Empirical Distributions of Laplacian Matrices of Large Dilute Random Graphs. Tiefeng Jiang 1

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1 Empirical Distributions of Laplacian Matrices of Large Dilute Random Graphs Tiefeng Jiang Abstract We study the spectral properties of the Laplacian matrices and the normalized Laplacian matrices of the Erdös-Rényi random graph G(n, p n ) for large n Although the graph is simple, we discover some interesting behaviors of the two Laplacian matrices In fact, under the dilute case, that is, p n (0, ) and np n, we prove that the empirical distribution of the eigenvalues of the Laplacian matrix converges to a deterministic distribution, which is the free convolution of the semicircle law and N(0, ) However, for its normalized version, we prove that the empirical distribution converges to the semi-circle law Introduction In graph theory, the Erdös-Rényi model G(n, p), named for Erdös-Rényi [7, 8, 9, 20], is a graph with n vertices; a potential edge between each pair of vertices has probability p, independently of the other edges Some properties of G(n, p) such as the diameters, the sizes and the giant components are known, see, eg, [5, 9, 7, 8, 9, 20] for details For the spectral properties of the graphs, one can see, eg, [9, 0,, 5, 27, 3] The graph G(n, p n ) corresponds to Bernoulli random variables {ξ (n) ; i < j n}, which are independent random variables with ξ (n) = ξ (n) ji and P (ξ (n) = ) = P (ξ (n) = 0) = p n for all i < j n The adjacency matrix A n and Laplacian matrix n are defined as follows: 0 ξ (n) 2 ξ (n) n ξ (n) 2 0 ξ (n) 2n A n = ξ (n) n ξ (n) n2 0 () and n = j ξ(n) j ξ (n) 2 ξ (n) n ξ (n) 2 ξ (n) n j 2 ξ(n) 2j ξ (n) 2n ξ (n) n2 j n ξ(n) nj (2) Supported in part by NSF #DMS , School of Statistics, University of Minnesota, 224 Church Street, MN55455, tjiang@statumnedu Key Words: random matrix, random graph, dilute graph, Laplacian matrix, normalized Laplacian matrix, spectral distribution, semi-circle law, free convolution AMS (2000) subject classifications: 05C80, 05C50, 5A52, 60B0

2 The normalized Laplacian matrix is defined by L n = I n D /2 n A n D /2 n, (3) where D n = ( n j i ξ(n) ) i n and D /2 n = (( n j i ξ(n) ) /2 ) i n with 0 /2 := 0 The matrix n is always non-negative definite, and the smallest eigenvalue is zero The Kirchhoff theorem [30] establishes the relationship between the number of the spanning trees of the graph and the eigenvalues of n The matrix L n is a different version of n It looks a bit complicated, however, the spectrum of L n is related to the graph discrepancy (or the graph invariant), and the second smallest eigenvalue of L n relates to the Cheeger constant and the rate convergence of random walks on the graph, see, eg, [8, 9] The spectral properties of the graph-relevant matrices such as A n, n and L n have applications in chemistry [4], where eigenvalues were connected to the stability of molecules They are also studied and used in theoretical physics and quantum mechanics, eg, [2, 22, 23, 24, 25, 36, 37] For graph G(n, p n ) there are two major asymptotic regimes: np n and np n c (0, ) The former corresponds to the dilute model, and the latter corresponds to the sparse model Of course, the dilute model consists of two important cases: p n p and p n 0 with np n There are several work on n Bauer and Golinelli [3] simulate the eigenvalues for the Erdös- Rényi random graph with p n p Assuming the entries {ξ (n) ξ ; i < j n < } in n in (2) are independent and identically distributed random variables with mean zero, Bryc, Dembo and Jiang [6] prove that the empirical distribution of the eigenvalues of n converges to the free convolution γ M of the semi-circular law and N(0, ) This results does not apply to the Erdös-Rényi random graph directly because the means of ξ s in the graph are not zero A result from Ding and Jiang [5] complements their result in [6] by showing that, under a general framework including the Erdös-Rényi random graph G(n, p n ) with p n p, the corresponding empirical distribution converges to γ M When ξ (n) s in (2) are deterministic, some results are given for the normalized Laplacian matrix L n in [8] In this paper, for the Erdös-Rényi random graph G(n, p n ) with np n, we study the empirical distributions of the eigenvalues of n and L n Before stating the main results, we review some notation Given an n n symmetric matrix M, let λ λ 2 λ n be the eigenvalues of M, we sometimes also write this as λ (M) λ 2 (M) λ n (M) for clarity The notation λ max = λ max (M), λ min = λ min (M) stand for the largest eigenvalue and the smallest eigenvalue of M, respectively The following is the situation we will consider in this paper Let {ξ (n) ; i < j < } be defined on the same probability space with ξ(n) ii = 0 and ξ (n) for all i, j n < For each n 2, assume {ξ (n) ; i < j n} are iid with P (ξ (n) 2 = ξ (n) ji = ) = P (ξ(n) 2 = 0) = p n (0, ) (4) For an n n matrix M, we use M = sup x Rn : x = Mx to denote its spectral norm, where x = x x2 n for x = (x,, x n ) R n The following are the main results of this paper 2

