Component sizes of the random graph outside the scaling window

0.16 MB·English

by  Asaf Nachmias

7 Component sizes of the random graph outside the
0
0 scaling window
2

n
a Asaf Nachmias and Yuval Peres∗
J

1
February 2, 2008
1

]
R
P Abstract
.
h We provide simple proofs describing the behavior of the largest
t
a component of the Erdo˝s-R´enyi random graph G(n,p) outside of the
m scaling window, p= 1+ǫ(n) where ǫ(n) 0 but ǫ(n)n1/3 .
n → →∞
[

2 1 Introduction
v
6
6 Consider the random graph G(n,p) obtained from the complete graph on
4 n vertices by retaining each edge with probability p and deleting each edge
0
with probability 1 p. We denote by the j-th largest component. Let
1 j
− C
6 ǫ(n)beanon-negative sequencesuchthatǫ(n) 0andǫ(n)n1/3 . The
0 following theorems, proved by Bollob´as [4] an→d L uczak [8] usin→g d∞ifferent
/
h methods, describe the behavior of the largest components when p is outside
t
a the “scaling-window”.
m
: Theorem 1 [Subcritical phase] If p(n)= 1−ǫ(n) then for any η > 0 and
v n
integer ℓ > 0 we have
i
X
r P |Cℓ| 1 >η 0,
a (cid:16)(cid:12)2ǫ(n)−2log(nǫ(n)3) − (cid:12) (cid:17) →
(cid:12) (cid:12)
as n . (cid:12) (cid:12)
→ ∞
Theorem 2 [Supercritical phase] If p(n) = 1+ǫ(n) then for any η > 0
n
we have
1
P |C | 1 > η 0,
(cid:16)(cid:12)2nǫ(n) − (cid:12) (cid:17) →
(cid:12) (cid:12)
∗Microsoft Research and U.C(cid:12). Berkeley. R(cid:12)esearch of both authors supported in part
by NSFgrants #DMS-0244479 and #DMS-0104073
1
and for any integer ℓ > 1 we have
ℓ
P |C | 1 > η 0,
(cid:16)(cid:12)2ǫ−2(n)log(nǫ3) − (cid:12) (cid:17) →
(cid:12) (cid:12)
as n . (cid:12) (cid:12)
→ ∞
The proofs of these theorems in [4] and [8] are quite involved and use
the detailed asymptotics from [14], [4] and [3] for the number of graphs on
k vertices with k + ℓ edges. The proofs we present here are simple and
require no hard theorems. The main advantage, however, of these proofs
is their robustness. In a companion paper [12] we use similar methods to
analyze critical percolation on a random regular graphs. In this case, the
enumerative methods employed in [4] and [8] are not available.
Thephasetransition intheErd˝os-R´enyi randomgraphsG(n,p)happens
when p = c. Namely, with high probability, if c > 1 then is linear in n,
n |C1|
and if c < 1 then is logarithmic in n. When c 1 the situation is more
1
delicate. In [9], L |uCcz|ak, Pittel and Wierman prov∼e that for p = 1+λn−1/3,
n
the law of n−2/3 converges to a positive non-constant distribution which
1
|C |
in [1] is identified as the longest excursion length of some Brownian motion
with variable drift. See [11] for a recent account of the case p = 1+λn−1/3
n
with simple proofs.
Thus, is not concentrated and is roughly of size n2/3 if p = 1+λn−1/3.
|C1| n
However, if ǫ(n) a sequence such that n1/3ǫ(n) and p = 1+ǫ(n) then as
→ ∞ n
stated in Theorems 1 and2, the size of the largest component in G(n,p)
1
|C |
isconcentrated. Insummary,G(n,p)hasascalingwindowoflengthn−1/3 in
which the percolation is “critical” in the sense that is not concentrated.
1
|C |
2 The exploration process
We recall an exploration process, due to Karp and Martin-Lo¨f (see [7] and
[10]), in which vertices will be either active, explored or neutral. After the
completion of step t 0,1,...,n we will have precisely t explored vertices
∈ { }
and the number of the active and neutral vertices is denoted by A and N
t t
respectively.
