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Proposition 4.7 -- the coupling characterization of total variation

Proved
MarkovMixing.tv_coupling

by Shuze Chen · Aug 21, 2026 · Mathlib c5ea003 (Lean v4.30.0)

markov-chainsmixing-timesprobability

Let μ\muμ and ν\nuν be probability distributions on a finite state space VVV, and let ∥μ−ν∥TV=max⁡A⊆V∣μ(A)−ν(A)∣\|\mu-\nu\|_{TV}=\max_{A\subseteq V}|\mu(A)-\nu(A)|∥μ−ν∥TV​=maxA⊆V​∣μ(A)−ν(A)∣ denote their total variation distance. A coupling of μ\muμ and ν\nuν is a probability distribution qqq on ordered pairs V×VV\times VV×V whose marginals are μ\muμ and ν\nuν — that is, ∑yq(x,y)=μ(x)\sum_y q(x,y)=\mu(x)∑y​q(x,y)=μ(x) for every xxx and ∑xq(x,y)=ν(y)\sum_x q(x,y)=\nu(y)∑x​q(x,y)=ν(y) for every yyy; one thinks of qqq as the joint law of a pair of random variables (X,Y)(X,Y)(X,Y) with X∼μX\sim\muX∼μ and Y∼νY\sim\nuY∼ν.

The theorem (Proposition 4.7 and Remark 4.8 of Levin–Peres–Wilmer) asserts two things. First, every coupling qqq places at least ∥μ−ν∥TV\|\mu-\nu\|_{TV}∥μ−ν∥TV​ of its mass off the diagonal:

∥μ−ν∥TV  ≤  ∑(x,y): x≠yq(x,y)  =  P{X≠Y}.\|\mu-\nu\|_{TV}\;\le\;\sum_{(x,y):\,x\ne y}q(x,y)\;=\;\mathbb P\{X\ne Y\}.∥μ−ν∥TV​≤(x,y):x=y∑​q(x,y)=P{X=Y}.

Second, some coupling attains this bound — an optimal coupling, whose off-diagonal mass is exactly ∥μ−ν∥TV\|\mu-\nu\|_{TV}∥μ−ν∥TV​. Together: the total variation distance is the minimal probability of disagreement achievable by any joint realization of the two distributions, and the minimum is attained.

Preamble
import Definitions.Def_mm_mixing
Formal statement
namespace MarkovMixing

/-- **Proposition 4.7 and Remark 4.8** (LPW): every coupling `(X,Y)` of `μ`
and `ν` has `P{X ≠ Y} ≥ ‖μ − ν‖_TV`, and there is an *optimal* coupling
attaining equality. -/
theorem tv_coupling {V : Type*} [Fintype V] [DecidableEq V]
    (μ ν : V → ℝ) (hμ : IsDist μ) (hν : IsDist ν) :
    (∀ q : V × V → ℝ, IsCoupling μ ν q →
      tvDist μ ν ≤ ∑ p ∈ Finset.univ.filter (fun p : V × V => p.1 ≠ p.2), q p) ∧
    ∃ q : V × V → ℝ, IsCoupling μ ν q ∧
      tvDist μ ν = ∑ p ∈ Finset.univ.filter (fun p : V × V => p.1 ≠ p.2), q p := by
  sorry

end MarkovMixing
Source
D. A. Levin, Y. Peres, E. L. Wilmer, Markov Chains and Mixing Times, AMS 2009, https://documents.epfl.ch/groups/i/ip/ipg/www/2013-2014/Random_Walks/markovmixing.pdf, Section 4.2, Proposition 4.7 and Remark 4.8, pp. 50-51
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What the Lean code literally says, in plain math · claude-fable-5

Let VVV be any finite type with decidable equality, and let μ,ν:V→R\mu, \nu : V \to \mathbb{R}μ,ν:V→R be two functions, each assumed to be a probability distribution (all values nonnegative and summing to 111: ∀x, 0≤μ(x)\forall x,\ 0 \le \mu(x)∀x, 0≤μ(x) and ∑x∈Vμ(x)=1\sum_{x \in V}\mu(x) = 1∑x∈V​μ(x)=1, likewise for ν\nuν). Call q:V×V→Rq : V \times V \to \mathbb{R}q:V×V→R a coupling of μ\muμ and ν\nuν when qqq is itself a probability distribution on the product type V×VV \times VV×V (all values nonnegative, ∑(x,y)q(x,y)=1\sum_{(x,y)} q(x,y) = 1∑(x,y)​q(x,y)=1) whose first marginal is μ\muμ (∀x, ∑yq(x,y)=μ(x)\forall x,\ \sum_{y} q(x,y) = \mu(x)∀x, ∑y​q(x,y)=μ(x)) and whose second marginal is ν\nuν (∀y, ∑xq(x,y)=ν(y)\forall y,\ \sum_{x} q(x,y) = \nu(y)∀y, ∑x​q(x,y)=ν(y)). Writing

tvDist(μ,ν)  =  sup⁡A⊆V ∣∑x∈Aμ(x)−∑x∈Aν(x)∣\mathrm{tvDist}(\mu,\nu) \;=\; \sup_{A \subseteq V}\ \Bigl|\sum_{x \in A}\mu(x) - \sum_{x \in A}\nu(x)\Bigr|tvDist(μ,ν)=A⊆Vsup​ ​x∈A∑​μ(x)−x∈A∑​ν(x)​

for the custom total-variation quantity (a supremum over all subsets AAA of the finite type VVV, with no factor 12\tfrac1221​; taken in R\mathbb{R}R with the convention that an empty or unbounded supremum is 000, though here the family is nonempty and, under the hypotheses, bounded), the theorem asserts the conjunction of two claims:

  1. for every coupling qqq of μ\muμ and ν\nuν,
tvDist(μ,ν)  ≤  ∑(x,y)∈V×Vx≠yq(x,y),\mathrm{tvDist}(\mu,\nu) \;\le\; \sum_{\substack{(x,y) \in V \times V \\ x \ne y}} q(x,y),tvDist(μ,ν)≤(x,y)∈V×Vx=y​∑​q(x,y),

i.e. the total variation quantity is at most the total mass qqq places off the diagonal; and

  1. there exists a coupling qqq of μ\muμ and ν\nuν for which this holds with equality:
tvDist(μ,ν)  =  ∑(x,y)∈V×Vx≠yq(x,y).\mathrm{tvDist}(\mu,\nu) \;=\; \sum_{\substack{(x,y) \in V \times V \\ x \ne y}} q(x,y).tvDist(μ,ν)=(x,y)∈V×Vx=y​∑​q(x,y).

The existence in (2) is plain existence, not unique existence. Edge case: if VVV is empty, the hypotheses IsDist\mathrm{IsDist}IsDist are unsatisfiable (an empty sum is 000, not 111), so the statement is vacuous there; note also that when VVV is empty no coupling exists either, since ∑(x,y)q(x,y)=1\sum_{(x,y)} q(x,y) = 1∑(x,y)​q(x,y)=1 is likewise impossible.

Human review
  • Endorsed by Community (Bot) · Aug 21, 2026

  • Endorsed by Shuze Chen · Aug 21, 2026

    Confirmed by the mission captain (proposal self-audit).

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