The Order of the Giant Component of Random Hypergraphs
Michael Behrisch⋆, Amin Coja-Oghlan⋆⋆, and Mihyun Kang
Humboldt-Universit¨at zu Berlin, Institut f¨ur Informatik, Unter den Linden 6, 10099 Berlin, Germany [email protected]
Abstract. We establish central and local limit theorems for the number of vertices in the largest com- n−1 ponent of a random d-uniform hypergraph Hd(n, p) with edge probability p = c/ d−1 , where −1 (d − 1) + ε 1 Introduction and Results A d-uniform hypergraph H = (V, E) consists of a set V of vertices and a set E of edges, which are subsets of V of cardinality d. Moreover, a vertex w is reachable in H from a vertex v if either v = w or there is a sequence e1,...,ek of edges such that v e1, w ek, and ei ei+1 = for i =1,...,k 1. Of course, reachability in H is an equivalence relation.∈ The∈ equivalence∩ classes6 are∅ the components of− H, and H is connected if there is only one component. Throughoutthe paper, we let V = 1,...,n beasetof n vertices. Moreover, if 2 d is a fixed integer { } ≤ and 0 p = p(n) 1 is a sequence, then we let Hd(n,p) signify a random d-uniform hypergraph with ≤ ≤ n vertex set V in which each of the d possible edges is present with probability p independently. We say that H (n,p) enjoys some property with high probability (w.h.p.) if the probability that H (n,p) has d d tends to 1 as n . If d =2, then theP H (n,p) model is identical with the well-known G(n,p) model ofP → ∞ d random graphs. In order to state some related results we will also need a different model Hd(n,m) of ran- dom hypergraphs, where the hypergraph is chosen uniformly at random among all d-uniform hypergraphs with n vertices and m edges. Since the pioneering work of Erd˝os and R´enyi [8], the component structure of random discrete struc- tures has been a central theme in probabilistic combinatorics. In the present paper, we contribute to this theme by analyzing the maximum order (Hd(n,p)) of a component of Hd(n,p) in greater detail. More precisely, establishing central and localN limit theorems for (H (n,p)), we determine the asymptotic N d distribution of (Hd(n,p)) precisely. Though such limit theorems are known in the case of graphs (i.e, d =2), they areN new in the case of d-uniform hypergraphs for d> 2. Indeed, to the best of our knowledge none of the argumentsknown for the graph case extends directly to the case of hypergraphs(d> 2). There- fore, we present a new, purely probabilistic proof of the central and local limit theorems, which, in contrast to most prior work, does not rely on involved enumerative techniques. We believe that this new technique is interesting in its own right and may have further applications. The giant component. In the seminal paper [8], Erd˝os and R´enyi proved that the number of vertices in the largest component of G(n,p) undergoes a phase transition as np 1. They showed that if np < 1 ε for an arbitrarily small ε > 0 that remains fixed as n , then all∼ components of G(n,p) consist− of O(ln n) vertices. By contrast, if np > 1+ ε, then G(n,p→) ∞has one giant component on a linear number Ω(n) of vertices, while all other components contain only O(ln n) vertices. In fact, in the case 1+ ε Central and local limit theorems. In terms of limit theorems, (1) provides a strong law of large numbers for (G(n,p)), i.e., it yields the probable value of (G(n,p)) up to fluctuations of order o(n). Thus, N N a natural question is whether we can characterize the distribution of (G(n,p)) (or (Hd(n,p))) more precisely; for instance, is it true that (G(n,p)) “converges to the normalN distribution”N in some sense? Our first result, which we will prove inN Section 5, shows that this is indeed the case. 1 Theorem 1. Let ((d 1)− , ) be a compact interval, and let 0 p = p(n) 1 be a sequence J ⊂ n 1− ∞ ≤ ≤ such that c = c(n) = d−1 p for all n. Furthermore, let 0 < ρ = ρ(n) < 1 be the unique solution to (2), and set − ∈ J d 1 ρ 1 ρ + c(d 1)(ρ ρ − ) n σ2 = σ(n)2 = − − − . (3) (1 c(d 1)ρd 1)2 − − − 1 Then σ− ( (H (n,p)) (1 ρ)n) converges in distribution to the standard normal distribution. N d − − Theorem 1 provides a central limit theorem for (H (n,p)); it shows that for any fixed numbers a 0 1 γσ (ν (1 ρ)n t)2 (5) P [ (Hd(n,p)) ν γσ] exp − − 2 − dt, |N − |≤ ∼ √2πσ γσ 2σ Z− i.e., we can estimate the probability that (H (n,p)) deviates from some value ν by at most γσ. However, N d it is impossible to derive from (4) or (5) the asymptotic probability that (Hd(n,p)) hits ν exactly. By contrast, our next theorem shows that for any integer ν such thatNν (1 ρ)n O(σ) we have | − − |≤ 1 (ν (1 ρ)n)2 P [ (Hd(n,p)) = ν] exp − − , (6) N ∼ √2πσ − 2σ2 1 n 1 provided that (d 1)− + ε − p = O(1). Note that (6) is exactly what we would obtain from (5) if − ≤ d 1 we were allowed to set δ = 1 σ(n,p−) 1 in that equation. Stated rigorously, the local limit theorem reads as 2 − follows. 1 Theorem 2. Let d 2 be a fixed integer. For any two compact intervals R, ((d 1)− , ), ≥ I ⊂ J ⊂ − ∞ and for any δ > 0 there exists n0 > 0 such that the following holds. Let p = p(n) be a sequence such that n 1 c = c(n) = d−1 p for all n, let 0 < ρ = ρ(n) < 1 be the unique solution to (2), and let σ be as − ∈ J 1 in (3). If n n and if ν is an integer such that σ− (ν (1 ρ)n) , then ≥ 0 − − ∈ I 1 δ (ν (1 ρ)n)2 1+ δ (ν (1 ρ)n)2 − exp − − P [ (Hd(n,p)) = ν] exp − − . √2πσ − 2σ2 ≤ N ≤ √2πσ − 2σ2 3 Related work. Since the work of Erd˝os and R´enyi [8], the component structure of G(n,p)= H2(n,p) has received considerable attention. Stepanov [19] provided central and local limit theorems for (G(n,p)), thereby proving the d =2 case of Theorems1 and 2. In order to establish these limit theorems,N he estimates the probability that a random graph G(n,p) is connected up to a factor 1+ o(1) using recurrence formulas for the number of connected graphs. Furthermore, Barraez, Boucheron, and Fernandez de la Vega [2] reproved the central limit theorem for (G(n,p)) via the analogy of breadth first search on a random graph and a Galton-Watson branching process.N In addition, a local limit theorem for (G(n,p)) can also be derived using the techniques of van der Hofstad and Spencer [9], or the enumerativeN results of either Bender, Canfield, and McKay [5] or Pittel and Wormald [15]. Moreover, Pittel [14] proved a central limit theorem for the largest component in the G(n,m) model of random graphs; G(n,m) is just a uniformly distributed graph with exactly n vertices and m edges. Indeed, Pittel actually obtained his central limit theorem via a limit theorem for the joint distribution of the number of isolated trees of a given order, cf. also Janson [10]. A comprehensive treatment of further results on the components of G(n,p) can be found in [11]. In contrast to the case of graphs, only little is known for d-uniform hypergraphs with d > 2; for the methods used for graphs do not extend to hypergraphs directly. Using the result [12] on the number of sparsely connected hypergraphs, Karo´nski and Łuczak [13] investigated the phase transition of Hd(n,p). They established (among other things) a local limit theorem for (Hd(n,m)) for m = n/d(d 1)+ l and 3 l ln n nN 1 1 − 1/3 1 n2 ln ln n which is similar to Hd(n,p) at the regime d−1 p = (d 1)− + ω, where n− ≪ ≤ 1/3 − − ≪ ω = ω(n) n− ln n/ ln ln n. These results were extended by Andriamampianina, Ravelomanana and ≪ 1/3 2/3 Rijamamy [1,16] to the regime l = o(n ) (ω = o(n− ) respectively). n 1 1 By comparison, Theorems 1 and 2 deal with edge probabilities p such that d−1 p = (d 1)− +Ω(1), n 1 1 − − i.e., − p is bounded away from the critical point (d 1) . Thus, Theorems 1 and 2 complement [1, d 1 − − n 1 − 1 13,16]. The only prior paper dealing with − p = (d 1) + Ω(1) is that of Coja-Oghlan, Moore, and d 1 − Sanwalani [7], where the authors computed the− expectation− and the variance of (H (n,p)) and obtained N d qualitative results on the component structure of Hd(n,p). In addition, in [7] the authors estimated the probability that Hd(n,p) or a uniformly distributed d-uniform hypergraph Hd(n,m) with n vertices and m edges is connected up to a constant factor. While in the present work we build upon the results on the component structure of Hd(n,p) from [7], the results and techniques of [7] by themselves are not strong enough to obtain a central or even a local limit theorem for (H (n,p)). N d Techniques and outline. The aforementioned prior work [1,12,13] on the giant component for random hypergraphs relies on enumerative techniques to a significant extent; for the basis [1,12,13] are results on the asymptotic number of connected hypergraphs with a given number of vertices and edges. By con- trast, in the present work we employ neither enumerative techniques nor results, but rely solely on proba- bilistic methods. Our proof methods are also quite different from Stepanov’s [19], who first estimates the asymptotic probability that a random graph G(n,p) is connected in order to determine the distribution of (Hd(n,p)). By contrast, in the present work we prove the local limit theorem for (Hd(n,p)) directly, therebyN obtaining “en passant” a new proof for the local limit theorem for random graphsN G(n,p), which may be of independent interest. Besides, the local limit theorem can be used to compute the asymptotic probability that G(n,p) or, more generally, Hd(n,p) is connected, or to compute the asymptotic number of connected