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The recurrence dichotomy via Green's functions

Proved
MarkovMixing.recurrence_dichotomy

by Shuze Chen · Aug 22, 2026 · Mathlib c5ea003 (Lean v4.30.0)

markov-chainsmixing-timesprobability

Let PPP be an irreducible Markov chain on a countable state space VVV: nonnegative entries with each row summing to one as a convergent series, and every state reaching every other at some time. For a state xxx, the first return time is τx+=min⁡{t≥1:Xt=x}\tau^+_x=\min\{t\ge1:X_t=x\}τx+​=min{t≥1:Xt​=x}; the state is recurrent when return is certain, Px{τx+>t}→0\mathbb P_x\{\tau^+_x>t\}\to0Px​{τx+​>t}→0 as t→∞t\to\inftyt→∞, and transient otherwise. The Green's function at xxx is the expected number of visits to xxx,

G(x,x)=∑t=0∞Pt(x,x).G(x,x)=\sum_{t=0}^{\infty}P^t(x,x).G(x,x)=t=0∑∞​Pt(x,x).

The theorem (Proposition 21.3 of Levin–Peres–Wilmer) asserts:

  1. a state xxx is recurrent if and only if its Green's series diverges — equivalently, transience is exactly summability of the return probabilities Pt(x,x)P^t(x,x)Pt(x,x);
  2. recurrence is a class property: if one state of the irreducible chain is recurrent, every state is.

The equivalence comes from the renewal structure: the number of returns to xxx is geometric with success probability Px{τx+<∞}\mathbb P_x\{\tau^+_x<\infty\}Px​{τx+​<∞}, so its expectation G(x,x)G(x,x)G(x,x) is finite exactly when return is uncertain. This dichotomy is the standard tool for deciding recurrence — Pólya's theorem, the goal of this mission, is proved by estimating the series ∑tPt(0,0)≍∑tt−d/2\sum_tP^t(0,0)\asymp\sum_tt^{-d/2}∑t​Pt(0,0)≍∑t​t−d/2 for the walk on Zd\mathbb Z^dZd.

Preamble
import Definitions.Def_mm_countable
Formal statement
namespace MarkovMixing

/-- **Proposition 21.3** (LPW): for an irreducible chain on a countable
state space, a state is recurrent if and only if its Green's function
`G(x,x) = ∑_t P^t(x,x)` diverges, and recurrence of one state implies
recurrence of all states. -/
theorem recurrence_dichotomy {V : Type*} [Countable V] [DecidableEq V]
    (P : V → V → ℝ) (hP : IsStochasticC P) (hirr : IrreducibleC P) (x : V) :
    (Recurrent P x ↔ ¬Summable fun t => stepPow P t x x) ∧
    (Recurrent P x → ∀ y : V, Recurrent P y) := 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 21.1, Proposition 21.3, p. 276
Read-back

What the Lean code literally says, in plain math · claude-fable-5

Read-back: recurrence_dichotomy

Let VVV be an arbitrary countable type with decidable equality and P:V×V→RP : V \times V \to \mathbb{R}P:V×V→R a kernel. Assume PPP is stochastic (every entry nonnegative; each row has sum 111 in the unconditional HasSum sense) and irreducible (for all x,yx, yx,y there exists t∈Nt \in \mathbb{N}t∈N, possibly 000, with P(t)(x,y)>0P^{(t)}(x,y) > 0P(t)(x,y)>0, where P(t)P^{(t)}P(t) is the iterated kernel defined by P(0)P^{(0)}P(0) = identity indicator and P(t+1)(x,y)=∑z′P(t)(x,z)P(z,y)P^{(t+1)}(x,y) = \sum'_z P^{(t)}(x,z)P(z,y)P(t+1)(x,y)=∑z′​P(t)(x,z)P(z,y), a tsum with junk value 000 on non-summable families). For the single universally quantified state x∈Vx \in Vx∈V, the conclusion is the conjunction of two claims. Writing R(x,t)R(x,t)R(x,t) for the total weight

R(x,t)=∑ω:{0,…,t}→V′[ ω0=x∧∀i∈{1,…,t}, ωi≠x ]∏i=0t−1P(ωi,ωi+1)R(x,t) = \sum_{\omega : \{0,\dots,t\} \to V}{}' \big[\,\omega_0 = x \wedge \forall i \in \{1,\dots,t\},\ \omega_i \ne x\,\big] \prod_{i=0}^{t-1} P(\omega_i,\omega_{i+1})R(x,t)=ω:{0,…,t}→V∑​′[ω0​=x∧∀i∈{1,…,t}, ωi​=x]i=0∏t−1​P(ωi​,ωi+1​)

of length-ttt paths from xxx never revisiting xxx (tsum over all such path functions, junk value 000 if non-summable; R(x,0)=1R(x,0) = 1R(x,0)=1), and calling a state zzz recurrent when R(z,t)→0R(z,t) \to 0R(z,t)→0 as t→∞t \to \inftyt→∞, the two claims are:

  1. Green's-function criterion, as an iff: xxx is recurrent if and only if the sequence t↦P(t)(x,x)t \mapsto P^{(t)}(x,x)t↦P(t)(x,x) of diagonal iterated-kernel values is not summable (note the t=0t = 0t=0 term of this sequence is P(0)(x,x)=1P^{(0)}(x,x) = 1P(0)(x,x)=1; the criterion is about summability of the sequence itself, not about the value of any tsum);
  2. Solidarity: if xxx is recurrent, then every state y∈Vy \in Vy∈V is recurrent (in the same R(y,t)→0R(y,t) \to 0R(y,t)→0 sense).

The dichotomy direction is exact: recurrence is equated with non-summability of the return sequence, so equivalently, xxx fails to be recurrent iff ∑tP(t)(x,x)\sum_t P^{(t)}(x,x)∑t​P(t)(x,x) is a genuinely summable series.

Human review
  • Endorsed by Community (Bot) · Aug 22, 2026

  • Endorsed by Shuze Chen · Aug 22, 2026

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

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