Growth of the Number of Spanning Trees of the Erdös-Rényi Giant Component
aa r X i v : . [ m a t h . P R ] M a y Growth of the Number of Spanning Treesof the Erd˝os-R´enyi Giant Component
Russell Lyons ∗† Ron Peled ‡§ Oded Schramm ¶
12 May 2008
Abstract
The number of spanning trees in the giant component of the random graph G ( n, c/n )( c >
1) grows like exp (cid:8) m (cid:0) f ( c ) + o (1) (cid:1)(cid:9) as n → ∞ , where m is the number of verticesin the giant component. The function f is not known explicitly, but we show thatit is strictly increasing and infinitely differentiable. Moreover, we give an explicitlower bound on f ′ ( c ). A key lemma is the following. Let PGW ( λ ) denote a Galton-Watson tree having Poisson offspring distribution with parameter λ . Suppose that λ ∗ > λ >
1. We show that
PGW ( λ ∗ ) conditioned to survive forever stochasticallydominates PGW ( λ ) conditioned to survive forever. Methods of enumeration of spanning trees in a finite graph G and relations to various areasof mathematics and physics have been investigated for more than 150 years. The number ofspanning trees is often called the complexity of the graph, denoted here by τ ( G ). The usualErd˝os-R´enyi model of random graphs, G ( n, p ), is a graph on n vertices, each pair of which isconnected by an edge with probability p , independently of other edges. Fix c >
1. It is wellknown that with probability approaching 1 as n → ∞ , the largest component of G ( n, c/n ) ∗ Partially supported by NSF grants DMS-0406017 and DMS-0705518 and Microsoft Research. † Department of Mathematics, Indiana University, Bloomington, IN 47405-5701. Email:[email protected]. ‡ Supported by Microsoft Research and NSF grant DMS-0605166. § Department of Statistics, UC Berkeley. Email: [email protected]. ¶ Microsoft Research. n , while the second largest component is of logarithmic size. (See,e.g., [ER60] or [Bol01].) The largest component is thus called the giant component andwill be denoted by G n = G n ( c ). As an example of a general theory, [Lyo05] proved thatthere is a number f ( c ) such that f ( c ) = lim n →∞ | V ( G n ) | log τ ( G n )in probability. In the same paper it was shown that f ( c ) > c >
1, that f (1 + ) = 0,and that f is continuous on [1 , ∞ ). [Lyo05] asked whether f is strictly increasing and realanalytic on (1 , ∞ ). Note that as c increases, both the number of trees τ ( G n ) as well as thenumber of vertices | V ( G n ) | increase, so that it is not clear which increase dominates. Herewe prove that f is strictly increasing and C ∞ ; prior to our work, it was not known even that f was non-decreasing.Let PGW ( c ) be the law of a rooted Galton-Watson tree ( T, o ) with Poisson( c ) offspringdistribution. Write PGW ∗ ( c ) for the law of PGW ( c ) conditioned on non-extinction. Some-times we also write this measure as PGW ∗ c . The event of extinction has probability q ( c ),which is well known to be the smallest positive solution of the equation q ( c ) = e − c (1 − q ( c )) . (1.1)Let p k ( x ; G ) denote the probability that simple random walk on a graph G started at avertex x is back at x after k steps. [Lyo05] proved that f ( c ) = Z (cid:16) log deg T ( o ) − X k ≥ k p k ( o ; T ) (cid:17) d PGW ∗ c ( T, o ) . (1.2) Theorem 1.1.
