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Imitation learning and DAgger

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

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

Follow-the-Leader regret for bounded strongly convex losses

Statement

Let d,N≥1 be integers, let K⊆Rd be nonempty and convex, and define ∥x−y∥2=(∑r=1d(xr−yr)2)1/2. Let G≥0 and μ>0. For each i=1,…,N, let Li:K→[0,1] be G-Lipschitz, meaning ∣Li(x)−Li(y)∣≤G∥x−y∥2, and μ-strongly convex, meaning

Li((1−λ)x+λy)≤(1−λ)Li(x)+λLi(y)−μ2λ(1−λ)∥x−y∥22

for all x,y∈K and λ∈[0,1]. For n=1,…,N, assume Fn:=∑i=1nLi attains a minimum on K. Let θ1∈K and, for n=1,…,N, let θn+1 be the unique minimizer of Fn. Define H0=0 and Hn=∑j=1n1/j for n≥1. Then

∑i=1NLi(θi)−min⁡θ∈K∑i=1NLi(θ)≤1+4G2μHN−1.

The minimizers are unique because Fn is nμ-strongly convex. The first policy θ1 may be any supplied point of K.

Statement references

Facts & Assumptions

Given: K, L1,…,LN, G, μ, the attained cumulative minimizers, and θ1 as in the Statement.

[A1]

Each Li takes values in [0,1], is G-Lipschitz and μ-strongly convex on the convex set K, and each Fn has an attained minimum.

Proof

technique · be-the-leader telescoping and a stability bound for consecutive minimizers
1.1givenA1algebra

Each Fn is nμ-strongly convex by summing the defining inequalities, so it has at most one minimizer; the assumed attainment therefore makes every θn+1 well-defined and unique. For any comparator v∈K, expand Li(θi+1)=Fi(θi+1)−Fi−1(θi+1) and telescope to obtain ∑i=1NLi(θi+1)=FN(θN+1)+∑i=1N−1(Fi(θi+1)−Fi(θi+2))≤FN(v), because θi+1 minimizes Fi and θN+1 minimizes FN.

2.1A1step 1.1algebra

For i≥2, apply the midpoint strong-convexity inequality for Fi−1 to θi and θi+1. Since θi minimizes Fi−1, this gives (i−1)μ4∥θi+1−θi∥22≤Fi−1(θi+1)−Fi−1(θi). Optimality of θi+1 for Fi=Fi−1+Li bounds the right side by Li(θi)−Li(θi+1), which is at most G∥θi+1−θi∥2 by Lipschitz continuity. If the distance is zero the loss difference is zero; otherwise divide by it to obtain Li(θi)−Li(θi+1)≤4G2/((i−1)μ).

3.1A1step 1.1step 2.1algebra∎

The first difference satisfies L1(θ1)−L1(θ2)≤1 because both values lie in [0,1]. Summing this with the bounds from step 2.1 yields ∑i=1N(Li(θi)−Li(θi+1))≤1+4G2μHN−1. For a minimizer v of FN, step 1.1 gives ∑iLi(θi+1)≤FN(v); adding that inequality proves the Statement.

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

Finite-horizon MDP policy occupancy is a probability distribution

Statement

Let a finite Markov decision process have a nonempty finite state set S and finite nonempty action sets A(s), with state transition probabilities P(s′∣s,a) obtained by summing its joint reward-transition law over rewards. Let T≥1 be an integer, let μ0:S→[0,1] satisfy ∑s∈Sμ0(s)=1, and let π(a∣s) be a stationary Markov policy. Define recursively

μt+1(s′)=∑s∈S∑a∈A(s)μt(s)π(a∣s)P(s′∣s,a),0≤t<T.

Then each μt is a probability vector on S, and the finite-horizon average occupancy

dπ(s)=1T∑t=0T−1μt(s)

is also a probability vector. This construction uses only finite sums and does not require an infinite-path probability extension.

Statement references

Finite Markov decision process with joint reward-transition law: the finite state/action sets, row-normalized joint reward-transition law, derived P(s′∣s,a), and stationary Markov policy.

The sum ∑i∈Sai over a finite index set, and its product form: finite-index sums are well-defined and independent of enumeration.

Laws of finite sums and finite products: finite sums preserve nonnegativity, add termwise, and commute with scalar multiplication.

∑i∈S∑j∈Taij=∑(i,j)∈S×Taij=∑j∈T∑i∈Saij for finite index sets S and T: finite sums over two finite index sets may be interchanged.

The principle of mathematical induction: a property proved at 0 and preserved by successors holds for every natural index.

Sutton and Barto, Reinforcement Learning: An Introduction: §3.6, printed pp. 67–69.

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: §2, PDF p. 1.

Facts & Assumptions

Given: A nonempty finite MDP state set S, finite nonempty action sets A(s), a normalized initial vector μ0, a stationary Markov policy π, and an integer horizon T≥1.

[A1]

μ0 is nonnegative with finite sum one, and T is a positive integer.

[L1]

At a successor step t<T, assume μt is nonnegative and ∑sμt(s)=1.

[F1]

Finite Markov decision process with joint reward-transition law: the joint masses p(s′,ρ∣s,a) are nonnegative and sum to one over (s′,ρ); P(s′∣s,a)=∑ρp(s′,ρ∣s,a); and each policy row π(⋅∣s) is a probability vector.

[F2]

The sum ∑i∈Sai over a finite index set, and its product form: sums over finite index sets are the enumeration-independent finite sums.

[F4]

Laws of finite sums and finite products: finite sums preserve nonnegative inequalities, add termwise, and satisfy ∑icai=c∑iai.

[F5]

The principle of mathematical induction: a property of t∈N holds for every t if it holds at zero and the successor step preserves it.

