and 1−Δ, so μ1=1, μ2=1−Δ, and (Δ1,Δ2)=(0,Δ). Each centered reward is identically 0, so its MGF is 1≤eλ2/2 for every real λ and [F4] applies.

2.1givenF1F3step 1.1algebra

The exploration means equal the deterministic rewards, so [F3] commits to arm 1. During exploration it pulls arm 2 exactly m times, each with gap Δ; all other pulls have gap 0. Therefore both the realized reward regret and its expectation equal mΔ.

3.1givenF4F5step 2.1algebra∎

Substituting Δ1=0 and Δ2=Δ into [F4] gives mΔ+(n−2m)Δe−mΔ2/4, which is at least mΔ because n−2m>0, Δ>0, and [F5] makes the exponential factor positive.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

Deterministic zero-reward arms have zero regret

Example

Fix any horizon n≥1 and take two arms with deterministic reward stacks Xs,1=Xs,2=0 for every s∈[n]. Then μ1=μ2=μ∗=0 and Δ1=Δ2=0. Every admissible policy has identically zero pseudo-regret and realized reward regret, so both expected regrets are zero.

Example references

Finite stochastic bandit model, gaps, and regret: the finite-arm means, gaps, pseudo-regret, and realized reward regret used in the example.

Facts & Assumptions

Given: The two-arm deterministic reward table in the example and the definitions of gaps and the two regret quantities.

[F1]

The arm means, gaps, pseudo-regret, and realized reward regret are those in the finite stochastic bandit model (Finite stochastic bandit model, gaps, and regret).

Verification

1.1givenF1algebra

Each arm's law is the point mass at 0, so both finite means equal 0. Hence μ∗=0 and Δ1=Δ2=0.

2.1givenF1step 1.1algebra

For every policy and every round, μ∗−μAt=0 and the observed reward Yt=0. The defining sums therefore give Rˉn=0 and Rnrew=0 pointwise.

3.1givenstep 2.1algebra∎

Taking expectations of the two identically zero random variables shows that expected pseudo-regret and expected reward regret both equal zero.

ExampleConstruction: AI-adaptedVerification: AI-generatedjudge pass (gpt-6-luna)Open item page →

A finite-mean polynomial-tail reward is not sub-Gaussian

Example

For p∈{2,3}, write ζp=∑j=1∞j−p, identifying each positive integer with its canonical real image. Let Ω={1,2,…} with its full power set, put pj=1/(ζ3j3), and define the probability measure Q=∑j=1∞pjδj by its countable weighted sum of Dirac measures. Let X:Ω→R be the inclusion map and let PX be its law. The reward has finite mean EQ[X]=∫Rx PX(dx)=ζ2/ζ3, but ∫Rexp⁡(λx) PX(dx)=+∞ for every λ>0. Thus this finite-mean reward lies outside the centered 1-sub-Gaussian model.

Example references

Finite stochastic bandit model, gaps, and regret: a stochastic arm law is a probability measure on the real Borel space with finite first absolute moment.

The zeta sums at 2 and 3 are finite: the two normalizing sums ζ2 and ζ3 are finite and positive.

Expectation of a countably valued nonnegative random variable is its weighted series: a countably valued nonnegative random variable has expectation equal to its weighted series.

Centered 1-sub-Gaussian reward: a 1-sub-Gaussian reward has a finite centered MGF bounded by exp⁡(λ2/2) for every real λ.

Probability measures and probability spaces: a probability measure has total mass one.

The Dirac set function at a point: the Dirac measure at a point assigns mass one to sets containing that point and zero otherwise.

A Dirac set function is a probability measure: each Dirac measure is a probability measure.

Nonnegative scalar multiples and countable weighted sums of measures: countable weighted sums of measures are defined pointwise by nonnegative extended sums.

Nonnegative scalar multiples and countable weighted sums of measures are measures: a countable nonnegative weighted sum of measures is a measure.

The real exponential function and the number e by a power series: the real exponential is defined by its all-real-convergent power series.

The exponential function is strictly increasing: the real exponential is continuous, hence Borel measurable.

The exponential series converges absolutely for every real argument: the exponential power series converges absolutely at every real argument.

A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum: partial sums of a nonnegative series are bounded above by its extended sum.

The factorial n! and the falling factorial nk‾, defined by recursion in N: the factorial is recursively defined and 4!=24 in the natural numbers.

Integer powers am: natural powers use recursive multiplication and negative powers of positive reals are reciprocals.

The canonical natural ι(n)=n⋅1F of a field: natural numbers in the real field are represented by their canonical images.