The function f is strictly increasing and C ∞ on (1 , ∞ ) . In fact, f ′ ( c ) > ( c − e − cq ( c ) c > for c > . From (1.2), it is not hard to see that for any ǫ >
0, we have that (cid:0) f ( c + ǫ ) − f ( c ) (cid:1) /ǫ ∼ /c as c → ∞ . Since ce − c = cq ( c ) e − cq ( c ) (1.3)and the function x xe − x is unimodal in (0 , ∞ ) and vanishes at 0 and ∞ , it follows thatlim c →∞ cq ( c ) = 0. Using this, we find that our lower bound for f ′ ( c ) in Theorem 1.1 has thesame asymptotic, 1 /c , as c → ∞ . We do not have any information on f ′ (1).A key lemma to prove Theorem 1.1 is the following:2 heorem 1.2. If c ′ > c ≥ , then PGW ∗ ( c ′ ) stochastically dominates PGW ∗ ( c ) . Here,
PGW ∗ (1) denotes the weak limit as c ↓ PGW ∗ ( c ); see [AP98], Lemma 23.We now recall what the stochastic domination referred to in the theorem means. If ( T, o )and ( T ′ , o ′ ) are rooted trees, we say that ( T, o ) dominates ( T ′ , o ′ ) if there is an isomorphismfrom T ′ to a subtree of T that takes o ′ to o . A probability measure on the collection ofrooted trees is said to stochastically dominate another probability measure on the collectionof rooted trees if they may be coupled so that the sample from the first measure a.s. dominatesthe sample from the second measure.Of course, PGW ( c ′ ) dominates PGW ( c ) when c ′ > c . It is the conditioning that makesTheorem 1.2 nontrivial. Indeed, the offspring distribution that has 1 or 3 children with prob-ability 1/2 each stochastically dominates the offspring distribution that has 0 or 3 childrenwith probability 1/2 each, but if we condition on survival, the domination does not persistsince conditioning does not change the former, but forces the latter to have 3 children of theroot. Let T n = T n ( λ ) be a PGW ( λ ) tree conditioned to have n vertices, where n ∈ N + ∪ {∞} .We consider the values of T n to be equivalence classes of rooted trees under isomorphismsthat preserve the root. It is easy to check that the distribution of T n does not depend on λ . It turns out that it is the same as the distribution obtained by forgetting the labels of auniform tree on n vertices with uniform root. Also, the probability that a PGW ( λ ) tree has k vertices is given by the Borel( λ ) distribution, namely,( λ e − λ ) k k k − λk ! . (2.1)These facts are well known and have a variety of proofs; see [Pit98] for some of them.[LW04] show the following: Theorem 2.1. T n +1 stochastically dominates T n for every n ∈ N + . More precisely, in their Theorem 4.1, for each d ≥ d, /d ). Taking alimit as d → ∞ gives Theorem 2.1. It is interesting to note that this is the same as sayingthat a uniformly rooted uniform tree on n + 1 vertices dominates a uniformly rooted uniformtree on n vertices. 3efine θ ( c ) as the survival probability of PGW ( c ), that is, θ ( c ) := 1 − q ( c ) . For µ > λ >
0, set α ( λ, µ ) := log e µ − µ − log e λ − λ > . Lemma 2.2.
For µ > λ > , we have α ( λ, µ ) < µ − λ and α (cid:0) λθ ( λ ) , µθ ( µ ) (cid:1) > λq ( λ ) − µq ( µ ) . Proof.
The first inequality states that x log (cid:0) ( e x − /x (cid:1) − x is decreasing for 1 < x < ∞ ,which in turn is a consequence of the inequality e x > x .By (1.1), we have e λθ ( λ ) − λθ ( λ ) = 1 − q ( λ ) λq ( λ ) θ ( λ ) = 1 λq ( λ ) . Therefore α (cid:0) λθ ( λ ) , µθ ( µ ) (cid:1) = log (cid:0) λq ( λ ) (cid:1) − log (cid:0) µq ( µ ) (cid:1) . (2.2)We next note that x xe − x is strictly increasing on (0 ,
1) and strictly decreasing on (1 , ∞ ).Recalling from (1.3) that λ exp( − λ ) = λq ( λ ) exp( − λq ( λ )) and using that λ >
1, we deducethat λq ( λ ) <
1. We similarly deduce that µq ( µ ) < λq ( λ ) since µ > λ . The second claimedinequality therefore follows from (2.2) and the fact that x log x − x is increasing on(0 , Q λ denote a Poisson( λ ) random variable and Q ∗ λ denote a random variable whosedistribution is the same as that of Q λ conditioned on Q λ > Lemma 2.3.
Let µ > λ > and set α = α ( λ, µ ) . Then Q ∗ µ stochastically dominates the sumof mutually independent copies of Q ∗ λ and Q α . Moreover, this does not hold for any larger α . Proof.