Fix an ordering of the vertices v ,...,v . In step t = 0 of the process,
1 n
{ }
we declare vertex v active and all other vertices neutral. Thus A = 1 and
1 0
N = n 1. In step t 1,...,n , if A > 0 let w be the first active
0 t−1 t
− ∈ { }
vertex; if A = 0, let w be the first neutral vertex. Denote by η the
t−1 t t
2
number of neutral neighbors of w in G(n,p), and change the status of these
t
vertices to active. Then, set w itself explored.
t
Denote by the σ-algebra generated by η ,...,η . Observe that
t 1 t
F { }
given the random variable η is distributed as Bin(N 1 ,p)
Ft−1 t t−1− {At−1=0}
and we have the recursions
N = N η 1 , t n, (1)
t t−1− t− {At−1=0} ≤
and
A +η 1, A > 0
At = (cid:26) ηt,−1 t− At−1 = 0, t n. (2)
t t−1
≤
As every vertex is either neutral, active or explored,
N = n t A , t n. (3)
t t
− − ≤
At each time j n in which A = 0, we have finished exploring a
j
≤
connected component. Hence the random variable Z defined by
t
t−1
Z = 1 ,
t {Aj=0}
Xj=1
counts thenumberof components completely explored by theprocessbefore
time t. Define the process Y by Y = 1 and
t 0
{ }
Y = Y +η 1.
t t−1 t
−
By(2)wehavethatY = A Z ,i.e. Y countsthenumberofactive vertices
t t t t
−
at step t minus the number of components completely explored before step
t.
At each step we marked as explored precisely one vertex. Hence, the
component of v has size min t 1 : A = 0 . Moreover, let t < t ...
1 t 1 2
{ ≥ }
be the times at which A = 0; then (t ,t t ,t t ,...) are the sizes of
tj 1 2 − 1 3 − 2
the components. Observe that Z = Z +1 for all t t +1,...,t .
t tj ∈ { j j+1}
Thus Y = Y 1 and if t t +1,...,t 1 then A > 0, and thus
tj+1 tj − ∈ { j j+1− } t
Y < Y . By induction we conclude that A = 0 if and only if Y < Y
tj+1 t t t s
for all s < t, i.e. A = 0 if and only if Y has hit a new record minimum
t t
{ }
at time t. By induction we also observe that Y = (j 1) and that for
tj − −
t t + 1,...t we have Z = j. Also, by our previous discussion
j j+1 t
∈ { }
for t t + 1,...t we have min Y = Y = (j 1), hence by
∈ { j j+1} s≤t−1 t tj − −
induction we deduce that Z = min Y +1. Consequently,
t s≤t−1 t
−
A = Y min Y +1. (4)
t t s
−s≤t−1
3
Lemma 3 For all p 2 there exists a constant c > 0 such that for any
≤ n
integer t > 0,
P N n 5t e−ct.
t
≤ − ≤
(cid:16) (cid:17)
Proof. Let α t be a sequence of i.i.d. random variables distributed
{ i}i=1
as Bin(n,p). It is clear that we can couple η and α so η α for all i, and
i i i i
≤
thus by (1)
t
N n 1 t α . (5)
t i
≥ − − −
Xi=1
The sum t α is distributed as Bin(nt,p) and p 2 so by Large Devia-
i=1 i ≤ n
tions (seeP[2] section A.14) we get that for some fixed c > 0
t
P α 3t e−ct,
i
≥ ≤
(cid:16)Xi=1 (cid:17)
which together with (5) concludes the proof. 2
3 The subcritical phase
Before beginning the proof of Theorem 1 we require some facts about pro-
cesses with i.i.d. increments. Fix some small ǫ > 0 and let p = 1−ǫ for some
m
integer m > 1. Let β be a sequence of random variables distributed as
j
{ }
Bin(m,p). Let W be a process defined by
t t≥0
{ }
W = 1, W = W +β 1.
0 t t−1 t
−
Let τ be the hitting time of 0,
τ = min W = 0 .
t
t { }
By Wald’s lemma we have that Eτ = ǫ−1. Further information on the
tail distribution of τ is given by the following lemma.
