Degree distribution of an inhomogeneous random intersection graph

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1 Degree distribution of an inhomogeneous random intersection graph Mindaugas Bloznelis Faculty of mathematics and informatics Vilnius University Vilnius, Lithuania Julius Damarackas Faculty of mathematics and informatics Vilnius University Vilnius, Lithuania Submitted: Oct 8, 2012; Accepted: Jul 11, 2013; Published: Jul 19, 2013 Mathematics Subject Classifications: 05C80, 05C07, 05C82 Abstract We show the asymptotic degree distribution of the typical vertex of a sparse inhomogeneous random intersection graph. Keywords: degree distribution; random graph; random intersection graph; power law 1 Introduction Let X 1,..., X m, Y 1,..., Y n be independent non-negative random variables such that each X i has the probability distribution P 1 and each Y j has the probability distribution P 2. Given realized values X = {X i } m i=1 and Y = {Y j } n j=1 we define the random bipartite graph H X,Y with the bipartition V = {v 1,..., v n }, W = {w 1,..., w m }, where edges {w i, v j } are inserted with probabilities p ij = min{1, X i Y j (nm) 1/2 } independently for each {i, j} [m] [n]. The inhomogeneous random intersection graph G(P 1, P 2, n, m) defines the adjacency relation on the vertex set V : vertices v, v V are declared adjacent (denoted v v ) whenever v and v have a common neighbour in H X,Y. The degree distribution of the typical vertex of the random graph G(P 1, P 2, n, m) has been first considered by Shang [21]. The proof of the main result of [21] contains gaps and the result is incorrect in the regime where m/n β (0, + ) as m, n +. We remark that this regime is of particular importance, because it leads to inhomogeneous graphs with the clustering property: the clustering coefficient P(v 1 v 2 v 1 v 3, v 2 v 3 ) is bounded away from zero provided that 0 < β < + and EX 3 1 <, EY 2 1 <, see [8]. Supported by the Research Council of Lithuania Grant MIP-053/2011. the electronic journal of combinatorics 20(3) (2013), #P3 1

2 The aim of the present paper is to show the asymptotic degree distribution in the case where m/n β for some β [0, + ]. We consider a sequence of graphs {G n = G(P 1, P 2, n, m)}, where m = m n + as n, and where P 1, P 2 do not depend on n. We denote a k = EX1 k, b k = EY1 k. By d(v) = d Gn (v) we denote the degree of a vertex v in G n (the number of vertices adjacent to v in G n ). We remark that for every n the random variables d Gn (v 1 ),..., d Gn (v n ) are identically distributed. In Theorem 1 below we show the asymptotic distribution of d(v 1 ). Theorem 1. Let m, n. (i) Assume that m/n 0. Suppose that that EX 1 <. Then P(d(v 1 ) = 0) = 1 o(1). (ii) Assume that m/n β for some β (0, + ). Suppose that EX 2 1 < and EY 1 <. Then d(v 1 ) converges in distribution to the random variable Λ 1 d = τ j, (1) j=1 where τ 1, τ 2,... are independent and identically distributed random variables independent of the random variable Λ 1. They are distributed as follows. For r = 0, 1, 2,..., we have P(τ 1 = r) = r + 1 P(Λ 2 = r + 1) and P(Λ i = r) = E e λ λr i i, i = 1, 2. (2) EΛ 2 r! Here λ 1 = Y 1 a 1 β 1/2 and λ 2 = X 1 b 1 β 1/2. (iii) Assume that m/n +. Suppose that EX 2 1 < and EY 1 <. Then d(v 1 ) converges in distribution to a random variable Λ 3 having the probability distribution Here λ 3 = Y 1 a 2 b 1. P(Λ 3 = r) = Ee λ λr 3 3, r = 0, 1,.... (3) r! Remark 2. The probability distributions P Λi of Λ i, i = 1, 2, 3, are Poisson mixtures. One way to sample from the distribution P Λi is to generate random variable λ i and then, given λ i, to generate Poisson random variable with the parameter λ i. The realized value of the Poisson random variable has the distribution P Λi. Remark 3. The asymptotic distributions (1) and (3) admit heavy tails. In the case (ii) we obtain a power law asymptotic degree distribution (1) provided that the heavier of the tails t P(τ 1 > t) and t P(Λ 1 > t) has a power law, see, e.g., [13]. Similarly, in the case (iii) we obtain a power law asymptotic degree distribution (3) provided that P 2 has a power law. Remark 4. Since the second moment a 2 does not show up in (1), (2) we expect that in the case (ii) the second moment condition EX 2 1 < is redundant and could perhaps be replaced by the weaker first moment condition EX 1 <. the electronic journal of combinatorics 20(3) (2013), #P3 2

3 Random intersection graphs have attracted considerable attention in the recent literature, see, e.g., [1, 2, 3, 9, 10, 12, 19]. Starting with the paper by Karoński et al [17], see also [22], where the case of degenerate distributions P 1 = P 1n, P 2 = P 2n depending on n was considered (i.e., P 1n (c n ) = P 2n (c n ) = 1, for some c n > 0), several more complex random intersection graph models were later introduced by Godehardt and Jaworski [14], Spirakis et al. [18], Shang [21]. The asymptotic degree distribution for various random intersection graph models was shown in [4, 5, 6, 7, 11, 15, 16, 20, 23]. 2 Proof Before the proof of Theorem 1 we introduce some notation and give two auxiliary lemmas. The event that the edge {w i, v j } is present in H = H X,Y is denoted w i v j. We denote I ij = I {wi v j }, I i = I i1, u i = m I ij, L = u i I i and remark that u i counts all neighbours of w i in H belonging to the set V \{v 1 }. Denote â k = m 1 Xi k, ˆbk = n 1 Yj k, 1 i m λ ij = X iy j, S XY = mn m i=1 n j=2 i=1 X i mn λ ij min{1, λ ij }, (4) and introduce the event A 1 = {λ i1 < 1, 1 i m}. We denote β n = m/n. By P 1 and E 1 we denote the conditional probability and conditional expectation given Y 1. By P and Ẽ we denote the conditional probability and conditional expectation given X, Y. By d T V (ζ, ξ) we denote the total variantion distance between the probability distributions of random variables ζ and ξ. In the case where ζ, ξ, X, Y are defined on the same probability space we denote by d T V (ζ, ξ) the total variation distance between the conditional distributions of ζ and ξ given X, Y. In the proof below we use the following simple fact about the convergence of a sequence of random variables {κ n }: κ n = o P (1), E sup κ n < Eκ n = o(1). (5) n