Consider some β ∈ (0 , µ − λ ), and let Z denote the sum of two mutually independentcopies of Q ∗ λ and Q β . For k ∈ N + , set a k := P (cid:2) Z = k (cid:3) = e − λ − β − e − λ k X j =1 λ j β k − j j ! ( k − j )! = e − β e λ − λ + β ) k − β k k ! , (2.3)4nd b k := P (cid:2) Q ∗ µ = k (cid:3) = 1 e µ − µ k k ! . In order for Q ∗ µ to dominate Z , it is necessary that a ≥ b . This translates and simplifies to β ≤ α . Now fix β = α ; note that α < µ − λ by Lemma 2.2. Let J be the set of k ∈ N + suchthat a k ≥ b k . We claim that there is a k ∈ R such that J = N + ∩ [1 , k ]. Before proving theclaim, we shall demonstrate that the lemma follows from it. Indeed, to prove domination itis sufficient to show that ∀ k ∈ N + ∞ X j = k b j ≥ ∞ X j = k a j . (2.4)This is clear if k > k , because b j > a j for j > k . On the other hand, if k ≤ k , then a j ≥ b j for j ≤ k , which gives k X j =1 b j ≤ k X j =1 a j . Now subtracting both sides from P j ∈ N + b j = 1 = P j ∈ N + a j gives (2.4).It remains to prove that J = N + ∩ [1 , k ] for some k ∈ R . Note that a k b k = ( e µ − e − β e λ − (cid:18)(cid:16) λ + βµ (cid:17) k − (cid:16) βµ (cid:17) k (cid:19) . Now think of the right-hand side as a function g ( k ) of positive real k . As such, it may bewritten in the form A ( B k − C k ), with constants A, B, C satisfying
A > > B >C >
0. We claim that g does not have any local minimum. Indeed, g ′ ( k ) = A ( B k log B − C k log C ) and g ′′ ( k ) = A (cid:0) B k (log B ) − C k (log C ) (cid:1) . Therefore, when g ′ ( k ) = 0, we have(log B ) / (log C ) = C k B − k and so g ′′ ( k ) = A (log C ) C k ( C k B − k − <
0. This verifiesthat g does not have a local minimum. Hence, the set of k ∈ (0 , ∞ ) such that g ( k ) ≥ β = α , this interval contains 1. This proves the claim, andcompletes the proof of the lemma.Given a rooted tree T with root o and a node v in T , let N ( v ) denote the cardinality ofthe set of nodes in the subtree of T corresponding to v , that is, the number of nodes in T that are not in the connected component of o in T \ { v } . Given a random rooted tree T , let n k = n k ( T ) denote the number of children v of the root satisfying N ( v ) = k .Let T ( λ ) denote a sample from PGW ( λ ). Note that the random variables (cid:0) n k ( T ( λ )) : k ∈ N + ∪ {∞} (cid:1) are independent Poisson random variables. It follows that the random variables (cid:0) n k ( T ∞ ( λ )) : k ∈ N + ∪ {∞} (cid:1) are independent, and n k ( T ∞ ( λ )) has the same law as n k ( T ( λ ))when k ∈ N + , while n ∞ ( T ∞ ( λ )) has the law of n ∞ ( T ( λ )) conditioned on being positive.5y (2.1), we have ∀ k ∈ N + E (cid:2) n k ( T ( λ )) (cid:3) = ( λ e − λ ) k k k − k ! . Observe that this is monotone decreasing in λ in the range λ ≥ n ∞ ( T ( λ )) is Poisson with parameter λ θ , we have X k ∈ N + E (cid:2) n k ( T ( λ )) (cid:3) = (1 − θ ) λ . If T and T ′ are rooted trees, we write T ≤ T ′ if there is an injective map i from thechildren of the root in T to the children of the root in T ′ such that N ( i ( v )) ≥ N ( v ) for everychild v of the root in T . If T and T ′ are random rooted trees, we write T ≤ L T ′ if T and T ′ may be coupled in such a way that T ≤ T ′ a.s. Observe that ≤ L is a partial order relation.Theorem 1.2 will follow easily from the following lemma. Lemma 2.4.
Let µ > λ > and let n ∈ N + . Then T n ≤ L T n +1 ≤ L T ∞ ( λ ) ≤ L T ∞ ( µ ) . Proof.
We start by proving T ∞ ( λ ) ≤ L T ∞ ( µ ). Let ( Z k : k ∈ N + ) and ( Z ′ k : k ∈ N + ) be inde-pendent Poisson random variables with E (cid:2) Z k (cid:3) = E (cid:2) n k ( T ( µ )) (cid:3) and E (cid:2) Z ′ k (cid:3) = E (cid:2) n k ( T ( λ )) (cid:3) − E (cid:2) n k ( T ( µ )) (cid:3) . (Recall that the latter is non-negative.) Let Z ′ := P k ∈ N + Z ′ k , which is aPoisson random variable. By the above, E (cid:2) Z ′ (cid:3) = (cid:0) − θ ( λ ) (cid:1) λ − (cid:0) − θ ( µ ) (cid:1) µ . By Lemma 2.2, we have α (cid:0) λθ ( λ ) , µθ ( µ ) (cid:1) ≥ λq ( λ ) − µq ( µ ) = E (cid:2) Z ′ (cid:3) . (2.5)Consequently, by Lemma 2.3, n ∞ ( T ∞ ( µ )) may be coupled to dominate n ∞ ( T ∞ ( λ )) plus anindependent copy of Z ′ . Thus, we may take n ∞ ( T ∞ ( λ )) independent from ( Z k : k ∈ N + ) and( Z ′ k : k ∈ N + ) and take n ∞ ( T ∞ ( µ )) independent from ( Z k : k ∈ N + ) such that n ∞ ( T ∞ ( µ )) ≥ Z ′ + n ∞ ( T ∞ ( λ )). For k ∈ N + we take n k ( T ∞ ( λ )) = Z k + Z ′ k , and n k ( T ∞ ( µ )) = Z k . Since Z ′ = P k ∈ N + Z ′ k , with these choices we have T ∞ ( λ ) ≤ T ∞ ( µ ). This proves that T ∞ ( λ ) ≤ L T ∞ ( µ ).The fact that T n ≤ L T n +1 follows from Theorem 2.1. Recall that the limit in law of T n as n → ∞ is the same as the limit in law of T ∞ ( λ ) as λ ց
1; see [AP98], Lemma 23. Let T ∞ (1) denote a random tree with this limit law. Then T n ≤ L T ∞ (1). By taking the limit as λ ′ ց T ∞ ( λ ′ ) ≤ L T ∞ ( λ ), we find that T ∞ (1) ≤ L T ∞ ( λ ). Thus, T n ≤ L T ∞ ( λ ) follows. Proof of Theorem 1.2.
