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Section 4.5 -- standard mixing-time inequalities

Proved
MarkovMixing.mixing_time_eps

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

markov-chainsmixing-timesprobability

Let PPP be an irreducible, aperiodic 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∈V∥Pt(x,⋅)−π∥TVd(t)=\max_{x\in V}\bigl\|P^t(x,\cdot)-\pi\bigr\|_{TV}d(t)=x∈Vmax​​Pt(x,⋅)−π​TV​

for the worst-case distance to stationarity. For a tolerance ε\varepsilonε, the mixing time is the first time this distance drops to ε\varepsilonε,

tmix(ε)=min⁡{t∈N:d(t)≤ε},tmix=tmix(1/4).t_{\mathrm{mix}}(\varepsilon)=\min\{t\in\mathbb N: d(t)\le\varepsilon\},\qquad t_{\mathrm{mix}}=t_{\mathrm{mix}}(1/4).tmix​(ε)=min{t∈N:d(t)≤ε},tmix​=tmix​(1/4).

The theorem asserts, for any tolerance 0<ε≤10<\varepsilon\le10<ε≤1, the two standard consequences of submultiplicativity (displays (4.34)–(4.36) of Levin–Peres–Wilmer). First, running the chain for ℓ\ellℓ blocks of length tmix(ε)t_{\mathrm{mix}}(\varepsilon)tmix​(ε) shrinks the distance geometrically:

d(ℓ⋅tmix(ε))≤(2ε)ℓfor every ℓ∈N.d\bigl(\ell\cdot t_{\mathrm{mix}}(\varepsilon)\bigr)\le(2\varepsilon)^{\ell}\qquad\text{for every }\ell\in\mathbb N.d(ℓ⋅tmix​(ε))≤(2ε)ℓfor every ℓ∈N.

Second, mixing to any tolerance costs only logarithmically many standard mixing times:

tmix(ε)  ≤  ⌈log⁡2ε−1⌉  tmix.t_{\mathrm{mix}}(\varepsilon)\;\le\;\bigl\lceil\log_2\varepsilon^{-1}\bigr\rceil\;t_{\mathrm{mix}}.tmix​(ε)≤⌈log2​ε−1⌉tmix​.

This is why the convention ε=1/4\varepsilon=1/4ε=1/4 is harmless: any other tolerance changes the mixing time by at most a logarithmic factor.

Preamble
import Definitions.Def_mm_mixing
import Mathlib.Analysis.SpecialFunctions.Log.Base
Formal statement
namespace MarkovMixing

/-- **§4.5, Eqs. (4.34)–(4.36)** (LPW): for an irreducible aperiodic chain,
`d(ℓ · t_mix(ε)) ≤ (2ε)^ℓ`, and consequently
`t_mix(ε) ≤ ⌈log₂ ε⁻¹⌉ · t_mix`. -/
theorem mixing_time_eps {V : Type*} [Fintype V] [DecidableEq V] [Nonempty V]
    (P : Matrix V V ℝ) (hP : IsStochastic P) (hirr : Irreducible P)
    (hap : Aperiodic P) (π : V → ℝ) (hπ : IsStationary P π)
    (ε : ℝ) (hε : 0 < ε) (hε1 : ε ≤ 1) :
    (∀ ℓ : ℕ, distStationary P π (ℓ * mixingTime P π ε) ≤ (2 * ε) ^ ℓ) ∧
    mixingTime P π ε ≤ ⌈Real.logb 2 ε⁻¹⌉₊ * tMix P π := 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 4.5, Eqs. (4.34)-(4.36), p. 55
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What the Lean code literally says, in plain math · claude-fable-5

