EDGE-STATISTICS ON LARGE GRAPHS

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1 EDGE-STATISTICS ON LARGE GRAPHS NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN Abstract. The inducibility of a graph H measures the maximum number of induced copies of H a large graph G can have. Generalizing this notion, we study how many induced subgraphs of fixed order and size l a large graph G on n vertices can have. Clearly, this number is n for every n, and l { 0, }. We conjecture that for every n, and 0 < l < this number is at most 1/e + o 1 n. If true, this would be tight for l {1, 1}. In support of our Edge-statistics conjecture we prove that the corresponding density is bounded away from 1 by an absolute constant. Furthermore, for various ranges of the values of l we establish stronger bounds. In particular, we prove that for almost all pairs, l only a polynomially small fraction of the -subsets of V G has exactly l edges, and prove an upper bound of 1/ + o 1 n for l = 1. Our proof methods involve probabilistic tools, such as anti-concentration results relying on fourth moment estimates and Brun s sieve, as well as graph-theoretic and combinatorial arguments such as Zyov s symmetrization, Sperner s theorem and various counting techniques. 1. Introduction Let be a positive integer and let G be a finite graph of order at least. Let A = A G, be chosen uniformly at random from all subsets of V G of size and let X G, = eg[a]. That is, X G, is the random variable counting the number of edges of G with both endpoints in A. Naturally, the above quantities can also be interpreted as densities rather than probabilities, and we shall frequently switch between these two perspectives. Given integers n and 0 l, let In,, l = max{pxg, = l : G = n}, that is, In,, l is the maximum density of induced subgraphs with vertices and l edges, taen over all graphs of order n. A standard averaging argument shows that In,, l is a monotone decreasing function of n. Consequently, we define ind, l := lim { n In,, l to be the edgeinducibility of and l. While this quantity is trivially 1 for l 0, } simply tae G to be a large empty or complete graph, respectively, it is natural to as how large can ind, l be for 0 < l <. This question is closely related to the problem of determining the inducibilities of fixed graphs, a concept which was introduced in 1975 by Pippenger and Golumbic [16]. For a graph H, let D H G denote the number of induced subgraphs of G that are isomorphic to H, and let I H n = max{d H G : G = n}. Again, the sequence {I H n/ n H } n= H is monotone decreasing and thus converges to a limit indh, the inducibility of H. Recently there has been a surge of interest in this area see, e.g., [4, 10, 17, 11]. Observe that both types of inducibility are invariant under taing complements, that is, ind, l = ind, l and indh = indh. Note also that indh ind H, eh. { Moreover, if H = and eh 1, } 1, then indh = ind, eh, as H is the unique up to isomorphism graph with vertices and eh edges. Consider a random graph G Gn, p, where p = 1. A straightforward calculation shows that the expected value of the number of -subsets of V G which span precisely one edge is about 1/e. This implies that ind, 1 1/e + o 1 as the o 1 notation suggests, Date: January 4, 019. The research of Noga Alon is supported in part by ISF grant 81/17 and by GIF grant G /016. The research of Dan Hefetz is supported by ISF grant 8/18. The research of Michael Krivelevich is supported in part by BSF grant and ISF grant 161/17. The research of Myhaylo Tyomyn is supported in part by ERC Starting Grant

2 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN we will often thin of as an asymptotic quantity and, in particular, we will assume to be sufficiently large wherever needed. In fact, we will see later that there are many constructions which achieve 1/e + o 1 as a lower bound for ind, 1. Another construction, achieving the same asymptotic value for l = 1 is the complete bipartite graph with the smaller part of size n/, so that ind, 1 indk 1, 1 1/e + o 1. In fact, it is nown [6] that indk 1, 1 = 1/e + o 1. Note that the o 1 term is necessary. For example, counting cherries in K n/,n/ shows that ind3, = indk 1, 3/4 in fact, it follows from Goodman s Theorem that ind3, 1 = ind3, = 3/4. Motivated by the aforementioned constructions as well as some additional data, we conjecture that the lower bound of 1/e is asymptotically tight. Conjecture 1.1 The Edge-statistics Conjecture. For every ε > 0 there exists 0 = 0 ε such that for all integers > 0 and 0 < l < we have ind, l 1/e + ε. For graph-inducibilities we mae an analogous conjecture, which would be implied by the Edge-statistics Conjecture. Conjecture 1. The Large Inducibility Conjecture. lim sup { indh : H / {K H, K H } } = 1/e. Our first theorem in this paper constitutes a first step towards proving Conjecture 1.1. It asserts that ind, l is bounded away from 1 by an absolute constant for every and 0 < l <. Theorem 1.3. There exists an ε > 0 such that for all positive integers and l which satisfy 0 < l < we have ind, l < 1 ε. For clarity of presentation, we do not give explicit bounds on ε and refer to Section 6 for a discussion. Note that { it is not hard to prove that for every positive integer we have ind, l = 1 if and only if l 0, }. Indeed, if 0 < l <, then ind, l < 1 4 is an easy consequence of Ramsey s Theorem and the aforementioned monotonicity of In,, l. With a bit more effort, this bound can be improved to 1. On the other hand, we do not see a simple argument that would upper bound ind, l away from 1 by an absolute constant as in Theorem 1.3. Note also that the related problem of minimizing graph-inducibilities has been extensively studied. In particular, Pippenger and Golumbic [16] showed that the inducibility of any -vertex graph is at least 1 + o 1!/. It follows that indh > 0 for every graph H and thus ind, l > 0 for every and l. We refer the reader to Section 6 for further discussion. For various ranges of values of l viewed as a function of we establish{ much better upper bounds than the one stated in Theorem 1.3. First, for every l satisfying min l, } l = ω, we prove an upper bound of 1/. Proposition 1.4. For every ε > 0 there exist Cε > 0 and 0 ε > 0 such that the following holds. Let and l be integers satisfying 0 and C l C. Then ind, l 1 + ε. Next, we prove Conjecture 1.1 almost everywhere. { In fact, we prove a much stronger statement, namely that for every l satisfying min l, } l = Ω the quantity ind, l is actually polynomially small in the right asymptotic behavior as can be seen by considering the random graph Gn, l/, which gives ind, l = Ω 1. Theorem 1.5. For every positive integers and l such that min{l, / l} = Ω we have ind, l = O 0.1. Lastly, we consider the case when l is fixed i.e., does not depend on. Here we prove an upper bound of 3/4. In the interesting sub-case l = 1, which corresponds to the inducibility of the one-edge graph equivalently, of K, the complete graph with one edge removed we prove a yet better bound of 1/.

