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ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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At the critical quadratic ULA step size, no invariant law exists

Example

Let U(x)=x2/2 on R and set γ=2 in ULA. Then the update is X′=−X+2Z for a fresh independent Z∼N(0,1). This Markov kernel has no invariant probability law.

Facts & Assumptions

Given: AC; the ULA update for U(x)=x2/2 at γ=2.

[A1]

AC is assumed in the standard-normal construction. The Axiom of Choice Standard normal and normal laws

[F1]

For U(x)=x2/2, ∇U(x)=x, so the ULA update at γ=2 is X′=−X+2Z, with fresh independent Z∼N(0,1). Unadjusted Langevin algorithm (ULA) Standard normal and normal laws

[F2]

Affine maps and sums of independent variables transform characteristic functions by φaX+b(t)=eitbφX(at) and by multiplication. Characteristic functions under affine maps and independent sums

[F3]

Every characteristic function has φ(0)=1, is continuous, and satisfies φ(−t)=φ(t)‾. Characteristic function of a real random variable Basic properties of characteristic functions

[F4]

The standard normal characteristic function is e−t2/2. Characteristic function of a normal law

Verification

Given: The update and facts above.

1.1A1F1F2F4given

Suppose for contradiction that an invariant probability μ exists and write φ for its characteristic function. Invariance of the kernel in [F1], together with the affine characteristic-function identity [F2] and the standard-normal characteristic function [F4], gives for every real t, φ(t)=φ(−t)e−2t2.

2.1F3step 1.1algebra∎

By [F3], ∣φ(−t)∣=∣φ(t)∣. For each t≠0, the displayed identity implies ∣φ(t)∣=∣φ(t)∣e−2t2, and e−2t2<1 forces φ(t)=0. Thus φ vanishes at every nonzero t, contradicting continuity at zero and φ(0)=1 from [F3]. No invariant probability can exist.

Depends on

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