3 THEOREM Suppose (4) holds with α n := (np n ( p n )) /2 as n, then, almost surely, F n (x) := n n I i= { } λ i ( n ) np n npn ( p n ) x, x R, (5) converges weakly to the free convolution γ M of the semi-circle law and N(0, ) It is known from [6] that the measure γ M is a non-random symmetric probability measure with smooth bounded density, doesn t depend on p n s and has an unbounded support Theorem 3 from [6] and Theorem 2 from [5] deal with similar problems by assuming ξ (n) s in (2) being arbitrary random variables of certain finite moments With truncations on ξ (n) s, the proofs of the two theorems from [6, 5] are essentially reduced to the case that ξ (n) s are bounded Here the situation is different: we will need to work on (ξ (n) p n )/ p n s, which are unbounded since p n 0 in one of the dilute cases mentioned earlier The truncation method used in [6, 5] does not apply here simply because the current random variables just take two values In fact, more involved analysis and combinatorics are used in the proof of Theorem If p n in Theorem is equal to a constant, we get the following result COROLLARY Suppose (4) holds with p n p (0, ) for all n 2 Then, almost surely, { } F n (x) := n λ i ( n ) np I x, x R, n np( p) i= converges weakly to the free convolution γ M of the semi-circle law and N(0, ) Take p n in Theorem to be a special dilute case: p n 0 with np n + as n Note that, under this condition, np n ( p n ) np n as n, we then easily have the following COROLLARY 2 Suppose (4) holds with p n 0 and np n as n, then, almost surely, F n (x) := n n i= { } λi ( n ) np n I x, x R, npn converges weakly to the free convolution γ M of the semi-circle law and N(0, ) Now we study the normalized Laplacian matrix L n in (3) THEOREM 2 Suppose (4) holds with sup{p n ; n 2} < If np n / log n as n, then, with probability one, F n (x) := n n i= ( ) npn I ( λ i (L n )) x, x R, p n converges weakly to the semi-circle law with density 2π 4 x2 I( x 2) 3

4 Fan, Lu and Vu [] derive a result for a weak form of L n when the graphs are of the given expected degrees, of which the Erdös-Rényi model is a special case The three authors prove that /2 the the empirical distribution of the eigenvalues of I n D n A D /2 n n converges to the semi-circle law in probability, where D n is the mathematical expectation of D n When p n does not depend on n, the above theorem obviously implies the following COROLLARY 3 Suppose (4) holds with p n p (0, ) for all n 2 Then, with probability one, F n (x) := n n i= ( ) np I p ( λ i(l n )) x, x R, converges weakly to the semi-circle law with density 2π 4 x2 I( x 2) If p n 0, then npn p n np n as n From Theorem 2 we immediately obtain a limiting result for the dilute Erdös-Rényi random graph COROLLARY 4 Suppose (4) holds with p n 0 and np n / log n as n, then, almost surely, F n (x) := n n I ( np n ( λ i (L n )) x), x R, i= converges weakly to the semi-circle law with density 2π 4 x2 I( x 2) In the general context of random graphs, the limits of the empirical distributions of the eigenvalues of A n, n and L n in (), (2) and (3) respectively are related to the three distributions: the Wigner s semi-circle law, the free convolution of the semi-circle law and N(0, ), and the Kesten- McKay law The last is the limit of the empirical distributions of the eigenvalues of A n for the random d-regular graphs, see [33] REMARK The above theorems study the limiting spectral distributions of n and L n for the dilute case There are some simulations and theoretical work on A n and n for the sparse case in [3, 26, 29, 32, 36, 37, 38], that is, p n = c/n for all n 2 and c is a constant It seems that the limits of the spectral distributions of A n and n are still not identified REMARK 2 Theorems and 2 are derived based on the Erdös-Rényi model The methods of the proofs could be carried to more complex models, for instance, the random geometric graphs [34], the weighted dilute random matrices [28], the power law random graphs, and the random graphs with given expected degrees [9] REMARK 3 There are a couple of books on the study of the spectral properties of the matrices generated from graphs, see, eg, [, 8, 9, 2, 3, 35] for reference The rest of the paper is organized as follows: the proof of Theorem is given in Section 2; the proof of Theorem 2 is given in Section 3 4