We remark that the condition E sup n κ n < (which assumes implicitly that all κ n are defined on the same probability space) can be replaced by a more restrictive condition that there exists a constant c > 0 such that κ n c, for all n. In the latter case we do not need all κ n to be defined on the same probability space. In particular, given a sequence of bivariate random vectors {(η n, θ n )} such that, for every n and m = m n random variables η n, θ n and {X i } m i=1, {Y j } n j=1 are defined on the same probability space, we have d T V (η n, θ n ) = o P (1) d T V (η n, θ n ) E d T V (η n, θ n ) = o(1). Here and below all limits are taken as n + (if not stated otherwise). the electronic journal of combinatorics 20(3) (2013), #P3 3

4 Lemma 5. Assume that EX 2 1 < and EY 1 <. We have as n + P(d(v 1 ) L) = o(1), (6) P(A 1 ) = 1 o(1), (7) S XY = o P (1), ES XY = o(1). (8) Proof. Proof of (6). We observe that the event A = {d(v 1 ) L} occurs in the case where for some 2 j n and some distinct i 1, i 2 [m] the event A i1,i 2,j = {w i1 v 1, w i1 v j, w i2 v 1, w i2 v j } occurs. From the union bound and the inequality P(A i1,i 2,j) m 2 n 2 Xi 2 1 Xi 2 2 Y1 2 Yj 2 we obtain P(A) = P A i1,i 2,j n 1ˆb2 Y1 2 Q X. (9) {i 1,i 2 } [m] Here Q X = m 2 {i 1,i 2 } [m] X2 i 1 X 2 i 2. We note that Q X and Y 2 1 are stochastically bounded and n 1ˆb2 = o P (1) as n +. Therefore, P(A) = o P (1). Now (5) implies (6). Proof of (7). Let A 1 denote the complement event to A 1. We have, by the union bound and Markov s inequality, P 1 (A 1 ) P 1 (λ i1 1) (nm) 1 Y1 2 EXi 2 = n 1 a 2 Y1 2. i [m] We conclude that P 1 (A 1 ) = o P (1), and now (5) implies P(A 1 ) = EP 1 (A 1 ) = o(1). Proof of (8). Since the first bound of (8) follows from the second one, we only prove the latter. Denote ˆX 1 = max{x 1, 1} and Ŷ1 = max{y 1, 1}. We observe that EX 2 1 <, EY 1 < implies i [m] lim E ˆX 1I 2 t + { ˆX1 >t} = 0, lim EŶ1I t + { Ŷ 1 >t} = 0. Hence one can find a strictly increasing function ϕ : [1, + ) [0, + ) with lim t + ϕ(t) = + such that E ˆX 2 1ϕ( ˆX 1 ) <, EŶ1ϕ(Ŷ1) <. (10) We remark that one can find ϕ which satisfies, in addition, the inequalities ϕ(t) < t and ϕ(st) ϕ(t)ϕ(s), s, t 1. (11) For this purpose we choose ϕ of the form ϕ(t) = e ψ(ln(t)), where ψ : [0, + ) [0, + ) is a differentiable concave function, which grows slowly enough to satisfy (10) and the first inequality of (11), and takes value 0 at the origin. We note that the second inequality of (11) follows from the concavity property of ψ. We remark, that (11) implies t/(st) = s 1 1/ϕ(s) ϕ(t)/ϕ(st), s, t 1. (12) the electronic journal of combinatorics 20(3) (2013), #P3 4

5 Let ŜXY be defined as in (4) above, but with λ ij replaced by ˆλ ij = ˆX i Ŷ j / mn. We note, that S XY ŜXY. Furthermore, from the inequalities min{1, ˆλ { ij } min 1, ϕ( ˆX i Ŷ j ) } ϕ( ϕ( ˆX i Ŷ j ) mn) ϕ( mn) ϕ( ˆX i )ϕ(ŷj) ϕ( mn) we obtain S XY ŜXY S XY /ϕ( nm), where S XY = (mn) 1 ( 1 i m ˆX 2 i ϕ( ˆX i ) ) ( Ŷ j ϕ(ŷj) We remark that (11) and (12) and imply the third and the first inequality of (13), respectively. Finally, the bound ES XY = o(1) follows from the inequality S XY SXY /ϕ( nm) and the fact that ESXY remains bounded as n, m +, see (10). In the proof of Theorem 1 we use the following inequality referred to as LeCam s inequality, see e.g., [24]. Lemma 6. Let S = I 1 + I I n be the sum of independent random indicators with probabilities P(I i = 1) = p i. Let Λ be Poisson random variable with mean p p n. The total variation distance between the distributions P S of P Λ of S and Λ sup P(S A) P(Λ A) = 1 P(S = k) P(Λ = k) p 2 i. (14) A {0,1,2... } 2 k 0 ). 1 i n Proof of Theorem 1. The case (i). Since P(d(v 1 ) L) = 1 it suffices to show that P(L > ε) = o(1), for any ε (0, 1). For this purpose we write P(L > ε) = EP 1 (L > ε) and prove that P 1 (L > ε) = o P (1), see (5). Here we estimate P 1 (L > ε) using the union bound and Markov s inequality, P 1 (L > ε) P 1 (I k = 1) E 1 λ k1 = m/ny 1 EX 1 = o P (1). The case (ii). Before the proof we introduce some notation. Given X, Y, let { lk } m k=1, l = 1, 2, 3, {ξ rk } m k=1, r = 1, 2, 3, 4, and {ξ 3k} m k=1, {ξ 4k} m k=1, (15) and {η k } m k=1 be sequences of conditionally independent (within each sequence) Poisson random variables with mean values Ẽξ 1k = p kj, Ẽξ 2k = λ kj, Ẽξ 3k = 1 b 1 X k, Ẽξ 4k = 1 b 1 X k, βn β Ẽ 1k = (λ kj p kj ), Ẽ 2k = 1 βn δ 2 X k, Ẽ 3k = 1 βn δ 3 X k, Ẽξ 3k = β 1/2 n bx k, Ẽξ 4k = β b 1 X k, Ẽη k = λ k1. (13) the electronic journal of combinatorics 20(3) (2013), #P3 5

6 Here we denote b := min{ˆb 1, b 1 }, δ 2 := ˆb 1 b, δ 3 := b 1 b and β := min{β 1/2, βn 1/2 }. We recall that Ẽ denotes the conditional expectation given X, Y. We assume that, given X, Y, the sequences {ξ 3k }m k=1, { 2k} m k=1, { 3k} m k=1 are conditionally independent, the sequences {I k } m k=1, {ξ 3k} m k=1 are conditionally independent, the sequences {ξ 1k} m k=1, { 1k} m k=1 are conditionally independent, and the sequence {η k } m k=1 is conditionally independent of the sequences (15). We assume, in addition, that for 1 k m ξ 2k := ξ 1k + 1k, ξ 3k := ξ 4k + (β 1/2 n β )b 1 X k, ξ 4k := ξ 4k + (β 1/2 β )b 1 X k. Furthermore, we suppose that, given X, Y 1, the sequence {η k } m k=1 is conditionally independent of the subgraph of H X,Y spanned by the vertices {v 2,..., v n } W. Next we introduce several random variables which are intermediate between L and d : L 0 = η k u k, L r = η k ξ rk, r = 1, 2, 3, 4, L 2 = Finally, we denote η k (ξ 3k + 2k ), L 3 = η k (ξ 3k + 3k ). f τ (t) = Ee itτ 1, fτ (t) = e itr 1 p r, p r = λ r 0 λ k1 I {ξ4k =r}, δ = ( f τ (t) 1) λ (f τ (t) 1)λ 1, f(t) = E 1 e itd, f(t) = Ēe itl 4. m λ = λ k1, k=1 Here Ē denotes the conditional expectation given X, Y and ξ 41,..., ξ 4m. We recall that E 1 denotes the conditional expectation given Y 1. Let us prove the statement (ii). In view of (6) the random variables d(v 1 ) and L have the same asymptotic distribution (if any). We shall show that L converges in distribution to (1). We proceed in two steps. Firstly, we approximate the distribution of L by that of L 4 using LeCam s inequality, see Lemma 6, and a coupling argument. Secondly, we prove the convergence of the characteristic functions Ee itl 4 Ee itd. Here and below i = 1 denotes the imaginary unit. Step 1. Let us show that L and L 4 have the same asymptotic probability distribution (if any). To this aim we prove that d T V (L, L 0 ) = o(1), d T V (L 0, L 1 ) = o(1), (16) E L 1 L 2 = o(1), L 2 L 3 = o P (1), E L 3 L 4 = o(1), (17) and observe that L 2 (respectively L 3 ) has the same distribution as L 2 (respectively L 3). Let us prove the first bound of (16). In view of (5) it suffices to show that d T V (L 0, L) = o P (1). In order to prove the latter bound we apply the inequality d T V (L 0, L)I A1 n 1 Y 2 1 â 2 (18) the electronic journal of combinatorics 20(3) (2013), #P3 6

7 shown below. We remark that (18) implies d T V (L 0, L) d T V (L 0, L)I A1 + I A1 n 1 Y 2 1 â 2 + I A1 = o P (1). Here n 1 Y1 2 â 2 = o P (1), because Y1 2 â 2 is stochastically bounded. Furthermore, the bound I A1 = o P (1) follows from (7). It remains to prove (18). We denote L k = k l=1 I lu l + m l=k+1 η lu l and write, by the triangle inequality, d T V (L 0, L) m k=1 d T V ( L k 1, L k ). Then we estimate d T V ( L k 1, L k ) d T V (η k, I k ) (nm) 1 Y1 2 Xk 2. Here the first inequality follows from the properties of the total variation distance. The second inequality follows from Lemma 6 and the fact that on the event A 1 we have p k1 = λ k1. Let us prove the second bound of (16). In view of (5) it suffices to show that d T V (L 0, L 1 ) = o P (1). We denote L k = k l=1 η lu l + m l=k+1 η lξ 1l and write, by the triangle inequality, m d T V (L 0, L 1 ) d T V (L k 1, L k). (19) Here k=1 d T V (L k 1, L k) d T V (η k u k, η k ξ 1k ) P(η k 0) d T V (u k, ξ 1k ). (20) Now, invoking the inequalities P(η k 0) = 1 e λ k1 λk1 and d T V (u k, ξ 1k ) n j=2 p2 kj, we obtain the desired bound from (8), (19) and (20) d T V (L 0, L 1 ) m λ k1 k=1 j=2 n p 2 kj Y 1 S XY = o P (1). and Let us prove the first bound of (17). We observe that L 2 L 1 = L 2 L 1 = η k 1k Ẽ η k 1k = λ k1 (λ kj 1)I {λkj >1} Y 1 S XY. Hence, we obtain E L 2 L 1 EY 1 ES XY = o(1), see (8). Let us prove the second bound of (17). It suffices to show that Ẽ L 2 L 3 = o P (1). We have Ẽ L 2 L 3 = (δ 2 + δ 3 )Y 1 â 2 = ˆb 1 b 1 Y 1 â 2 = o P (1). (21) In the last step we used the fact that Y 1 â 2 = O P (1) and ˆb 1 b 1 = o P (1). the electronic journal of combinatorics 20(3) (2013), #P3 7

8 The third bound of (17) is straightforward 1 E L 3 L 4 1 b1 βn β Eη k X k = βn b 2 1 β 1 a 2 = o(1). Step 2. Here we show that for every real t we have E (t) = o(1), where (t) = e itl 4 e itd. To this aim we prove that for any realized value Y 1 there exists a positive constant c = c(t, Y 1 ) such that for every 0 < ε < 0.5 we have lim sup E( (t) Y 1 ) < c ε. (22) n,m + Clearly, (22) implies E( (t) Y 1 ) = o(1). This fact together with the simple inequality (t) 2 yields the desired bound E (t) = o(1), by Lebesgue s dominated convergence theorem. We fix 0 < ε < 0.5 and prove (22). Introduce event D = { â 1 a 1 < ε min{1, a 1 }} and let D denote the complement event. Furthermore, select a number T > 1/ε such that P(τ 1 T ) < ε. By c 1, c 2,... we denote positive numbers which do not depend on n, m. We observe that, given Y 1, the conditional distribution of d is the compound Poisson distribution with the characteristic function f(t) = e λ 1(f τ (t) 1). Similarly, given X, Y and ξ 41,..., ξ 4m, the conditional distribution of L 4 is the compound Poisson distribution with the characteristic function f(t) = e λ( f τ (t) 1). In the proof of (22) we exploit the convergence λ λ 1 and f τ (t) f τ (t). In what follows we assume that m, n are so large that β 2β n 4β. Let us show (22). For this purpose we write E( (t) Y 1 ) = E 1 (t) = I 1 + I 2, I 1 = E 1 (t)i D, I 2 = E 1 (t)i D and estimate I 1 and I 2. We have I 2 2P 1 (D) = 2P(D) = o(1), by the law of large numbers. Now we show that I 1 c ε + o(1). Combining the identity E 1 (t) = E 1 f(t)(e δ 1) with the inequalities f(t) 1 and e s 1 s e s we obtain Here we estimated e δ e 8λ 1 =: c 1 using the inequalities I 1 E 1 δ e δ I D c 1 E 1 δ I D. (23) δ 2 λ + 2λ 1, λ = Y1 β 1/2 n â 1 3λ 1. We remark that the last inequality holds for β n 2β provided that the event D occurs. We shall show below that E 1 δ I D (c 2 + λ 1 c 3 + λ 1 c 4 )ε + o(1). (24) This inequality together with (23) yields the desired upper bound for I 1. the electronic journal of combinatorics 20(3) (2013), #P3 8

9 Let us prove (24). We write and estimate δ = ( f τ (t) 1)( λ λ 1 ) + ( f τ (t) f τ (t))λ 1 δ 2 λ λ 1 + λ 1 f τ (t) f τ (t). Now, from the inequalities â 1 a 1 < ε and β n 2β we obtain λ λ 1 Y 1 â 1 a βn 1/2 + Y 1 a 1 βn 1/2 β 1/2 2Y 1 β 1/2 ε + o(1). Hence, E 1 λ λ 1 I D c 2 ε + o(1), where c 2 = 2Y 1 β 1/2. We complete the proof of (24) by showing that E 1 f τ (t) f τ (t) I D (c 3 + c 4 )ε + o(1). (25) In order to prove (25) we split f τ (t) f τ (t) = r 0 e itr ( p r p r ) = R 1 R 2 + R 3, and estimate separately the terms R 1 = e itr p r, R 2 = e itr p r, r T r T R 3 = 0 r<t e itr ( p r p r ). Here we denote p r = P(τ 1 = r). The upper bound for R 2 follows by the choice of T R 2 r T p r = P(τ 1 T ) < ε. Next, combining the identity p r = (â 1 m) 1 X ki {ξ4k =r} with the inequalities R 1 (â 1 m) 1 X k I {ξ4k =r} = (â 1 m) 1 X k I {ξ4k T } (26) r T and â 1 1 I D 2a 1 1, we estimate E 1 R 1 I D 2a 1 1 E 1 (X k I {ξ4k T }) 2a 1 1 T 1 E 1 (X k ξ 4k ) = 2a 1 1 a 2 b 1 β 1/2 ε. Hence, we obtain E 1 R 1 I D c 4 ε, where c 4 = 2a 1 1 a 2 b 1 β 1/2. Now we estimate R 3. We denote p r = (â 1 /a 1 ) p r and observe that the inequality â 1 a εa 1 implies â 1 a ε and e itr ( p r p r) ε p r ε. 0 r T 0 r T In the last inequality we use the fact that the probabilities { p r } r 0 sum up to 1. It follows now that R 3 I D ε + p r p r. 0 r T the electronic journal of combinatorics 20(3) (2013), #P3 9

