aa r X i v : . [ m a t h . P R ] M a r ON MULTIVARIATE STRONG RENEWAL THEOREM
Zhiyi ChiDepartment of Statistics, University of ConnecticutStorrs, CT 06269, USA. E-mail: [email protected] 1, 2018
Abstract
This paper takes the so-called probabilistic approach to the Strong Renewal Theorem (SRT)for multivariate distributions in the domain of attraction of a stable law. A version of the SRTis obtained that allows any kind of lattice-nonlattice composition of a distribution. A generalbound is derived to control the so-called “small- n contribution”, which arises from randomwalk paths that have a relatively small number of steps but make large cumulative moves.The asymptotic negligibility of the small- n contribution is essential to the SRT. Applications ofthe SRT are given, including some that provide a unified treatment to known results but withsubstantially weaker assumptions. Keywords and phrases.
Renewal, regular variation, infinitely divisible, large deviations.
For a probability distribution F on R d , the Strong Renewal Theorem (SRT) is said to hold if | x | d A ( | x | ) U ( x + E ) → g ( x/ | x | ) u ( E ) (1.1)uniformly in a certain sense as | x | → ∞ , where U = P ∞ n =1 F ∗ n is the renewal measure with F ∗ n the n -fold convolution of F , A ( · ) > , ∞ ), E is some “nice” set, x + E denotes { x + y : y ∈ E } , g ( · ) { ω ∈ R d : | ω | = 1 } , and finally, u is a nonzero σ -finitemeasure on R d ; see Theorem 2.3 for precise explanation. The definition extends the one in [37]that only considers F on Z d . In [37], x stays in Z d and E = { } . However, in general, x can takeany value in R d and E has to depend on the lattice-nonlattice composition of F .There are two main approaches to the SRT. One is based on Fourier analysis of the renewalmeasure [8, 14, 15, 17, 35, 37]. The other is the so-called probabilistic approach [3, 5, 6, 10, 11, 31,36, 37]. It is based on the realization that the two partial sums that comprise the renewal measure, X n ≥ A ( δ | x | ) F ∗ n ( x + E ) and X n n ” and “small- n ” contributions,respectively. In general, the big- n contribution can be dealt with using Local Limit Theorems(LLTs), essentially yielding the limit (1.1) provided it exists [6, 14, 17, 31, 37]. In contrast, withoutadditional conditions, the small- n contribution often fails to converge, hence ruling out the exis-tence of the limit [17, 36, 37]. Recently, to control the small- n contribution when d = 1, integralcriteria were proposed [5, 6]. This paper extends the idea in [5, 6] to the multivariate case. As inprevious works, it investigates the SRT for F in the domain of attraction without centering of a nondegenerate stable law. By definition, there are a n ∈ R such that F ∗ n ( a n d x ) weakly convergesto an α -stable law not concentrated in any linear manifold of dimension d −
1. Denote this by1 ∈ D ( α ). To establish the SRT for F , the small- n contribution is approached by analyzingvarious subsets of random walk paths, in particular components of the paths at different scales.In addition to being quite easily applicable, the resulting SRT gives a unified treatment to manyknown results, sometimes with substantially weaker assumptions.It should be remarked that for d >
1, a more general type of stability can be defined, namelyoperator-stability (cf. [30]). Characterizations of domain of attraction for operator-stability aswell as the corresponding LLTs are known (cf. [9, 20–22, 27]). Since operator-stability has foundapplications, e.g., in the study on the ladder height and ladder epoch of random walks in R [12, 19],it is of interest to consider the related SRT. This topic is beyond the scope of the paper.During the revision of the paper, sufficient and necessary conditions for the SRT in the univari-ate case were announced [4]. The key to the new result is a new local large deviation bound for F ∗ n . An extension to the multivariate case will be interesting in future work.Section 2 presents the main result of the paper, which is a multivariate SRT in Theorem 2.3.The SRT is preceded by a result on the lattice-nonlattice composition of a distribution, which is anissue unique to the multivariate case and has to addressed in order to formulate the SRT properly.Applications of the SRT are also presented in the section. The proof of Theorem 2.3 is outlinedin Section 3. It is shown in this section that the theorem is a consequence of Theorems 3.1 and3.2 that deal with the big- n and small- n contributions, respectively. As a preparation for theirproofs, Section 4 derives bounds for the L´evy concentration and local large deviation of F ∗ n . ThenTheorems 3.1 and 3.2 are proved in Sections 5–6, respectively. Section 7 collects proofs of minorresults on the SRT. The lattice-nonlattice composition is proved in Appendix A.The rest of this section fixes notation. For a , b ∈ R , denote a ∨ b = max( a, b ), a ∧ b = min( a, b ),and a + = a ∨
0. For x ∈ R d , denote by | x | its Euclidean norm and k x k = max i | t i | its sup-norm,where t i are the coordinates of x . Denote B d = { x ∈ R d : | x | ≤ } , S d − = { x ∈ R d : | x | = 1 } ,and I d = [0 , d . For Λ ⊂ R and D , E ⊂ R d , denote aD = { ay : y ∈ D } , Λ x = { λx : λ ∈ Λ } , x + D = { x + y : y ∈ D } , and D + E = { y + z : y ∈ D, z ∈ E } . Denote M ∈ Λ m × d if M is an m × d matrix of elements in Λ, and M D = { M y : y ∈ D } . Denote by diag( a , . . . , a n ) the diagonal matrixwith the i -th diagonal element being a i , and Id n the n × n identity matrix. For a linear subspace V of R d , denote by π V the projection onto V . If f ∈ L ( R d ), denote b f ( t ) = R e i h t,x i f ( x ) d x .For functions f and g , f ( x ) = O ( g ( x )), f ( x ) ≪ g ( x ), and g ( x ) ≫ f ( x ) all mean | f ( x ) | ≤ C | g ( x ) | for some constant C >
0, and f ( x ) ≍ g ( x ) means g ( x ) ≪ f ( x ) ≪ g ( x ). If C depends on param-eters a , . . . , a k , when it is necessary to emphasize the dependence, denote f ( x ) = O a ,...,a k ( g ( x )), f ( x ) ≪ a ,...,a n g ( x ), or g ( x ) ≫ a ,...,a n f ( x ). By f ( x ) = o a ,...,a k ( g ( x )) as x → ∞ it means there is afunction M ( ǫ ) = M ( ǫ ; a , . . . , a k ), such that | f ( x ) | ≤ ǫ | g ( x ) | for all x ≥ M ( ǫ ). It is well known that the SRT, in particular, the big- n contribution involved, has to be handleddifferently for lattice distributions and nonlattice ones [1, 18, 28, 32, 33]. Recall that if a distributionis concentrated on a + Γ for some a ∈ R d and lattice Γ ⊂ R d , then the distribution as well as anyrandom variable following it is said to be lattice. By definition, Γ is an additive subgroup of R d with no cluster points. For d >
1, a complication is that a distribution may be jointly lattice andnonlattice, so it is necessary to first know its lattice-nonlattice composition in order to establishthe SRT. The lattice-nonlattice composition of a nondegenerate distribution is characterized bythe next result that will be proved in Appendix A. Recall that two integers are coprime if their2reatest common divisor is 1, and a , . . . , a n ∈ R are rationally independent if for m , . . . , m n ∈ Z , P m i a i ∈ Z ⇐⇒ all m i = 0 ([25], p. 51). Proposition 2.1.
Let X be a nondegenerate random variable in R d . Denote by ϕ X ( u ) = E [ e i h u,X i ] its characteristic function. Then there exist a linear subspace V ⊂ R d , a nonsingular matrix T ∈ R d × d , and integers ≤ ν ≤ r ≤ d and q ≥ with the following properties.(1) π V ( X ) is lattice, and | ϕ X (2 πv ) | < for v ∈ R d \ V .(2) | ϕ T X (2 πu ) | = 1 ⇐⇒ u ∈ Z r × { } , where is the zero vector in Z d − r .(3) Let T X = (
Y, Z ) with Y ∈ R r and Z ∈ R d − r . Then P { Y ∈ β + Z r } = 1 for β =(0 , . . . , , β ν , . . . , β r ) , where β ν = p/q ∈ Q with ≤ p < q being coprime and β ν +1 , . . . , β r ∈ (0 , \ Q are rationally independent.Furthermore, V , r , ν , and q with above properties are unique, and r = dim( V ) .Remark. (1) In the decomposition, β ν = 0 ⇐⇒ p = 0 and q = 1.(2) It was claimed in [33] that according to p. 64–75 of [31], X can always be linearly transformedinto a nondegenerate ξ ∈ R d , such that for some β ∈ R r , ϕ ξ (2 πu ) = exp[2 π i h β, v i ] if u = ( v, ∈ Z r × { } and | ϕ ξ (2 πu ) | < X is assumed to be Z d -valued in [31], so it isunclear how the claim was obtained. Moreover, the SRT requires detailed information about β , sothe claimed transformation is insufficient.Denote l = (0 , . . . , , β ν ) , w = ( β ν +1 , . . . , β r ) . (2.1)By Proposition 2.1, if Y is partitioned as ( L, W ) ∈ R ν × R r − ν so that T X = (
Y, Z ) = (
L, W, Z ) , (2.2)then ( L, W ) and Z are the lattice and nonlattice components of X , respectively. Meanwhile, L and ( W, Z ) are the arithmetic and nonarithmetic components, respectively. The dimensions of thecomponents are unique, and the number q ∈ N such that DL is Z ν -valued is unique, where D = diag(1 , . . . , , q ) ∈ Z ν × ν . (2.3)However, Proposition 2.1 does not say β ν , . . . , β r are unique.The SRT for an aperiodic random walk is studied in [37]. A random variable ξ ∈ Z ν is said tobe aperiodic if for any nonrandom t ∈ R ν , P {h t, ξ i ∈ Z } = 1 ⇐⇒ t ∈ Z ν , and is said to be stronglyaperiodic if for any nonrandom t ∈ R ν and c ∈ R , P {h t, ξ i ∈ c + Z } = 1 ⇐⇒ t ∈ Z ν and c ∈ Z (cf. [31], T7.1, P7.8). The following result will be proved in Appendix A. Recall that a matrix K ∈ Z ν × ν has an inverse in Z ν × ν ⇐⇒ | det K | = 1, in which case K − is det K times the adjugateof K . Proposition 2.2. (1) Let ξ ∈ Z ν be nondegenerate. Then ξ is aperiodic ⇐⇒ there are K ∈ Z ν × ν with | det K | = 1 and coprime integers ≤ p < q , such that ξ = K − ( Dζ + pe ν ) , where ζ is strongly aperiodic, D isdefined in (2.3) , and e ν = (0 , . . . , , is the ν -th standard base vector of Z ν .(2) For the L and D in (2.2) – (2.3) , L − β ν e ν is strongly aperiodic and DL is aperiodic. .2 A sufficient condition for the SRT The lattice-nonlattice composition in Proposition 2.1 allows the SRT to be formulated properly.The next SRT is the main result. It also implies certain property of the limiting law involved.Recall that a nondegenerate stable law has an infinitely differentiable density with all derivativesvanishing at ∞ ([29], Example 28.2). Theorem 2.3 (SRT) . Let F ∈ D ( α ) be a distribution on R d with < α ∨ < d . Let ψ be thedensity of the limiting stable law of F ∗ n ( a n d x ) , where a n > is a sequence of norming constants.Let A ( s ) be any function regularly varying at ∞ such that A ( a n ) /n → as n → ∞ . Define ̺ s ( ω ) = αq − Z /s ψ ( uω ) u d − α − d u, ω ∈ S d − , s > . (2.4) Let T ∈ R d × d be nonsingular such that T X = (
L, W, Z ) as in (2.2) . Fixing Υ ∈ Z ν × ν with | det Υ | = 1 , define ∆ h = ( D − Υ I ν ) × ( hI d − ν ) , h > , (2.5) where D is given in (2.3) . Define K ( t, a, η, h ) = Z | z | <η | t | F ( t − z + hI d ) e −| z | /a d z (2.6) for t ∈ R d , a > , r > , and h > . Define κ = ⌊ d/α ⌋ . (2.7) Then, if there are θ ∈ (0 , /κ ) and η > such that lim δ → lim s →∞ s d A ( s ) X n ≤ A ( δs ) na − dn sup | t | >θs K ( t, a n , η, h ) = 0 , (2.8) then the following convergence holds lim s →∞ sup ω ∈ S d − (cid:12)(cid:12)(cid:12)(cid:12) s d A ( s ) U ( sω + T − ∆ h ) − h d − ν ̺ ( ω ) | det T | (cid:12)(cid:12)(cid:12)(cid:12) = 0 (2.9) and moreover, lim δ → sup ω ∈ S d − | ̺ ( ω ) − ̺ δ ( ω ) | = 0 . (2.10) Remark.
Eq. (2.9) makes clear what the uniform convergence in (1.1) means and will be referredto as the SRT. For α = 2, since ψ is a normal density, the uniform convergence in (2.10) holdswithout assuming (2.8). However, for α ∈ (0 ,
2) and d >
1, it may fail to hold. Indeed, followingupon Example 5-B of [37], given 1 ≤ k < d , if ψ is the density of ( ξ , . . . , ξ d ), where ξ i ∈ R are i.i.d.symmetric α -stable with α ≤ ( d − k ) / ( k + 1) and density g , then for any ω ∈ S d − with at most k nonzero coordinates, ψ ( sω ) s d − α − = g ( ω s ) · · · g ( ω d s ) s d − α − ≫ s d − ( k +1)(1+ α ) ≫ s − as s → ∞ ,giving ρ ( ω ) = ∞ .The condition (2.8) is often easy to check. From Theorem 2.3, the following SRT follows. When F is concentrated on Z d , the same result was established in [37] but with a very different argument.Unlike [37], the proof given in Section 7 applies to F with any lattice-nonlattice composition.4 heorem 2.4. If ≤ d/ < α ∈ (0 , (so d = 2 or 3), then the SRT (2.9) holds for any F ∈ D ( α ) and (2.10) holds for any nondegenerate α -stable law on R d . The above result does not cover α = 2. For this case, the next result provides weaker conditionsthan [31], P26.1, and [35]. Proposition 2.5.
Let F ∈ D (2) and X ∼ F . Denote q X ( s ) = P {| X | > s } . Then the SRT (2.9) holds for F in each of the following cases.(1) d = 3 .(2) d = 4 and q X ( s ) Z s u − A ( u ) d u = o ( A ( s ) /s ) , s → ∞ . (3) d ≥ and q X ( s ) = o ( s − d ) . For d = 3, the SRT for X ∼ F ∈ D (2) is established in [31], P26.1, under the condition σ = E | X | < ∞ . For d = 4, the SRT is established in [35] under the condition E | X | (ln | X | ) + < ∞ ,which implies σ < ∞ . However, if σ < ∞ , then by the Central Limit Theorem, A ( s ) /s ≍ q X ( s ) = o (1 / ( s ln s )), which is a weaker condition. In Example 2.6 below,it is shown that even σ < ∞ is not necessary. Finally, for d ≥
5, the SRT is established in [35]under the condition E | X | d − < ∞ . Clearly, the condition in (3) is weaker. In Example 2.8, it willbe seen that the condition and even σ < ∞ is not necessary. Example 2.6.
Let X ∈ R be spherically symmetric with q X ( s ) ≍ / ( s ln s ). Put V X ( s ) = E [ | X | {| X | ≤ s } ]. By V X ( s ) ≍ R s uq X ( u ) d u ≍ R s ( u ln u ) − d u ≍ ln ln s , E | X | = ∞ . On theother hand, since A ( s ) ∼ s /V X ( s ) ≍ s / ln ln s , then q X ( s ) = o (1 /A ( s )) and so X ∈ D (2) (cf.(3.3) and [28], Th. 4.1). Meanwhile, R s u − A ( u ) d u ≍ ln s/ (ln ln s ) . As a result, the condition inProposition 2.5(2) is satisfied and the SRT holds.Next consider a multivariate version of a result in [5, 6]. Define φ ( x ) = | x | d F ( x + hI d ) A ( | x | ) . The classical condition sup φ < ∞ for d = 1 played a critical role in several works [10, 36, 37]. Theorem 2.7.
Let α ∈ (0 , ∩ (0 , d/ . Suppose there are T ≥ and η > such that sup ω Z | z | <ηs [ φ ( sω − z ) − T ] + d z = o ( A ( s ) ) , s → ∞ , (2.11) then for any θ > and δ > , X n ≤ A ( δs ) na − dn sup | t |≥ θs K ( t, a n , η, h ) = [ o (1) + δ α ] A ( s ) s d , s → ∞ (2.12) and consequently the SRT (2.9) holds for F and (2.10) for the limiting stable law of F ∗ n ( a n d x ) . Example 2.8.
As an application of Theorem 2.7, it can be shown that for d ≥
5, the conditionin Proposition 2.5(3), i.e., q X ( s ) = o ( s − d ), is not necessary for the SRT. Indeed even E | X | < ∞ is not necessary. Let X have density f ( x ) = c (1 + | x | ) − d − , where c > V X ( s ) = E [ | X | {| X | ≤ s } ]. Then V X ( s ) = R | x |≤ s | x | f ( x ) d x ≍ ln s as s → ∞ , so E | X | = ∞ .However, by s q X ( s ) = s R | x |≥ s f ( x ) d x = O (1), the law of X is in D (2) ([28], Th. 4.1). Moreover,since A ( s ) ∼ s /V X ( s ) (cf. (3.3)), φ ( x ) ≍ h | x | d f ( x ) A ( | x | ) is bounded. Then by Theorem 2.7, theSRT holds. 5 xample 2.9. Let ξ ∈ R be symmetric α -stable with α ∈ (0 ,
2) and X = ( X , . . . , X d ) with X i i.i.d. ∼ ξ . From the remark for Theorem 2.3, if α ≤ ( d − /
2, then ̺ ( e i ) = ∞ . Therefore, both(2.9) and (2.10) fail to hold. The goal here is to show that if α > ( d − / d ≤ θ > < η < / (10 d ). Given t = ( t , . . . , t d ), put x = t i , where | t i | = max | t j | .Then | x | ≥ | t | /d and so K ( t, a, η, h ) = Z | z | <η | t | d Y j =1 P { X i ∈ t j − z j + hI } e −| z | /a d z ≤ Z | z i | <ηd | x | d Y j =1 P { X i ∈ t i − z i + hI } e −| z i | /a d z = h d − Z | u | <ηd | x | P { ξ ∈ x − u + hI } e −| u | /a d u (2.13)For x ∈ R , P { ξ ∈ x + hI } ≪ h | x | − α − . On the other hand, for | t | ≫
1, if | x | ≥ | t | /d and | u | ≤ ηd | x | < | x | /
10, then | x − u | ≍ | t | . Then K ( t, a, η, h ) ≪ h | t | − α − Z ∞−∞ e −| u | /a d u ≪ a | t | − α − . Since a n ∼ n /α and A ( t ) ∼ t α , X n ≤ A ( δs ) na − dn sup | t | >θs K ( t, a n , η, h ) ≪ h s − α − X n ≤ A ( δs ) n − ( d − /α ≪ δ α − d +1 A ( s ) /s d . Then by Theorem 2.3, the SRT holds for X .As seen earlier, for a strictly α -stable distribution G on R d with α ∈ (0 , ̺ δ to ̺ may fail to hold if d >
1, where ̺ δ is defined in (2.4). As an application ofTheorem 2.3, a sufficient condition for the uniform convergence will be provided next. Let ψ bethe density of G . By Theorem 14.10 in [29], b ψ ( t ) = exp[ − C E f α ( h ξ, t i ) + i h τ, t i { α = 1 } ] , t ∈ R d , where C > τ ∈ R d are constants, ξ ∈ S d − with E ξ = 0 if α = 1, and for θ ∈ R , f α ( θ ) = | θ | α [1 − i tan( πα/ θ ) { α = 1 } + i(2 /π )sgn( θ ) ln | θ | { α = 1 } ]. Conversely, for any C > τ ∈ R d , and ξ on S d − , provided E ξ = 0 if α = 1, the RHS is the characteristic function ofa nonconstant strictly stable distribution. Proposition 2.10. If ξ has a bounded density with respect to the spherical measure on S d − , then (2.10) holds for G , i.e., sup | ̺ δ − ̺ | → as δ → . Finally, it is of interest to infer properties of an ID distribution from its L´evy measure (cf. [1],8.2.7; [16], XVII.4; [13]). This is the motivation of the following result.
Proposition 2.11.
Let α ∈ (0 , and F ∈ D ( α ) be ID with L´evy measure ν . Define A ν ( s ) =1 /ν ( R d \ sB d ) and φ ν ( x ) = | x | d ν ( x + hI d ) A ν ( | x | ) . If condition (2.11) is satisfied with φ and A replaced with φ ν and A ν , respectively, then the SRT holds for F . Outline of proof
Several facts about distributions in the domain of attraction will be needed. For random variable X in R d , for s > u ∈ R d , denote q X ( s ) = P {| X | > s } , c X ( s ) = E [ X {| X | ≤ s } ] ,m X ( s, u ) = E [ h u, X i {| X | ≤ s } ] , V X ( s ) = E [ | X | {| X | ≤ s } ] . For x , x , . . . ∈ R d , denote S ( x ) = 0 and S n ( x ) = S n − ( x ) + x n , n ≥
1. For 0 < α ≤
2, denote F ∈ D ( α ) if there are a n ∈ R and b n ∈ R d , such that for X , X , . . . i.i.d. ∼ F , S n ( X ) /a n − b n weakly converges to an α -stable law that is nondegenerate ([29], Def. 24.16). See [28], Th. 4.1–4.2,for necessary and sufficient conditions for F ∈ D ( α ). In particular, if X ∼ F ∈ D ( α ), then V X ∈ R − α , (3.1)where R θ denotes the class of functions that are regularly varying at ∞ with exponent θ , and q X ( s ) = [2 /α − o (1)] V X ( s ) /s , as s → ∞ . (3.2)Let A be any function such that A ( s ) ∼ s /V X ( s ) as s → ∞ . (3.3)Then for any sequence a n such that A ( a n ) /n → n → ∞ , S n ( X ) /a n − b n D → µ for suitable b n , (3.4)where µ is the aforementioned stable law. Define a = 1. By definition, F ∈ D ( α ) if (3.4) holdswith b n = 0, in which case µ is strictly stable ([2], § F ∈ D ( α ), F ∈ D ( α ) ⇐⇒ ( n/a n ) c X ( a n ) converges as n → ∞ . (3.5)Proofs of the above facts are readily available for the univariate case ([1]) but not so for themultivariate case. For convenience, their proofs are given in the Appendix B. As noted in Section 1, the probabilistic approach to the SRT deals with the big- n and small- n contributions (1.2) in different ways. Theorem 2.3 is a consequence of the following results. Theorem 3.1 (Big- n contribution) . Let F ∈ D ( α ) , A , ψ , and ̺ s be as in Theorem 2.3. Let X ∼ F and T ∈ R d × d be nonsingular such that T X = (
L, W, Z ) as in (2.2) . Give δ > and h > ,define ∆ h as in (2.5) and r δ,h ( sω ) = s d A ( s ) X n ≥ A ( δs ) F ∗ n ( sω + T − ∆ h ) . Then as s → ∞ , sup ω ∈ S d − (cid:12)(cid:12)(cid:12)(cid:12) r δ,h ( sω ) − h d − ν ̺ δ ( ω ) | det T | (cid:12)(cid:12)(cid:12)(cid:12) = o δ,h (1) . (3.6)7 heorem 3.2 (Small- n contribution) . Let F ∈ D ( α ) and A be as in Theorem 2.3. Define κ as in (2.7) . Then given < θ < /κ , η > , and ǫ > , for < δ ≪ θ,η,ǫ , s ≫ θ,η,ǫ,δ,h , and n ≤ A ( δs ) , sup ω ∈ S d − F ∗ n ( sω + hI d ) ≪ h na − dn sup | t | >θs K ( t, a n , η, h ) + ǫ n ( s ) (3.7) with ǫ n ( s ) > satisfying X n ≤ A ( δs ) sup r ≥ s ǫ n ( r ) ≪ ǫA ( s ) /s d . (3.8) In particular, X n ≤ A ( δs ) sup | x |≥ s F ∗ n ( x + hI d ) ≪ h X n ≤ A ( δs ) na − dn sup | t | >θs K ( t, a n , η, h ) + ǫA ( s ) /s d . It is clear that the big- n contribution r δ,h ( sω ) depends on the the lattice-nonlattice compositionof X . In contrast, in dealing with the small- n contribution, the lattice-nonlattice composition isunimportant. Once the two theorems are proved, Theorem 2.3 immediately follows from the nextresult, which itself has some application; see Example 3.4. Proposition 3.3.
Let F ∈ D ( α ) , A , ψ , and ̺ s be as in Theorem 2.3. Then, if the small- n contribution is asymptotically negligible, i.e, lim δ → lim s →∞ sup ω ∈ S d − s d A ( s ) X n ≤ A ( δs ) F ∗ n ( sω + hI d ) = 0 , (3.9) then (2.9) and (2.10) hold. Conversely, if (2.9) and (2.10) hold, then (3.9) holds.Remark. (1) If X ∈ Z d is aperiodic, then Theorem 3.1 is implied by [37], Eq. (3.6). When X is notstrongly aperiodic, the proof in [37] relies on approximating X by a strongly aperiodic one; also see[31], P26.1. However, by Proposition 2.2, for some L and D as in (2.2)–(2.3) and some Υ ∈ Z d × d with det Υ = ± T X = L with T = D − Υ. Letting ∆ d = D − Υ I d , U ( x + I d ) = U ( x + T − ∆ d ).Then, without approximation, Theorem 3.1 leads to Eq. (3.6) in [37].(2) In [37], for aperiodic X , it is shown that (3.9) combined with (2.10) implies (2.9). However,by Proposition 3.3, (3.9) implies both (2.9) and (2.10).(3) Proposition 3.3 is weaker than Proposition A of [6] which essentially states that for d = 1, | x | U ( x + ∆ h ) /A ( | x | ) converges as x → ±∞ ⇐⇒ (3.9) holds, and if either happens the limit mustbe h̺ . However, the argument for that result does not apply to d > Proof of Proposition 3.3.
Since ψ is bounded and d > α , sup ω ̺ s ( ω ) < ∞ for each s > ̺ s ( ω ) ↑ ̺ ( ω ) as s ↓
0. Clearly, r δ,h ( sω ) ≥ δ . By (3.6), for s ≫
1, sup ω r ,h ( sω ) < ∞ . Then from r ,h ( sω ) − r ,h ( sω ) = ( s d /A ( s )) P n
0, there is η >
0, such that for any 0 < δ ≤ η and s ≫ η ≤ sup ω | r ,h ( sω ) − r δ,h ( sω ) | ≤ sup ω | r ,h ( sω ) − r η,h ( sω ) | ≤ h d − ν ǫ . By Theorem 3.1, for s ≫ δ ω | r δ,h ( sω ) − h d − ν ̺ δ ( ω ) | ≤ h d − ν ǫ . Combining the inequalities, sup ω | r ,h ( sω ) − h d − ν ̺ δ ( ω ) | ≤ h d − ν ǫ . As the inequality holds for η and δ , sup ω | ̺ δ ( ω ) − ̺ η ( ω ) | ≤ ǫ . Letting δ → ǫ >
0, there is δ > s →∞ sup ω | r ,h ( sω ) − r δ,h ( sω ) | < ǫ , so (3.9) holds if hI d therein isreplaced with ∆ h . Since hI d can be covered by a finite number of z + ∆ h , then (3.9) follows.8s an application of Proposition 3.3, consider a classical example on multivariate SRT given in[37]. The following formulas will be used in Example 3.4 and in the proofs in Section 7. First, itcan and will always be assumed without loss of generality that A is strictly increasing and A (0) = 0 , A ′ ( s ) ≍ A ( s ) /s for s > . (3.10)Then given β , for s ≫
1, by change of variable and A ′ ( s ) ≍ A ( s ) /s ,˜ A β ( s ) := X n ≤ A ( s ) na − βn ≪ Z A ( s )1 u d u ( A − ( u )) β ≪ Z sa A ( u ) d uu β +1 ≪ ( A ( s ) s − β if α > β/ O (1) if α < β/ . (3.11) Example 3.4.
Consider the following modified version of Example 5-A in [37]. Let d >
1. Let ξ ∈ Z \ { } , such that for k ∈ N , P { ξ = k } = P { ξ = − k } = ( ck − − d/ ln k if k
6∈ { n : n ≥ } ck − d/ /b k otherwisewhere c > b k ≫
1. Let X = ( X , . . . , X d ), with X i i.i.d. ∼ ξ . Then X ∈ D ( α ) with α = d/
2. The limiting stable density is ψ ( u ) = g ( u ) · · · g ( u d ), with g the univariatesymmetric α -stable density. Then for any ω ∈ S d − , since it has at least one coordinate withabsolute value ≥ /d , ψ ( sω ) ≪ sup /d ≤ a ≤ g ( as ) ≪ s − α − , giving ψ ( sω ) s d − α − ≪ s d − α − = s − for s ≫
1. As a result, (2.10) holds.On the other hand, by [37], if d = 2 , b k ≪ k , then the SRT fails to hold for X . Itwill be shown next that if d = 2 ,
3, then the SRT (2.9) holds ⇐⇒ ln k = o ( b k ) as k → ∞ , and if d = 4, then the SRT (2.9) holds ⇐⇒ (ln k ) = o ( b k ).Without loss of generality, let h = 1, so hI = [0 , b ( z ) be a function such that b ( z ) ≡ b k in each [ k, k + 1). First, let d = 2 or 3. It suffices to show that if ln k = o ( b k ), then the SRT (2.9)holds. By [37], X ∼ − X ∈ D ( α ) with α = d/ ∈ (0 , α ∈ (0 , P { ξ > s } ≤ q X ( s ) = P {| X | > s } ≤ d P {| ξ | > s/ √ d } ≍ P { ξ > s } ≍ s − d/ ln s, s ≫ , it follows that A ( s ) ∼ /q X ( s ) ∼ Cs d/ / ln s for some constant C > θ > < η < / (10 d ). Then the bound (2.13) still holds, i.e., for t = ( t , . . . , t d ),letting x = t i with | t i | = max | t j | , K ( t, a, η, h ) ≤ Z | u | <ηd | x | P { ξ ∈ x − u + I } e −| u | /a d u. For each u with | u | < ηd | x | , x − u + I has exactly one k ∈ Z . For | t | ≫ | k | ≍ | x − u | ≍ | x | ≍ | t | .If k
6∈ {± n : n ≥ } , then P { ξ ∈ x − u + I } = c | k | − − d/ ln | k | ≍ | t | − − d/ ln | t | = q X ( | t | ) / | t | , andlikewise, if k ∈ {± n : n ≥ } , then P { ξ ∈ x − u + I } ≍ | t | − d/ /b ( t ) ≪ q X ( | t | ) / ( b ( | t | ) ln | t | ). It canbe seen that the set of u ∈ ( − ηd | x | , ηd | x | ) with x − u + I containing one k ∈ {± n : n ≥ } is asingle interval of length at most 1. Then K ( t, a, η, h ) ≪ q X ( | t | ) | t | Z e −| u | /a d u + q X ( | t | ) b ( | t | ) ln | t | ≪ aq X ( | t | ) | t | + q X ( | t | ) b ( | t | ) ln | t | . Since q X ( s ) / (ln s ) ≍ A ( s ) /s d , if ln s = o ( b ( s )) as s → ∞ , then for s ≫ X n ≤ A ( δs ) na − dn sup | t | >θs K ( t, a n , η, h ) ≪ ˜ A d − ( δs ) sA ( s ) + o ( A ( s ) /s d ) ˜ A d ( δs ) . (3.12)9iven δ >
0, for s ≫ δ
1, by (3.11) ˜ A d − ( δs ) ≪ A ( δs ) / ( δs ) d − ≪ δs/ (ln s ) , while˜ A d ( δs ) ≪ Z δsa A ( u ) d uu d +1 ≪ Z a A ( u ) d uu d +1 + Z ∞ d uu (ln u ) < ∞ . Then the LHS in (3.12) is O ( δ (ln s ) − /A ( s )) + o ( A ( s ) /s d ) = O ( δA ( s ) /s d ). Since δ is arbitrary,then by Theorem 2.3, the SRT holds.Next let d = 4. Then E | X | = ∞ . However, as s → ∞ , s q X ( s ) ≍ ln s , V X ( s ) = Z s u P {| X | ∈ d u } = 2 Z s u [ q X ( u ) − q X ( s )] d u ≍ (ln s ) , and for t = ( t , . . . , t ), m X ( s, t ) = X i =1 t i E [ X i {| X | ≤ s } ] + X i = j t i t j E [ X i X j {| X | ≤ s } = | t | V X ( s ) /d. Then by Theorem 4.1 of [28], X ∼ − X ∈ D (2) and A ( s ) ∼ Cs / (ln s ) as s → ∞ for someconstant C >
0. The bound on K just above (3.12) still holds. Then X n ≤ A ( δs ) na − n sup | t | >θs K ( t, a n , η, h ) ≪ q X ( s ) ˜ A ( δs ) s + q X ( s ) ˜ A ( δs ) b ( s ) ln s . (3.13)Given δ >
0, for s ≫ δ
1, as A ( s ) = s / (ln s ) , by similar calculation as in the case d = 2 or 3,˜ A ( δs ) ≪ δs/ (ln s ) and ˜ A ( δs ) = O (1). Then the LHS of (3.13) is s − A ( s ) O ( δ/ ln s +(ln s ) /b ( s )).Thus, if (ln s ) = o ( b ( s )), then (2.9) holds. Conversely, if b ( s ) / (ln s )
6→ ∞ , then by similarargument as in [37], (3.9) cannot hold. Since (2.10) holds, by Proposition 3.3, the SRT (2.9)cannot hold.
For a random variable X ∈ R d , define Q X ( h ) = sup x ∈ R d P { X ∈ x + hI d } , h ≥ . The function is a special case of L´evy concentration function of multivariate random variables,which has been studied before ([24, 38]). The purpose here is to show the following result.
Lemma 4.1.
There is an absolute constant c d > , such that Q X ( h ) ≤ c d [(1 /a ) ∨ h ] d Z k t k≤ a | ϕ X ( t ) | d t, a > . Proof.
The argument follows the one on p. 22–26 of [26]. Let f be a probability density on R d suchthat f ( x ) = f ( − x ) and b f ∈ L . For y ∈ R d and a >
0, by applying Fourier inversion formula tothe density of X + a − Y , where Y has density f and is independent of X , Z f ( ax ) P { X ∈ y + d x } = 1(2 πa ) d Z e i h t,y i b f ( t/a ) ϕ X ( t ) d t ≤ πa ) d Z | b f ( t/a ) ϕ X ( t ) | d t.
10n the other hand, Z f ( ax ) P { X ∈ y + d x } ≥ Z k x k≤ h/ f ( ax ) P { X ∈ y + d x }≥ inf k x k≤ ah/ f ( x ) · P { X ∈ y + [ − h/ , h/ d } . As a result, Q X ( h ) ≤ (2 πa ) − d sup k x k≤ ah/ f ( x ) Z | b f ( t/a ) ϕ X ( t ) | d t. (4.1)Now for x = ( x , . . . , x d ), let f ( x ) = Q f ( x i ), where f ( y ) = 3 / (8 π )[sin( y/ / ( y/ for y ∈ R \{ } and f (0) = 3 / (8 π ). Then for t = ( t , . . . , t d ), b f ( t ) = Q b f ( t i ), where b f ( t i ) = | t i | ≥ , − | t i | ) if | t i | ∈ [1 / , − t i + 6 | t i | if | t i | ≤ / . See p. 25 [26]. Let c d = (2 π ) − d sup k x k≤ / /f ( x ). Then by (4.1), for a ≤ /h , Q X ( h ) ≤ c d (1 /a ) d Z k t k≤ a | ϕ X ( t ) | d t. On the other hand, for a ≥ /hc d h d Z k t k≤ a | ϕ X ( t ) | d t ≥ c d h d Z k t k≤ /h | ϕ X ( t ) | d t ≥ Q X ( h ) , where the second inequality follows from the previous display. Combining the two displays thenfinishes the proof. The following bounds will be used in the proof of Theorem 3.2.
Proposition 4.2.
Let F ∈ D ( α ) and X, X , X , . . . be i.i.d. ∼ F . Put M = 0 and for n ≥ , put M n = max {| X | , . . . , | X n |} . Then there are s > , C > both only depending on { F, A } , suchthat for all x ∈ R d , s ≥ s , h > , and n ≥ , P { S n ( X ) ∈ x + hI d , M n ≤ s } ≪ h ( s − d + a − dn ) e −| x | /s + Cn/A ( s ) . The bound is a multivariate generalization of the local large deviation bounds in [5, 7, 10].Letting s = ∞ , the bound yields Q S n ( X ) ( h ) ≪ h a − dn , which can also be derived from the LLT ofStone (cf. Proposition 5.1). The case n = 0 is included in Proposition 4.2 only for convenience,which follows by noting P { S ( X ) ∈ x + hI d } = {− x ∈ hI d } ≤ {| x | ≤ h √ d } . Let n ≥ Lemma 4.3.
There is s > such that inf ω ∈ S d − E [ h ω, X − X i {| X | ∨ | X | ≤ s / (2 √ d ) } ] > . s > ω ∈ S d − and s ≥ s , Z ( s, ω ) := E [ e h ω,X i /s {| X | ≤ s } ] ∈ (0 , e ) . Define the following probability measure concentrated in sB d , G s,ω (d x ) = Z ( s, ω ) − e h ω,x i /s {| x | ≤ s } F (d x ) . Let x = rv , where r = | x | and v ∈ S d − . Let Y, Y , Y , . . . be i.i.d. ∼ G s,v . Then P { S n ( X ) ∈ x + hI d , M n ≤ s } = [ Z ( s, v )] n E [ e −h v,S n ( Y ) i /s { S n ( Y ) ∈ x + hI d } ] . The function Z ( s, v ) on the RHS has the following property. Lemma 4.4. [ln Z ( s, ω )] + ≪ /A ( s ) for s ≥ s and ω ∈ S d − . Thus, there is a constant C = C ( F, A ) > Z ( s, ω ) ≤ e C/A ( s ) for s ≥ s and ω ∈ S d − .On the other hand, since h v, y i − h v, x i ≥ −| y − x | ≥ −√ dh for y ∈ x + hI d , for s ≥ s , e −h v,S n ( Y ) i /s { S n ( Y ) ∈ x + hI d } ≪ h e −h v,x i /s { S n ( Y ) ∈ x + hI d } = e − r/s { S n ( Y ) ∈ x + hI d } . As a result, P { S n ( X ) ∈ x + hI d , M n ≤ s } ≤ [ Z ( s, v )] n e − r/s P { S n ( Y ) ∈ x + hI d }≤ [ Z ( s, v )] n e − r/s Q S n ( Y ) ( h ) ≪ e − r/s + Cn/A ( s ) Q S n ( Y ) ( h ) . (4.2)By Lemma 4.1, if W = Y − Y , then ϕ W ( t ) = | ϕ Y ( t ) | > Q S n ( Y ) ( h ) ≤ c d ( s ∨ h ) d Z k t k≤ /s | ϕ S n ( Y ) ( t ) | d t ≪ h Z k t k≤ /s ϕ W ( t ) n/ d t ≪ h s − d + Z /s< | t |≤√ d/s ϕ W ( t ) n/ d t. (4.3)By x ≤ e − (1 − x ) , ϕ W ( t ) = E cos h t, W i ≤ e − (1 − E cos h t,W i ) / . By 1 − cos x ≫ x for | x | ≤ − E cos h t, W i ≥ E [(1 − cos h t, W i ) {|h t, W i| ≤ } ] ≫ E [ h t, Y − Y i {|h t, Y i i| ≤ / , i = 1 , } ]= Z ( s, v ) − E [ h t, X − X i e h v,X + X i /s {|h t, X i i| ≤ / , | X i | ≤ s, i = 1 , } ] . Given t ∈ R d with 1 /s < | t | ≤ √ d/s , let ω = t/ | t | and L = | t | − /
2. Then1 − E cos h t, W i ≫ | t | E [ h ω, X − X i {| X i | ≤ L, i = 1 , } ] . (4.4)Since X and X are i.i.d. ∼ X , E [ h ω, X − X i {| X i | ≤ L } ] = 2 m X ( L, ω ) P {| X | ≤ L } − |h ω, c X ( L ) i| . Since L ≥ s / (2 √ d ), by Lemma 4.3, the infimum of the LHS over ω ∈ S d − is positive. In particular m X ( L, ω ) ≥ inf ω ∈ S d − , s ≥ s / (2 √ d ) m X ( s, ω ) >
0. Assume the following is true for now.12 emma 4.5. m X ( s, ω ) ≍ s /A ( s ) for s ≥ s and ω ∈ S d − . It follows that m X ( L, ω ) P {| X | ≤ L } ≍ | t | − /A (1 / | t | ) for | t | ≤ √ d/s with L = | t | − /
2. Onthe other hand, by (3.5), |h ω, c X ( L ) i| ≤ | c X ( L ) | ≪ | t | − /A (1 / | t | ) . This combined with last twodisplays yields that, for 1 /s < | t | ≤ √ d/s , 1 − E cos h t, W i ≫ /A (1 / | t | ), and so for some c > { F, A } , ϕ W ( t ) ≤ exp {− c/A (1 / | t | ) } . Then Z /s< | t |≤√ d/s ϕ W ( t ) n/ d t ≤ Z /s< | t |≤√ d/s exp (cid:26) − cnA (1 / | t | ) (cid:27) d t ≪ Z √ d/s /s y d − exp (cid:26) − cA ( a n ) A (1 /y ) (cid:27) d y. By Potter’s Theorem ([1], Th. 1.5.6), A ( a n ) /A (1 /y ) ≫ ( a n y ) α/ ∧ ( a n y ) α/ for n ≥ y ≤√ d/s . Combining this with the above display and e − ( x ∧ y ) ≤ e − x + e − y , there is C = C ( F, A ) > Z /s< | t |≤√ d/s ϕ W ( t ) n/ d t ≪ Z √ d/s /s y d − ( e − C ( a n y ) α/ + e − C ( a n y ) α/ ) d y. On the other hand, for any b > q > Z ∞ /s y d − e − b ( a n y ) q d y = a − dn Z ∞ a n /s y d − e − by q d y ≪ b,q a − dn . The above three displays combined with (4.2) and (4.3) then prove Proposition 4.2.
Proof of Lemma 4.3.
Put µ ( ω, s ) = E [ h ω, X − X i {| X | ∨ | X | ≤ s/ } ]. Since F ∈ D ( α ) isnondegenerate, for each ω ∈ S d − , µ ( ω, ∞ ) >
0, so by monotone convergence, there is s ( ω ) > µ ( ω, s ( ω )) >
0. Fixing ω , by continuity of the mapping v → µ ( v, s ( ω )), there is r ( ω ) > µ ( v, s ( ω )) ≥ µ ( ω, s ( ω )) / v ∈ [ ω + r ( ω ) B d ] ∩ S d − . Since S d − is compact, there area finite number of ω i ∈ S d − , such that S d − is covered by the union of ω i + r ( ω i ) B d . Then it iseasy to see that s = max i s ( ω i ) has the asserted property. Proof of Lemma 4.4. By Z ( s, ω ) = E [ e h ω,X i /s {| X | ≤ s } ] and ln x ≤ x − x > Z ( s, ω ) ≤ E [ e h ω,X i /s {| X | ≤ s } ] − ≤ E [( e h ω,X i /s − {| X | ≤ s } ] = I ( s, ω ) + h ω, s − c X ( s ) i , (4.5)where I ( s, ω ) = E [( e h ω,X i /s − − h ω, X i /s ) {| X | ≤ s } ]. By | e z − − z | ≤ cz for | z | ≤
1, where c > s ≥ s , | I ( s, ω ) | ≪ s − m X ( s, ω ) ≤ s − V X ( s ) ∼ /A ( s ). On theother hand, from (3.5), sup ω |h ω, s − c X ( s ) i| ≪ /A ( s ). By (4.5), the proof is complete. Proof of Lemma 4.5.
Since for all s ≥ s and ω ∈ S d − , 0 < m X ( s, ω ) ≤ V X ( s ) ≍ s /A ( s ), itsuffices to show that for s ≥ s , inf ω ∈ S d − m X ( s, ω ) ≫ s /A ( s ). For u , v ∈ R d , by |h u, X i −h v, X i | = |h u − v, X ih u + v, X i| ≤ | u − v || u + v || X | , | m X ( s, u ) − m X ( s, v ) | ≤ | u − v || u + v | V X ( s ) . In particular, for u, v ∈ S d − , | m X ( s, u ) − m X ( s, v ) | ≤ | u − v | V X ( s ). Then, by the compactnessof S d − , it suffices to show that given ω ∈ S d − , for s ≥ s , m X ( s, ω ) ≫ s /A ( s ). First, let α = 2.13et Σ be the covariance matrix of the limit normal distribution and put b ( u ) = h u, Σ u i . By [28],Th. 4.1, m X ( s, u ) /m X ( s, v ) → b ( u ) /b ( v ). Letting v = e i , it follows that m X ( s, u ) /V X ( s ) = m X ( s, u ) / X i m X ( s, e i ) → b ( u ) / X i b ( e i ) , (4.6)which together with (3.3) leads to the desired result. Now let α ∈ (0 , m X ( s, ω ) = Z z ∈ [0 ,x ] , v ∈ S d − z h ω, v i P {| X | ∈ d z, X/ | X | ∈ d v } = 2 Z z ∈ [0 ,s ] , v ∈ S d − h ω, v i (cid:18)Z z x d x (cid:19) P {| X | ∈ d z, X/ | X | ∈ d v } = 2 Z x ∈ [0 ,s ] , v ∈ S d − h ω, v i x P { x ≤ | X | ≤ s, X/ | X | ∈ d v } d x Let s → ∞ . By Th. 4.2 of [28] and Th. 14.10 of [29], there is a finite nonzero measure γ on S d − ,such that for any measurable E ⊂ S d − , P {| X | ≥ s, X/ | X | ∈ E } /q X ( s ) → γ ( E ) /γ ( S d − ). Thenstandard argument based on Riemann sum approximation to the integral over v ∈ S d − yields m X ( s, ω ) = 2 (cid:20)Z h ω, v i γ (d v ) + o (1) (cid:21) Z s x [ q X ( x ) − q X ( s )] d x = [ c + o (1)] s (2 − α ) A ( s ) (cid:20)Z h ω, v i γ (d v ) + o (1) (cid:21) , where c = c ( F, A ) > S n ( X ) /a n is nondegenerate,by Lemma 3.1 of [28], R h ω, v i γ (d v ) >
0. Then the proof is complete. n contribution This section proves Theorem 3.1. The following LLT will be used.
Proposition 5.1 (Stone [33]) . Let Y ∈ R r and Z ∈ R d − r and constant vector β ∈ R r be as inProposition 2.1(3). Let ξ i = ( Y i , Z i ) , i ≥ , be i.i.d. ∼ ( Y, Z ) . Let a n → ∞ and d n = ( b n , c n ) ∈ R r × R d − r such that S n ( ξ ) /a n − d n weakly converges to a stable law with density ψ ( y, z ) . Denote Λ n = ( nβ + Z r ) × R d − r . Then as n → ∞ , sup ( y,z ) ∈ Λ n , h ≤ (cid:12)(cid:12)(cid:12)(cid:12) a dn P { S n ( Y ) = y, S n ( Z ) ∈ z + hI d − r } − h d − r ψ (cid:18) ya n − b n , za n − c n (cid:19)(cid:12)(cid:12)(cid:12)(cid:12) → . The proof consists of two steps. First, Theorem 3.1 is proved assuming that no transform of X is needed to reveal its lattice-nonlattice composition. That is, in (2.2), one can let T = Id d so that X = ( L, W, Z ) . (5.1)Then the general case is proved with some uniform convergence argument. Proof of Theorem 3.1 under (5.1) . In the following, x and sω with s ≥ ω ∈ S d − will beused interchangeably. Writing x = ( u, z ) with u ∈ R ν , for any k ∈ N , F ∗ n ( x + ∆ h ) is equal to thesum of F ∗ n ( x a + ∆ h/k ) over a ∈ { , , . . . , k − } d − ν , where x a = ( u, z + ha/k ). Thus, if (3.6) holdsfor h ∈ (0 , h >
0. So without loss of generality, let h ∈ (0 , M > δ ∨
