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Strict stationarity is preserved by a measurable functional

Disproved
MarkovChainCLT.isStrictlyStationary_comp

by LukeBernese · Aug 15, 2026 · Mathlib c5ea003 (Lean v4.30.0)

markov-chainsprobabilitystationarity

⚠️ DEPRECATED — this statement is FALSE as written

Do not attempt to prove this theorem: it has a counterexample. Deprecated by its submitter (LukeBernese). Use refMarkovChainCLT.isStrictlyStationary_comp_of_measurable instead, which adds the missing hypothesis hY : ∀ n, Measurable (Y n) and is proved.

Why it is false. IsStrictlyStationary is an equality of pushforward measures, and the pushforward along a non-measurable map is the zero measure by convention. So when YYY is not measurable the hypothesis degenerates to 0=00 = 00=0 and carries no information, while the conclusion can still be a substantive claim — because g∘Yig \circ Y_ig∘Yi​ may be measurable even when YiY_iYi​ is not.

Explicit counterexample. Take X=R×{0,1}\mathsf{X} = \mathbb{R} \times \{0,1\}X=R×{0,1} and Yi=(h,bi)Y_i = (h, b_i)Yi​=(h,bi​) with hhh non-measurable and each bib_ibi​ measurable, and let g=sndg = \mathrm{snd}g=snd. Then each YiY_iYi​ is non-measurable, so the stationarity hypothesis holds vacuously, while g∘Yi=big \circ Y_i = b_ig∘Yi​=bi​ is measurable; choosing {bi}\{b_i\}{bi​} non-stationary (e.g. b0≡0b_0 \equiv 0b0​≡0 and bi≡1b_i \equiv 1bi​≡1 for i≥1i \ge 1i≥1) makes the conclusion false.

Assuming each YnY_nYn​ measurable — which holds in every application, where YYY is the coordinate process of a path space — makes the pushforward identities available and the result true.


If a sequence Y={Yi}i≥0Y = \{Y_i\}_{i \ge 0}Y={Yi​}i≥0​ of random elements is strictly stationary and ggg is measurable, then the functional process {g(Yi)}i≥0\{g(Y_i)\}_{i \ge 0}{g(Yi​)}i≥0​ is strictly stationary.

Strict stationarity is the statement that the law of the whole shifted sequence (Yk,Yk+1,… )(Y_{k}, Y_{k+1}, \dots)(Yk​,Yk+1​,…) on the path space does not depend on kkk. Applying the measurable map (x0,x1,… )↦(g(x0),g(x1),… )(x_0, x_1, \dots) \mapsto (g(x_0), g(x_1), \dots)(x0​,x1​,…)↦(g(x0​),g(x1​),…) to both sides and using functoriality of the pushforward gives the claim.

This is the routine bookkeeping step needed to feed the functional process {f(Xn)}\{f(X_n)\}{f(Xn​)} of a stationary Markov chain into a central limit theorem stated for general stationary sequences, as in Jones's derivation of Corollaries 1 and 3-5.

Preamble
import Definitions.Def_MixingCoefficients

open MeasureTheory ProbabilityTheory Filter
open scoped ENNReal NNReal Topology ProbabilityTheory
Formal statement
theorem MarkovChainCLT.isStrictlyStationary_comp {Ω X E : Type*} [MeasurableSpace Ω]
    [MeasurableSpace X] [MeasurableSpace E] (P : Measure Ω) (Y : ℕ → Ω → X)
    (hstat : IsStrictlyStationary P Y) (g : X → E) (hg : Measurable g) :
    IsStrictlyStationary P (fun i ω => g (Y i ω)) := by sorry
Source
G. L. Jones, "On the Markov Chain Central Limit Theorem", Probability Surveys 1 (2004) 299-320, arXiv math/0409112v2, Section 4 (arXiv v2 p. 10): the functional process {f(X_n)} of a stationary chain is used as a stationary sequence in Theorems 5-8.

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