Probab. Theory Relat. Fields 117, 419–447 (2000)
Digital Object Identifier (DOI) 10.1007/s004400000057
Harry Kesten · Zhong-Gen Su
Asymptotic behavior of the critical probability
for ρ-percolation in high dimensions
c Springer-Verlag 2000
Received: 7 January 1999 / Published online: 14 June 2000 – Abstract. We consider oriented bond or site percolation on ⺪d+ . In the case of bond percolation we denote by Pp the probability measure on configurations of open and closed
bonds which makes all bonds of ⺪d+ independent, and for which Pp {e is open} = 1 −
Pp {e is closed} = p for each fixed edge e of ⺪d+ . We take X(e) = 1 (0) if e is open
(respectively, closed). We say that ρ-percolation occurs for some given 0 < ρ ≤ 1, if
there exists an oriented
infinite path v0 = 0, v1 , v2 , . . ., starting at the origin, such that
P
lim inf n→∞ (1/n) ni=1 X(ei ) ≥ ρ, where ei is the edge {vi−1 , vi }. [MZ92] showed that
there exists a critical probability pc = pc (ρ, d) = pc (ρ, d, bond) such that there is a.s.
no ρ-percolation for p < pc and that Pp {ρ-percolation occurs} > 0 for p > pc . Here
we find limd→∞ d 1/ρ pc (ρ, d, bond) = D1 , say. We also find the limit for the analogous
quantity for site percolation, that is D2 = limd→∞ d 1/ρ pc (ρ, d, site). It turns out that for
ρ < 1, D1 < D2 , and neither of these limits equals the analogous limit for the regular d-ary
trees.
1. Introduction
Let G be an infinite connected graph with edge set E and vertex set V. As usual in
percolation, we allow each edge to be in one of two states, which we call open and
closed. The configurations of open and closed states are chosen randomly, so that
all edges are independent and so that
P {e is open} = 1 − P {e is closed} = p
(1.1)
for some p ∈ [0, 1]. We denote the corresponding probability measure on the configurations of open and closed edges by Pp , and use Ep to denote expectation with
respect to Pp . We set
n
1 if e is open
X(e) =
0 if e is closed.
[MZ92] introduced so-called ρ-percolation. Following their paper we say that
ρ-percolation occurs if there exists an infinite path on G which starts at a fixed
H. Kesten, Z. Su∗ : Department of Mathematics, White Hall, Cornell University, Ithaca,
NY 14853 USA. e-mail: kesten@math.cornell.edu
∗
Current address: Department of Mathematics, Hangzhou University, Zhejiang, 310028
P. R. China. e-mail: zgsu@mail.hz.zj.cn
Mathematics Subject Classification (1991): Primary 60K35; Secondary 82B43
Key words and phrases: ρ-percolation – Critical probability – Second moment method
420
H. Kesten, Z. Su
distinguished vertex v0 and successively traverses the edges e1 , e2 , . . . without
using any edge twice, and such that
n
lim inf
n→∞
1X
X(ei ) ≥ ρ.
n
(1.2)
i=1
If G is a directed graph, then we of course require in this definition that e1 , e2 , . . .
be a directed path on G. In this paper we shall mainly work with ⺪d+ viewed as a
directed graph, in which each edge is oriented from a vertex v to v + ξi for some
1 ≤ i ≤ d, where ξi is the positive i-th coordinate vector. For this graph we always take the origin for the distinguished vertex v0 . [MZ92] introduced the critical
probability
pc = pc (ρ, G) = pc (ρ, G, bond)
:= sup{p : Pp {ρ-percolation occurs} = 0}.
(1.3)
Since Pp {ρ-percolation occurs} is increasing we automatically have
Pp {ρ-percolation occurs} > 0 when p > pc .
(1.4)
It is also obvious that for ρ ≤ 1
pc (ρ, G) ≤ critical probability for ordinary bond percolation on G.
(1.5)
A simple Peierls argument (see Lemma 1 below) shows further that for G = oriented
⺪d+ , and ρ > 0, pc (ρ, ⺪d+ ) > 0. Thus in any dimension d ≥ 2,
0 < pc (ρ, ⺪d ) ≤ pc (ρ, ⺪d+ ) < 1 for 0 < ρ ≤ 1.
(1.6)
In this paper we prove the following theorem about the behavior of pc for oriented
ρ-percolation on ⺪d+ for large d.
Theorem 1. For 0 < ρ ≤ 1
lim d 1/ρ pc (ρ, ⺪d+ , bond) = D1 = D1 (ρ) :=
d→∞
h 1 − ρ i(1−ρ)/ρ
ρe
(1.7)
(D1 is to be interpreted as 1 when ρ = 1).
We also consider the anologous problem for oriented site percolation on ⺪d+ . In
this case we define for a vertex v the random variable Y (v) = 1(0) if v is open (respectively, closed), and we say that ρ-percolation occurs if there exists an oriented
path v0 = 0, v1 , v2 , . . . on ⺪d+ such that
n
1X
Y (vi ) ≥ ρ.
lim inf
n→∞ n
i=1
The corresponding critical probability is denoted by pc (ρ, ⺪d+ , site). The following
theorem holds.
ρ-percolation in high dimensions
421
Theorem 2. For 0 < ρ ≤ 1
lim d 1/ρ pc (ρ, ⺪d+ , site) = D2 = D2 (ρ) :=
d→∞
θ 1/ρ
,
eθ − 1
(1.8)
where θ is the unique solution of
1
θeθ
=
θ
e −1
ρ
(1.9)
(also D2 = 1 when ρ = 1).
One can also consider ρ-percolation on the unoriented graph G = ⺪d . Again
one may consider bond and site percolation. We denote the critical percolation
for this kind of ρ-percolation by pc (ρ, ⺪d , bond) and pc (ρ, ⺪d , site), respectively.
Note that our notation does not explicitly indicate whether we consider oriented or
unoriented percolation; we merely indicate the graph G on which we are working.
If G = ⺪d+ , then we tacitly assume that this is oriented in the positive direction.
Since an oriented path on ⺪d+ is also an unoriented path on ⺪d we can use Theorems
1 and 2 to obtain bounds on pc for the unoriented case.
Theorem 3. For 0 < ρ ≤ 1
2−1/ρ D1 ≤ lim inf d 1/ρ pc (ρ, ⺪d , bond)
d→∞
≤ lim sup d 1/ρ pc (ρ, ⺪d , bond) ≤ D1 ,
(1.10)
d→∞
and
2−1/ρ D2 ≤ lim inf d 1/ρ pc (ρ, ⺪d , site)
d→∞
≤ lim sup d 1/ρ pc (ρ, ⺪d , site) ≤ D2 .
(1.11)
d→∞
Remark 1. It is clear that occurrence of ordinary percolation (that is the occurrence
of an infinite self-avoiding path all of whose edges, respectiveley vertices, are open)
implies the occurrence of 1-percolation (that is, ρ-percolation with ρ = 1). Thus,
if pc (⺪d , bond) and pc (⺪d , site) denote the critical probabilities for ordinary bond
and site percolation on ⺪d , then
pc (1, ⺪d , bond) ≤ pc (⺪d , bond) and pc (1, ⺪d , site) ≤ pc (⺪d , site).
(1.12)
In fact, it follows from Theorem 5 of [Lee93] that both inequalities here are equalities. Since it is known that
2dpc (⺪d , bond) → 1 and 2dpc (⺪d , site) → 1 as d → ∞
(1.13)
([Kes90, HS90, Gor91, BK94]), it follows from Theorem 3 and the fact that D1 =
D2 = 1 when ρ = 1, that
2dpc (1, ⺪d , bond) → 1 and 2dpc (1, ⺪d , site) → 1 as d → ∞.
(1.14)
422
H. Kesten, Z. Su
Remark 2. If G is a rooted regular d-ary tree, then pc (ρ, G, bond) and pc (ρ, G, site)
are known exactly. (A rooted regular d-ary tree is a tree in which each vertex has
d + 1 neighbors, and each edge is oriented away from a distinguished vertex which
is called the root. Thus there are d outgoing edges incident to each vertex, except
for the root which is incident to d + 1 outgoing edges.) For such trees [MZ92]
already showed that these critical probabilities for the bond and site problem are
the same, and both are equal to the solution for p of the equation
p ρ (1 − p)1−ρ d = ρ ρ (1 − ρ)1−ρ .
(1.15)
Simple asymptotic analysis therefore shows that for these trees
lim d 1/ρ pc (ρ, G, bond) = lim d 1/ρ pc (ρ, G, site)
d→∞
d→∞
= D0 = D0 (ρ) := ρ(1 − ρ)(1−ρ)/ρ .
