Chapter 4

Coupling

Coupling is a quintessial probabilistic technique with a wide range of applications.The idea behind the coupling method is that, to compare two probability measuresµ and ⌫, it is sometimes useful to construct a joint probability space with marginalsµ and ⌫. For instance, in the classical application of coupling to the convergenceof Markov chains (Theorem 1.22), one simultaneously constructs two copies of aMarkov chain—one of which is at stationarity—and shows that they can be madeto coincide after a random amount of time called the coupling time.

In this chapter, we discuss a number of applications of the coupling method indiscrete probability, including applications of the important concept of stochasticdomination and its connection to correlation inequalities.

4.1 Background

Throughout this section, we will denote by µZ the law of random variable Z.

4.1.1 Basic definitions

A formal defintion of coupling follows. Recall that for measurable spaces (S1

,S1

)(S

2

,S2

), we can consider the product space (S1

⇥ S2

,S1

⇥ S2

) where

S1

⇥ S2

:= {(s1

, s2

) : s1

2 S1

, s2

2 S2

}

is the Cartesian product of S1

and S2

, and S1

⇥ S2 is the smallest �-field S1

⇥ S2

containing the rectangles A1

⇥A2

for all A1

2 S1

and A2

2 S2

.

Version: September 9, 2015Modern Discrete Probability: An Essential ToolkitCopyright c� 2015 Sebastien Roch

148

Definition 4.1 (Coupling). Let µ and ⌫ be probability measures on the same mea-surable space (S,S). A coupling of µ and ⌫ is a probability measure � on the

couplingproduct space (S ⇥ S,S ⇥ S) such that the marginals of � coincide with µ and ⌫,i.e.,

�(A⇥ S) = µ(A) and �(S ⇥A) = ⌫(A), 8A 2 S.Similarly, for two random variables X and Y taking values in (S,S), a couplingof X and Y is a joint variable (X 0, Y 0) taking values in (S⇥S,S ⇥S) whose lawis a coupling of the laws of X and Y . Note that, under this definition, X and Yneed not be defined on the same probability space—but X 0 and Y 0 do need to. Wealso say that (X 0, Y 0) is a coupling of µ and ⌫ if the law of (X 0, Y 0) is a couplingof µ and ⌫.

We give a few examples.

Example 4.2 (Coupling of Bernoulli variables). Let X and Y be Bernoulli randomvariables with parameters 0 q < r 1 respectively. That is, P[X = 0] = 1� qand P[X = 1] = q, and similarly for Y . Here S = {0, 1} and S = 2S .

- (Independent coupling) One coupling of X and Y is (X 0, Y 0) where X 0 d

= X

and Y 0 d

= Y are independent. Its law is

⇣

P[(X 0, Y 0) = (i, j)]⌘

i,j2{0,1}=

✓

(1� q)(1� r) (1� q)rq(1� r) qr

◆

.

- (Monotone coupling) Another possibility is to pick U uniformly at randomin [0, 1], and set X 00 = {Uq} and Y 00 = {Ur}. The law of coupling(X 00, Y 00) is Then (X 00, Y 00) is a coupling of X and Y with law

⇣

P[(X 00, Y 00) = (i, j)]⌘

i,j2{0,1}=

✓

1� r r � q0 q

◆

.

J

Example 4.3 (Bond percolation: monotonicity). Let G = (V,E) be a countablegraph. Denote by Pp the law of bond percolation on G with density p. Let x 2 Vand assume 0 q < r 1. Using the coupling in the previous example on eachedge independently produces a coupling of Pq and Pr. More precisely:

- Let {Ue}e2E be independent uniforms on [0, 1].

- For p 2 [0, 1], let Wp be the set of edges e such that Ue p.

149

Thinking of Wp as specifying the open edges in the percolation process on G underPp, we see that (Wq,Wr) is a coupling of Pq and Pr with the property that P[Wq ✓Wr] = 1. Let C(q)

x and C(r)x be the open clusters of x under Wq and Wr respectively.

Because C(q)x ✓ C(r)

x ,

✓(q) := Pq[|Cx| = +1]

= P[|C(q)x | = +1] P[|C(r)

x | = +1]

= Pr[|Cx| = +1]

= ✓(r),

as claimed in Section 2.1.3. JExample 4.4 (Biased random walk on Z). For p 2 [0, 1], let (S(p)

t ) be nearest-neighbor random walk on Z started at 0 with probability p of jumping to the rightand probability 1� p of jumping to the left. Assume 0 q < r 1. Using againthe coupling of Bernoulli variables above we produce a coupling of S(q) and S(r).

- Let (X 00i , Y

00i )i be an infinite sequence of i.i.d. monotone Bernoulli couplings

with parameters q and r respectively.

- Define (Z(q)i , Z(r)

i ) := (2X 00i � 1, 2Y 00

i � 1). Note that P[2X 001

� 1 = 1] =P[X 00

1

= 1] = q and P[2X 001

� 1 = �1] = P[X 001

= 0] = 1� q.

- Let S(q)t =

P

it Z(q)i and S(r)

t =P

it Z(r)i .

Then (S(q)t , S(r)

t ) is a coupling of (S(q)t , S(r)

t ) such that S(q)t S(r)

t for all n almostsurely. So for all y and all t

P[S(q)t y] = P[S(q)

t y] � P[S(r)t y] = P[S(r)

t y].

J

4.1.2 . Harmonic functions on lattices and trees

Let (Xt) be a Markov chain on a (finite or) countable state space V with transitionmatrix P and let Px be the law of (Xt) started at x.⇤ Recall that a function h :V ! R is P -harmonic on V (or harmonic for short) if

h(x) =X

y2VP (x, y)h(y), 8x 2 V.

We first give a coupling-based criterion for harmonic functions to be constant.⇤Requires: Section 3.3.1.

150

Lemma 4.5 (Coupling and bounded harmonic functions). If, for all y, z 2 V , thereis a coupling ((Yt), (Zt)) of Py and Pz such that

limt

P[Yt 6= Zt] = 0,

then all bounded harmonic functions on V are constant.

Proof. Let h be bounded and harmonic on V with supx |h(x)| = M < +1. Lety, z be any points in V . By harmonicity, (h(Yt)) and (h(Zt)) are martingales and,in particular,

E[h(Yt)] = E[h(Y0

)] = h(y) and E[h(Zt)] = E[h(Z0

)] = h(z).

So by Jensen’s inequality and the boundedness assumption

|h(y)�h(z)| = |E[h(Yt)]� E[h(Zt)]| E |h(Yt)� h(Zt)| 2M P[Yt 6= Zt]! 0.

So h(y) = h(z).

Harmonic functions on Zd Consider simple random walk on Zd, d � 1. In thatcase, we show that all bounded harmonic functions are constant.

Theorem 4.6 (Bounded harmonic functions on Zd). All bounded harmonic func-tions on Zd are constant.

Proof. Clearly, h is harmonic with respect to simple random walk if and only if itis harmonic with respect to lazy simple random walk. Let Py and Pz be the lawsof lazy simple random walk on Zd started at y and z. We construct a coupling((Yt), (Zt)) = ((Y (i)

t )i2[d], (Z(i)t )i2[d]) of Py and Pz as follows: at time t, pick a

coordinate I 2 [d] uniformly at random, then

• if Y (I)t = Z(I)

t then do nothing with probability 1/2 and otherwise pickW 2 {�1,+1} uniformly at random, set Y (I)

t+1

= Z(I)t+1

:= Z(I)t + W and

leave the other coordinates unchanged;

• if instead Y (I)t 6= Z(I)

t , pick W 2 {�1,+1} uniformly at random, and withprobability 1/2 set Y (I)

t+1

:= Y (I)t +W and leave Zt and the other coordinates

of Yt unchanged, or otherwise set Z(I)t+1

:= Z(I)t + W and leave Yt and the

other coordinates of Zt unchanged.

151

It is straightforward to check that ((Yt), (Zt)) is indeed a coupling of Py and Pz .To apply the previous lemma, it remains to bound P[Yt 6= Zt].

The key is to note that, for each coordinate i, the difference (Y (i)t � Z(i)

t ) isitself a random walk on Z started at y(i) � z(i) with holding probability 1 � 1

d—until it hits 0. Simple random walk on Z is irreducible and recurrent. The holdingprobability does not affect the type of the walk, as can be seen for instance fromthe characterization in terms of effective resistance. So (Y (i)

t �Z(i)t ) hits 0 in finite

time with probability 1. Hence, letting ⌧ (i) be the first time Y (i)t � Z(i)

t = 0, wehave P[Y (i)

t 6= Z(i)t ] P[⌧ (i) > t]! P[⌧ (i) = +1] = 0.

By a union bound,

P[Yt 6= Zt] X

i2[d]P[Y (i)

t 6= Z(i)t ]! 0,

as desired.

Harmonic functions on Td On trees, the situation is different. Let Td be theinfinite d-regular tree with root ⇢. For x 2 Td, we let Tx be the subtree, rooted atx, of descendants of x.

Theorem 4.7 (Bounded harmonic functions on Td). For d � 3, let (Xt) be simplerandom walk on Td and let P be the corresponding transition matrix. Let a be aneighbor of the root and consider the function

h(x) = Px[Xt 2 Ta for all but finitely many t].

Then h is a non-constant, bounded P -harmonic function on Td.

Proof. The function h is clearly bounded and by the usual one-step trick

h(x) =X

y⇠x

1

dPy[Xt 2 T

0

for all but finitely many t] =X

y

P (x, y)h(y),

so h is P -harmonic.Let b 6= a be a neighbor of the root. The key of the proof is the following

lemma.

Lemma 4.8.q := Pa[⌧⇢ = +1] = Pb[⌧⇢ = +1] > 0.

152

Proof. The second equality follows by symmetry. To see that q > 0, let (Zt)be simple random walk on Td started at a until the walk hits 0 and let Lt be thegraph distance between Zt and the root. Then (Lt) is a biased random walk onZ started at 1 jumping to the right with probability 1 � 1

d and jumping to the leftwith probability 1

d . The probability that (Lt) hits 0 in finite time is < 1 because1� 1

d > 2 when d � 3.

Note thath(⇢)

✓

1� 1

d

◆

(1� q) < 1.

Indeed if on the first step the random walk started at ⇢ moves away from a, anevent of probability 1 � 1

d , then it must come back to ⇢ in finite time to reach Ta.Similarly, by the strong Markov property,

h(a) = q + (1� q)h(⇢).

Since h(⇢) 6= 1 and q > 0, this shows that h(a) > h(⇢).

4.2 Coupling inequality

In the examples above, we used coupling to prove monotonicity statements. Cou-pling is also useful to bound the distance between probability measures.

4.2.1 Bounding the total variation distance via coupling

Let µ and ⌫ be probability measures on (S,S). Recall the definition of the totalvariation distance

kµ� ⌫kTV

:= supA2S

|µ(A)� ⌫(A)|.

Lemma 4.9 (Coupling inequality). Let µ and ⌫ be probability measures on (S,S).For any coupling (X,Y ) of µ and ⌫,

kµ� ⌫kTV

P[X 6= Y ].

Proof. For any A 2 S ,

µ(A)� ⌫(A) = P[X 2 A]� P[Y 2 A]

= P[X 2 A, X = Y ] + P[X 2 A, X 6= Y ]

� P[Y 2 A, X = Y ]� P[Y 2 A, X 6= Y ]

= P[X 2 A, X 6= Y ]� P[Y 2 A, X 6= Y ]

P[X 6= Y ],

153

and, similarly, ⌫(A)� µ(A) P[X 6= Y ]. Hence

|µ(A)� ⌫(A)| P[X 6= Y ].

Here is a simple example.

Example 4.10 (A coupling of Poisson random variables). Let X ⇠ Poi(�) andY ⇠ Poi(⌫) with � > ⌫. Recall that a sum of independent Poisson is Poisson. Thisfact leads to a natural coupling: let Y ⇠ Poi(⌫), Z ⇠ Poi(� � ⌫) independentlyof Y , and X = Y + Z. Then (X, Y ) is a coupling and

kµX � µY kTV

P[X 6= Y ] = P[Z > 0] = 1� e�(��⌫) �� ⌫.

J

In fact, the inequality in Lemma 4.9 is tight. For simplicity, we prove this inthe finite case only.

Lemma 4.11 (Maximal coupling). Assume S is finite and let S = 2S . Let µ and ⌫be probability measures on (S,S). Then,

kµ� ⌫kTV

= inf{P[X 6= Y ] : coupling (X,Y ) of µ and ⌫}.

Proof. We construct a coupling which achieves equality in the coupling inequality.Such a coupling is called a maximal coupling.

maximalcoupling

Let A = {x 2 S : µ(x) > ⌫(x)}, B = {x 2 S : µ(x) ⌫(x)} and

p :=X

x2Sµ(x) ^ ⌫(x), ↵ :=

X

x2A[µ(x)� ⌫(x)], � :=

X

x2B[⌫(x)� µ(x)].

First, two lemmas. See Figure 4.1 for a proof by picture.

Lemma 4.12.X

x2Sµ(x) ^ ⌫(x) = 1� kµ� ⌫k

TV

.

154

Figure 4.1: Proof by picture that: 1� p = ↵ = � = kµ� ⌫kTV

.

Proof. We have

2kµ� ⌫kTV

=X

x2S|µ(x)� ⌫(x)|

=X

x2A[µ(x)� ⌫(x)] +

X

x2B[⌫(x)� µ(x)]

=X

x2Aµ(x) +

X

x2B⌫(x)�

X

x2Sµ(x) ^ ⌫(x)

= 2�X

x2Bµ(x)�

X

x2A⌫(x)�

X

x2Sµ(x) ^ ⌫(x)

= 2� 2X

x2Sµ(x) ^ ⌫(x).

Lemma 4.13.X

x2A[µ(x)� ⌫(x)] =

X

x2B[⌫(x)� µ(x)] = kµ� ⌫k

TV

= 1� p.

Proof. The first equality is immediate. The second equality follows from the sec-ond line in the proof of the previous lemma.

155

The maximal coupling is defined as follows:

- With probability p, pick X = Y from �min

where

�min

(x) :=1

pµ(x) ^ ⌫(x), x 2 S.

- Otherwise, pick X from �A where

�A(x) :=µ(x)� ⌫(x)

1� p, x 2 A,

and, independently, pick Y from

�B(x) :=⌫(x)� µ(x)

1� p, x 2 B.

Note that X 6= Y in that case because A and B are disjoint.