hypergraphs with a given number of vertices and edges (cf. Section 6). Hence, the general approach taken in the present work is actually converse to the prior ones [1,12,13,19]. The proof of Theorem 1 makes use of Stein’s method, which is a general technique for proving central limit theorems [18]. Roughly speaking, Stein’s result implies that a sum of a family of dependent random variables converges to the normal distribution if one can bound the correlations within any constant-sized subfamily sufficiently well. The method was used by Barbour, Karo´nski, and Ruci´nski [3] in order to prove that in a random graph G(n,p), e.g., the number of tree components of a given (bounded) size is asymptotically normal. To establish Theorem 1, we extend their techniques in two ways. – Instead of dealing with the number of vertices in trees of a given size, we apply Stein’s method to the total number n (Hd(n,p)) of vertices outside of the giant component; this essentially means that we need to sum− over N all possible (hyper)tree sizes up to about ln n. 4 – Since we are dealing with hypergraphs rather than graphs, we are facing a somewhat more complex situation than [3], because the fact that an edge may involve an arbitrary number d of vertices yields additional dependencies. The main contribution of this paper is the proof of Theorem 2. To establish this result, we think of the edges of Hd(n,p) as being added in two “portions”. More precisely, we first include each possible edge with probability p1 = (1 ε)p independently, where ε> 0 is small but independent of n (and denote the resulting random hypergraph− by H ); by Theorem 1, the order (H ) of the largest component of H is 1 N 1 1 asymptotically normal. Then, we add each possible edge that is not present in H1 with a small probability p2 εp and investigate closely how these additional random edges attach further vertices to the largest component∼ of H . Denoting the number of these “attached” vertices by , we will show that the conditional 1 S distribution of given the value of (H1) satisfies a local limit theorem. Since p1 and p2 are chosen such that each edgeS is present with probabilityN p after the second portion of edges has been added, this yields the desired result on (Hd(n,p)). The analysis of theN conditional distribution of involves proving that is asymptotically normal. To show this, we employ Stein’s method once more. InS addition, in order to showS that satisfies a local limit S theorem, we prove that the number of isolated vertices of H1 that get attached to the largest component of H1 by the second portion of random edges is binomially distributed. Since the binomial distribution satisfies a local limit theorem, we thus obtain a local limit theorem for . S Our proof of Theorem 2 makes use of some results on the component structure of Hd(n,p) derived in [7]. For instance, we employ the results on the expectation and the variance of (Hd(n,p)) from that paper. Furthermore, the analysis of given in the present work is a considerable extensionN of the argument used in [7], which by itself would justS yield the probability that attains a specific value s up to a constant factor. S The main part of the paper is organized as follows. After making some preliminaries in Section 2, we outlinethe proofof Theorem2 in Section 3.In thatsection we explain in detail how Hd(n,p) is generated in two “portions”. Then, in Section 4 we analyze the random variable , assuming the central limit theorem for . Further, Section 5 deals with the proof of Theorem 1 and theS proof of the central limit theorem for S via Stein’s method; the reason why we defer the proof of Theorem 1 to Section 5 is that we can S use basically the same argument to prove the asymptotic normality of both (Hd(n,p)) and . Finally, Section 6 contains some concluding remarks, e.g., on the use of the presentN results to derive furtherS limit theorems and to solve enumerative problems. 2 Preliminaries Throughout the paper, we let V = 1,...,n . If d 2 is an integer and V ,...,V V , then we let { } ≥ 1 k ⊂ d(V1,...,Vk) signify the set of all subsets e V of cardinality d such that e Vi = for all i. We omit Ethe subscript d if it is clear from the context. ⊂ ∩ 6 ∅ If H is a hypergraph, then we let V (H) denote its vertex set and E(H) its edge set. We say that a set S V (H) is reachable from T V (H) if each vertex s S is reachable from some vertex t T . Further,⊂ if V (H) V = 1,...,n⊂ , then the subsets of V can∈ be ordered lexicographically; hence,∈ we can define the largest⊂ component{ of H} to be the lexicographically first component of order (H). N We use the O-notation to express asymptotic estimates as n . Furthermore, if f(x1,...,xk,n) is a function that depends not only on n but also on some further→ parameters ∞ x from domains D R i i ⊂ (1 i k),and if g(n) 0 is another function, then we say that the estimate f(x1,...,xk,n)= O(g(n)) holds≤ uniformly≤ in x ,...,x≥ if the following is true: if and D , D , are compact sets, then there 1 k Ij j Ij ⊂ j exist numbers C = C( 1,..., k) and n0 = n0( 1,..., k) such that f(x1,...,xk,n) Cg(n) for all I kI I I | |≤ n n0 and (x1,...,xk) j=1 j . We define uniformity analogously for the other Landau symbols Ω, Θ,≥ etc. ∈ I We shall make repeatedQ use of the following Chernoff bound on the tails of a binomially distributed variable X = Bin(ν, q) (cf. [11, p. 26] for a proof): for any t> 0 we have t2 P [ X E(X) t] 2exp . (7) | − |≥ ≤ −2(E(X)+ t/3) 5 Moreover, we employ the following local limit theorem for the binomial distribution (cf. [6, Chapter 1]). Proposition 3. Suppose that 0 p = p(n) 1 is a sequence such that np(1 p) as n . Let X = Bin(n,p). Then for any sequence≤ x = x≤(n) of integers such that x np−= o(→np ∞(1 p))→2/3 ∞, | − | − 1 (x np)2 P [X = x] (2πnp(1 p))− 2 exp − as n . ∼ − −2p(1 p)n → ∞ − Furthermore, we make use of the following theorem, which summarizes results from [7, Section 6] on the component structure of Hd(n,p). 1 n 1 − Theorem 4. Let p = c d−1 . − 1 1. If there is a fixed c < (d 1)− such that c = c(n) c , then 0 − ≤ 0 2 2 100 P (H (n,p)) 3(d 1) (1 (d 1)c )− ln n 1 n− . N d ≤ − − − 0 ≥ − 1 2. Suppose that c0 > (d 1)− is a constant, and that c0 c = c(n) = o(ln n) as n . Then the transcendental equation− (2) has a unique solution 0 <ρ ≤= ρ(c) < 1, which satisfies → ∞ ρn 1 p for some number c0′ > 0 that depends only on c0. Moreover, o(1) E [ (Hd(n,p))] (1 ρ)n n , | N − − |≤ d 1 ρ 1 ρ + c(d 1)(ρ ρ − ) n Var( (H (n,p))) − − − . N d ∼ (1 c(d 1)ρd 1)2 − − − 100 Furthermore, with probability 1 n− there is precisely one component of order (1+o(1))(1 ρ)n in H (n,p), while all other components≥ − have order ln2 n. In addition, − d ≤ 0.51 100 P (H (n,p)) E( (H (n,p))) n n− . |N d − N d |≥ ≤ n 1 Finally, the following result on the component structure of Hd(n,p) with average degree d−1 p < 1 − (d 1)− below the threshold has been derived in [7, Section 6] via the theory of branching processes. − 1 k Proposition 5. There exists a function q : (0, (d 1)− ) [0, 1] R 0, (ζ, ξ) q(ζ, ξ)= ∞ qk(ζ)ξ − × → ≥ 7→ k=1 whose coefficients ζ qk(ζ) are differentiable such that the following holds. Suppose that 0 p = 7→ n 1 1 P ≤ p(n) 1 is a sequence such that 0 < d−1 p = c = c(n) < (d 1)− ε for an arbitrarily small ε> 0 ≤ − − − that remains fixed as n . Let P (c, k) denote the probability that in Hd(n,p) some fixed vertex v V lies in a component of order→ ∞k. Then ∈ 2/3 2 P (c, k)=(1+ o(n− ))q (c) for all 1 k ln n. k ≤ ≤ Furthermore, for any fixed ε> 0 there is a number 0 <γ = γ(ε) < 1 such that k 1 q (c) γ for all 0 3 Proof of Theorem 2 n 1 Throughout this section, we assume that c = c(n) = − p for some compact interval d 1 ∈ J J ⊂ ((d 1) 1, ). Moreover, we let R be some fixed compact− interval, and ν denotes an integer such − that−(ν (1∞ρ)n)/σ . All asymptoticsI ⊂ are understood to hold uniformly in c and (ν (1 ρ)n)/σ. − − ∈ I − − 6 3.1 Outline n 1 1 Let ε = ε( ) > 0 be independent of n and small enough so that (1 ε) d−1 p > (d 1)− + ε. Set p = (1 Jε)p. Moreover, let p be the solution to the equation p +−p p−p = p; then− p εp. We 1 − 2 1 2 − 1 2 2 ∼ expose the edges of Hd(n,p) in four “rounds” as follows. n R1. As a first step, we let H1 be a random hypergraphobtained by including each of the d possible edges with probability p1 independently. Let G denote the largest component of H1. R2. Let H be the hypergraph obtained from H by adding each edge e H that lies completely outside 2 1 6∈ 1 of G (i.e., e V G) with probability p2 independently. R3. Obtain H ⊂by adding\ each possible edge e H that contains vertices of both G and V G with 3 6∈ 1 \ probability p2 independently. R4. Finally, include each possible edge e H such that e G with probability p independently. 6∈ 1 ⊂ 2 Here the 1st round corresponds to the first portion of edges mentioned in Section 1, and the edges added in the 2nd–4th round correspond to the second portion. Note that for each possible edge e V the probability ⊂ n 1 that e is actually present in H is p + (1 p )p = p, hence H = H (n,p). Moreover, as − p > 4 1 − 1 2 4 d d 1 1 (d 1) 1 + ε by our choice of ε, Theorem 4 entails that w.h.p. H has exactly one largest component− of − 1 linear− size Ω(n) (the “giant component”). Further, the edges added in the 4th round do not affect the order of the largest component, i.e., (H4)= (H3). In order to analyze the distributionN of N(H (n,p)), we first establish central limit theorems for (H )= N d N 1 G and (H3) = (H4) = (Hd(n,p)), i.e., we prove that (centralized and normalized versions of)| | (HN) and (HN ) are asymptoticallyN normal. Then, we investigate the number of vertices = N 1 N 3 S (H3) (H1) that get attached to G1 during the 3rd round. We shall prove that given that G = n1, N − N 2 | | S is locally normal with mean µ + (n1 µ1)λ and variance σ independent of n1. Finally, we combine S − S S these results to obtain the local limit theorem for (Hd(n,p)) = (H3)= (H1)+ . n 1 n 1 N N N S Let c1 = d−1 p1 and c3 = d−1 p. Moreover, let 0 < ρ3 < ρ1 < 1 signify the solutions to the − − d 1 transcendental equations ρ = exp c (ρ − 1) and set for j =1, 3 j j j − d 1 ρ 1 ρ +c (d 1)(ρ ρ − ) n 2 j − j j − j − j µj = (1 ρj)n, σj = d 1 (cf. Theorem 4). − (1 c (d 1)ρ − )2 − j − j The following proposition, which we will prove in Section 5, establishes a central limit theorem for both (H ) and (H ) and thus proves Theorem 1. N 1 N 3 Proposition 6. ( (Hj ) µj )/σj converges in distribution to the standard normal distribution for j = 1, 3. N − With respect to the distribution of , we will establish the following local limit theorem in Section 4. S Proposition 7. Suppose that n µ n0.6. | 1 − 1|≤ 1. The conditional expectation of given that G = n1 satisfies E( 1 = n1)= µ + λ (n1 µ1)+ S | | S|N S S − o(√n), where µ = Θ(n) and λ = Θ(1) are independent of n1. S S 0.6 2. There is a constant C > 0 such that for all s satisfying µ + λ (n1 µ1) s n we have 1 | S S − − | ≤ P [ = ν = n ] Cn− 2 . S |N1 1 ≤ 3. If s is an integer such that µ + λ (n1 µ1) s O(√n), then | S S − − |≤ 2 1 (µ + λ (n1 µ1) s) P [ = s 1 = n1] exp S S − − , S |N ∼ √2πσ − 2σ2 S S where σ = Θ(√n) is independent of n1. S Since = + , Propositions 6 and 7 yield N3 N1 S µ3 = µ1 + µ + o(√n). (10) S Combining Propositions 6 and 7, we derive the following formula for P [ 3 = ν] in Section 3.2. Recall that we are assuming that ν is an integer such that (ν µ)/σ = (ν µ )/σN . − − 3 3 ∈ I 7 Corollary 8. Letting z = (ν µ )/σ , we have − 3 3 2 2 1 ∞ x 1 σ1 σ3 P [ 3 = ν] exp (x (1 + λ ) z dx. (11) N ∼ 2πσ − 2 − 2 · S σ − · σ S Z−∞ " S S # Proof of Theorem 2. Integrating the right hand side of (11), we obtain an expression of the form 1 (ν κ)2 P [ 3 = ν] exp − , (12) N ∼ √2πτ − 2τ 2 2 where κ, τ = Θ(n). Therefore, on the one hand ( 3 µ3)/σ3 converges in distribution to the normal dis- tribution with mean κ µ and variance (τ/σ )2. OnN the− other hand, Proposition 6 states that ( µ )/σ − 3 3 N3 − 3 3 converges to the standard normal distribution. Consequently, κ µ3 = o(τ) and τ σ3. Plugging these 1 1| − | 2 2 ∼ estimates into (12), we obtain P [ 3 = ν] exp (ν µ3) σ− . Since 3 = (Hd(n,p)), N ∼ √2πσ3 − 2 − 3 N N this yields the assertion. ⊓⊔ 3.2 Proofof Corollary8 Let α > 0 be arbitrarily small but fixed as n , and let C′ = C′(α) > 0 be a large enough number → ∞ depending only on α. Set J = n1 ZZ : n1 µ1 C′√n , let J ′ = n1 ZZ : C′√n< n1 µ1 0.6 { ∈ | 0.6− |≤ } { ∈ | − |≤ n , and J ′′ = n ZZ : n µ >n . Then letting } { 1 ∈ | 1 − 1| } Ψ = P [ = n ] P [ = ν n = n ] , for X J, J ′, J ′′ X N1 1 S − 1|N1 1 ∈{ } n1 X X∈ we have P [ 3 = ν]= ΨJ + ΨJ′ + ΨJ′′ , and we shall estimate each of the three summands individually. N 0.51 100 Since Theorem 4 implies that P µ >n n− , we conclude that |N1 − 1| ≤ 100 Ψ ′′ P [ J′′] n− . (13) J ≤ N1 ∈ ≤ 2 Furthermore, as σ1 = O(n), Chebyshev’s inequality implies that 2 2 1 P [ J ′] P µ > C′√n σ C′− n− < α/C′, (14) N1 ∈ ≤ |N1 − 1| ≤ 1 provided that C′ is large enough. Hence, combining (14) with the second part of Proposition 7, we obtain C αC 1/2 ΨJ′ P [ 1 J ′] < αn− , (15) ≤ N ∈ · √n ≤ C′√n where once we need to pick C′ sufficiently large. σ1 To estimate the contribution of n1 J, we split J into subintervals J1,...,JK of length between 2C′ σ1 ∈ and C′ . Moreover, let Ij be the interval [(min Jj µ1)/σ1, (max Jj µ1)/σ1]. Then Proposition 6 implies that − − 1 α 2 1+ α 2 − exp( x /2)dx P [ 1 = n1] exp( x /2)dx (16) √2π Ij − ≤ N ≤ √2π Ij − n1 Jj Z X∈ Z for each 1 j K. Furthermore, Proposition 7 yields ≤ ≤ 2 1 (ν n1 µ λ (n1 µ1)) P [ = ν n1 1 = n1] exp − − S − S − . S − |N ∼ √2πσ − 2σ2 S S for each n1 J. Hence, choosing C′ sufficiently large, we can achieve that for all n1 Jj and all x Ij the bound ∈ ∈ ∈ 2 2 (1 + α) (ν µ1 σ1x µ λ (n1 µ1)) P [ = ν n1 1 = n1] exp − − − S − S − S − |N ≤ √2πσ − 2σ2 S S 2 2 (10) (1 + α) 1 σ σ exp (x (1 + λ ) 1 z 3 (17) ∼ √2πσ −2 · S σ − · σ ! S S S 8 holds. Now, combining (16) and (17), we conclude that K Ψ = P [ = n ] P [ = ν n = n ] J N1 1 S − 1|N1 1 j=1 n1 Jj X X∈ K (1 + α)3 x2 1 σ σ 2 exp (x (1 + λ ) 1 z 3 dx ≤ 2πσ − 2 − 2 · S σ − · σ j=1 Ij " # S X Z S S 2 2 1+4α ∞ x 1 σ σ exp (x (1 + λ ) 1 z 3 dx. (18) ≤ 2πσ − 2 − 2 · S σ − · σ S Z−∞ " S S # Analogously, we derive the matching lower bound 2 2 1 4α ∞ x 1 σ1 σ3 ΨJ − exp (x (1 + λ ) z dx. (19) ≥ 2πσ − 2 − 2 · S σ − · σ S Z−∞ " S S # Finally, combining (13), (15), (18), and (19), and remembering that P [ 3 = ν] = ΨJ + ΨJ′ + ΨJ′′ , we obtain the assertion, because α> 0 can be chosen arbitrarily small if n getsN sufficiently large. 4 The Conditional Distribution of S Throughout this section, we keep the notation and the assumptions from Section 3. In addition, we let G V be a set of cardinality n such that n µ n0.6. ⊂ 1 | 1 − 1|≤ 4.1 Outline The goal of this section is to prove Proposition 7. Let us condition on the event that the largest component of H1 is G. To analyze the conditional distribution of , we need to overcome the problem that in H1 the edges in the set V G do not occur independently anymoreS once we condition on G being the largest \ component of H1. However, we will see that this conditioning is “not very strong”. To this end, we shall compare with an “artificial” random variable , which models the edges contained in V G as mutually S SG \ independent objects. To define G, we set up random hypergraphs Hj,G, j = 1, 2, 3, in three “rounds” as follows. S n n1 R1’. The vertex set of H1,G is V = 1,...,n , and each of the −d possible edges e V G is present in H with probability p {independently.} ⊂ \ 1,G 1 R2’. Adding each possible edge e V G not present in H with probability p independently yields ⊂ \ 1,G 2 H2,G. R3’. Obtain H from H by including each possible edge e incident to both G and V G with proba- 3,G 2,G \ bility p2 independently. The process R1’–R3’ relates to the process R1–R4 from Section 3.1 as follows. While in H1 the edges in V G are mutually dependent, we have “artificially” constructed H in such a way that the edges \ 1,G outside of G occur independently. Then, H2,G and H3,G are obtained similarly as H2 and H3, namely by including further edges inside of V G and crossing edges between G and V G with probability p . \ \ 2 Letting SG denote the set of vertices in V G that are reachable from G, the quantity G = SG now corresponds to . In contrast to R1–R4, the\ process R1’–R3’ completely disregards edgesS inside| | of G, S because these do not affect G. The following lemma, which we will prove in Section 4.3 shows that G is indeed a very good approximationS of , so that it suffices to study . S S SG 9 Lemma 9. For any ν ZZ we have P [ = ν (H )= n ] P [ = ν] n− . ∈ | S | N 1 1 − SG |≤ As a next step, we investigate the expectation of G.While thereis no needto compute E( G) precisely, we do need that E( ) depends on n µ linearlyS. The corresponding proof can be found inS Section 4.4. SG 1 − 1 9 Lemma 10. We have E( G) = µ + λ (n1 µ1)+ o(√n), where µ = Θ(n) and λ = Θ(1) do not S S S − S S depend on n1. Furthermore, we need that the variance of G is essentially independent of the precise value of n1. This will be proven in Section 4.5. S Lemma 11. We have Var( )= O(n). Moreover, if G′ V is another set such that µ G′ = o(n), SG ⊂ | 1 − | || then Var( ) Var( ′ ) = o(n). | SG − SG | To show that satisfies a local limit theorem, the crucial step is to prove that for numbers s and t SG such that s is “close” to t the probabilities P [ G = s], P [ G = t] are “almost the same”. More precisely, the following lemma, proven in Section 4.2, holds.S S 0.6 Lemma 12. For every α> 0 there is β > 0 such that for all s,t satisfying s E( G) , t E( G) n and s t βn1/2 we have | − S | | − S |≤ | − |≤ 10 10 (1 α)P [ = s] n− P [ = t] (1 + α)P [ = s]+ n− . − SG − ≤ SG ≤ SG 1/2 Moreover, there is a constant C > 0 such that P [ = s] Cn− for all integers s. SG ≤ 2 Letting G0 = 1,..., µ1 , we define σ = Var( G0 ) and obtain a lower bound on σ as an immediate consequence of{ Lemma⌈ 12.