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Theorem 9.12 -- Rayleigh's monotonicity law

Proved
MarkovMixing.rayleigh_monotonicity

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

markov-chainsmixing-timesprobability

Let ccc and c′c'c′ be two networks on the same finite vertex set — symmetric nonnegative conductance functions, each with strictly positive total conductance c(x)=∑yc(x,y)c(x)=\sum_yc(x,y)c(x)=∑y​c(x,y) at every vertex and an irreducible associated walk P(x,y)=c(x,y)/c(x)P(x,y)=c(x,y)/c(x)P(x,y)=c(x,y)/c(x). For distinct vertices a≠za\ne za=z, the effective resistance Rc(a↔z)R_c(a\leftrightarrow z)Rc​(a↔z) is defined through the voltage W(x)=Px{τa<τz}W(x)=\mathbb P_x\{\tau_a<\tau_z\}W(x)=Px​{τa​<τz​} and the current ∥I∥=∑yc(a,y)[W(a)−W(y)]\|I\|=\sum_yc(a,y)[W(a)-W(y)]∥I∥=∑y​c(a,y)[W(a)−W(y)] as Rc(a↔z)=∥I∥−1R_c(a\leftrightarrow z)=\|I\|^{-1}Rc​(a↔z)=∥I∥−1.

The theorem (Rayleigh's Monotonicity Law, Theorem 9.12 of Levin–Peres–Wilmer) asserts: if c′(x,y)≤c(x,y)c'(x,y)\le c(x,y)c′(x,y)≤c(x,y) on every edge — resistances are only increased — then

Rc(a↔z)  ≤  Rc′(a↔z).R_c(a\leftrightarrow z)\;\le\;R_{c'}(a\leftrightarrow z).Rc​(a↔z)≤Rc′​(a↔z).

Decreasing conductances can only increase effective resistance. Deceptively simple, this is one of the most-used facts of the theory: it lets one bound resistances in a complicated network by deleting edges (setting conductances to zero) or by comparison with a tractable subnetwork, and through the commute-time identity it transfers to monotonicity statements for hitting times.

Preamble
import Definitions.Def_mm_network
Formal statement
namespace MarkovMixing

/-- **Theorem 9.12, Rayleigh's Monotonicity Law** (LPW): decreasing
conductances (increasing resistances) can only increase the effective
resistance: if `c' ≤ c` edgewise, then `R_c(a↔z) ≤ R_{c'}(a↔z)`. -/
theorem rayleigh_monotonicity {V : Type*} [Fintype V] [DecidableEq V]
    (c c' : V → V → ℝ) (hc : IsConductance c) (hc' : IsConductance c')
    (hpos : ∀ x : V, 0 < vertexConductance c x)
    (hpos' : ∀ x : V, 0 < vertexConductance c' x)
    (hirr : Irreducible (networkWalk c)) (hirr' : Irreducible (networkWalk c'))
    (hle : ∀ x y : V, c' x y ≤ c x y) (a z : V) (haz : a ≠ z) :
    effectiveResistance c a z ≤ effectiveResistance c' a z := 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 9.4, Theorem 9.12, p. 123
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What the Lean code literally says, in plain math · claude-fable-5

Let VVV be a finite type with decidable equality, and let c,c′:V×V→Rc, c' : V \times V \to \mathbb{R}c,c′:V×V→R be two functions each satisfying nonnegativity and symmetry — c(x,y)≥0c(x,y) \ge 0c(x,y)≥0, c(x,y)=c(y,x)c(x,y) = c(y,x)c(x,y)=c(y,x), and likewise for c′c'c′ (self-loop values c(x,x)c(x,x)c(x,x) are permitted). Write c(x)=∑y∈Vc(x,y)c(x) = \sum_{y \in V} c(x,y)c(x)=∑y∈V​c(x,y) and c′(x)=∑y∈Vc′(x,y)c'(x) = \sum_{y \in V} c'(x,y)c′(x)=∑y∈V​c′(x,y) for the vertex conductances, and assume both are strictly positive at every vertex: c(x)>0c(x) > 0c(x)>0 and c′(x)>0c'(x) > 0c′(x)>0 for all x∈Vx \in Vx∈V. Let PPP and P′P'P′ be the transition matrices P(x,y)=c(x,y)/c(x)P(x,y) = c(x,y)/c(x)P(x,y)=c(x,y)/c(x) and P′(x,y)=c′(x,y)/c′(x)P'(x,y) = c'(x,y)/c'(x)P′(x,y)=c′(x,y)/c′(x), and assume each is irreducible in the sense that 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 (respectively (P′t)(x,y)>0(P'^t)(x,y) > 0(P′t)(x,y)>0; since P0P^0P0 is the identity, the case x=yx = yx=y is automatic). Assume the pointwise comparison c′(x,y)≤c(x,y)c'(x,y) \le c(x,y)c′(x,y)≤c(x,y) for all x,y∈Vx, y \in Vx,y∈V, and fix two vertices a≠za \ne za=z. For a conductance function bbb, the effective resistance Rb(a,z)\mathcal{R}_b(a,z)Rb​(a,z) is defined as follows: the voltage vb(x)v_b(x)vb​(x) is the probability that the Markov chain with transition matrix b(x,y)/b(x)b(x,y)/b(x)b(x,y)/b(x) started at xxx reaches aaa strictly before zzz — concretely vb(x)=∑t=0∞∑ω∏i=0t−1b(ωi,ωi+1)b(ωi)v_b(x) = \sum_{t=0}^{\infty} \sum_{\omega} \prod_{i=0}^{t-1} \frac{b(\omega_i,\omega_{i+1})}{b(\omega_i)}vb​(x)=∑t=0∞​∑ω​∏i=0t−1​b(ωi​)b(ωi​,ωi+1​)​, summed over paths ω:{0,…,t}→V\omega : \{0,\dots,t\} \to Vω:{0,…,t}→V with ω0=x\omega_0 = xω0​=x, ωt=a\omega_t = aωt​=a, ωi≠a\omega_i \ne aωi​=a for i<ti < ti<t, and ωi≠z\omega_i \ne zωi​=z for all i≤ti \le ti≤t (so vb(a)=1v_b(a) = 1vb​(a)=1, vb(z)=0v_b(z) = 0vb​(z)=0) — the current strength is Ib=∑y∈Vb(a,y) (vb(a)−vb(y))I_b = \sum_{y \in V} b(a,y)\,\bigl(v_b(a) - v_b(y)\bigr)Ib​=∑y∈V​b(a,y)(vb​(a)−vb​(y)), and Rb(a,z)=Ib−1\mathcal{R}_b(a,z) = I_b^{-1}Rb​(a,z)=Ib−1​, with the Lean convention that the reciprocal of 000 is 000 (so if the current strength for either network were 000, the corresponding effective resistance would be the junk value 000). Under all of these hypotheses, the theorem asserts the single inequality

Rc(a,z)≤Rc′(a,z),\mathcal{R}_c(a,z) \le \mathcal{R}_{c'}(a,z),Rc​(a,z)≤Rc′​(a,z),

i.e. the effective resistance between aaa and zzz computed from the larger conductance function ccc is less than or equal to the effective resistance computed from the smaller conductance function c′c'c′ (non-strict inequality, for this one fixed pair of vertices).

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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