5 2 Proof of Theorem Given n 2, let Γ n = {(i, j); j < i n} be a graph Any element (i, j) Γ n is called a vertex, and i and j are called indices We say vertices a = (i, j ) and b = (i 2, j 2 ) form an edge and denote it by a b, if one of i and j is identical to one of i 2 and j 2 For convenience of notation, from now on, we write a = (a +, a ) for any a Γ n Of course, a + > a Given a, b Γ n, define an n n matrix, if i = a +, j = b +, or i = a, j = b ; Q a,b [i, j] =, if i = a +, j = b, or i = a, j = b + ; 0, otherwise Recall (4) Set η (n) = ξ(n) p n pn ( p n ) (2) for all i j n and n 2, and Ψ n = j η(n) j η (n) 2 η (n) n η (n) 2 η (n) n j 2 η(n) 2j η (n) 2n η (n) n2 j n η(n) nj (22) With this and the above notation, we rewrite Ψ n as follows Ψ n = a Q a,a, (23) where η (n) a a Γ n η (n) = η (n) a + a for a = (a +, a ) Γ n Let t a,b = tr(q a,b ) We summarize some facts from [6] in the following lemma LEMMA 2 Let a, b Γ n The following assertions hold: (i) t a,b = t b,a 2, if a = b;, if a b and a = b or a + = b + ; (ii) t a,b =, if a b and a = b + or a + = b ; 0, otherwise (iii) Q a,b Q c,d = t b,c Q a,d Therefore, tr(q a,a Q a2,a 2 Q ar,a r ) = r j= t a j,a j+, where a,, a r Γ n, and a r+ = a Given n 2, we call π = (a,, a r ) a circuit of length r in Γ n if a i Γ n for i n and a a r a For such a circuit, set η (n) π = ( r ) r t aj,a j+ j= j= η (n) a j (24) 5

6 with a r+ = a From (23) and (iii) of Lemma 2, we know tr(ψ r n) = ( ) r π η (n) π and E tr(ψ r n) = ( ) r π Eη (n) π (25) where the sum is taken over all circuits of length r in Γ n DEFINITION 2 We say that a circuit π = (a a r a ) of length r in Γ n is vertexmatched if for each i =,, r there exists some j i such that a j = a i ; we say π has a match of order 3 if some value is repeated at least three times among {a j, j =,, r} Clearly, by independence, the only possible non-zero terms in (25) come from vertex-matched circuits For x 0, denote by x the integer part of x The following two lemmas will be used later LEMMA 22 (Proposition 40 from [6]) Fix r N Let N denote the number of vertex-matched circuits in Γ n with vertices having at least one match of order 3 Then N = O(n (r+)/2 ) as n Recall Γ n = {(i, j); j < i n} for n 2 Let m, r, s,, s m be positive integers with r m and s + + s m = r Define A m,n,r by {a a 2 a r a : a i Γ n for i n; there exist pairwise different a,, a m {a,, a r } such that a i appears s i times in the sequence a a 2 a r for i m} (26) LEMMA 23 Let m, r, s,, s m be positive integers with r m and s + + s m = r Then, there exists a constant C r depending only on r such that A m,n,r C r n m+ for all n 2 Proof First, denote by B the total number of ways to assign m different items, say, ω,, ω m, into a sequence of length r such that ω i appears exactly s i times for each i m Then B = ( ) r s,,s l r! Given one of such assignment, recalling (26), the number of choices of a = (a +, a ) Γ n is at most n 2 Then, each time a new vertex a i = (a+ i, a i ) appears in the circuit sequence, one index from {a + i, a i } is free, while the other is identical to one of {a+, a } s which appear in the circuit sequence earlier than a i Thus, the total number of the choices of a i is bounded by 2r n + 2 r n = 2 r n It follows that the total number of the circuits corresponding to the assignment is no more than n 2 (2 r n) m 2 r2 n m+ Take C r = 2 r2 r! The conclusion then follows by combining the two steps Recall vertex a = (a +, a ) Γ n has two end points a + and a DEFINITION 22 Fix integer n 2 Let π, π 2, π 3, π 4 be four circuits of length r in Γ n, we say they are matched if each vertex of any one of these circuits is either self-matched, that is, there is another vertex of the same circuit with the same end points, or is cross-matched, that is, there is a vertex of the other circuit with the same end points (or both) Given l N, let Q l be the number of the above matched quadruples satisfying: (a) for any i =, 2, 3, 4, there is j i such that π i and π j share a common vertex; (b) the total number of different vertices in the four circuits is l 6

7 LEMMA 24 For fixed l, r N with l 2r, there exists a constant C r not depending on n, such that Q l C r n l+2 for all n 2 Proof Write π = a a 2 a r a ; (27) π 2 = b b 2 b r b ; π 3 = c c 2 c r c ; π 4 = d d 2 d r d For i j, denote by π i π j if π i and π j have at least one crossed match Up to the permutations of π, π 2, π 3, π 4, all matched quadruples satisfying (a) above the statement of the lemma can be divided into three cases: (i) π π 2 and π 3 π 4 ; (ii) π π 2, π π 3 and π 2 π 4 ; (iii) π π 2, π π 3 and π π 4 (By permutations we mean that the circuits π, π 2, π 3, π 4 in (i), (ii) and (iii) can be π i, π i2, π i3, π i4 for any permutation {i, i 2, i 3, i 4 } of {, 2, 3, 4}) Thus, to prove the lemma, it suffices to show that the total number of the circuits in each of the three cases are no more than Cn l+2 for some constant C not depending on n To do so, since each of the four circuits has length r, it is enough to show the total number of the vertices in the four circuits is bounded by Cn l+2 with C not depending on n Case (i): π π 2 and π 3 π 4 Suppose there are l different vertices among circuits π and π 2, then there are l 2 = l l 0 different vertices in π 3 and π 4 that do not appear in π and π 2 Since π π 2, recalling (27), say, (a +, a ) = a b Then, there are at most n(n )/2 ways to assign {,, n} to a + and a with a+ > a Now assign such numbers to other l vertices, there are at most 2n assignment for each of a new vertex since two distinct neighbors are connected So the total number of vertices in π and π 2 is no more than n 2 (2n) l 2 l n l+ Similarly, there are at most 2 l 2 n l2+ different vertices in circuits π 3 and π 4 So the total number of vertices in circuits π, π 2, π 3, π 4 is bounded by 2 l n l+ 2 l2 n l2+ = 2 l n l+2 Cases (ii) and (iii) The difference between the two cases and case (i) is that the four circuits here are all connected, while (π, π 2 ) and (π 3, π 4 ) in case (i) may not have a common vertex at all For i =, 2, 3, 4, let u i be the number of the distinct vertices in π i that do not appear in π j for j < i Then u + u 2 + u 3 + u 4 = l We proceed to assign (i, j) s with i > j to the vertices in π first, then π 2, π 3, π 4 consecutively By the same consideration as that in (i), the total number of possible vertices in π is at most n u+ Since for any i = 2, 3, 4, there exists j =, 2, 3 such that π j π i, then the total number of different vertices in π k is at most (2n) u k for k = 2, 3, 4 Therefore, the total number of the distinct vertices in the four circuits is at most n u + (2n) u2 (2n) u3 (2n) u 4 2 l n l+ LEMMA 25 Let m, n N Let X,, X n be non-negative, independent random variables Set g(a,, a n ) = E (X a Xa n n ) < for non-negative integers a,, a n with 0 0 = Then m g(a,, a n ) g(a j,, a nj ) j= 7

8 where {a ; i n, j m} are nonnegative integers satisfying m j= a = a i for all i n Proof Given random variable Y 0 and non-decreasing functions f(x) and g(x) on [0, ) If E f(y ) <, E g(y ) < and E f(y )g(y ) <, then, by the covariance inequality, we have that E(f(Y )g(y )) Ef(Y ) Eg(Y ) Thus, for non-negative integers b, b,, b m with b = b + + b m, we obtain E(X b i ) E(X b i for any i n By independence, n g(a,, a n ) = E (X ai i ) and i= ) E(X b 2+ +b n i ) n j= m g(a j,, a nj ) = j= E(X b j i ) (28) n m i= j= E ( ) X a i From (28), E (X a i i ) m j= E ( X a ) m i since j= a = a i for all i =, 2,, n Then the conclusion follows Recall from (25) that n r/2 E tr(ψ r n) = ( ) r n r/2 π Eη(n) π where the sum is taken over all circuits of length r in Γ n We next show that some terms in the sum are negligible LEMMA 26 Suppose the conditions in Theorem hold Fix r N Let W n = n r/2 π Λ n Eη (n) π, where Λ n is the set of vertex matched circuits of order 3 and of length r in Γ n Then W n 0 as n Proof Take a circuit in Λ n : a a 2 a r a Let a,, a m {a,, a r } be pairwise different vertices with a i appearing s i 2 times in the sequence a a 2 a r for i m with s + + s m = r and max i m s i 3 Evidently, m (r )/2 By (24) and independence, Eη (n) π = ( r ) m t aj,a j+ E j= (η (n) a i i= ) s i 2 r m i= E( η (n) a s i ), i where the last inequality is from (ii) of Lemma 2 Now, let Z Ber(p n ) and q n = p n Notice p s i n q n + p n q s i n 2p n q n, we obtain from (2) that E( η (n) a s i ) is identical to i ( ) si Z pn E pn q n = ps i n q n + p n q s i n (p n q n ) s i/2 2 (p n q n ) s i/2 (29) Using s + + s m = r to see that Eη π (n) 2 r 2 m (p n q n ) m (r/2) Let C m,r be the number of solutions of the equation s + + s m = r with integer s i 2 for each i m Now we classify the sum π Λ n Eη π (n) into the sums corresponding to m =, 2,, (r )/2 By Lemma 23, the m-th sum is bounded by C m,r C r n m+ 2 m+r (p n q n ) m (r/2) It follows that W n C n r/2+ m,r C r n m+ 2 m+r (p n q n ) m (r/2) m (r )/2 ( ) C (np n q n ) m (r/2) = O 0 npn q n m (r )/2 8

9 as n since np n q n by assumption, where C above is a constant not depending on n Let U n be a symmetric matrix of form j Y j Y 2 Y n Y 2 j 2 U n = Y 2j Y 2n Y n Y n2 j n Y nj where {Y ; i < j < } are iid standard normal random variables not depending on n (20) LEMMA 27 Suppose the conditions in Theorem hold Let Ψ n and U n be as in (22) and (20) respectively Then (i) lim n (ii) lim n for any integer k E tr(ψ2k nk+/2 n ) = 0; n k+ ( E tr(ψ 2k n ) E tr(u 2k n ) ) = 0 Proof (i) As in (25), E tr(ψ 2k n ) = π Eη(n) π where the sum is taken over all circuits of length 2k in Γ n in which each vertex appearing at least twice Notice r = 2k is odd, one vertex has to appear at least three times The result then follows from Lemma 26 (ii) Recall (20) For a circuit a a r a with a i Γ n for all i r, similar to notation η π (n) in (24), we define Taking r = 2k, we have Y (n) π E tr(ψ 2k n ) E tr(u 2k n ) = π = ( r ) r t aj,a j+ Y aj (2) j= (E η (n) π j= E Y (n) π ) E η π (n) + E Y π (n) + (E η π (n) E Y π (n) ) π A π A π A 2 := I + I 2 + I 3, (22) where A denotes the set of the vertex-matched circuits with match of order 3, and A 2 denotes the set of the vertex-matched circuits in Γ n such that there are exact k distinct matches Now we analyze the three terms one by one By Lemma 26, I = o(n k+ ) as n Observe that each vertex of any circuit in A 2 matches exactly two times From the independence assumption and that E η (n) 2 = E Y 2 = for all i < j n and n 2, we know E η π (n) = E Y π (n) = for all π A 2 This gives I 3 = 0 Now we turn to estimate I 2 Recalling (2), for π A, 2k j= Y a j has the form of w s= V l s s, where V,, V w are iid N(0, )-distributed random variables, and l 2,, l w 2 with w s= l s = 2k Thus, by the Hölder 9

10 inequality, E 2k j= Y a j E( N(0, ) 2k ) From (ii) of Lemma 2, E Y π (n) 2 2k E( N(0, ) 2k ) It follows that I 2 2 2k E( N(0, ) 2k ) A = O(n k ) as n by Lemma 22 In summary, I + I 2 + I 3 = o(n k+ ) as n, which together with (22) concludes (ii) LEMMA 28 Suppose the conditions in Theorem hold Let Ψ n be as in (22) Then for any integer r lim n n r/2+ (tr(ψr n) Etr(Ψ r n)) = 0 as Proof By (24) and (25), tr(ψ r n) Etr(Ψ r n) = ( ) r π (η (n) π Eη (n) π ) where the sum is taken over all circuits of length r in Γ n, and η (n) π = r r t aj,a j+ j= j= for π = (a,, a r ) We claim that, to prove the lemma, it is enough to show as n In fact, if this holds, by the Markov inequality, P η (n) a j E tr(ψ r n) Etr(Ψ r n) 4 = O(n 2r+2 ) (23) ( tr(ψ r n n) Etr(Ψ r n) ) ϵ E tr(ψ r n) Etr(Ψ r n) 4 ( ) r/2+ n 2r+4 ϵ 4 = O n 2 as n for any ϵ > 0 Thus, the sum of the left hand side of the above over all n 2 is finite, the Borel-Cantelli lemma then gives the desired conclusion Now, = E tr(ψ r n) Etr(Ψ r n) 4 { (η π (n) π,π 2,π 3,π 4 E Eη π (n) )(η π (n) 2 Eη π (n) 2 )(η π (n) 3 } Eη π (n) 3 )(η π (n) 4 Eη π (n) 4 ) (24) where the sum is taken over all circuits π j, j =, 2, 3, 4 of length r each For every n 2, with random variables η a (n) s independent and of mean zero (recalling (2)), if a circuit π k has a vertex that is not matched with any other vertex in the four circuits, then Eη π (n) k = 0 and E 4 (η π (n) j j= Eη π (n) j ) = E{η π (n) k (η π (n) j j k Eη (n) π j )} = 0 Further, if one of the circuits, say π, is only self-matched, ie, has no cross-matched vertex, then obviously E 4 (η π (n) j j= Eη π (n) j ) = E(η π (n) Eη (n) π ) E 4 (η π (n) j j=2 Eη (n) π j ) = 0 0

11 Therefore, it suffices to take the sum in (24) over all matched quadruples of circuits on π, π 2, π 3, π 4 of length r (see Definition 22) with the property: for any i =, 2, 3, 4, there is j i such that π i and π j share the same vertex Recalling the definition of Q l in Lemma 24, since the quadruples are matched, we know l 2r, and by this lemma, Q l C r n l+2 for all l 2r, where C r is a constant depending on r only Thus, by (24), E tr(ψ r n) Etr(Ψ r n) 4 where h(π, π 2, π 3, π 4 ) = E 4 j= (η(n) π j 2r l= (π,π 2,π 3,π 4 ) Q l h(π, π 2, π 3, π 4 ), (25) Eη π (n) j ) Expand the expectation, there are 6 terms, and the absolute value of each is bounded by E 4 j= η(n) π j by Lemma 25 Thus, by (25), E tr(ψ r n) Etr(Ψ r n) 4 6C r 2r n l+2 l= max (π,π 2,π 3,π 4) Q l 4 E η π (n) j j= (26) Given l, let Z be a Ber(p n )-distributed random variable Then, for each (π, π 2, π 3, π 4 ) Q l, by independence, E 4 j= η (n) π j max l i= ( ) mi Z pn E (27) pn q n where the maximum is over all m,, m l subject to the constraint m + + m l = 4r By (29), the above is controlled by l i= ( ) mi Z pn E pn q n It follows from (26) and (27) that l i= 2 (p n q n ) = 2 l m i/2 (p n q n ) 2r l E tr(ψ r n) Etr(Ψ r n) 4 2r 2 l n l+2 6C r (p n q n ) 2r l l= 2r (6C r 2 2r )n 2r+2 (np n q n ) 2r l = O(n 2r+2 ) l= as n since np n q n by assumption This gives (23) For an n n symmetric matrix M, let F M be the empirical distribution of the eigenvalues of M LEMMA 29 Suppose α n := (np n ( p n )) /2 as n Let Ψ n be as in (22) Then, as n, with probability one, F Ψ n/ n converges weakly to the free convolution γ M of the semi-circle law and N(0, )

12 Proof Proposition A3 from [6] says that γ M is a symmetric distribution and uniquely determined by its moments Thus, to prove the theorem, it is enough to show that n r/2+ tr(ψr n) = n tr(n /2 Ψ n ) r = x r df n /2 Ψ n x r dγ M as (28) as n for any integer r First, since γ M is symmetric, we know x 2k dγ M = 0 for any integer k Second, recalling U n in (20), Proposition 43 in [6] says that n E tr((n /2 U n ) 2k ) x 2k d γ M as n for any k Consequently, the two facts together with Lemmas 27 and 28 yield (28) R Proof of Theorem Let µ n = p n and σ n = p n ( p n ) Recalling n in (2), η (n) in (2) and Ψ n in (22), we know n = σ n Ψ n + (nµ n )I n }{{} µ n J n (29) n, where I n is the n n identity matrix, and J n is the n n matrix with all of its entries equal to By Lemma 29, with probability one, as n Therefore, F Ψ n/ n converges weakly to γ M F ( n, nµ n I n )/ nσ n = F Ψ n/ n converges weakly to γ M (220) almost surely as n By (29) and the rank inequality (see Lemma 22 from [2]), F ( n nµ n I n )/ nσ n F ( n, nµ n I n )/ nσ n ( ) n rank n n, nσn n rank (J n) 0, (22) n as n, where f := sup x R f(x) is the supremum norm of a bounded, measurable function f(x) defined on R Therefore, (220) and (22) lead to that, as n, n n I i= ( λ i ( n ) np n npn ( p n ) x ) = F ( n nµ n I n )/ nσ n (x) converges weakly to γ M with probability one The proof is complete 2