Canonical naturals are positive and strictly increasing: the canonical image of each positive natural is positive.

Every complete ordered field is Archimedean: the natural numbers are unbounded in the real field.

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

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

Monotonicity and nonnegative homogeneity of the nonnegative integral: positive finite scalars factor out of nonnegative Lebesgue integrals.

Law or distribution of a random element: the law of a random element is its pushforward probability measure.

The law of a random element is a probability measure: the law of a random element is a probability measure.

Change of variables for expectation: for nonnegative measurable g, E[g(X)]=∫g dPX.

Tor Lattimore and Csaba Szepesvári, Bandit Algorithms: §5.3, PDF pp. 86–88 / printed pp. 77–79; the local discrete construction and divergence proof are supplied here.

Facts & Assumptions

Given: The two convergent sums ζ2,ζ3, the positive-integer sample space and weights in the Example, the inclusion random variable X, and the centered 1-sub-Gaussian definition.

[F1]

The series ζ2 and ζ3 converge and are at least one (The zeta sums at 2 and 3 are finite).

[F3]

If a nonnegative measurable random variable is constant on each cell of a countable measurable partition, its expectation is the series of those values weighted by cell probabilities; a finite series gives integrability (Expectation of a countably valued nonnegative random variable is its weighted series).

[F4]

The inclusion X is a measurable random element, its law is a probability measure, and expectation of a nonnegative measurable function of X equals its integral against that law (Random elements and real random variables, Law or distribution of a random element, The law of a random element is a probability measure, Change of variables for expectation).

[F6]

The natural numbers are unbounded in R (Every complete ordered field is Archimedean).

[F8]

A centered 1-sub-Gaussian reward has finite MGF bounded by exp⁡(λ2/2) for every real λ (Centered 1-sub-Gaussian reward).

[F9]

A positive finite scalar factors out of a nonnegative Lebesgue integral (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[F10]

A stochastic arm law is a probability measure on the real Borel space with finite first absolute moment, and its mean is the integral of the identity function (Finite stochastic bandit model, gaps, and regret).

Verification

1.1givenF1F2F4constructchoosealgebra

By [F1], ζ3 is finite and positive. For pj=1/(ζ3j3), the finite partial sum of the weights is SN/ζ3, where SN=∑j=1Nj−3→ζ3. Given ε>0, choose N with ∣SN−ζ3∣<εζ3; then ∣SN/ζ3−1∣<ε, so ∑jpj=1. By [F2], the Dirac weighted sum Q is a measure and Q(Ω)=∑jpj=1, so Q is a probability measure. The inclusion map from a full-power-set space is measurable, and [F4] makes its law PX a probability measure.

1.2givenF1F3F4F5F6algebrachoose

Fix λ>0. The variable exp⁡(λX) is measurable on the full power set and equals exp⁡(λj) on {j}. The function x↦exp⁡(λx) is Borel measurable by continuity in [F5]. By [F3] and [F4], its expectation and the MGF under PX are ∑j≥1pjexp⁡(λj). By [F5], pjexp⁡(λj)≥cj, where c=λ4/(24ζ3)>0. Thus the Nth partial sum is at least c∑j=1Nj≥cN. For any real M, [F6] gives an integer N>M/c, so these partial sums are unbounded and the nonnegative extended sum is +∞. Hence ∫exp⁡(λx) PX(dx)=+∞.

2.1givenF1F3F4step 1.1choosealgebra

Apply [F3] to the singleton partition Ω=⋃j≥1{j}: EQ[X]=∑j≥1jpj. Its Nth partial sum is TN/ζ3, where TN=∑j=1Nj−2→ζ2. Given ε>0, choose N with ∣TN−ζ2∣<εζ3; then ∣TN/ζ3−ζ2/ζ3∣<ε. Thus EQ[X]=ζ2/ζ3<∞, so [F3] makes X integrable. By [F4], this equals ∫x PX(dx), the finite arm mean in the bandit model.

3.1givenF4F7F8F9F10step 2.1step 1.2algebra∎

Let μ=EQ[X], finite by step 2.1 and equal to the arm-law mean by [F4, F10]. Pointwise, [F7] gives exp⁡(λ(X−μ))=exp⁡(−λμ)exp⁡(λX), with positive finite scalar exp⁡(−λμ). By [F9] and step 1.2, the centered MGF is still +∞. The change-of-variables formula [F4] transfers this centered MGF to PX. This contradicts the finite bound required by [F8], so the arm law is not 1-sub-Gaussian.

Sources