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ULA's invariant-law bias under smoothness and radial drift

Statement

Assume AC (The Axiom of Choice) and the hypotheses of Unadjusted Langevin algorithm (ULA). In addition, let $U \in C^{2} (R^{d})$ , let $\nabla U$ be $L$ -Lipschitz for $L > 0$ , and suppose there are a global minimizer $x_{⋆}$ , $ρ > 0$ , $α \in (1, 2]$ , and $M_{ρ} \geq 0$ such that $⟨ \nabla U (x), x - x_{⋆} ⟩ \geq ρ ∥ x - x_{⋆} ∥^{α} whenever ∥ x - x_{⋆} ∥ \geq M_{ρ} .$ Assume $0 < Z_{U} = \int e^{- U (x)} d x < \infty$ and normalize $U (x_{⋆}) = 0$ ; subtracting a constant leaves the target and ULA kernel unchanged and only rescales the weight by a constant. Put $π_{U} (d x) = Z_{U}^{- 1} e^{- U (x)} d x$ and $V (x) = e^{U (x) / 2}$ . For every $0 < γ < L^{- 1}$ , the ULA kernel has a unique invariant probability $π_{γ}$ , both $π_{U}$ and $π_{γ}$ have finite $V^{1 / 2}$ moments, and $∥ π_{U} - π_{γ} ∥_{V^{1 / 2}} \leq C γ^{1 / 2},$ where the finite source-dependent constant $C$ is independent of $γ$ . The norm is the weighted total-variation norm of Weighted total variation for probability measures.

Facts & Assumptions

Given: AC and the potential, target, weights and step size in the Statement.

[A1]

AC is inherited through the Brownian/Gaussian, Ito, change-of-measure and Hahn/Jordan interfaces. The Axiom of Choice

[F1]

The ULA kernel $R_{γ}$ is that of Unadjusted Langevin algorithm (ULA). It has a unique invariant $π_{γ}$ with finite $V$ moment for every $0 < γ < L^{- 1}$ . With $γ_{0} = (2 L)^{- 1}$ , constants $M_{V}, M_{W}, M_{g} < \infty$ independent of the step satisfy $π_{γ} V \leq M_{V}$ for $γ \leq γ_{0}$ , $π_{γ} W \leq M_{W}$ for all allowed steps, and $∥ \nabla U (x) ∥^{2} + ∥ x - x_{⋆} ∥^{2} \leq M_{g} V (x)$ , where $W = V^{1 / 2}$ . ULA invariant laws have uniform exponential moments on the needed step ranges

[F2]

The Langevin semigroup leaves $π_{U}$ invariant, has $π_{U} V < \infty$ , satisfies $P_{t} V \leq V + B_{V}$ for a fixed finite $B_{V}$ , and contracts zero-mass differences in the W norm: $∥ μ P_{t} - ν P_{t} ∥_{W} \leq C_{W} e^{- v_{W} t} ∥ μ - ν ∥_{W}$ , with $C_{W} < \infty$ and $v_{W} > 0$ . Radial drift gives weighted geometric contraction of Langevin diffusion

[F3]

From any finite-second-moment initial probability, the endpoint relative entropy of an m-step Euler interpolation against the Langevin endpoint satisfies $D (Law (X_{m}) ∥ Law (Y_{m γ})) \leq (L^{2} / 4) \sum_{k = 0}^{m - 1} (γ^{3} E ∥ \nabla U (X_{k}) ∥^{2} / 3 + d γ^{2})$ . Pathwise relative-entropy error of the Langevin Euler interpolation

[F4]

For probabilities with finite V moments, weighted Pinsker with $W^{2} = V$ gives $∥ μ - ν ∥_{W}^{2} \leq 2 (μ V + ν V) D (μ ∥ ν)$ . A weighted total-variation bound from relative entropy

[F5]

Weighted total variation is the supremum over Borel tests $∣ f ∣ \leq W$ and obeys the triangle inequality; it is at most $μ W + ν W$ for two probabilities with finite W moments. Weighted total variation for probability measures

Proof

Given: The hypotheses in the Statement, with $W = e^{U / 4}$ and $γ_{0} = (2 L)^{- 1}$ .