⌉} S S S Corollary 13. We have σ = Ω(√n). S Proof. By Lemma 12 there exists a number 0 <β< 0.01 independent of n such that for all integers s,t satisfying s E( ) , t E( ) √n and s t β√n we have | − SG | | − SG |≤ | − |≤ 2 10 P [ = t] P [ = s] n− . (20) SG ≥ 3 SG − 2 2 Set γ = β /64 and assume for contradiction that σ < γn/2. Moreover, suppose that G = G0 = S 1 1,..., µ1 . Then Chebyshev’s inequality entails that P G E( G) √γn 2 . Hence, there { ⌈ ⌉} |S − S |≥1 ≤ exists an integer s such that s E( ) γn and P [ = s] 1 (γn) 2 . Therefore, due to (20) we G √ G 2 − 1 1| − S | ≤ S ≥ 2 have P [ = t] (γn)− 2 for all integers t such that s t β√n. Thus, recalling that γ = β /64, SG ≥ 4 | − | ≤ we obtain β√n This contradiction shows 1 P [ G s β√n] = t: t s β√n P [ G = t] 4√γn > 1. ≥ |S − |≤ | − |≤ S ≥ that σ2 γn/2. S ≥ P ⊓⊔ Using the above estimates of the expectation and the variance of G and invoking Stein’s method once more, in Section 5 we will show the following. S 0.66 Lemma 14. If n1 µ1 n , then ( G E( G))/σ is asymptotically normal. | − |≤ S − S S Proof of Proposition 7. The first part of the proposition follows readily from Lemmas 9 and 10. Moreover, the second assertion follows from Lemma 12. Furthermore, we shall establish below that 1 (s E( ))2 P [ G = s] exp − S for any integer s such that s E( G) = O(√n). (21) S ∼ √2π − 2σ2 | − S | S 2 2 2 2 This claim implies the third part of the proposition. For (s E( )) σ− (µ + λ (n1 µ1)) σ− by Lemma 10 and Corollary 13, and P [ = s = n ] P [− =S s] byS Lemma∼ S 9. S − S S |N1 1 ∼ SG To prove (21) let α> 0 be arbitrarily small but fixed. Since σ2 = Θ(n) by Lemma 11 and Corollary 13, S 0.6 Lemma 12 entails that for a sufficiently small β > 0 and all s,t satisfying s E( G) , t E( G) n and s t βσ we have | − S | | − S |≤ | − |≤ S 10 10 (1 α)P [ = s] n− P [ = t] (1 + α)P [ = s]+ n− . (22) − SG − ≤ SG ≤ SG 10 Now, suppose that s is an integer such that s E( G) O(√n), and set z = (s E( G))/σ . Then Lemma 14 implies that | − S | ≤ − S S z+β 1 α 2 β 2 P [ G s βσ ] − exp( x /2)dx (1 2α) exp( z /2), (23) |S − |≤ S ≥ √2π z β − ≥ − √2π − Z − provided that β is small enough. Furthermore, (22) yields that 10 P [ G s βσ ]= P [ G = t] βσ ((1 + α)P [ G = s]+ n− ) |S − |≤ S S ≤ S S t: t s βσ | −X|≤ S 9 (1 + α)βσ P [ G = s]+ n− , (24) ≤ S S because σ = O(√n) by Lemma 11. Combining (23) and (24), we conclude that S 2 1 2α 1 2 9 1 4α (s E( G)) P [ G = s] − exp( z /2) n− − exp − S . S ≥ 1+ α · √2πσ − − ≥ √2πσ − 2σ2 S S S 2 1+4α (s E( G)) − S2 Since analogous arguments yield the matching upper bound P [ G = s] √ exp 2σ , S ≤ 2πσS − S and because α> 0 may be chosen arbitrarily small, we obtain (21). ⊓⊔ Next we will prove Lemma 12 which provides the central locality argument while the more technical proofs of Lemma 9, 10 and 11 are deferred to the end of this section. 4.2 ProofofLemma 12 1 Since the assertion is symmetric in s and t, it suffices to prove that P [ G = s] (1 α)− P [ G = s]+ 10 S ≤ − S n− . Let = E(H ) E(H ) be the (random) set of edges added during R3’. We split into three F 3,G \ 2,G F subsets: let 1 consist of all e such that either e G 2 or e contains a vertex that belongs to a component ofF V G of order ∈2 F. Moreover, is the| \ set| of ≥ all edges e that contain a vertex \ ≥ F2 ∈F\F1 of V G that is also contained in some other edge e′ . Finally, = ( ); thus, all edges \ ∈F1 F3 F\ F1 ∪F2 e 3 connect d 1 vertices in G with a vertex v V G that is isolated in H2,G + 1 + 2, see Figure 1 for∈F an example. Hence,− H = H + + +∈ \. F F 3,G 2,G F1 F2 F3 1 F 2 F 3 1 F F 1 F V G \ G Fig. 1. The three kinds of edges (black) which attach small components to G. The edges of H2,G are depicted in grey. The (3-uniform) edges are depicted as circular arcs spanned by the three vertices contained in the corresponding edge. As a next step, we decompose G into two contributions corresponding to 1 2 and 3. More precisely, we let big be the numberS of vertices in V G that are reachable from GFin H∪F + F+ and SG \ 2,G F1 F2 11 iso big set G = G G . Hence, if we let signify the set of all isolated vertices of H2,G + 1 + 2 in the S S − Siso W F F set V G, then G equals the number of vertices in that get attached to G via the edges in 3. We\ can determineS the distribution of iso precisely.W For if v , then each edge e containingF v SG ∈ W and exactly d 1 vertices of G is present with probability p2 independently. Therefore, the probability − n1 ( 1) that v gets attached to G is 1 (1 p2) d− . In fact, these events occur independently for all v . Consequently, − − ∈ W n n iso 1 iso 1 = Bin , 1 (1 p )(d−1) , µ = E( )= (1 (1 p )(d−1))= Ω( ), (25) SG |W| − − 2 iso SG |W| − − 2 |W| 1 d where the last equality sign follows from the fact that p2 εp1 = Θ(n − ). big iso ∼ Hence, G = G + G features a contribution that satisfies a local limit theorem, namely the bino- S S iso S mially distributed G . Thus, to establish the locality of G (i.e., Lemma 12), we are going to prove that “inherits” the localityS of iso. To this end, we need toS bound , thereby estimating µ = E( iso). SG SG |W| iso SG 1 10 Lemma 15. We have P (n n ) exp( c) 1 n− . |W| ≥ 2 − 1 − ≥ − The proof of Lemma 15 is just a standard application of Azuma’s inequality, cf. Section 4.6. Further, let M be the set of all triples (H, F1, F2) such that 11 M1. P [ G = s H2,G = H, 1 = F1, 2 = F2] n− , and M2. givenS that H| = H, F = F , andF = F≥, the set has size 1 (n n ) exp( c)= Ω(n). 2,G F1 1 F2 2 W ≥ 2 − 1 − Lemma 16. If s t β√n for some small enough β = β(α) > 0, then P [ G = t (H2,G, 1, 2) M] (1 α)P [ =| −s (|≤H , , ) M]. S | F F ∈ ≥ − SG | 2,G F1 F2 ∈ big Proof. Let (H, F1, F2) M, and let b be the value of G given that H2,G = H, 1 = F1 and 2 = F2. Then given that this event∈ occurs, we have = s iff Siso = s b. As (H, F , F )F M, we concludeF that SG SG − 1 2 ∈ n1 M1 ( 1) 11 P [ = s H = H, = F , = F ]=P Bin , 1 (1 p ) d− = s b n− . SG | 2,G F1 1 F2 2 |W| − − 2 − ≥ h i 0.6 Therefore, the Chernoff bound (7) implies that s b µiso n . Furthermore, since we assume that | − − | ≤ n1 1/2 ( 1) t s βn for some small β = β(α) > 0 and as µiso = (1 (1 p2) d− ) Ω(n) due to M2, Proposition| − |≤ 3 entails that |W| − − ≥ n1 n1 P Bin , 1 (1 p )(d−1) = t b (1 α)P Bin , 1 (1 p )(d−1) = s b . |W| − − 2 − ≥ − |W| − − 2 − h i h i Thus, the assertion follows from (25). ⊓⊔ Proof of Lemma 12. By Lemmas 15 and 16, we have 1 P [ = s] P [ = s (H , , ) M] P [(H , , ) M]+(1 α)− P [ = t] SG ≤ SG | 2,G F1 F2 6∈ 2,G F1 F2 6∈ − SG M1, M2 11 1 1 10 n− + P [ = o(n)]+(1 α)− P [ = t] (1 α)− P [ = t]+ n− , ≤ |W| − SG ≤ − SG as claimed. ⊓⊔ 4.3 ProofofLemma 9 Let signify the event that G is the largest component of H . Given that occurs, the edges in H G LG 1 LG 3 − do not occur independently anymore. For if G occurs, then H1 G does not contain a component on more than G vertices. Nonetheless, the followingL lemma shows− that if E (V ) (G) is a set of edges such| that| the hypergraph H(E) = (V, E (V G)) does not feature a⊂ “big” E component,\E then the ∩E \ dependence of the edges is very small. In other words, the probability that the edges E are present in H3 is very close to the probability that these edges are present in the “artificial” model H3,G, in which edges occur independently. 12 Lemma 17. For any set E (V ) (G) such that (H(E)) ln2 n we have ⊂E \E N ≤ 10 P [E(H ) (G)= E ]=(1+ O(n− ))P [E(H )= E] . 3 \E | LG 3,G Before getting down to the proof of Lemma 17, we first show how it implies Lemma 9. As a first step, we 2 derive that it is actually quite unlikely that either H3 G or H3,G G features a component on ln n vertices. − − ≥ 2 2 10 Corollary 18. We have P (H G) > ln n , P (H G) > ln n = O(n− ). N 3 − |LG N 3,G − 2 10 Proof. Theorem 4 implies that P (H3,G G) > ln n = O(n− ), because H3,G simply is a random nNn1 − n µ1 1 hypergraph H (n n ,p), and − p − p < (d 1) by (8). Hence, Lemma 17 yields that d 1 d 1 d 1 − − 2 − ∼10 − − 2 10 P (H G) ln n (1 O(n− ))P (H G) ln n 1 O(n− ). N 3 − ≤ |LG ≥ − N 3,G − ≤ ≥ − ⊓⊔ Proof of Lemma 9. Let denote the set of all subsets E (V ) (G ) such that in the hypergraph As ⊂ E \E (V, E) exactly s vertices in V G are reachable from G. Moreover, let s signify the set of all E s such that (H(E)) ln2 n. Then\ B ∈ A N ≤ P [ = s ]=P[E(H ) (G) ] , and P [ = s]=P[E(H ) ] . (26) S |LG 3 \E ∈ As|LG SG 3,G ∈ As Furthermore, by Corollary 18 2 10 P [E(H ) (G) ] P (H G) > ln n = O(n− ), (27) 3 \E ∈ As \Bs|LG ≤ N 3 − |LG 2 10 P [E(H ) ] P (H G) > ln n = O(n− ). (28) 3,G ∈ As \Bs ≤ N 3,G − Combining (26), (27), and (28), we conclude that 10 P [ = s ] = P[E(H ) (G) ]+ O(n− ) S |LG 3 \E ∈Bs|LG Lemma 17 10 10 = P [E(H ) ]+ O(n− )=P[ = s]+ O(n− ), 3,G ∈Bs SG thereby completing the proof. ⊓⊔ Thus, the remaining task is to prove Lemma 17. To this end, let (E) denote the event that (V H1 E \ G) E(H1)= E. Moreover, let 2(E) signify the event that (V G) E(H2) E(H1)= E (i.e., E is the∩ set of edges added during R2).H Further, let (E) be the eventE that\ ∩(G, V G\) E(H )= E (i.e., E H3 E \ ∩ 3 consists of all edges added by R3). In addition, define events 1,G(E), 2,G(E), 3,G(E) analogously, with H , H , H replaced by H , H , H . Finally, let Hdenote theH event thatHG is a component of 1 2 3 1,G 2,G 3,G CG H1. In order to prove Lemma 17, we establish the following. Lemma 19. Let E (V G), E (V G) E , and E (G, V G). Moreover, suppose that 1 ⊂E \ 2 ⊂E \ \ 1 3 ⊂E \ 2 3 10 3 (H(E )) ln n. Then P (E ) =(1+ O(n− ))P (E ) . N 1 ≤ i=1 Hi i |LG i=1 Hi,G i h i h i Proof. Clearly, V V 3 P [ (E ) (E ) (E )]P[ (E ) ] P (E ) = H2 2 ∧ H3 3 |LG ∧ H1 1 H1 1 ∧ LG . (29) Hi i |LG P [ ] "i=1 # G ^ L Furthermore, since R2 and R3 add edges independently of the 1st round with probability p2, and because the same happens during R2’ and R3’, we have P [ (E ) (E ) (E )] = P [ (E ) (E ) (E )] . (30) H2 2 ∧ H3 3 |LG ∧ H1 1 H2,G 2 ∧ H3,G 3 |H1,G 1 2 Moreover, given that 1(E1) occurs, H1 G has no component on more than ln n vertices. Hence, G is the largest componentH of H iff G is a− component; that is, given that (E ) occurs, the events 1 H1 1 G and G are equivalent. Therefore, P [ G 1(E1)] = P [ G 1(E1)]. Further, whether or not G Lis a componentC of H is independent ofL the∧ edges H containedC in V∧ H G, and thus P [ (E )] = 1 \ CG ∧ H1 1 13 P [ G] P [ 1(E1)]. Hence, as