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Finite-time weak order two of BAOAB

Statement

Assume the model data of Underdamped Langevin equation and canonical Gibbs density, V∈C8, and DjV is globally bounded for 2≤j≤8. Let T>0, N≥1, h=T/N≤1, and let Z0 be a common F0-measurable R2d-valued initial state with E∣Z0∣6<∞, independent of σ(Wt−W0:0≤t≤T); initialize both processes at this state. Let ϕ∈Cb8(R2d). Run BAOAB by taking each fresh Gaussian vector to be the standardized exact OU integral on that grid interval from the same Brownian motion that drives the exact SDE. If ZNB is its state after N steps and ZT is the exact SDE state, then ∣Eϕ(ZNB)−Eϕ(ZT)∣≤CT,ϕ,Z0h2. The order is a finite-time weak order statement under the displayed smoothness and moment assumptions; it gives no invariant-measure bias or ergodicity conclusion.

Facts & Assumptions

Given: The hypotheses above and a uniform grid tn=nh on [0,T]; the initial state is common, F0-measurable, and independent of σ(Wt−W0:0≤t≤T), and the BAOAB OU draws use the prescribed Brownian coupling.

[F1]

Underdamped Langevin equation and canonical Gibbs density: Cb8 means bounded continuous derivatives through order eight, and the total generator is L=A+B+O with the three operators in Symmetric BAOAB one-step weak expansion.

[F2]

Symmetric BAOAB one-step weak expansion: for test functions ψ∈Cb6 with a common bound on derivatives through order six, ∣Khψ(z)−Phψ(z)∣≤Ch3(1+∣z∣3) uniformly under the stated model assumptions. The lemma's general CP6 branch instead permits an exponent depending on the test-function growth.

[F3]

Global strong well-posedness for Lipschitz Itô SDEs: the exact solution exists uniquely and globally for the globally Lipschitz, linearly growing drift.

[F4]

Multidimensional Ito formula for Brownian-driven processes: It^o's formula supplies the generator moment estimate for smooth Lyapunov functions and the short-time semigroup expansion.

[F5]

Dominated convergence: dominated pointwise convergence permits differentiation limits and integral limits under the stated integrable bounds.

[F6]

Fatou's lemma: the integral of a nonnegative pointwise limit is at most the limit inferior of the integrals.

[A1]

The Axiom of Choice: AC is inherited through [F2]–[F6]; the finite-grid argument uses no further choice.

Proof

Given: The assumptions above.

1.1F1F2F3F4F6givenalgebra

Put Ws(z)=(1+∣z∣2)s for integer s≥1. The drift b has linear growth and bounded derivatives of positive order, and G is constant, so ∣LWs(z)∣≤CsWs(z). Applying [F4] to the solution stopped on radius R, taking expectations and applying Gronwall gives EWs(Zt∧τR)≤eCstEWs(Z0); Fatou’s lemma [F6] as R→∞ gives uniformly bounded exact moments through order six on [0,T]. Each B or A stage has the form z+hδ(z) with ∣δ(z)∣≤C(1+∣z∣), hence the mean-value formula and Young’s inequality give Ws(z+hδ(z))≤(1+Ch)Ws(z)+Ch. In the O stage p becomes ap+η; conditional terms linear in η vanish and terms of degree j≥2 have moments at most Cjhj/2, so Young’s inequality gives the same stage bound. Composing five stages and iterating over Nh=T yields max⁡n≤NEWs(ZnB)≤CT for s≤3.

2.1F1F2F3F5step 1.1

For 0≤t≤T, the solution flow is differentiable through order six in its initial state. The first derivative solves J˙=Db(Zt)J; each higher derivative solves the same linear equation with forcing a finite sum of products of lower flow derivatives and bounded derivatives of b. Induction and Gronwall bound these derivatives uniformly. Differentiating Ptϕ(z)=Eϕ(Ztz) under expectation is justified by bounded derivatives of ϕ and the flow bounds, so Ptϕ has bounded derivatives through order six uniformly. Together with the Markov bound ∥Ptϕ∥∞≤∥ϕ∥∞, these derivative estimates give a common Cb6 bound for ψn=PT−tn+1ϕ. Apply precisely the bounded-test branch of [F2], whose cubic weight is uniform in n,N; no generic polynomial-growth exponent is substituted.

3.1A1F2F3F5F6step 1.1step 2.1algebra∎

Let Kh be the BAOAB transition and Ph the exact transition. Inserting one exact transition at a time and using the Markov property gives [A1, F2, F3, F5, F6, step 1.1, step 2.1, algebra] Eϕ(ZNB)−Eϕ(ZT)=∑n=0N−1E[(Kh−Ph)PT−tn+1ϕ(ZnB)]. By [F2] and step 2.1, each summand has absolute value at most Ch3E(1+∣ZnB∣3)≤CTh3 using the s=2 moment bound in step 1.1. Summing N=T/h terms gives CTNh3=CTTh2. The result is finite-time weak order two and says nothing about invariant bias.

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