1.1A1F1F2algebrachoose

By [F1], for each allowed step the unique invariant probability $π_{γ}$ exists and has finite V moment, hence finite W moment since $1 \leq W \leq V$ . By [F2], $π_{U} V < \infty$ , so its W moment is finite too. Choose a fixed $T_{0} > 0$ so that $C_{W} e^{- v_{W} T_{0}} \leq 1 / 2$ , possible because $v_{W} > 0$ . All these constants depend only on the potential and its hypotheses.

2.1A1F1F3step 1.1algebraconstruct

First suppose $0 < γ \leq γ_{0}$ . Put $m = ⌈ T_{0} / γ ⌉$ and $T = m γ$ , so $T_{0} \leq T < T_{0} + γ \leq T_{0} + γ_{0}$ . Start Euler with law $μ = π_{γ}$ , independently of its Brownian increments; every grid-time law is $μ$ by invariance. Its second moment is finite by [F1], since $∥ x ∥^{2} \leq 2 ∥ x - x_{⋆} ∥^{2} + 2 ∥ x_{⋆} ∥^{2}$ . Moreover $E ∥ \nabla U (X_{k}) ∥^{2} \leq M_{g} M_{V}$ at every grid time. Apply [F3] to the diffusion with the same initial law. Euler's endpoint law is $μ$ and diffusion's is $μ P_{T}$ , giving $D (μ ∥ μ P_{T}) \leq \frac{L^{2}}{4} m (\frac{γ^{3}}{3} M_{g} M_{V} + d γ^{2}) \leq K_{D} γ, K_{D} = \frac{L^{2}}{4} (T_{0} + γ_{0}) (d + \frac{γ_{0}}{3} M_{g} M_{V}) .$ The initial Brownian/Gaussian construction uses only the inherited AC assumption.

3.1F1F2F4F5step 1.1step 2.1algebra

By [F1], $μ V \leq M_{V}$ ; by [F2], $(μ P_{T}) V \leq M_{V} + B_{V}$ . Weighted Pinsker [F4] and step 2.1 therefore imply $∥ μ - μ P_{T} ∥_{W} \leq K_{E} \sqrt{γ}, K_{E} = \sqrt{2 (2 M_{V} + B_{V}) K_{D}} .$ Since $π_{U} P_{T} = π_{U}$ , the contraction [F2], the choice of $T_{0}$ , and $T \geq T_{0}$ give $∥ μ - π_{U} ∥_{W} \leq K_{E} \sqrt{γ} + ∥ μ P_{T} - π_{U} P_{T} ∥_{W} \leq K_{E} \sqrt{γ} + \frac{1}{2} ∥ μ - π_{U} ∥_{W} .$ The last norm is finite by step 1.1, so subtraction yields $∥ π_{γ} - π_{U} ∥_{W} \leq 2 K_{E} \sqrt{γ}$ throughout the small-step range.

4.1F1F5step 1.1step 3.1algebra∎

If $γ_{0} < γ < L^{- 1}$ , [F1] and [F5] give $∥ π_{γ} - π_{U} ∥_{W} \leq M_{W} + π_{U} W \leq (M_{W} + π_{U} W) γ_{0}^{- 1 / 2} \sqrt{γ}$ . Together with step 3.1, this proves the stated bound with $C = \max {2 K_{E}, (M_{W} + π_{U} W) γ_{0}^{- 1 / 2}} < \infty,$ independent of $γ$ over the entire specified open interval. Step 1.1 supplies both finite weight moments and uniqueness/existence for every such step.

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Sources

Alain Durmus and Éric Moulines, Non-asymptotic convergence analysis for the Unadjusted Langevin Algorithm, Theorem 10 and proof §4.6 (standard reference, not scraped)