13 3 Proof of Theorem 2 For any two probability measures µ and ν on R, define d BL (µ, ν) = sup{ fdµ fdν : f + f L }, (3) where f = sup x R f(x), f L = sup x y f(x) f(y) / x y It is well known (see, eg, Section 3 from [6]) that d BL (, ) is called the bounded Lipschitz metric, which characterizes the weak convergence of probability measures Given n n real and symmetric matrices M and M 2, let ˆµ(M i ) be the empirical measure of the eigenvalues (the spectral measure) of M i for i =, 2, that is, ˆµ(M i ) = n n δ λj (M i ), i =, 2, j= where δ x is the point mass probability measure at x We have (see, eg, (26) from [6]) d 2 BL(ˆµ(M ), ˆµ(M 2 )) n tr((m M 2 ) 2 ) (32) A similar result is the Difference Inequality in Lemma 23 from [2]: L 3 (F M, F M 2 ) n tr(m M 2 ) 2 (33) where F Mi is the empirical cumulative distribution function of the eigenvalues of M i for i =, 2 For each n 2, recall from (4) that {ξ (n) ; i < j n} are iid Ber(p n)-distributed random variables, and ξ (n) ii = 0 and ξ (n) = ξ (n) ji for all i j n Define r n,i = n j= ξ (n), r n,i = r n,i (n )p n, a n = max i n { r n,i } and b n = min i n {r n,i} (34) With this notation, we have the following lemma LEMMA 3 Let A n and L n be as in () and (3), respectively Set L n = I n ((n )p n ) A n Then for all n 2 n tr( L n L n ) 2 2a4 n + 8n 2 p 2 na 2 n n(n ) 4 p 4 nb 2 n i j n Proof If b n = 0, the conclusion holds trivially Now, without loss of generality, assume that b n > 0 for all n 2 Note that L n L n = A n + diag(r /2 n, (n )p,, r /2 n,n ) A n diag(r /2 n,,, r /2 n ( ) = ( + )ξ (n) (n )p n rn,i rn,j 3 n n ξ (n) n,n )

14 Thus, using (ξ (n) )2 = ξ (n), we get Now tr( L n L n ) 2 = i,j n ( ) 2 ξ (n) (n )p (35) n rn,i rn,j ( ) 2 = ( r n,i r n,j (n )p n ) 2 (n )p n rn,i rn,j (n ) 2 p 2 (36) nr n,i r n,j Use formula x y = x y /( x + y) x y / y for x = r n,i r n,j and y = (n ) 2 p 2 n to obtain that the right hand side of (36) is dominated by From the definition of r n,i, we see that (r n,i r n,j (n ) 2 p 2 n) 2 (n ) 4 p 4 (r n,ir n,j (n ) 2 p 2 n) 2 nr n,i r n,j (n ) 4 p 4 nb 2 n r n,i r n,j (n ) 2 p 2 n = ( r n,i + (n )p n )( r n,j + (n )p n ) (n ) 2 p 2 n = r n,i r n,j + (n )p n ( r n,i + r n,j ) a 2 n + 2(n )p n a n It follows from the inequality (x + y) 2 2x 2 + 2y 2 for any x, y R that (r n,i r n,j (n ) 2 p 2 n) 2 (a 2 n + 2(n )p n a n ) 2 2a 4 n + 8n 2 p 2 na 2 n Thus, collecting all the assertions above, we see that the coefficient of ξ (n) bounded by in the sum in (35) is 2a 4 n + 8n 2 p 2 na 2 n (n ) 4 p 4 nb 2 n for all i < j n and n 2 This yields the desired inequality We need the following result LEMMA 32 Let ϵ,, ϵ n be iid random variables with P (ϵ = ) = P (ϵ = 0) = p (0, ) Then for all n and δ (0, ) P ( n i= ϵ i np ) δ 2 e (np)δ2 /4 When p is a constant, that is, p does not depend on n, the above is a simple application of the Chernoff bound (see, eg, p27 from [4]) In fact, the above result captures the behavior when p = p n goes to 0 LEMMA 33 Assume (4) holds Let a n and b n be as in (34) If np n / log n, then as n a n np n 0 as, b n np n as and n 2 p n i j n ξ (n) as 4