Proof

1.1A1

At t=0, μ0 is nonnegative and has sum one by [A1].

1.2L1F1F2F3F4algebra

Fix t<T and assume [L1]. Every term in the recursion for μt+1(s′) is nonnegative by [F1]. For each admissible (s,a), [F1] and finite-sum interchange [F3] give ∑s′∈SP(s′∣s,a)=∑s′∈S∑ρ∈Rp(s′,ρ∣s,a)=1. Interchange the finite state sums and factor constants using [F2]–[F4]: ∑s′∈Sμt+1(s′)=∑s′∈S∑s∈S∑a∈A(s)μt(s)π(a∣s)P(s′∣s,a)=∑s∈Sμt(s)∑a∈A(s)π(a∣s)∑s′∈SP(s′∣s,a)=∑s∈Sμt(s)∑a∈A(s)π(a∣s)=∑s∈Sμt(s)=1. The calculation and nonnegativity prove that μt+1 is a probability vector.

2.1F5step 1.1step 1.2

Step 1.1 proves the base case and step 1.2 proves the successor case, so [F5] gives nonnegative normalized vectors μt for every 0≤t≤T.

3.1A1F2F4step 2.1algebra∎

Every term in dπ(s) is nonnegative. Summing over s and using the finite-sum scaling law [F4] gives ∑sdπ(s)=T−1∑t=0T−1∑sμt(s)=T−1∑t=0T−11=1, because there are exactly T indices and T≥1. Hence dπ is a probability vector.

DefinitionDefinition: Literature-sourcedProof: Not applicableOpen item page →

Finite-horizon imitation learning and expert-surrogate loss

Definition

Fix a finite Markov decision process with nonempty finite state set S and finite nonempty action set A(s) at each state, a positive integer horizon T, and an initial probability vector μ0 on S. A policy is a stationary Markov policy π(a∣s); a deterministic expert is a function π∗:S→⋃s∈SA(s) with π∗(s)∈A(s). For a policy π, let μtπ and dπ be the finite-horizon state and average-occupancy probability vectors defined by the finite recursion in the preceding lemma.

Let ct(s,a)∈[0,1] be the task cost at time t for 0≤t<T. Define the total expected task cost by

J(π)=∑t=0T−1∑s∈Sμtπ(s)∑a∈A(s)π(a∣s)ct(s,a).

Define the expert cost-to-go recursively by VT∗(s)=0 and, for 0≤t<T,

Qt∗(s,a)=ct(s,a)+∑s′∈SP(s′∣s,a)Vt+1∗(s′),Vt∗(s)=Qt∗(s,π∗(s)).

For a learner policy πθ indexed by a parameter θ in a nonempty set K, define its expert-action disagreement probability at s by e(s,θ)=∑a∈A(s):a≠π∗(s)πθ(a∣s). An expert surrogate loss is a function ℓ:S×K→[0,1] satisfying ℓ(s,θ)≥e(s,θ) for every s∈S and θ∈K. The average surrogate loss under a policy's own state occupancy is Es∼dπθ[ℓ(s,θ)]=∑sdπθ(s)ℓ(s,θ).

Behavior cloning fits a learner policy using expert labels on states sampled from dπ∗. DAgger (Dataset Aggregation) is the interactive alternative: at iteration i, it rolls out a mixture πi=βiπ∗+(1−βi)π^i, where π^i=πθi is the current learner with parameter θi∈K and βi∈[0,1] is the expert-use probability at each decision, queries the expert on the visited states, aggregates those labeled states with prior data, and updates the learner. For a positive integer N of iterations, in the infinite-sample analysis define Li(θ)=∑sdπi(s)ℓ(s,θ). When the minimum is attained, define ϵN=min⁡θ∈KN−1∑i=1NLi(θ) and the learner's average regret γN=N−1∑i=1NLi(θi)−ϵN. Fixed demonstrations do not in general provide expert labels for every learner-visited state; if they cover those states, their labels can be reused without new expert interaction.

Definition references

Finite Markov decision process with joint reward-transition law: the finite state/action sets, joint reward-transition masses, derived state transition probabilities, and stationary Markov policies.

Finite-horizon MDP policy occupancy is a probability distribution: finite recursion gives probability vectors μtπ and their normalized average occupancy dπ without an infinite-path construction.

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

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: §2, PDF pp. 1–2.

Bagnell, Boots, and Choudhury, Modern Adaptive Control and Reinforcement Learning, Chapter 6: PDF pp. 1–5 / printed pp. 53–58.

Duisterhof, Truong, and Agrawal, Algorithmic Foundations of Interactive Learning, Lecture 9: PDF pp. 50–53 / note pages 46–47.

AlgorithmAlgorithm: Literature-sourcedProof: Not applicableOpen item page →

Dataset Aggregation (DAgger)

Definition

Let a finite-horizon imitation-learning problem have the finite MDP, initial state distribution, positive integer horizon T, deterministic expert π∗, learner class {πθ:θ∈K}, and expert surrogate loss ℓ(s,θ) defined previously. Let N≥1 and m≥1 be integers giving the number of training rounds and trajectories per round, let β1,…,βN∈[0,1] be expert-use probabilities, supply an initial parameter θ1∈K, and supply a deterministic selector for attained minimizers of the cumulative losses. Require that each cumulative training-loss function below attains a minimum on K; when the minimizer is unique the selector returns it.

Initialize the aggregated training dataset as the empty multiset D0=∅.

For each round i=1,…,N:

  1. Set π^i=πθi. For the rollout policy, at each decision time of every trajectory, independently draw an expert-use coin with success probability βi. At current state s, if the coin succeeds, execute π∗(s); otherwise sample an action from π^i(⋅∣s) and execute it. This per-decision stochastic policy is πi=βiπ∗+(1−βi)π^i.