This follows by repeatedly applying Lemma 2.4 at each nodeof the T ∞ ( λ ) tree, while keeping the corresponding couplings appropriately conditionallyindependent. 6 Return Probabilities
A general result on monotonicity [Lyo07], combined with Theorem 1.2 implies the mono-tonicity claim in Theorem 1.1. Here, we analyze in more detail the expression (1.2) in orderto gain an explicit lower bound on the derivative of f ( c ), which that general result does notsupply.Our main aim in this section is to prove the following result: Theorem 3.1.
The expression
Z X k ≥ k p k ( o ; T ) d PGW ∗ c ( T, o ) is monotonic decreasing in c > . In light of (1.2) and Theorem 1.2, this implies the monotonicity claim in Theorem 1.1and will lead to an explicit lower bound on the derivative in Section 5. It also implies thefollowing lower bound for f ( c ) itself: f ( c ) ≥ X k ≥ e − c c k (1 − q ( c ) k ) log kθ ( c ) k ! − X k ≥ e − log(1 + k ) k ! ≥ . To see this, note first that by, say, (2.3), we have that
PGW ∗ c [deg T ( o ) = k ] = e − c c k (1 − q ( c ) k ) θ ( c ) k ! . Second, recall that lim c ↓ f ( c ) = 0. Therefore, Theorem 3.1 and (1.2) imply that f ( c ) ≥ Z log deg T ( o ) d PGW ∗ c ( T, o ) − Z log deg T ( o ) d PGW ∗ ( T, o ) ≥ , and this equals the above expression by the well-known form of PGW ∗ (1) ([AP98], Corollary3). This lower bound should be compared to the trivial upper bound f ( c ) ≤ X k ≥ e − c c k (1 − q ( c ) k ) log kθ ( c ) k ! . To prove Theorem 3.1, let V ( s, T, o ) := P k ≥ p k ( o ; T ) s k . Since Z E [ V ( s, T, o )] − s ds = Z X k ≥ k p k ( o ; T ) d PGW ∗ c ( T, o ) , Theorem 3.1 will be a consequence of the following result:7 heorem 3.2.
For all s ∈ (0 , , the expectation R V ( s, T, o ) d PGW ∗ c ( T, o ) is decreasing in c > . Fix µ > λ > T and T ′ have the distributions PGW ∗ ( λ ) and PGW ∗ ( µ ), respec-tively. Let X count the number of visits to the root in a random walk on the tree T startedfrom the root in which at each step the walker has probability 1 − s to die, independent ofthe other steps (note that X ≥ X ′ be the same for awalk on T ′ . Because V ( s, T, o ) = E [ X ], Theorem 3.2 follows from: Theorem 3.3. X stochastically dominates X ′ . In words, larger trees have fewer returns of simple random walk for this model. We shallneed a technical lemma for the proof.
Lemma 3.4.
Fix integers a ≥ and b ≥ . Let F be a convex increasing function on [0 , ∞ ) . Let X , . . . , X a , Y , . . . , Y b be independent non-negative random variables. If each X i stochastically dominates each Y j , then E F a + b a X i =1 X i + b X i =1 Y i !! ≤ E F a a X i =1 X i !! . Proof.
Define an auxiliary random vector A with a + b coordinates to be uniformly chosenamong the N := (cid:18) a + ba (cid:19) vectors containing exactly a values equal to 1 /a and b zeroes.Condition on the X ’s and Y ’s and consider the random variable R := F E A a X i =1 A i X i + b X i =1 A i + a Y i !! , where E A denotes expectation over A . On the one hand, we have R = F a + b a X i =1 X i + b X i =1 Y i !! . (3.1)On the other hand, by Jensen’s inequality (since the X ’s and Y ’s are non-negative) R ≤ E A F a X i =1 A i X i + b X i =1 A i + a Y i ! = N − N X i =1 M i , (3.2)8here each M i is a random variable of the form F a a X k =1 X i k + 1 a b X k =1 Y j k ! ;here, ( X i k ) a k =1 is a subset of a of the X ’s, ( Y j k ) b k =1 is a subset of b of the Y ’s, a ≤ a, b ≤ b and a + b = a . Taking now expectation over the X ’s and Y ’s and using that each X i stochastically dominates each Y j , we get E ( M i ) ≤ E F a a X k =1 X k ! (3.3)since the X i ’s and Y j ’s are independent, non-negative and each M i is increasing in each ofthe random variables. Putting (3.1), (3.2) and (3.3) together, we get E F a + b a X i =1 X i + b X i =1 Y i !! ≤ E F a a X i =1 X i ! , proving the lemma. Proof of Theorem 3.3.
