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✓ 4 results · all verified · 2 also independently AI-judged

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Langevin splittings and invariant measure error — Examples

1 · Prerequisites

2 · Summary

Summary

These examples compute the exact BAOAB harmonic position variance and its separate kinetic bias, exhibit a step above the proven harmonic stability limit where no invariant law exists, and show why a constant potential on the line has no normalizable Gibbs probability. The final comparison gives the published overdamped ULA variance beside the underdamped BAOAB harmonic covariances; their different state variables and transition maps keep the conclusions separate.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passOpen item page →

Harmonic BAOAB thermostat: exact position variance and kinetic bias

Example

Take a one-dimensional harmonic oscillator with mass $m = 2$ , stiffness $k = 8$ , inverse temperature $β = 1$ , friction $γ = 1$ , and BAOAB step $h = 1 / 2$ . The stability threshold is $h < 2 \sqrt{m / k} = 1$ . The example concerns the exact stationary covariance of the discrete BAOAB chain, not Monte Carlo error from a finite trajectory.

Facts & Assumptions

Given: $m = 2$ , $k = 8$ , $β = 1$ , $γ = 1$ , $h = 1 / 2$ , and the harmonic BAOAB theorem.

[F1]

Exact invariant covariance and stability threshold for scalar harmonic BAOAB: when $0 < h < 2 \sqrt{m / k}$ , $Var (Q) = 1 / (β k)$ and $Var (P) = (m / β) (1 - h^{2} k / (4 m))$ .

[A1]

The Axiom of Choice: AC is inherited through [F1]; the parameter substitution and arithmetic use no additional choice.

Verification

Given: $m = 2$ , $k = 8$ , $β = 1$ , $γ = 1$ , $h = 1 / 2$ , and the harmonic BAOAB theorem.

[F1] Exact invariant covariance and stability threshold for scalar harmonic BAOAB: when $0 < h < 2 \sqrt{m / k}$ , $Var (Q) = 1 / (β k)$ and $Var (P) = (m / β) (1 - h^{2} k / (4 m))$ .

[A1] The Axiom of Choice: AC is inherited through [F1]; the parameter substitution and arithmetic use no additional choice.

Given: The parameter values in the Example.

1.1F1givenalgebra

Here $2 \sqrt{m / k} = 2 \sqrt{2 / 8} = 1$ , so $h = 1 / 2$ is in the strict stable range of [F1]. The dimensionless quantity is $r = h^{2} k / (4 m) = (1 / 4) 8 / 8 = 1 / 4$ .

2.1A1F1step 1.1algebra∎

The discrete stationary position variance is $1 / (β k) = 1 / 8$ , exactly the canonical Gibbs position variance. The momentum variance is $(2 / 1) (1 - 1 / 4) = 3 / 2$ , whereas the continuous canonical value is $m / β = 2$ , so the momentum bias is $- 1 / 2 = - h^{2} k / (4 β)$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

An unstable harmonic BAOAB step has no invariant probability

Example

For the one-dimensional harmonic oscillator with $m = k = β = γ = 1$ , choose the BAOAB step $h = 5 / 2$ . The method's affine Gaussian transition is unstable, so this discrete chain has no invariant probability.

Facts & Assumptions

Given: $m = k = β = γ = 1$ , $h = 5 / 2$ , and the harmonic BAOAB theorem.

[F1]

Exact invariant covariance and stability threshold for scalar harmonic BAOAB: for the scalar harmonic chain, no invariant probability exists when $h > 2 \sqrt{m / k}$ .

[A1]

The Axiom of Choice: AC is inherited through [F1]; the threshold comparison uses no additional choice.

Verification

Given: $m = k = β = γ = 1$ , $h = 5 / 2$ , and the harmonic BAOAB theorem.

[F1] Exact invariant covariance and stability threshold for scalar harmonic BAOAB: for the scalar harmonic chain, no invariant probability exists when $h > 2 \sqrt{m / k}$ .

[A1] The Axiom of Choice: AC is inherited through [F1]; the threshold comparison uses no additional choice.

Given: The parameters in the Example.

1.1F1givenalgebra

The threshold is $2 \sqrt{m / k} = 2$ , and $5 / 2 > 2$ . Thus the strict unstable case of [F1] applies.

2.1A1F1step 1.1∎

Therefore the BAOAB transition at this step size has no invariant probability law. This example does not infer failure at the boundary $h = 2$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-luna)Open item page →

A constant potential has no canonical Gibbs probability on the line

Example

On $R$ , take $V (q) = 0$ , any fixed mass $m > 0$ , friction $γ > 0$ , and inverse temperature $β > 0$ . The configurational partition integral diverges, so the canonical Gibbs expression does not define a probability. The momentum factor is normalizable but cannot repair the infinite configurational factor.