  2. Run m trajectories of length T from the specified initial distribution using πi. At every visited state Sijt, for 1≤j≤m and 0≤t<T, query and record the expert label π∗(Sijt). Append these labeled pairs to Di−1, with multiplicity, to form Di. The expert label is recorded for training and is not executed unless the mixture in step 1 selected the expert.

  3. Define the empirical round loss on every candidate parameter by

    L^i(θ)=1mT∑j=1m∑t=0T−1ℓ(Sijt,θ).

  4. Update by Follow-the-Leader on the cumulative empirical loss: set θi+1 to the supplied selector's choice from arg⁡min⁡θ∈K∑r=1iL^r(θ).

The finite-sample output is the sequence π^1,…,π^N and the aggregated labeled multiset DN. The population-loss idealization replaces each empirical L^i by Li(θ)=∑sdπi(s)ℓ(s,θ); it is used only for the infinite-sample theorem. The procedure requires an interactive expert capable of labeling learner-visited states.

Definition references

Finite-horizon imitation learning and expert-surrogate loss: specifies the finite MDP, normalized state occupancy, deterministic expert, learner class, task/surrogate losses, and DAgger round losses.

Online decision protocol: the learner selects its current policy from earlier feedback before collecting current-round trajectories and labels.

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: Algorithm 3.1, §3, PDF p. 3.

Bagnell, Boots, and Choudhury, Modern Adaptive Control and Reinforcement Learning, Chapter 6: PDF pp. 4–6 / printed pp. 57–59.

Duisterhof, Truong, and Agrawal, Algorithmic Foundations of Interactive Learning, Lecture 9: PDF pp. 51–53 / note pages 47–48.

LemmaStatement: AI-adaptedProof: AI-generatedOpen item page →

Expert mixing changes finite-horizon state occupancy by at most the mixing mass

Statement

Let a finite-horizon MDP have positive integer horizon T, and let π∗ and π^ be stationary Markov policies. For a stationary policy π, define its state transition matrix by Pπ(s,s′)=∑a∈A(s)π(a∣s)P(s′∣s,a); for a state vector μ, define (μPπ)(s′)=∑s∈Sμ(s)Pπ(s,s′). Fix β∈[0,1] and let πβ(a∣s)=βπ∗(a∣s)+(1−β)π^(a∣s), which is the per-decision expert/learner mixture. For every function g:S→[0,1], its finite-horizon average occupancies satisfy

∣∑s∈Sdπβ(s)g(s)−∑s∈Sdπ^(s)g(s)∣≤1−(1−β)T≤min⁡{1,Tβ}.

Here dπ=T−1∑t=0T−1μtπ is the normalized average of the state marginals from the common initial distribution.

Statement references

Finite-horizon imitation learning and expert-surrogate loss: defines the finite-horizon state recursion, normalized average occupancy, and the DAgger per-decision policy mixture.

The principle of mathematical induction: finite recursion and Bernoulli inequality use natural-number induction.

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: Lemma 4.1 and proof, §4.2, PDF p. 4.

Duisterhof, Truong, and Agrawal, Algorithmic Foundations of Interactive Learning, Lecture 9: PDF pp. 50–53 / note pages 46–47.

Facts & Assumptions

Given: The finite MDP, initial distribution, horizon, policies, and mixture parameter in the Statement.

[F1]

Finite-horizon imitation learning and expert-surrogate loss: for each policy π, μt+1π=μtπPπ and dπ=T−1∑t<Tμtπ.

[F2]

Finite-horizon MDP policy occupancy is a probability distribution: every μtπ and dπ is a probability vector, with nonnegative coordinates.

[F3]

The principle of mathematical induction: a property holding at zero and preserved by successors holds for all natural indices.

Proof

1.1F1F2F3algebraconstruct

Put q=1−β. The transition matrix of the mixed policy is Pπβ=qPπ^+βPπ∗ by finite-sum distributivity. For each t≤T, we show μtπβ=qtμtπ^+rt, where rt has nonnegative coordinates and total mass 1−qt. At t=0, take r0=0. If the formula holds at t<T, the recursion gives μt+1πβ=qt+1μt+1π^+qtβμtπ^Pπ∗+rtPπβ⏟rt+1. The two terms defining rt+1 have nonnegative coordinates, and their total masses are qtβ and 1−qt, so rt+1 has mass 1−qt+1. Induction proves the decomposition for every t≤T.

2.1F1F2step 1.1algebraconstruct

Since 0≤q≤1, for each t<T one has qt≥qT. Thus rt′:=rt+(qt−qT)μtπ^ is nonnegative and has mass 1−qT. Averaging over t=0,…,T−1 gives dπβ=qTdπ^+r′, where r′=T−1∑t<Trt′ has mass 1−qT. If 1−qT>0, normalize r′ to a probability vector ν; then for g:S→[0,1] the difference of expectations is (1−qT)(∑sν(s)g(s)−∑sdπ^(s)g(s)), whose absolute value is at most 1−qT. If 1−qT=0, the occupancies are equal and the same bound is zero.

3.1F3step 2.1algebra∎

The Bernoulli inequality qT=(1−β)T≥1−Tβ follows by induction: it is equality at T=0, and multiplying qn≥1−nβ by q≥0 gives qn+1≥(1−nβ)(1−β)=1−(n+1)β+nβ2≥1−(n+1)β. Also 1−qT≤1, so 1−qT≤min⁡{1,Tβ}. Combining this with step 2.1 proves the Statement.

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

Expert-surrogate loss bounds finite-horizon task-cost increase

Statement

Let a finite-horizon MDP, horizon T≥1, initial distribution, deterministic expert π∗, learner policy πθ, task cost J, expert cost-to-go Qt∗, expert value Vt∗(s)=Qt∗(s,π∗(s)), occupancy dπθ, and surrogate ℓ(s,θ) be as defined previously. Suppose u≥0 and

Qt∗(s,a)−Vt∗(s)≤u

for every time 0≤t<T, state s∈S, and admissible action a∈A(s), and suppose ℓ(s,θ)≥∑a≠π∗(s)πθ(a∣s) for every state. Then

J(πθ)−J(π∗)≤uT∑s∈Sdπθ(s)ℓ(s,θ).

The one-step cost-to-go bound is required at all states and times, including states not reached by the expert.

Statement references

Finite-horizon imitation learning and expert-surrogate loss: defines finite-horizon task cost, expert cost-to-go recursion, normalized state occupancy, and the surrogate's expert-action-disagreement domination condition.

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: Theorem 2.2 and its complete hybrid-policy proof, §2.2, PDF pp. 2–3.

Duisterhof, Truong, and Agrawal, Algorithmic Foundations of Interactive Learning, Lecture 9: equations (9.72)–(9.74), PDF p. 54 / note page 48; used for terminology only.

Facts & Assumptions

Given: The finite MDP and policy definitions in the Statement, a learner parameter θ, u≥0, the uniform one-step expert cost-to-go upper bound, and the statewise surrogate disagreement upper bound.

[F1]

Finite-horizon imitation learning and expert-surrogate loss: J(π) is the finite sum of expected per-time costs, dπ=T−1∑t<Tμtπ, and Qt∗(s,a)=ct(s,a)+∑s′P(s′∣s,a)Vt+1∗(s′) with Vt∗(s)=Qt∗(s,π∗(s)).

[F2]

For any sequence of Markov policies (ρt)t=0T−1, define μ0ρ=μ0, recursively set μt+1ρ(s′)=∑sμtρ(s)∑a∈A(s)ρt(a∣s)P(s′∣s,a), and define J(ρ)=∑t=0T−1∑sμtρ(s)∑a∈A(s)ρt(a∣s)ct(s,a). For a stationary policy this agrees with J in [F1].

[A1]

Qt∗(s,a)−Vt∗(s)≤u for every t,s,a.

[A2]

ℓ(s,θ)≥∑a≠π∗(s)πθ(a∣s) for every s.

Proof

1.1givenF1F2constructalgebra

For k=0,…,T, let ρt(k)=πθ for t<k and ρt(k)=π∗ for t≥k. By [F2], these time-dependent policies have finite-horizon costs J(ρ(k)), with J(ρ(0))=J(π∗) and J(ρ(T))=J(πθ). Telescoping gives J(πθ)−J(π∗)=∑k=0T−1(J(ρ(k+1))−J(ρ(k))).

2.1F1F2step 1.1algebra

The two hybrid policies in the kth difference have the same state distribution μkπθ at time k, because both use πθ for the first k decisions. From then on, one takes the learner action at time k and the other takes the expert action, and both follow the expert afterward. The cost recursion [F2] and the definition of Qk∗ therefore give J(ρ(k+1))−J(ρ(k))=∑s∈Sμkπθ(s)∑a∈A(s)πθ(a∣s)(Qk∗(s,a)−Vk∗(s)).

3.1A1A2step 2.1algebra

For a=π∗(s) the cost-to-go difference in step 2.1 is zero by the definition of Vk∗. For every other action it is at most u by [A1]. Therefore each inner sum in step 2.1 is at most u∑a≠π∗(s)πθ(a∣s)≤uℓ(s,θ) by [A2].

4.1F1step 1.1step 3.1algebra∎

Sum step 3.1 over k=0,…,T−1 and use dπθ=T−1∑k<Tμkπθ from [F1]. The result is J(πθ)−J(π∗)≤u∑k<T∑sμkπθ(s)ℓ(s,θ)=uT∑sdπθ(s)ℓ(s,θ), as claimed.

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

DAgger no-regret reduction under strongly convex surrogate losses

Statement

Consider the finite-horizon imitation-learning setting with deterministic expert π∗, positive integer horizon T, and expert surrogate ℓ(s,θ)∈[0,1] defined previously. Let d≥1 be an integer and let K⊆Rd be nonempty and convex, with ∥x−y∥2=(∑r=1d(xr−yr)2)1/2. Assume there are constants μ>0 and G≥0 such that, uniformly in s, ℓ(s,⋅) is G-Lipschitz, meaning ∣ℓ(s,x)−ℓ(s,y)∣≤G∥x−y∥2, and is μ-strongly convex on K; that is,

ℓ(s,(1−λ)x+λy)≤(1−λ)ℓ(s,x)+λℓ(s,y)−μ2λ(1−λ)∥x−y∥22

for all x,y∈K and λ∈[0,1]. Run the population-loss version of DAgger for a positive integer N rounds with exact FTL updates, and assume each cumulative population loss attains its minimum on K. Let βi∈[0,1] satisfy β1≤1 and βi≤(1−α)i−1 for 2≤i≤N, where α∈(0,1] is fixed independently of T and N. Suppose u≥0 satisfies Qt∗(s,a)−Vt∗(s)≤u for every time t, state s, and admissible action a.

Let Li(θ)=∑sdπi(s)ℓ(s,θ) be the true round-i loss under the mixed rollout policy, and let ϵN=min⁡θ∈KN−1∑i=1NLi(θ), with the minimum attained. Define H0=0, Hn=∑j=1n1/j, let qT be the least nonnegative integer with T≤2qT, and let kα be the least positive integer with kαα≥1. Then some learned policy π^∈{π^1,…,π^N} satisfies

J(π^)≤J(π∗)+uT[ϵN+1+4G2μHN−1+kα(qT+2)N].