15 Proof We first prove the last limit Since ξ (n) = ξ (n) ji assertion in the lemma, it is enough to prove that 2 Z n := n(n )p n i<j n ξ (n) for all i < j n, to show the third as as n Taking δ (0, ), since np n / log n, we obtain from Lemma 32 that P ( Z n δ) = O(n 2 ) as n This shows that n 2 P ( Z n δ) <, which implies Z n as by the Borel-Cantelli lemma Recall that r n,i in (34) is a sum of n iid Ber(p n )-distributed random variables for each i n Also, r n,i = r n,i (n )p n, a n = max i n { r n,i } and b n = min i n {r n,i } By Lemma 32 again, use condition np n / log n to get ( ) ( a n n j=2 P δ n P ξ ) j δ (n )p n (n )p n n 2 for any δ (0, ) as n (depending on δ) is sufficiently large It follows that n 2 P (a n/(n )p n δ) is finite for any δ (0, ) By the Borel-Cantelli lemma again, we have a n np n 0 as (37) as n Finally, from (37), b n = min i n{r n,i (n )p n } a n 0 as (n )p n (n )p n (n )p n as n This leads to b n /(np n ) as as n Proof of Theorem 2 Review from Lemma 3 that L n = I n ((n )p n ) A n, and ˆµ( L n ) and ˆµ(L n ) are the empirical measures of the eigenvalues of L n and L n, respectively By Lemma 3, for n 2 Set d 2 BL(ˆµ(L n ), ˆµ( L n )) n tr(l n L n ) 2 2a4 n + 8n 2 p 2 na 2 n n(n ) 4 p 4 nb 2 n a n = (np n )u n, b n = (np n )v n and i j n ξ (n) i j n = n 2 p n w n ξ (n) (38) for n 2 By Lemma 33, u n 0 as, v n as and w n as as n Therefore, noting np n, we obtain as n Therefore, 2a 4 n + 8n 2 p 2 na 2 n n(n ) 4 p 4 nb 2 n i j n ξ (n) = 2n4 p 4 nu 4 n + 8n 4 p 4 nu 2 n n 3 (n ) 4 p 6 nvn 2 n 2 p n w n (2u4 n + 8u 2 ( ) n)w n (np n )vn 2 = o as np n ( ) n tr(l n L n ) 2 = o np n 5 as

16 as n It is easy to see from (32) that ( ˆµ(γL n + ρi n ), ˆµ(γ L ) n + ρi n ) d 2 BL γ 2 n tr( L n L n ) 2 for n 2 and any two real numbers γ and ρ The above two assertions together with (38) conclude that ( ) γ d 2 2 BL(ˆµ(γ n L n + ρ n I n ), ˆµ(γ n Ln + ρ n I n )) = o n np n as (39) as n for any two sequences of constants {γ n, n } and {ρ n, n } Notifying L n = I n ((n )p n ) A n, it is clear that (n )p n ( λ i ( L n )) = λ i (A n ) (30) for i =, 2,, n From the given condition, α n := (np n ( p n )) /2 By corollary 2 from [5], with probability one, ˆµ(A n /α n ) converges weakly to the semi-circle law µ cir with density 2π 4 x2 I( x 2) Thus, by (30), almost surely, n ( ) (n )pn I ( λ i ( L n )) x, x R, n i= converges weakly to µ cir as n Observe that (n )p n (n )p n = α n (np n ( p n )) npn /2 p n α n as n By a trivial manipulation, we obtain that, with probability one, n ( ) npn I ( λ i ( L n )) x, x R, (3) n p n i= converges weakly to µ cir as n Taking γ n = ρ n = np n /( p n ), we have npn λ i (γ n Ln + ρ n I n ) = ( λ i ( L)) p n for i n Thus, by (3), with probability one, ˆµ(γ n Ln + ρ n I n ) converges weakly to µ cir as n Equivalently, d BL ( ˆµ(γ n Ln + ρ n I n ), µ cir ) 0 as (32) as n By the condition that sup{p n ; n 2} <, we know o(γ 2 n/(np n )) = o() as n Since d BL (, ) is a metric, by the triangle inequality, we finally conclude from (39) and (32)that i= d BL (ˆµ(γ n L n + ρ n I n ), µ cir ) 0 as as n Equivalently, with probability one, n ( ) npn I ( λ i (L n )) x, x R, n p n converges weakly to µ cir as n Acknowledgement The author thanks Dr Xue Ding for very helpful discussions on the proof of Theorem 6

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