The marginal law of X at x 2 S is

p �min

(x) + (1� p) �A(x) = µ(x),

and similarly for Y . Finally P[X 6= Y ] = 1� p = kµ� ⌫kTV

.

Remark 4.14. A proof of this result for general Polish spaces can be found in [dH,Section 2.5].

We return to our coupling of Bernoulli variables.

Example 4.15 (Coupling of Bernoulli variables (continued)). Let X and Y beBernoulli random variables with parameters 0 q < r 1 respectively. That is,P[X = 0] = 1 � q and P[X = 1] = q, and similarly for Y . Here S = {0, 1}and S = 2S . Let µ and ⌫ be the laws of X and Y respectively. To construct themaximal coupling as above, we note that

p :=X

x

µ(x) ^ ⌫(x) = (1� r) + q, 1� p = ↵ = � := r � q,

A := {0}, B := {1},

(�min

(x))x=0,1 =

✓

1� r

(1� r) + q,

q

(1� r) + q

◆

, �A(0) := 1, �B(1) := 1.

156

The law of the maximal coupling (X 000, Y 000) is given by

⇣

P[(X 000, Y 000) = (i, j)]⌘

i,j2{0,1}=

✓

p �min

(0) (1� p) �A(0)�B(1)0 p �

min

(1)

◆

=

✓

1� r r � q0 q

◆

,

which coincides with the monotone coupling. J

Poisson approximation Here is a classical application of coupling. Let X1

, . . . , Xn

be independent Bernoulli random variables with parameters p1

, . . . , pn respec-tively. We are interested in the case where the pis are “small.” Let Sn :=

P

inXi.We approximate Sn with a Poisson random variable Zn as follows: let W

1

, . . . ,Wn

be independent Poisson random variables with means �1

, . . . ,�n respectively anddefine Zn :=

P

inWi. We choose �i = � log(1� pi) so as to ensure

(1� pi) = P[Xi = 0] = P[Wi = 0] = e��i .

Note that Zn ⇠ Poi(�) where � =P

in �i.

Theorem 4.16 (Poisson approximation).

kµSn � Poi(�)kTV

1

2

X

in

�2

i .

Proof. We couple the pairs (Xi,Wi) independently for i n. Let

W 0i ⇠ Poi(�i) and X 0

i = W 0i ^ 1.

Because �i = � log(1 � pi), (X 0i,W

0i ) is a coupling of (Xi,Wi). Let S0

n :=P

inX0i and Z 0

n :=P

inW0i . Then (S0

n, Z0n) is a coupling of (Sn, Zn). By the

coupling inequality

kµSn � µZnkTV

P[S0n 6= Z 0

n] X

in

P[X 0i 6= W 0

i ] =X

in

P[W 0i � 2]

=X

in

X

j�2

e��i�ji

j!X

in

�2

i

2

X

`�0

e��i�`i

`!=X

in

�2

i

2.

157

Maps reduce total variation distance The following lemma will be useful.

Lemma 4.17 (Mappings). Let X and Y be random variables taking values in(S,S), let h be a measurable map from (S,S) to (S0,S 0), and let X 0 := h(X) andY 0 := h(Y ). It holds that

kµX0 � µY 0kTV

kµX � µY kTV

.

Proof. It holds that

supA02S0

�

�P[X 0 2 A0]� P[Y 0 2 A0]�

� = supA02S0

�

�P[h(X) 2 A0]� P[h(Y ) 2 A0]�

�

= supA02S0

�

�P[X 2 h�1(A0)]� P[Y 2 h�1(A0)]�

�

= supA2S

|P[X 2 A]� P[Y 2 A]| .

4.2.2 . Erdos-Renyi graphs: degree sequence

Let Gn ⇠ Gn,pn be an Erdos-Renyi graph with pn := �n and � > 0. For i 2 [n],

let Di(n) be the degree of vertex i and define

Nd(n) :=nX

i=1

{Di(n)=d}.

Theorem 4.18 (Erdos-Renyi graphs: degree sequence).

1

nNd(n)!p

fd := e���d

d!, 8d � 1.

Proof. We proceed in two steps:

1. we use the coupling inequality (Lemma 4.9) to show that the expectation of1

nNd(n) is close to fd;

2. we use Chebyshev’s inequality (Theorem 2.13) to show that 1

nNd(n) is closeto its expectation.

We prove each step as a lemma below.

Lemma 4.19 (Convergence of the mean).

1

nEn,pn [Nd(n)]! fd, 8d � 1.

158

Proof. Note that the Di(n)s are identically distributed (but not independent) so1

nEn,pn [Nd(n)] = Pn,pn [D1

(n) = d]. Moreover D1

(n) ⇠ Bin(n � 1, pn). LetSn ⇠ Bin(n, pn) and Zn ⇠ Poi(�). By the Poisson approximation

kµSn�µZnkTV

1

2

X

in

(� log(1� pn))2 =

1

2

X

in

✓

�

n+O(n�2)

◆

2

=�2

2n+O(n�2).

We can couple D1

(n) and Sn as (P

in�1

Xi,P

inXi) where the Xis are i.i.d. Bernoulliwith parameter �

n . By the coupling inequality

kµD1

(n) � µSnkTV

P

2

4

X

in�1

Xi 6=X

in

Xi

3

5 = P[Xn = 1] =�

n.

By the triangle inequality for total variation distance,

1

2

X

d�0

|Pn,pn [D1

(n) = d]� fd| �+ �2/2

n+O(n�2).

Therefore,�

�

�

�

1

nEn,pn [Nd(n)]� fd

�

�

�

�

2�+ �2

n+O(n�2)! 0.

Lemma 4.20 (Concentration around the mean).

Pn,pn

�

�

�

�

1

nNd(n)� 1

nEn,pn [Nd(n)]

�

�

�

�

� "

�

2�+ 1

"2n, 8d � 1, 8n.

Proof. By Chebyshev’s inequality, for all " > 0

Pn,pn

�

�

�

�

1

nNd(n)� 1

nEn,pn [Nd(n)]

�

�

�

�

� "

�

Varn,pn [1

nNd(n)]

"2.

159

Note that

Varn,pn

1

nNd(n)

�

=1

n2

8

<

:

En,pn

2

4

0

@

X

in

{Di(n)=d}

1

A

2

3

5� (nPn,pn [D1

(n) = d])2

9

=

;

=1

n2

n

n(n� 1)Pn,pn [D1

(n) = d,D2

(n) = d]

+ nPn,pn [D1

(n) = d]� n2Pn,pn [D1

(n) = d]2o

1

n+n

Pn,pn [D1

(n) = d,D2

(n) = d]� Pn,pn [D1

(n) = d]2o

.

We bound the second term using a neat coupling argument. Let Y1

and Y2

beindependent Bin(n� 2, pn) and let X

1

and X2

be independent Ber(pn). Then theterm in curly bracket above is equal to

P[(X1

+ Y1

, X1

+ Y2

) = (d, d)]� P[(X1

+ Y1

, X2

+ Y2

) = (d, d)]

P[(X1

+ Y1

, X1

+ Y2

) = (d, d), (X1

+ Y1

, X2

+ Y2

) 6= (d, d)]

= P[(X1

+ Y1

, X1

+ Y2

) = (d, d), X2

+ Y2

6= d]

= P[X1

= 0, Y1

= Y2

= d, X2

= 1] + P[X1

= 1, Y1

= Y2

= d� 1, X2

= 0]

2�

n.

So Varn,pn⇥

1

nNd(n)⇤ 2�+1

n .

Combining the lemmas concludes the proof of Theorem 4.18.

4.3 Stochastic domination

In comparing two probability measures, a natural relationship to consider is the no-tion of “domination.” To see why this might be useful, let (Xi)

ni=1

be independentZ+

-valued random variables with

P[Xi � 1] � p,

and let S =Pn

i=1

Xi be their sum. Consider also a random variable

S⇤ ⇠ Bin(n, p).

160

Then it is intuitively clear that one should be able to obtain some bounds on S bystudying S⇤ instead—which may be easier. Indeed, in some sense, S “dominates”S⇤, that is, S should have a tendency to be bigger than S⇤. One expects morespecifically that

P[S > x] � P[S⇤ > x].

Coupling provides a handy characterization of this notion, as we detail in this sec-tion.

In particular we study an important special case known as positive association.In that case a measure “dominates itself” in the following sense: conditioning oncertain events makes other events more likely. This concept, which is formalizedin Section 4.3.4, has numerous applications in discrete probability.

4.3.1 Definitions

We start with the case of real random variables.

Ordering of real random variables For real random variables, stochastic dom-ination is defined as follows. See Figure 4.2 for an illustration.

Definition 4.21 (Stochastic domination). Let µ and ⌫ be probability measures onR. The measure µ is said to stochastically dominate ⌫, denoted µ ⌫ ⌫, if for all

stochasticdomination

x 2 Rµ⇥

(x,+1)⇤ � ⌫

⇥

(x,+1)⇤

.

A real random variable X stochastically dominates Y , denoted by X ⌫ Y , if thelaw of X dominates the law of Y .

Example 4.22 (Bernoulli vs. Poisson). Let X ⇠ Poi(�) be Poisson with mean� > 0 and let Y be a Bernoulli trial with success probability p 2 (0, 1), i.e.,P[Y = 1] = 1 � P[Y = 0] = p. In order for X to stochastically dominate Y , itsuffices to have

P[X > `] � P[Y > `], 8` � 0.

This is always true for ` � 1 since P[X > `] > 0 but P[Y > `] = 0. So it remainsto consider the case ` = 0. We have

1� e�� = P[X > 0] � P[Y > 0] = p,

if and only if� � � log(1� p).

J

161

Figure 4.2: The law of X , represented here by its cumulative distribution functionFX in red, stochastically dominates the law of Y , in orange. The construction of amonotone coupling, (X, Y ) := (F�1

X (U), F�1

Y (U)) where U is uniform in [0, 1],is also depicted.

162

Note that stochastic domination does not require X and Y to be defined onthe same probability space. The connection to coupling arises from the followingcharacterization.

Theorem 4.23 (Coupling and stochastic domination). The real random variableX stochastically dominates Y if and only if there is a coupling (X, Y ) of X andY such that

P[X � Y ] = 1. (4.1)

We refer to (X, Y ) as a monotone coupling of X and Y . monotonecouplingProof. One direction is clear. Suppose there is such a coupling. Then for all x 2 R

P[Y > x] = P[Y > x] = P[X � Y > x] P[X > x] = P[X > x].

For the other direction, define the cumulative distribution functions FX(x) =P[X x] and FY (x) = P[Y x]. Assume X ⌫ X 0. The idea of the proofis to use the following standard way of generating a real random variable. Recall(e.g. [Dur10, Section 1.2]) that

Xd

= F�1

X (U), (4.2)

where U is a [0, 1]-valued uniform random variable and

F�1

X (u) := inf{x 2 R : FX(x) � u},

is a generalized inverse. Indeed F�1

X (u) > x precisely when u > FX(x). Itis natural to construct a coupling of X and Y by simply using the same uniformrandom variable U in this representation, i.e., we define X = F�1

X (U) and Y =F�1

Y (U). See Figure 4.2. By (4.2), this is a coupling of X and Y . It remainsto check (4.1). Because FX(x) FY (x) for all x by definition of stochasticdomination, by the definition of the generalized inverse,

P[X � Y ] = P[F�1

X (U) � F�1

Y (U)] = 1,

as required.

Example 4.24. Returning to the example in the first paragraph of Section 4.3, let(Xi)

ni=1

be independent Z+

-valued random variables with P[Xi � 1] � p andconsider their sum S :=

Pni=1

Xi. Further let S⇤ ⇠ Bin(n, p). Write S⇤ as thesum

Pni=1

Yi where (Yi) are independent {0, 1}-variables with P[Yi = 1] = p. Tocouple S and S⇤, first set (Yi) := (Yi) and S⇤ :=

Pni=1

Yi. Let Xi be 0 wheneverYi = 0. Otherwise, i.e. if Yi = 1, generate Xi according to the distribution of

163

Xi conditioned on {Xi � 1}, independently of everything else. By constructionXi � Yi a.s. for all i and as a result

Pni=1

Xi =: S � S⇤ a.s. or S ⌫ S⇤ bythe previous theorem. That implies for instance that P[S > x] � P[S⇤ > x] aswe claimed earlier. A special case of this argument gives the following useful factabout binomials

n � m, q � p =) Bin(n, q) ⌫ Bin(m, p).

J

Example 4.25 (Poisson distribution). Let X ⇠ Poi(µ) and Y ⇠ Poi(⌫) with µ >⌫. Recall that a sum of independent Poisson is Poisson (use moment-generatingfunctions or see e.g. [Dur10, Exercise 2.1.14]). This fact leads to a natural cou-pling: let Y ⇠ Poi(⌫), Z ⇠ Poi(µ � ⌫) independently of Y , and X = Y + Z.Then (X, Y ) is a coupling and X � Y a.s. because Z � 0. Hence X ⌫ Y . J

We record two useful consequences of Theorem 4.23.

Corollary 4.26. Let X and Y be real random variables with X ⌫ Y and letf : R ! R be a non-decreasing function. Then f(X) ⌫ f(Y ) and furthermore,provided E|f(X)|,E|f(Y )| < +1, we have that

E[f(X)] � E[f(Y )].

Proof. Let (X, Y ) be the monotone coupling of X and Y whose existence is guar-anteed by Theorem 4.23. Then f(X) � f(Y ) a.s. so that, provided the expecta-tions exist,

E[f(X)] = E[f(X)] � E[f(Y )] = E[f(Y )],

and furthermore (f(X), f(Y )) is a monotone coupling of f(X) and f(Y ). Hencef(X) ⌫ f(Y ).

Corollary 4.27. Let X1

, X2

be independent random variables. Let Y1

, Y2

be inde-pendent random variables such that Xi ⌫ Yi, i = 1, 2. Then

X1

+X2

⌫ Y1

+ Y2

.

Proof. Let (X1

, Y1

) and (X2

, Y2

) be independent, monotone couplings of (X1

, Y1

)and (X

2

, Y2

) (on the same probability space). Then

X1

+X2

⇠ X1

+ X2

� Y1

+ Y2

⇠ Y1

+ Y2

.