10 Furthermore, observing that E 1 p r = a 1 E 1 X k I {ξ4k =r} = p r, for 1 k m, we obtain E 1 p r p r 2 = m 1 E 1 a 1 X k I {ξ4k =r} p r 2 m 1 a 2 1 EX 2 k. Hence, E 1 p r p r = O(m 1/2 ). We conclude that E 1 R 3 I D ε + O( T m 1/2 ) = ε + o(1), thus completing the proof of (25). The case (iii). Before the proof we introduce several random variables. Given ε (0, 1), denote I k = I, γ {Xk βn 1/2 b 1 <ε} k = X k βn 1/2 b 1 I k, 1 k m, and γ = λ k1 γ k. Given X, Y, let Ĩ1,..., Ĩm be conditionally independent Bernoulli random variables with success probabilities P(Ĩk = 1) = 1 P(Ĩk = 0) = γ k. We assume that, given X, Y, the sequences {I k } m k=1, {Ĩk} m k=1 and {ξ 3k} m k=1 (see (15)) are conditionally independent. Furthermore, introduce random variables L 5 = I k ξ 3k, L 6 = I k I kξ 3k, L 7 = I k Ĩ k and let L 8 be a random variable with the distribution P(L 8 = r) = Ee γ γ r /r!, for r = 0, 1,.... We remark that the probability distribution of L 8 is a Poisson mixture, i.e., the Poisson distribution with random parameter γ. We note that by (16) and the first two bounds of (17), the random variables L and L 3 have the same asymptotic distribution (if any). Now we prove that L 3 converges in distribution to Λ 3. For this purpose we show that for any ε (0, 1) d T V (L 3, L 5 ) = o(1), E L 5 L 6 = o(1), (27) d T V (L 6, L 7 ) a 2 b 2 1ε, d T V (L 7, L 8 ) = o(1), (28) Ee itl 8 Ee itλ 3 = o(1). (29) Let us prove (27). The first bound of (27) is obtained in the same way as the first bound of (16). To show the second bound of (27) we write Ẽ L 5 L 6 = (1 I k)ẽikẽξ 3k Y 1 b 1 m 1 (1 I k)xk 2 and obtain E L 5 L 6 = EẼ L 5 L 6 b 2 1EX1I 2 = o(1). {X1 βn 1/2 b 1 ε} We note that the right hand side tends to zero since β n +. the electronic journal of combinatorics 20(3) (2013), #P3 10

11 Let us prove the first inequality of (28). Proceeding as in (19), (20) and using the identity Ĩk = ĨkI k we write d T V (L 6, L 7 ) I P(I k k 0) d T V (ξ 3k, Ĩk). Next, we estimate I d k T V (ξ 3k, Ĩk) γk 2, by LeCam s inequality (14), and invoke the inequality P(I k 0) λ k1. We obtain d T V (L 6, L 7 ) I kλ k1 γk 2 ε I kλ k1 γ k εy 1 b 1 â 2. Here we estimated γ 2 k εγ k. Now the inequalities d T V (L 6, L 7 ) E d T V (L 6, L 7 ) a 2 b 2 1ε imply the first relation of (28). Let us prove the second relation of (28). d T V (L 7, L 8 ) = o P (1). For this purpose we write d T V (L 7, L 8 ) I A1 dt V (L 7, L 8 ) + I A1, In view of (5) it suffices to show that where I A1 = o P (1), see (7), and estimate using LeCam s inequality (14) I A1 dt V (L 7, L 8 ) I A1 P 2 (I k Ĩ k = 1)I k = I A1 λ 2 k1γk 2 b 2 1Y1 2 m 1 â 4 = o P (1). In the last step we used the fact that EX 2 1 < implies m 1 â 4 = o P (1). Finally, we show (29). We write ẼeitL 8 = e γ(eit 1) and observe that Y 1 b 1 a 2 γ = o P (1). (30) Furthermore, since for any real t the function z e z(eit 1) is bounded and uniformly continuous for z 0, we conclude that (30) implies the convergence Ee itl 8 = Ee γ(eit 1) Ee Y 1b 1 a 2 (e it 1) = Ee itλ 3. It remains to prove (30). We write Y 1 b 1 a 2 γ = Y 1 b 1 (a 2 â 2 ) + Y 1 b 1 â 2 γ and note that a 2 â 2 = o P (1), by the law of large numbers, and 0 E(Y 1 b 1 â 2 γ) = b 2 1EX1I 2 = o(1). {X1 βn 1/2 b 1 ε} Acknowledgements We thank the anonymous referee for remarks and suggestions that have improved the presentation. the electronic journal of combinatorics 20(3) (2013), #P3 11

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