Facts & Assumptions

Given: $V (q) = 0$ for every $q \in R$ , $m > 0$ , $γ > 0$ , $β > 0$ , and the canonical normalizer in Underdamped Langevin equation and canonical Gibbs density.

[F1]

Underdamped Langevin equation and canonical Gibbs density: a canonical Gibbs probability requires a finite positive normalizing integral $Z = \int e^{- β H (q, p)} d q d p$ .

[A1]

The Axiom of Choice: AC is inherited from the model-definition interface [F1]; the finite-rectangle lower bound uses no further choice.

Verification

Given: $V (q) = 0$ for every $q \in R$ , $m > 0$ , $γ > 0$ , $β > 0$ , and the canonical normalizer in Underdamped Langevin equation and canonical Gibbs density.

[F1] Underdamped Langevin equation and canonical Gibbs density: a canonical Gibbs probability requires a finite positive normalizing integral $Z = \int e^{- β H (q, p)} d q d p$ .

[A1] The Axiom of Choice: AC is inherited from the model-definition interface [F1]; the finite-rectangle lower bound uses no further choice.

Given: The data in the Example.

1.1givenalgebra

For every $R > 0$ , $\int_{- R}^{R} e^{- β V (q)} d q = \int_{- R}^{R} 1 d q = 2 R$ . These integrals are unbounded as $R \to \infty$ , so $\int_{R} e^{- β V (q)} d q = \infty$ .

2.1A1F1givenalgebra∎

On the finite rectangle $[- R, R] \times [- 1, 1]$ the phase-space integrand is at least $e^{- β / (2 m)}$ , so its integral is at least $4 R e^{- β / (2 m)}$ . These lower bounds diverge as $R \to \infty$ , hence the full normalizer $Z$ is infinite and the expression cannot be normalized to a probability.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passOpen item page →

Overdamped ULA and underdamped BAOAB have different harmonic variance errors

Example

For overdamped ULA targeting $N (0, 1)$ , take step $η = 1$ . For underdamped BAOAB, take the harmonic oscillator $m = k = β = γ = 1$ and step $h = 1$ . The two invariant laws concern different state variables and transition kernels; their variance formulas are juxtaposed only to make that distinction explicit.

Facts & Assumptions

Given: The ULA Gaussian theorem, the harmonic BAOAB theorem, $η = 1$ , and $m = k = β = γ = 1$ , $h = 1$ .

[F1]

The Gaussian ULA invariant law has variance $2 / (2 - γ)$ : for the standard normal target and $0 < η < 2$ , ULA has invariant variance $2 / (2 - η)$ .

[F2]

Exact invariant covariance and stability threshold for scalar harmonic BAOAB: for $0 < h < 2 \sqrt{m / k}$ , BAOAB has $Var (Q) = 1 / (β k)$ and $Var (P) = (m / β) (1 - h^{2} k / (4 m))$ .

[A1]

The Axiom of Choice: AC is inherited through [F1] and [F2]; the side-by-side substitutions use no additional choice.

Verification

Given: The ULA Gaussian theorem, the harmonic BAOAB theorem, $η = 1$ , and $m = k = β = γ = 1$ , $h = 1$ .

[F1] The Gaussian ULA invariant law has variance $2 / (2 - γ)$ : for the standard normal target and $0 < η < 2$ , ULA has invariant variance $2 / (2 - η)$ .

[F2] Exact invariant covariance and stability threshold for scalar harmonic BAOAB: for $0 < h < 2 \sqrt{m / k}$ , BAOAB has $Var (Q) = 1 / (β k)$ and $Var (P) = (m / β) (1 - h^{2} k / (4 m))$ .

[A1] The Axiom of Choice: AC is inherited through [F1] and [F2]; the side-by-side substitutions use no additional choice.

Given: The parameters in the Example.

1.1F1givenalgebra

Since $0 < η = 1 < 2$ , [F1] gives the ULA invariant variance $2 / (2 - 1) = 2$ , compared with the target variance $1$ .

2.1A1F2givenalgebra∎

For BAOAB, $h = 1 < 2 \sqrt{m / k} = 2$ and $r = h^{2} k / (4 m) = 1 / 4$ . Thus [F2] gives position variance $1$ and momentum variance $1 (1 - 1 / 4) = 3 / 4$ , compared with the canonical momentum variance $1$ . The ULA variance describes a single overdamped coordinate; the BAOAB pair describes both position and momentum. Neither result supplies the other's invariant-law claim.

Langevin splittings and invariant measure error — Examples

1 · Prerequisites

2 · Summary

Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Harmonic BAOAB thermostat: exact position variance and kinetic bias

Example

Facts & Assumptions

Verification

An unstable harmonic BAOAB step has no invariant probability

Example

Facts & Assumptions

Verification

A constant potential has no canonical Gibbs probability on the line

Example

Facts & Assumptions

Verification

Overdamped ULA and underdamped BAOAB have different harmonic variance errors

Example

Facts & Assumptions

Verification

Sources