Lecture 22 - Dec. 1, 2008
EE 87021: Advanced Topics in Random Wireless Networks
Scribes: Martin Haenggi and Sundaram Vanka
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Executive Summary
We begin with a proof of the integral form of Russo’s formula. The effect of occupation probability p on
cluster sizes is then investigated. For some critical probablity pc , this is studied for the subcritical (p < pc )
and supercritical (p > pc ) phases, and also at the critical threshold p = pc .
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Proof of Integral Form of Russo’s Formula
Corollary 1 [Integral form of Russo’s formula] Let A be an increasing event, which depends on the state
of finitely many edges of L, and let N (A) denote the number of edges that are pivotal for A. Then for any
0 6 p1 < p2 6 1,
Z p2
1
Pp2 (A) = Pp1 (A) exp
Ep (N (A) | A)dp .
p1 p
Proof:
X
d
Pp (A) =
P(e is pivotal for A)
dp
e∈E
1X
P({e is pivotal for A} ∩ {e is open})
=
p
e∈E
1X
=
P({e is pivotal for A} ∩ A)
p
e∈E
1X
P(e is pivotal for A | A)Pp (A)
=
p
e∈E
=
1
E(N (A) | A)Pp (A) .
p
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3.1
Critical Threshold for Bond Percolation on L
Subcritical Phase: Exponential decrease of the radius of the mean cluster size
Let S(n) be the diamond of radius n (i.e., the ball of radius n with the Manhattan distance). Let An =
{0 ↔ ∂S(n)} be the event that there exists an open path connecting the origin to any vertex lying on
the surface of S(n). Defining the radius of the origin’s cluster C by rad C = maxx∈C |x|1 , we see that
An = {rad C > n}.
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Theorem 2 If p < pc , there exists ψ(p) > 0 such that
P(rad C > n) = P(An ) < exp(−nψ(p)) .
3.2
Supercritical Phase: Uniqueness of the Infinite Open Cluster
Theorem 3 If p > pc , then
P(there exists exactly one infinite open cluster) = 1 .
Proof: (Sketch.) Trivial if p ∈ {0, 1}, so assume 0 < p < 1. Let N be the number of infinite open
clusters, and let B be a finite set of vertices and EB the set of edges of L joining pairs of vertices in B. We
write NB (0) for the number of infinite open clusters when all edges in EB are closed, and NB (1) for when
they are open. Finally, MB denotes the number of infinite open clusters which intersect B (have a common
vertex with B).
Since the sample space Ω = {0, 1}E is a product space with a natural family of translations, P is a
product measure on Ω, and N is translation-invariant on Ω, it is a.s. constant, i.e.
∃k ∈ N ∪ {0, ∞}
s.t. P(N = k) = 1 .
Naturally the value of k depends on p.
We show first that k ∈ {0, 1, ∞}. Suppose to the contrary that 2 6 k < ∞. Since every configuration
on EB has a strictly positive probability, it follows by the almost sure constantness of N that
P(NB (0) = k) =
P({NB = k} ∩ {all edges of S(n) are closed})
=1
P({all edges of S(n) are closed})
The same holds for the case when all edges in B are open, so
P(NB (0) = NB (1) = k) = 1 .
Now NB (0) = NB (1) iff B intersects at most one infinite open cluster, since if it intersected two or more,
then it would matter whether B’s edges are open or not: if they were open the clusters would be joined.
Therefore
P(MB > 2) = 0 .
Now take B = S(n) to be the diamond of radius n, such that S(n) → Z2 as n → ∞. Hence
0 = lim P(MS(n) > 2) = P(N > 2) ,
n→∞
in contradiction with P(NB = k) = 1 for some k > 2.
Suppose next that k = ∞. This assumption also leads to a contradiction if the mean number of trifurcations in a diamond S(n) is calculated. A node x is called a trifurcation if it lies in an infinite open cluster,
has (open) degree 3, and the deletion of x and its three open edges splits the infinite cluster into exactly
three disjoint infinite clusters and no finite clusters. If Tx is the event that x is a trifurcation, then it can be
shown (under the assumption that k = ∞) that for each x ∈ Z2 , P(Tx ) > 0. So the number of trifurcations
in S(n) grows like |S(n)| ∼ n2 . On the other hand, by construction, each trifurcation in S(n) is connected
to a unique vertex on ∂S(n) which grows only linearly in n. This is the contradiction.
Note that bifurcations would not be sufficient, since if they lie on the same line, bifurcations would not give
rise to an additional point on ∂S(n).
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3.3
Critical Threshold
Lemma 4 [Lower bound]
θ(1/2) = 0 .
Proof: Proceed (again) by contradiction. Suppose that θ(1/2) > 0. Consider the square B(n) =
[−n, n]2 and let Ai , i ∈ {t, b, l, r} be the event that some vertex on the top,bottom,left,right side of B(n)
belongs to an infinite open path of L that uses no other vertex of B(n). These are four increasing events of
equal probability whose union is the event that some vertex of ∂B(n) belongs to the infinite cluster. Since
we assume θ(1/2) > 0, there is a.s. an infinite cluster, and hence as n → ∞,




 [



i
P1/2 
A (n) → 1


i∈{t,b,l,r}

|
{z
}
A(n)
Now we are using the 4-th root trick (and the fact that the 4 probabilities are all equal) to obtain
1/4
P1/2 (Ai (n)) > 1 − 1 − P1/2 (A(n))
so that
P(Ai (n)) → 1
as n → ∞. Therefore there is n0 large enough such that
P(Ai (n0 )) > 7/8 .
Now consider the dual box Bd (n) defined as
Bd (n) = {(i + 1/2, j + 1/2) | (i, j) ∈ B(n)} ,
and let Aid (n) be the event that some vertex on the respective side of Bd (n) belongs to an infinite closed path
of Ld that uses no other vertex of Bd (n). Each edge of Ld is closed with probability 1/2, so P1/2 (Aid (n)) =
P1/2 (Ai (n)) for all n, and, in particular,
P1/2 (Aid (n0 )) > 7/8 .
Consider the event A = Al (n0 ) ∩ Ar (n0 ) ∩ Atd (n0 ) ∩ Abd (n0 ) that there exist infinite open paths of L
connecting to the left and right sides of B(n0 ) without using any other vertex of B(n0 ) and that there exist
infinite closed path connecting to some vertex on the top and bottom sides of Bd (n0 ) without using any
other vertex of Bd (n0 ). Using the union bound,
P1/2 (A) = 1 − P Āl (n0 ) ∪ Ār (n0 ) ∪ Ātd (n0 ) ∪ Ābd (n0 )
> 1 − P1/2 (Āl (n0 )) + P1/2 (Ār (n0 )) + P1/2 (Ātd (n0 )) + P1/2 (Ābd (n0 ))
> 1/2
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since P1/2 (Āi (n0 )) < 1/8. If A occurs there must be two infinite open clusters in L\B(n0 ), one containing
the infinite open path connected to the left side of B(n) and the other containing the infinite open path
connected to the right side of B(n). These two infinite open clusters must be disjoint because they are
separated by two infinite closed path in Ld \ B(n0 ).
The same reasoning implies that there must be two disjoint infinite closed clusters in Ld \ Bd (n0 ), one
containing an infinite path starting from the top and the other from the bottom, and separated by the two
infinite open paths of L \ B(n0 ). Now since θ(1/2) > 0, the uniqueness theorem says that there is exactly
one infinite open cluster. Therefore, there must be a left-right crossing within B(n) which forms a barrier
to any top-bottom closed crossing of Bd (n). As a consequence there must be at least two disjoint infinite
closed clusters in Ld , but the probability that there are two infinite closed clusters in Ld is the same as the
probability that there are two infinite open clusters in L which is zero.
We have thus reached a contradiction, which means that P1/2 (A) has to be zero. The assumption
θ(1/2) > 0 cannot be valid, which establishes the result.
This implies that pc > 1/2. The next lemma is the main step in showing the converse pc 6 1/2.
Lemma 5 [Crossing the square for p = 1/2] Let LR(n) be the event that there is a left-right crossing of
the rectangle R(n) = [0, n + 1] × [0, n]. Then P1/2 (LR(n)) = 1/2 for all n.
Proof: The rectangle R(n) is the subgraph of L having vertex set [0, n + 1] × [0, n] and edge set
comprising all edges of L joining pairs of vertices in S(n) except those joining pairs (i, j), (k, l) with
either i = k = 0 or i = k = n + 1, i.e., edges in the left or right sides. Let Rd (n) be the subgraph of Ld
having vertex set {(i + 1/2, j + 1/2) | 0 6 i 6 n, −1 6 j 6 n} and all edges inherited from Ld except
for the one along the top and bottom sides. The two subgraphs are obtained by a 90 degree rotation.
Let us consider the two following events: LR(n) is the event that there exists an open path of R(n)
joining a vertex n the left side of R(n) to a vertex on the right side, and TBd (n) is the event that there exists
a closed path of Rd (n) joining a vertex on the top to a vertex at the bottom.
If LR(n) ∩ TBd (n) 6= ∅ there is a left-right open path in R(n) crossing a top-bottom closed path in
Rd (n). But at the crossing of these two path an open edge of L would cross a closed edge of Ld which is
impossible. Hence LR(n) ∩ TBd (n) = ∅. On the other hand, either LR(n) or TBd (n) must occur. Let D
be the set of vertices that are reachable from the left side of R(n) by an open path. Suppose that LR(n)
does not occur. Then there exists a top-bottom closed path of Ld crossing only edges of R(n) contained in
the edge boundary of D, and so TBd (n) occurs. Consequently, LR(n) and TBd (n) form a partition of the
sample space Ω and
P(LR(n)) + P(TBd (n)) = 1 .
Since R(n) and Rd (n) are isomorphic, flipping each edge of Ld yields that Pp (TBd (n)) = P1−p (LR(n)),
thus
Pp (LR(n)) + P1−p (LR(n)) = 1 .
Inserting p = 1/2 proves the lemma.
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4
Main Take-Aways
For bond percolation over L, with an occupation probability p and a critical probabilty pc :
• If p > pc , then almost surely there exists exactly one infinite open cluster.
• At the critical threshold (i.e., p = pc ), there exists no infinite open cluster.
• In the sub-critical phase (p < pc ): the mean cluster size is finite. In other words, almost surely an
infinite open cluster does not exist.
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