3 Theorem 1.6. For every fixed positive integer l we have EDGE-STATISTICS ON LARGE GRAPHS 3 ind, l o 1. Moreover, for l = 1 we have ind, o 1. Our results are summarized in the following table. For various ranges of l /4, it states the best nown upper bound on ind, l. Note that for l /4 the table can be extended symmetrically. [ l = l 1 const. [ω1, O] ω, o ] [ Ω, /4 ] ind, l 1/ 3/4 1 ε 1/ O Notation. Throughout this paper, log stands for the natural logarithm, unless explicitly stated otherwise. For positive integers n we denote by n the falling factorial 1 i=0 n i. The symmetric difference of two sets A and B, denoted by A B, is A \ B B \ A. Our graph-theoretic notation is standard and follows that of [5]. In particular, we use the following. For a graph G, let V G and EG denote its sets of vertices and edges respectively, and let G = V G and eg = EG. The complement of G, denoted by G, is the graph with vertex set V G and edge set V G \ EG. For a set S V G, let G[S] denote the graph with vertex set S and edge set {uv EG : u, v S}. For disjoint sets S, T V G, let G[S, T ] denote the bipartite graph with parts S and T and edge-set {uv EG : u S, v T }. For a set S V G and a vertex v V G, let N G v, S = {u S : uv EG} denote the neighbourhood of v in S and let d G v, S = N G v, S denote the degree of v into S. We abbreviate N G v, V G under N G v and d G v, V G under d G v; we refer to the former as the neighbourhood of v in G and to the latter as the degree of v in G. The maximum degree of a graph G is G = max{d G u : u V G}. Often, when there is no ris of confusion, we omit the subscript G from the notation above. The rest of this paper is organized as follows. In Section we establish a number of facts and lemmas which will be useful later on when we will upper bound ind, l; we then prove Proposition 1.4. In Sections 3 and 4 we prove Theorems 1.3 and 1.5 respectively. Moving on to the fixed l regime, in Section 5 we establish some additional tools and prove Theorem 1.6. In Section 6 we conclude the paper with several remars and open problems.. Preliminaries Recall that A = A G, is the set chosen uniformly at random from all subsets of V G of size and X G, = eg[a]. To simplify notation we abbreviate X G, to X whenever there is no ris of confusion. Our first lemma provides a useful global-local criterion for handling edgeinducibilities. Lemma.1. Let and l be positive integers satisfying 0 < l < and let a = ind, l. Let n be a sufficiently large integer and let G be a graph on n vertices which attains In,, l. Then, for every vertex v V G, we have PX = l v A = a + o n 1. Proof. The main idea of the proof is the same as in the proof of Lemma.4 from [10]. Doublecounting yields a + o n 1 = PX = l = 1 PX = l, v A = 1 PX = l v A Pv A = 1 n v V G v V G PX = l v A = 1 n v V G v V G PX = l v A. Let v + and v be the vertices with the largest and the smallest value of PX = l v A, respectively. Let the graph G be obtained from G by Zyov s symmetrization [18], i.e., remove

4 4 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN v and add a twin copy of v + instead say, the two copies of v + are not connected by an edge in G. Then PX = l PX G, = l PX = l PX = l, v A + PX = l, v + A PX = l, v A, v + A = PX = l n P X = l v A + n P X = l v + A 1 nn 1 P X = l v +, v A = PX = l + n PX = l v + A PX = l v A On, where the first inequality follows from our assumption that G maximizes PX G, = l. Therefore PX = l v + A PX = l v A = On 1. Hence, PX = l v + A a + On 1 and PX = l v A a On 1. Since a > 0 as remared in the introduction, this concludes the proof of the lemma. For two vertices v, w V G we now consider a subset of V G \ {v, w} of size 1, chosen uniformly at random among all such subsets. Our next lemma shows that, assuming that ind, l is large, with a significant probability v and w will have the same degree into this set. Lemma.. Let and l be positive integers satisfying 0 < l <, let a = ind, l, and suppose that a > 1/. Let n be a sufficiently large integer and let G be a graph on n vertices which attains In,, l. Then for any two vertices v, w V G we have Proof. By Lemma.1 we have and, symmetrically, P e G v, A G\{v,w}, 1 = e G w, A G\{v,w}, 1 > a 1 o n 1. PX = l v A, w / A = PX = l v A + o1 = a + o n 1..1 PX = l w A, v / A = a + o n 1.. Setting G = G \ {v, w} and B = A G\{v,w}, 1, identities.1 and. imply that PX G, 1 + e G B, v = l = a + o n 1 and PX G, 1 + e G B, w = l = a + o n 1. Let L v respectively, L w denote the event X G, 1 + e G B, v = l respectively, X G, 1 + e G B, w = l. Then Pe G B, v = e G B, w PL v L w = PL v + PL w PL v L w > a 1 o n 1. Lemma.3. For every 1/ < a < 1 there exists C = Ca > 0 for which the following holds. Suppose that and l are positive integers with 0 < l < such that a = ind, l. Suppose that n is sufficiently large and that G is a graph on n vertices which attains In,, l. Then for any two vertices v, w V G we have Moreover, as a 1, inequality.3 holds with C 0. N G v N G w < Cn/..3 Note that the second part of the statement of Lemma.3 i.e., the one referring to a 1 is only needed to prove Theorem 1.3 by contradiction, and is otherwise vacuous, as it contradicts Theorem 1.3.

5 EDGE-STATISTICS ON LARGE GRAPHS 5 Proof. We can assume to be larger than any given absolute constant, since, in the first statement, the small values of can be accommodated for by adjusting C, and the second statement implicitly assumes, since a = ind, l < 1 for any given 0 < l <, and taes only finitely many values when is bounded. Fix two arbitrary vertices v, w V G and let P = N G v \ w N G w, R = N G w \ v N G v and Q = V G \ {v, w} P R, Let B = A G\{v,w}, 1. It then follows by Lemma. that P B P = B R > a 1 o n 1..4 Suppose that P R cn/ for some absolute constant c > 0. Then PB P R o11 c/ 1 = 1 e c + o 1 = Ω 1..5 In particular, if P R = ωn/, then PB P R = 1 o 1. Therefore, using.4 we obtain P B P = B R B P R P B P = B R B P R = PB P R a 1 o1 = = a 1 o o 1 Similarly, if a = 1 o 1 and P R = Ωn/, then.4 implies that P B P = B R = 1 o 1. Therefore P B P = B R B P R P B P = B R B P R = PB P R = PB P R o 1 PB P R.5 = 1 o 1..7 Note that the above argument does not mae any use of the graph structure of G. Indeed, the situation at hand can be viewed as an urn model, in which we have a large urn filled with P pin balls and R red balls, and we draw a fixed but otherwise arbitrary number 1 s 1 of balls from the urn uniformly at random, without replacement. We would lie to upper bound the probability of drawing equally many pin balls and red balls. To this end, we first prove the following auxiliary claim. Claim.4. For every integer 1 t 1/ we have P B P = B R = t 3 P B P R = t. 4 Moreover, for every ε > 0 there exists t 0 = t 0 ε such that for every t t 0 we have P B P = B R = t ε P B P R = t. Proof. Fix some 1 t 1/. By the log-concavity of the binomial coefficients we have P R t t P + R. t Moreover, using the fact that t is much smaller than P + R as t < and P + R cn/ by assumption, and thus can be assumed to be sufficiently large, straightforward calculations show that P + R 3 t 4 P + R t.