(1.16)
For ρ = 1, D0 = D1 = D2 = 1. Also for ordinary oriented percolation on ⺪d+ , the
critical probabilities pc (⺪d+ , bond) and pc (⺪d+ , site) are asymptotically equivalent
to 1/d (see [CD83]). Therefore, both oriented percolation and oriented 1-percolation on ⺪d+ and a d-ary tree have asymptotically equivalent critical probabilities.
Also ordinary unoriented percolation on ⺪d is in this sense similar to percolation
on a 2d-ary tree, by virtue of (1.13). It is somewhat surprising that this kind of
similarity with trees for high d does not extend to ρ < 1. In fact for ρ < 1 one can
see that D0 < D1 < D2 . Indeed, D0 < D1 follows immediately from
∞
e1−ρ <
X
1
1
=
=
(1 − ρ)n .
ρ
1 − (1 − ρ)
n=0
That D1 < D2 for ρ < 1 is best seen from the following expressions for D1 , D2 :
h
h
i
i−1
−λ −1
= inf θ (ρ−1)/ρ eθ
D1 = inf eλ(1−ρ)/ρ ee
θ>0
(1.17)
h λ∈⺢
i−1 h
i−1
e−λ
λ/ρ
D2 = inf e
= inf θ −1/ρ eθ − 1
.
e −1
λ∈⺢
θ>0
These representations, together with
θ (ρ−1)/ρ eθ =
∞
∞
X
X
θ n−1/ρ
θ n−1/ρ
>
= θ −1/ρ eθ − 1 ,
(n − 1)!
n!
n=1
0 < ρ < 1,
n=1
show that D1 < D2 . The expressions in (1.17) can of course be verified directly
from the definitions of D1 , D2 in (1.7), (1.8), but this is rather artificial. These expressions arise more naturally from large deviation estimates in the course of Peierls
arguments used to find a lower bound for pc (ρ, ⺪d+ , bond) and pc (ρ, ⺪d+ , site) (see
Lemma 1).
Remark 3. ρ-percolation seems to have received little attention so far. We only
know of one reference where it appears, somewhat implicitly, to wit [BF86]. Section 6.1 of this reference raises the question whether for d = 2 and a given value of
p (denoted by α rather than p in [BF86]), there exists outside any square a circuit at
least half of whose edges are open. This is essentially equivalent to asking whether
p > pc (1/2, ⺪d , bond) (or may be, whether p ≥ pc (1/2, ⺪d , bond)).
ρ-percolation in high dimensions
423
Idea of the proof. In the next section we shall only prove Theorem 1. The proof
of Theorem 2 is entirely analogous and will be left to the reader. The left hand
inequalities in (1.10) and (1.11) follow from simple Peierls arguments similar to
Lemma 1 below, while the right hand inequalities are immediate from Theorems 1
and 2.
For the proof of Theorem 1 we introduce the following notation. If e00 = (a 00 , b00 )
and e0 = (a 0 , b0 ) are oriented edges, then e0 ≺ e00 means that there exists an oriented
path on ⺪d+ from b0 to a 00 . In this case we say that e0 precedes e00 , or e00 is a successor
of e0 . For an oriented edge e = (a, b) we write
kek =
d
X
ai .
(1.18)
i=1
For ε > 0 we further introduce the following random variable:
bn (ε) = number of sequences e1 , . . . , en for which each ei , 1 ≤ i ≤ n,
N
n(1 + ε)
.
(1.19)
is open and ei ≺ ei+1 , 1 ≤ i ≤ n − 1, and ken k ≤
ρ
We define
µbxc
bxc!
= P {a Poisson variable with mean µ takes the value bxc}.
Pµ (x) = e−µ
It is not hard to show (see Lemma 1) that for all ε > 0, d ≥ 1, n ≥ 1,
(1 + ε)
bn (ε)
n − n ≤ Ep N
p n d n end Pnd
ρ
(1−ρ+ε)/ρ h
−d in
dρ
1
≤ pn d n 1 +
1−
1−ρ+ε
1 + dρ
(1.20)
1−ρ+ε
bn (ε) > 0 for all large n, it
Since ρ-percolation can occur only if for all ε > 0, N
follows that
−d
(1−ρ+ε)/ρ dρ
1
≥ 1. (1.21)
1−
pc (ρ, ⺪d+ )d lim 1 +
ε↓0
1−ρ+ε
1 + dρ
1−ρ+ε
Consequently, for 0 < ρ ≤ 1,
1
ρ −(1−ρ)/ρ 1 − ρ d
+
1−
d 1/ρ pc (ρ, ⺪d+ ) ≥
d
1−ρ
1 − ρ + dρ
h 1 − ρ i(1−ρ)/ρ
1
=
+ O( ).
ρe
d
(1.22)
Of course this is stronger than
lim inf d 1/ρ pc (ρ, ⺪d+ ) ≥
d→∞
h 1 − ρ i(1−ρ)/ρ
ρe
.
(1.23)
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H. Kesten, Z. Su
For an inequality in the opposite direction, we first deal with rational ρ and make
sure that our estimates are explicit enough to later allow for approximating an irrational ρ by rational numbers. When ρ = r/s for integers r, s, we introduce for
bnr . (Roughly speaking Nn is still a sum
n ≥ 1 a quantity Nn which is similar to N
bnr , but now the
of weights over the same sequences e1 , . . . , enr as counted in N
sequences are not all given weight 1.) It will be the case that Nn > 0 implies that
there exists an oriented path of length nr/ρ = ns which has at least nr open edges.
In addition the events {Nn > 0} will be decreasing in n. Therefore, if
Pp {Nn > 0} ≥ C1 > 0 for large n,
(1.24)
then
Pp {ρ-percolation occurs} ≥ P {Nn > 0 for all large n} ≥ C1 .
(1.25)
We then follow the method of [CD83]. We use the so-called “second moment method" to prove that (1.24) holds. More specifically, we shall show that for η > 0 there
exists some constant C2 = C2 (η) < ∞ such that for
1 − ρ (1−ρ)/ρ
p > (1 + η)
ρe
d −1/ρ ,
(1.26)
it holds that
Ep [Nn ]2 ≤ C2 [Ep Nn ]2 .
Schwarz’ inequality then implies that
Pp {Nn > 0} ≥
[Ep Nn ]2
1
≥
.
2
C2
Ep [Nn ]
(1.27)
Thus (1.26) implies that percolation occurs. Theorem 1 then follows by letting η
tend to 0.
2. Proof of Theorem 1
The proof will be broken down in a number of lemmas. We will use Ci to denote
various constants strictly between 0 and ∞ whose precise value is not important
for us. The same symbol Ci in different equations may have different values.
bn (ε) (as defined in (1.19)) satisfies (1.20).
Lemma 1. N
Proof. Note that if e0 = (a 0 , b0 ) is an oriented edge, then the number of edges
e00 = (a 00 , b00 ) which succeed e0 and satisfy ke00 k = ke0 k + k is
dk
−d
d +k−2
k−1
, k ≥ 1.
≥
d
= (−1) d
k−1
k−1
(k − 1)!
Therefore, the number of sequences of oriented edges e1 , . . . , en with ei ≺ ei+1 ,
ke1 k = k1 and kei+1 k = kei k + ki+1 for 1 ≤ i ≤ n − 1 is
n Y
d + ki − 2
n
.
d
ki − 1
i=1
ρ-percolation in high dimensions
425
The expected number of such sequences with all ei open is therefore
n Y
d + ki − 2
n n
p d
ki − 1
i=1
and
n Y
d + ki − 2
ki − 1
X
bn (ε) = pn d n
EN
k1 +···+kn ≤n(1+ε)/ρ i=1
ki ≥1
X
≥ pn d n
d
n
Y
P
(ki −1)
i=1
k1 +···+kn ≤n(1+ε)/ρ
ki ≥1
X
= p n d n end
1
(ki − 1)!
Pnd (`)
`≤n(1+ε)/ρ−n
≥ p n d n end Pnd (n(1 + ε)/ρ − n).
(2.1)
This proves the lower bound in (1.20).
For the upper bound in (1.20) we use a standard large deviation estimate, valid
for any λ ≥ 0 (compare [Var84], Section 3).
i
n ∞
hX
Y
X
n
d + ki − 2
λn(1+ε)/ρ
−λk d + k − 2
≤e
e
ki − 1
k−1
k1 +···+kn ≤n(1+ε)/ρ i=1
ki ≥1
k=1
= eλn(1+ε)/ρ
h
= eλ(1+ε)/ρ
∞
hX
e−λ(`+1) (−1)`
`=0
in
e−λ
.
(1 − e−λ )d
−d
`
i
n
(2.2)
The right hand inequality in (1.20) follows by taking
λ = log 1 +
d
.