164

Example 4.28 (Binomial vs. Poisson). A sum of n Poisson variables with mean� is Poi(n�). A sum of n Bernoulli trials with success probability p is Bin(n, p).Using Example 4.22 and Corollary 4.27, we get

� � � log(1� p) =) Poi(n�) ⌫ Bin(n, p). (4.3)

The following special case will be useful later. Let 0 < ⇤ < 1 and let m be aninteger. Then

⇤

m� 1� ⇤

m� ⇤=

m

m� ⇤� 1 � log

✓

m

m� ⇤

◆

= � log

✓

1� ⇤

m

◆

,

where we used that log x x � 1 for all x 2 R. So, setting � := ⇤

m�1

, p := ⇤

mand n := m� 1 in (4.3), we get

⇤ 2 (0, 1) =) Poi(⇤) ⌫ Bin

✓

m� 1,⇤

m

◆

. (4.4)

J

Ordering on partially ordered sets The definition of stochastic dominationhinges on the totally ordered nature of R. It also extends naturally to posets. Let(X ,) be a poset, i.e., for all x, y, z 2 X : poset

- [Reflexivity] x x,

- [Antisymmetry] if x y and y x then x = y,

- [Transitivity] if x y and y z then x z.

For instance the set {0, 1}F is a poset when equipped with the relation x y ifand only if xi yi for all i 2 F . Equivalently the subsets of F , denoted by 2F ,form a poset with the inclusion relation. (A totally ordered set satisfies in additionthat, for any x, y, we have either x y or y x.)

Let F be a �-field over the poset X . An event A 2 F is increasing if x 2 Aincreasing event,function

implies that any y � x is also in A. A function f : X ! R is increasing if x yimplies f(x) f(y).

Definition 4.29 (Stochastic domination for posets). Let (X ,) be a poset and letF be a �-field on X . Let µ and ⌫ be probability measures on (X ,F). The measureµ is said to stochastically dominate ⌫, denoted by µ ⌫ ⌫, if for all increasing

stochasticdomination forposets

A 2 Fµ(A) � ⌫(A).

An X -valued random variable X stochastically dominates Y , denoted by X ⌫ Y ,if the law of X dominates the law of Y .

165

As before, a monotone coupling (X, Y ) of X and Y is one which satisfies monotonecoupling forposets

X � Y a.s.

Example 4.30 (Monotonicity of the percolation function). We have already seenan example of stochastic domination in Section 2.1.3. Consider bond percolationon the d-dimensional lattice Ld. Here the poset is the collection of all subsetsof edges, specifying the open edges, with the inclusion relation. Recall that thepercolation function is given by

✓(p) := Pp[|C0| = +1],

where C0

is the open cluster of the origin. We argued in Section 2.1.3 that ✓(p) isnon-decreasing by considering the following alternative representation of the per-colation process under Pp: to each edge e, assign a uniform [0, 1]-valued randomvariable Ue and declare the edge open if Ue p. Using the same Ues for twodifferent p-values, p

1

< p2

, gives a monotone coupling of the processes for p1

and p2

. It follows immediately that ✓(p1

) ✓(p2

), where we used that the event{|C

0

| = +1} is increasing. J

The existence of a monotone coupling is perhaps more surprising for posets.We prove the result in the finite case only, which will be enough for our purposes.

Theorem 4.31 (Strassen’s theorem). Let X and Y be random variables takingvalues in a finite poset (X ,) with the �-field F = 2X . Then X ⌫ Y if and onlyif there exists a monotone coupling (X, Y ) of X and Y .

Proof. One direction is clear. Suppose there is such a coupling. Then for all in-creasing A

P[Y 2 A] = P[Y 2 A] = P[X � Y 2 A] P[X 2 A] = P[X 2 A].

The proof in the other direction relies on the max-flow min-cut theorem (The-orem 1.9). To see the connection with flows, let µX and µY be the laws of X andY respectively, and denote by ⌫ their joint distribution under the desired coupling.Noting that we want ⌫(x, y) = 0 if x y, the marginal conditions on the couplingread

X

yx

⌫(x, y) = µX(x), 8x 2 X ,

andX

x�y

⌫(x, y) = µY (y), 8y 2 X .

166

Figure 4.3: Construction of a monotone coupling through the max-flow represen-tation for independent Bernoulli pairs with parameters r (on the left) and q < r(on the right). Edge labels indicate capacity. Edges without labels have infinitecapacity. The colored edges depict a suboptimal cut. The blue and orange verticescorrespond respectively to the sets A⇤ and Z⇤ for this cut. The capacity of the cutis r2 + r(1� r) + (1� q)2 + (1� q)q = r + (1� q) > r + (1� r) = 1.

These equations can be interpreted as flow-conservation constraints. Considerthe following directed graph. There are two vertices, (w, 1) and (w, 2), for each el-ement w in X with edges connecting each (x, 1) to those (y, 2)s with x � y. Theseedges have capacity +1. In addition there is a source a and a sink z. The sourcehas a directed edge of capacity µX(x) to (x, 1) for each x 2 X and, similarly, each(y, 2) has a directed edge of capacity µY (y) to the sink. The existence of a mono-tone coupling will follow once we show that there is a flow of strength 1 between aand z. Indeed, in that case, all edges from the source and all edges to the sink are atcapacity. If we let ⌫(x, y) be the flow on edge h(x, 1), (y, 2)i, the constraints abovethen impose the conservation of the flow on the vertices (X ⇥ {1}) [ (X ⇥ {2}).Hence the flow between X ⇥ {1} and X ⇥ {2} yields the desired coupling. SeeFigure 4.3.

By the max-flow min-cut theorem (Theorem 1.9), it suffices to show that aminimum cut has capacity 1. Such a cut is of course obtained by choosing alledges out of the source. So it remains to show that no cut has capacity less than 1.

167

This is where we use the fact that µX(A) � µY (A) for all increasing A. Becausethe edges between X ⇥ {1} and X ⇥ {2} have infinite capacity, they cannot beused in a minimum cut. So we can restrict our attention to those cuts containingedges from a to A⇤ ⇥ {1} and from Z⇤ ⇥ {2} to z for subsets A⇤, Z⇤ ✓ X . Wemust have

A⇤ ◆ {x 2 X : 9y 2 Zc⇤, x � y},

to block all paths of the form a ⇠ (x, 1) ⇠ (y, 2) ⇠ z with x and y as above. Infact, for a minimum cut, we further have

A⇤ = {x 2 X : 9y 2 Zc⇤, x � y},

as adding an x not satisfying this property is redundant. See Figure 4.3. In particu-lar A⇤ is increasing: if x

1

2 A⇤ and x2

� x1

, then 9y 2 Zc⇤ such that x1

� y and,since x

2

� x1

� y, the same y works for x2

. Observe that, because y � y, the setA⇤ also includes Zc⇤. If it were the case that A⇤ 6= Zc⇤, then we could construct acut with lower or equal capacity by fixing A⇤ and setting Z⇤ := Ac⇤: because A⇤ isincreasing, any y 2 A⇤ \Z⇤ is such that paths of the form a ⇠ (x, 1) ⇠ (y, 2) ⇠ zwith x � y are cut by x 2 A⇤. Hence, for a minimum cut, we can assume that infact A⇤ = Zc⇤. The capacity of the cut is

µX(A⇤) + µY (Z⇤) = µX(A⇤) + 1� µY (A⇤) = 1 + (µX(A⇤)� µY (A⇤)) � 1,

where the term in parenthesis is nonnegative by assumption and the fact that A⇤ isincreasing. That concludes the proof.

Remark 4.32. Strassen’s theorem holds more generally on Polish spaces with a closedpartial order. See e.g. [Lin02, Section IV.1.2] for the details.

The proof of Corollary 4.26 immediately extends to:

Corollary 4.33. Let X and Y be X -valued random variables with X ⌫ Y andlet f : X ! R be an increasing function. Then f(X) ⌫ f(Y ) and furthermore,provided E|f(X)|,E|f(Y )| < +1, we have that

E[f(X)] � E[f(Y )].

Ordering of Markov chains Stochastic domination also arises in the context ofMarkov chains. We begin with an example.

Example 4.34 (Lazier chain). Consider a random walk (Xt) on the network N =((V,E), c) where V = {0, 1, . . . , n} and i ⇠ j if and only |i � j| 1 (includingself-loops). Let N 0 = ((V,E), c0) be a modified version of N on the same graph

168

where for all i c(i, i) c0(i, i). That is, if (X 0t) is random walk on N 0, then (X 0

t) is“lazier” than (Xt) in that it is more likely to stay put. Assume that both (Xt) and(X 0

t) start at i0

and define Ms := maxtsXt and M 0s := maxtsX 0

t. Since (X 0t)

“travels less” than (Xt) the following claim is intuitively obvious:

Claim 4.35.Ms ⌫M 0

s.

We prove this by producing a monotone coupling. First set (Xt) := (Xt). We thengenerate (X 0

t) as a “sticky” version of (Xt). That is, (X 0t) follows exactly the same

transitions as (Xt) (including the self-loops), but at each time it opts to stay whereit currently is, say j, for an extra time step with probability

c0(j, j)� c(j, j)P

i⇠j c0(i, j)

,

which is in [0, 1] by assumption. Marginally, (X 0t) is a random walk on N 0 because

by construction

c0(j, j)P

i⇠j c0(i, j)

=c0(j, j)� c(j, j)P

i⇠j c0(i, j)

+

P

i⇠j c(i, j)P

i⇠j c0(i, j)

!

c(j, j)P

i⇠j c(i, j),

and for i 6= j with i ⇠ j

c0(i, j)P

i⇠j c0(i, j)

=

P

i⇠j c(i, j)P

i⇠j c0(i, j)

!

c(i, j)P

i⇠j c(i, j),

since c0(i, j) = c(i, j). This coupling satisfies

cMs := maxts

Xt � maxts

X 0t =: cM 0

s, a.s.

because (X 0t)ts visits a subset of the states visited by (Xt)ts. In other words

(cMs, cM 0s) is a monotone coupling of (Ms,M 0

s) and this proves the claim. JThe previous example involved an asynchronous coupling of the chains. Often,

a simpler step-by-step approach is possible.

Definition 4.36 (Stochastic domination for Markov kernels). Let P and Q be tran-sition matrices on a finite or countable poset (X ,). The transition matrix Q issaid to stochastically dominate the transition matrix P if

stochasticdomination forMarkov chains

x y =) P (x, ·) � Q(y, ·). (4.5)

If the above condition is satisfied for P = Q, we say that P is stochasticallymonotone.

stochasticmonotonicity

169

The equivalent of Strassen’s theorem in this case is the following theorem,which we prove in the finite case only again.

Theorem 4.37 (Strassen’s theorem for Markov kernels). Let (Xt) and (Yt) beMarkov chains on a finite poset (X ,) with transition matrices P and Q respec-tively. Assume that Q stochastically dominates P . Then for all x

0

y0

there is acoupling (Xt, Yt) of (Xt) started at x

0

and (Yt) started at y0

such that a.s.

Xt Yt, 8t.Furthermore, if the chains are irreducible and have stationary distributions ⇡ andµ respectively, then ⇡ � µ.

Observe that, for a step-by-step monotone coupling to exist, it is not generallyenough for the weaker condition P (x, ·) � Q(x, ·) to hold for all x, as shouldbe clear from the proof. Also you should convince yourself that the chains inExample 4.34 do not in general satisfy (4.5). (Which pairs x, y cause problems?)

Proof of Theorem 4.37. Let

W := {(x, y) 2 X ⇥ X : x y}.For all (x, y) 2 W , let R((x, y), ·) be the joint law of a monotone coupling ofP (x, ·) and Q(y, ·). Such a coupling exists by Strassen’s theorem and Condi-tion (4.5). Let (Xt, Yt) be a Markov chain on W with transition matrix R startedat (x

0

, y0

). By construction, Xt Yt for all t a.s. That proves the first half of thetheorem.

For the second half, let A be increasing in X . Then, by the ergodic theorem forMarkov chains (e.g. [Dur10, Exercise 6.6.4]),

⇡(A) 1

t

X

st

ˆXs2A 1

t

X

st

ˆYs2A ! µ(A), a.s.

where we used that Xs 2 A implies Ys 2 A because Xs Ys and A is increasing.This proves the claim by definition of stochastic domination.

An example of application of this theorem is given in the next subsection.

4.3.2 . Ising model on Zd: extremal measures

Consider the d-dimensional lattice Ld. Let ⇤ be a finite subset of vertices in Ld

and define X := {�1,+1}⇤, which is a poset when equipped with the relation� �0 if and only if �i �0

i for all i 2 ⇤. For shorthand, we occasionally write +

170

and � instead of +1 and �1. For ⇠ 2 {�1,+1}Ld , recall that the (ferromagnetic)Ising model on ⇤ with boundary conditions ⇠ and inverse temperature � is the

boundaryconditions,inversetemperature,spins

probability distribution over spin configurations � 2 X given by

µ⇠�,⇤(�) :=

1

Z⇤,⇠(�)

e��H⇤,⇠(�),

whereH

⇤,⇠(�) := �X

i⇠ji,j2⇤

�i�j �X

i⇠ji2⇤,j /2⇤

�i⇠j ,

is the Hamiltonian andHamiltonian

Z⇤,⇠(�) :=

X

�2Xe��H

⇤,⇠(�),

is the partition function. (Warning: it is easy to get confused with the � signs thatpartitionfunction

cancel out in the exponent.) For the all-(+1) and all-(�1) boundary conditionswe write respectively µ+

�,⇤(�) and µ��,⇤(�). In this section, we show that these

two measures are extremal in the following sense. For all boundary conditions⇠ 2 {�1,+1}Ld :

Claim 4.38.µ+

�,⇤ ⌫ µ⇠�,⇤ ⌫ µ�

�,⇤.

In words, because the ferromagnetic Ising model favors spin agreement, the all-(+1) boundary condition tends to produce more +1s which in turn makes increas-ing events more likely.

The idea of the proof is to use Theorem 4.37 with a suitable Markov chain.

Stochastic domination In this context, vertices are often referred to as sites.site

Recall that the single-site Glauber dynamics of the Ising model is the Markov chainon X which, at each time, selects a site i 2 ⇤ uniformly at random and updates thespin �i according to µ⇠

�,⇤(�) conditioned on agreeing with � at all sites in ⇤\{i}.Specifically, for � 2 {�1,+1}, i 2 ⇤, and � 2 X , let �i,� be the configuration �with the state at i being set to �. Then, letting n = |⇤|, because the Ising measurefactorizes, the transition matrix of the Glauber dynamics is simply

Q⇠�,⇤(�,�

i,�) :=1

n· e��S

⇠i (�)

e��S⇠i (�) + e�S

⇠i (�)

,

171

whereS⇠i (�) :=

X

j⇠ij2⇤

�j +X

j⇠ij /2⇤

⇠j .