6 6 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN We conclude that P B P = B R = t = P t R t n P R 1 t n = 3 P B P R = t. 4 The second statement can be proved analogously; we omit the details. P + R n P R t 1 t n 1 Coming bac to the proof of Lemma.3, using Claim.4 with a = 1 o 1 we have Therefore P B P = B R > 0 = 1/ t=1 3 1/ 4 P B P R = t 3 4 t=1 P B P = B R = t P B P R > 0. P B P = B R B P R 3 4, contrary to.7. Similarly, for any a > 1/, if P R = ωn/, we obtain that for some t = ta we have P B P = B R B P R > t < 1 a 1, contrary to.6. This concludes the proof of Lemma.3. Under closer inspection, Lemma.3 has the following immediate consequence. Lemma.5. For every 1/ < a < 1 there exists C = Ca > 0 for which the following holds. Suppose that and l are positive integers with 0 < l < such that a = ind, l. Suppose that n is sufficiently large and that G is a graph on n vertices which attains In,, l. Suppose that eg n /. Then G < Cn/. Moreover, as a 1, the above holds with C 0. Proof. We prove the first statement the second can be proven analogously. It follows from Lemma.3 that dv dw = On/ holds for any v, w V G. Let v be a vertex of minimum degree in G. Put U := N G v and W := N G v. Observe that U W since we assumed that dv = δg and eg n /. Suppose for a contradiction that U = ωn/. We double-count the edges of the bipartite graph G[U, W ]. Applying Lemma.3 to v and u for every u U, we derive that d G[U,W ] u = On/. In particular eg[u, W ] = O U n = O W n. On the other hand, applying Lemma.3 to v and w for every w W, yields n n n n d G[U,W ] w U O ω O = ω. Therefore n W ω eg[u, W ] = O W n which is clearly a contradiction, and thus δg = On/. Since, moreover, dv dw = On/ holds for any v, w V G by Lemma.3, we conclude that G = On/ as claimed. From now on, we denote eg by m. For every e EG, let X e be the indicator random variable for the event e EG[A], that is, X e = 1 if both endpoints of e are in A and X e = 0 otherwise. Observe that X = e EG X e. Putting µ = EX we have µ = e EG This has the following consequence. 1 EX e = m nn 1 = m 1 + O1/n..8 n,

7 EDGE-STATISTICS ON LARGE GRAPHS 7 Lemma.6. Let and l be positive integers satisfying 0 < l < and let a = ind, l. Let n be a sufficiently large integer and let G be a graph with n vertices and m edges which attains In,, l. Then In particular, if a = 1 o 1, then m 1 o 1al n. m 1 o 1 n. Proof. Suppose for a contradiction that m < 1 εal n / for some constant ε > 0. It then follows by.8 that µ < 1 ε/al. On the other hand, since X 0 and by the choice of G we have µ l PX = l = 1 o1al, a contradiction. Combining the above facts, we can immediately prove Proposition 1.4. Proof of Proposition 1.4. Suppose for a contradiction that there exist ε > 0 and an integer l = l such that C l C for some large C > 0, and ind, l = a > 1/ + ε. Let G be a graph attaining In,, l, where n is sufficiently large. By symmetry we may assume that eg n /. Then, on the one hand, by Lemma.5 we have G = On/ entailing eg = On /. On the other hand, Lemma.6 implies that eg 1/ o1l n C 3 n, which is a contradiction for large enough C. We conclude that a 1/ + o1 as claimed. 3. Proof of Theorem 1.3 Our goal in this section is to prove Theorem 1.3. Since, as remared in the introduction, ind, l is never identically 1 assuming 0 < l <, it suffices to prove the theorem for large values of. Hence, suppose now, for a contradiction, that the assertion of Theorem 1.3 is false for arbitrarily large values of. That is, for every ε > 0 there exist arbitrarily large values of such that for some 0 < l < there will be arbitrarily large values of n and graphs G on n vertices for which PX G, = l > 1 ε. We may assume that G maximizes PX G, = l over all n-vertex graphs and, by symmetry, that eg n /. We would lie to calculate the variance of X G,. Using Lemma.5 and our assumption that is sufficiently large we obtain EX = E e EG X e = µ + e,f EG e f= EX e X f + e,f EG e f =1 EX e X f = µ + m 4 n O 1/n + S 3 n O 1/n, where S = v V G dv. Therefore, VarX = EX µ = m n m n 4 + m 4 n 4 + S 3 n 3 = m n + S 3 n m n O 1/n = m n + S 3 n 3 m 43 n o O 1/n Since, by Lemma.5 we have m = on /, and, consequently, 4m 3 /n 4 = om /n, we can write VarX = 1 + o1 m n + S 3 n

8 8 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN We shall need the following anti-concentration inequality from [3], involving the fourth moment of a random variable. Lemma 3.1 [3]: Lemma 3.i. Let Y be a real random variable and suppose that its first, second and fourth moments satisfy EY = 0, E Y = σ > 0 and E Y 4 bσ 4 for some constant b > 0. Then PY > 0 1 4/3 b and PY < 0 1 4/3 b. The aim in the rest of this section is to prove the following lemma, which directly implies Theorem 1.3. Lemma 3.. There exists a constant b > 0 such that E X G, µ 4 b VarX G,. 3. Having established this, we conclude the proof Theorem 1.3 as follows. By Lemma 3. the random variable Y := X µ satisfies the assumptions of Lemma 3.1. Since the event {X = l} is disjoint either from the event {Y > 0} or from the event {Y < 0}, using Lemma 3.1 we obtain PX = l 1 1 < 1 ε, contrary to our assumption that PX = l > 1 ε. 4/3 b Proof of Lemma 3.. First let us expand the right hand side of 3.. VarX 3.1 = 1 + o1 m n + S 3 n 3 = Θ m ms n4 n 5 + S 6 n Now, putting p := /n, the left hand side of 3. can be written as 4 E X µ 4 = E X e p = Cove, f, g, h, 3.4 e EG e,f,g,h EG 4 where, by abuse of notation, Cove, f, g, h := E X e px f px g px h p. 3.5 The following simple technical claim will play a crucial role in our proof of Lemma 3.. Claim 3.3. For every 4-tuple e, f, g, h EG 4 we have Cove, f, g, h = OEX e X f X g X h. Proof. Fix some 4-tuple e, f, g, h EG 4. Let B be some subset of {e, f, g, h}. Let H B denote the graph spanned by the edges in B and let t = V H B. Let H B be a graph obtained from H B by deleting one of its edges together with any of its endpoints if its degree in H is 1, adding two new vertices, and connecting them by an edge. Let t denote the number of vertices of H B ; clearly t t. Since n, a straightforward calculation show that E p = E t = O n t a EH B EH B X a a EH B X a t = O = O E Therefore, for every B {e, f, g, h} we have p 4 B E X a = OEX e X f X g X h. We conclude that as claimed. a B n t Cove, f, g, h = OEX e X f X g X h, a EH B X a.