Lemma 4 There exists constant C ,C ,c ,c > 0 such that for all T > ǫ−2
1 2 1 2
we have
P(τ T) C1 ǫ−2T−3/2e−(ǫ2−c21ǫ3)T ,
≥ ≤
(cid:16) (cid:17)
and
P(τ T) c1 ǫ−2T−3/2e−(ǫ2+c22ǫ3)T .
≥ ≥
(cid:16) (cid:17)
Furthermore,
Eτ2 = O(ǫ−3).
4
We will use the following proposition due to Spitzer (see [13]).
Proposition 5 Let a ,...,a Z satisfy k−1a = 1. Then there is
0 k−1 ∈ i=0 i −
precisely one j 0,...,k 1 such that for Pall r 0,...,k 2
∈ { − } ∈ { − }
r
a 0.
(j+i) mod k
≥
Xi=0
Proof of Lemma 4. By Proposition 5, P(τ = t) = 1P(W = 0). As
t t
t β is distributed as a Bin(mt,p) random variable we have
j=1 j
P
mt
P(W = 0) = pt−1(1 p)m−(t−1).
t
(cid:18)t 1(cid:19) −
−
Replacingt 1withtintheaboveformulaonlychangesitbyamultiplicative
−
constantwhichisalways between1/2and2. Astraightforwardcomputation
using Stirling’s approximation gives
1 tm 1 ǫ t(m−1)
P(W = 0) = Θ t−1/2(1 ǫ)t 1+ 1 − . (6)
t
− m 1 − m
n (cid:16) (cid:17) (cid:16) (cid:17) o
−
m m−1
Denote q = (1 ǫ) 1+ 1 1 1−ǫ , then
− m−1 − m
(cid:16) (cid:17) (cid:16) (cid:17)
1
P(τ T)= P(τ = t)= P(W = 0) = Θ t−3/2qt .
t
≥ t
Xt≥T Xt≥T (cid:16)Xt≥T (cid:17)
This sum can be bounded above by
qT
T−3/2 qt =T−3/2 ,
1 q
Xt≥T −
and below by
2T qT(1 qT)
t−3/2qt (2T)−3/2 − .
≥ 1 q
Xt=T −
Observe that as m we have that q tends to (1 ǫ)eǫ. By expanding eǫ
→ ∞ −
we find that
ǫ2 ǫ2
q = (1 ǫ)(1+ǫ+ )+Θ(ǫ3)= 1 +Θ(ǫ3).
− 2 − 2
5
Using this and the previous bounds on P(τ T) we get the first assertion
≥
of the Lemma.
The second assertion follows from the following computation. By (6) we
have that for some constant C > 0
Eτ2 = t2P(τ =t) = tP(W = 0) C √tqt.
t
≤
Xt≥1 Xt≥1 Xt≥1
Thus, by direct computation (or by [6], section XIII.5, Theorem 5)
1 3/2
Eτ2 O = O(ǫ−3).
≤ 1 q
(cid:16) (cid:17)
−
2
Proof of Theorem 1. We begin with an upper bound. Recall that com-
ponent sizes are t t for some j > 0 where t are record minima of the
j+1 j j
−
process Y . For a vertex v denote by C(v) the connected component of
t
{ }
G(n,p) which contains v. We first bound P(C(v ) > T) where
1
| |
T = 2(1+η)ǫ−2log(nǫ3).
Recall that C(v ) = min Y = 0 . Couple Y with a process W as
1 t t t t
| | { } { } { }
in Lemma 4, which has increments distributed as Bin(n,p) 1 such that
−
Y W for all t. Define τ as in Lemma 4. As p = 1−ǫ and T > ǫ−2, by
t ≤ t n
Lemma 4 we have
P(τ > T) Cǫ(nǫ3)−(1+η)log(nǫ3)−3/2,
≤
for some fixed C > 0. Our coupling implies that P(C(v ) > T) P(τ >
1
| | ≤
T). Denote by X the number of vertices v such that C(v) > T. If > T
1
| | |C |
then X > T. Also, for any two vertices v and u by symmetry we have that
C(v) and C(u) are identically distributed. We conclude that
| | | |
EX nP(C(v ) > T)
1
P( > T) P(X >T) = | |
1
|C | ≤ ≤ T T
C nǫ(nǫ3)−(1+η)(1−C2ǫ)log−3/2(nǫ3)
1 (nǫ3)−η(1−C2ǫ)+C2ǫ 0.
