Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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.

Discounted occupancy change is bounded by average policy TV

Statement

Let a finite discounted CMDP and two stationary policies π,π′ be given, with normalized discounted occupancies dπ,dπ′. Define

TV⁡s(π′,π):=12∑a∈A(s)∣π′(a∣s)−π(a∣s)∣

and ∥x∥1:=∑s∈S∣x(s)∣ for a state vector. Then

∥dπ′−dπ∥1≤2γ1−γ∑s∈Sdπ(s)TV⁡s(π′,π).

Facts & Assumptions

Given: A finite CMDP, 0≤γ<1, stationary policies π,π′, and their state-mass vectors and normalized discounted occupancies.

[F1]

Finite discounted constrained Markov decision process: S and each A(s) are finite, p(s′,ρ∣s,a) is a normalized joint law, P(s′∣s,a):=∑ρp(s′,ρ∣s,a) is a probability distribution, and π(a∣s) is a probability distribution over A(s).

[F2]

Normalized discounted state occupancy: dσ=(1−γ)∑t≥0γtμtσ and the column transition matrix is Kσ(s′,s)=∑a,ρσ(a∣s)p(s′,ρ∣s,a).

[F3]

Normalized discounted occupancy is a probability distribution: every μtσ and dσ is a probability distribution.

Proof

1.1F1F4F5givenalgebra

For σ∈{π,π′}, every entry of Kσ is nonnegative and each column sums to one by [F1]. Thus Kσt is column-stochastic for every t, and its induced matrix 1-norm ∥A∥1:=max⁡s∑s′∣A(s′,s)∣ is one. Every entry of Kσt lies in [0,1], so each of the finitely many coordinate series ∑t=0∞γt(Kσt)(s′,s) converges by comparison with [F4]; their limits give convergence of the matrix partial sums in this finite-dimensional norm. The matrix series Gσ:=∑t=0∞γtKσt therefore exists; its finite partial sums satisfy (I−γKσ)∑t=0NγtKσt=I−γN+1KσN+1, whose remainder norm tends to zero by [F5]. The same identity with multiplication on the other side shows Gσ=(I−γKσ)−1.

1.2F1F3algebra

By [F1], for each coordinate s′ of (Kπ′−Kπ)dπ, the triangle inequality gives ∑s′∣((Kπ′−Kπ)dπ)(s′)∣≤∑s,a,s′dπ(s)∣π′(a∣s)−π(a∣s)∣P(s′∣s,a)=2∑sdπ(s)TV⁡s(π′,π), because ∑s′P(s′∣s,a)=1.

2.1F1F2step 1.1algebra

The recurrence in [F1] and the matrix definition in [F2] give μtσ=Kσtμ, hence dσ=(1−γ)Gσμ. The inverse identity Gπ′−Gπ=Gπ′[(I−γKπ)−(I−γKπ′)]Gπ therefore yields dπ′−dπ=γGπ′(Kπ′−Kπ)dπ.

3.1F4step 2.1step 1.2algebra∎

Since ∥Gπ′∥1≤∑t≥0γt=1/(1−γ) by [F4] and column-stochasticity, combine step 2.1 with step 1.2 to obtain the claimed 1-norm bound.

Depends on

Used by

Dependency tree · two levels

32 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources