Almost sure limit theorems for random allocations

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1 Almost sure limit theorems for random allocations István Fazekas and Alexey Chuprunov Institute of Informatics, University of Debrecen, P.O. Box, 400 Debrecen, Hungary, and Department of Math. Stat. and Probability, Chebotarev Inst. of Mathematics and Mechanics, Kazan State University, Universitetskaya 7, Kazan, Russia, Abstract. Almost sure limit theorems are presented for random allocations. A general almost sure limit theorem is proved for arrays of random variables. It is applied to obtain almost sure versions of the central limit theorem for the number of empty boxes when the parameters are in the central domain. Almost sure versions of the Poisson limit theorem in the left domain are also proved. AMS 000 subject classification: 60F05, 60F5, 60C05. Key words and phrases. Almost sure central limit theorem, random allocation.. Introduction Let n balls be placed successively and independently into urns. Let µ 0 n, ) denote the number of empty urns. There are several theorems concerning the limit laws of µ 0 n, ) when the parameters belong to certain domains see, e.g., Weiss 958), Rényi 96), Békéssy 963), and the monograph Kolchin, Sevast yanov and Chistyakov 978)). It is known that if n, in the central domain, then the limit of the standardized µ 0 n, ) is standard normal. The same is true in the left intermediate domain. In the left domain the limit of the appropriately centralized µ 0 n, ) is Poisson distribution. In this paper we obtain almost sure a.s.) versions of the above limit theorems for µ 0 n, ). The general form of the a.s. limit theorem is the following. Let ζ n, n, be a sequence of random elements defined on the probability space Ω, A, P). Supported by the Hungarian Foundation of Scientific Researches under Grant o. OTKA T0336/000 and Grant o. OTKA T03658/000. The research was partially realized while this author was visiting Institute of Informatics, University of Debrecen, Debrecen, Hungary.

2 I. Fazekas, A. Chuprunov A.s. limit theorems state that n d k δ ζk ω) µ, as n, for almost every ω Ω,.) D n k= where δ x is the unit mass at point x and µ denotes weak convergence to the probability measure µ. In the simplest form of the a.s. CLT ζ n = X + +X n )/ n, where X, X,..., are i.i.d. real random variables with mean 0 and variance, d k = /k, D n = log n, and µ is the standard normal law 0, ); see Berkes 998) for an overview. In Section, we consider an appropriate representation of µ 0 n, ) in terms of independent, uniformly distributed random variables in order to handle the dependence structure inside the array µ 0 n, ), n, =,,.... As µ 0 n, ) depends on two parameters, there are different settings of the a.s. limit theorem. The simplest is when we choose a sequence n, n)) in the domain considered. Then we obtain classical type a.s. limit theorems. These are Theorems.7,.8 and.3 in the central, in the left intermediate, and in the left domains, respectively. On the other hand, we can consider sections of our domains. If T n denotes the section considered, then we obtain limit theorems of the form D n k,k) T n d kk δ ζkk ω) µ, as n,, for almost every ω Ω..) To prove the above type theorems, we present a general a.s. limit theorem Theorem.). This result is an extension of known general a.s. limit theorems see, e.g., Fazekas and Rychlik 00)). Then we use our general result to get Theorems.5 and.6 in the central domain, Theorem.9 in the left intermediate domain, and Theorem. in the left domain. The most interesting result of these is Theorem.. The reason is that in the left domain the parameter of the original limit Poisson distribution depends on the subsequence considered. Therefore the limit distribution in our Theorem. is constructed from different Poisson distributions. The phenomenon that there is no ordinary limit distribution for the whole sequence) but the almost sure limit theorem is valid was described e.g. in Berkes, Csáki and Csörgő 999) and in Berkes, Csáki, Csörgő and Megyesi 00). In our paper the situation is similar. To obtain the limit, we described the appropriate accompanying Poisson distributions for µ 0 n, ) in the left domain Theorem.0). Finally, we obtain a.s. limit theorems when n and converge to infinity independently. These two-index limit theorems are Theorem.5 in the central domain and Theorem. in the left domain. These are easy consequences of the previous results, so we do not need multiindex a.s. limit thorems described in Fazekas and