All other transitions have probability 0.This chain is clearly irreducible. It is also reversible with respect to µ⇠

�,⇤.Indeed, for all � 2 X and i 2 ⇤, letting

S⇠6=i(�) := H

⇤,⇠(�i,+) + S⇠

i (�) = H⇤,⇠(�

i,�)� S⇠i (�),

we have

µ⇠�,⇤(�

i,�)Q⇠�,⇤(�

i,�,�i,+) =e��S⇠

6=i(�)e��S⇠i (�)

Z⇤,⇠(�)

· e�S⇠i (�)

n[e��S⇠i (�) + e�S

⇠i (�)]

=e��S⇠

6=i(�)

nZ⇤,⇠(�)[e��S⇠

i (�) + e�S⇠i (�)]

=e��S⇠

6=i(�)e�S⇠i (�)

Z⇤,⇠(�)

· e��S⇠i (�)

n[e��S⇠i (�) + e�S

⇠i (�)]

= µ⇠�,⇤(�

i,+)Q⇠�,⇤(�

i,+,�i,�).

In particular µ⇠�,⇤ is the stationary distribution of Q⇠

�,⇤.

Claim 4.39.

⇠0 � ⇠ =) Q⇠0

�,⇤ stochastically dominates Q⇠�,⇤. (4.6)

Proof. Because the Glauber dynamics updates a single site at a time, establishingstochastic domination reduces to checking simple one-site inequalities:

Lemma 4.40. To establish (4.6), it suffices to show that, for all � ⌧ ,

Q⇠�,⇤(�,�

i,+) Q⇠0

�,⇤(⌧, ⌧i,+). (4.7)

Proof. Assume (4.7) holds. Let A be increasing in X and let � ⌧ . Then, for thesingle-site Glauber dynamics, we have

Q⇠�,⇤(�, A) = Q⇠

�,⇤(�, A \B�), (4.8)

whereB� := {�i,� : i 2 ⇤, � 2 {�1,+1}},

172

and similarly for ⌧ , ⇠0. Moreover, because A is increasing and ⌧ � �,

�i,� 2 A =) ⌧ i,� 2 A, (4.9)

and�i,� 2 A =) �i,+ 2 A. (4.10)

Letting

I±�,A := {i 2 ⇤ : �i,� 2 A}, I+�,A := {i 2 ⇤ : �i,+ 2 A},

and similarly for ⌧ , we have by (4.7), (4.8), (4.9), and (4.10),

Q⇠�,⇤(�, A) = Q⇠

�,⇤(�, A \B�)

=X

i2I+�,A

Q⇠�,⇤(�,�

i,+) +X

i2I±�,A

h

Q⇠�,⇤(�,�

i,�) +Q⇠�,⇤(�,�

i,+)i

X

i2I+�,A

Q⇠0

�,⇤(⌧, ⌧i,+) +

X

i2I±�,A

1

n

X

i2I+⌧,A

Q⇠0

�,⇤(⌧, ⌧i,+) +

X

i2I±⌧,A

h

Q⇠0

�,⇤(⌧, ⌧i,�) +Q⇠0

�,⇤(⌧, ⌧i,+)

i

= Q⇠0

�,⇤(⌧, A),

as claimed.

Returning to the proof of Claim 4.39, observe that

Q⇠�,⇤(�,�

i,+) =1

n· e�S

⇠i (�)

e��S⇠i (�) + e�S

⇠i (�)

=1

n· 1

e�2�S⇠i (�) + 1

,

which is increasing in S⇠i (�). Now � ⌧ and ⇠ ⇠0 imply that S⇠

i (�) S⇠0

i (⌧).That proves the claim.

Finally:

Proof of Claim 4.38. Combining Theorem 4.37 and Claim 4.39 gives Claim 4.38.

Observe that we have not used any special property of the d-dimensional lattice.Indeed Claim 4.38 in fact holds for any countable, locally finite graph with positivecoupling constants.

173

Thermodynamic limit To be written. See [RAS, Theorem 9.13].

4.3.3 . Random walk on trees: speed

To be written. See [LP, Proposition 13.3 and Exercise 13.1].†

4.3.4 Correlation inequalities: FKG and Holley’s inequalities

A special case of stochastic domination is positive associations. In this section,we restrict ourselves to posets of the form {0, 1}F for F finite. We begin with anexample.

Example 4.41 (Erdos-Renyi graphs: positive associations). Consider an Erdos-Renyi graph G ⇠ Gn,p. Let E = {{x, y} : x, y 2 [n], x 6= y}. Think of G astaking values in the poset ({0, 1}E ,) where a 1 indicates that the correspondingedge is present. In fact observe that the law of G, which we denote as usual by Pn,p,is a product measure on {0, 1}E . The event A that G is connected is increasingbecause adding edges cannot disconnect a connected graph. So is the event Bof having a chromatic number larger than 4. Intuitively then, conditioning on Amakes B more likely. Indeed the occurence of A tends to be accompanied witha larger number of edges which in turn makes B more probable. This is a moregeneral phenomenon. That is, for any non-empty increasing events A and B, wehave:

Claim 4.42.Pn,p[B |A] � Pn,p[B]. (4.11)

Or, put differently, the conditional measure Pn,p[ · | A] stochastically dominatesthe unconditional measure Pn,p[ · ]. This is a special case of what is known asHarris’ inequality (proved below). Note that (4.11) is equivalent to Pn,p[A \ B] �Pn,p[A]Pn,p[B], i.e., to the fact that A and B are positively correlated. J

More generally:

Definition 4.43 (Positive associations). Let µ be a probability measure on {0, 1}Fwhere F is finite. Then µ is said to have positive associations, or is positively

positiveassociations

associated, if for all increasing functions f, g : {0, 1}F ! R

µ(fg) � µ(f)µ(g),

†Requires: Section 2.2.3.

174

whereµ(h) :=

X

!2{0,1}Fµ(!)h(!).

In particular, for any increasing events A and B it holds that

µ(A \B) � µ(A)µ(B),

i.e., A and B are positively correlated.positivecorrelationRemark 4.44. Note that positive associations is concerned only with increasing events.

See Remark 4.63.

Remark 4.45. A notion of negative associations, which is a somewhat more delicate con-cept, was defined in Remark 3.109. See also [Pem00].

Let µ be positively associated. Note that if A and B are decreasing, i.e. theirdecreasingevent

complements are increasing, then

µ(A \B) = 1� µ(Ac [Bc)

= 1� µ(Ac)� µ(Bc) + µ(Ac \Bc)

� 1� µ(Ac)� µ(Bc) + µ(Ac)µ(Bc)

= µ(A)µ(B).

Similarly, if A is increasing and B is decreasing, we have µ(A\B) µ(A)µ(B).Harris’ inequality states that product measures on {0, 1}F have positive asso-

ciations. We prove a more general result known as the FKG inequality. For twoconfigurations !, !0 in {0, 1}F , we let ! ^ !0 and ! _ !0 be the coordinatewiseminimum and maximum of ! and !0.

Theorem 4.46 (FKG inequality). Let X = {0, 1}F where F is finite. Suppose µ isa positive probability measure on X satisfying the FKG condition

FKG condition

µ(! _ !0)µ(! ^ !0) � µ(!)µ(!0), 8!,!0 2 X . (4.12)

This property is also known as log-convexity or log-supermodularity. We call sucha measure an FKG measure. Then µ has positive associations.

FKG measure

Remark 4.47. Strict positivity is not in fact needed [FKG71]. The FKG condition isequivalent to a strong form of positive associations. See Exercise 4.4.

175

Note that product measures satisfy the FKG condition with equality. Indeed ifµ(!) is of the form

Q

f2F µf (!f ) then

µ(! _ !0)µ(! ^ !0) =Y

f

µf (!f _ !0f ) µf (!f ^ !0

f )

=Y

f :!f=!0f

µf (!f )2

Y

f :!f 6=!0f

µf (!f )µf (!0f )

=Y

f :!f=!0f

µf (!f )µf (!0f )

Y

f :!f 6=!0f

µf (!f )µf (!0f )

= µ(!)µ(!0).

So the FKG inequality applies, for instance, to bond percolation and Erdos-Renyigraphs. The pointwise nature of the FKG condition also makes it relatively easy tocheck for measures, such as that of the Ising model, which are defined explicitlyup to a normalizing constant.

Example 4.48 (Ising model on Zd: checking FKG). Consider again the settingof Section 4.3.2. Of course we work on the space X := {�1,+1}⇤ rather than{0, 1}F . Fix ⇤ ✓ Ld, ⇠ 2 {�1,+1}Ld and � > 0.

Claim 4.49. The measure µ⇠�,⇤ satisfies the FKG condition and therefore has pos-

itive associations.

Intuitively, taking the maximum or minimum of two configurations tends to in-crease spin agreement and therefore leads to a higher likelihood. By taking loga-rithms in the FKG condition, one sees that proving the claim boils down to check-ing an inequality for each term in the Hamiltonian. For �,�0 2 X , let ⌧ = � _ �0

and ⌧ = � ^ �0. When i 2 ⇤ and j /2 ⇤ such that i ⇠ j, we have

⌧ i⇠j + ⌧ i⇠j = (⌧ i + ⌧ i)⇠j = (�i + �0i)⇠j = �i⇠j + �0

i⇠j . (4.13)

For i, j 2 ⇤ with i ⇠ j, note first that the case �j = �0j reduces to the previous

calculation, so we assume �i 6= �0i and �j 6= �0

j . Then

⌧ i⌧ j + ⌧ i⌧ j = (+1)(+1) + (�1)(�1) = 2 � �i�j + �0i�

0j ,

since 2 is the largest value the rightmost expression ever takes. We have shown that

H⇤,⇠(⌧) +H

⇤,⇠(⌧) H⇤,⇠(�) +H

⇤,⇠(�0),

which implies the claim.

176

Again, we have not used any special property of the lattice and the same resultholds for countable, locally finite graphs with positive coupling constants. Notehowever that in the anti-ferromagnetic case, i.e., if we multiply the Hamiltonian by�1, the above argument does not work. Indeed there is no reason to expect positiveassociations in that case. J

The FKG inequality in turn follows from a more general result known as Hol-ley’s inequality.

Theorem 4.50 (Holley’s inequality). Let X = {0, 1}F where F is finite. Supposeµ1

and µ2

are positive probability measures on X satisfying

µ2

(! _ !0)µ1

(! ^ !0) � µ2

(!)µ1

(!0), 8!,!0 2 X . (4.14)

Then µ1

� µ2

.

Before proving Holley’s inequality, we check that it indeed implies the FKGinequality. See Exercise 4.1 for an elementary proof in the independent case, i.e.,of Harris’ inequality.

Proof of Theorem 4.46. Assume that µ satisfies the FKG condition and let f , g beincreasing functions. Because of our restriction to positive measures in Holley’sinequality, we will work with positive functions. This is done without loss of gen-erality. Indeed, note that f and g are increasing if and only if f 0 := f�f(0)+1 > 0and g0 := g � g(0) + 1 > 0 are increasing and that, moreover,

µ(f 0g0)� µ(f 0)µ(g0) = µ([f 0 � µ(f 0)][g0 � µ(g0)])= µ([f � µ(f)][g � µ(g)])

= µ(fg)� µ(f)µ(g).

In Holley’s inequality, we let µ1

:= µ and define the positive probability mea-sure

µ2

(!) :=g(!)µ(!)

µ(g).

We check that µ1

and µ2

satisfy the conditions of Holley’s inequality. Note that!0 !_!0 for any ! so that, because g is increasing, we have g(!0) g(!_!0).

177

Hence, for any !, !0,

µ1

(!)µ2

(!0) = µ(!)g(!0)µ(!0)

µ(g)

= µ(!)µ(!0)g(!0)µ(g)

µ(! ^ !0)µ(! _ !0)g(! _ !0)

µ(g)

= µ1

(! ^ !0)µ2

(! _ !0),

where on the third line we used the FKG condition satisfied by µ.So Holley’s inequality implies that µ

2

⌫ µ1

. Hence, since f is increasing, byCorollary 4.33

µ(f) = µ1

(f) µ2

(f) =µ(fg)

µ(g),

and the theorem is proved.

Proof of Theorem 4.50. We use Theorem 4.37. This is similar to what was donein Section 4.3.2. This time we couple Metropolis-like chains. For x 2 X and � 2{0, 1}, we let xi,� be x with coordinate i set to �. We write x ⇠ y if kx� yk

1

= 1.Let n = |F |.

For ↵,� > 0 small enough, the following transition matrix over X is irre-ducible and reversible w.r.t. its stationary distribution µ

2

: for all i 2 F , y 2 X ,

Q(yi,0, yi,1) =↵

n{�} ,

Q(yi,1, yi,0) =↵

n

⇢

�µ2

(yi,0)

µ2

(yi,1)

�

,

Q(y, y) = 1�X

z⇠y

Q(y, z).

Let P be similarly defined for µ1

with the same values of ↵ and �. For reasons thatwill be clear below, the value of � is chosen so that the sum of the two expressionsin brackets above is smaller than 1 for all y, i. The value of ↵ is then chosen sothat P (x, x), Q(y, y) � 0 for all x, y. Reversibility follows immediately from thefirst two equations. We call the first transition above an upward transition and the

upward/downwardtransition

second one a downward transition.By Theorem 4.37, it remains to show that Q stochastically dominates P . That

is, for any x y, we want to show that P (x, ·) � Q(y, ·). We produce a monotone

178

coupling (X, Y ) of these two distributions. Our goal is never to perform an upwardtransition in x simultaneously with a downward transition in y. Observe that

µ1

(xi,0)

µ1

(xi,1)� µ

2

(yi,0)

µ2

(yi,1)(4.15)

by taking ! = yi,0 and !0 = xi,1 in Condition (4.14).The coupling works as follows. Fix x y. With probability 1 � ↵, set

(X, Y ) := (x, y). Otherwise, pick a coordinate i 2 F uniformly at random.There are several cases to consider depending on the values of xi, yi (with xi yiby assumption):

- (xi, yi) = (0, 0): With probability �, perform an upward transition in both,i.e., set X := xi,1 and Y := yi,1. With probability 1 � �, set (X, Y ) :=(x, y) instead.

- (xi, yi) = (1, 1): With probability � µ2

(yi,0)µ2

(yi,1), perform a downward transition

in both, i.e., set X := xi,0 and Y := yi,0. With probability

�

✓

µ1

(xi,0)

µ1

(xi,1)� µ

2

(yi,0)

µ2

(yi,1)

◆

,

perform a downward transition in x only, i.e., set X := xi,0 and Y := y.Note that (4.15) guarantees that the previous step is well-defined. With theremaining probability, set (X, Y ) := (x, y) instead.

- (xi, yi) = (0, 1): With probability �, perform an upward transition in x

only, i.e., set X := xi,1 and Y := y. With probability � µ2

(yi,0)µ2

(yi,1), perform

a downward transition in y only, i.e., set X := x and Y := yi,0. With theremaining probability, set (X, Y ) := (x, y) instead. (This is where we usethe odd choice of �.)

By construction, this coupling satisfies X Y a.s. An application of Theo-rem 4.37 concludes the proof.

Example 4.51 (Ising model: extremality revisited). Holley’s inequality gives an-other proof of Claim 4.38. To see this, just repeat the calculations of Example 4.48,where now (4.13) is replaced with an inequality. See Exercise 4.2. J

179

4.3.5 . Erdos-Renyi graph: Janson’s inequality and application to thecontainment problem

Let G = (V,E) ⇠ Gn,p be an Erdos-Renyi graph. Repeating the computations ofSection 2.2.2 (or see Claim 2.21), we see that the property of being triangle-freehas threshold n�1. That is, the probability that G contains a triangle goes to 0 or1 as n ! +1 depending on whether p ⌧ n�1 or p � n�1 respectively. In thissection, we investigate what happens at the threshold. From now on, we assumethat p = �/n for some � > 0 not depending on n.

For any subset S of three distinct vertices of G, let BS be the event that S formsa triangle in G. So

" := Pn,p[BS ] = p3 ! 0. (4.16)

Let Xn =P

S2(V3

) BSbe the number of triangles in G. By the linearity of

expectation, the expected number of triangles is

En,pXn =

✓

n

3

◆

p3 =n(n� 1)(n� 2)

6

✓

�

n

◆

3

! �3

6,

as n ! +1. If the events {BS}S were mutually independent, Xn would bebinomially-distributed and the event that G is triangle-free would have probability

Y

S2(V3

)

Pn,p[BcS ] = (1� p3)(

n3

) ! e��3/6. (4.17)

In fact, by the Poisson approximation to the binomial (e.g. [Dur10, Theorem 3.6.1]),we would have that the number of triangles converges weakly to Poi(�3/6).

In reality, of course, the events {BS} are not mutually independent. Observehowever that, for most pairs S, S0, the events BS and BS0 are in fact pairwiseindependent. That is the case whenever |S \ S0| 1, i.e., whenever the edgesconnecting S are disjoint from those connecting S0. Write S ⇠ S0 if S 6= S0 arenot independent, i.e. if |S \ S0| = 2. The expected number of (unordered) pairsS ⇠ S0 both forming a triangle is

� :=1

2

X

S,S0(V3

)S⇠S0

Pn,p[BS \BS0 ] =1

2

✓

n

3

◆

(n� 3)p5 = ⇥(n4p5)! 0. (4.18)

Given that the events {BS} are “mostly” independent, it is natural to expect thatXn behaves asymptotically as it does in the independent case. Indeed we prove:

Claim 4.52.Pn,p[Xn = 0]! e��3/6.

180

Remark 4.53. In fact, Xn

d! Poi(�3/6). See Exercises 2.11 and 4.5.

The FKG inequality immediately gives one direction. Recall that Pn,p, as aproduct measure over edge sets, satisfies the FKG condition and therefore has pos-itive associations by the FKG inequality. Moreover the events Bc

S are decreasingfor all S. Hence

Pn,p

h

T

S2(V3

)BcS

i

�Y

S2(V3

)

Pn,p[BcS ]! e��3/6,

by (4.17). As it turns out, the FKG inequality also gives a bound in the otherdirection. This is known as Janson’s inequality, which we state in a more generalcontext.

Janson’s inequality Let X := {0, 1}F where F is finite. Let Bi, i 2 I , bea finite collection of events of the form Bi := {! 2 X : ! � �(i)} for some�(i) 2 X . Think of these as “bad events” corresponding to a certain subset ofcoordinates being set to 1. By definition, the Bis are increasing. Assume P is apositive product measure on X . Write i ⇠ j if �(i)

r = �(j)r = 1 for some r and

note that Bi is independent of Bj if i ⌧ j. Set

� :=X

{i,j}i⇠j

P[Bi \Bj ].

Theorem 4.54 (Janson’s inequality). Let X , P, {Bi}i2I and � be as above. As-sume further that there is " > 0 such that P[Bi] " for all i 2 I . Then

Y

i2IP[Bc

i ] P[\i2IBci ] e

�

1�"

Y

i2IP[Bc

i ].

Before proving the theorem, we show that it implies Claim 4.52. We havealready shown in (4.16) and (4.18) that " ! 0 and � ! 0. Janson’s inequalityimmediately implies the claim by (4.17).

Proof of Theorem 4.54. The lower bound follows from the FKG inequality.In the other direction, the first step is somewhat clear. We apply the chain rule

to obtain

P[\i2IBci ] =

mY

i=1

P[Bci | \j2[i�1]

Bcj ].

181

The rest is clever manipulation. W.l.o.g. assume I = [m]. For i 2 [m], let N(i) :={` 2 [m] : ` ⇠ i} and N<(i) := N(i) \ [i � 1]. Note that Bi is independent of{B` : ` 2 [i� 1]\N<(i)}. Then

P[Bi | \j2[i�1]

Bcj ] =

Ph

Bi \⇣

\j2[i�1]

Bcj

⌘i

P[\j2[i�1]

Bcj ]

�Ph

Bi \⇣

\j2[i�1]

Bcj

⌘i

P[\j2[i�1]\N<(i)Bcj ]

= Ph

Bi \�\j2N<(i)B

cj

�

�

�

�

\j2[i�1]\N<(i) Bcj

i

= Ph

Bi

�

�

�

\j2[i�1]\N<(i) Bcj

i

⇥Ph

\j2N<(i)Bcj

�

�

�

Bi \�\j2[i�1]\N<(i)B

cj

�

i

= P [Bi]

⇥Ph

\j2N<(i)Bcj

�

�

�

Bi \�\j2[i�1]\N<(i)B

cj

�

i

,

By a union bound the second term on the last line is

Ph

\j2N<(i)Bcj

�

�

�

Bi \�\j2[i�1]\N<(i)B

cj

�

i

� 1�X

j2N<(i)

Ph

Bj

�

�

�

Bi \�\j2[i�1]\N<(i)B

cj

�

i

� 1�X

j2N<(i)

Ph

Bj

�

�

�

Bi

i

,

where the last line follows from the FKG inequality applied to the product measureP[ · |Bi] (on {0, 1}F 0 with F 0 := {` 2 [m] : �(i)

` = 0}). Combining the last threedisplays and using 1 + z ez , we get

P[\i2IBci ]

mY

i=1

2

4P [Bci ] +

X

j2N<(i)

P [Bi \Bj ]

3

5

mY

i=1

P [Bci ]

2

41 +1

1� "

X

j2N<(i)

P [Bi \Bj ]

3

5

mY

i=1

P [Bci ] exp

0

@

1

1� "

X

j2N<(i)

P [Bi \Bj ]

1

A .

182

By the definition of �, we are done.

4.3.6 . Percolation on Z2: RSW theory and a proof of Harris’ theorem

Consider bond percolation on the two-dimensional lattice L2. Recall that the per-colation function is given by

✓(p) := Pp[|C0| = +1],

where C0

is the open cluster of the origin. We known from Example 4.30 that ✓(p)is non-decreasing. Let

pc(L2) := sup{p � 0 : ✓(p) = 0},

be the critical value. We proved in Section 2.1.3 that there is a non-trivial transition,i.e., pc(L2) 2 (0, 1). In fact we showed that pc(L2) 2 [1/3, 2/3] (see Exercise 2.2).

Our goal in this section is to use the FKG inequality to improve this further to:

Theorem 4.55 (Harris’ theorem).

✓(1/2) = 0.

Or, put differently, pc(L2) � 1/2.

Remark 4.56. This bound is tight, i.e., in fact pc

(L2) = 1/2. The other direction, knownas Kesten’s theorem, is postponed to Section 7.3.2 where an additional ingredient is intro-duced, Russo’s formula.

Several proofs of Harris’ theorem are known. A particularly elegant one issketched in Exercise 6.1. Here we present a proof that uses an important tool inpercolation theory, the RSW lemma, an application of the FKG inequality.

Harris’ theorem To motivate the RSW lemma, we start with the proof of Harris’theorem.

Proof of Theorem 4.55. Fix p = 1/2. We use duality. Consider the L2 annulus

Ann(`) := [�3`, 3`]2\[�`, `].

The existence of a closed dual cycle inside Ann(`), which we denote by Od

(`),prevents the possibility of an infinite open self-avoiding path from the origin in theprimal lattice L2. See Figure 4.4. That is,

183

Figure 4.4: Top: the event Od

(`). Bottom: the event O#

d

(`).

184

P1/2[|C0| = +1]

KY

k=0

{1� P1/2[Od

(3k)]}, (4.19)

for all K, where we took powers of 3 to make the annuli disjoint and thereforeindependent.

To prove the theorem, it suffices to show that there is a constant c⇤ > 0 suchthat, for all `, P

1/2[Od

(`)] � c⇤. Then the r.h.s. of (4.19) tends to 0 as K ! +1.To simplify further, thinking of Ann(`) as a union of four rectangles [�3`,�`) ⇥[�3`, 3`], [�3`, 3`]⇥ (`, 3`], etc., it suffices to consider the event O#

d

(`) that eachone of these rectangles contains a closed dual self-avoiding path connecting itstwo shorter sides. (More precisely, for the first rectangle above, the path connects[�3`+1/2,�`�1/2]⇥{3`�1/2} to [�3`+1/2,�`�1/2]⇥{�3`+1/2} andstays inside the rectangle, etc.) See Figure 4.4. By symmetry the probability thatsuch a path exists is the same for all four rectangles. Denote it by ⇢`. Moreoverthe event that such a path exists is increasing so, although the four events are notindependent, we can apply the FKG inequality. Hence, because O#

d

(`) ✓ Od

(`),we finally get the bound

P1/2[Od

(`)] � ⇢4` .

The RSW lemma and some symmetry arguments, both of which are detailedbelow, imply that there is some c > 0 such that, for all `:

Lemma 4.57.⇢` � c.

That concludes the proof.

It remains to prove Lemma 4.57. We first state the RSW lemma.

RSW theory We have reduced the proof of Harris’ theorem to bounding theprobability that certain closed paths exist in the dual lattice. To be consistent withthe standard RSW notation, we switch to the primal lattice and consider open paths.We also let p take any value in (0, 1).

Let Rn,↵(p) be the probability that the rectangle

B(↵n, n) := [�n, (2↵� 1)n]⇥ [�n, n],

has an open self-avoiding path connecting its left and right sides with the pathremaining inside the rectangle. Such a path is called an (open) left-right crossing.

left-right/top-bottomcrossing

The event that a left-right crossing exists in a rectangle B is denoted by LR(B).We similarly define the event, TB(B), that a top-bottom crossing exists in B. In

185

essence, the RSW lemma says this: if there is a significant probability that a left-right crossing exists in the square B(n, n), then there is a significant probabilitythat a left-right crossing exists in the rectangle B(3n, n). More precisely, here is aversion of the theorem that will be enough for our purposes. (See Exercise 4.6 fora generalization.)

Lemma 4.58 (RSW lemma). For all n � 2 (divisible by 4) and p 2 (0, 1),

Rn,3(p) � 1

4Rn,1(p)

11Rn/2,1(p)12. (4.20)

The r.h.s. of (4.20) depends only on the probability of crossing a square from leftto right. By a duality argument, at p = 1/2, it turns out that this probability is atleast 1/2 independently of n. Before presenting a proof of the RSW lemma, wedetail this argument and finish the proof of Harris’ theorem.

Proof of Lemma 4.57. The point of (4.20) is that, if Rn,1(p) is bounded away from0 uniformly in n, so is the l.h.s. By the argument in the proof of Harris’ theorem,this then implies that an open self-avoiding cycle exists in Ann(n) with a probabil-ity bounded away from 0 as well. Hence to prove Lemma 4.57 it suffices to give alower bound on Rn,1(1/2). It is crucial that this bound not depend on the “scale,”n. As it turns out, a simple duality-based symmetry argument does the trick. Thefollowing fact about L2 is a variant of the contour lemma (Lemma 2.8). Its proof issimilar and Exercise 4.7 asks for the details (the “if” direction being the non-trivialimplication).

Lemma 4.59. There is an open left-right crossing in the primal rectangle [0, n +1]⇥ [0, n] if and only if there is no closed top-bottom crossing in the dual rectangle[1/2, n+ 1/2]⇥ [�1/2, n+ 1/2].

By symmetry, when p = 1/2, the two events in Lemma 4.59 have equal probability.So they must have probability 1/2 because they form a partition of the space ofoutcomes. By monotonicity, that implies Rn,1(1/2) � 1/2 for all n. The RSWlemma then implies the required bound.

The proof of the RSW lemma involves a clever choice of event that relates theexistence of crossings in squares and rectangles.

Proof of Lemma 4.58. There are several steps in the proof.

186

Figure 4.5: Illustration of the implication LR(B01

) \ TB(B01

\ B02

) \ LR(B02

) ✓LR(B(3n, n)).

Step 1: it suffices to bound Rn,3/2(p) We first reduce the proof to finding abound on Rn,3/2(p). Let B0

1

:= B(2n, n) and B02

:= [n, 5n] ⇥ [�n, n]. Notethat B0

1

[ B02

= B(3n, n) and B01

\ B02

= [n, 3n] ⇥ [�n, n]. Then we have theimplication

LR(B01

) \ TB(B01

\B02

) \ LR(B02

) ✓ LR(B(3n, n)).

See Figure 4.5. Each event on the l.h.s. is increasing so the FKG inequality gives

Rn,3(p) � Rn,2(p)2Rn,1(p).

A similar argument over B(2n, n) gives

Rn,2(p) � Rn,3/2(p)2Rn,1(p).

Combining the two, we have proved:

Lemma 4.60 (Proof of RSW: step 1).

Rn,3(p) � Rn,3/2(p)4Rn,1(p)

3. (4.21)

187

Step 2: bounding Rn,3/2(p) The heart of the proof is to bound Rn,3/2(p) usingan event involving crossings of squares. Let

B1

:= B(n, n) = [�n, n]⇥ [�n, n],B

2

:= [0, 2n]⇥ [�n, n],B

12

:= B1

\B2

= [0, n]⇥ [�n, n],S := [0, n]⇥ [0, n].

Let �1

be the event that there are paths P1

, P2

, where P1

is a top-bottom crossingof S and P

2

is an open self-avoiding path connecting the left side of B1

to P1

andstays inside B

1

. Similarly let �02

be the event that there are paths P 01

, P 02

, where P 01

is a top-bottom crossing of S and P 02

is an open self-avoiding path connecting theright side of B

2

to P 01

and stays inside B2

. Then we have the implication

�1

\ LR(S) \ �02

✓ LR(B(3n/2, n)).

See Figure 4.6. By symmetry Pp[�1

] = Pp[�02

]. Moreover, the events on thel.h.s. are increasing so by the FKG inequality:

Lemma 4.61 (Proof of RSW: step 2).

Rn,3/2(p) � Pp[�1

]2Rn/2,1(p). (4.22)

Step 3: bounding Pp[�1

] It remains to bound Pp[�1

]. That requires several ad-ditional definitions. Let P

1

and P2

be top-bottom crossings of S. There is a naturalpartial order over such crossings. The path P

1

divides S into two subgraphs: [P1

}which includes the left side of S (including edges on the left incident with P

1

butnot those edges on P

1

itself) and {P1

] which includes the right side of S (and P1

itself). Then we write P1

� P2

if {P1

] ✓ {P2

]. Assuming TB(S) holds, onealso gets the existence of a unique rightmost crossing. Roughly speaking, take the

rightmostcrossing

union of all top-bottom crossings of S as sets of edges; then the “right boundary”of this set is a top-bottom crossing P ⇤

S such that P ⇤S � P for all top-bottom cross-

ings P of S. (We accept as a fact the existence of a unique rightmost crossing. SeeExercise 4.7 for a related construction.)

Let IS be the set of self-avoiding (not necessarily open) paths connecting thetop and bottom of S and stay inside S. For P 2 IS , we let P 0 be the reflection ofP through the x-axis in B

12

\S and we let PP 0 be the union of P and P 0. Define

[ PP 0 } to be the subgraph of B1

to the left of PP 0 (including edges on the left incident

with PP 0 but not those edges on P

P 0 itself). Let LR+

�

[ PP 0 }�

be the event that thereis a left-right crossing of [ PP 0 } ending on P , i.e., that there is an open self-avoidingpath connecting the left side of B

1

and P that stays within [ PP 0 }. See Figure 4.6.

188

Figure 4.6: Top: illustration of the implication �1

\ LR(S) \ �02

✓LR(B(3n/2, n))1. Bottom: the event LR+

�

[ PP 0 }� \ {P = P ⇤

S}; the dashed pathis the mirror image of the rightmost top-bottom crossing in S; the pink region isthe complement in B

1

of the set [ PP 0 }. Note that, because on the bottom figurethe left-right path must stay within [ PP 0 }, the configuration shown in the top fig-ure where the purple left-right path “travels behind” the top-bottom crossing of Scannot occur.

189

Note that the existence of a left-right crossing of B1

implies the existence of anopen self-avoiding path connecting the left side of B

1

to PP 0 . By symmetry we then

get

Pp

⇥

LR+

�

[ PP 0 }�⇤ � 1

2Pp[LR(B1

)] =1

2Rn,1(p). (4.23)

Now comes a subtle point. We turn to the rightmost crossing of S—for two reasons.First, by uniqueness, {P = P ⇤

S}P2IS forms a partition of TB(S). Second, therightmost crossing has a Markov-like property. Observe that, for P 2 IS , the eventthat {P = P ⇤

S} depends only the bonds in {P ]. In particular it is independent ofthe bonds in [ PP 0 }, e.g. of the event LR+

�

[ PP 0 }�

. Hence

Pp

⇥

LR+

�

[ PP 0 }� |P = P ⇤

S

⇤

= Pp

⇥

LR+

�

[ PP 0 }�⇤

.

With (4.23), we get

Pp[�1

] �X

P2ISPp[P = P ⇤

S ]Pp

⇥

LR+

�

[ PP 0 }� |P = P ⇤

S

⇤

� 1

2Rn,1(p)

X

P2ISPp[P = P ⇤

S ]

=1

2Rn,1(p)Pp[TB(S)]

=1

2Rn,1(p)Rn/2,1(p).

We have proved:

Lemma 4.62 (Proof of RSW: step 3).

Pp[�1

] � 1

2Rn,1(p)Rn/2,1(p). (4.24)

Step 4: putting everything together Combining (4.21), (4.22) and (4.24) givesthat

Rn,3(p) � Rn,3/2(p)4Rn,1(p)

3

� [Pp[�1

]2Rn/2,1(p)]4Rn,1(p)

3

�"

✓

1

2Rn,1(p)Rn/2,1(p)

◆

2

Rn/2,1(p)

#

4

Rn,1(p)3.

Collecting the terms concludes the proof of the RSW lemma.

Remark 4.63. This argument is quite subtle. In that respect, it is instructive to read theremark after [Gri97, Theorem 9.3].

190

4.4 Couplings of Markov chains

As we have seen, coupling is useful to bound total variation distance. A naturalapplication is mixing as we show in this section.

4.4.1 Bounding the mixing time via coupling

Let P be an irreducible, aperiodic Markov transition matrix on the finite state spaceV with stationary distribution ⇡. Recall that, for a fixed 0 < " < 1/2, the mixingtime of P is

tmix

(") := min{t : d(t) "},where

d(t) := maxx2VkP t(x, ·)� ⇡k

TV

.

It will be easier to work with

d(t) := maxx,y2V

kP t(x, ·)� P t(y, ·)kTV

.

The quantities d(t) and d(t) are related in the following way.

Lemma 4.64.d(t) d(t) 2d(t), 8t.

Proof. The second inequality follows from an application of the triangle inequality.For the first inequality, note that by definition of the total variation distance

kP t(x, ·)� ⇡kTV

= supA✓V

|P t(x,A)� ⇡(A)|

= supA✓V

�

�

�

�

�

�

X

y2V⇡(y)[P t(x,A)� P t(y,A)]

�

�

�

�

�

�

supA✓V

X

y2V⇡(y)|P t(x,A)� P t(y,A)|

X

y2V⇡(y)

(

supA✓V

|P t(x,A)� P t(y,A)|)

X

y2V⇡(y)kP t(x, ·)� P t(y, ·)k

TV

maxx,y2V

kP t(x, ·)� P t(y, ·)kTV

.

191

A coupling of Markov chains with transition matrix P is a Markov chain (Xt, Yt)on V ⇥ V such that both (Xt) and (Yt) are Markov chains with transition matrixP . For our purposes, the following special type of coupling will suffice.

Definition 4.65 (Markovian coupling). A Markovian coupling of P is a MarkovMarkoviancoupling

chain (Xt, Yt)t on V ⇥ V with transition matrix Q satisfying:

- (Markovian coupling) For all x, y, x0, y0 2 V ,X

z0

Q((x, y), (x0, z0)) = P (x, x0),

X

z0

Q((x, y), (z0, y0)) = P (y, y0).

We say that a Markovian coupling is coalescing if further:coalescing

- (Coalescing) For all z 2 V ,

x0 6= y0 =) Q((z, z), (x0, y0)) = 0.

Let (Xt, Yt) be a coalescing Markovian coupling of P . By the coalescingcondition, if Xs = Ys then Xt = Yt for all t � s. That is, once (Xt) and (Yt)meet, they remain together. Let ⌧

coal

be the coalescence time (also called couplingcoupling!coalescencetime

time), i.e.,⌧coal

:= inf{t � 0 : Xt = Yt}.The key to the coupling approach to mixing times is the following immediate con-sequence of the coupling inequality (Lemma 4.9). For any starting point (x, y),

kP t(x, ·)� P t(y, ·)kTV

P(x,y)[Xt 6= Yt] = P

(x,y)[⌧coal > t]. (4.25)

Combining (4.25) and Lemma 4.64, we get the main result of this section.

Theorem 4.66 (Bounding the mixing time: coupling method). Let (Xt, Yt) be acoalescing Markovian coupling of an irreducible transition matrix P on a finitestate space V with stationary distribution ⇡. Then

d(t) maxx,y2V

P(x,y)[⌧coal > t].

In particular

tmix

(") inf�

t � 0 : P(x,y)[⌧coal > t] ", 8x, y .

We give a few simple examples in the next subsection.

192

4.4.2 . Markov chains: mixing on cycles, hypercubes, and trees

In this section, we consider lazy simple random walk on various graphs. By this welazy walk

mean that the walk stays put with probability 1/2 and otherwise picks an adjacentvertex uniformly at random. In each case, we construct a coupling to bound themixing time. As a reference, we compare our upper bounds to the diameter-basedlower bound derived in Section 2.3.4. Recall that, by Claim 2.47, for a finite,reversible Markov chain with stationary distribution ⇡ and diameter � we have thelower bound

tmix

(") = ⌦

�2

log(n _ ⇡�1

min

)

!

,

for " > 0, where ⇡min

is the smallest value taken by ⇡.

Cycle Let (Zt) be lazy simple random walk on the cycle of size n, Zn :={0, 1 . . . , n � 1}, where i ⇠ j if |j � i| = 1 (mod n). For any starting pointsx, y, we construct a coalescing Markovian coupling (Xt, Yt) of this chain. Set(X

0

, Y0

) := (x, y). At each time, flip a fair coin. On heads, Yt stays put and Xt

moves one step, the direction of which is uniform at random. On tails, proceedsimilarly with the roles of Xt and Yt reversed. Let Dt be the clockwise distancebetween Xt and Yt. Observe that, by construction, (Dt) is simple random walk on{0, . . . , n} and ⌧

coal

= ⌧D{0,n}, the first time (Dt) hits {0, n}.We use Markov’s inequality, Theorem 2.4, to bound P

(x,y)[⌧D{0,n} > t]. Denote

by D0

= dx,y the starting distance. By Wald’s second equation (e.g. [Dur10,Example 4.1.6]),

E(x,y)

h

⌧D{0,n}i

= dx,y(n� dx,y).

Applying Theorem 4.66 and Markov’s inequality we get

d(t) maxx,y2V

P(x,y)[⌧coal > t]

maxx,y2V

E(x,y)

h

⌧D{0,n}i

t

= maxx,y2V

dx,y(n� dx,y)

t

n2

4t,

or:

193

Claim 4.67.tmix

(") n2

4".

By our diameter-based lower bound on mixing in Section 2.3.4, this boundgives the correct order of magnitude in n up to logarithmic factors. Indeed, thediameter is � = n/2 and ⇡

min

= 1/n so that Claim 2.47 gives

tmix

(") � n2

64 log n,

for n large enough. Exercise 4.9 sketches a tighter lower bound.

Hypercube Let (Zt) be lazy simple random walk on the n-dimensional hyper-cube Zn

2

:= {0, 1}n where i ⇠ j if ki � jk1

= 1. We denote the coordinates ofZt by (Z(1)

t , . . . , Z(n)t ). The coupling (Xt, Yt) started at (x, y) is the following.

At each time t, pick a coordinate i uniformly at random in [n], pick a bit value bin {0, 1} uniformly at random independently of the coordinate choice. Set bothi coordinates to b, i.e., X(i)

t = Y (i)t = b. This is equivalent to performing the

Glauber dynamics chain on an empty graph. Because of the way the updating isdone, the chains stay put with probability 1/2 at each time as required. Clearlythe chains coalesce when all coordinates have been updated at least once. Thefollowing standard bound on the coupon collector problem is what is needed toconclude.

Lemma 4.68 (Coupon collecting). Let ⌧coll

be the time it takes to update eachcoordinate at least once. Then, for any c > 0,

P [⌧coll

> dn log n+ cne] e�c.

Proof. Let Bi be the event that the i-th coordinate has not been updated by timedn log n+ cne. Then

P[⌧coll

> dn log n+ cne] X

i

P[Bi]

=X

i

✓

1� 1

n

◆dn logn+cne

n exp

✓

�n log n+ cn

n

◆

= e�c.

194

Applying Theorem 4.66, we get

d(dn log n+ cne) maxx,y2V

P(x,y)[⌧coal > dn log n+ cne]

P[⌧coll

> dn log n+ cne] e�c.

Hence for c := c" > 0 large enough:

Claim 4.69.tmix

(") dn log n+ c"ne.Again we get a quick lower bound using our diameter-based result from Sec-

tion 2.3.4. Here � = n and ⇡min

= 1/2n so that Claim 2.47 gives

tmix

(") � n2

12 log n+ (4 log 2)n= ⌦(n),

for n large enough. So the upper bound we derived above is off at most by alogarithmic factor in n. In fact:

Claim 4.70.tmix

(") � 1

2n log n�O(n).

Proof. For simplicity, we assume that n is odd. Let Wt be the number of 1s, orHamming weight, at time t. Let A be the event that the Hamming weight is n/2.

Hamming weightTo bound the mixing time, we use the fact that for any z

0

d(t) � kP t(z0

, ·)� ⇡kTV

� |P t(z0

, A)� ⇡(A)|. (4.26)

Under the stationary distribution, the Hamming weight is equal in distribution toa Bin(n, 1/2). In particular the probability that a majority of coordinates are 0 is1/2. That is, ⇡(A) = 1/2.

On the other hand, let (Zt) start at z0

= 1, where 1 is the all-1 vector. LetUt be the number of updated coordinates up to time t in the Glauber dynamicsrepresentation of the chain discussed above. We use Chebyshev’s inequality (The-orem 2.13) to bound the probability of event A at time t. We first need to computethe expectation and variance of Wt. Observe that, conditioned on Ut, the Ham-ming weight Wt is equal in distribution to Bin(Ut, 1/2) + (n�Ut) as the updated

195

coordinates are uniform and the other ones are 1. Thus we have

E[Wt] = E[E[Wt |Ut]]

= E

1

2Ut + (n� Ut)

�

= n� 1

2n

"

1�✓

1� 1

n

◆t#

=n

2

"

1 +

✓

1� 1

n

◆t#

, (4.27)

where on the third line we used that E[Ut] = nh

1� �

1� 1

n

�ti

by linearity ofexpectation, and

Var[Wt] = E[Var[Wt |Ut]] + Var[E[Wt |Ut]]

=1

4E [Ut] +

1

4Var[Ut]. (4.28)

It remains to compute Var[Ut]. Let I(i)t be 1 if coordinate i has not been updatedup to time t and 0 otherwise. Note that for i 6= j

Cov[I(i)t , I(j)t ] = E[I(i)t I(j)t ]� E[I(i)t ]E[I(j)t ]

=

✓

1� 2

n

◆t

�✓

1� 1

n

◆

2t

=

✓

1� 2

n

◆t

�✓

1� 2

n+

1

n2

◆t

0,

that is, I(i)t and I(j)t are negatively correlated, while

Var[I(i)t ] = E[(I(i)t )2]� (E[I(i)t ])2 E[I(i)t ] =

✓

1� 1

n

◆t

.