9 EDGE-STATISTICS ON LARGE GRAPHS 9 We now distinguish between several cases, depending on how the four edges e, f, g, h are arranged. Suppose first that e, f, g, h are pairwise distinct. Let H be the four-edge graph spanned by e, f, g and h. Note that H is a subgraph of G, but not necessarily an induced one. The options for H are i H = 4K ; ii H = K + K 1, ; iii H = K 1, ; iv H = K + K 1,3 ; v H = K + P 4 ; vi H = K + K 3 ; vii H = K 1,4 ; viii H = K 1,3 + a 3-star with one edge subdivided into two; ix H = P 5 ; x H = C 4 ; xi H = K + 3 a triangle with one pendant edge. There are also several degenerate cases, in which two of the edges e, f, g, h coincide; we will deal with these cases at the end of the proof. We claim that in each of the above listed cases, after some cancellation, the respective contributions can be bounded from above using the terms appearing on the right hand side of 3.3. As a helpful piece of notation, we denote the number of unlabelled not necessarily induced copies of H in G by NH. Case i. H = 4K. Given a fixed 4-tuple e, f, g, h forming H, it follows by 3.5 that Cove, f, g, h = EX e X f X g X h 4EX e X f X g p + 6EX e X f p 4EX e p 3 + p 4 = 1 n 8 [ ] 1 + O 1/n = 1 n 8 [ O 6 ] 6 = O. n 8 Taing the sum over all such tuples e, f, g, h, we obtain an overall contribution of at most m 4 6 O n 8 = O m 4 n 4 = O V arx, where the penultimate equality holds since m = On / follows from Lemma.5 and the last equality holds by 3.3. Case ii. H = K + K 1,. Let e and f denote the two edges that share a vertex. It follows by 3.5 and a straightforward calculation that Cove, f, g, h = [ EX e X f X g X h EX e X f X g p EX e X f X h p + EX e X f p ] 1 + O 1/n = 1 n 7 [ ] 1 + O 1/n = 1 n 7 [ O 5 ] = 1 n O 5 < 0. Thus, we can ignore such tuples e, f, g, h as their contribution is negative.

10 10 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN Case iii. H = K 1,. A straightforward calculation shows that Cove, f, g, h = 1+o1EX e X f X g X h = O 6 /n 6. Moreover, the total number of such 4-tuples e, f, g, h is at most 4! S. Hence, Cove, f, g, h = O S 6 n 6 = O V arx, e,f,g,h =H where the last equality holds by 3.3. Case iv. H = K + K 1,3. It follows from Claim 3.3 that Cove, f, g, h = O 6 /n 6. Note that a copy of K 1,3 in G can be viewed as a copy of P 3 with an additional edge attached to its middle vertex. This implies that NK 1,3 = ONP 3 G = OSn/, where the last equality holds by Lemma.5. Therefore, Cove, f, g, h = O ms 5 /n 5 = O V arx, e,f,g,h =H where the last equality holds by 3.3. Case v. H = K + P 4. This case is analogous to Case iv with P 4 in place of K 1,3. Since a copy of P 4 in G can be viewed as a copy of P 3 with another vertex connected by an edge to one of its end vertices, we have NP 4 = ONP 3 G = OSn/, where the last equality holds by Lemma.5. We conclude that Cove, f, g, h = O m NP 4 6 /n 6 = O ms 5 /n 5 = O V arx, e,f,g,h =H where the last equality holds by 3.3. Case vi. H = K + K 3. This is similar to cases iv and v. It follows from Claim 3.3 that Hence, e,f,g,h =H Cove, f, g, h = O Cove, f, g, h = O 5 /n 5. m NK 3 5 n 5 = O ms 5 /n 5 = O V arx, where the second equality holds since NK 3 = OS and the last equality holds by 3.3. Case vii. H = K 1,4. It follows from Claim 3.3 that Note that NK 1,4 = O v V G e,f,g,h =H Cove, f, g, h = O dv 4 = O G v V G NK 1,4 5 n dv n = O S = OmS, where the penultimate equality holds by Lemma.5 and the last equality holds by Lemma.6. We conclude that Cove, f, g, h = O ms 5 n 5 = O V arx, e,f,g,h =H where the last equality holds by 3.3.

11 EDGE-STATISTICS ON LARGE GRAPHS 11 Case viii. H = K 1,3 +. This is very similar to Case vii with K 1,3 + in place of K 1,4. Since a copy of K 1,3 + in G can be viewed as a copy of P 3 with two leaves attached to one of its end vertices, it follows that N K 1,3 + = O S G = OmS. Therefore e,f,g,h =H Cove, f, g, h = O N K 1,3 + 5 n 5 = O where the last equality holds by 3.3. ms 5 n 5 = O V arx, Case ix. H = P 5. This is again very similar to Case vii with P 5 in place of K 1,4. Since a copy of P 5 in G can be viewed as a copy of P 3 with a leaf attached to each of its end vertices, it follows that Therefore e,f,g,h =H Cove, f, g, h = O where the last equality holds by 3.3. NP 5 = O S G = OmS. NP 5 5 n 5 = O Case x. H = C 4. It follows from Claim 3.3 that e,f,g,h =H Cove, f, g, h = O ms 5 n 5 = O V arx, NC 4 4 n 4. It is evident that NC 4 = ONP 4. Since, moreover, we can view P 4 as P 3 with an additional edge appended to one of its leaves, it follows that NP 4 = O NP 3 G = O S G = O S n, where the last equality holds by Lemma.5. Therefore Cove, f, g, h = O NC 4 4 = O e,f,g,h =H = O n 4 ms 5 n 5 S 3 n 3 = O V arx, where the penultimate equality holds by Lemma.6 and the last equality holds by 3.3. Case xi. H = K 3 +. Using the calculations made in Case x we obtain N K 3 + = ONP4 = O S n. It thus follows by Claim 3.3 that Cove, f, g, h = O N K = O e,f,g,h =H where the last equality holds by 3.3. n 4 S 3 n 3 = O ms 5 n 5 = O V arx,