3 Almost sure limit theorems for random allocations 3 Rychlik 003)). The main technical tools in this paper are the strong law of large numbers in Lemma 3. and the inequalities in Lemma 4... Main results A general almost sure limit theorem. Recently, several papers are devoted to the background and to general forms of the a.s. CLT, see e.g. Major 000), Berkes and Csáki 00), Fazekas and Rychlik 00), Móri and Székely 003), Fazekas and Chuprunov 003). In this paper, the general result is Theorem. which is an extension of the basic result in Fazekas and Rychlik 00). Then we apply this theorem to prove a.s. versions of some limit theorems for random allocations. Let {α k)} and {α k)} be given integer valued sequences with α k) α k) <, for k. Let B, ϱ) be a complete separable metric space and let ζ ki, α k) i α k), k, be an array of random elements in B. Let µ ζ denote the distribution of the random element ζ. Let log + x = log x, if x and log + x = 0, if x <. Theorem.. Assume that there exist C > 0, ε > 0; an increasing sequence of positive numbers c n with lim n c n =, c n+ /c n = O); and B-valued random elements ζlj ki, for k, i, l, j, k < l, α k) i α k), α l) j α l), such that the random elements ζ ki and ζlj ki are independent for k < l and for any i, j; and { )} E{ϱζ lj, ζlj ki cl +ε) ) } C log + log +,.) c k for k < l and any i, j. Let 0 d k logc k+ /c k ), assume that k= d k =. Assume that d k = α k) i=α k) for each k, with nonnegative numbers d ki. Let D n = n k= d k. Then for any probability distribution µ on the Borel σ-algebra of B the following two statements are equivalent n α k) D n k= i=α k) d ki d ki δ ζki ω) µ, as n, for almost every ω Ω ;.) n α k) D n k= i=α k) d ki µ ζki µ, as n..3)

4 4 I. Fazekas, A. Chuprunov In the special case, when α k) = α k) for each k, our Theorem. is the same as Theorem. in Fazekas and Rychlik 00). We remark that the same universal approach was used in Fazekas and Rychlik 00) as in Berkes and Csáki 00). Multiindex versions of a.s. limit theorems were obtained in Fazekas and Rychlik 003). However, as the weights there are of product type, we can not apply those results for domains like {k, i) : α k) i α k), k }. The importance of condition.3) is demonstrated in Berkes, Csáki and Csörgő 999), Berkes, Csáki, Csörgő and Megyesi 00), where examples are given in which a.s. limit theorems are true, however the ordinary limit theorems are not valid. Remark.. following Theorem. remains valid if condition.) is replaced by the E{ϱζ lj, ζ ki lj ) } C for k < l and for any i, j, where β > 0. ck c l ) β,.4) Remark.3. If conditions.) or.4) are valid only for < k 0 k < l, then Theorem. resp. Remark. remain valid. To prove this, one has to apply the statements for the array ζ ki, α k) i α k), k = k 0, k 0 +,.... Random allocations. Let ξ, ξ j, j, be independent random variables uniformly distributed on [0, ]. Let n,, i = i = [ i, i ), i. Define the following events connected with the subdivision of the interval [0, ) into the subintervals i A i = A i n, ) = j {,...,n} {ξ j / i }. Let θ i = θ i n, ) be the indicator of A i n, ) and µ 0 n, ) = θ i. We consider the intervals i, i =,...,, as a row of boxes. Random variables ξ j, j =,,..., are realizations of ξ. Each realization of ξ we treat as a random allocation of one ball into the boxes. The equality θ i = means that at a random allocation of n balls into boxes the ith box is empty. Moreover, µ 0 n, ) is the number of empty boxes. We will use the notation α = n. It is known see Kolchin et al. 978), Ch., Sec.) that the following limit realitions.5) and.6) hold if n, such that α = o). For the expectation we have Eµ 0 n, ) = ) n = e α α ) α + α)e α e α + O i=.5)

