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Proposition 6.10 -- the strong stationary time bound

Proved
MarkovMixing.strong_stationary_bound

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

markov-chainsmixing-timesprobability

Let PPP be an irreducible Markov chain on a finite state space VVV with stationary distribution π\piπ. Write Pt(x,⋅)P^t(x,\cdot)Pt(x,⋅) for the distribution at time ttt started at xxx, ∥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)∣ for the total variation distance, and d(t)=max⁡x∥Pt(x,⋅)−π∥TVd(t)=\max_x\|P^t(x,\cdot)-\pi\|_{TV}d(t)=maxx​∥Pt(x,⋅)−π∥TV​ for the worst-case distance to stationarity. A strong stationary time τ\tauτ for the chain started at xxx is a randomized stopping rule that stops in finite time almost surely, with the stopped state distributed exactly as π\piπ and independent of the stopping time.

The theorem (Proposition 6.10 of Levin–Peres–Wilmer) asserts: if a single stopping rule is a strong stationary time from every starting state, then for every time ttt

d(t)  ≤  max⁡x∈V Px{τ>t}.d(t)\;\le\;\max_{x\in V}\,\mathbb P_x\{\tau>t\}.d(t)≤x∈Vmax​Px​{τ>t}.

To bound the mixing time of a chain it therefore suffices to construct one strong stationary time and control its tail uniformly in the start.

Preamble
import Definitions.Def_mm_stopping
Formal statement
namespace MarkovMixing

/-- **Proposition 6.10** (LPW): if `τ` is a strong stationary time from every
starting state, then `d(t) ≤ max_x P_x{τ > t}`. -/
theorem strong_stationary_bound {V : Type*} [Fintype V] [DecidableEq V]
    [Nonempty V] (P : Matrix V V ℝ) (hP : IsStochastic P) (hirr : Irreducible P)
    (π : V → ℝ) (hπ : IsStationary P π)
    (s : ∀ t : ℕ, (Fin (t + 1) → V) → ℝ)
    (hs : ∀ x : V, IsStrongStationaryTime P π x s) (t : ℕ) :
    distStationary P π t ≤ ⨆ x : V, stopTailProb P x s t := 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 6.4, Proposition 6.10, p. 79
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What the Lean code literally says, in plain math · claude-fable-5

Let VVV be a finite nonempty type with decidable equality, and let PPP be a V×VV \times VV×V real matrix assumed to be: stochastic (P(x,y)≥0P(x,y) \ge 0P(x,y)≥0 for all x,yx,yx,y, and each row sums to 111) and irreducible in the sense that for every pair of states x,yx, yx,y there is some natural number t≥0t \ge 0t≥0 with (Pt)(x,y)>0(P^t)(x,y) > 0(Pt)(x,y)>0 (with t=0t = 0t=0 allowed, so x=yx = yx=y is always witnessed by the identity P0P^0P0). Let π:V→R\pi : V \to \mathbb{R}π:V→R be assumed stationary for PPP: π≥0\pi \ge 0π≥0 pointwise, ∑yπ(y)=1\sum_y \pi(y) = 1∑y​π(y)=1, and πP=π\pi P = \piπP=π as a row vector. Let sss be a family assigning to each time ttt and each trajectory ω=(ω0,…,ωt)∈Vt+1\omega = (\omega_0, \dots, \omega_t) \in V^{t+1}ω=(ω0​,…,ωt​)∈Vt+1 a real number st(ω)s_t(\omega)st​(ω), and for a start xxx define

σtx(y)  =  ∑ω:{0,…,t}→Vω0=x, ωt=y(∏i=0t−1P(ωi,ωi+1))(∏u=0t−1(1−su(ω0,…,ωu)))st(ω),\sigma^x_t(y) \;=\; \sum_{\substack{\omega : \{0,\dots,t\} \to V \\ \omega_0 = x,\ \omega_t = y}} \left(\prod_{i=0}^{t-1} P(\omega_i, \omega_{i+1})\right) \left(\prod_{u=0}^{t-1}\bigl(1 - s_u(\omega_0,\dots,\omega_u)\bigr)\right) s_t(\omega),σtx​(y)=ω:{0,…,t}→Vω0​=x, ωt​=y​∑​(i=0∏t−1​P(ωi​,ωi+1​))(u=0∏t−1​(1−su​(ω0​,…,ωu​)))st​(ω),

the probability, starting from xxx, of tracing a path that ends at yyy at time ttt, not stopping at any earlier time, and stopping at time ttt under the randomized rule sss. The hypothesis on sss is that it is a strong stationary time simultaneously for every starting state x∈Vx \in Vx∈V; for each xxx this means the conjunction of: (i) 0≤st(ω)≤10 \le s_t(\omega) \le 10≤st​(ω)≤1 for all t,ωt, \omegat,ω; (ii) ∑t=0∞∑yσtx(y)=1\sum_{t=0}^{\infty} \sum_y \sigma^x_t(y) = 1∑t=0∞​∑y​σtx​(y)=1, where the countable sum over ttt takes the junk value 000 if the family is not summable (so summability with sum 111 is being asserted); and (iii) σtx(y)=(∑zσtx(z)) π(y)\sigma^x_t(y) = \bigl(\sum_z \sigma^x_t(z)\bigr)\,\pi(y)σtx​(y)=(∑z​σtx​(z))π(y) for every ttt and yyy. Under these hypotheses the theorem asserts, for every natural number ttt (including t=0t = 0t=0):

sup⁡x∈V  sup⁡A⊆V∣ ∑y∈A(Pt)(x,y)−∑y∈Aπ(y) ∣  ≤  sup⁡x∈V(1−∑u=0t∑y∈Vσux(y)).\sup_{x \in V}\; \sup_{A \subseteq V} \left|\, \sum_{y \in A} (P^t)(x,y) - \sum_{y \in A} \pi(y) \,\right| \;\le\; \sup_{x \in V} \left( 1 - \sum_{u=0}^{t} \sum_{y \in V} \sigma^x_u(y) \right).x∈Vsup​A⊆Vsup​​y∈A∑​(Pt)(x,y)−y∈A∑​π(y)​≤x∈Vsup​​1−u=0∑t​y∈V∑​σux​(y)​.

The left-hand side is the worst-case (over starting states xxx) distance at time ttt between the row y↦(Pt)(x,y)y \mapsto (P^t)(x,y)y↦(Pt)(x,y) of the ttt-th matrix power and π\piπ, where the distance used is the supremum over all finite subsets AAA of VVV (the empty set included, contributing 000) of the absolute difference in mass — the supremum of an absolute value, not a half-ℓ1\ell^1ℓ1 norm. The right-hand side is the supremum over starting states xxx of one minus the total probability of having stopped at some time u≤tu \le tu≤t (the inner sum over uuu runs from 000 through ttt inclusive), i.e. the worst-case probability of not yet having stopped by time ttt; nothing in the statement constrains this expression to lie in [0,1][0,1][0,1] beyond what follows from the hypotheses. Both outer suprema are over the finite nonempty index set VVV. No aperiodicity or reversibility assumption appears.

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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