≤ 2(1+η)ǫ−2log(nǫ3) ≤ →
We now turn to prove a lower bound. Write
T = 2(1 η)ǫ−2log(nǫ3),
−
6
and define the stopping time
ηǫn
γ = min t : N n .
t
{ ≤ − 8 }
Recall that t aretimesinwhichA = 0andalsoY isarecordminimum
{ j} tj tj
(j)
for Y . For each integer j let W be a process with increments dis-
{ t} { t }
tributedasBin(n ηǫn,p)wherethestartingpointisW(j) = Y = (j 1).
− 8 0 tj − −
(j)
Note that if t < γ then we can couple Y and W such that
j+1 { t} { t }
Y W for all t [t ,t ]. Define the stopping times τ by
tj+t ≥ t ∈ j j+1 { j}
(j)
τ = min t :W = j .
j { t − }
Take
N = ǫ−1(nǫ3)(1−η8) .
l m
We will prove that with high probability t < γ and that there exists
N
k < k < ... < k < N such that τ > T. Note that these two events
1 2 ℓ ki
imply that > T. Indeed, by Lemma 3 we have
ℓ
|C |
ηǫn
P γ e−cǫn. (7)
≤ 40 ≤
(cid:16) (cid:17)
Byboundingtheincrementsof Y abovebyvariablesdistributedasBin(n,p)
t
{ } −
1 we learn by Wald’s Lemma (see [5]) that E[t t ] ǫ−1, hence
j+1 j
EtN ǫ−2(nǫ3)(1−η8). We conclude that − ≤
≤
P(tN > ηǫn) 40ǫ−2(nǫ3)(1−η8) = 40(nǫ3)−η8 , (8)
40 ≤ ηǫn η
which goes to 0 as ǫn−1/3 tends to . In Lemma 4 take m = n ηǫn and
∞ − 8
(1−ǫ)(1−ηǫ) 1−(1+η)ǫ
note that p = 8 8 , and so Lemma 4 gives that for any j
m ≥ m
P(τj >T) c1ǫ(nǫ3)−(1+η8)2(1−η)(1+c2ǫ)log−3/2(ǫ3n) ǫ(nǫ3)−(1−η4).
≥ ≥
Let X be the number of j N such that τ > T. Then we have
j
≤
EX Nǫ(nǫ3)−(1−η4) C(nǫ3)η8 ,
≥ ≥ → ∞
hence by Large Deviations (see [2], section A.14) for any fixed integer ℓ > 0
we have
η
P X < ℓ e−c(nǫ3)8 ,
≤
(cid:16) (cid:17)
7
for some fixed c> 0. By our previous discussion, this together with (7) and
(8) gives
(nǫ3)−η8
P( < T) O .
ℓ
|C | ≤ η
(cid:16) (cid:17)
2
4 The supercritical phase
In this section we denote ξ = η 1. We first prove some Lemmas.
t t
−
Lemma 6 If p = 1+ǫ then for all t < 3ǫ(n)n
n
EA = O(ǫt+√t), (9)
t
and
EZ = O(ǫt+√t). (10)
t
Proof. Write T = 3ǫn. We will use (4). First observe that as η can
t
always be bounded above by a Bin(n,p) random variable we can bound
Eξ ǫ for all t and hence EY ǫt. Denote by τ the stopping time
t t
≤ ≤
τ = min t :N n 15ǫn . By definition of η we have
t t
{ ≤ − }
E[ξ ]= pN p1 1.
t | Ft−1 t−1− {At−1=0}−
As N is a decreasing sequence, we deduce that as long as t < τ, we
t
{ }
have E[ξ ] > Dǫ for D > 0 large enough. Hence, the process
t t−1
| F −
Dǫj Y t∧τ isasubmartingaleforanyt. ByDoob’smaximalL2 inequality
{ − j}j=0
we have
E[max(Dǫj Y )2] 4E[(Dǫ(t τ) Y )2]. (11)
j t∧τ
j≤t∧τ − ≤ ∧ −
For any j < τ the random variable η can be stochastically bounded
j
from below by a Bin(n 15ǫn,p) random variable and above by a Bin(n,p)
−
random variable. Hence for any k < j < τ we have
E[ξ Dǫ ] = O(ǫ),
j k
(cid:12) − |F (cid:12)
(cid:12) (cid:12)
and so (cid:12) (cid:12)
E[(ξ Dǫ)(ξ Dǫ)] = O(ǫ2).
