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A uniform second-moment bound is inherited by the Gaussian limit

Proved
MarkovChainCLT.gaussian_variance_le_of_tendstoInDistribution

by LukeBernese · Aug 16, 2026 · Mathlib c5ea003 (Lean v4.30.0)

central-limit-theoremgaussianprobabilitysecond-momentweak-convergence

A uniform second-moment bound passes to the Gaussian limit. If Yn→N(0,v)Y_n \to N(0,v)Yn​→N(0,v) in distribution and E[Yn2]≤B\mathbb E[Y_n^2] \le BE[Yn2​]≤B for every nnn, then v≤Bv \le Bv≤B.

Why this is not automatic. Convergence in distribution says nothing about moments: they can jump up in the limit only if mass escapes, and can certainly fail to converge. What is true is the lower-semicontinuity direction, E[Y2]≤lim inf⁡nE[Yn2]\mathbb E[Y^2] \le \liminf_n \mathbb E[Y_n^2]E[Y2]≤liminfn​E[Yn2​], and that is exactly what is needed to transfer a uniform bound to the limit.

Where it is used. In the Markov chain central limit theorem for a square-integrable observable one truncates, proves the theorem for each bounded piece fKf_KfK​ obtaining a limit variance vKv_KvK​, and must then show (vK)(v_K)(vK​) converges. The route goes through the characteristic function, ∫cos⁡ dN(0,vK)=e−vK/2\int\cos\,dN(0,v_K) = e^{-v_K/2}∫cosdN(0,vK​)=e−vK​/2, whose Cauchy property transfers back to vKv_KvK​ only if the vKv_KvK​ stay bounded — otherwise e−vK/2e^{-v_K/2}e−vK​/2 could tend to 000 with vK→∞v_K\to\inftyvK​→∞. This lemma supplies exactly that boundedness: the O(n)O(n)O(n) variance bound for partial sums gives E[(TnK)2]≤4N∥fK∥L2(π)2\mathbb E[(T_n^K)^2] \le 4N\|f_K\|_{L^2(\pi)}^2E[(TnK​)2]≤4N∥fK​∥L2(π)2​ uniformly in nnn, hence vK≤4N∥fK∥L2(π)2v_K \le 4N\|f_K\|_{L^2(\pi)}^2vK​≤4N∥fK​∥L2(π)2​.

Proof. Truncate the square: φM(x)=min⁡(x2,M)\varphi_M(x) = \min(x^2, M)φM​(x)=min(x2,M) is bounded and continuous, so convergence in distribution gives ∫φM dN(0,v)=lim⁡nE[φM(Yn)]≤sup⁡nE[Yn2]≤B\int\varphi_M\,dN(0,v) = \lim_n \mathbb E[\varphi_M(Y_n)] \le \sup_n\mathbb E[Y_n^2] \le B∫φM​dN(0,v)=limn​E[φM​(Yn​)]≤supn​E[Yn2​]≤B, using φM≤x2\varphi_M \le x^2φM​≤x2. Letting M→∞M\to\inftyM→∞, dominated convergence — with dominating function x2x^2x2, integrable for a Gaussian — gives ∫x2 dN(0,v)≤B\int x^2\,dN(0,v) \le B∫x2dN(0,v)≤B. Finally the second moment of N(0,v)N(0,v)N(0,v) is its variance, since the mean vanishes, and that is vvv.

Preamble
import Mathlib.MeasureTheory.Function.ConvergenceInDistribution
import Mathlib.Probability.Distributions.Gaussian.Real
import Mathlib.Probability.Moments.Variance
import Mathlib.MeasureTheory.Integral.Bochner.Set

open Filter MeasureTheory ProbabilityTheory
open scoped ENNReal NNReal Topology
Formal statement
theorem MarkovChainCLT.gaussian_variance_le_of_tendstoInDistribution {Ω : Type*}
    [MeasurableSpace Ω] (μ : Measure Ω) [IsProbabilityMeasure μ]
    (Y : ℕ → Ω → ℝ) (v : ℝ≥0) (B : ℝ) (hY : ∀ n, Measurable (Y n))
    (hclt : TendstoInDistribution Y atTop (id : ℝ → ℝ) (fun _ => μ) (gaussianReal 0 v))
    (hint : ∀ n, Integrable (fun ω => (Y n ω) ^ 2) μ)
    (hbd : ∀ n, ∫ ω, (Y n ω) ^ 2 ∂μ ≤ B) :
    (v : ℝ) ≤ B := by sorry
Source
P. Billingsley, Convergence of Probability Measures, 2nd ed., Wiley 1999, Section 3 (uniform integrability and moment convergence); P. Billingsley, Probability and Measure, 3rd ed., Wiley 1995, Theorem 25.11; G. L. Jones, "On the Markov Chain Central Limit Theorem", Probability Surveys 1 (2004) 299-320.

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