Then, writing n� Ut as the sum of these indicators, we have

Var[Ut] = Var[n� Ut] =nX

i=1

Var[I(i)t ] + 2X

i<j

Cov[I(i)t , I(j)t ] n

✓

1� 1

n

◆t

.

196

Plugging this back into (4.28), we get

Var[Wt] n

4

"

1�✓

1� 1

n

◆t#

+n

4

✓

1� 1

n

◆t

=n

4.

For t↵ = 1

2

n log n� (log 2↵)n with ↵ > 0, by (4.27),

E[Wt↵ ] =n

2+ et↵(�1/n+⇥(1/n2

)) =n

2+ ↵pn+ o(1),

where we used that by a Taylor expansion, for |z| 1/2, log (1� z) = �z +⇥(z2). Fix 0 < " < 1/2. By Chebyshev’s inequality (Theorem 2.13), for t↵ =1

2

n log n� (log 2↵)n and n large enough,

P[Wt↵ n/2] P[|Wt↵ � E[Wt↵ ]| � (↵/2)pn] n/4

(↵/2)2n 1

2� ",

for ↵ large enough. By (4.26), that implies d(t↵) � " and we are done.

The proof of Claim 4.70 relies on a “distinguishing statistic.” Recall from Lemma 4.17that for any random variables X , Y and map h it holds that kµh(X)

� µh(Y )

kTV

kµX � µY kTV

, where µZ is the law of Z. The map used in the proof of the claimis the Hamming weight. In essence, we gave a lower bound on the total varia-tion distance between the laws of the Hamming weight at stationarity and underP t(z

0

, · ). See Exercise 4.10 for a more general instantiation of the distinguishingstatistic approach.

Remark 4.71. The upper bound in Claim 4.69 is indeed off by a factor of 2. See [LPW06,Theorem 18.3] for an improved upper bound and a discussion of the so-called cutoff phe-nomenon. The latter refers to the fact that for all 0 < " < 1/2 it can be shown that

cutoff

limn!+1

t(n)

mix

(")

t(n)

mix

(1� ")= 1,

where t(n)

mix

(") is the mixing time on the n-dimensional hypercube. In words, for large n,the total variation distance drops from 1 to 0 in a short time window. See Exercise 6.2 fora necessary condition for cutoff.

b-ary tree Let (Zt) be lazy simple random walk on the `-level rooted b-ary tree,bTb,`. The root, 0, is on level 0 and the leaves, L, are on level `. All vertices havedegree b + 1, except for the root which has degree b and the leaves which havedegree 1. Hence the stationary distribution is

⇡(x) :=�(x)

2(n� 1),

197

where n is the number of vertices and �(x) is the degree of x. We constructa coupling (Xt, Yt) of this chain started at (x, y). Assume w.l.o.g. that x is nofurther from the root than y, which we denote by x 4 y. The coupling has twostages:

- In the first stage, at each time, flip a fair coin. On heads, Yt stays put and Xt

moves one step chosen uniformly at random among its neighbors. Similarly,on tails, reverse the roles of Xt and Yt and proceed in the same manner. Dothis until Xt and Yt are on the same level.

- In the second stage, i.e., once the two chains are on the same level, at eachtime first let Xt move as a lazy random walk on bTb,`. Then let Yt move inthe same direction as Xt, i.e., if Xt moves closer to the root, so does Yt andso on.

By construction, Xt 4 Yt for all t. The key observation is the following. Let ⌧⇤

be the first time (Xt) visits the root after visiting the leaves. By time ⌧⇤, the twochains have necessarily met: because Xt 4 Yt, when Xt reaches the leaves, so doesYt; after that time, the coupling is in the second stage so Xt and Yt remain on thesame level; in particular, when Xt reaches the root, so does Yt. Hence ⌧

coal

⌧⇤.Intuitively, the mixing time is indeed dominated by the time it takes to reach theroot from the worst starting point, a leaf. See Figure 4.7 and the correspondinglower bound argument.

To estimate P(x,y)[⌧

⇤ > t], we use Markov’s inequality (Theorem 2.4), forwhich we need a bound on E

(x,y)[⌧⇤]. We note that E

(x,y)[⌧⇤] is less than the mean

time for the walk to go from the root to the leaves and back. Let Lt be the levelof Xt and let N be the corresponding network (where the conductances are equalto the number of edges on each level of the tree). In terms of Lt, the quantity weseek to bound is the mean of ⌧

0,`, the commute time of the chain (Lt) between thestates 0 and `. By the commute time identity (Theorem 3.103),

E0

[⌧0,`] = cN R(0$ `), (4.29)

wherecN = 2

X

e={x,y}2Nc(e) = 4(n� 1),

where we simply counted the number of edges in bTb,` and the factor of 4 accountsfor self-loops. Using network reduction techniques, we computed the effectiveresistance R(0 $ `) in Examples 3.84 and 3.85—without self-loops. Of courseadding self-loops does not affect the effective resistance as we can use the same

198

voltage and current. So, ignoring them, we get

R(0$ `) =`�1

X

j=0

r(j, j + 1) =`�1

X

j=0

b�(j+1) =1

b· b

�` � 1

b�1 � 1,

which implies1

b R(0$ `) 1

b� 1 1.

Finally, applying Theorem 4.66 and Markov’s inequality and using (4.29), we get

d(t) maxx,y2V

P(x,y)[⌧

⇤ > t]

maxx,y2V

E(x,y)[⌧

⇤]t

E0

[⌧0,`]

t

4n

t,

or:

Claim 4.72.tmix

(") 4n

".

This time the diameter-based bound is far off. We have � = 2` = ⇥(log n)and ⇡

min

= 1/2(n� 1) so that Claim 2.47 gives

tmix

(") � (2`)2

12 log n+ 4 log(2(n� 1))= ⌦(log n),

for n large enough. Here is a better lower bound. Intuitively the mixing timeis significantly greater than the squared diameter because the chain tends to bepushed away from the root: going from the leaves on one side of the root to theleaves on the other typically takes time linear in n. Formally let x

0

be a leaf of bTb,`

and let A be the set of vertices “on the other side of root (inclusively),” i.e., verticeswhose graph distance from x

0

is at least `. See Figure 4.7. Then ⇡(A) � 1/2 bysymmetry. We use the fact that

kP t(x0

, ·)� ⇡kTV

� |P t(x0

, A)� ⇡(A)|,to bound the mixing time. We claim that, started at x

0

, the walk typically takes timelinear in n to reach A. Consider again the level Lt of Xt. Using the expression forthe effective resistance above, we have

P`[⌧0 < ⌧+` ] =1

c(`)R(0$ `)=

1

2b`· b� 1

1� b�`=

b� 1

2b` � 2= O

✓

1

n

◆

.

199

Figure 4.7: Setup for the lower bound on the mixing time on a b-ary tree. (Hereb = 2.)

Hence, started from the leaves, the number of excursions back to the leaves neededto reach the root for the first time is geometric with success probability O(n�1).Each such excursion takes time at least 2. So P t(x

0

, A) is bounded above bythe probability that at least one such excursion was successful among the first t/2attempts. That is,

P t(x0

, A) 1� �

1�O�

n�1

��t/2<

1

2� ",

for all t ↵"n with ↵" > 0 small enough and

kP↵"n(x0

, ·)� ⇡kTV

� |P↵"n(x0

, A)� ⇡(A)| > ".

We have proved that tmix

(") � ↵"n.

4.4.3 Path coupling

Path coupling is a method for constructing Markovian couplings from “simpler”couplings. The building blocks are one-step couplings starting from pairs of initialstates that are close in some dissimilarity graph.

200

Let (Xt) be an irreducible Markov chain on a finite state space V with transi-tion matrix P and stationary distribution ⇡. Assume that we are given a dissimi-larity graph H

0

= (V0

, E0

) on V0

:= V with edge weights w0

: E0

! R+

. Thisdissimilaritygraph, pathmetric

graph need not have the same edges as the transition graph of (Xt). We extend w0

to the path metric

w0

(x, y) := inf

(

m�1

X

i=0

w0

(xi, xi+1

) : x = x0

, . . . , xm = y is a path in H0

)

,

where the infimum is over all paths connecting x and y in H0

. We call a pathachieving the infimum a minimum-weight path. Let

minimum-weightpath, weighteddiameter

�0

:= maxx,y

w0

(x, y),

be the weighted diameter of H0

.

Theorem 4.73 (Path coupling method). Assume that

w0

(u, v) � 1,

for all {u, v} 2 E0

. Assume further that there exists 2 (0, 1) such that:

- (Local couplings) For all x, y with {x, y} 2 E0

, there is a coupling (X⇤, Y ⇤)of P (x, ·) and P (y, ·) satisfying the following contraction property

E[w0

(X⇤, Y ⇤)] w0

(x, y). (4.30)

Thend(t) �

0

t,

or

tmix

(") ⇠

log�0

+ log "�1

log �1

⇡

.

Proof. The crux of the proof is to extend (4.30) to arbitrary pairs of vertices.

Claim 4.74 (Global coupling). For all x, y 2 V , there is a coupling (X⇤, Y ⇤) ofP (x, ·) and P (y, ·) such that (4.30) holds.

201

Iterating the coupling in this claim immediately implies the existence of a coa-lescing Markovian coupling (Xt, Yt) of P such that

E(x,y)[w0

(Xt, Yt)] = E(x,y) [E[w0

(Xt, Yt) |Xt�1

, Yt�1

]]

E(x,y) [w0

(Xt�1

, Yt�1

)]

· · · t E

(x,y)[w0

(X0

, Y0

)]

= tw0

(x, y)

t�0

.

By assumption, {x6=y} w0

(x, y) so that by the coupling inequality and Lemma4.64, we have

d(t) d(t) maxx,y

P(x,y)[Xt 6= Yt] max

x,yE(x,y)[w0

(Xt, Yt)] t�0

,

which implies the theorem. It remains to prove Claim 4.74.

Proof of Claim 4.74. Fix x0, y0 2 V such that {x0, y0} is not an edge in the dissim-ilarity graph H

0

. The idea is to combine the local couplings on a minimum-weightpath between x0 and y0 in H

0

. Let x0 = x0

⇠ · · · ⇠ xm = y0 be such a path. For alli = 0, . . . ,m�1, let (Z⇤

i,0, Z⇤i,1) be a coupling of P (xi, ·) and P (xi+1

, ·) satisfyingthe contraction property (4.30). Set Z(0) := Z⇤

0,0 and Z(1) := Z⇤0,1. Then itera-

tively pick Z(i+1) according to the law P[Z⇤i,1 2 · |Z⇤

i,0 = Z(i)]. By induction oni, (X⇤, Y ⇤) := (Z(0), Z(m)) is then a coupling of P (x0, ·) and P (y0, ·). Formally,define the transition matrix

Ri(z(i), z(i+1)) := P[Z⇤

i,1 = z(i+1) |Z⇤i,0 = z(i)],

and observe thatX

z(i+1)

Ri(z(i), z(i+1)) = 1, (4.31)

andX

z(i)

P (xi, z(i))Ri(z

(i), z(i+1)) = P (xi+1

, z(i+1)), (4.32)

by construction of the coupling (Z⇤i,0, Z

⇤i,1). See Figure 4.8. The law of the full

coupling (Z(0), . . . , Z(m)) is

P[(Z(0), . . . , Z(m)) = (z(0), . . . , z(m))]

= P (x0

, z(0))R0

(z(0), z(1)) · · ·Rm�1

(z(m�1), z(m)).

202

Figure 4.8: Coupling of P (x0, ·) and P (y0, ·) constructed from a sequence of localcouplings (Z⇤

0,0, Z⇤0,1), . . . , (Z

⇤0,m�1

, Z⇤0,m�1

).

Using (4.31) and (4.32) inductively gives

P[X⇤ = z(0)] = P[Z(0) = z(0)] = P (x0

, z(0)),

P[Y ⇤ = z(m)] = P[Z(m) = z(m)] = P (xm, z(m)),

as required.By the triangle inequality for w

0

, the coupling (X⇤, Y ⇤) satisfies

E[w0

(X⇤, Y ⇤)] = Eh

w0

(X(0), X(m))i

m�1

X

i=0

Eh

w0

(X(i), X(i+1))i

m�1

X

i=0

w0

(xi, xi+1

)

= w0

(x0, y0),

where, on the third line, we used (4.30) for adjacent pairs and the last line followsfrom the fact that we chose a minimum-weight path.

That concludes the proof of the theorem.

203

We illustrate the path coupling method in the next two subsections. See Exer-cise 4.11 for an optimal transport perspective on the path coupling method.

4.4.4 . Ising model: Glauber dynamics at high temperature

Let G = (V,E) be a finite, connected graph with maximal degree �. DefineX := {�1,+1}V . Recall that the (ferromagnetic) Ising model on V with inversetemperature � is the probability distribution over spin configurations � 2 X givenby

µ�(�) :=1

Z(�)e��H(�),

whereH(�) := �

X

i⇠j

�i�j ,

is the Hamiltonian andHamiltonianZ(�) :=

X

�2Xe��H(�),

is the partition function. In this context, recall that vertices are often referred to aspartitionfunction

sites. The single-site Glauber dynamics of the Ising model is the Markov chain on

siteX which, at each time, selects a site i 2 V uniformly at random and updates thespin �i according to µ�(�) conditioned on agreeing with � at all sites in V \{i}.Specifically, for � 2 {�1,+1}, i 2 V , and � 2 X , let �i,� be the configuration �with the state at i being set to �. Then, letting n = |V |, the transition matrix of theGlauber dynamics is

Q�(�,�i,�) :=

1

n· e��Si(�)

e��Si(�) + e�Si(�)=

1

n

⇢

1

2+

1

2tanh(��Si(�))

�

, (4.33)

whereSi(�) :=

X

j⇠i

�j .

All other transitions have probability 0. Recall that this chain is irreducible andreversible with respect to µ� . In particular µ� is the stationary distribution of Q� .