In particular, the additive term beyond uTϵN is at most 1 whenever N≥uT(1+4G2μHN−1+kα(qT+2)). For fixed α,μ,G and a valid scale u≥1, the sufficient iteration count is O~(uT) because HN−1 and qT grow logarithmically. If a valid one-step bound has u<1, it may be enlarged to 1 for this iteration-count corollary; the exact inequality above remains valid with the original u. This is an infinite-sample statement: expert labels must be obtained by interaction on states visited under πi.

Statement references

Finite-horizon imitation learning and expert-surrogate loss: defines task cost, finite-horizon state occupancy, surrogate disagreement domination, mixed rollout losses, and ϵN.

Dataset Aggregation (DAgger): population-loss FTL updates, expert/learner mixture, and interactive expert labels.

Follow-the-Leader regret for bounded strongly convex losses: exact cumulative FTL regret bound 1+(4G2/μ)HN−1.

Expert mixing changes finite-horizon state occupancy by at most the mixing mass: a bounded state loss changes by at most min⁡(1,Tβi) between the mixed and learner occupancies.

Expert-surrogate loss bounds finite-horizon task-cost increase: task-cost increase is at most uT times the learner's own-occupancy surrogate loss.

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: Theorems 3.1–3.2 and §4.2, PDF pp. 3–4.

Bagnell, Boots, and Choudhury, Modern Adaptive Control and Reinforcement Learning, Chapter 6: PDF pp. 4–7 / printed pp. 57–60.

Duisterhof, Truong, and Agrawal, Algorithmic Foundations of Interactive Learning, Lecture 9: PDF pp. 51–54 / note pages 47–48.

Facts & Assumptions

Given: The finite-horizon MDP and expert definitions, uniform bounded strongly convex Lipschitz surrogate, exact population FTL updates, geometric expert-use schedule, attained minimum defining ϵN, and one-step expert cost-to-go bound in the Statement.

[F1]

Finite-horizon imitation learning and expert-surrogate loss: dπ is the normalized average state occupancy; Li(θ)=∑sdπi(s)ℓ(s,θ); the surrogate upper-bounds action disagreement; and J(π) is finite-horizon expected task cost.

[F2]

Dataset Aggregation (DAgger): at round i, DAgger rolls out πi=βiπ∗+(1−βi)π^i, obtains expert labels on visited states, and sets θi+1 to an attained minimizer of the cumulative losses.

[F3]

Follow-the-Leader regret for bounded strongly convex losses: for the stated bounded μ-strongly convex G-Lipschitz losses, ∑iLi(θi)−NϵN≤1+(4G2/μ)HN−1.

[F4]

Expert mixing changes finite-horizon state occupancy by at most the mixing mass: for every g:S→[0,1], the absolute difference between its mixed-policy and learner-policy occupancy expectations is at most min⁡(1,Tβi); in particular, either expectation is at most the other plus this error.

[F5]

Expert-surrogate loss bounds finite-horizon task-cost increase: J(π^i)−J(π∗)≤uT∑sdπ^i(s)ℓ(s,θi).

[A1]

For every state s, ℓ(s,⋅) is G-Lipschitz and μ-strongly convex on the nonempty convex parameter set K, uniformly in s, with 0≤ℓ≤1.

[A2]

Every cumulative population loss attains its minimum on K, and the minimum defining ϵN is attained.

[A3]

The expert-use schedule satisfies 0≤β1≤1 and 0≤βi≤(1−α)i−1 for 2≤i≤N, with fixed 0<α≤1.

[A4]

The expert cost-to-go gap satisfies Qt∗(s,a)−Vt∗(s)≤u for every time, state, and admissible action, with u≥0.

Proof

1.1A1A2F1F2F3algebra

Since Li is the finite probability-weighted average of the statewise losses, averaging the uniform strong-convexity inequality and the G-Lipschitz inequality in [A1] over dπi shows that each Li is μ-strongly convex, G-Lipschitz, and valued in [0,1]. The exact FTL updates [F2], attained minima [A2], and bound [F3] therefore give 1N∑i=1NLi(θi)≤ϵN+1+4G2μHN−1N.

1.2A3F2algebra

Let q=1−α and k=kα. Since (1−α)(1+α)=1−α2≤1, q≤(1+α)−1. For every integer n≥1, (1+α)n≥1+nα: it is equality at n=1, and multiplying the inductive inequality by 1+α>0 gives (1+α)n+1≥1+(n+1)α+nα2≥1+(n+1)α. Thus (1+α)k≥1+kα≥2 and qk≤1/2. Partition indices i into blocks i=jk+r+1 with j≥0 and 0≤r<k. On block j, [A3] gives min⁡(1,Tβi)≤min⁡(1,Tqjk)≤min⁡(1,T2−j). Hence ∑i=1Nmin⁡(1,Tβi)≤k∑j=0∞min⁡(1,T2−j)≤k(qT+2), because the first qT+1 terms are at most 1 and the remaining geometric tail is T∑j=qT+1∞2−j=T2−qT≤1.

1.3F3algebraconstruct

For every positive integer n, let qn be the least nonnegative integer with n≤2qn. If N=1, then HN−1=0≤1+qN. If N≥2, group HN−1 into the initial term 1 and blocks 2j−1<r≤2j, 1≤j≤qN. Each block has at most 2j−1 terms, each at most 2−(j−1), so HN−1≤1+qN. These dyadic lengths grow logarithmically in their arguments.