It is enough to show that for each integer M ≥
2, we have P ( X ≥ M ) ≥ P ( X ′ ≥ M ) . (3.4)We couple the two trees according to the coupling given in the preceding section, in theproof of Lemma 2.4 and Theorem 1.2, for T ∞ ( λ ) and T ∞ ( µ ). It is then enough to show theinequality (3.4) conditioned on the number of subtrees of each size that the roots of T and T ′ have (the variables n k ( T ) and n k ( T ′ )). Henceforth we always condition on these values.Denote N F := P ∞ k =1 n k ( T ) , N I := n ∞ ( T ) , N ′ F := P ∞ k =1 n k ( T ′ ) and N ′ I := n ∞ ( T ′ ). Accord-ing to the coupling, we have N F ≥ N ′ F and d := N F + N I ≤ N ′ F + N ′ I =: d ′ . We constructour coupling of T and T ′ to have the following properties: T is a rooted subtree of T ′ ; thechildren of the root are ordered; the first N ′ F children of the root lie in T and have finitesubtrees, all pairwise equal in T and T ′ ; the next N F − N ′ F children lie in T and have finitesubtrees; and the next N I children lie in T . Given the sizes of the subtrees, recall that thepairs of coupled subtrees of the children of the root are independent, even including the d ′ − d left-over subtrees of T ′ , and that all N ′ I of the infinite subtrees of T ′ are i.i.d.For each 1 ≤ j ≤ N F , suppose that the random walk enters in its first step the j th finitesubtree of T . Let P F j be the probability that continuing this random walk, we ever return to9he root. Similarly, define for 1 ≤ j ≤ N I the probability P I j to return from the j th infinitesubtree of T . Define analogously Q F j and Q I j on T ′ . Thus, P F j = Q F j for j ≤ N ′ F because ofthe coupling.The inequality (3.4) can now be written as follows: E sd N F X j =1 P F j + N I X j =1 P I j !! M − ≥ E sd ′ N ′ F X j =1 Q F j + N ′ I X j =1 Q I j M − . We prove this inequality in two steps. First we observe that E sd N F X j =1 P F j + N I X j =1 P I j !! M − ≥ E sd N ′ F X j =1 Q F j + d − N ′ F X j =1 Q I j M − ;this is because if the walk entered a branch of T ′ that contains a branch of T , then certainlyits probability ever to return to the root is smaller in T ′ than it is in T (by coupling thewalks). Now we may use Lemma 3.4 to get that E sd N ′ F X j =1 Q F j + d − N ′ F X j =1 Q I j M − ≥ E sd ′ N ′ F X j =1 Q F j + N ′ I X j =1 Q I j M − since each Q I j is stochastically dominated by each Q F j (again, by coupling the walks on thecoupled subtrees). This proves the theorem. Let ¯ p k = ¯ p k ( c ) := R p k ( o ; T ) d PGW ∗ c ( T, o ). We shall prove
Theorem 4.1.
For each k ≥ , ¯ p k ( c ) is real analytic in c > and there exists β > suchthat for c > , k ≥ , and n ≥ , we have (cid:12)(cid:12)(cid:12)(cid:12) ∂ n ¯ p k ∂c n (cid:12)(cid:12)(cid:12)(cid:12) ≤ A n n ! k βn e − ak / , where the constants A, a > depend only on c and are bounded from 0 and infinity for c inevery compact subinterval of (1 , ∞ ) . emark . We obtain β = 1 in the proof, but this could be reduced further.An immediate corollary is Corollary 4.3. f ( c ) is C ∞ for c ∈ (1 , ∞ ) . To prove Theorem 4.1, we shall prove
Theorem 4.4.
For each k ≥ , ¯ p k can be analytically continued to the domain Ω k := { x + iy | x ∈ (1 , ∞ ) , | y | ≤ ak − β } for some β > and satisfies for c ∈ Ω k | ¯ p k ( c ) | ≤ Ae − ak / , where A, a > depend only on x = Re ( c ) and are bounded from 0 and infinity for x in everycompact subinterval of (1 , ∞ ) . Theorem 4.1 is an immediate corollary by Cauchy estimates. To see this, for each c > C of radius r := min( ak − β , c − ) around c . Then (cid:12)(cid:12)(cid:12)(cid:12) ∂ n ¯ p k ∂c n (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) n !2 πi I C ¯ p k ( z )( z − c ) n +1 dz (cid:12)(cid:12)(cid:12)(cid:12) ≤ An ! e − ak / r n ≤ AA ′ n ! k βn e − ak / , where A ′ := max (cid:16) a − n , (cid:0) c − (cid:1) − n (cid:17) .In the rest of the section, we prove Theorem 4.4. We start by quoting a known resultconcerning a priori bounds on ¯ p k ; see [Pia98, Theorem 2]. Theorem 4.5.