12 1 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN The degenerate cases. We now consider the cases where some edge of {e, f, g, h} appears more than once. In those cases, the corresponding graph H has at most three edges. xii H = 3K xiii H = K + K 1, xiv H = K 1,3 xv H = P 4 xvi H = K 3 xvii H = K 1, xviii H = K xix H = K Case xii. H = 3K. Without loss of generality, suppose that e, f, g are independent and that h = e. Since, clearly X e = X e, we obtain Cove, f, g, h = EX e p X f px g p = EX e X f X g pex e X f pex e X g + p EX e 1 + O 1/n = 1 n O 1/n = 45 + O 4 n 6 < 0. Thus, the total contribution of all such tuples is negative. Case xiii. H = K + K 1,. It follows from Claim 3.3 that Cove, f, g, h = O mnk 1, 5 n 5 = O ms 5 n 5 e,f,g,h =H where the last equality holds by 3.3. Case xiv. H = K 1,3. It follows from Claim 3.3 that Cove, f, g, h = O e,f,g,h =H = O = O V arx, NK 1,3 4 = O G NK 1, 4 n 4 S 3 n 3 = O n 4 ms 5 n 5 = O V arx, where the third equality holds by Lemma.5, the fourth equality holds by Lemma.6, and the last equality holds by 3.3. Case xv. H = P 4. It follows from Claim 3.3 that Cove, f, g, h = O e,f,g,h =H NP 4 4 n 4 = O S 3 n 3 = O ms 5 n 5 = O V arx, where the second equality was proved in Case x and the last equality holds by 3.3. Case xvi. H = K 3. It follows from Claim 3.3 that Cove, f, g, h = O NK 3 3 n 3 = O S 3 n 3 e,f,g,h =H where the last equality holds by 3.3. = O ms 5 n 5 = O V arx,

13 e,f,g,h =H EDGE-STATISTICS ON LARGE GRAPHS 13 Case xvii. H = K 1,. It follows from Claim 3.3 that Cove, f, g, h = O NK 1, 3 n 3 = O S 3 n 3 where the last equality holds by 3.3. Case xviii. H = K. It follows from Claim 3.3 that Cove, f, g, h = O NK 4 n 4 = O e,f,g,h =H where the last equality holds by 3.3. Case xix. H = K. It follows from Claim 3.3 that Cove, f, g, h = O m n = O e,f,g,h =H = O ms 5 n 5 = O V arx, m 4 n 4 = O V arx, m 4 n 4 = O V arx, where the last equality holds by 3.3. To conclude, we have considered every possible case and in each one we have shown that its respective contribution to E X µ 4 is of order of magnitude O V arx. Since the number of cases is constant, this completes the proof of Lemma 3. and thus also of Theorem Very small inducibility for almost all l Our { aim in this section is to prove Theorem 1.5. Let and l be positive integers for which min, } l = Ω. Since, ind, l is defined as lim n In,, l and the latter sequence is monotone decreasing, it suffices to show that I,, l = O 0.1. Suppose then that we have a -vertex graph G = V, E, in which we are selecting a -vertex set A V uniformly at random. As in the proofs in previous sections, we may assume that G maximizes PX G, = l amongst all -vertex graphs. Claim 4.1. min{eg, eg} = Ω. Proof. Since G = and G contains an induced subgraph of order and size l = Ω as, clearly, I,, l > 0, we have that eg = Ω. The statement for G follows analogously. A pair of distinct vertices {u, v} V will be called distinguished if N G u N G v = Θ. Let D = DG be the set of all distinguished pairs in G. Claim 4.. D = Θ. Proof. It is evident that D = O. To see that D = Ω, note that Nu Nv = du + dv Nu Nv {u,v} V {u,v} V = V 1 dv dv v V v V [ = 1eG ] dv. v V By a theorem of Ahlswede and Katona [1], for a graph H of given order and size the quantity, which corresponds to the number of cherries K1,, is maximized when either H or its v dv

14 14 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN complement is a union of a clique and an independent set. In either case it follows by Claim 4.1 and a straightforward calculation that Nu Nv N H u N H v = Ω 3. {u,v} V {u,v} V H Since the maximum degree of G is O, we conclude that DG = Ω, as claimed. Claim 4.3. For any graph F with vertices and Ω edges, and any integer t satisfying t +, we have P X F,t ef = o 1. 5 Proof. Let B V F be a random set of size t and let B = V F \ B be its complement. Fix an arbitrary vertex v V. It follows by standard tail estimates for the hypergeometric distribution that P dv, B dv, B > 0.9 = o in fact, the right hand side is exponentially small in a positive constant power of, see for example [1, Theorem.10]. Hence, by the union bound, with probability 1 o 1, it holds that dv, B dv, B 0.9 for every v V F. Summing over all vertices in B, we obtain ef [B] ef [B, B] v B dv, B dv, B 1.9 = oef. Similarly, summing over all vertices in B, we obtain Hence, with probability 1 o 1 we have ef [B] ef [B, B] = oef. ef [B] = ef [B] + oef = 1 ef [B, B] + oef = 1/4 + o1 ef. In particular, X F,t = ef [B] ef /5 holds with probability 1 o 1. Now, let us sample the set A in two steps as follows. We first sample a set A 0 V of size 0. uniformly at random, then we sample a set A 1 V \ A 0 of size 0. in a manner which will be specified later, and finally we set A = A 0 A 1. Note that A 1 will be sampled in a way which will ensure that the resulting set A will indeed be chosen uniformly amongst all subsets of V G of size. Given A 0, a distinguished pair {u, v} is said to be bad if du, A 0 dv, A 0 0.4, and good otherwise. Our next claim shows that, with sufficiently high probability, most distinguished pairs are good. Claim 4.4. With probability 1 O 0.1 there are at most D /6 bad pairs. Proof. Fix some distinguished pair {u, v} and let X uv = 1 if {u, v} is bad and X uv = 0 otherwise. Let Y uv = A 0 Nu Nv and let s = Nu Nv ; recall that {u, v} is a distinguished pair and thus s = Θ. Then s PX uv = 1 = PX uv = 1 Y uv = t PY uv = t = t=0 0.9s t=ε PX uv = 1 Y uv = t PY uv = t + O 1, 4.1 where the first equality holds by the law of total probability and the second equality holds for a sufficiently small constant ε > 0 since, by standard tail estimates for the hypergeometric distribution, we have that PY uv ε = O 1 and PY uv 0.9s = O 1.

