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Theorem 11.2 -- the Matthews upper bound on cover times

Proved
MarkovMixing.matthews_upper

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 n=∣V∣n=|V|n=∣V∣ states. Write Ex(τy)\mathbb E_x(\tau_y)Ex​(τy​) for the expected number of steps to reach yyy from xxx, and define the two extremal quantities

thit=max⁡x,y∈V Ex(τy),tcov=max⁡x∈V Ex(τcov),t_{\mathrm{hit}}=\max_{x,y\in V}\ \mathbb E_x(\tau_y),\qquad t_{\mathrm{cov}}=\max_{x\in V}\ \mathbb E_x(\tau_{\mathrm{cov}}),thit​=x,y∈Vmax​ Ex​(τy​),tcov​=x∈Vmax​ Ex​(τcov​),

where τcov\tau_{\mathrm{cov}}τcov​ is the cover time — the first time the chain has visited every state — so tcovt_{\mathrm{cov}}tcov​ is the worst-case expected time to see the whole state space.

The theorem (the Matthews method, Theorem 11.2 of Levin–Peres–Wilmer) asserts:

tcov  ≤  thit(1+12+⋯+1n).t_{\mathrm{cov}}\;\le\;t_{\mathrm{hit}}\Bigl(1+\frac12+\cdots+\frac1n\Bigr).tcov​≤thit​(1+21​+⋯+n1​).

Covering costs at most a harmonic-sum factor ≈log⁡n\approx\log n≈logn beyond the worst single hitting time. The proof is a striking randomization trick: reveal the states in uniformly random order and bound the expected extra time to collect each new one; the kkk-th freshly revealed state is the last of kkk to be visited with probability 1/k1/k1/k.

Preamble
import Definitions.Def_mm_network
Formal statement
namespace MarkovMixing

/-- **Theorem 11.2, the Matthews method** (LPW): for an irreducible chain on
`n` states, `t_cov ≤ t_hit (1 + 1/2 + ⋯ + 1/n)`. -/
theorem matthews_upper {V : Type*} [Fintype V] [DecidableEq V] [Nonempty V]
    (P : Matrix V V ℝ) (hP : IsStochastic P) (hirr : Irreducible P) :
    coverTimeMax P ≤
      hitTimeMax P * ∑ k ∈ Finset.Icc 1 (Fintype.card V), (1 : ℝ) / k := 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 11.2, Theorem 11.2, pp. 143-144
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What the Lean code literally says, in plain math · claude-fable-5

Read-back: matthews_upper

Let VVV be a finite, nonempty type (with decidable equality) and let PPP be a V×VV \times VV×V matrix of real numbers, 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: ∑yP(x,y)=1\sum_y P(x,y) = 1∑y​P(x,y)=1 for every xxx) and "irreducible" in the following sense: for every pair x,y∈Vx, y \in Vx,y∈V there exists a natural number t≥0t \ge 0t≥0 with (Pt)(x,y)>0(P^t)(x,y) > 0(Pt)(x,y)>0 — the exponent t=0t = 0t=0 is allowed, and P0P^0P0 is the identity, so the case x=yx = yx=y holds trivially. For a length-ttt vertex sequence ω=(ω0,…,ωt)\omega = (\omega_0, \dots, \omega_t)ω=(ω0​,…,ωt​) let its weight be ∏i=0t−1P(ωi,ωi+1)\prod_{i=0}^{t-1} P(\omega_i, \omega_{i+1})∏i=0t−1​P(ωi​,ωi+1​) (the empty product 111 when t=0t=0t=0). Define three quantities by unfolding the custom definitions in the statement. (i) For x,y∈Vx, y \in Vx,y∈V, the "expected hitting time" is h(x,y)=∑t=0∞ut(x,y)h(x,y) = \sum_{t=0}^{\infty} u_t(x,y)h(x,y)=∑t=0∞​ut​(x,y), where ut(x,y)u_t(x,y)ut​(x,y) is the total weight of all length-ttt sequences starting at ω0=x\omega_0 = xω0​=x that avoid yyy at every index including index 000; probabilistically utu_tut​ is the chance the walk from xxx has not visited yyy through time ttt, so h(x,y)h(x,y)h(x,y) is the expected first-hitting time of yyy from xxx — and in the diagonal case x=yx = yx=y every term is 000 (the start point itself violates avoidance), so h(x,x)=0h(x,x) = 0h(x,x)=0, not an expected return time. (ii) The quantity written thit=sup⁡(x,y)∈V×Vh(x,y)t_{\mathrm{hit}} = \sup_{(x,y) \in V \times V} h(x,y)thit​=sup(x,y)∈V×V​h(x,y) is the supremum of hhh over all ordered pairs, diagonal pairs included (harmless, since they contribute 000). (iii) The "expected cover time from xxx" is C(x)=∑t=0∞rt(x)C(x) = \sum_{t=0}^{\infty} r_t(x)C(x)=∑t=0∞​rt​(x), where rt(x)r_t(x)rt​(x) is the total weight of all length-ttt sequences starting at xxx for which some vertex y∈Vy \in Vy∈V is avoided at every index 0,…,t0, \dots, t0,…,t; probabilistically rt(x)r_t(x)rt​(x) is the chance that the walk from xxx has not yet visited every vertex by time ttt, so C(x)C(x)C(x) is the expected time to cover VVV; and tcov=sup⁡x∈VC(x)t_{\mathrm{cov}} = \sup_{x \in V} C(x)tcov​=supx∈V​C(x). All infinite series here are Lean tsums, which take the value 000 whenever the series fails to be summable, and the suprema are real-number suprema over finite nonempty index sets (hence genuine maxima). With n=∣V∣≥1n = |V| \ge 1n=∣V∣≥1 the cardinality of VVV, the theorem asserts the single non-strict inequality

tcov  ≤  thit⋅∑k=1n1k,t_{\mathrm{cov}} \;\le\; t_{\mathrm{hit}} \cdot \sum_{k=1}^{n} \frac{1}{k},tcov​≤thit​⋅k=1∑n​k1​,

i.e. the largest expected cover time over all starting states is at most the largest expected pairwise hitting time multiplied by the harmonic number Hn=1+12+⋯+1nH_n = 1 + \tfrac12 + \dots + \tfrac1nHn​=1+21​+⋯+n1​ (here the sum runs over the integers kkk from 111 to nnn inclusive, so H1=1H_1 = 1H1​=1 when VVV has a single element, in which case both tcovt_{\mathrm{cov}}tcov​ and thitt_{\mathrm{hit}}thit​ are 000 and the claim is trivially true). No aperiodicity, reversibility, or stationarity assumption appears — only stochasticity and the irreducibility condition above.

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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