(1 + ε)/ρ − 1
t
u
As pointed out in Section 1, Lemma 1 (that is, inequality (1.20)) implies (1.21)
and therefore (1.23). The next lemma is a preparation for the definition of the
random variable Nn . We assume that
r0
with r0 , s0 ∈ {1, 2, . . .} and (r0 , s0 ) = 1.
(2.3)
ρ=
s0
Lemma 2. For all η > 0, 0 < σ < 1, there exists an integer m0 = m0 (η, σ ) ≥ 1
and integers α` ≥ 0, ` ≥ 1, such that
X
`α` = m0 s0 ,
(2.4)
`≥1
426
H. Kesten, Z. Su
and such that for all d ≥ 1 and σ ≤ ρ ≤ 1 − σ
oi
0 r0 n 0 r0
h X(α) mY
h X(α) mY
1/(m0 r0 )
d ki i1/(m0 r0 )
d + ki − 2
≥
d
ki − 1
(ki − 1)!
i=1
i=1
h ρe i(1−ρ)/ρ
≥ (1 − η)d 1/ρ
, (2.5)
1−ρ
P
where the sum (α) runs over all choices of k1 , k2 , . . . , km0 r0 ≥ 1 which satisfy
ki = ` for exactly α` values of i ∈ {1, 2, . . . , m0 r0 }.
(2.6)
Note that m0 in this lemma depends on r0 , s0 only through σ . The αi will depend
on r0 and s0 , though. We also point out that for any k1 , . . . , km0 r0 which appears in
P(α)
it holds that
m
0 s0
X
α` = m0 r0 and
`=1
m
0 r0
X
ki =
i=1
X
`α` = m0 s0 = m0 r0 /ρ.
(2.7)
`≥1
Proof of Lemma 2. The first inequality in (2.5) was already observed in the beginning of Lemma 1 and is trivial anyway. The second inequality is probably standard
knowledge among the experts on large deviations, but we give here a selfcontained
proof.
P
We shall write k,1≤ki ≤A for the sum over all k1 , . . . , kmr0 which satisfy 1 ≤
Pmr
ki ≤ A, 1 ≤ i ≤ mr0 and 1 0 ki = ms0 . We further introduce
φA (θ) :=
φ∞ (θ) :=
A
X
k=1
∞
X
k=1
eθk
,
(k − 1)!
eθk
θ
= eθ +e .
(k − 1)!
Then
mr0
mr0 X Y
d ki
eθ ki
ms0 −ms0 θ
=d e
φA (θ)
(ki − 1)!
φA (θ )(ki − 1)!
k,1≤ki ≤A i=1
k,1≤ki ≤A i=1
mr
(A)
= d ms0 e−ms0 θ φA (θ) 0 P {Tmr
= ms0 },
(2.8)
0
X
mr0
Y
where
(A)
Tmr
0
for a sequence
(A)
Xi ,
=
mr0
X
i=1
(A)
Xi
i ≥ 1, of i. i. d. random variables with distribution
(A)
P {Xi
= k} =
eθk
,
φA (θ)(k − 1)!
1 ≤ k ≤ A.
ρ-percolation in high dimensions
427
We shall choose θ = θA as the solution of
0 (θ)
φA
s0
1
=
=
φA (θ)
r0
ρ
(2.9)
(we shall soon see that this solution is unique when A ≥ 2). We similarly define
θ∞ to be the solution of
0 (θ)
1
φ∞
= .
φ∞ (θ)
ρ
Explicit calculation shows that
e θ∞ =
1−ρ
.
ρ
(2.10)
We claim that
1 − ρ (1−ρ)/ρ
as A → ∞,
e
ρ
θA → θ∞ and φA (θA ) → φ∞ (θ∞ ) =
(2.11)
uniformly in r0 , s0 with σ ≤ r0 /s0 ≤ 1 − σ . To see this we note first that |θ∞ | =
| log(1 − ρ)/ρ| ≤ C1 (σ ) and that
00 (θ)φ (θ) − φ 0 (θ) 2
0 (θ )
φ∞
d φ∞
∞
∞
≥ C2 (σ ) for |θ | ≤ C1 + 1.
=
2
dθ φ∞ (θ )
φ∞ (θ)
0 (·)/φ (·) is strictly increasing, and even
Therefore φ∞
∞
0 (θ − ε)
φ 0 (θ∞ )
s0
φ∞
∞
≤ −C2 ε + ∞
= −C2 ε + ,
φ∞ (θ∞ − ε)
φ∞ (θ∞ )
r0
0 (θ + ε)
0 (θ )
φ
s
φ∞
∞
∞
0
≥ C2 ε + ∞
= C2 ε +
φ∞ (θ∞ + ε)
φ∞ (θ∞ )
r0
(2.12)
0 (·)/φ (·) is strictly increasing when A ≥ 2, so that θ is uniquely
for ε ≤ 1. Also φA
A
A
determined by (2.9) as we claimed. Moreover,
0 (θ)
φA
φ 0 (θ)
→ ∞
uniformly on |θ | ≤ C1 + 1.
φA (θ)
φ∞ (θ)
Together with (2.12) this easily implies our claim (2.11).
Now note that for θ = θA ,
(A)
P {1 ≤ Xi
≤ A} = 1
(2.13)
and
(A)
P {Xi
(A)
= 1} ≥ C3 (σ, A) > 0, P {Xi
as well as
(A)
E{Xi } =
= 2} ≥ C3 (σ, A) > 0,
0 (θ )
φA
s0
A
= .
φA (θA )
r0
(2.14)
428
H. Kesten, Z. Su
The local central limit theorem therefore implies that there exists an m1 = m1 (σ, A)
such that for all m ≥ m1
(A)
(A)
(A)
P {Tmr
= ms0 } = P {Tmr
− ETmr
= 0}
0
0
0
1
1
≥ √
.
≥ q
A π mr0
(A)
πmr0 var(Xi )
The fact that m1 only depends on σ, A under the conditions (2.13), (2.14), follows
from any of the standard proofs of the local central limit theorem (see for instance
[GK54], Section 49).
Combining this with (2.8) we see that for θ = θA , and m ≥ m1 (σ, A),
h
mr0
Y
X
k1 +···+kmr0 =ms0 i=1
1≤ki ≤A
h
i1/(mr0 )
1
d ki i1/(mr0 )
≥ d 1/ρ e−θA /ρ φA (θA ) √
.
(ki − 1)!
A π mr0
(2.15)
Clearly we can find an m2 = m2 (σ, A) ≥ m1 (σ, A) such that for m ≥ m2 ,
h
e−θA /ρ φA (θA )
i1/(mr0 )
1
η
≥ (1 − )e−θA /ρ φA (θA ).
√
A πmr0
3
Next we take A = A0 (η, σ ) so large that for A = A0 this is at least
(1 −
2η h ρe i(1−ρ)/ρ
2η −θ∞ /ρ
)e
)
φ∞ (θ∞ ) = (1 −
.
3
3 1−ρ
This then yields for m2 = m2 (σ, A0 (η, σ )) that
h
m
2 r0
Y
X
k1 +···+km2 r0 =m2 s0 i=1
1≤ki ≤A0
2η 1/ρ h ρe i(1−ρ)/ρ
d ki i1/(m2 r0 )
≥ (1 −
.
)d
(ki − 1)!
3
1−ρ
(2.16)
The sum in the left hand side here contains only terms with all ki ≤ A0 . One
therefore has for m = qm2 that
h
X
qm
2 r0
Y
k1 +···+kqm2 r0 =qm2 s0
1≤ki ≤A0
i=1
≥
h
X
d ki i1/(qm2 r0 )
(ki − 1)!
m
2 r0
Y
k1 +···+km2 r0 =m2 s0 i=1
1≤ki ≤A0
≥ (1 −
d ki iq/(qm2 r0 )
(ki − 1)!
2η 1/ρ h ρe i(1−ρ)/ρ
)d
.
3
1−ρ
(2.17)
ρ-percolation in high dimensions
429
On the other hand, any summand in the left hand side of (2.17) which has exactly
β` of the ki equal to ` equals
P
d
`
A0 h
`β` Y
`=1
iβ`
1
.
(` − 1)!
One can therefore write the expression inside the square brackets in the left hand
side of (2.17) as
X X(β)
β
where
P
d
`
A0 h
`β` Y
`=1
P
β
0 h
iβ`
iβ`
X X(β) Y
1
1
= d qm2 s0
,
(` − 1)!
(` − 1)!
A
β
`=1
(2.18)
stands for the sum over all β1 , . . . , βA0 ≥ 0 which satisfy
A0
X
`β` = qm2 s0
(2.19)
`=1
P
and (β) is the sum over all k1 , . . . , kqm2 r0 with exactly β` of the ki equal to `.