each edge in E1 is present in H1 as well as in H1,G with probability p1 independently,C H we obtain E1 (V G) E1 P [ (E )] = P [ ] p| |(1 p )E \ −| | = P [ ] P [ (E )] . (31) LG ∧ H1 1 CG 1 − 1 CG H1,G 1 Combining (29), (30), and (31), we obtain 3 P [ ] 3 P (E ) = CG P (E ) . (32) Hi i |LG P [ ] · Hi,G i "i=1 # G "i=1 # ^ L ^ 10 Since by Theorem 4 with probability 1 n− the random hypergraph H1 = Hd(n,p1) has precisely ≥P[ G−] 10 one component of order Ω(n), we get C =1+ O(n ). Hence, (32) implies the assertion. P[ G] − L ⊓⊔ Proof of Lemma 17. For any set E (V ) (G) let (E) denote the set of all decompositions ⊂ E \ E F (E1, E2, E3) of E into three disjoint sets such that E1, E2 (V G) and E3 (G, V G). If (H(e)) ln2 n, then Lemma 19 implies that ⊂ E \ ⊂ E \ N ≤ 3 P [E(H ) (G)= E ]= P (E ) 3 \E |LG Hi i |LG (E1,E2,E3) (E) "i=1 # X∈F ^ 3 10 10 = (1+ O(n− )) P (E ) =(1+ O(n− ))P [E(H )= E] , Hi,G i 3,G (E1,E2,E3) (E) "i=1 # X∈F ^ as claimed. ⊓⊔ 4.4 ProofofLemma 10 Recall that SG signifies the set of all vertices v V G that are reachable from G in H3,G, so that = S . Letting denote the component of H∈ that\ contains v V , we have SG | G| Cv 2,G ∈ n n1 − E( )= P [v S ]= P [v S = k] P [ = k] (33) SG ∈ G ∈ G||Cv| |Cv| v V G v V G k=1 ∈X\ ∈X\ X n n1 n µ1 1 Since H2,G is just a random hypergraph Hd(n n1,p), and because d− 1 p d− 1 p < (d 1)− 2 − − 10 ∼ − − by (8), Theorem 4 entails that (H ) ln n with probability 1 n− . Therefore, (33) yields N 2,G ≤ ≥ − E( )= o(1) + P [v S = k] P [ = k] . (34) SG ∈ G||Cv| |Cv| v V G 1 k ln2 n ∈X\ ≤ X≤ n 1 n µ1 To estimate P [v SG v = k], let z = z(n1) = (n1 µ1)/σ1, ξ0 = exp p2 d−1 d− 1 , and ∈ ||C | − − − − − n µ1 n n1 n µ1 h h n µ1 ii ξ(z)= ξ0 1+ zσ1p2 d− 2 . Additionally, let ζ(z)= d− 1 p d− 1 p zσ 1 d−2 p . − − ∼ − − − h i 2 k 1 Lemma 20. For all 1 k ln n we have P [v S = k]=1 ξ(z) + O(n− polylog n). ≤ ≤ ∈ G | |Cv| − · Proof. Suppose that = k but v S . This is the case iff in H there occurs no edge that is incident |Cv| 6∈ G 3,G to both G and v. Letting (G, (v)) denote the set of all possible edges connecting G and v, we shall prove below thatC E C C n n µ1 zσ1 n µ1 d 2 (G, ) = k − + − + O(n − polylog n) |E Cv | d 1 − d 1 d 1 d 2 · − − − − d 1 = O(n − polylog(35)n). · 14 By construction every edge in (G, ) occurs in H with probability p independently. Therefore, E Cv 3,G 2 (G, v) 1 P [v S = k] = (1 p )|E C | = (1+ O(n− polylog n)) exp [ p (G, ) ] 6∈ G||Cv| − 2 · − 2|E Cv | (35) 1 k = (1+ O(n− polylog n))ξ(z) , · hence the assertion follows. Thus, the remaining task is to prove (35). As a first step, we show that n n k n n n n k (G, ) = − − 1 + − 1 − . (36) |E Cv | d − d − d d n n k n n1 For there are d possible edges in total, among which −d contain no vertex of v, −d contain n n1 k C no vertex of G, and − − contain neither a vertex of nor of G; thus, (36) follows from the in- d v C n n k 1 clusion/exclusion formula. Furthermore, as k = O(polylog n), we have − = (1+ O(n d d − n n n1 n n1 k 1 n n−1 · polylog n))k d 1 and −d − d − = (1+ O(n− polylog n))k d− 1 . Thus (36) yields − − · − 1 n n n1 (G, (v)) = (1+ O(n− polylog n))k − . (37) |E C | · d 1 − d 1 − − n n1 n µ1 n n1 d 2 As n1 = µ1 + zσ1, we have d− 1 = d− 1 zσ1 d− 2 + O(n − polylog n), so that (35) follows from (37). − − − − · ⊓⊔ k Let q(ζ, ξ) = k∞=1 qk(ζ)ξ be the function from Proposition 5. Combining (34) with Proposition 5 and Lemma 20, we conclude that P E( )= o(n1/2)+ q((n n )p, ξ(z))(n n )= o(n1/2)+ q(ζ(z), ξ(z))(n n ). (38) SG − 1 − 1 − 1 ∂q ∂q Since q is differentiable (cf. Proposition 5), we let ∆ζ = ∂ζ (ζ(0), ξ(0)) and ∆ξ = ∂ξ (ζ(0), ξ(0)). As 1/2 ζ(z) ζ(0), ξ(z) ξ(0) = O(n− ), we get − − 1/2 q(ζ(z), ξ(z)) q(ζ(0), ξ(0)) = (ζ(z) ζ(0))∆ + (ξ(z) ξ(0))∆ + o(n− ) − − ζ − ξ n µ1 1/2 = zσ − [ξ ∆ p ∆ p]+ o(n− ). (39) 1 d 2 0 ξ 2 − ζ − n µ1 Finally, let µ = (n µ1)q(ζ(0), ξ(0)) and λ = q(ζ(0), ξ(0)) (d 1) [εξ0∆ξ ∆ζ ] d− 1 p. Then S − S − − − − combining (38) and (39), we see that E( G)= µ + zσ1λ + o(√n), as desired. S S S 4.5 ProofofLemma 11 Remember that SG denotes the set of all “attached” vertices, and Nv,G the order of the component of v V G in the graph H . ∈ \ 2,G The following lemma provides an asymptotic formula for Var( ). SG Lemma 21. Let r = P [N = i v S ] and r¯ = P [N = i v S ] for any vertex v G,i v,G ∧ ∈ G G,i v,G ∧ 6∈ G ∈ V G. Moreover, set r = L r , R = L ir , R¯ = L ir¯ for L = ln2 n . In \ G i=1 G,i G i=1 G,i G i=1 G,i addition, let αG =1 G /n and − | | P P P 2 d 2 RG d 2 ¯2 1 αG− ¯ ΓG = (1 RG)(RG rG) + ((d 1)c 1) + RG + (d 1)(1 αG− )εcRG + − d 1 RG. (40) − − − − rG − − 1 α − − G Then Var( ) α2 Γ n + α r (1 r )n. SG ∼ G G G G − G 15 Before we get down to the proof of Lemma 21, we observe that it implies Lemma 11. Proof of Lemma 11. By Theorem 4 part 2 together with Lemma 9 we know that with probability at least 8 2 k 1 n− there are no components of order > ln n inside of V G. Let q(ζ, ξ) = k∞=1 qk(ζ)ξ be the function− from Proposition 5, and let ξ(z) be as in Lemma 20. Then\ Proposition 5 and Lemma 20 entail that for all v V G P ∈ \ n G n G r = q − | | p ξ(( G µ )/σ ), r¯ q − | | p (1 ξ(( G µ )/σ )). G,i i d 1 | |− 1 1 G,i ∼ i d 1 − | |− 1 1 − − n G i By (9) there exists a number 0 <γ< 1 such that qi d−|1 | p γ . Since 0 ξ(( G µ1)/σ1) 1, − ≤ ≤ | |− ≤ i ¯ this yields rG,i, r¯G,i γ . Hence, RG, RG = O(1), so that Lemma 21 implies Var( G)= O(n). ≤ 0.9 n G n G′ S Finally, if G′ V satisfies G′ G n , then d−|1 | p −|d 1 | p = O( G G′ )/n, because ⊂ || |−| || ≤ | − − − | | |−| | 1 d n G n G′ p = O(n − ). Therefore, qi d−|1 | p qi −|d 1 | p = O ( G G′ )/n, because the function | − − − | | | − | | n µ1 ζ qi(ζ) is differentiable. Similarly, as ξ(z)=ξ 0(1 + zσ 1p2 d− 2 ) for some fixed ξ0 = Θ(1), we have 7→ − ξ(( G µ )/σ ) ξ(( G′ µ )/σ ) = O( G G′ )/n. Consequently, r r ′ = O( G G′ )/n | | |− 1 1 − | |− 1 1 | | |−| | | G,i − G ,i| | |−| | and r¯ r¯ ′ = O( G G′ )/n, and thus | G,i − G ,i| | | − | | 0.1 r r ′ , R R ′ , R¯ R¯ ′ = O( G G′ )/n = O(n− ). | G − G | | G − G | | G − G | | | − | | Hence, Lemma 21 implies that Var( ) Var( ′ ) = o(n). | SG − SG | ⊓⊔ The remaining task is to establish Lemma 21. Keeping G fixed, in the sequel we constantly omit the subscript G in order to ease up the notation; thus, we write α instead of αG etc. As a first step, we compute P(v, w S) r2. Setting ∈ − L S = [P [N = j w S w C ,N = i, v S] P [N = j w S]] 1 w ∧ ∈ | 6∈ v v ∈ − w ∧ ∈ i,j=1 X P [w C N = i, v S] P [N = i v S] , × 6∈ v| v ∈ v ∧ ∈ L S = (1 r) P [w C N = i, v S] P [N = i v S] , 2 − ∈ v| v ∈ v ∧ ∈ i=1 X we have P(v, w S) r2 = S + S . ∈ − 1 2 To compute S2, observe that whether w Cv depends only on Nv, but not on the event v S. ∈ 1 ∈ n 2 n 1 − i 1 Therefore, P [w Cv Nv = i, v S]=P[w Cv Nv = i]= i −2 i −1 = n− 1 , because given that ∈ n0 | 1 ∈ ∈ | − − n0 −2 N = i, there are − ways to choose the set C V G, while there are − ways to choose C v i 1 v i 2 v in such a way that w − C . As a consequence, ⊂ \ − ∈ v L 1 r 1 r S − (i 1)P [N = i v S]= − (R r). 2 ∼ n 1 − v ∧ ∈ n 1 − i=1 − X − With respect to S1, we let P (i, j)=P[N = j w C ,N = i] , 1 w | 6∈ v v P (i, j)=P[w S N = j, w C ,N = i, v S] , 2 ∈ | w 6∈ v v ∈ so that S = [P (i, j)P (i, j) P [N = j w S]] P [w C N = i, v S] P [N = i v S] 1 1 2 − w ∧ ∈ 6∈ v| v ∈ v ∧ ∈ i,j X [P (i, j)P (i, j) P [N = j] P [w S N = j]] P [N = i v S] . ∼ 1 2 − w ∈ | w v ∧ ∈ i,j X 16 1 ((d 1)c 1)ij+i 2 Lemma 22. We have − − − . P1(i, j)P [Nw = j] =1+ n n1 + O(n− polylog n) − · Proof. This argument is similar to the one used in the proof of Lemma 41 in [7]. Remember that if we restrict our view on H to the set V G the hypergraphis similar to a H (n n ,p). In order to estimate 3,G \ d − 1 S1, we observe that P [N = j in H (n n ,p) N = i, w C ]=P[N = j in H (n n ,p) C ] . (41) w d − 1 | v 6∈ v w d − 1 \ v Given that Nv = i, Hd(n,p) Cv is distributed as a random hypergraph Hd(n n1 i,p). Hence, the probability that N = j in H \(n,p) C equals the probability that a given vertex− of −H (n n i,p) w d \ v d − 1 − belongs to a component of order j. Therefore, we can compare P [Nw = j in Hd(n n1,p) Cv] and n n1 i 1 − \ P [Nw = j in Hd(n n1,p)] as follows: in Hd(n n1 i,p) there are − j −1 − ways to choose the set − n n1 i n n1 −i j − j − C j . Moreover, there are − − − − − possible edges connecting the chosen set C w \{ } d − d − d w with V Cw, and as Cw is a component, none of these edges is present. Since each such edge is present with probability\ p independently, the probability that there is no C -V C edge equals w \ w n−n1−i (n−n1−i−j) (j) (1 p)( d )− d − d . − n n1 1 By comparison, in Hd(n n1,p) there are −j 1− ways to choose the vertex set of Cw. Further, there − − are n n1 n n1 j j possible edges connecting and , each of which is present with −d − d − d Cw V Cw − − n n1 i n n1 i j \ n n1 n n1 j probability p independently. Thus, letting γ = − − − − − − − − , we obtain d − d − d − d 1 P [N = j in H (n n ,p) C ] n n i 1 n n 1 − w d − 1 \ v = − 1 − − − 1 − (1 p)γ. (42) P [N = j in H (n n ,p)] j 1 j 1 − w d − 1 − − Concerning the quotient of the binomial coefficients, we have 1 n n1 i 1 n n1 1 − i(j 1) 2 − − − − − = exp − + O(n− polylog n) . (43) j 1 j 1 − n n · − − − 1 n n1 (n n1 i)d+(n n1 j)d (n n1 i j)d Moreover, − − − − − − − − − . Expanding the falling factorials, we get γ = d (n n1) 1 − d − h i d 2 2 2 n n1 2 (i + j (i + j) ) 3 n n1 d 3 γ = − − + O(n− polylog n) = − ij + O(n − polylog n). d (n n )2 · − d 2 · " − 1 # − (44) Plugging (43) and (44) into (42), we obtain P [N = j in H (n n ,p) C ] i(j 1) n−n1 d−3 w d 1 v 2 ( 2 )ij+O(n polylog n) − \ = exp − + O(n− polylog n) (1 p)− d− · P [N = j in H (n n ,p)] − n n · − w d − 1 − 1 i(j 1) n n1 2 = exp − + − ijp + O(n− polylog n) − n n d 2 · 1 − 1 − 2 =1+(n n )− [((d 1)c 1)ij + i]+ O(n− polylog n). − 1 − − · Therefore, by (41) P [Nw = j Nv = i, w Cv] P [Nw = j in Hd(n n1,p)] | 6∈ − − 1 2 =P[N = j in H (n n ,p)] n− [((d 1)c 1)ij + i]+ O(n− polylog n)(45). w d − 1 − − · ⊓⊔ 2 1 αd− d 2 Lemma 23. Setting γ = − d−1 and γ = (d 1)(1 α − )εc, we have P (i, j) 1 P[v S Nv=i](1 α ) 2 2 1 ∈ | − − 2 − − P [w S N = j]= n− P [w S N = j] (jγ ijγ )+ O(n− polylog n). ∈ | w 6∈ | w 1 − 2 · 17 Proof. Let be the event that N = j, w C , N = i, and v S. Moreover, let be the event F w 6∈ v v ∈ Q that in H3 there exists an edge incident to the three sets Cv, Cw, and G simultaneously, so that P2(i, j)= P [ ]+P[w S , ] P [ ] . Q|F ∈ |¬Q F ¬Q|F To bound P [w S , ] P [w S N = j], we condition on the event that C and C are fixed ∈ |¬Q F − ∈ | w v w disjoint sets of sizes i and j. Let Q′ signify the probability that Cw is reachable from G in H3,G, and let Q denote the probability that C is reachable from G in H , and that the event occurs. Then Q′ w 3,G ¬Q corresponds to P [w S Nw = j] and Q to P [w S , ], so that our aim is to estimate Q Q′. As there are (G, C ) ∈ |(G, C , C ) possible edges∈ |¬Q that joinF C and G but avoid C , each of− which is |E v |−|E v w | v w present in H3,G with probability p2 independently, we have (G,Cv) (G,Cv,Cw) (G,Cw) Q =1 (1 p )|E |−|E |, while Q′ =1 (1 p )|E |. − − 2 − − 2 Therefore, (G,Cw) (G,Cv,Cw) Q Q′ = (1 p )|E | 1 (1 p )−|E | − − 2 − − 2 h i n n0 (1 Q′) (1 exp[p (G, C , C ) ]) ij(Q′ 1) p . ∼ − − 2|E v w | ∼ − d 2 − d 2 2 − − n 1 As d−1 p2 εc, we thus get − ∼ d 2 1 P [w , ] P [w S N = j] ij(P [w S N = j] 1)(d 1)(1 α − )εcn− . (46) ∈ S|¬Q F − ∈ | w ∼ ∈ | w − − − With respect to P [ ], we let signify the number of edges joining C and G. Given that occurs, Q|F K v F n n0 d 1 is asymptotically Poisson with mean λi = i d 1 d 1 p2 i(1 α − )εc. Moreover, given that K − − − ∼ − k (G,Cv,Cw) = k, the probability that one of these k edgesh hits C is (ki) E , and thus w (Cv,G) K P ∼ E 1 d 2 n n n n − 1 α − (k) jk 0 0 jk(d 1) − . P ∼ d 2 − d 2 d 1 − d 1 ∼ − 1 αd 1 − − − − − − Consequently, exp( λ ) jkλk j(1 αd 2) i i − (47) P [ ] − (k) − d 1 . Q|F ∼ 1 exp( λi) k! P ∼ n(1 exp( λi))(1 α ) k 1 − − − X≥ − − − Combining (46) and (47), we obtain the assertion. ⊓⊔ Thus, L nS P [v S N = i] 1 ∼ ∈ ∧ v i=1 X L [((d 1)c 1)ij + i] P [w S N = j]+P[w S N = j] [γ j + γ ij] × − − ∈ ∧ w 6∈ ∧ w 1 2 j=1 X 2 N d 2 R 1 α − = ((d 1)c 1) + R + γ R¯2 + − P [N = i] R¯ − − r 2 1 αd 1 v i=1 − − 2 X d 2 R d 2 2 1 α − = ((d 1)c 1) + R + (d 1)(1 α − )εcR¯ + − R.¯ − − r − − 1 αd 1 − − Hence, letting Γ be as defined by (40)we have P [v, w S] P [v S] P [w S]) Γ/n. Consequently, Var(S) αΓ n + α2r(1 r)n. ∈ − ∈ ∈ ∼ ∼ − 18 4.6 ProofofLemma 15 n1−1 n ( 1 ) ( 1) The probability that a vertex v V G is isolated in H3,G is at least (1 p) d− (1 p2) d− n1 1 n ∈ \ − − ∼ exp( p d −1 εp d 1 ) exp( c). Therefore, − − − − ≥ − E( ) (1 o(1)) exp( c)(n n ). (48) |W| ≥ − − − 1 To show that is concentrated about its mean, we employ the following version of Azuma’s inequality (cf. [11, p. 38]).|W| K Lemma 24. Let Ω = i=1 Ωi be a product of probability spaces. Moreover, let X : Ω R be a random variable that satisfies the following Lipschitz condition. → Q If two tuples ω = (ωi)1 i K ,ω′ = (ωi′)1 i K Ω differ only in their j’th components for ≤ ≤ ≤ ≤ ∈ (49) some 1 j K, then X(ω) X(ω′) 1. ≤ ≤ | − |≤ 2 Then P [ X E(X) t] 2 exp( t ), provided that E(X) exists. | − |≥ ≤ − 2K Using Lemma 24, we shall establish the following. Corollary 25. Let Y be a random variable that maps the set of all d-uniform hypergraphs with vertex set V to [0,n]. Assume that Y satisfies the following condition. Let H be a hypergraph,and let e (V ). Then Y (H) Y (H +e) , Y (H) Y (H e) 1. (50) ∈E | − | | − − |≤ Then P Y (H ) E(Y (H )) n0.66 exp( n0.01). | 3,G − 3,G |≥ ≤ − K Proof. In order to apply Lemma 24, we need to decompose H3,G into a product i=1 Ωi of probability spaces. To this end, consider an arbitrary decomposition of the set (V ) of all possible edges into sets 0.1 E Q 1 K so that K n and E(E(H3,G) j ) n for all 1 j K; such a decomposition exists, E ∪···∪E ≤ ∩E ≤ n ≤ ≤ because the expected number of edges of H3,G is d p = O(n). Now, let Ωe be a Bernoulli experiment with success probability p for each e (V G),≤ resp. with success probability p for e (G, V G). ∈ E \ 2 ∈ E \ Then setting Ω = Ω , we obtain a product decomposition H = K Ω . i e i e 3,G i=1 i ∈E In addition, construct for each hypergraph H with vertex set V another hypergraph H∗ by removing Q 0.1 Q from H all edges e i such that E(H) i 4n (1 i K). Since E(H3,G) i is the sum of ∈E | ∩E |≥ ≤ ≤ | ∩E | 0.1 two binomially distributed variables, the Chernoff bound (7) implies that P [ E(H3,G) i] 4n ) exp( n0.05). As K n, this entails | ∩E |≥ ≤ − ≤ 0.05 0.04 P H = H∗ K exp( n ) exp( n ),sothat (51) 3,G 6 3,G ≤ − ≤ − E(Y (H )) E(Y (H∗ )) 1 [because 0 Y n]. (52) | 3,G − 3,G |≤ ≤ ≤ 1 0.1 As a next step, we claim that Y ∗(H) = 4 n− Y (H∗) satisfies the Lipschitz condition (49). For by construction modifying (i.e., adding or removing) an arbitrary number of edges belonging to a single 0.1 factor i can affect at most 4n edges of H∗. Hence, (50) implies that Y ∗(H) satisfies (49). Therefore, LemmaE 24 entails that 0.63 0.52 0.02 P Y (H3∗,G) E(Y (H3∗,G)) n P Y ∗(H3,G) E(Y ∗(H3,G)) n exp( n ). | − |≥ ≤ | − |≥ ≤ − (53) Finally, combining (51), (52), and (53), we conclude that 0.64 0.63 P Y (H ) E(Y (H )) n P Y ∗(H) E(Y ∗(H)) n +P H = H∗ | 3,G − 3,G |≥ ≤ | − |≥ 3,G 6 3,G exp( n0.01), ≤ − thereby completing the proof. ⊓⊔ Finally, since /d satisfies (50), Lemma 15 follows from Corollary 25 and (48). |W| 19 5 Normality via Stein’s Method In this section we will use Stein’s Method to prove that (Hd(n,p)) as well as G tend (after suitable normalization) in distribution to the normal distribution.ThisN proofsProposition6 asS well as Theorem1 and Lemma 14. First we will define a general setting for using Stein’s Method with random hypergraphs which defines some conditions the random variables have to fulfill. Then we show in two lemmas (Lemma 28 and Lemma 29) that the random variables corresponding to (Hd(n,p)) and G do indeed comply to the conditions and last but not least a quite technical part will showN how to derive theS limiting distribution from the conditions. 5.1 Stein’s method for random hypergraphs Let bethe set ofall subsetsofsize d of V = 1,...,n ,andlet bethe powerset of . Moreover, let 0 E { } H E ≤ pe 1 for each e , and define a probability distribution on by letting P [H]= e H pe e H 1 ≤ ∈E H ∈ · ∈E\ − pe. That is H can be considered a random hypergraph with ”individual” edge probabilities. ∈ H Q Q Furthermore, let be a family of subsets of V , and let (Yα)α be a family of random variables. A ∈A Remember that for Q V we set (Q)= e : e Q = . We say that Yα is feasible if the following holds. ⊂ E { ∈E ∩ 6 ∅} For any two elements H, H′ such that H (α)= H′ (α) we have Y (H)= Y (H′). ∈ H ∩E ∩E α α That means Yα is feasible if its value depends only on edges having at least one endpoint in α. In addition, S S set Yα (H)= Yα(H (S)) (H , α , S V , S α = ). Thus Yα (H) is the value of Yα after removing all edges incident\E with S∈. HWe define∈ A ⊂ ∩ ∅ Y = Y , µ = E [Y ], σ2 = Var [Y ], X = (Y µ )/σ (54) α α α α α − α α X∈A 1 Yβ if α β = , Zα = Zαβ, where Zαβ = σ− α ∩ 6 ∅ (55) × Yβ Yβ if α β = , β − ∩ ∅ X∈A α α α β V = Y /σ + (Y Y ∪ )/σ, and (56) αβ γ γ − γ γ:β∩γ6=∅ γ:β∩γ=∅ ∧αX∩γ=∅ ∧αX∩γ=∅ δ = E X Z2 + (E [ X Z V ]+E[ X Z ]E [ Z + V ]) . (57) | α| α | α αβ αβ | | α αβ| | α αβ | α α,β X∈A X∈A The following theorem was proven for graphs (i.e. d =2) in [3]. The argument used there carries over to the case of hypergraphs without essential modifications. Thus for the sake of brevity we omit a detailed proof of this result. Y E[Y ] Theorem 26. Suppose that all Yα are feasible. If δ = o(1) as n , then −σ converges to the standard normal distribution. → ∞ Now the following lemma states that a number of conditions on the expectations of the product of up to S three random variables Yα will suffice for δ = o(1). The conditions are identical for both statements we want to prove and we will prove that they are fulfilled in both cases in the next two sections while the proof of the lemma itself is deferred to the end of the section. Lemma 27. Let k = O(polylog n) and let (Yα)α be a feasible family such that 0 Yα k for all α . If the following six conditions are satisfied,∈A then δ = o(1) as n . ≤ ≤ ∈ A → ∞ Y1. We have E(Y ), Var(Y )= Θ(n), and β :β α= µβ = O(E(Y )/n polylog n)= O(polylog n).for any α ∈A ∩ 6 ∅ · Y2. Let α,β,γ∈ A be distinct elements of .P Then A Y (Y Y α)Y α =0 if α β = , (58) α β − β β ∩ ∅ Y Y =0 if α β = , (59) α β ∩ 6 ∅ (Y Y α)Y α = (Y Y α)Y =0 if α β = α γ = = β γ. (60) β − β γ β − β γ ∩ ∩ ∅ 6 ∩ 20 Y3. For all α, β we have E(Y Y α) k2µ . γ:γ β= ,γ α= β γ ≤ β Y4. If α, β are disjoint, then∩ 6 ∅ ∩ ∅ ∈ A P E [YαYβ]= O(µαµβ polylog n), (61) µ · E Y Y α = O( β polylog n), (62) | β − β | n · µ µ E Y Y Y α = O( α β polylog n). (63) α| β − β | n · Y5. If α,β,γ are pairwise disjoint, then ∈ A α α β µβ µγ E Y Y Y ∪ = O( polylog n), (64) β| γ − γ | n · α α α β µβ µγ E Y Y Y Y ∪ = O( polylog n), (65) | β − β | · | γ − γ | n2 · α α α β µαµβµγ E Y Y Y Y Y ∪ = O( polylog n), (66) α| β − β | · | γ − γ | n2 · µ µ µ E Y Y Y α Y Y α = O( α β γ polylog n), (67) α| β − β | · | γ − γ | n2 · µ µ E (Y Y α)(Y Y α) = O( α β polylog n). (68) | β − β γ − γ | n2 · Y6. If α,β,γ satisfy α β = α γ = , then ∈ A ∩ ∩ ∅ β β γ µγ E Y Y ∪ = O( polylog n). (69) | α − α | n · 5.2 Conditions for the normality of N (Hd(n, p)) In this section we will prove the properties Y1–Y6 defined in Lemma 27 for the case of the normality of (H (n,p)). N d Let k = O(polylog n) and let = α V : 1 α k . Moreover, for A V with A α = let IA =1 if α is a component of HA ({A),⊂ and 0 otherwise.