15 EDGE-STATISTICS ON LARGE GRAPHS 15 Assume without loss of generality that Nu \ Nv Nv \ Nu ; observe that by the definition of a distinguished pair it thus follows that r := Nu \ Nv = Θ. Let Z uv = A 0 Nu \ Nv. Observe that for any ε t s we have PX uv = 1 Y uv = t t+ 0.4 / i=t 0.4 / PZ uv = i Y uv = t. 4. Note that Z uv Y uv = t is a hypergeometric random variable with parameters s, t and r. Since, moreover, EZ uv = rt/s = Θ, a straightforward calculation shows that for every t 0.4 / i t / we have PZ uv = i Y uv = t = O 1/. 4.3 Combining 4.1, 4., and 4.3 we obtain PX uv = 1 = O Now, let X = {u,v} D X uv be the random variable which counts the number of bad pairs in D. It follows by 4.4 that EX = EX uv = O 1.9. {u,v} D Applying Marov s inequality to X we conclude that PX D /6 = O 0.1. Claim 4.5. With probability 1 O 0.1 the set A 0 will have the following property: there exist disjoint sets X, Y V \ A 0 such that X = Y = Θ and dx, A 0 dy, A holds for every x X and y Y. Proof. Recall that D = Θ holds by Claim 4.. By considering an auxiliary graph F with V F = V and EF = {uv : {u, v} D} to which we apply Claim 4.3, we infer that, with probability 1 o 1, at least D /5 of the distinguished pairs are disjoint from A 0. On the other hand, it follows by Claim 4.4 that, with probability 1 O 0.1, at most D /6 of the distinguished pairs are bad. We conclude that, with probability 1 O 0.1, there is a set D D V \A 0 of size c, for a constant c > 0, such that du, A 0 dv, A for every {u, v} D. Let u 1,... u + 0. be an ordering of the vertices of V \ A 0 by non-increasing order of degrees into A 0, that is, du i, A 0 du j, A 0 for every 1 i < j Let q = c/3, let X = {u 1,..., u q }, and let Y = {u + 0. q+1,..., u + 0.}. Observe that X Y = and X = Y = q = Θ. It thus remains to prove that dx, A 0 dy, A holds for every x X and y Y. Suppose for a contradiction that there exist x X and y Y such that dx, A 0 dy, A 0 < 0.4. Since, by definition, dx, A 0 du, A 0 dy, A 0 for every u V \ A 0 X Y, it follows that {u, v} X Y for every {u, v} D. Therefore X Y c = D + X Y V \ A 0 X Y which is clearly a contradiction. qq q c/3 < c, Let A 0 be a set chosen randomly as described above. Let X and Y be disjoint subsets of V \ A 0 such that X = Y = c for some c > 0, and dx, A 0 dy, A holds for every x X and y Y ; such sets exist with probability 1 O 0.1 by Claim 4.5. Now, we choose a random set A 1 V \ A 0 of size 0. as follows. First, we choose 0. pairwise-disjoint vertex pairs from V \ A 0 uniformly at random. Then, from each such pair, we choose uniformly at random exactly one element to be in A 1 ; all choices being mutually independent. Claim 4.6. Let A 1 be chosen randomly as described above and let A = A 0 A 1. Then, for any integer l, we have Pe G A = l = O 0.1.

16 16 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN Proof. Let M denote the set of all 0. randomly chosen pairs. Let M XY denote the set of chosen pairs which have one element in X and one in Y, and let m = M XY. We claim that m = Ω 0. with probability 1 e Ω0.. Indeed, consider choosing the pairs which constitute M one by one. Suppose that we have already chosen j pairs for some 0 j < 0., and now choose the j + 1st pair. Let M j denote the set of all vertices in the union of these j pairs; clearly M j = O 0.. Let T j+1 = 1 if the j + 1st pair has one element in X and one in Y, and T j+1 = 0 otherwise. Then PT j+1 = 1 = X \ M j Y \ M j c 3. Note that m = 0. j=1 T j. Let Z Bin 0., c /3 and observe that, by the above calculation, m stochastically dominates Z. Hence, using standard bounds on the tail of the binomial distribution, we conclude that Pm < c 0. /6 PZ < c 0. /6 PZ < EZ/ < e c 0. /4. Let {x 1, y 1 }, {x, y },..., {x m, y m } be the elements of M XY, where {x 1,..., x m } X and {y 1,..., y m } Y. Fix any choice of one element from every pair in M \ M XY. With any choice of one element from every pair in M XY, we associate a binary vector z = z 1,..., z m in a natural way, namely, for every 1 i m, z i = 1 if we chose x i to be in A 1 and z i = 0 if we chose y i. For each such vector z, we denote the resulting random -subset of V G by A z. For two such vectors z and w, we say that z > w if z i w i for every 1 i m and z w. We claim that e G A z > e G A w whenever z > w. Indeed, let 1 i m be an index for which z i = 1 and w i = 0. Then e G A z e G A w dx i, A 0 dy i, A 0 e G A w \ A > 0. It follows that, for any integer l, the elements of { z {0, 1} m : e G A z = l} form an anti-chain. Hence, by Sperner s Theorem we conclude that Pe G A = l m m/ m = O 1/ m = O 0.1. Claim 4.6 implies that I,, l = O 0.1. Thus indeed ind, l = O 0.1 and the proof of Theorem 1.5 is complete. Note that the proof of Claim 4.6 resembles and was inspired by the Littlewood-Offord problem and its solution by Erdős [7]. 5. When l is fixed: Proof of Theorem 1.6 Our aim in this section is to show that, under certain natural conditions, X G, exhibits a Poisson-lie behaviour. The following proposition maes this precise. Proposition 5.1. Suppose that 1 l n are integers and that G is a graph with n vertices and m edges. Suppose that lim m = µ for some constant µ = µl > 0 and that n := G = o n/. Then PX G, = l = 1 + o 1e µ µl l!. Proof. Throughout this proof the o -notation will solely refer to o ; hence we omit the subscript for readability. By Brun s Sieve see, e.g., Theorem[ in [], in order to prove X that PX G, = l = 1 + o1e µ µl l!, it suffices to show that E ] r = 1 + o1 µr r! holds for every fixed integer 1 r rl.

17 EDGE-STATISTICS ON LARGE GRAPHS 17 Let e 1,..., e m be an arbitrary ordering of the edges of G. For every 1 j m, let X j = 1 if both endpoints of e j are in the random -set A, and X j = 0 otherwise. Observe that X G, = m j=1 X j. In particular, EX G, = m EX j = m j=1 n n = m 1 nn 1. [ X ] It thus follows by the definition of µ that E 1 = EX = 1 + o1µ. Now, fix some positive integer r. Let 1 j 1 <... < j r m be arbitrary indices. Assume first that e j1,..., e jr form a matching in G. Then PX j1 = 1... X jr = 1 = n r r n = 1 + o1 r. n Moreover, the number of ways to choose indices 1 j 1 <... < j r m for which e j1,..., e jr form a matching in G is 1 + o1 m r trivially, it is at most m r. Indeed, by our assumption on, this number is at least r 1 1 m t 1 r! r! t=0 r 1 t=0 m tm/n = mr r! m 1 + o1e r /n = 1 + o1, r where the last equality holds since, by assumption, n is sufficiently large with respect to and r. Next, assume that e j1,..., e jr do not form a matching in G. Let H be the graph whose edges are e j1,..., e jr and whose vertices are the endpoints of e j1,..., e jr. Let C 1,..., C t denote the connected components of H, and let c i = C i for every 1 i t. Then PX j1 = 1... X jr = 1 = n t i=1 c i t i=1 c i n = 1 + o1 t i=1 c i. n Assume without loss of generality that c 1... c t. Since H is not a matching, we must have c 1 3. For all integers a and b a 1, the number of ways to choose b edges of G that form a connected component on a vertices, is at most ma 1! a a b a+1 = { Om if a = m o [ n/ a ] if a > Indeed, we begin by choosing an arbitrary edge of G in m ways, then, one by one, we choose a additional edges to form a tree on a vertices in at most 3... a 1 = a 1! a ways, and then we choose the remaining b a + 1 edges such that their endpoints are among the a vertices of the tree. Hence, the total number of ways to choose r edges of G that form a graph H consisting of connected components of orders c 1... c t, where c 1 3, is at most t [ n ] [ ci n ] t m o = m t i=1 o c i t. i=1