2.1A2F1F2F4step 1.1algebra

For each round, apply the absolute-value estimate [F4] to g(s)=ℓ(s,θi)∈[0,1], using its direction ∑sdπ^i(s)g(s)≤Li(θi)+min⁡(1,Tβi). Averaging over i and using step 1.1 yields 1N∑i=1N∑sdπ^i(s)ℓ(s,θi)≤ϵN+1+4G2μHN−1N+1N∑i=1Nmin⁡(1,Tβi).

3.1F2step 1.2step 2.1algebra

By step 1.2 and step 2.1, the average own-occupancy surrogate loss of the finite sequence π^1,…,π^N is at most ϵN+1+4G2μHN−1+kα(qT+2)N. At least one member π^i has own-occupancy surrogate loss no larger than this average, because otherwise every term in the finite average would be strictly larger than the average.

4.1A4F5step 3.1algebra

Apply the one-step assumption [A4] through [F5] to that index i and substitute the bound from step 3.1. This gives the stated task-cost inequality.

5.1A1A3A4step 1.3step 4.1algebra∎

If N≥uT(1+4G2μHN−1+kα(qT+2)), then the additive term in the Statement is at most 1. To verify the iteration order, set uˉ=max⁡(1,u) and B=uˉT≥1; let b=⌈B⌉ and let Q be the least integer with b+T+2≤2Q. Choose an integer C≥1 and set N=⌈CbQ⌉. Since b≤2Q and Q≤2Q, N≤(C+1)22Q, so the least qN with N≤2qN satisfies qN≤2Q+qC+1. By step 1.3 and qT≤Q, the bracket is at most 1+4G2μ(2Q+qC+1+1)+kα(Q+2). Set c=4G2/μ, a=2c+kα, b0=c, and b1=1+c+2kα; the bracket bound equals aQ+b0qC+1+b1. Choose an integer r≥2 so large that r(r−1)/2≥a+b0(r+1)+b1; such an r exists because the left side is quadratic with positive leading coefficient and the right side is affine in r. The binomial theorem gives C:=2r≥r(r−1)/2, while C+1≤2r+1 gives qC+1≤r+1. Hence C≥a+b0qC+1+b1. Since Q≥1 and a,b0,b1≥0, CQ≥(a+b0qC+1+b1)Q≥aQ+b0qC+1+b1; as b≥B, this gives N≥CbQ≥B(aQ+b0qC+1+b1). Thus N≥uˉT times the bracket, which also implies the required inequality with u. Moreover Q=O(log⁡(B+2)), so N=O(Blog⁡(B+2))=O~(uˉT). When u≥1, this is O~(uT); if u<1, replacing it by uˉ=1 is a valid upper-bound normalization for the iteration count. The estimate retains the hindsight optimum ϵN and assumes no realizability condition π∗∈Π.

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

Finite-sample DAgger surrogate-loss guarantee

Statement

Assume the finite-horizon imitation-learning setting with deterministic expert π∗, positive integer horizon T, expert surrogate ℓ(s,θ)∈[0,1], learner policies πθ, and nonempty convex parameter set K⊆Rd defined previously. Let d≥1 be an integer and use ∥x−y∥2=(∑r=1d(xr−yr)2)1/2. Suppose for every state s, ℓ(s,⋅) is G-Lipschitz and μ-strongly convex on K, with constants G≥0 and μ>0, where Lipschitz means ∣ℓ(s,x)−ℓ(s,y)∣≤G∥x−y∥2 and strong convexity means ℓ(s,(1−λ)x+λy)≤(1−λ)ℓ(s,x)+λℓ(s,y)−μ2λ(1−λ)∥x−y∥22 for x,y∈K and λ∈[0,1]. Assume also that θ↦πθ(a∣s) is Borel measurable for every state-action pair.

Run finite-trajectory DAgger for positive integers N,m, with per-decision expert-use probabilities βi∈[0,1] satisfying β1≤1 and βi≤(1−α)i−1 for 2≤i≤N, where α∈(0,1] is fixed. Let π^i=πθi and πi=βiπ∗+(1−βi)π^i. In round i, let Sijt, 1≤j≤m, 0≤t<T, be the states of trajectory j, and define ℓˉij(θ)=T−1∑t=0T−1ℓ(Sijt,θ), L^i(θ)=m−1∑j=1mℓˉij(θ), and ϵ^N=min⁡θ∈KN−1∑i=1NL^i(θ). Assume this minimum and every cumulative empirical-loss minimum are attained, and DAgger uses exact Follow-the-Leader updates.

Assume the samples are observed in a filtered probability space with nested sigma-algebras Fi,j, where F1,0=F0, Fi,0=Fi−1,m for i>1, and Fi,j−1⊆Fi,j. Suppose θi is Fi,0-measurable, each trajectory state Sijt is Fi,j-measurable, and the conditional rollout law satisfies E[ℓˉij(θi)∣Fi,j−1]=Li(θi) almost surely, where Li(θ)=∑sdπi(s)ℓ(s,θ) is the true round-i loss under the mixed rollout policy. Assume the Axiom of Choice for the conditional-expectation and martingale results used below. Let δ∈(0,1), define H0=0 and Hn=∑r=1n1/r, let qT be the least nonnegative integer with T≤2qT, and let kα be the least positive integer with kαα≥1. Then with probability at least 1−δ, some i∈{1,…,N} satisfies

∑sdπ^i(s)ℓ(s,θi)≤ϵ^N+1+4G2μHN−1+kα(qT+2)N+2ln⁡(1/δ)mN.

For each fixed m≥1, α∈(0,1], μ>0, and G≥0, there are constants C,C′>0 depending only on these parameters such that if δ∈(0,1/2] and N≥CT2ln⁡(1/δ), the right side is at most ϵ^N+C′/T. This is a surrogate-loss guarantee with interactive expert labels; it does not assert a finite-sample task-cost bound.