5 Almost sure limit theorems for random allocations 5 and for the variance we have D n, = D µ 0 n, ) = ) ) n + ) n ) n = = e α + α)e α) + O α + α)e α e α + ))..6) As in the theory of random allocations the roles of n and are fixed, we shall use the following notation for twodimensional indices: n, ), k, K). Let S n = µ 0n, ) Eµ 0 n, ) D n, be the standardized variable, where n, ). Almost sure limit theorems for random allocations in the central domain. If n, such that 0 < α n α <, where α and α are some constants, then it is said that n, in a central domain. In a central domain we have the following central limit theorem. Theorem A. Let 0 < α < α <. If n, such that α = n [α, α ], then S n γ. Here and in the following γ denotes the standard normal law. The proof of Theorem A can be found in Weiss 958) and Rényi 96). See also the monograph Kolchin et al. 978), Ch., Sec.3. Consider almost sure versions of Theorem A. Theorem.4. Let 0 < α < α < and Q n ω) = Then, as n, we have log α log α ) log n k n {K : α k K α } Q n ω) γ, for almost every ω Ω. kk δ S kk ω). In the above theorem the limit was considered for n and the indices of the summands were in a fixed central domain). The following theorem is a two-index limit theorem, i.e. n and. The relation of n and could be arbitrary

6 6 I. Fazekas, A. Chuprunov however, as the indices of the summands are in a fixed central domain, we assume that n, ) is in the central domain considered. Theorem.5. Let 0 < α < α < and Q n ω) = log α log α ) log n k n Then, as n, so that α n α, we have {K : K, α k K α } Q n ω) γ, for almost every ω Ω. kk δ S kk ω). The following theorem is a version of Theorem.4, here we shall use the weights k. Theorem.6. Let 0 < α < α < and Q nω) = α α Then, as n, it holds that ) log n k n {K : α k K α } Q nω) γ, for almost every ω Ω. k δ S kk ω). ow consider an almost sure version of Theorem A in ordinary form. We will assume that in n, ) the number depends on n: = n). In this case S nn) will be denoted by S n. Theorem.7. Let 0 < α < α <. Suppose that α n n) α for all n. Let Then, as n, one has Q 0) n ω) = log n n k= k δ S k ω). Q 0) n ω) γ for almost every ω Ω. Almost sure limit theorems for random allocations in the left intermediate domain. If n, such that α 0 and α then it is said that n, in the left intermediate domain. Theorem B. Let n, in the left intermediate domain. Then S n γ.

7 Almost sure limit theorems for random allocations 7 For the proof of Theorem B see Rényi 96) and Kolchin et al. 978). Ch., Sec.3. We have the following almost sure version of Theorem B. Theorem.8. Suppose that n such that n, n)) belongs to the left intermediate domain. Then, as n, one has Q 0) n ω) γ for almost every ω Ω. The following theorem is similar to Theorem.4, as we consider summands with indices in a domain. Theorem.9. Let / < α < α < be fixed. Let Q g) n ω) = Then, as n, we have log α log α ) log n k n {K : k α K k α } Q g) n ω) γ, for almost every ω Ω. kk log K δ S kk ω). In the left intermediate domain we could not obtain a two-index version i.e. a result like Theorem.5) of the a.s. limit theorem. Almost sure limit theorems for random allocations in the left domain. Let 0 < λ < be fixed. If n, such that α 0 and α then it is said that n, in a left domain. Theorem C. λ, Let n, in the left domain. Then we have µ 0 n, ) n) πλ), where πλ) is Poisson distribution with the parameter λ. For the proof of Theorem C see Békéssy 963) or Kolchin et al. 978), Ch., Sec.4. In the left domain for different subsequences of pairs n, ) the parameters of the limit Poisson distributions can be different. However, we can find appropriate

8 8 I. Fazekas, A. Chuprunov accompanying Poisson law to µ 0 n, ) n). Let πλ, k) = λk k! e λ denote the kth element of the Poisson law. Theorem.0. Let 0 < λ < λ < be fixed. Then P µ 0 n, ) n) = m ) ) n π, m 0.7) λ. The above conver- for each fixed m {0,,...}, if n so that λ n gence is uniform in satisfying λ n λ. The following theorems can be considered as almost sure versions of Theorem C. Theorem.. Let 0 < λ < λ < be fixed. Let T n be the following domain in { } T n = k, K) : k n, λ k K λ. Let and let Then, as n, Q p) n ω) = Z kk = µ 0 k, K) K k) λ λ ) log n k,k) T n K 3/ δ Z kk ω)..8) Q p) n ω) µ τ, for almost every ω Ω, where τ is a random variable with distribution Pτ = l) = a l l! = l! λ λ λ λ x l e x dx, l = 0,,.... The following theorem is a two-index almost sure limit theorem in the left domain. The setting is similar to the one in Theorem.5. Theorem.. Let 0 < λ < λ < be fixed. Let Z kk be defined in Theorem.. Let T n, be the following domain in Let T n, = Q p) n,ω) = { } k, K) : k n, K, λ k K λ If n, so that λ n λ, then λ λ ) log n k,k) T n, K 3/ δ Z kk ω)..9) Q p) n,ω) µ τ, for almost every ω Ω,.

9 Almost sure limit theorems for random allocations 9 where τ is a random variable with distribution defined in Theorem.. The following theorem in an a.s. limit theorem in classical form. Theorem.3. Suppose that n such that n, n)) belongs to the left domain. Let Z n p) = µ 0 n, n)) n) n) and Q p0) n ω) = n log n k δ. Z p) k ω) Then, as n, one has k= Q p0) n ω) πλ) for almost every ω Ω. 3. Proof of the general a.s. limit theorem Theorem. and Remark. are formally two-index statements. However, we do not need a multiindex strong law as in Fazekas and Rychlik 003)) or a strong law for particular domains see, e.g. Gut 983)). To prove Theorem. and Remark. we will use the same strong law of large numbers that was used in ordinary a.s. limit theory. Lemma 3.. Fazekas and Rychlik 00), Lemma.3.) Let η i, i, be uniformly bounded random variables. Let F n = D n n k= d kη k, where {d k } is a nonnegative sequence with k= d k = and D n = n k= d k. Assume that there exist C > 0, ε > 0, an increasing sequence of positive numbers c n with lim n c n =, c n+ /c n = O) such that { )} cl +ε) E{η k η l } C log + log +, 3.) c k for k < l. Assume that 0 d k logc k+ /c k ), k =,,.... Then lim F n = 0 a.s. 3.) n The proof of the next lemma follows from that of Theorem.3.3 in Dudley 989). Let B, ϱ) be a complete separable metric space. Let BLB) be the space of the Lipschitz continuous bounded functions g : B R with g BL = g + g L <, where g is the sup norm and g L = sup x y gx) gy). ϱx, y)

10 0 I. Fazekas, A. Chuprunov Lemma 3.. Let µ be a finite Borel measure on B. Then there exists a countable set M BLB) depending on µ) such that for any sequence of finite Borel measures µ n, n, on B we have: µ n µ, n, if and only if for each g M gx)dµ n x) gx)dµx), n. B B Proof of Theorem...3) =.). Let µ be fixed. Let M be the countable set of functions from Lemma 3. that determines the convergence to µ. Let g M. Define the random variable η k as η k = d k α k) i=α k) d ki [gζ ki ) Egζ ki )], if d k 0 and η k = 0 if d k = 0 k ). Let K be a constant with gx) K and gx) gy) Kϱx, y), x, y B. Then for k < l, using the independence of ζ ki lj and ζ ki, = E{η k η l } = E α l) d l α k) d k i=α k) j=α l) d ki d l α k) d k i=α k) d ki [gζ ki ) Egζ ki )] 3.3) d lj [gζ lj ) gζlj ki ) + gζlj ki ) Egζ lj )] = α l) j=α l) d lj E { [gζ ki ) Egζ ki )][gζ lj ) gζlj ki )] }. As E [gζki ) Egζ ki )][gζ lj ) gζlj ki )] KE gζlj ) gζlj ki ) { )} KE{Kϱζ lj, ζlj ki ) K} 4K cl +ε) C log + log +, c k therefore we have E{η k η l } α k) d k i=α k) d ki d l α l) j=α l) { )} 4K cl +ε) C log + log +. c k So, by Lemma 3., we obtain B gx)d D n n k= i=α k) { )} d lj 4K cl +ε) C log + log + c k α k) d ki δ ζki ω)) x) 3.4)