Since there are at most [1 + qm2 s0 ]A0 choices for β satisfying (2.19), there is a
choice of β for which
#
"
A0 h
iβ` 1/(qm2 r0 )
X(β) Y
1
(` − 1)!
`=1
2η 1/ρ h ρe i(1−ρ)/ρ
)d
≥ [1 + qm2 s0 ]−A0 /(qm2 r0 ) (1 −
. (2.20)
3
1−ρ
Finally, we can choose q = q(η, σ, A0 ) so that
[1 + qm2 s0 ]−A0 /(qm2 r0 ) (1 −
2η
) ≥ (1 − η).
3
We take m0 = qm2 and write α` for the value of β` for which (2.20) holds. Then
(2.4) holds by (2.19), and (2.5) is immediate from (2.20), and m0 only depends on
η and σ by construction.
t
u
We now take m0 and α` so that (2.4) and (2.5) hold. We further take
r
r = m0 r0 , s = m0 s0 , so that ρ = .
s
(2.21)
To define Nn we introduce marked paths of length ns. A marked path of length ns
is a pair (v, k) such that
v = {0, v1 , v2 , . . . , vns }
is an oriented path on ⺪d+ which starts at the origin (the vi here are vertices of ⺪d+ ),
and
k = (k1 , . . . , knr )
430
H. Kesten, Z. Su
is a sequence of nr integers ≥ 1 such that
for t = 1, . . . , n and ` ≥ 1, there are exactly
α` values of i ∈ ((t − 1)r, tr] with ki equal to `.
(2.22)
We define Nn as the number of marked paths (v, k) of length ns for which
all the edges (vKi −1 , vKi ), with Ki = k1 + · · · + ki , are open.
(2.23)
We shall call the edge (vi−1 , vi ) the i-th edge of the path v. We think of k as ‘marking’ the Ki -th edge for 1 ≤ i ≤ nr. Then Nn counts the number of marked paths
all of whose marked edges are open. Note that by virtue of (2.22) and (2.4)
X
`α` = m0 s0 = s and hence Ktr = ts, 1 ≤ t ≤ n. (2.24)
Ktr − K(t−1)r =
`≥1
Thus the edges whose number isP
a multiple of s are always marked. Also, the
number of Ki in ((j − 1)s, j s] is ` α` = m0 r0 = r.
The number of possible marks, that is the number of sequences k1 , . . . , knr
which satisfy (2.22) is
in
h
r!
Q
,
(2.25)
`≥1 α` !
because one can arrange r objects, α` of which belong to class `, and with objects in
the same class being indistinguishable, in exactly r!/α1 !α2 ! . . . ways. Let us write
K = Kn for the collection of possible marks. Its cardinality is given by (2.25). It
follows that
in
h
r!
.
(2.26)
Ep Nn = p nr d ns Q
`≥1 α` !
Next we start on an upper bound for Ep Nn2 . It is standard (see [CD83, Kes90])
that this second moment can be written as
XXXX
p2nr−L(v,k,w,m) ,
(2.27)
Ep Nn2 =
P
v
k
w
m
P
P
where v and w run over all oriented paths of length ns, while k and m run
over the class Kn . Moreover, if m = (m1 , . . . , mnr ) and Mi = m1 + · · · + mi ,
and v0 = w0 = 0 = the origin, then
P
L(v, k, w, m) := number of i ∈ [1, ns] for which vi−1 = wi−1 , vi = wi ,
(2.28)
and i = Kx = My for some x, y ≤ nr;
In words, L(v, k, w, m) is the number of edges common to v and w which are
marked in both k and m.
It is convenient to think of v, w as a pair of independent paths, each chosen
uniformly from the oriented paths of length ns. Let ⺠ = ⺠n denote the probability
measure which picks pairs of paths in this way and let ⺕ = ⺕n be expectation with
respect to ⺠. Then (2.27) can be rewritten as
XX
p−L(v,k,w,m) .
(2.29)
Ep Nn2 = p2nr d 2ns ⺕
k
m
ρ-percolation in high dimensions
431
Since the details of the estimate of the right hand side of (2.29) are somewhat
technical we first give an outline of the argument. The principal idea is to use the
edges with a number divisible by s as a kind of regeneration points. As observed
above, these edges are marked in all marked paths and Lemma 5 will show that
the right hand side of (2.29) is not much larger than essentially the same sum, but
with the expectation restricted only to pairs of paths v, w which for some q have
their edges numbered s, 2s, . . . , qs in common, and without common edges with
number > (q + 1)s. Once we have Lemma 5 the estimation of (2.29) reduces to
estimating a similar expression, but now with the expectation only over paths v, w
of length s which have ≤r marked edges in common (including the s-th edge).
More precisely the quantity U defined in (2.69) has to be estimated. This is done in
Lemma 6, basically by finding the worst possible configuration of common marked
edges.
Lemma 5 will be proven by estimating the ‘probability cost’ of restricting the
expectation in the right hand side of (2.29) to pairs v, w which have the edges numbered s, 2s, . . . , qs in common. This cost turns out to be controllable if we already
know that the gaps between the common edges of v and w are not too large. The
next lemma prepares for this by comparing the right hand side of (2.29) to the same
expression, but with the expectation restricted to pairs v, w whose common marked
edges do not skip a complete block of edges numbered ts + 1, ts + 2, . . . , (t + 1)s
(for some t). This comparison involves a series ε1 of ratios of random walk probabilities which is estimated in Lemma 4 by some general combinatorial and analytical
tools.
We shall write St for vt − wt , where vt (wt ) is the t-th point in v (w, respectively). If v and w are chosen according to the measure ⺠, then v and w are independent
oriented simple random walks on Zd+ and {St } is the difference between two independent simple random walks. Without loss of generality we may assume that these
random walks are running for all time instead of just till time ns. We can therefore
drop the subscript n from ⺠ and ⺕.
Lemma 3. Let
ηq = sup
⺠{Si+(q−1)s = 0}
i<2s
and
ε1 =
⺠{Si = 0}
∞
X
ηq .
q=2
Then
Ep Nn2 ≤ p2nr d 2ns
∞
X
j =0
p−j (1 + ε1 )j
X
⺕W (j, k, m, v, w),
(2.30)
k,m
where W (j, k, m, v, w) = Wn (j, k, m, v, w) is the number of choices of j edges
which are marked in k and m and which are common to v and w, and which are
such that ‘no block of the form (ts, (t + 1)s] is skipped by these common edges’.
432
H. Kesten, Z. Su
More precisely, if v, w, k, m are given, then W (j, k, m, v, w) equals the number
of j -tuples 1 ≤ i1 < i2 < · · · < ij ≤ ns such that for z = 1, . . . , j ,
viz −1 = wiz −1 , viz = wiz , and iz = Kx = My for some x, y ≤ ns,
(2.31)
and such that (with i0 = 0)
(t − 1)s < iz−1 ≤ ts implies iz ≤ (t + 1)s.
(2.32)
Proof. We write iz ∈ K to indicate that iz = Kx for some 1 ≤ x ≤ nr, and i ∈ K
to indicate that i1 , . . . , ij ∈ K. We define iz ∈ M and i ∈ M similarly.
Now if two sets of marks k, m are given, and some i1 < · · · < ij are such
that i ∈ K ∩ M, then the requirement that i1 , i2 , . . . , ij should be counted in
L(v, k, w, m) is equivalent to the conditions
viz −1 = wiz −1 , viz = wiz ,
1≤z≤j
(see (2.31)). We therefore obtain from (2.29) that
Ep Nn2 ≤ p2nr d 2ns
= p 2nr d 2ns
∞
X
∞
X
j =0
p−j
j =0
= p 2nr d 2ns
∞
X
XX X
p−j
k
XX X
p−j ⺠{Siz −1 = 0 and Siz = 0, 1 ≤ z ≤ j }
m i∈K∩M
k
X
p−j
⺠{viz −1 = wiz −1 and viz = wiz , 1 ≤ z ≤ j }
m i∈K∩M
j h
X X Y
i
d −1 ⺠{Siz −1 − Siz−1 = 0}
i1 <i2 <···<ij k:i∈K m:i∈M z=1
j =0
(2.33)
(the term with j = 0 is taken to be p 2nr d 2ns ).
Now define t (z) ≥ 1 for z = 1, . . . , j , as a function of i1 < i2 < · · · < ij , by
(t (z) − 1)s < iz−1 ≤ t (z)s,
and set
q(z) = t (z + 1) − t (z) (with t(1) = 0).