Statement references

Finite-horizon imitation learning and expert-surrogate loss: defines finite-horizon state occupancy, the mixed rollout policy, and the true round loss.

Dataset Aggregation (DAgger): defines finite-trajectory DAgger, empirical round loss, and exact FTL updates using interactive expert labels.

Follow-the-Leader regret for bounded strongly convex losses: supplies the empirical FTL regret bound 1+(4G2/μ)HN−1.

Expert mixing changes finite-horizon state occupancy by at most the mixing mass: bounds the absolute change in a bounded state-loss expectation under expert mixing by min⁡(1,Tβi).

Azuma-Hoeffding inequality: supplies the bounded-increment martingale tail bound under the stated Axiom of Choice assumption.

Filtration and filtered probability space: defines the nested information sigma-algebras used to reveal samples.

Adapted and integrable stochastic process: defines adaptedness and integrability of the trajectory-loss increments.

Martingale difference sequence: defines the centered adapted increments.

Martingales and martingale differences correspond: converts the centered increments to a martingale.

Martingale submartingale and supermartingale: defines the martingale property used by Azuma–Hoeffding.

Conditional expectation as an ae class: fixes the almost-sure class convention for conditional means.

The Axiom of Choice: states the axiom assumed by the conditional-expectation and Azuma interfaces.

Ross, Gordon, and Bagnell, A Reduction of Imitation Learning and Structured Prediction to No-Regret Online Learning: Theorem 4.2 and complete proof, §4.2, PDF pp. 4–5.

Bagnell, Boots, and Choudhury, Modern Adaptive Control and Reinforcement Learning, Chapter 6: PDF pp. 4–7 / printed pp. 57–60.

Duisterhof, Truong, and Agrawal, Algorithmic Foundations of Interactive Learning, Lecture 9: PDF pp. 51–54 / note pages 47–48; cited for conventions, not for the incomplete final cost-proof sketch.

Facts & Assumptions

Given: The finite-horizon MDP, expert and surrogate definitions, bounded strongly convex Lipschitz loss, exact empirical DAgger updates, nested sample filtration, conditional rollout-mean identity, attained empirical minima, geometric expert-use schedule, confidence parameter, and Axiom of Choice in the Statement.

[F1]

Finite-horizon imitation learning and expert-surrogate loss: dπ is the normalized average state occupancy and Li(θ)=∑sdπi(s)ℓ(s,θ) is the true round loss.

[F2]

Dataset Aggregation (DAgger): L^i(θ)=1mT∑j=1m∑t=0T−1ℓ(Sijt,θ) and θi+1 minimizes ∑r=1iL^r.

[F3]

Follow-the-Leader regret for bounded strongly convex losses: for bounded, G-Lipschitz, μ-strongly convex round losses with attained cumulative minima, empirical FTL regret is at most 1+(4G2/μ)HN−1.

[F4]

Expert mixing changes finite-horizon state occupancy by at most the mixing mass: for every g:S→[0,1], the absolute difference between its mixed-policy and learner-policy occupancy expectations is at most min⁡(1,Tβi).

[F5]

Azuma-Hoeffding inequality: under AC, martingale increments with predictable endpoints of width at most ck satisfy P(Mn−M0≥t)≤exp⁡(−2t2/∑kck2).

[F6]

Martingale difference sequence: a martingale difference is integrable, adapted, and has zero conditional mean given the preceding sigma-algebra.

[F7]

Martingales and martingale differences correspond: an adapted integrable martingale-difference sequence sums, from any integrable initial value, to a martingale.

[F8]

Filtration and filtered probability space: a filtered probability space has nested sigma-algebras Fn⊆Fn+1.

[F9]

Adapted and integrable stochastic process: adaptedness means time-n measurability with respect to Fn, and integrability means finite E∣Xn∣ at each time.

[F10]

Martingale submartingale and supermartingale: an integrable adapted process is a martingale when E[Mn+1∣Fn]=Mn almost surely.

[F11]

Conditional expectation as an ae class: conditional expectations are unique almost-sure equivalence classes of measurable integrable versions.

[F12]

The Axiom of Choice: AC states that every family of nonempty sets has a choice function.

[F13]

Basic algebra and order properties of conditional expectation: conditional expectation is linear on integrable random variables.

[F14]

Conditioning a known variable and an independent variable: if an integrable variable is measurable with respect to the conditioning sigma-algebra, its conditional expectation is itself almost surely.

[F15]

Arithmetic and lattice operations preserve measurability whenever they are defined: finite sums and products of real-valued Borel-measurable functions are Borel measurable.

[F16]

The principle of mathematical induction: a property of positive integer indices that holds at the first index and is preserved by successors holds for every positive integer index.

[A1]

For every state, ℓ(s,⋅) is valued in [0,1], G-Lipschitz, and μ-strongly convex on the nonempty convex set K.

[A2]

All cumulative empirical-loss minima and the minimum defining ϵ^N are attained; the algorithm uses exact FTL updates.

[A3]

The per-sample conditional mean identity E[ℓˉij(θi)∣Fi,j−1]=Li(θi) holds almost surely, with θi measurable before round i and trajectory states measurable when revealed.

[A4]

0≤β1≤1 and 0≤βi≤(1−α)i−1 for i≥2, with fixed 0<α≤1.

[A5]

For the rate corollary, 0<δ≤1/2, while the main probability bound only requires 0<δ<1.