≤ | | ≤ Further,} set Y A = α ⊆IA. We briefly∩ write∅ α \E α | | · α Iα = Iα∅ and Yα = Yα∅. Then (Yα)α is a feasible family, because whether α is a component or not only depends on the presence of edges that∈A contain at least one vertex of α. Let (S) denote the even that the subhypergraph of H induced on S V is connected. If Iα =1, then (α) occurs.C Moreover, H contains no edges joining α and V α, i.e., H⊂ (α, V α) = . Since each Cedge occurs in H with probability p independently, we thus obtain\ ∩E \ ∅ (α,V α) P [I =1]=P[ (α)](1 p)|E \ |. (70) α C − Furthermore, observe that α , A B V α : IA =1 IB =1. (71) ∀ ∈ A ⊂ ⊂ \ α → α Proof of Y1: We know from Theorem 1 that E [Y ]= Θ(n) and Var [Y ]= Θ(n). To see that µ = O(E [Y ]/n polylog n), β · β :β α= ∈AX∩ 6 ∅ note that µβ := E [Yβ] depends only on the size of β. Thus with µb = µβ for an arbitrary set β of k k n size b we have E [Y ] = β µβ = b=1 β∈A µβ = b=1 b µb while β :β α= µβ = ∈A |β|=b ∈A ∩ 6 ∅ k k n = b=1 β∩α6 ∅ µβ b=1Pk b 1 µb. P P P P |β|=b ≤ − P ProofP of Y2: (58):P Suppose that Iα = 1. Then H features no edge that contains a vertex in α and a α α vertex in β. If in addition Iβ =1, then we obtain that Iβ =1 as well. Hence, Yβ = Yβ . (59): This just means that any two components of H are either disjoint or equal. 21 (60): To show that Y (Y Y α)=0, assume that I = 1. Then γ is a component of H, so that β γ β − β γ cannot be a component, because γ = β but γ β = ; hence, Iβ =0. Furthermore, if γ is a component of 6 ∩ 6 ∅α α α H, then γ is also a component of H (α), so that Iγ =1. Consequently, Iβ =0. Thus, Yβ = Yβ =0. In order to prove that Y α(Y Y\Eα)=0, suppose that Iα =1. Then Iα =0, because the intersecting γ β − β γ β sets β,γ cannot both be componentsof H (α). Therefore,we also have Iβ =0;forif β were a component of H, then β would also be a component\E of H (α). Hence, also in this case we obtain Y = Y α =0. \E β β Proof of Y3: Suppose that Iβ = 1, i.e., β is a component of H. Then removing the edges (α) from H may cause β to split into several components B ,...,B . Thus, if Y β > 0 for some γ Esuch that 1 l γ ∈ A γ β = , then γ is one of the components B1,...,Bl. Since l β k, this implies that given Iβ =1 ∩ 6 ∅ α 2 ≤ | |≤ we have the bound γ:γ β= ,γ α= Yγ k . Hence, we obtain Y3. The following lemma∩ which6 ∅ ∩ gives∅ a description≤ of the limited dependence between the random vari- P ables Iα and Iβ for disjoint α and β together with the fact that P [Iα =1]= O(µα) implies Y4–Y6. Lemma 28. Let 0 l, r 2, and let α ,...,α ,β ,...,β be pairwise disjoint. Moreover, let ≤ ≤ 1 l 1 r ∈ A A1,...,Ar,B1,...,Br V be sets such that Ai Bi V βi and Bi 2k for all 1 i r, and assume that r B A⊂= . Then ⊂ ⊂ \ | | ≤ ≤ ≤ i=1 i \ i ∅ lT r l r Aj Bj r P , Iα =1 I = I O(n− polylog n) P [Iα = 1] P Iβ =1 . i ∧ βj 6 βj ≤ · i j i=1 j=1 j=1 j=1 ^ ^ Y Y Proof. Since (71) entails that IAj = IBj IBj =1 IAj =0, we have βj 6 βj ↔ βj ∧ βj Aj Bj Aj Bj P i, j : Iα =1 I = I =P i, j : Iα =1 I =0 I =1 . (72) ∀ i ∧ βj 6 βj ∀ i ∧ βj ∧ βj h i h i We shall bound the probability on the right hand side in terms of mutually independent events. If I = 1 and IBj = 1 for all i, j, then the hypergraphs induced on α and β are connected, i.e., αi βj i j the events (αi) and (βj) occur. Note that these events are mutually independent, because (αi) (resp. (β )) onlyCdepends onC the presence of edges e (α ) (V α ) (resp. e (β ) (V C β )). C j ∈E i \E \ i ⊂E j \E \ j Furthermore, if αi is a component,then in H there occur no edges joining αi and V αi; in other words, H (α , V α )= . However, these events are not necessarily independent, because\ (α , V α ) may ∩E i \ i ∅ E 1 \ 1 contain edges that are incident with vertices in α2. Therefore, we consider the sets r (α )= α ′ β B , (α )= (α , V α ) ( (α )), F i i ∪ j ∪ j D i E i \ i \E F i i′=i j=1 [6 [ l r (β )= α β ′ B ′ , (β )= (β , V β ) ( (β )). F j i ∪ j ∪ j D j E j \ j \E F j i=1 j′=j j′ =1 [ [6 [ Bj Then Iα = 1 (resp. I = 1) implies that (αi) H = (resp. (βj ) H = ). Moreover, since i βj D ∩ ∅ D ∩ ∅ the sets (αi) and (βj ) are pairwise disjoint, the events (αi) H = , (βi) H = are mutually independent.D D D ∩ ∅ D ∩ ∅ Finally, we need to express the fact that IAj = 0 but IBj = 1. If this event occurs, then H contains βj βj an edge connecting β with B A , i.e., H (β ,B A ) = . Thus, let denote the event that j j \ j ∩E j j \ j 6 ∅ Q H (βj ,Bj Aj ) = for all 1 j r. ∩EThus, we obtain\ 6 ∅ ≤ ≤ Aj Bj P i, j : Iα =1 I =0 I =1 ∀ i ∧ βj ∧ βj h l ri P ( (α ) ( (α ) H = )) ( (β ) ( (β ) H = )) ≤ C i ∧ D i ∩ ∅ ∧ C j ∧ D j ∩ ∅ ∧Q i=1 j=1 ^ ^ l r = P [ (α )] P [ (α ) H = ] P [ (β )] P [ (β ) H = ] P [ ] . (73) C i D i ∩ ∅ × C j D j ∩ ∅ × Q i=1 j=1 Y Y 22 We shall prove below that (αi,V αi) (βj ,V βj ) P [ (αi) H = ] (1 p)|E \ |, P [ (βj ) H = ] (1 p)|E \ |, (74) D ∩ ∅ ∼ − r D ∩ ∅ ∼ − P [ ]= O(n− polylog n). (75) Q · Combining (70) and (72)–(75), we then obtain the assertion. To establish (74), note that by definition (α ) (α , V α ). Therefore, D i ⊂E i \ i (αi) (αi,V αi) P [ (α ) H = ]=(1 p)|D | (1 p)|E \ |. (76) D i ∩ ∅ − ≥ − On the other hand, we have αi , (αi) = O(polylog n), and thus (αi, (αi)) αi (αi) n d 2 | | |F | 1 d |E F | ≤ | |·|F | · d 2 = O(n − polylog n). Hence, as p = O(n − ), we obtain − · (αi) (αi,V αi) (αi, (αi)) P [ (α ) H = ]=(1 p)|D | (1 p)|E \ |−|E F | D i ∩ ∅ − ≤ − (αi,V αi) d 2 (αi,V αi) (1 p)|E \ | exp(p O(n − polylog n)) (1 p)|E \ |. (77) ∼ − · · ∼ − (αi,V αi) Combining (76) and (77), we conclude that P [ (α ) H = ] (1 p)|E \ |. As the same argu- D i ∩ ∅ ∼ − ment applies to P [ (βj ) H = ], we thus obtain (74). Finally, we proveD (75).∩ If r =1∅ , then H contains an edge of (β ,B A ). Since E 1 1 \ 1 d 2 d 2 (β ,B A ) β B A n − = O(n − polylog n), |E 1 1 \ 1 | ≤ | 1| · | 1 \ 1| · · and because each possible edge occurs with probability p independently, the probability of this event is d 2 1 P [ ] O(n − polylog n)p = O(n− polylog n), as desired. QNow,≤ assume· that r =2. Then H features· edges e (β ,B A ) (j =1, 2). j ∈E j j \ j 1st case: e1 = e2. In this case, e1 contains a vertex of each of the four sets β1, β2, B1 A1, B2 A2. d 4 2 d \4 \ Hence, the number of possible such edges is n − β B A = O(n − polylog n). ≤ j=1 | j | · | j \ j | · Consequently, the probability that such an edge occurs in H is O(nd 4 polylog n)p = O(n 3 Q − − polylog n). ≤ · · d 2 d 2 2nd case: e1 = e2. There are βj Bj Aj n − = O(n − polylog n) ways to choose ej (j =1, 2). 6 ≤ | |·| \ |· · d 2 2 2 Hence, the probability that such edges e1,e2 occur in H is O(n − polylog n)p = O(n− polylog n). ≤ · · Thus, in both cases we obtain the bound claimed in (75). ⊓⊔ 5.3 Conditions for the normality of SG In this section we will prove the properties Y1–Y6 defined in Lemma 27 for the case of the normality of G. S Consider a set G V of size n . Let be the set of all subsets α V G of size α k. Moreover, ⊂ 1 A ⊂ \ | |≤ let pe = p for e V G, pe = p2 for e (G, V G), and pe =0 if e G. ⊂ \ A ∈E \ ⊂ A For A V and A α = set Iα =1 if α is a component of H (A G). Moreover, let Jα =1 if (H (A))⊆ (G, α)∩= . Further,∅ let KA = IAJ A and Y A = α K\EA. Then∪ \E ∩E 6 ∅ α α α α | | α P [Kα =1]= Ω(P [Iα = 1]). (78) Proof of Y1: Using Lemma 10 we have E [Y ]= Θ(n) and using Lemma 11 we have Var [Y ]= Θ(n). The proof of the rest of Y1 is analogous to the proof of Y1 in the case of (H (n,p)). N d Proof of Y2: (58): Suppose that Kα =1. Then Iα =1, so that H (G) has no α-β-edges. Hence, if α \E α also Kβ =1, then β is a component of H (G) as well. Thus, Kβ =1, so that Yβ = Yβ . (59): If K =1, then α is a componentof\EH (G). Since any two components of H (G) are either α \E \E disjoint or equal, we obtain Iβ =0, so that Yβ =0 as well. (60): To show that Y (Y Y α)=0, assume that K = 1. Then I = 1, i.e., γ is a component of γ β − β γ γ H (G). Since β = γ but β γ = , we conclude that Iβ = 0. Furthermore, if γ is a component of H \E(G), then γ is also6 a component∩ 6 of∅ H (G α), whence Iα =0. Consequently, Y = Y α =0. \E \E ∪ β β β 23 In order to prove that Y α(Y Y α)=0, suppose that Kα = 1. Then Kα = 1. Therefore, Iα = 0, γ β − β γ γ β because the intersecting sets β,γ cannot both be components of H (α). Thus, we also have Iβ =0; for if β were a component of H, then β would also be a component of\EH (α). Hence, also in this case we α \E obtain Yβ = Yβ =0. Proof of Y3: Suppose that Kβ =1. Then Iβ =1, i.e., β is a component of H (G). Then removing \E β the edges α from H (G) may cause β to split into several components B1,...,Bl. Thus, if Yγ > 0 for some γE such that\E γ β = , then γ is one of the components B ,...,B . Since l β k, this ∈ A ∩ 6 ∅ 1 l ≤ | |≤ implies that given Iβ =1 we have the bound Y α k2. γ ≤ γ:γ β= ,γ α= ∩ 6 X∅ ∩ ∅ Hence, we obtain Y3. Similar to Lemma 28 the following lemma on the limited dependence of Kα and Kβ for disjoint α and β implies Y4–Y6. Lemma 29. Let 0 l, r 2, and let α ,...,α ,β ,...,β be pairwise disjoint. Moreover, let ≤ ≤ 1 l 1 r ∈ A A1,...,Ar,B1,...,Br V be sets such that Ai Bi V βi and Bi O(1) for all 1 i r, and assume that r B A⊂= . Then ⊂ ⊂ \ | |≤ ≤ ≤ i=1 i \ i ∅ l Tr l r Aj Bj r P Kα =1 K = K O(n− polylog n) P [Kα = 1] P Kβ =1 . i ∧ βj 6 βj ≤ · i j i=1 j=1 j=1 j=1 ^ ^ Y Y Aj Bj Aj Bj Aj Bj Aj Proof. Let = P i, j : Kα =1 K = K . If K = K , then either I = I or I = P ∀ i ∧ βj 6 βj βj 6 βj βj 6 βj βj IBj =1 and J Aj =hJ Bj . Therefore, letting = ji: IAj = IBj and ¯ = 1,...,r , we obtain βj βj 6 βj J { βj 6 βj } J { } \ J l Aj Bj Aj Aj Bj P Iα =1 I = I I =1 J = J . (79) P ≤ i ∧ βj 6 βj ∧ βj ∧ βj 6 βj i=1 j j ¯ ^ ^∈J ^∈J Aj Bj Aj Now, the random variables Iα , I , and I are determined just by the edges in (G), while J i βj βj E\E βj and J Bj depend only on the edges in (G). Hence, as the edges in (G) and in (G) occur in H βj independently, (79) yields E E\E E l Aj Aj Bj Aj Bj P Iα =1 I =1 I = I P J = J . (80) P ≤ i ∧ βj ∧ βj 6 βj · βj 6 βj i=1 j ¯ j j ¯ ^ ^∈J ^∈J ^∈J Furthermore, Lemma 28 entails that l l r Aj Aj Bj P Iα =1 I =1 I = I O(n−|J| polylog n) P [Iα = 1] P Iβ =1 . i ∧ βj ∧ βj 6 βj ≤ · · i · j i=1 j ¯ j i=1 j=1 ^ ^∈J ^∈J Y Y (81) In addition, we shall prove below that Aj Bj ¯ P J = J O(n−|J| polylog n). (82) βj 6 βj ≤ · j ¯ ^∈J r l r Plugging (81) and (82) into (80), we get O(n− polylog n) i=1 P [Iαi = 1] j=1 P Iβj =1 , so that the assertion follows from (78). P ≤ · · · Thus, the remaining task is to establish (82). Let us first dealQ with the case ¯ Q= 1. Let j ¯. Aj Bj Aj Bj |J | ∈ J If J = J , then J = 1 and J = 0, because Aj Bj . Thus, βj is connected to G via an βj 6 βj βj βj ⊂ 24 edge that is incident with Aj Bj ; that is, H (βj ,Bj Aj ) = . Since there are (βj ,Bj Aj ) d 2 d 2 \ ∩E \ 6 ∅ |E \ | ≤ β B n − = O(n − polylog n) such edges to choose from, and because each such edge is present | j | · | j | · · 1 d Aj Bj with probability p2 = O(n − ), we conclude that P J = J P [H (βj ,Bj Aj ) = ] βj 6 βj ≤ ∩E \ 6 ∅ ≤ d 2 1 O(n − polylog n)p2 = O(n− polylog n), whenceh we obtain (82).i · · Aj Bj Finally, suppose that ¯ =2. If J = J for j =1, 2, then there occur edges ej H (βj ,Bj |J | βj 6 βj ∈ ∩E \ Aj ) (j =1, 2). 1st case: e1 = e2. In this case e1 = e2 is incident with all four sets βj ,Bj Aj (j = 1, 2). Hence, as the d 4 2 d 4 \ numberof such edgesis n − j=1 βj Bj Aj O(n − polylog n) and each such edge occurs ≤ 1 d | |·| \ |≤ · d 4 with probability p2 = O(n − ), the probability that the 1st case occurs is O(n − polylog n)p2 = 3 Q · O(n− polylog n). · d 2 d 2 2nd case: e1 = e2. There are βj Bj Aj n − O(n − polylog n) ways to choose ej for j =1, 2, 6 ≤ | |·| \ |· ≤ 1 d · each of which is present with probability p2 = O(n − ) independently. Hence, the probability that the d 2 2 2 second case occurs is bounded by O(n − polylog n)p O(n− polylog n). · 2 ≤ · Thus, the bound (82) holds in both cases. ⊓⊔ 5.4 ProofofLemma 27 All we need to show is that the conditions defined in Lemma 27 imply that δ as defined by (57) tends to 0. 3 We will do so by proving that each of the three summands contributing to δ is O(σ− E [Y ] polylog n). Together with condition Y1, stating that E [Y ], σ2 = Θ(n), this implies the statement. We formulate· one lemma for each summand, bounding the expectations using conditions Y1–Y6. The proof of the lemmas are mainly long and technical computations then. 2 3 Lemma 30. α E Xα Zα = O(σ− E [Y ] polylog n) ∈A | | · Proof. Let P 2 2 S = E Y Y , S = E µ Y , 1 α β 2 α β α β:α β= α β:α β= X∈A X∩ 6 ∅ X∈A X∩ 6 ∅ 2 2 S = E Y (Y Y α) , S = E µ (Y Y α) . 3 α β − β 4 α β − β α β:α β= α β:α β= X∈A X∩ ∅ X∈A X∩ ∅ Since X = (Y µ )/σ (Y + µ )/σ, (54) entails that α α − α ≤ α α 2 2 2 3 α E X Z 2σ− E (Y + µ ) Y + (Y Y ) | α| α ≤ α α β β − β β:α β= β:α β= X∩ 6 ∅ X∩ ∅ 3 2σ− (S + S + S + S ). ≤ 1 2 3 4 Therefore, it suffices to show that S = O(E(Y ) polylog n) for j =1, 2, 3, 4. j · Regarding S1, we obtain the bound (59) S = E [Y Y Y ] k2 E [Y ] O(E [Y ] polylog n). 1 α β γ ≤ α ≤ · α β:α β= γ:α γ= α X∈A X∩ 6 ∅ X∩ 6 ∅ X∈A With respect to S2, note that due to (59) and (61) we have E [YβYγ ] kµβ if β = γ, E [YβYγ]=0 if β = γ but β γ = , and E [Y Y ]= O(µ µ polylog n) if β γ = ≤. Consequently, 6 ∩ 6 ∅ β γ β γ · ∩ ∅ S2 = µα E [YβYγ] α β:α β= γ:α γ= X∈A X∩ 6 ∅ X∩ 6 ∅ Y1 µ O(µ µ polylog n) O(E(Y ) polylog n). (83) ≤ α β γ · ≤ · α β:α β= γ:α γ= X∈A X∩ 6 ∅ X∩ 6 ∅ 25 Concerning S3, we obtain S = E Y (Y Y α)(Y Y α) 3 α β − β γ − γ α β:α β= γ:α γ= X∈A X∩ ∅ X∩ ∅ (67), (60) 2 O(µ µ µ n− polylog n) ≤ α β γ · α β:α β= γ:α γ= X∈A X∩ ∅ X∩ ∅ Y1 2 3 O(n− polylog n)E(Y ) O(E(Y ) polylog n). ≤ · ≤ · To bound S4, we note that for disjoint α, β and γ disjoint from α the conditions (62), (59), and (68) yield ∈ A ∈ A µβ O( n polylog n) if β = γ α α · E (Yβ Yβ )(Yγ Yγ ) = 0 if β = γ,β γ = | − − | µβ µγ 6 ∩ 6 ∅ O( 2 polylog n) if β γ = . n · ∩ ∅ Therefore, µ µ µ E (Y Y α)(Y Y α) O( β γ polylog n)+ O( β polylog n) | β − β γ − γ | ≤ n2 · n · β:α β= γ:α γ= β γ β X∩ ∅ X∩ ∅ X∈A X∈A X∈A O(E(Y )2/n2 polylog n)+ O(E(Y )/n polylog n) ≤ · · = O(polylog n). α α Hence, we obtain S4 α µα β:α β= γ:α γ= E (Yβ Yβ )(Yγ Yγ ) O(E [Y ] polylog n). ≤ ∈A ∩ 6 ∅ ∩ 6 ∅ − − ≤ · P P P h i ⊓⊔ 3 Lemma 31. α β E [ XαZαβVαβ ]= O(σ− E [Y ] polylog n) ∈A ∈A | | · P P α Proof. Let S1 = β:α β= E [ XαYβVαβ ] and S2 = β:α β= E Xα(Yβ Yβ )Vαβ . Then the def- ∩ 6 ∅ | | ∩ ∅ − 1 inition (55) of ZαβPyields that α β E [ XαZαβPVαβ ] σ−h (S1 + S2) Hence, it sufficesi to show 2 ∈A ∈A | | ≤ that S ,S = O(σ− E [Y ] polylog n). 1 2 · P P To bound S1, we note that YαYβ = 0 if α β = but α = β by (59), and that Vαβ = 0 if α = β by the definition (56) of V . Thus, if α β = ,∩ then6 ∅ 6 αβ ∩ 6 ∅ (54) 1 1 E [ X Y V ] σ− E [(Y + µ ) Y V ] σ− µ E [ Y V ]. (84) | α β αβ| ≤ α α | β αβ| ≤ α | β αβ | Furthermore, Y7 T (α, β) = E Y Y α k2µ . (85) 1 | β γ ≤ β γ:γ β= ,γ α= ∩ 6 X∅ ∩ ∅ (64) α α β µβ µγ T (α) = E Y Y Y ∪ O( polylog n) 2 β| γ − γ | ≤ n · β:α β= γ:β∩γ=∅ β:α β= γ:β∩γ=∅ X∩ 6 ∅ ∧αX∩γ=∅ X∩ 6 ∅ ∧αX∩γ=∅ 1 O(n− polylog n) µ µ ≤ · γ β γ β:α β= X∈A X∩ 6 ∅ Y1 1 O(n− E(Y ) polylog n)= O(polylog n). (86) ≤ · Combining (84)–(86), we get (56) 1 2 S σ− µ E [ Y V ] σ− µ T (α)+ T (α, β) 1 ≤ α | β αβ | ≤ α 2 1 α β:α β= α β:α β= X∈A X∩ 6 ∅ X∈A X∩ 6 ∅ Y1 2 2 2 O(σ− polylog n) E(Y )+ k µ O(σ− E(Y ) polylog n) ≤ · β ≤ · β:α β= X∩ 6 ∅ 26 To bound S , let α, β be disjoint. As X (Y + µ )/σ, we obtain 2 ∈ A α ≤ α α α 1 α E X (Y Y )V σ− E (Y + µ )(Y Y )V α β − β αβ ≤ α α β − β αβ (56), (60) 2 α α σ− E (Yα + µα)(Yβ Y )Y ≤ − β β 2 α α α β +σ− E (Y + µ )(Y Y )(Y Y ∪ ) α α β β γ γ γ:β∩γ=∅ − − ∧αX∩γ=∅ 2 σ− (T + T + T + T ), ≤ 1 2 3 4 where T =E Y (Y Y α)Y α ,T = µ E (Y Y α)Y α , 1 α β − β β 2 α β − β β α α α β α α α β T3 = E Yα(Yβ Y )(Y Y ∪ ) ,T4 = µα E (Yβ Y )(Y Y ∪ ) . − β γ − γ − β γ − γ γ:β∩γ=∅ γ:β∩γ=∅ ∧αX∩γ=∅ ∧αX∩γ=∅ Now, T1 =0 by (58). Moreover, bounding T2 by (62), T3 by (66) and T4 by (65), we obtain µ µ µ µ µ σ2E X (Y Y α)V O( α β polylog n)+ O( α β γ polylog n) α β − β αβ ≤ n · n2 · γ:β∩γ=∅ ∧αX∩γ=∅ µ µ = O( α β polylog n). n · 2 µαµβ 1 2 2 2 Thus, (87) yields S2 σ− β:α β= O( n polylog n)= O(n− σ− E(Y ) polylog n)= O(σ− E(Y ) polylog n), as desired.≤ ∩ ∅ · · · P ⊓⊔ 3 Lemma 32. α β E [ XαZαβ ]E [ Zα + Vαβ ]= O(σ− E [Y ] polylog n) ∈A ∈A | | | | · Proof. SincePσX PY + µ , | α|≤ α α 1 E [ X Z ]E [ Z + V ] σ− µ E [ Z ](E [ Z ]+E[ V ]) + | α αβ| | α αβ| ≤ α | αβ| | α| | αβ| α β α β X∈A X∈A X∈A X∈A E [Y Z ](E [ Z ]+E[ V ]) .(87) α | αβ| | α| | αβ| Furthermore, we have the three estimates (55) σE [ Z ] σ E [ Z ] = µ + E Y Y α | α| ≤ | αβ| β β − β β β:α β= β:α β= X∈A X∩ 6 ∅ X∩ ∅ (62), Y1 1 O(n− µ polylog n)= O(polylog n), (88) ≤ β · β X∈A (56) α α α β σE [ V ] E Y + E Y Y ∪ | αβ | ≤ γ γ − γ γ:β∩γ6=∅ γ:β∩γ=∅ ∧αX∩γ=∅ ∧αX∩γ=∅ Y1 (69), 1 = O(n− µ polylog n) O(polylog n), (89) γ · ≤ γ X∈A (55) σE [Y Z ] = E [Y Y ]+ E Y Y Y α α | αβ| α β α β − β β β:α β= β:α β= X∈A X∩ 6 ∅ X∩ ∅ (59), (63) µ µ = kµ + α β = O(µ polylog n). (90) α n α · β:α β= X∩ ∅ 27 Now, (88)–(90) yield 2 µ E [ Z ](E [ Z ]+E[ V ]) = O(σ− polylog n) µ α | αβ| | α| | αβ | · α α β α X∈A X∈A X∈A 2 = O(σ− E [Y ] polylog n), (91) · 2 E [Y Z ](E [ Z ]+E[ V ]) = O(σ− polylog n) µ α | αβ| | α| | αβ | · α α β α X∈A X∈A X∈A 2 = O(σ− E [Y ] polylog n). (92) · Combining (87), (91), and (92), we obtain the assertion. ⊓⊔ Finally, Lemma 27 is an immediate consequence of Lemmas 30–32. 6 Conclusion Using a purely probabilistic approach, we have established a local limit theorem for (Hd(n,p)). This result has a number of interesting consequences, which we derive in a follow-up paperN [4]. Namely, via Fourier analysis the univariate local limit theorem (Theorem 2) can be transformed into a bivariate one that describes the joint distribution of the order and the number of edges of the largest component. Furthermore, since given its number of vertices and edges the largest component is a uniformly distributed connected graph, this bivariate limit theorem yields an asymptotic formula for the number of connected hypergraphs with a given number of vertices and edges. Thus, we can solve an involved enumerative problem (“how many connected hypergraphs with ν vertices and µ edges exist?”) via a purely probabilistic approach. 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