Fix a z0 and consider a choice of i1 , . . . , ij for which q(z0 ) ≥ 2 and consider the
contribution to the right hand side of (2.33) given by
X X
k:i∈K m:i∈M
Then take
iz0
=
d −j
j
Y
⺠{Siz −1 − Siz−1 = 0}.
(2.34)
z=1
iz
iz − (q(z0 ) − 1)s
if z ≤ z0 − 1
if z ≥ z0 .
(2.35)
ρ-percolation in high dimensions
433
Note that this definition still gives i10 < i20 < · · · < ij0 . Let us compare (2.34) to the
contribution
j
Y
X X
d −j
⺠{Siz0 −1 − Si 0 = 0}.
(2.36)
z−1
k:i0 ∈K m:i0 ∈M
z=1
As a first step for this comparison we prove that
X
X
1=
1,
(2.37)
k:i0 ∈K
k:i∈K
or, in words, the number of marks which contain {i1 , . . . , inr } is the same as the
0 }. To see this, recall that K = ts.
number of marks which contain {i10 , . . . , inr
tr
Thus if iz , x < z ≤ y, are the values iz in (ts, (t + 1)s], then for the ku with
ts < ku ≤ (t + 1)s, the restriction that i1 , . . . , inr ∈ K amounts to the condition
that
exactly α` of the ku with tr < u ≤ (t + 1)r equal `,
and for each x < z ≤ y there is some q for which
q
X
ku .
iz − ts = Kq − Ktr =
u=tr+1
These conditions depend only on the iz + s − diz /ses, that is, on the difference
between iz and the last multiple of s which is <iz . These conditions for the different
blocks (ts, (t + 1)s] are independent of each other. Since the change from the i to
the i 0 only ignores the blocks (ts, (t + 1)s], t (z0 ) ≤ t < t (z0 + 1) − 1, which
contain no iz , and shifts the blocks (ts, (t +1)s], t ≥ t (z0 +1)−1, by a multiple of
s, the collection of values iz − s − diz /ses, 1 ≤ z ≤ j , is the same as the collection
of values iz0 − s − diz0 /ses. This implies (2.37).
0
for all z 6= z0 and
Now note that iz − iz−1 = iz0 − iz−1
iz0 − iz0 −1 = iz0 0 − iz0 0 −1 + (q(z0 ) − 1)s.
Also
iz0 0 − 1 − iz0 0 −1 = iz0 − (q(z0 ) − 1)s − 1 − iz0 −1
≤ t (z0 + 1)s − (q(z0 ) − 1)s − (t (z0 ) − 1)s − 1 = 2s − 1.
An immediate consequence of (2.37) and these facts is that
X X
d −j
⺠{Siz −1 − Siz−1 = 0}
z=1
k:i∈K m:i∈M
=
j
Y
⺠{Siz0 −1 − Siz0 −1 = 0} X
⺠{Siz0 −1 − Si 0
z
0
≤ ηq(z0 )
0 −1
X
X
k:i0 ∈K
m:i0 ∈M
This is our basic estimate.
= 0}
d −j
k:i0 ∈K
j
Y
z=1
X
d −j
m:i0 ∈M
j
Y
z=1
⺠{Siz0 −1 − Si 0
z−1
⺠{Siz0 −1 − Si 0
= 0}.
z−1
= 0}
(2.38)
434
H. Kesten, Z. Su
As a consequence of (2.38) we obtain that
i
j h
X X Y
X
p 2nr d 2ns
d −1 ⺠{Siz −1 − Siz−1 = 0}
i1 <i2 <···<ij k:i∈K m:i∈M z=1
q(z0 )≥2
≤
X
ηq p2nr d 2ns
i
d −1 ⺠{Siz −1 − Siz−1 = 0} .
j h
X X Y
X
i1 <i2 <···<ij k:i∈K m:i∈M z=1
q(z0 )=0 or 1
q≥2
In turn this says
p
d
i
j h
X X Y
X
2nr 2ns
d −1 ⺠{Siz −1 − Siz−1 = 0}
i1 <i2 <···<ij k:i∈K m:i∈M z=1
≤ [1 + ε1 ] p2nr d 2ns
X
×
i
d −1 ⺠{Siz −1 − Siz−1 = 0} .
j h
X X Y
(2.39)
i1 <i2 <···<ij k:i∈K m:i∈M z=1
q(z0 )=0 or 1
In the same way we can successively add the conditions q(z) = 0 or 1, 1 ≤ z ≤ j ,
at the expense of having to insert a factor (1 + ε1 )j into the left hand side of (2.39).
We therefore obtain
p
d
2nr 2ns
i
j h
X X Y
X
d −1 ⺠{Siz −1 − Siz−1 = 0}
i1 <i2 <···<ij k:i∈K m:i∈M z=1
≤ [1 + ε1 ]j p2nr d 2ns
i
d −1 ⺠{Siz −1 − Siz−1 = 0} . (2.40)
j h
X X Y
X
×
i1 <i2 <···<ij
q(z)=0
k:i∈K m:i∈M z=1
or 1 for 1≤z≤j
But the condition q(z) = 0 or 1 is equivalent to (2.32), and
X
i1 <i2 <···<ij
q(z)=0
j h
Y
i
d −1 ⺠{Siz −1 − Siz−1 = 0}
z=1
or 1 for 1≤z≤j
is precisely ⺕W (j, k, m, v, w). The lemma therefore follows from (2.40) and (2.33).
t
u
The next lemma estimates ε1 (d, s). The estimates are basically the same as in
[CD83], but it is necessary for us to keep track carefully of the dependence on d
and s.
ρ-percolation in high dimensions
435
Lemma 4. There exists a d0 < ∞ (independent of s) such that for d ≥ d0 and
s 4 ≤ d it holds that
(3s)!
ε1 (d, s) ≤ 3 s ≤ d −s/5 .
(2.41)
d
Proof. For any t ≤ i + (q − 1)s we have
⺠{Si+(q−1)s = 0} = ⺠{vi+(q−1)s = wi+(q−1)s }
≤ max ⺠{vi+(q−1)s = u} ≤ max ⺠{vt = u}.
u∈⺪d+
u∈⺪d+
P
coordinate vector. If u = d`=1 γ` ξ` ∈ ⺪d+ for some
Recall that ξ` denotes the `-th P
non-negative
integers γ` with ` γ` 6= t, then ⺠{vt = u} = 0. On the other hand,
P
if ` γ` = t, then there are at most t! ways to order t steps, γ` of which are in the
`-th direction. Therefore
⺠{St = 0} ≤ max ⺠{vt = u} ≤
u∈⺪d+
t!
.
dt
(2.42)
In the other direction we have the almost trivial bound
⺠{Si = 0} ≥ Pmin ⺠{vi =
γ` =i
X
γ` ξ` } ≥
1
.
di
(2.43)
It follows that for any t ≤ i + (q − 1)s,
P{Si+(q−1)s = 0}
≤ t!d i−t .
P{Si = 0}
(2.44)
We shall use the estimate (2.44) for q ≤ 2d/4 . First we observe that for i ≤ 2s
and (` + 2)s ≤ d/2
[(` + 2)s]s
(i + `s)!d −`s
≤
≤ 2−s .
ds
(i + (` − 1)s)!d −(`−1)s
Thus for i ≤ 2s, (q + 2)s ≤ d/2,
P{Si+(q−1)s = 0}
≤ i + (q − 1)s !d −(q−1)s
P{Si = 0}
q−1
(i + s)! Y
(i + `s)!d −`s
=
ds
i
+
(`
−
1)s
!d −(`−1)s
`=2
≤ 2−(q−2)s
Therefore
bd/(2s)c−2
X
q=2
ηq ≤
(i + s)!
(3s)!
≤ 2−(q−2)s s .
ds
d
X (3s)!
q≥2
ds
2−(q−2)s ≤
2(3s)!
.
ds
(2.45)
436
H. Kesten, Z. Su
Moreover, by (2.44) for t = [bd/(2s)c−2]s, one has for i ≤ 2s and (q +2)s > d/2
P{Si+(q−1)s = 0}
(3s)!
≤ 2−(bd/(2s)c−5)s s .
P{Si = 0}
d
This yields, for d ≥ some suitable d0 ,
d/4
b2
Xc
ηq ≤ 26s · 2−d/4
q=bd/(2s)c−1
(3s)!
(3s)!
≤
s
d
2d s
(2.46)
(recall s 4 ≤ d).
For q > 2d/4 we follow [CD83]. Let R = b i + (q − 1)s /dc. Then
X
⺠{Si+(q−1)s = 0} ≤ Pmax ⺠{vRd =
γ` ξ` }
γ` =Rd
=
Pmax
γ` =Rd
(Rd)!
(Rd)!
d −Rd =
γ1 ! · · · γd !