For each k ≥ , we have ¯ p k ( c ) ≤ Ae − ak / for c > , where A, a > dependonly on c and are bounded from 0 and infinity for c in every compact subinterval of (1 , ∞ ) . We remark that in [Pia98, Theorem 2] the boundedness of the constants is not claimed,just that constants exist for every c >
1, but this implies the theorem since ¯ p k is a continuousfunction of c .We now fix a compact subinterval I ⊆ (1 , ∞ ) and shall work only with c = x + iy suchthat x ∈ I . All the constants A, a > I and it is understoodthat their value may change from line to line: A may increase, while a may decrease.We record for later use the well-known structure of the PGW ∗ ( c ) distribution, as was alsodiscussed in Section 2. Lemma 4.6.
The
PGW ∗ ( c ) distribution is a 2-type Galton-Watson distribution, with thetypes called I and F (for “infinite” and “finite”). For vertices of type I the number of type Ichildren is distributed as Q ∗ c (1 − q ( c )) and of type F children as Q cq ( c ) . Vertices of type F haveonly type F children, the number of which is distributed as Q cq ( c ) .
11e next introduce the notion of the trace of a random walk path . This is all the in-formation about a path on a tree that starts at its root. The trace includes the followinginformation per step:1. whether the step is up or down (up is away from the root);2. if the step is up, whether it is to a type-I child or to a type-F child and which suchchild is it (e.g., the first type-I child, the second type-F child, etc.).We denote by W k the set of all traces that have exactly k steps and end at the root. Given W ∈ W k , let l I ( W ) , l F ( W ) be the number of distinct vertices of types I and F, respectively,that the trace visits, so that l I ( W ) + l F ( W ) ≤ k . Given a tree T , the trace W may be feasibleon T or not: It is feasible if and only if all the vertices that W visits exist in T (e.g., if onthe first step, W moves to the third type-F child of the root, then the root of T must haveat least 3 type-F children). Let d I ( W ) := ( d F1 , d I1 , . . . , d F l I ( W ) , d I l I ( W ) ) be the minimum requirednumber of children of type F and type I from each of the vertices of type I that W visitsin order for the walk to be feasible. Here, the subscript i indicates the i th distinct vertexof type I visited by W . Similarly, let d F ( W ) := ( ˜ d F1 , . . . , ˜ d F l F ( W ) ) be the minimum requirednumber of children of type F from the vertices of type F that W visits. Given two vectors e and d of the same length, we write e (cid:23) d if each coordinate of e is greater than or equal tothe corresponding coordinate of d . Finally, denote by p c ( W ) the probability under PGW ∗ ( c )to sample a feasible tree for W and then to sample W as a simple random walk path oflength k on that tree. From all the above discussion, we have¯ p k ( c ) = X W ∈W k p c ( W ) = X W ∈W k X e I (cid:23) d I ( W ) e F (cid:23) d F ( W ) p c ( e I , e F ) p ( W, e I , e F ) , (4.1)where p c ( e I , e F ) is the probability to sample a tree in which the vertices that W passesthrough have exactly the prescribed number of children e I , e F of each type, and p ( W, e I , e F )is the conditional probability, given e I and e F , to sample W as a simple random walk pathon the tree. We emphasize that p ( W, e I , e F ) does not depend on c , while p c ( e I , e F ) is thesame for all W that satisfy e I (cid:23) d I ( W ) and e F (cid:23) d F ( W ).Since c and q are analytically related for c ∈ (1 , ∞ ) by (1.1) (which can be rewritten as c = − log( q )1 − q ), there is a unique extension of q ( c ) to an analytic function of c for Re c > | Im c | ≤ κ (Re c ) for some continuous function κ : (1 , ∞ ) → (0 , ∞ ). (In fact, one can extendit much further, but we shall not need that.) Hence, the same holds for p c ( e I , e F ). We shalluse the same notations for the original functions as for these extensions, and likewise forsimilar functions below. 12ote that to prove Theorem 4.4, it is enough to show that for c = x + iy with x ∈ I and | y | ≤ κ ( x ) k − β , the sum (4.1) converges uniformly and is bounded by A exp( − ak / ). Denoteby max( e ) the maximal element of e . We continue with Lemma 4.7. If c = x + iy with x ∈ I and | y | ≤ κ ( x ) , then | p c ( e I , e F ) | ≤ p x ( e I , e F ) e Ak (max( e I )+max( e F )+1) | y | . Proof.