In this section we give an upper bound on the mixing time, tmix

("), of Q�

using path coupling. We say that the Glauber dynamics is fast mixing if tmix

(") =fast mixing

O(n log n). We first make a simple observation:

Claim 4.75 (Glauber dynamics: lower bound on mixing).

tmix

(") = ⌦(n), 8� > 0.

204

Proof. We use a coupon collecting argument. Let � be the all-(�1) configurationand let A be the set of configurations where at least half of the sites are +1. Then,by symmetry, µ�(A) = µ�(Ac) = 1/2 where we assumed for simplicity that n isodd. By definition of the total variation distance

d(t) � kQt�(�, ·)� µ�(·)kTV

� |Qt�(�, A)� µ�(A)| = |Qt

�(�, A)� 1/2|.

So it remains to show that by time c n, for c > 0 small, the chain is unlikely to havereached A. That happens if, say, fewer than a third of the sites have been updated.Using the notation of Example 2.15, we are seeking a bound on Tn,n/3, i.e., thetime to collect n/3 coupons out of n. We can write this random variable as asum of n/3 independent geometric variables Tn,n/3 =

Pn/3i=1

⌧n,i, where E[⌧n,i] =�

1� i�1

n

��1 and Var[⌧n,i] =�

1� i�1

n

��2. Hence

E[Tn,n/3] =

n/3X

i=1

✓

1� i� 1

n

◆�1

= nnX

j=2n/3+1

j�1 = ⇥(n), (4.34)

and

Var[Tn,n/3] n/3X

i=1

✓

1� i� 1

n

◆�2

= n2

nX

j=2n/3+1

j�2 = ⇥(n). (4.35)

So by Chebyshev’s inequality (Theorem 2.13)

P[|Tn,n/3 � E[Tn,n/3]| � "n] Var[Tn,n]

("n)2! 0,

by (4.34) and (4.35). Taking " > 0 small enough and n large enough, we haveshown that, for t c"n for some c" > 0,

Qt�(�, A) 1/3,

which proves the claim.

Remark 4.76. In fact, Ding and Peres proved that tmix

(") = ⌦(n log n) for any graph onn vertices [DP11]. In Claim 4.70, we treated the special case of the empty graph, whichis equivalent to lazy random walk on the hypercube.

In our main result, we show that the Glauber dynamics of the Ising model is fastmixing when the inverse temperature � is small enough as a function of the maxi-mum degree.

205

Claim 4.77 (Glauber dynamics: fast mixing at high temperature).

� < ��1 =) tmix

(") = O(n log n).

Proof. We use path coupling. Let H0

= (V0

, E0

) where V0

:= X and {�,!} 2 E0

if 1

2

k� � !k1

= 1 with unit w0

-weights on all edges. (To avoid confusion, wereserve the notation ⇠ for adjacency in G.) Let {�,!} 2 E

0

differ at coordinate i.We construct a coupling (X⇤, Y ⇤) of Q�(�, ·) and Q�(!, ·). We first pick the samecoordinate i⇤ to update. If i⇤ is such that all its neighbors in G have the same statein � and !, i.e., if �j = !j for all j ⇠ i⇤, we update X⇤ from � according to theGlauber rule and set Y ⇤ := X⇤. Note that this includes the case i⇤ = i. Otherwise,i.e. if i⇤ ⇠ i, we proceed as follows. From the state �, the probability of updatingsite i⇤ to state � 2 {�1,+1} is given by the expression in brackets in (6.6), andsimilarly for !. Unlike the previous case, we cannot guarantee that the update isidentical in both chains. However, in order to minimize the chances of increasingthe distance between the two chains, we pick a uniform-[�1, 1] variable U and set

X⇤i⇤ :=

(

+1, if U tanh(�Si⇤(�))

�1, o.w.

and

Y ⇤i⇤ :=

(

+1, if U tanh(�Si⇤(!))

�1, o.w.

We set X⇤j := �j and Y ⇤

j := !j for all j 6= i⇤. The expected distance between X⇤

and Y ⇤ is then

E[w0

(X⇤, Y ⇤)] = 1� 1

n|{z}

(a)

+1

n

X

j⇠i

1

2|tanh(�Sj(�))� tanh(�Sj(!))|

| {z }

(b)

, (4.36)

where (a) corresponds to i⇤ = i in which case w0

(X⇤, Y ⇤) = 0 and (b) correspondsto i⇤ ⇠ i in which case w

0

(X⇤, Y ⇤) = 2 with probability 1

2

| tanh(�Si⇤(�)) �tanh(�Si⇤(!))| by our coupling, and w

0

(X⇤, Y ⇤) = 1 otherwise. To bound (b),we note that for j ⇠ i

|tanh(�Sj(�))� tanh(�Sj(!))| = tanh(�(s+ 2))� tanh(�s), (4.37)

wheres := Sj(�) ^ Sj(!).

206

The derivative of tanh is maximized at 0 where it is equal to 1. So the r.h.s. of (4.37)is 2�. Plugging this back into (4.36), we get

E[w0

(X⇤, Y ⇤)] 1� 1� ��

n exp

✓

�1� ��

n

◆

= w0

(�,!),

where

:= exp

✓

�1� ��

n

◆

< 1,

by assumption. The diameter of H0

is �0

= n. By Theorem 4.73,

tmix

(") ⇠

log�0

+ log "�1

log �1

⇡

=

⇠

n(log n+ log "�1)

1� ��

⇡

,

which implies the claim.

Remark 4.78. A slighlty more careful analysis shows that the condition � tanh(�) < 1 isenough for the claim to hold. See [LPW06, Theorem 15.1].

4.4.5 . From approximate sampling to approximate counting

To be written. See [MU05, Section 10.3] and [LPW06, Sections 14.3 and 14.4].See also [JS97].

4.5 Further applications

4.5.1 . Voter model on the complete graph and on the line: duality

To be written. See [AF, Section 14.3] and [Ald13, Lig].

Exercises

Exercise 4.1 (Harris’ inequality: alternative proof). We say that f : Rn ! Ris nondecreasing if it is nondecreasing in each variable while keeping the othervariables fixed.

• (Chebyshev’s association inequality) Let f : R ! R and g : R ! R benondecreasing and let X be a real random variable. Show that

E[f(X)g(X)] � E[f(X)]E[g(X)].

[Hint: Consider the quantity (f(X) � f(X 0))(g(X) � g(X 0)) where X 0 isan independent copy of X .]

207

• (Harris’ inequality) Let f : Rn ! R and g : Rn ! R be nondecreasingand let X = (X

1

, . . . , Xn) be independent real random variables. Show byinduction on n that

E[f(X)g(X)] � E[f(X)]E[g(X)].

Exercise 4.2. Provide the details for Example 4.51.

Exercise 4.3 (FKG: sufficient conditions). Let X := {0, 1}F where F is finite andlet µ be a positive probability measure on X . We use the notation introduced in theproof of Holley’s inequality (Theorem 4.50).

a) To check the FKG condition, show that it suffices to check that, for all x y 2 X and i 2 F ,

µ(yi,1)

µ(yi,0)� µ(xi,1)

µ(xi,0).

[Hint: Write µ(! _ !0)/µ(!) as a telescoping product.]

b) To check the FKG condition, show that it suffices to check (4.12) only forthose !,!0 2 X such that k! � !0k

1

= 2 and neither ! !0 nor !0 !.[Hint: Use a).]

Exercise 4.4 (FKG and strong positive association). Let X := {0, 1}F where F isfinite and let µ be a positive probability measure on X . For ⇤ ✓ F and ⇠ 2 X , let

X ⇠⇤

:= {!⇤

⇥ ⇠⇤

c : !⇤

2 {0, 1}⇤},where !

⇤

⇥ ⇠⇤

c agrees with ! on coordinates in ⇤ and with ⇠ on coordinates inF\⇤. Define the measure µ⇠

⇤

over {0, 1}⇤ as

µ⇠⇤

(!⇤

) :=µ(!

⇤

⇥ ⇠⇤

c)

µ(X ⇠⇤

).

That is, µ⇠⇤

is µ conditioned on agreeing with ⇠ on F\⇤. The measure µ is saidto be strongly positively associated if µ⇠

⇤

(!⇤

) is positively associated for all ⇤ andstrong positiveassociation

⇠. Prove that the FKG condition is equivalent to strong positive association. [Hint:Use Exercise 4.3 as well as the FKG inequality.]

Exercise 4.5 (Triangle-freeness: a second proof). Consider again the setting ofSection 4.3.5.

a) Let et be the minimum number of edges in a t-vertex union of k not mutuallyvertex-disjoint triangles. Show that, for any k � 2 and k t < 3k, it holdsthat et > t.

208

b) Use Exercise 2.11 to give a second proof of the fact that P[Xn = 0] !e��3/6.

Exercise 4.6 (RSW lemma: general ↵). Let Rn,↵(p) be as defined in Section 4.3.6.Show that for all n � 2 (divisible by 4) and p 2 (0, 1)

Rn,↵(p) �✓

1

2

◆

2↵�2

Rn,1(p)6↵�7Rn/2,1(p)

6↵�6.

Exercise 4.7 (Primal and dual crossings). Modify the proof of Lemma 2.8 to proveLemma 4.59.

Exercise 4.8 (Square-root trick). Let µ be an FKG measure on {0, 1}F where F isfinite. Let A

1

and A2

be increasing events with µ(A1

) = µ(A2

). Show that

µ(A1

) � 1�p

1� µ(A1

[A2

).

Exercise 4.9 (Mixing on cycles: lower bound). Let (Zt) be lazy, simple randomwalk on the cycle of size n, Zn := {0, 1 . . . , n � 1}, where i ⇠ j if |j � i| = 1(mod n).

a) Let A = {n/2, . . . , n�1}. By coupling (Zt) with lazy, simple random walkon Z, show that

P↵n2

(n/4, A) <1

2� ",

for ↵ � ↵" > 0. [Hint: Use Kolmogorov’s maximal inequality (e.g. [Dur10,Theorem 2.5.2]).]

b) Deduce thattmix

(") � ↵"n2.

Exercise 4.10 (Lower bound on mixing: distinguishing statistic). Let X and Ybe random variables on a finite state space S. Let h : S ! R be a measurablereal-valued map. Assume that

E[h(Y )]� E[h(X)] � r�,

where r > 0 and �2 := max{Var[h(X)],Var[h(Y )]}. Show that

kµX � µY kTV

� 1� 8

r2.

[Hint: Consider the interval on one side of the midpoint between E[h(X)] andE[h(Y )].]

209

Exercise 4.11 (Path coupling and optimal transport). Let V be a finite state spaceand let P be an irreducible transition matrix on V with stationary distribution ⇡.Let w

0

be a metric on V . For probability measures µ, ⌫ on V , let

W0

(µ, ⌫) := inf {E[w0

(X,Y )] : (X,Y ) is a coupling of µ and ⌫} ,

be the so-called transportation metric or Wasserstein distance between µ and ⌫.Wassersteindistancea) Show that W

0

is a metric. [Hint: Proof of Claim 4.74.]

b) Assume that the conditions of Theorem 4.73 hold. Show that for any proba-bility measures µ, ⌫

W0

(µP, ⌫P ) W0

(µ, ⌫).

c) Use a) and b) to prove Theorem 4.73.

Notes

Bibliographic Remarks

Section 4.1 The coupling method is generally attributed to Doeblin [Doe38]. Thestandard reference on coupling is [Lin02]. See that reference for a history of cou-pling and a facsimile of Doeblin’s paper. See also [dH]. Section 4.2.2 is based on[vdH14, Section 5.3] and Section 4.1.2 is based on [Per, Section 6].

Section 4.3 Strassen’s theorem is due to Strassen [Str65]. Harris’ inequality isdue to Harris [Har60]. The FKG inequality is due to Fortuin, Kasteleyn, and Gini-bre [FKG71]. A “four-function” version of Holley’s inequality, which also ex-tends to distributive lattices, was proved by Ahlswede and Daykin [AD78]. Seee.g. [AS11, Section 6.1]. An exposition of submodularity and its connectionsto convexity can be found in [Lov83]. For more on Markov random fields, seee.g. [RAS]. Section 4.3.5 follows [AS11, Sections 8.1, 8.2, 10.1]. Janson’s inequal-ity is due to Janson [Jan90]. Boppana and Spencer [BS89] gave the proof presentedhere. For more on Janson’s inequality, see [JLR11, Section 2.2]. The presentationin Section 4.3.6 follows closely [BR06b, Sections 3 and 4]. See also [BR06a,Chapter 3]. Broadbent and Hammersley [BH57, Ham57] initiated the study of thecritical value of percolation. Harris’ theorem was proved by Harris [Har60] andKesten’s theorem was proved two decades later by Kesten [Kes80], confirmingnon-rigourous work of Sykes and Essam [SE64]. The RSW theorem was obtained

210

independently to Russo [Rus78] and Seymour and Welsh [SW78]. The proof wegave here is due to Bollobas and Riordan [BR06b]. Another short proof of a versionof the RSW theorem for critical site percolation on the triangular lattice was givenby Smirnov. See e.g. [Ste]. The type of “scale invariance” seen in the RSW theoremplays a key role in the contemporary theory of critical two-dimensional percolationand of two-dimensional lattice models more generally. See e.g. [Law05, Gri10a].

Section 4.4 The material in Section 4.4 borrows heavily from [LPW06, Chap-ters 5, 14, 15] and [AF, Chapter 12]. Aldous [Ald83] was the first author to makeexplicit use of coupling to bound total variation distance to stationarity of finiteMarkov chains. The link between couplings of Markov chains and total varia-tion distance was also used by Griffeath [Gri75] and Pitman [Pit76]. The proofof Claim 4.70 is partly based on [LPW06, Proposition 7.13]. See also [DGG+00]and [HS07] for alternative proofs. Path coupling is due to Bubley and Dyer [BD97].The optimal transport perspective on the path coupling method in Exercise 4.11 isfrom [LPW06, Chapter 14]. For more on optimal transport, see e.g. [Vil09]. Themain result in Section 4.4.4 is taken from [LPW06, Theorem 15.1]. For more back-ground on the so-called “critical slowdown” of the Glauber dynamics of Ising andPotts models on various graphs, see [CDL+12, LS12]. The connection betweensampling and counting was first considered by Jerrum, L. Valiant and V. Vazi-rani [JVV86]. For more on this topic, see e.g. [Sin93].

Section 4.5 For much more on interacting particle systems, see [Lig85].

211