(R!)d d Rd
≤ C1 eC2 d/R R (1−d)/2 d 1/2 (2π )−d/2 .
(2.47)
In the last inequality we used Stirling’s formula (see [WW52], Section 12.33); C1
and C2 are universal constants. Together with (2.43) this gives for d ≥ some d0
(independent of s) and q > 2d/4
ηq ≤ C3 d 2s+d/2 (qs)(1−d)/2 π −d/2 .
2d/4
Summing over q >
and taking into account that
large enough and d ≥ d0
X
ηq ≤ π −d/2 .
s4
(2.48)
< d now yields for d0
q>2d/4
Together with (2.45) and (2.46) this implies the lemma.
t
u
Lemma 5. Let W ∗ (j, k, m, v, w) = Wn∗ (j, k, m, v, w) be the number of j -tuples
1 ≤ i1 < · · · < ij ≤ ns which satisfy (2.31) and (2.32), and for which there exists
a q ≤ n such that s, 2s, . . . , qs each equal one of the iz , and ij ≤ (q + 1)s. Then,
⺕
∞
XX
p−j (1 + ε1 )j W (j, k, m, v, w)
k,m j =0
≤⺕
∞
XX
p−j (1 + ε1 + 28s p)j W ∗ (j, k, m, v, w).
(2.49)
k,m j =0
Proof. For the same reasons as given for (2.33) we have
o
nP
∞ P
⺕
p−j (1 + ε1 )j W (j, k, m, v, w)
j =0 k,m
∞
P P
=
k,m j =0
p−j (1 + ε1 )j
P
i1 <i2 <···<ij ∈K∩M
q(z)=0
i
d −1 ⺠{Siz −1 − Siz−1 = 0} .
j h
Q
z=1
or 1 for 1≤z≤j
(2.50)
ρ-percolation in high dimensions
437
We now want to compare the right hand side here with the same expression but
with the added restriction on the iz that there exists a q for which
s, 2s, . . . , qs each equal one of the iz and ij ≤ (q + 1)s.
(2.51)
Consider a choice of i1 < · · · < ij in the right hand side of (2.50). Let q0 be such
that q0 s < ij ≤ (q0 + 1)s. In order to obtain a summand which also satisfies (2.51)
we add the indices s, 2s, . . . , q0 s to the set i1 , . . . , ij ; of course if ts is already
equal to some i` , then we do not have to add ts. Let i10 < it0 < · · · < ij0 0 = ij
be the set {i1 , . . . , ij } ∪ {s, . . . , q0 s} arranged in increasing order. We shall say
that this sequence i10 , . . . , ij0 0 is associated to i1 , . . . , ij . For given i1 , . . . , ij the
associated sequence is unique. Note also that i0 ∈ K ∩ M, because ts ∈ K ∩ M for
ts ≤ ij ≤ ns, by virtue of (2.24). If
iz−1 < ts < iz
(2.52)
and the point ts is added to the set of i’s, then the factor
d −1 ⺠{Siz −1 − Siz−1 = 0}
(2.53)
in the right hand side of (2.50) has to be replaced by
p −1 (1 + ε1 )d −1 ⺠{Sts−1 − Siz−1 = 0}d −1 ⺠{Siz −1 − Sts = 0};
(2.54)
the extra factor p −1 (1 + ε1 ) comes in because we have added one more point,
so that j should be changed to j + 1. The ratio of the expression (2.53) and the
expression (2.54) is
p(1 + ε1 )−1
d −1 ⺠{Siz −1 − Siz−1 = 0}
.
d −1 ⺠{Siz−1 − Sts = 0}d −1 ⺠{Sts−1 − Siz−1 = 0}
(2.55)
We claim that for a, b = 1, 2, . . . it holds that
⺠{Sa+b = 0}
≤ 22(a+b) .
⺠{Sa = 0}⺠{Sb = 0}
(2.56)
To see this we first note that
X
⺠{Sa+b = 0} = ⺠{va+b = wa+b } =
` γ` =a+b
X
=
P
γ` :
Now for each sequence γ1 , γ2 , . . . , γd with
ative integers σ` , τ` satisfying
`=1
σ` = a,
d
X
`=1
⺠{va+b =
P
γ` :
d
X
P
2
γ` ξ` }
`=1
d −(a+b)
` γ` =a+b
` γ`
d
X
(a + b)! 2
.
γ1 ! · · · γd !
= a + b we can choose nonneg-
τ` = b and σ` + τ` = γ` .
(2.57)
438
H. Kesten, Z. Su
For instance, if m is such that
X
γ` < a ≤
`<m
then we can take
(
σ` =
X
γ` ,
`≤m
γ` P
a − `<m γ`
0
if ` < m
if ` = m
if ` > m,
and τ` = γ` − σ` . It is further clear that each collection σ` , τ` which satisfies
(2.57) is selected for at most one sequence γ` , because γ` = σ` + τ` . In addition,
if γ` = σ` + τ` , then γ` ! ≥ σ` !τ` !. It follows that
X
d −(a+b)
P
γ` :
` γ` =a+b
X
≤
=
X
P
σ` :
(a + b)! 2
γ1 ! · · · γd !
P
` σ` =a τ` :
d −(a+b)
` τ` =b
X
(a + b)! 2
a!b!
P
σ` :
P
σ` :
d −a
` σ` =a
X
≤ 22(a+b)
⺠{va =
` σ` =a
2
(a + b)!
σ1 ! · · · σd !τ1 ! · · · τd !
X
2
a!
σ1 ! · · · σd !
d
X
τ` :
d −b
` τ` =b
X
2
σ` ξ` }
`=1
P
P
τ` :
2
b!
τ1 ! · · · τd !
⺠{vb =
` τ` =b
d
X
2
τ` ξ` }
`=1
= 22(a+b) ⺠{Sa = 0}⺠{Sb = 0}.
(2.58)
This finally proves our claim (2.56).
As a special case of (2.56) we have the rather crude bound
⺠{Siz −1 − Siz−1 = 0} ≤ 22(iz −1−iz−1 ) ⺠{S1 = 0}P {Siz −2−iz−1 = 0}
= 22(iz −1−iz−1 ) d −1 P {Siz −2−iz−1 = 0}.
This bound, together with (2.56), shows that (2.55) is at most
p(1 + ε1 )−1 24(iz −iz−1 ) .
Because (2.50) only contains contributions from sequences i1 < · · · < ij with
q(z − 1) = 0 or 1 and q(z) = 0 or 1, or, equivalently, satisfying (2.32), (2.52)
forces
ts − iz−1 ≤ s and iz − ts ≤ s.
For such i1 , . . . , ij we see that (2.55) is at most
(1 + ε1 )−1 p28s .
(2.59)
ρ-percolation in high dimensions
439
In (2.59) we have given an upper bound for the factor by which the right hand
side of (2.50) is multiplied if we add one point ts to i1 , . . . , ij . If i10 , . . . , ij0 0 is
formed from i1 , . . . , ij by adding j 0 − j points of the form ts, then
p −j (1 + ε1 )j
j h
Y
d −1 ⺠{Siz −1 − Siz−1 = 0}
i
z=1
j0 h
i
p28s j 0 −j
Y
−j 0
j0
0
p (1 + ε1 )
= 0} .
d −1 ⺠{Siz0 −1 − Siz−1
≤
1 + ε1
(2.60)
z=1
However, we must compare not just a contribution coming from a single sequence
i1 , . . . , ij with the contribution for the associated i10 , . . . , ij0 0 , because the same sequence i 0 , . . . , ij0 0 is associated to many sequences i1 , . . . , ij . A given i10 , . . . , ij0 0
which contains s, 2s, . . . , q0 s and satisfies ij0 0 ≤ (q0 + 1)s, is associated with any
i1 , . . . , ij which differs from i10 , . . . , ij0 0 by a subset of {s, 2s, . . . , q0 s}. There are
q0
j0 − j
possibilities for such i1 , . . . , ij (when j is fixed). The sum of
p
−j
i
j h
Y
j
d −1 ⺠{Siz −1 − Siz−1 = 0}
(1 + ε1 )
z=1
over all the sequences which have the same associated sequence i10 , . . . , ij0 0 is
therefore at most
p28s j 0 −j
q0
j0 − j
1 + ε1
times the contribution
0
p
−j 0
j0
(1 + ε1 )
z=1
coming from the sequence
∞
X
X
i
j h
Y
0
= 0}
d −1 ⺠{Siz0 −1 − Siz−1
i10 , . . . , ij0 0 .