Using the structure Lemma 4.6, we know that p c ( e I , e F ) = l I ( W ) Y i =1 p c ( e I , i ) l F ( W ) Y i =1 p c ( e F , i )(the values l I ( W ) and l F ( W ) are implicit in e I and e F as their lengths), where p c ( e I , i ) = P ( Q ∗ c (1 − q ( c )) = e I i ) P ( Q cq ( c ) = e F i ) ,p c ( e F , i ) = P ( Q cq ( c ) = ˜ e F i ) . More explicitly, denoting j := e I i , m := e F i and n := ˜ e F i and abbreviating q := q ( c ), we have p c ( e I , i ) = e − c (1 − q ) ( c (1 − q )) j j !(1 − e − c (1 − q ) ) e − cq ( cq ) m m ! ,p c ( e F , i ) = e − cq ( cq ) n n ! . We have in the first case | p c ( e I , i ) | ≤ e − x j ! m ! | c | j + m | − q | j | q | m | − e − c (1 − q ) | . (4.2)We know that when x ∈ I and | y | ≤ κ ( x ), we have | q ( x + iy ) − q ( x ) | ≤ A | y | since q is ananalytic function of c . Hence | c | j + m = x j + m (cid:12)(cid:12)(cid:12)(cid:12) y x (cid:12)(cid:12)(cid:12)(cid:12) j + m ≤ x j + m e Ay ( j + m ) , | − q ( x + iy ) | j ≤ (1 − q ( x )) j (cid:18) A | y | − q ( x ) (cid:19) j ≤ (1 − q ( x )) j e Aj | y | , | q ( x + iy ) | m ≤ q ( x ) m (cid:18) A | y | q ( x ) (cid:19) m ≤ q ( x ) m e Am | y | , | − e − ( x + iy )(1 − q ( x + iy )) | ≤ − e − x (1 − q ( x )) + A | y | ≤ − e − x (1 − q ( x )) (1 + A | y | ) ≤ e A | y | − e − x (1 − q ( x )) . Substituting back into (4.2), we get | p c ( e I , i ) | ≤ p x ( e I , i ) e A ( j + m +1) | y | . This bound was for p c ( e I , i ), but we also obtain analogously that | p c ( e F , i ) | ≤ p x ( e F , i ) e A ( n +1) | y | .Hence | p c ( e I , e F ) | ≤ p x ( e I , e F ) e A ( l I ( W )+ l F ( W ))(max( e I )+max( e F )+1) | y | ≤ p x ( e I , e F ) e Ak (max( e I )+max( e F )+1) | y | . To continue, say that a vertex of a tree is L -big if it has either exactly L type-I childrenor exactly L type-F children or both. Let E k,L be the event that if we sample a tree and doa simple random walk on it (from the root), then the walk returns to the root after exactly k steps and visits an L -big vertex along the way but does not visit an M -big vertex alongthe way for any M > L . We observe that X W ∈W k X e I (cid:23) d I ( W ) e F (cid:23) d F ( W )max(max( e I ) , max( e F ))= L p x ( e I , e F ) p ( W, e I , e F ) = PGW ∗ x ( E k,L ) ≤ Ake − aL log L , (4.3)where the last inequality follows since there are no more than k vertices along any path W andsince the tails of a Poisson( c ) random variable decay as Ae − aL log L , even when conditionedto be at least 1.Thus, we find that if c = x + iy with x ∈ I and | y | ≤ κ ( x ) /k , then from Lemma 4.7 (for k large enough as a function of I ), we have | ¯ p k ( c ) | ≤ X W ∈W k X e I (cid:23) d I ( W ) e F (cid:23) d F ( W ) | p c ( e I , e F ) | p ( W, e I , e F ) ≤ X W ∈W k ∞ X L =1 e A ( L +1) X e I (cid:23) d I ( W ) e F (cid:23) d F ( W )max(max( e I ) , max( e F )))= L p x ( e I , e F ) p ( W, e I , e F ) ≤ e Aδk / X W ∈W k X L ≤ δk / ( · · · ) | {z } ( C ) + X W ∈W k X L>δk / e AL ( · · · ) | {z } ( D ) .
14y (4.3), we have ( D ) ≤ Ak X L>δk / e AL − aL log L ≤ Ae − aδk / log k , and by Theorem 4.5, we have( C ) ≤ e Aδk / ¯ p k ( x ) ≤ Ae ( Aδ − a ) k / ≤ Ae − ak / , where the last inequality follows by taking δ small enough (as a function of I ). Puttingeverything together, we get | ¯ p k ( c ) | ≤ Ae − ak / . The calculation was made for k large enough as a function of I , but the inequality will betrue for smaller k as well by taking A large enough. This completes the proof of Theorem4.4. By Theorem 3.1 and (1.2), we have f ′ ( c ) ≥ ddc Z log deg T ( o ) d PGW ∗ c ( T, o ) = ddc X k ≥ e − c c k (1 − q ( c ) k ) log kθ ( c ) k ! . Although this lower bound appears to be a fairly simple expression, the presence of thelogarithm makes it hard to evaluate. For that reason, it seems desirable to have a moreexplicit lower bound.Write r k ( c ) for the probability that the root has degree k under the PGW ∗ ( c ) distribution.We seek a lower bound for X k ≥ r ′ k ( c ) log k = X k ≥ r ′ k ( c ) log + k = X k ≥ s ′ k ( c )[log + ( k + 1) − log + k ]= X k ≥ s ′ k ( c ) log k + 1 k > X k ≥ s ′ k ( c ) 1 k + 1 , where s k ( c ) := P j>k r j ( c ) and we have used Lemma 2.4 for the fact that s ′ k ( c ) ≥ PGW ∗ ( c ) is stochastically increasing in c ). Now thedegree of the root has the same law as X c := Q ∗ cθ ( c ) + Q cq ( c ) . Let c > δ >