11 Almost sure limit theorems for random allocations = D n n k= i=α k) gx)d B D n α k) n α k) k= i=α k) d ki µ ζki ) x) = d ki [gζ ki ω)) Egζ ki )] = n d k η k ω) 0, D n as n, for almost all ω Ω. By.3), the second term in 3.4) converges to B gx)dµx). Therefore, as the set M is countable, for almost all ω Ω, for all g M we have B gx)d D n n α k) k= i=α k) d ki δ ζki ω) ) x) as n. By Lemma 3., this implies.3) =.)..) =.3). Define the following measures. µ n = D n n α k) k= i=α k) d ki µ ζki, µ n,ω = D n n k= B gx)dµx), α k) k= i=α k) d ki δ ζki ω). Let A be a continuity set of µ: µ A) = 0. The expectation of µ n,ω A) is µ n A), i.e. Ω µ n,ωa)dpω) = µ n A). ow,.) means that lim n µ n,ω A) = µa), for almost every ω. Take expectation in this relation, use dominated convergence theorem to obtain lim n µ n A) = µa). So we obtained.3). 4. Proofs for central and left intermediate domains We will assume that i a i = 0 and i a i =. For fixed k, n,, k < n, introduce the following notation. ζ n = µ 0 n, ) Eµ 0 n, ) 4.) and ζ k n = ) k n i= j=k+ I {ξj / i } ) n k. 4.) The proof of our a.s. limit theorems will be based on the following inequalities. Lemma 4.. Let k < n and be fixed. Then we have Eζ n ζn) k k ) n k 4.3)

12 I. Fazekas, A. Chuprunov and Eζ n ζ k n) kn. 4.4) Proof. Let F kn be the σ-algebra generated by ξ k+,..., ξ n. First we show that our choice is optimal in the sense that ζn k is the conditional expectation E{ζ n F kn }. n E{ζ n F kn } = E I {ξj / i } ) n Fkn = = i= j= n k I {ξj / i } E I {ξj / i } F kn i= j=k+ j= n = I {ξj / i } i= j=k+ ) k ) n = ζn k. ) n = ow we can use the general equality E [η E{η F}] = Eη E [E{η F}]. Therefore Eζ n ζn) k = n = E ) k n E = U V. Here I {ξj / i } i= j= I {ξj / i } i= j=k+ n U = E I{ξ j / i } + n E I {ξj / i }I {ξj / l } = i= j= i l j= = ) n + ) ) n. Similarly Therefore V = ) k [ ) n k + ) ) n k ]. U V = ) n [ ) k ] + 4.5) + ) ) n k [ ) k ) k ] = A + B. Using the inequality where 0 a b <, l, we obtain 0 A b l a l lb a)b l, 4.6) ) n [ )] k ) n k k,