p−j (1 + ε1 )j
j =0 i1 <i2 <···<ij has
associated sequence
i10 ,...,ij0 0
≤p
−j 0
i
j h
Y
d −1 ⺠{Siz −1 − Siz−1 = 0}
z=1
j0
j0 i
X
p28s ` Y h −1
q0
0
(1 + ε1 )
= 0}
d ⺠{Siz0 −1 − Siz−1
`
1 + ε1
j0
`=0
0
It follows that
≤ p −j (1 + ε1 + 28s p)j
z=1
0
j0 h
Y
z=1
i
0
= 0}
d −1 ⺠{Siz0 −1 − Siz−1
(2.61)
440
H. Kesten, Z. Su
(because q0 ≤ j 0 ). Finally we sum both sides of (2.61) over i10 , . . . , ij0 0 ∈ K ∩ M
and over all k, m. Since each sequence i1 , . . . , ij has a unique i10 , . . . , ij0 0 associated
to it, we see from (2.50) that the sum of the left side is precisely
⺕
∞ X
nX
o
p−j (1 + ε1 )j W (j, k, m, v, w) .
j =0 k,m
The sum of the right side of (2.61) is
∞
XX
X
k,m j 0 =0
i10 <···<ij0 0 ∈K∩M,
0
p−j (1 + ε1 + 28s p)j
0
for some q, ij0 0 ≤(q+1)s and
{s,2s,...,qs}⊂{i 0 ,...,ij0 0 }
0
×
z=1
=⺕
i
j h
Y
0
= 0}
d −1 ⺠{Siz0 −1 − Siz−1
∞
XX
p−j (1 + ε1 + 28s p)j W ∗ (j, k, m, v, w).
(2.62)
k,m j =0
t
u
This proves (2.49).
The inequalities (2.30), (2.49) and the equality (2.26) together show that
−2
Ep Nn2 Ep Nn
∞
i−2n X X
h
r!
≤ Q
⺕
p−j (1 + ε1 + 28s p)j W ∗ (j, k, m, v, w).
`≥1 α` !
(2.63)
k,m j =0
We shall now show that the right hand side of (2.63) is bounded in n for suitable p.
Lemma 6. Let η > 0 and let σ, m0 and α` be as in Lemma 2 such that (2.5) holds.
If
(2.64)
p(1 + ε1 + 28s p)−1 rs2s ≤ η
and
h 1 − ρ i(1−ρ)/ρ
d 1/ρ p ≥ (1 − η)−1−2/r (1 + ε1 + 28s p)
ρe
= (1 − η)−1−2/r (1 + ε1 + 28s p)D1 ,
(2.65)
then there exists a C4 = C4 (η, p, r, s, α1 , . . . , αs ) such that
h
Q
r!
`≥1 α` !
∞
i−2n X X
⺕
k,m j =0
p−j (1 + ε1 + 28s p)j W ∗ (j, k, m, v, w) ≤ C4 . (2.66)
ρ-percolation in high dimensions
441
Proof. We begin with an estimate for the left hand side of (2.62). We decompose
the sum over j 0 and over the permissible i10 , . . . , ij0 0 according to the minimal value
of q for which ij0 0 ≤ (q + 1)s, that is, q + 1 = dij0 0 /se. If dij0 0 /se = q0 + 1, then
the conditions on i10 , . . . , ij0 0 are that each ts, 1 ≤ t ≤ q0 is one of the iz0 , and
further that each iz0 ∈ K ∩ M. The conditions on k, m are just (2.22). All these are
conditions which apply separately to the Kx , My and iz0 in each block ((t − 1)s, ts].
More precisely the values Kx − (t − 1)s for which Kx ∈ ((t − 1)s, ts] are restricted
by (2.22), and similarly for the My . Moreover, the iz0 in this same block have to lie in
K ∩ M ∩ ((t − 1)s, ts], or equivalently, iz0 − (t − 1)s has to equal some Kx − (t − 1)s
and some My − (t − 1)s. However, the values of the Kx , My in the different blocks
do not influence each other. For dij0 0 /se = q0 + 1 the contribution to the left hand
side of (2.62) can therefore be written as a product of n contributions, each coming
from a separate block ((t − 1)s, ts]. The first q0 factors corresponding to the blocks
((t − 1)s, ts], 1 ≤ t ≤ q0 , are all equal to
r
X X
U :=
X
p−j (1 + ε1 + 28s p)j
k1 ,...,kr j =1 i1 <···<ij ∈K∩M
m1 ,...,mr
ij =s
×
i
d −1 ⺠{Siz −1 − Siz−1 = 0} ,
j h
Y
(2.67)
z=1
kr = s)
where this time the ki have to satisfy (2.22) with t = 1 (and automaticallyP
while the restrictions onP
i1 , . . . , ij are that each iz must equal some Kx = xi=1 ki
y
and equal some My = i=1 mi with x, y ≤ r, and that ij = s. We may start the
sum over j at j = 1, because s has to be one of the iz ; this is not an important point
though.
The block (q0 s, (q0 + 1)s] plays a special role, because (q0 + 1)s does not have
to equal one of the iz0 . The factor corresponding to this block is therefore
r
X X
V :=
X
p−j (1 + ε1 + 28s p)j
k1 ,...,kr j =0 i1 <···<ij ∈K∩M
m1 ,...,mr
ij ≤s
×
i
d −1 ⺠{Siz −1 − Siz−1 = 0} .
j h
Y
(2.68)
z=1
V differs from U only in that the condition ij = s has been relaxed to ij ≤ s and
that now j = 0 is possible; the former condition just expresses that the iz which
are >q0 s must be ≤(q0 + 1)s, that is, they can exceed q0 s by no more than s.
Finally, each of the blocks ((t − 1)s, ts], q0 + 1 < t ≤ n contains no indices
iz0 so that the factor corresponding to any of these blocks is
i2
h
X
r!
1= Q
.
`≥1 α` !
k,m
442
H. Kesten, Z. Su
This is so, because the number of choices for k satisfying (2.22) with t = 1 is
exactly
r!
Q
,
`≥1 α` !
by (2.25), and the number of choices for m is the same.
We conclude that the left hand side of (2.62) equals
n−1
X
h
r!
UqV Q
i2(n−q−1)
`≥1 α` !
q=0
.
(2.66) is therefore equivalent to
h
Q
n−1
i−2n X
r!
`≥1 α` !
h
r!
UqV Q
i2(n−q−1)
`≥1 α` !
q=0
≤ C4 .
V is a finite sum and is therefore always finite. Thus, it suffices for (2.66) to prove
that under the conditions (2.64) and (2.65), it holds that
U=
r
X X
X
p−j (1 + ε1 + 28s p)j
k1 ,...,kr j =1 i1 <···<ij ∈K∩M
m1 ,...,mr
ij =s
×
i
j h
Y
d −1 ⺠{Siz −1 − Siz−1 = 0}
z=1
h
r!
≤ (1 − η) Q
i2
`≥1 α` !
.
(2.69)
To prove this we look separately at the contributions to the left hand side of (2.69)
corresponding to different values of j . The main contribution will be the one corresponding to the maximal value of j , that is, the contribution coming from the
terms with j = r. Since there are only r values of Kx and of My , these can only
have r common values if Kt = Mt for t = 1, . . . , r, that is if k = m. To get r
values iz in K ∩ M we must also have iz = Kz = Mz , 1 ≤ z ≤ r. Thus, the terms
corresponding to j = r contribute to U
p −r (1 + ε1 + 28s p)r
i
r h
X Y
d −1 ⺠{SKz −1 − SKz−1 = 0}
k1 ,...,kr z=1
≤ p −r (1 + ε1 + 28s p)r
= p −r (1 + ε1 + 28s p)r
r
X Y
(Kz − 1 − Kz−1 )!
(see (2.42))
d Kz −Kz−1
k1 ,...,kr z=1
r
X Y
k1 ,...,kr z=1
(kz − 1)!
d kz
ρ-percolation in high dimensions
443
= p−r (1 + ε1 + 28s p)r d −s Q
s
Y
r!
`≥1 α` ! `=1
α`
(` − 1)!
(2.70)
(by (2.6) and (2.25)).
In the same way we find that the contributions to U of the terms with j < r is
at most
r−1
X X
X
p
−j
j −s
(1 + ε1 + 2 p) d
8s
k1 ,...,kr j =1 i1 <···<ij ∈K∩M
m1 ,...,mr
ij =s
j
Y
(iz − iz−1 − 1)! (2.71)
z=1
(where i0 is interpreted as 0). By dropping the condition that i1 , . . . , ij ∈ M and
summing over all possibilites for m1 , . . . , mr , we see that (2.71) is at most
Q
r−1
X X
r!
X
`≥1 α` ! k1 ,...,kr j =1 i1 <···<ij ∈K
ij =s
×
j
Y
p−j (1 + ε1 + 28s p)j d −s
(iz − iz−1 − 1)!