Let VVV be a finite, nonempty set and let PPP be a V×VV \times VV×V matrix of real numbers that is stochastic, meaning every entry satisfies P(x,y)≥0P(x,y) \ge 0P(x,y)≥0 and every row sums to 111 (∑yP(x,y)=1\sum_y P(x,y) = 1∑y​P(x,y)=1 for each xxx). Assume PPP is irreducible in the sense that for every ordered pair of states x,yx, yx,y there exists a natural number t≥0t \ge 0t≥0 with (Pt)(x,y)>0(P^t)(x,y) > 0(Pt)(x,y)>0 (note t=0t = 0t=0 is allowed, so for x=yx = yx=y this condition is automatic since P0P^0P0 is the identity). Assume PPP is aperiodic in the following custom sense: for every state xxx, the quantity per⁡(x):=sup⁡{d∈N:d divides every t∈Rx}\operatorname{per}(x) := \sup\{d \in \mathbb{N} : d \text{ divides every } t \in R_x\}per(x):=sup{d∈N:d divides every t∈Rx​} equals 111, where Rx={t≥1:(Pt)(x,x)>0}R_x = \{t \ge 1 : (P^t)(x,x) > 0\}Rx​={t≥1:(Pt)(x,x)>0} is the set of return times of xxx; here the supremum is taken in N\mathbb{N}N, with the convention that the supremum of an unbounded or empty-bounded situation is the junk value 000 (in particular, if Rx=∅R_x = \varnothingRx​=∅ the divisor set is all of N\mathbb{N}N and per⁡(x)=0≠1\operatorname{per}(x) = 0 \ne 1per(x)=0=1, so the aperiodicity hypothesis forces every state to have at least one return time). Let π:V→R\pi : V \to \mathbb{R}π:V→R be stationary for PPP, meaning π\piπ is a probability distribution (π(x)≥0\pi(x) \ge 0π(x)≥0 for all xxx and ∑xπ(x)=1\sum_x \pi(x) = 1∑x​π(x)=1) and the row vector π\piπ satisfies πP=π\pi P = \piπP=π. Finally, let ε\varepsilonε be a real number with 0<ε≤10 < \varepsilon \le 10<ε≤1.

Define the total-variation-style distance between two functions μ,ν:V→R\mu, \nu : V \to \mathbb{R}μ,ν:V→R as

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

the supremum over all (finite) subsets AAA of VVV of the absolute difference of the masses of AAA (note this is the sup of ∣μ(A)−ν(A)∣|\mu(A) - \nu(A)|∣μ(A)−ν(A)∣, without the factor 12\tfrac1221​ sometimes used for total variation). Define the distance-to-stationarity at time t∈Nt \in \mathbb{N}t∈N as

d(t)=sup⁡x∈Vtv⁡((Pt)(x,⋅), π),d(t) = \sup_{x \in V} \operatorname{tv}\bigl((P^t)(x, \cdot),\, \pi\bigr),d(t)=x∈Vsup​tv((Pt)(x,⋅),π),

the worst case over starting states xxx of the distance between row xxx of PtP^tPt and π\piπ. Define the mixing time at level δ\deltaδ as

tmix(δ)=inf⁡{t∈N:d(t)≤δ},t_{\mathrm{mix}}(\delta) = \inf\{t \in \mathbb{N} : d(t) \le \delta\},tmix​(δ)=inf{t∈N:d(t)≤δ},

an infimum over natural numbers, with the convention that the infimum of the empty set is 000 — so if d(t)d(t)d(t) never drops to δ\deltaδ or below, tmix(δ)t_{\mathrm{mix}}(\delta)tmix​(δ) is 000, not ∞\infty∞. Write tmix=tmix(1/4)t_{\mathrm{mix}} = t_{\mathrm{mix}}(1/4)tmix​=tmix​(1/4) for the mixing time at the fixed level 14\tfrac1441​.

Under all these hypotheses, the theorem asserts the conjunction of two claims:

  1. For every natural number ℓ≥0\ell \ge 0ℓ≥0,
d(ℓ⋅tmix(ε))≤(2ε)ℓ,d\bigl(\ell \cdot t_{\mathrm{mix}}(\varepsilon)\bigr) \le (2\varepsilon)^{\ell},d(ℓ⋅tmix​(ε))≤(2ε)ℓ,

where ℓ⋅tmix(ε)\ell \cdot t_{\mathrm{mix}}(\varepsilon)ℓ⋅tmix​(ε) is the product of natural numbers. (For ℓ=0\ell = 0ℓ=0 this reads d(0)≤1d(0) \le 1d(0)≤1; note the base of the exponential is 2ε2\varepsilon2ε, not ε\varepsilonε, and 2ε2\varepsilon2ε may be as large as 222 since ε\varepsilonε is only assumed ≤1\le 1≤1, in which case the bound grows with ℓ\ellℓ.)

tmix(ε)≤⌈log⁡2(ε−1)⌉⋅tmix,t_{\mathrm{mix}}(\varepsilon) \le \bigl\lceil \log_2(\varepsilon^{-1}) \bigr\rceil \cdot t_{\mathrm{mix}},tmix​(ε)≤⌈log2​(ε−1)⌉⋅tmix​,

where ⌈⋅⌉\lceil \cdot \rceil⌈⋅⌉ is the ceiling taken as a natural number (any nonpositive real value is clamped to 000). In particular, when ε=1\varepsilon = 1ε=1 the right-hand side is 000 and the claim becomes tmix(1)=0t_{\mathrm{mix}}(1) = 0tmix​(1)=0; and log⁡2\log_2log2​ here is the real base-2 logarithm, which takes the junk value 000 at 000.

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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