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The Gaussian ULA invariant law has variance $2 / (2 - γ)$

Statement

Assume AC, let $U (x) = x^{2} / 2$ on $R$ , and let $π = N (0, 1)$ . For every $0 < γ < 2$ , the one-dimensional ULA kernel of Unadjusted Langevin algorithm (ULA) has the invariant probability $π_{γ} = N ⁣ (0, \frac{2}{2 - γ}) .$ Every invariant probability with finite second moment has variance $2 / (2 - γ)$ . Thus its invariant variance is not the target variance $1$ . No uniqueness claim is made for invariant laws without a finite second moment.

Facts & Assumptions

Given: AC, the standard Gaussian target, and the ULA update with $U (x) = x^{2} / 2$ .

[A1]

AC is assumed in the standard-normal construction and characteristic-function uniqueness. The Axiom of Choice

[F1]

For this potential, $\nabla U (x) = x$ and the ULA update is $X^{'} = (1 - γ) X + \sqrt{2 γ} Z$ , where $Z \sim N (0, 1)$ is independent of the current state. Unadjusted Langevin algorithm (ULA) The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case

[F2]

The normal-law convention is Standard normal and normal laws; the characteristic function of $N (m, v)$ is $\exp (i m t - v t^{2} / 2)$ , and its mean and variance are $m$ and $v$ . Characteristic function of a normal law Moments, variance, and covariance on a probability space

[F3]

Affine transforms and finite independent sums multiply the corresponding characteristic functions. Characteristic functions under affine maps and independent sums

[F4]

Equal characteristic functions determine equal Borel probability laws. Uniqueness of a law from its characteristic function

[F5]

For independent square-integrable variables, expectations factor and finite linear combinations commute with expectation. Expectations factor over finite products of independent random variables The Lebesgue integral is linear on $L^{1} (μ)$

[F6]

A square-integrable variable is integrable by Cauchy–Schwarz on a probability space. Cauchy-Schwarz for random variables

Proof

Given: AC, $0 < γ < 2$ , and the ULA transition in [F1].

1.1A1F1F2F3F4algebra

Set $v_{γ} = 2 / (2 - γ) > 0$ and let $X \sim N (0, v_{γ})$ be independent of $Z \sim N (0, 1)$ . By [F2]–[F3], the updated state has characteristic function $φ_{X^{'}} (t) = \exp ⁣ (- \frac{(1 - γ)^{2} v_{γ} t^{2}}{2}) \exp (- γ t^{2}) = \exp ⁣ (- \frac{((1 - γ)^{2} v_{γ} + 2 γ) t^{2}}{2}) .$ Direct algebra gives $(1 - γ)^{2} v_{γ} + 2 γ = v_{γ}$ , so [F2] and [F4] imply $X^{'} \sim N (0, v_{γ})$ . This proves that $π_{γ}$ is invariant.

2.1F1F2F5F6algebra∎

Let $X$ have any invariant law with finite second moment, independently of the fresh standard normal $Z$ . By [F6], $X$ is integrable, and [F5] gives $Var (X^{'}) = (1 - γ)^{2} Var (X) + 2 γ,$ because $Var (Z) = 1$ and the cross-covariance of the independent variables is zero. Invariance gives $Var (X^{'}) = Var (X)$ , so $Var (X) = \frac{2 γ}{1 - (1 - γ)^{2}} = \frac{2}{2 - γ} .$ Since $γ > 0$ , this exceeds $1$ .

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Sources

Alain Durmus and Éric Moulines, Non-asymptotic convergence analysis for the Unadjusted Langevin Algorithm, §3.3, Gaussian-target calculation (standard reference, not scraped)
Gautam Iyer, Lecture 20: Langevin Monte Carlo, course notes, §3.1 (standard reference, not scraped)

The Gaussian ULA invariant law has variance 2/(2−γ)2/(2−γ)