(2.72)
z=1
Let us fix k1 , . . . kr , and hence K1 , . . . Kr for the moment. Then in the above sum
i1 , i2 , . . . , ij have to be chosen from K1 , . . . , Kr . We can choose these j indices
by first choosing j + 1 indices and then dropping one of them. (We cannot drop
the highest one, though; that has to stay equal to s.) Since there are at most r possibilities for each iz , each possible choice of j indices results from at most r sets of
j + 1 indices by this procedure. Moreover, if i1 < i2 < · · · < ij is obtained from
i1 < · · · < it < i < it+1 < · · · < ij by discarding the index i, then
j
Y
(iz − iz−1 − 1)! =
z=1
t
Y
(it+1 − it − 1)!
(iz − iz−1 − 1)!
(i − it − 1)!(it+1 − i − 1)!
z=1
×(i − it − 1)!(it+1 − i − 1)!
j
Y
(iz − iz−1 − 1)!
z=t+2
≤ s2s
t
Y
(iz − iz−1 − 1)! (i − it − 1)!(it+1 − i − 1)!
z=1
×
j
Y
(iz − iz−1 − 1)! .
z=t+2
From these observations it follows that under (2.64),
X
i1 <···<ij ∈K
ij =s
p−j (1 + ε1 + 28s p)j d −s
j
Y
(iz − iz−1 − 1)!
z=1
444
H. Kesten, Z. Su
≤ p(1 + ε1 + 28s p)−1 rs2s
X
p−j −1 (1 + ε1 + 28s p)j +1 d −s
i1 <···<ij +1 ∈K
ij +1 =s
×
jY
+1
(iz − iz−1 − 1)!
z=1
X
≤η
p−j −1 (1 + ε1 + 28s p)j +1 d −s
i1 <···<ij +1 ∈K
ij +1 =s
jY
+1
(iz − iz−1 − 1)!.
(2.73)
z=1
Iteration of the last inequality and summing over j now shows that the sum in (2.72)
is at most equal to η/(1 − η) times the expression in the second member of (2.70).
Consequently,
U ≤ (1 − η)−1 p−r (1 + ε1 + 28s p)r d −s Q
s
Y
r!
`≥1 α` ! `=1
α`
(` − 1)!
.
However, by our choice of the α` (see (2.5)) we have
Q
r!
s h
Y
`≥1 α` ! `=1
s h
iα`
Y
r!
d ` iα`
1
ds
= Q
(` − 1)!
(` − 1)!
`≥1 α` !
`=1
h ρe i(1−ρ)s
≥ (1 − η)r d s
.
1−ρ
It follows that
U ≤ (1 − η)−r−1 p−r (1 + ε1 + 28s p)r d −s
h ρe i−(1−ρ)s h
i2
r!
Q
. (2.74)
1−ρ
`≥1 α` !
Thus, by virtue of (2.65), (2.69), and hence (2.66), holds.
t
u
Lemma 7. For every 0 < ρ ≤ 1 there exists a sequence of ordered pairs of integers
{r0 (d), s0 (d)}d≥2 which has the following properties:
1 ≤ r0 (d) ≤ s0 (d), (r0 (d), s0 (d)) = 1,
r0 (d)
≥ ρ;
s0 (d)
(2.75)
with ζ = (16m0 ρ log 2)−1
s0 (d) ≤ ζ log d for
and
0≤
d ≥ some d1 = d1 (ρ);
1
r0 (d)
− ρ = o(
).
s0 (d)
log d
(2.76)
(2.77)
ρ-percolation in high dimensions
445
Proof. This is a well known fact from elementary number theory. If ρ is rational, then we simply take r0 , s0 relatively prime and such that ρ = r0 /s0 . Then
(2.75)–(2.77) trivially hold.
If ρ is irrational, then we take r0 and s0 relatively prime such that r0 (d)/s0 (d)
is the smallest rational number ≥ρ with denominator s0 (d) ≤ ζ log d. In other
words, r0 (d)/s0 (d) is the nearest number to the right of ρ in the Farey series of
order bζ log dc (see [HW54], Chapter 3). If r 0 /s 0 = r 0 (d)/s 0 (d) is the rational number in this Farey series which is immediately to the left of ρ, then (see [HW54],
Theorems 28, 30) s 0 + s0 (d) ≥ ζ log d and
0≤
r0
r0
1
2
r0
−ρ ≤
− 0 = 0 ≤
.
s0
s0
s
s s0
ζ log d min(s 0 , s0 )
Since both s 0 and s0 tend to ∞ with d, because ρ is irrational, (2.77) follows. (2.75)
and (2.76) hold by construction.
t
u
Proof of Theorem 1. The case ρ = 1 follows from the fact that
lim inf dpc (1, ⺪d+ , bond) ≥ 1
d→∞
(see (1.21)–(1.23)), in conjunction with the result of [CD83] that
lim dpc (⺪d+ , bond) = 1,
d→∞
where pc (⺪d+ , bond) is the critical probability for ordinary oriented bond percolation on ⺪d+ . (Recall that pc (1, ⺪d+ , bond) ≤ pc (⺪d+ , bond) just as in (1.12).)
From now on we take 0 < ρ < 1 and we fix σ such that 0 < σ ≤ ρ ≤ 1 − 2σ .
Now let 0 < η ≤ 1/2 and choose m0 = m0 (η, σ ) and α` as in Lemma 2, so that
(2.4) and (2.5) hold. Finally, let p = p(d) be such that
(1 + 2η)5 D1 ≤ d 1/ρ p ≤ 2D1 .
(2.78)
Now take r0 (d), s0 (d) with the properties (2.75)–(2.77) and define
ρ
b= ρ
b(d) =
r0 (d)
,
s0 (d)
r = r(d) = m0 r0 (d),
s = s(d) = m0 s0 (d).
(2.79)
We claim that for any such p(d), (2.64) and (2.65) with ρ replaced by ρ
b hold for
all large d. To see this, note first that 1 ≤ s = s(d) ≤ m0 ζ log d, so that
ε1 (d, s) ≤ d −1/5 → 0
(d → ∞),
by virtue of (2.41). Moreover,
ρ
b− ρ =
1
r0
− ρ = o(
),
s0
log d
and consequently also
0≤
1
1
1
− = o(
).
ρ
ρ
b
log d
446
H. Kesten, Z. Su
This implies that for large d
ρ
ρ
d 1/b
≤ d 1/ρ ≤ (1 + η)d 1/b
.
In addition
28s p ≤ d 8m0 ζ log 2 2D1 d −1/ρ → 0
(d → ∞).
Thus, for sufficiently large d,
1 + ε1 + 28s p ≤ 1 + η.
(2.80)
The inequality (2.65) with ρ replaced by ρ
b now holds, because
(1 − η)−1−2/r (1 + ε1 + 28s p)D1 ≤ (1 − η)−3 (1 + η)D1
ρ
≤ (1 + 2η)4 D1 ≤ (1 + 2η)−1 d 1/ρ p ≤ d 1/b
p.
(2.64) is obvious for similar reasons, because also
prs2s ≤ [m0 ζ (log d)]2 d m0 ζ log 2−1/ρ 2D1 → 0
(d → ∞).
We can now apply (2.63) and Lemma 6. This yields that for all large n
Pp {Nn > 0} ≥
[Ep Nn ]2
1
≥
.
C4
Ep Nn2
From the definition (2.23) we see that the event {Nn+1 > 0} is the event that the
(n + 1) blocks ((t − 1)s, ts], 1 ≤ t ≤ n + 1, each contain r marks which satisfy
(2.22) and (2.23). This immediadetly implies that the same holds for the first n
blocks corresponding to t = 1, . . . , n. In other words, {Nn+1 > 0} ⊂ {Nn > 0}.
We therefore also have
Pp {Nn > 0 for all n} ≥
1
C4
(2.81)
as soon as (2.64) and (2.65) are satisfied. But {Nn > 0 for all n} implies that there
exists an oriented path with r open edges in each block ((t − 1)s, ts], so that ρ
bpercolation occurs. Since ρ
b ≥ ρ, also ρ-percolation occurs. The only restriction
on p(d) for this was (2.78). We have therefore proven that
lim sup d 1/ρ pc (ρ, ⺪d+ , bond) ≤ (1 + 2η)5 D1 ,
d→∞
for any η with (1 + 2η)5 < 2.
t
u
Acknowledgements. The work of H.K. was supported by an NSF Grant to Cornell University. Z.S. was supported by a Fellowship from the China Scholarship Council, which he used
to visit the Mathematics Department at Cornell University for the year 1998. The research
for this article was carried out during this year, and Z.S. thanks the Mathematics Department
at Cornell for its hospitality.
ρ-percolation in high dimensions
447
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