18 18 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN For every 1 t r, let L t denote the set of all integer vectors c 1,..., c t such that c 1... c t, c 1 3, and t i=1 c i < r. Combining everything together we conclude that [ ] X m r E = 1 + o1 1 + o1 r r n r [ n ] t + m t i=1 o c t i t i=1 c i 1 + o1 n t=1 c 1,...,c t L t = 1 + o1 1 m r [ r m t ] r! n + o n t=1 c 1,...,c t L t [ ] µ r = 1 + o1 r! + oµr = 1 + o1 µr r!, where the penultimate equality holds since r, t and µ > 0 are fixed. The first consequence of Proposition 5.1 is that it provides a whole plethora of constructions demonstrating that ind, 1 1/e + o 1. Indeed, let n be a sufficiently large integer and let G be any graph with n vertices, 1 + o 1 n / edges, and maximum degree o n/. Then, G satisfies the assumptions of Proposition 5.1 with µ = 1, implying that ind, 1 1/e + o 1. This rich family of constructions includes, in particular, the random graphs G n, 1 with high probability which were mentioned in the introduction, and the pairwise disjoint union of cliques, on n 1 vertices each. Moreover, by combining Proposition 5.1 with some of our previous arguments as well as some new ones, we can prove Theorem The 1/-bound for l = 1. Let be a sufficiently large integer, let a = ind, 1, and let G be a graph on n vertices, attaining the maximum density of induced one-edged graphs, that is, PX G, = 1 = a + o n 1. Our aim is to show that either a 1/ + o 1, or that the conditions of Proposition 5.1 are satisfied. In the latter case we will be done, as applying Proposition 5.1 would imply a 1/e + o 1. Claim 5.. G 10n/. Proof. Let v be a vertex of maximum degree in G and suppose for a contradiction that dv > 10n/. It follows by Lemma.1 and the discussion succeeding the proof of Proposition 5.1 that PX G, = 1 v A = a + o n 1 1/e + o 1. However, a straightforward calculation which uses the Poisson-approximation shows that PX G, > 1 v A P A N G v > 1 v A > 1 e 10 10e 10 + o 1 which is an obvious contradiction. > 1 PX G, = 1 v A, Claim 5.3. If G = o n/, then a 1/e + o 1. Proof. Recall that eg = Ω n / holds by Lemma.6. Assume first that eg 0n /. By compactness we may assume that m = 1+o 1µ n for some constant 0 < µ 0, whereby the conditions of Proposition 5.1 are satisfied. Applying it yields a + o n 1 = PX G, = 1 = 1 + o 1µe µ 1/e + o 1. Assume then that eg > 0n /. Let G be an arbitrary subgraph of G with 0n / edges. Since G G = o n/, the graph G satisfies the conditions of Proposition 5.1 with µ = 0. Therefore a + o n 1 = PX G, = 1 PX G, 1 = 1e 0 + o 1 1/e + o 1.

19 EDGE-STATISTICS ON LARGE GRAPHS 19 By Claims 5. and 5.3 and by compactness, we may assume that G = cn/ for some constant c > 0. Let v be a vertex of maximum degree in G, let Q = N G v, and let R = V G\Q {v}. By the Poisson-approximation of the hypergeometric distribution, with probabilities of approximately e c and ce c, respectively, A will contain exactly 0 or 1 vertices from Q. Therefore, invoing Lemma.1, we obtain a + o n 1 = PX G, = 1 v A = PX G, = 1 v A and A Q = 0 P A Q = 0 v A + PX G, = 1 v A and A Q = 1 P A Q = 1 v A 1 + o 1 PX G[R], 1 = 1 e c + PX G[R], = 0 ce c. 5.1 Claim 5.4. For any graph H and any positive integers and t we have PX H, 1 = t t PX H, = t. Proof. Sample uniformly at random a vertex set A 1 V H of size 1, using the following two steps. First sample vertices uniformly at random and without replacement, obtaining a set A V H. Then, choose a vertex u A uniformly at random and put A 1 = A \ {u}. Observe that, if in the first step we sampled a t-edge graph, then the probability to destroy it in the second step is at most t/. Therefore PX H, 1 = t PeG[A 1 ] = t eg[a ] = t PeG[A ] = t t Claim 5.5. For any graph H and any positive integers and t we have PX H, 1 = t PX H, = t + t +. PX H, = t. Proof. Consider the same two-step sampling as in the proof of Claim 5.4. Note that, if in the first step, the set A contains more than t edges, then the probability to obtain a t-edge graph in the second step is at most t + / this is because, for every s 1, a graph with t + s edges contains at most t + vertices of degree s. Thus, PX H, 1 = t = PX H, = i PX H, 1 = t eg[a ] = i i=0 PX H, = t + i=t+1 PX H, = t + t + PX H, = t + t +. PX H, = i PX H, 1 = t eg[a ] = i i=t+1 PX H, = i Let b denote PX G[R], = 1. Then, applying Claim 5.4 to 5.1, we obtain Observe that, if c 1, then a 1 + o 1 PX G[R], 1 = 1 e c + PX G[R], = 0 ce c 1 + o 1 PX G[R], = 1 e c + PX G[R], = 0 ce c 1 + o 1 be c + 1 bce c. a 1 + o 1 be c + 1 bce c 1 + o 1 bce c + 1 bce c 1/e + o 1.

20 0 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN Assume then that c 1. Observe that, in this case, be c +1 bce c is an increasing function in b. Moreover, b 1 + o 1PX G[R], = o 1a, where the first inequality holds by Claim 5.5 applied twice, and the second inequality holds since a = ind, 1 and R is large. Hence, This implies that 1 + o 1a be c + 1 bce c ae c + 1 ace c. a 1 + o 1 1 e c + ce c = 1 + o c 1 e c 1 + c 1/ + o 1. ce c 5.. The 3/4-bound for a fixed l. We use a similar, but slightly more technical, argument to the one used in the case l = 1. Let l be a positive integer and let l. Let a = ind, l, and note that a is bounded away from 0 by a constant εl. This can be seen, for instance, by considering an appropriate random graph. Our aim is to prove that a 3/4 + o 1. Let G be a graph on n vertices, attaining the maximum density of induced l-edged graphs, that is, PX G, = l = a + o n 1. Using analogous arguments to the ones used to prove Claims 5. and 5.3, we may assume that G = cn/ for some constant c > 0. Let v be a vertex of maximum degree in G, let Q = N G v, and let R = V G \ Q {v}. By the Poisson-approximation of the hypergeometric distribution, for each fixed t, with probability of approximately e c c t /t!, the set A will contain exactly t vertices from Q. It thus follows by Lemma.1 that a + o n 1 = PX G, = l v A l = PX G, = l v A and A Q = t P A Q = t v A t=0 1 + o 1 PX G[R], 1 = l e c + l t=1 PX G[R], 1 t l t ct t! e c. 5. Let b denote PX G[R], l 1 = l. Then, applying Claim 5.4 to 5. a constant depending only on l number of times, we obtain l a 1 + o 1 PX G[R], 1 = l e c + PX G[R], 1 t l t ct t! e c t=1 l 1 + o 1 PX G[R], l 1 = l e c + PX G[R], l 1 l t ct t! e c. t=1 l 1 + o 1 be c c t + 1 b t! e c. Assume first that c < log. In this case we have l c t t! < ec 1 < 1 = 1, t=1 t=1 which implies that be c + 1 b l t=1 e c c t /t! is an increasing function in b. Now observe that b 1 + o 1PX G[R], = l 1 + o 1a, where the first inequality holds by Claim 5.5 applied l + 1 times, and the second inequality holds since a = ind, l and R is large. Hence, 1 + o 1a be c + 1 b l t=1 c t t! e c ae c + 1 a l t=1 c t t! e c.