j k
− −
8
We conclude that as long as t < τ
t t
E[(Dǫt Y )2] 2 E[(ξ Dǫ)(ξ Dǫ)]+ E[(ξ Dǫ)2] = O(ǫ2t2+t).
t j k j
− ≤ − − −
Xk<j Xj=1
Lemma 3 implies that for n large enough,
1
P N n 15ǫn e−3cǫn , (12)
T ≤ − ≤ ≤ n2
(cid:16) (cid:17)
and as N is a decreasing sequence we deduce that P(τ T) n−2.
t
{ } ≤ ≤
Hence for any t T
≤
E[(Dǫt Y )2] E[(Dǫ(t τ) Y )21 ]+O(n2)P(t τ)
t t∧τ {t<τ}
− ≤ ∧ − ≥
= O(ǫ2t2+t).
We deduce by (11) and Jensen inequality that for any t T
≤
E[min(Y Dǫj)] = O(ǫt+√t),
j
j≤t −
hence E[min Y ] = O(ǫt+√t) and so by (4) we obtain (9). Inequality
j≤t j
(10) follows immediately from the relation Z = A Y . 2
t t t
−
Lemma 7 If p = 1+ǫ then for all t < 3ǫ(n)n
n
EN =n(1 p)t+O(ǫ2n), (13)
t
−
and
t
Eξ =ǫ +O(ǫ2). (14)
t
− n
Proof. Observe that by (1) we have that
E[N ] = (1 p)N (1 p)1 .
t |Ft−1 − t−1− − {At−1=0}
By iterating this relation we get that EN = n(1 p)t + O(EZ ) which
t t
−
by Lemma 6 yields (13) (observe that for t = 3ǫn we have ǫt > √t by our
assumption on ǫ). Since
E[ξ ] = pN p1 1,
t | Ft−1 t−1− {At−1=0}−
by taking expectations and using (13) we get
9
1+ǫ
Eξ = (1+ǫ)(1 )t 1+O(ǫ2)
t
− n −
t
= (1+ǫ)(1 (1+ǫ)t/n) 1+O(ǫ2)= ǫ +O(ǫ2),
− − − n
where we used the fact that (1 x)t = 1 tx+O(t2x2). 2
− −
Proof of Theorem 2. Write T = 3ǫn and ξ∗ = E[ξ ]. The process
j j |Fj−1
t
M = Y ξ∗,
t t− j
Xj=1
is a martingale. By Doob’s maximal L2 inequality we have that
E(maxM2)) 4EM2.
t≤T t ≤ T
AsM hasorthogonalincrementswithboundedsecondmomentweconclude
t
that EM2 = O(T), hence, by Jensen’s inequality we have
T
t
E max Y ξ∗ O(√T)= O(√ǫn). (15)
h t≤T (cid:12)(cid:12) t−Xj=1 j(cid:12)(cid:12)i ≤
(cid:12) (cid:12)
As ξ∗ = pN p1 1 by (3) we have
j j−1− {Aj−1=0}−
E ξ∗ Eξ = pE A +1 EA E1 .
| j − j| | j−1 {Aj−1=0}− j−1− {Aj−1=0}|
By the triangle inequality and Lemma 6 we conclude that for all j T
≤
E ξ∗ Eξ p O(ǫj + j),
| j − j| ≤ ·
p
and hence for any t T
≤
E ξ∗ Eξ p O(ǫt2+t3/2) O(ǫ3n).
| j − j| ≤ · ≤
hXj≤t i
By the triangle inequality we get
t
E max (ξ∗ Eξ ) O(ǫ3n). (16)
h t≤T (cid:12)(cid:12)Xj=1 j − j (cid:12)(cid:12)i ≤
(cid:12) (cid:12)
Using the triangle inequality, (15), (16) and Markov inequality gives
10