13 Almost sure limit theorems for random allocations 3 0 B ) ) n k [ ) ) ] k ) n k k. So we obtain 4.3). To prove 4.4), we use 4.5) in the following form: Eζ n ζ k n) = { [ ) n ) k+n ] + [ + ) ) n ) k ) n k ]} { [ ) k ) n k ) n ]} = X Y. ow we expand the terms in X into finite geometrical series. Moreover 0 X = = 0 Y = k i=0 k i=0 k i=0 ) i+n k i=0 ) i ) n i = [ ) i+n ) i ) n i ] ) i n i nk kk ) nk. ) n k [ ) k ) k ] k nk. In the last steps we applied 4.6). Therefore Eζ n ζn) k = X Y kn, so we obtain 4.4). The proof is complete. Proof of Theorem.4. Let ζ kk = S kk. For k < n let ζn kk = ζn/d k n, where ζn k is defined in 4.). We will show that ζn kk satisfies the conditions of Theorem.. ζn kk and ζ kk are independent for k < n. In the central domain C D n, where C depends only on α and α, see.6). Therefore, by Lemma 4., we have E ) ζ n ζn kk k D n C k α k C n. Therefore d k = c k is an appropriate choice for any positive constant c. However, as d k = k {K : k K k } α α K k log α log α ),

14 4 I. Fazekas, A. Chuprunov the above choice is possible. So, in Theorem., we can put D n = log α log α ) log n. By Theorem A, log α log α ) log n as n. So we can apply Theorem.. k n {K : α k K α } kk µ S kk γ, Proof of Theorem.5. Consider Q n from Theorem.4 and Q n. Their difference is Q n ω) Q n ω) = log α log α ) log n k n {K : K>, α k K α } kk δ S kk ω). As the summands are probability measures, we can confine attention to the weights. However, a direct calculation shows that k n {K : K>, α k K α } kk clog α log α ). Therefore, when for a fixed ω we have Q n ω) γ, as n, then Q n ω) γ, as n,. Proof of Theorem.6. It is the same as the proof of Theorem.4. The only difference is that for the weights now we have d k = and D n = α α ) log n. {K : k α K k α } k ), k α α Proof of Theorem.7. It is similar to the proof of Theorem.4. Let ζn kk and ζ kk be the same as in Theorem.4. Choose in Theorem. α n) = α n) = n) for every n. Proof of Theorem.8. Let k, n so that k < n and let = n). Let ζ kk = S kk and ζn kk = ζn/d k n, where ζn k is defined in 4.). Using the representation of D n in.6), we see that lim α 0 n D n e α α = lim =. α 0 + α)e α Therefore, there exists C > 0 not depending on n, such that n n. Consequently, by Lemma 4., E ) ζn kk ζ kk apply Theorem. with α n) = α n) = n). kn D n D n C for all C k. So we can n

15 Almost sure limit theorems for random allocations 5 ζ kk n Proof of Theorem.9. Let k, n so that k < n. Let ζ kk = S kk and = ζn/d k n, where ζn k is defined in 4.). Then the appropriate independence conditions are satisfied. Like in the proof of Theorem.8, E ) ζn kk ζ kk C k. n ow d k = kk log K k log α log α ), D n = k n {K : k α K k So we can apply Theorem.. α } {K : k α K k α } kk log K log α log α ) log n. 5. Proofs for the left domain Proof of Theorem.0. First let be fixed. Following Kolchin et al. 978), Ch., Sec., let δ = and for l =,..., let δ l be the number of steps after filling in l boxes untill filling in l boxes. Then the variables δ l are independent with geometric distribution: for each l =, 3,...,. ) m l P δ l = m) = l ), m =,,..., Let ν k = δ + δ δ k be the minimal number of steps untill filling in k boxes. Then P ν k n) = P µ 0 n, ) k ). 5.) As the variables δ l are independent with geometric distribution, we can obtain the generating function of ν k k. k ϕ,k x) = x k k Ex δ l = l ) / l ) l= l= x. [ k log ϕ,k x) = log l ) log l )] l= x = k = l ) l l= + O + l ) l x O x = = x { kk ) x } + k )kk ) + O. 6