0. Define g ( c, δ ) := α (cid:0) cθ ( c ) , ( c + δ ) θ ( c + δ ) (cid:1) − [ cq ( c ) − ( c + δ ) q ( c + δ )] .
15y Lemma 2.2, we have g ( c, δ ) >
0. By Lemma 2.3, we have that X c + δ stochasticallydominates Q ∗ cθ ( c ) + Q α ( cθ ( c ) , ( c + δ ) θ ( c + δ )) + Q ( c + δ ) q ( c + δ ) , which has the same distribution as X c + Y c,δ , where Y c,δ := Q g ( c,δ ) is independent of X c .Therefore, s k ( c + δ ) − s k ( c ) = P (cid:2) X c + δ > k (cid:3) − P (cid:2) X c > k (cid:3) ≥ P (cid:2) X c + Y c,δ > k (cid:3) − P (cid:2) X c > k (cid:3) . It follows that s ′ k ( c ) ≥ r k ( c ) β ( c ) , where β ( c ) := lim δ → g ( c, δ ) /δ . By (2.2) and (1.3), we have that g ( c, δ ) = log (cid:0) cq ( c ) e − cq ( c ) (cid:1) − log (cid:0) ( c + δ ) q ( c + δ ) e − ( c + δ ) q ( c + δ ) (cid:1) = log (cid:0) ce − c (cid:1) − log (cid:0) ( c + δ ) e − ( c + δ ) (cid:1) = log c − c − log( c + δ ) + c + δ . Therefore, β ( c ) = 1 − c . Thus, we obtain f ′ ( c ) > X k ≥ r k ( c ) β ( c ) k + 1 = e − c β ( c ) θ ( c ) Z (cid:16) e cs − e csq ( c ) (cid:17) ds = (cid:18) − c (cid:19) (cid:18) − e − c cθ ( c ) − e − cθ ( c ) − e − c cq ( c ) θ ( c ) (cid:19) = ( c − e − cq ( c ) c > . This completes the proof of Theorem 1.1.
A number of questions suggest themselves in light of our results, some of which arose inconversation with Itai Benjamini. 16. Given two finite graphs H and G , say that H G if there is a coupling of uniformvertices Y of H and X of G such that there is an isomorphism ϕ of the componentof Y in H to a subgraph of G such that ϕ ( Y ) = X . Let G ( n, M ) denote the randomgraph on n vertices with M edges. Write G ∗ ( n, M ) for the union of all components of G ( n, M ) that have the maximum number of edges (the maximum being taken over allthe components of G ( n, M ); for large M , there is likely to be only one such component).One very strong finitary version of Theorem 1.2 would say that G ∗ ( n, M ) G ∗ ( n, M +1)for M < (cid:0) n (cid:1) . Does this hold?2. Consider a ( d + 1)-regular tree and p > p > /d . Let T ( p ) denote the component ofthe root under Bernoulli( p ) percolation conditioned on the event that this component isinfinite. Does T ( p ) stochastically dominate T ( p )? Unpublished work of Erik Bromanand the first author here shows that for the (slightly different) case of d -ary trees, thisholds for d = 2 , G be a transitive graph, especially such as Z d , and p > p > p c ( G ),where p c ( G ) is the critical probability for Bernoulli (bond or site) percolation on G .Fix o ∈ G and let G ( p ) denote the component of o given that it is infinite. Does G ( p )stochastically dominate G ( p )? If this holds, then there is a weak limit of G ( p ) as p ↓ p c ( G ), which could be called the incipient infinite cluster. (It is conjectured thatthere is no infinite component at p c ( G ); see [BS96].) Such a limit is not known to existin Z d for d ≥
3, although another incipient infinite cluster has been constructed for d ≥
19 by [vdHJ04].4. Again, if G is a transitive graph, o ∈ G , and n ≥
1, let T n denote a uniformly chosenrandom subtree of G rooted at o and with n vertices. Is T n T n +1 ?5. Let ρ and ρ be two Galton-Watson measures on rooted trees. If k ≥ ρ ρ ,then is it necessarily the case that R p k ( o ; T ) dρ ( T, o ) ≥ R p k ( o ; T ) dρ ( T, o )?We thank Yuval Peres for several conversations.
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