Proof

1.1A1A2F2F3algebra

For any realized sample sequence, L^i is a finite average of functions ℓ(Sijt,⋅), so [A1] makes it [0,1]-valued, G-Lipschitz, and μ-strongly convex on K. By [F2] and [A2], the iterates are exact FTL iterates and the hindsight minimum is attained. Applying [F3] samplewise gives 1N∑i=1NL^i(θi)≤ϵ^N+1+4G2μHN−1N.

1.2A1A3F1F6F8F9F11F13F14F15construct

Flatten the pairs (i,j) in round-major order. Put G0=F1,0 and, for r=(i−1)m+j with 1≤i≤N and 1≤j≤m, put Gr=Fi,j; for every r>mN, put Gr=GmN. Within each round the sigma-algebras are nested, and at round boundaries Fi,0=Fi−1,m, so (Gr)r≥0 is an infinite filtration. Define D(i−1)m+j=Li(θi)−ℓˉij(θi) for the sample indices and set Dr=0 for r>mN. For each fixed round i, the policy probabilities are Borel functions of θ, the finite occupancy recursion uses only finite sums and products, and ℓ(s,⋅) is continuous by its Lipschitz property. Thus [F15] makes Li(θ) Borel, so Li(θi) is measurable with respect to G(i−1)m=Fi,0 and hence with respect to each Gr−1 in round i. Since the state set is finite and each statewise loss is continuous, (s,θ)↦ℓ(s,θ) is measurable; the revealed states and preceding measurable parameter therefore make each ℓˉij(θi) measurable with respect to Gr. By [A1] and the probability-occupancy property in [F1], both Li(θi) and ℓˉij(θi) lie in [0,1], so each sample increment is integrable and in [−1,1]; the zero extension has the same properties. For r=(i−1)m+j≤mN, [A3], [F13], and [F14] give E[Dr∣Gr−1]=Li(θi)−Li(θi)=0. For r>mN, [F14] gives E[0∣Gr−1]=0. Thus D is an infinite adapted martingale-difference sequence by [F6], using the filtration and process conventions [F8, F9, F11].

1.3A4F16algebrainduction

Put q=1−α and k=kα. For every integer n≥1, (1+α)n≥1+nα: equality holds for n=1, and if it holds for n, then multiplication by 1+α>0 gives (1+α)n+1≥(1+nα)(1+α)=1+(n+1)α+nα2≥1+(n+1)α. Thus [F16] gives (1+α)k≥1+kα≥2. Also q(1+α)=1−α2≤1, so q≤(1+α)−1 and qk≤1/2. For block j≥0, let Ij={jk+1,…,(j+1)k}. If j=0, [A4] gives β1≤1=q0 and βi≤qi−1≤1 for the other indices; if j≥1, every i∈Ij satisfies βi≤qi−1≤qjk. Thus each block contributes at most kmin⁡(1,T2−j). Let J=⌈N/k⌉−1, so the blocks j=0,…,J cover all sample indices. Their terms with j≤qT contribute at most qT+1 inside the sum over blocks. If J≥qT+1, the identity 2−j=2−(j−1)−2−j telescopes to ∑j=qT+1J2−j=2−qT−2−J≤2−qT; if J<qT+1, this tail is empty. Hence the remaining block contribution is at most T2−qT≤1, and ∑i=1Nmin⁡(1,Tβi)≤kα(qT+2).

2.1A3F5F7F10F12step 1.2algebra

By [F7], M0=0 and Mr=∑k=1rDk form a martingale on the infinite filtration (Gr). Its restriction to r=0,…,mN is the finite martingale required by [F5]. Each of its first mN increments has predictable endpoints −1 and 1, of width 2, so [F5] and the AC hypothesis [F12] give P(MmN≥t)≤exp⁡(−t2/(2mN)). Taking t=2mNln⁡(1/δ) makes this probability at most δ. Since MmN=m∑i=1N(Li(θi)−L^i(θi)), outside that event 1N∑iLi(θi)≤1N∑iL^i(θi)+2ln⁡(1/δ)mN.

3.1A2step 1.1step 2.1algebra

On the event from step 2.1, combine its mixed-policy loss bound with step 1.1 to obtain 1N∑iLi(θi)≤ϵ^N+1+4G2μHN−1N+2ln⁡(1/δ)mN.

4.1A1F1F4step 2.1step 3.1step 1.3algebra

Apply both directions of the absolute occupancy estimate [F4] with g(s)=ℓ(s,θi), giving ∑sdπ^i(s)ℓ(s,θi)≤Li(θi)+min⁡(1,Tβi). Average over i and use step 3.1 and step 1.3. At least one of the N own-occupancy losses is no larger than their average, so on the event of step 2.1 some i satisfies the displayed inequality in the Statement. Its probability is at least 1−δ.

5.1A5step 4.1algebra∎

For the rate claim, set L=ln⁡(1/δ)≥ln⁡2 and choose a fixed C≥3/ln⁡2, so N0=CT2L≥3>e. With c=4G2/μ, for N≥2 we have HN−1=1+∑r=2N−11/r≤1+∑r=2N−1∫r−1rdx/x=1+ln⁡(N−1)≤1+ln⁡N, with the sum empty when N=2. The derivative of x↦[1+c(1+ln⁡x)]/x is (c−1−cln⁡x)/x2≤0 for x≥e, so for N≥N0 the FTL term is at most [1+c(1+ln⁡N0)]/N0=O((1+ln⁡T)/T2), since L≥ln⁡2 and (1+ln⁡L)/L is bounded on [ln⁡2,∞). Also qT≤1+log⁡2T, so the mixing term is O((1+ln⁡T)/T2), while the Azuma term is at most 2/(mC)/T. As 1+ln⁡T≤T for T≥1, the two former terms are O(1/T) as well. This proves the stated rate on the same event as step 4.1.

4 · Examples, counterexamples and false statements

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