21 This implies that 1 + o 1a EDGE-STATISTICS ON LARGE GRAPHS 1 e c l t=1 ct /t! 1 e c + e c l t=1 ct /t! = l t=1 ct /t! e c 1 + l t=1 ct /t! < l t=1 ct /t! l t=1 ct /t! = 1 < 3 4. It thus remains to consider the case c log. Recall that d G v = G = cn/ and let w be a vertex of minimum degree in G. Claim 5.6. d G w = δg = o n/. Proof. Suppose that δg αn/ for some constant α = αl > 0. Let G be a subgraph of G obtained by selecting each edge of G independently at random with probability β/, where β = βl, α is a sufficiently large constant. Since n, with high probability all degrees in G will be between 1 o1αβn/ and 1+o1cβn/, and by compactness we may assume that eg = 1 + o 1µn / for some constant µ = µl, α, β > αβ/3 > 0. Hence G satisfies the conditions of Proposition 5.1, implying that the distribution of X G, is approximately Poisson with parameter µ. However, since β was chosen to be sufficiently large, this means that A G,, and a fortiori A G, will induce more than l edges with probability 1 o 1. Now, let B = A G\{v,w}, 1 denote a subset of V G \ {v, w} of size 1, chosen uniformly at random among all such subsets. Assuming that a > 1/ as otherwise we are done, we apply Lemma. to v and w to infer P e G v, B = e G w, B > a 1 o n 1. However, since d G w = o n/ holds by Claim 5.6 and d G v log n/ holds by assumption, a straightforward calculation shows that Pe G w, B = 0 = 1 o 1 and Pe G v, B = o 1e log = 1/ + o 1. It follows that 1/ + o 1 > a 1 o n 1, implying a 3/4 + o 1, as claimed. 6. Concluding remars Bounds on ε in Theorem 1.3. While, for clarity of presentation, we did not mae an effort to calculate ε explicitly, it is not difficult to see that this constant is not too small. when is sufficiently large, the value ε = 1/100 is certainly sufficient, and with some care, one can improve it to ε = 1/10. On the other hand, looing at some of our arguments, it is evident that in order to go below 1/ for every sufficiently large and every 0 < l <, one would need new ideas. Upper bounds for fixed l. It is not hard to see that our argument in the proof of Theorem 1.6 gives in fact better bounds than 1/ + o1 for l = 1 and 3/4 + o1 in the general case. Tae for example the case l = 1. For G = cn/ we obtain c a 1 + o1 e c 1 + c, where the right hand side attains its maximum of 1/ when c is close to zero. On the other hand, when G = on/ Proposition 5.1 Brun s sieve implies that a 1/e + o1. Interpolating between these two arguments shows that indeed ind, 1 < 1/ Ω1. Strengthening Conjecture 1.1 in some ranges. Theorem 1.5 demonstrates that the assertion of Conjecture 1.1 holds, with room to spare, for almost all values of and l. In particular, ind, l tends to 0 with as long as l and l are quadratic in. We believe that in the above statement quadratic can be replaced with super-linear. { Conjecture 6.1. For all pairs, l satisfying min o1. l, l } = ω, we have ind, l =

22 NOGA ALON, DAN HEFETZ, MICHAEL KRIVELEVICH, AND MYKHAYLO TYOMKYN On the other hand, our construction for l = 1 can be straightforwardly extended to show that for any fixed integer C we have ind, C C = Ω1. Note also that the argument appearing in the proof of Theorem 1.5 can be applied in a wider { range, namely, it can be used to prove that ind, l is polynomially small in whenever min l, } l = Ω δ for some explicit constant δ > 0. It would be interesting to determine the correct power of in various ranges. For instance, when l and l are quadratic in, Theorem 1.5 yields ind, l = O 0.1, whereas, for some values of l = Θ, we have ind, l = Ω 1/ tae for instance l = /4 and G = K n/,n/. We believe that the latter bound is tight. { Conjecture 6.. For all pairs, l satisfying min l, } l = Ω, we have ind, l = O 1/. Minimum edge-inducibility. A natural counterpart to Conjecture 1.1 would be to determine the asymptotic value of { } η := min ind, l : 0 l, as tends to. Note that this question has been well-studied in the setting of graphinducibilities [10, 11, 16, 17]. In particular, it is nown [16, 17] that min{indh : H = } = 1 + o1! = e +o1. This is in star contrast with the lower bound of η = Ω1/ for edge-inducibilities, achieved by the random graphs G n, l/. 3/4 as a general upper bound. As noted in the introduction, ind, l < 1 for every positive integer and every 0 < l <. It seems plausible that in fact ind, l 3/4 for all such pairs, l. Note that, if true, this bound would be tight since, as noted in the introduction, ind3, 1 = ind3, = 3/4. Such a result would simultaneously improve Theorems 1.3 and 1.6. Hypergraphs. The concepts of both graph- and edge-inducibility extend naturally to r-uniform hypergraphs. Since the lower bound constructions of 1/e for l = 1 and for the star K 1, 1 extend to higher uniformities as well, it maes sense to as if Conjectures 1.1 and 1. would also hold in this more general setting. Needless to say that we expect these questions to be difficult. Further questions. The constant of 1/e, being closely related to the Poisson distribution, maes an appearance in many combinatorial and probabilistic setups. In pariticular, we would lie to draw the reader s attention to the interesting conjectures of Feige [8] and Rudich see [13]. Addendum In the period when this paper was under review, our results have been extended in several directions. In particular, The Edge-Statistics Conjecture Conjecture 1.1 was proved in the superlinear regime by Kwan, Sudaov and Tran [14]. In the sublinear regime the conjecture was proved by Fox and Sauermann [9] and, independently, by Martinsson, Mousset, Noever and Trujić [15]. Thus, the results of [14], combined with either [9] or [15] prove the assertion of Conjecture 1.1. Some questions which remain open can be found in these papers. Acnowledgement We than the anonymous referee for carefully reading our paper and for providing helpful remars.