16 6 I. Fazekas, A. Chuprunov Denote by ϱ,k the generating function of the Poisson distribution with parameter k. Then log ϱ,kx) = k x ). The difference of the logarithms of the two generating functions is log ϱ,k x) log ϕ,k x) = k { x } + k )kk ) x ) + O 6 0, as k so that 0 < M k M < and the convergence is uniform for these values of. Therefore, as k, ϱ,k x) ϕ,k x) 0, uniformly in x < and in satisfying 0 < M k M <. Using the Cauchy integral formula, p l = πi Gz) dz, C r zl+ where p l is the lth term of a distribution, while G is its differentiable) generating function, we obtain that ) k Pν k k = m) π, m 0 5.) for each fixed m {0,,...}, if k so that 0 < M k M < moreover, the above convergence is uniform in satisfying this condition. The above argument is a version of the proof of Theorem 6 in Kolchin et al. 978), Ch., Sec.. ow we shall use the ideas in Kolchin et al. 978), Ch., Sec.4. Substituting k = n m into 5.), where m is fixed, we obtain P ν n m n m) m ) = P µ 0 n, ) n) m ). 5.3) As m is fixed and n satisfies the appropriate boundedness condition, we can apply 5.). Therefore, as n, P ν n m n m) = m ) ) n m) π, m 0 uniformly in. By 5.3), the last relation implies P µ 0 n, ) n) = m ) ) n m) π, m 0 uniformly in, as n. Finally, for the parameters of the Poisson laws, when n, then n m) n 0, uniformly in. So we obtain.7).

17 Almost sure limit theorems for random allocations 7 Proof of Theorem.. We check conditions of Theorem.. Let ζ kk = Z kk = µ 0 k, K) K k). Let ζn kk = ζn k + Eµ 0 n, ) n) for k < n, where ζn k is defined in Lemma 4.. Then the independence conditions in Theorem. are satisfied. Moreover, E ) ζn kk ζ n kn We see that Moreover n k= d k = is an appropriate choice. {K : λ k K λ } {K : λ k K λ } λ k n, if λ n λ. K λ 3/ λ ) k. K λ 3/ λ ) log n = D n ow, we deal with the limit distribution. Let l and ε > 0 be fixed. Then, using Theorem.0, for appropriate n ε we have [ P µ D n k,k) T n K 3/ 0 k, K) K k) = l ) )] k π K, l D n k,k) T nε K + 3/ D n k,k) T n\t nε K ε ln n ε 3/ ln n + ε ε, if n is large enough. Here On the other hand logn /λ ) A D n k,k) T n K 3/ π A = D n {K n } λ {K n λ } {k : λ k K λ } = l! λ λ ) K ) k K, l A + B. K 3/ l! {k : λ k K λ } By the dominated convergence theorem, as K, {k : λ k K λ } k K logn /λ ) log n k K ) l ) k e k K K K ) l e k K = ) l ) e k K. K λ λ x l e x dx.

18 8 I. Fazekas, A. Chuprunov Therefore, the limit of A is lim n l! Moreover, λ λ ) = a l l! logn /λ ) log n lim n 0 B = D n { n λ K n λ } lim A = n logn /λ ) logn /λ ) {K n λ } {k : λ k K λ } {K n λ } K K = a l l!. ) k l e k K 3/ K l! K λ λ x l e x dx = λ λ ) logn /λ ) logn /λ ) log n logn /λ ) logn /λ ) { n λ K n λ } as n. So the above calculations show that as n. is D n K {k : λ k K λ } k,k) T n K 3/ P µ 0 k, K) K k) = l ) a l l!, K 0, Proof of Theorem.. The difference of the coefficients in.8) and.9) 0 D n k,k) T n K 3/ D n D n { n λ K n λ } {k : λ k K λ } 5.4) k,k) T n, K3/ K 3/ 0, when n, as we showed in the proof of Theorem.. ow.8) implies.9). Proof of Theorem.3. We check the conditions of Theorem.. Let ζ kk = Z kk = µ 0 k, K) K k). Let ζn kk = ζn k + Eµ 0 n, ) n) for k < n, where ζn k is defined in Lemma 4.. Then the independence conditions in Theorem n. are satisfied. Since lim n = λ, there exists C > 0 not depending on n, such that n C for all n. Consequently, E ) ζ kk n ζ n kn C k. n

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