1. Put J s,δ,M = [ A ( δs ) , A ( M s )) ∩ N and B δ,M ( x, h ) = X n ∈ J s,δ,M F ∗ n ( x + ∆ h ) . Then r δ,h ( x ) = s d B δ,M ( x, h ) /A ( s ) + r M,h ( x ). First, by Proposition 5.1, F ∗ n ( x + ∆ h ) ≪ a − dn . Thenby A − ∈ R /α and A − ( t ) = [1 + o (1)] a n for t ∈ [ n − , n + 1] as n → ∞ , r M,h ( x ) ≪ s d A ( s ) X n ≥ A ( Ms ) a − dn ≪ s d A ( s ) Z ∞ A ( Ms ) d t ( A − ( t )) d . By change of variable t = A ( su ) and A ′ ( x ) = [ α + o (1)] A ( x ) /x as x → ∞ , Z ∞ A ( Ms ) d t ( A − ( t )) d ≪ Z ∞ M A ( su ) d us d u d +1 ≪ A ( s ) s d Z ∞ M u α − d − d u. Since d > α , the above two displays show that r M,h ( x ) ≪ R ∞ M u α − d − d u is arbitrarily small if M is large enough. Also, since ψ is bounded, sup ω ∈ S d − R /M ψ ( uω ) u d − α − d u is arbitrarily small aswell. Therefore, to show (3.6), it only remains to show s d B δ,M ( x, h ) A ( s ) = αq − h d − ν Z /δ /M ψ ( uω ) u d − α − d u + o δ,M,h (1) . (5.2)Let x = ( u, w, z ) ∈ R ν × R r − ν × R d − r . Put Λ = D − Υ I ν . Then { S n ( X ) ∈ x + ∆ h } = { S n ( L ) ∈ u + Λ , S n ( W ) ∈ w + hI r − ν , S n ( Z ) ∈ z + hI d − r } . Let l = (0 , . . . , , β ν ) = (0 , . . . , , p/q ) and w = ( β ν +1 , . . . , β r ) be as in (2.1). As L ∈ l + Z ν , S n ( L ) ∈ u + Λ only if ( u + Λ) ∩ ( nl + Z ν ) = ∅ . Clearly, nl + Z ν ⊂ Z l + Z ν . Since p and q arecoprime, D ( Z l + Z ν ) = Z ν . Then by Υ − Z ν = Z ν ,( u + Λ) ∩ ( Z l + Z ν ) = D − [( Du + Υ I ν ) ∩ Z ν ] = D − Υ[(Υ − Du + I ν ) ∩ Z ν ]has exactly one element l and there is a unique number among 0 , . . . , q −
1, denoted κ ( u ), suchthat l ∈ κ ( u ) l + Z ν . Then l ∈ nl + Z ν ⇐⇒ [ n − κ ( u )] l ∈ Z ν ⇐⇒ q | n − κ ( u ). It follows that( u + Λ) ∩ ( nl + Z ν ) = ∅ ⇐⇒ l ∈ nl + Z ν ⇐⇒ n ∈ κ ( u ) + q Z . (5.3)Thus S n ( L ) ∈ u + Λ only if n ∈ κ ( u ) + q Z . Next, since W ∈ w + Z r − ν , S n ( W ) ∈ w + hI r − ν onlyif ( w + hI r − ν ) ∩ ( nw + Z r − ν ) = ∅ . As a result, F ∗ n ( x + ∆ h ) > n ∈ R x , where R x = { n ∈ N : n ∈ κ ( u ) + q Z , ( w + hI r − ν ) ∩ ( nw + Z r − ν ) = ∅} . Then by Proposition 5.1, as s → ∞ , for n ≥ A ( δs ), F ∗ n ( x + ∆ h ) = X y =(˜ u, ˜ w ): ˜ u ∈ ( u +Λ) ∩ ( nl + Z ν )˜ w ∈ ( w + hI r − ν ) ∩ ( nw + Z r − ν ) P { S n ( Y ) = y, S n ( Z ) ∈ z + hI d − r } = a − dn h d − r [ ψ ( x/a n ) + o δ (1)] { n ∈ R x } , h < w + hI r − ν contains at most one point in nw + Z r − ν . Let m and m be the first and last integers in [ A ( δs ) , A ( M s ) + 1). For s ≫ δ,M m ≍ δ A ( s ) and m − m ≍ δ,M A ( s ). Fix integers m = N < N < . . . < N k < N k +1 = m with k = k ( s ), such that as s → ∞ ,min i ( N i +1 − N i ) → ∞ , min i ( N i +1 − N i ) ≍ max i ( N i +1 − N i ) = o δ,M ( A ( s )) . Then by A − ∈ R /α , max i ( a N i +1 − a N i ) = o δ,M ( s ). It follows that uniformly for N i ≤ n < N i +1 , a n = a N i (1 + o δ,M (1)) (5.4)and by the continuity of ψ and s/a n = O δ (1), ψ ( x/a n ) = ψ ( x/a N i ) + o δ,M (1) . (5.5)Combining the above displays, by ψ being bounded, F ∗ n ( x + ∆ h ) = h d − r a − dN i ψ ( x/a N i ) { n ∈ R x } + o δ,M,h ( a − dn ) . Let g i ( x ) be the cardinality of R x ∩ { N i , N i + 1 , . . . , N i +1 − } . Then B δ,M ( x, h ) = N X i =1 X N i ≤ n 1, for 1 ≤ i ≤ k and N i ≤ n < N i +1 ,( w + hI r − ν ) ∩ ( nw + Z r − ν ) = ∅ ⇐⇒ e π i( nβ ν + j − w j ) ∈ Γ for every 1 ≤ j ≤ r − ν, where Γ is the arc { e π i z : 0 ≤ z < h } of the unit circle S . Let b i = ⌈ ( N i − κ ( u )) /q ⌉ , c i = ⌈ ( N i +1 − κ ( u )) /q ⌉ − . For s ≫ δ,M N i − κ ( u ) > A ( δs ) − q > 0, so b i ≥ 1. Therefore, by (5.3), for N i ≤ n < N i +1 , n ∈ κ ( u ) + q Z ⇐⇒ n = κ ( u ) + qk with b i ≤ k ≤ c i . Then g i ( x ) = c i − b i X k =0 r − ν Y j =1 { e π i(( κ ( u )+( b i + k ) q ) β ν + j − w j ) ∈ Γ } = c i − b i X k =0 r − ν Y j =1 { θ j e π i kτ j ∈ Γ } , (5.7)where θ j = e π i(( κ ( u )+ b i q ) β ν + j − w j ) and τ j = qβ ν + j . Then θ := ( θ , . . . , θ r − ν ) ∈ K := ( S ) r − ν . Define Hz = ( z e π i τ , . . . , z r − ν e π i τ r − ν ) , z ∈ C r − ν . Under the Euclidean norm, H is an isometry and H K = K . Since τ , . . . , τ r − ν are rationallyindependent, for any θ ∈ K , { H n θ } n ≥ is dense in K ([25], p. 158). Then the pair ( K , H ) is strictlyergodic ([25], Prop. 4.2.15), and the normalized Lebesgue measure on K is the unique probabilitymeasure invariant under H . By Prop. 4.2.8 of [25] followed by dominated convergence, as n → ∞ ,1 n n X k =0 r − ν Y j =1 { θ j e π i kτ j ∈ Γ } = 1 n n X k =0 { H k θ ∈ Γ r − ν } → h r − ν uniformly in θ ∈ K .16hen by (5.7), as s → ∞ , for 1 ≤ i ≤ s , g i ( x ) = [1 + o (1)]( c i − b i ) h r − ν = [1 + o δ,M (1)] q − ( N i +1 − N i ) h r − ν , which together with (5.6) and another application of (5.4) and (5.5) yields B δ,M ( x, h ) = [1 + o δ,M (1)] q − h d − ν k X i =1 a − dN i ψ ( x/a N i )( N i +1 − N i ) + o δ,M,h (1) X n ∈ J x,δ,M a − dn = q − h d − ν X n ∈ J x,δ,M a − dn ψ ( x/a n ) + o δ,M,h (1) X n ∈ J x,δ,M a − dn . Since [ A − ( t )] − d ψ ( x/A − ( t )) = a − dn [ ψ ( x/a n ) + o δ (1)] for n ≥ A ( δs ) and t ∈ [ n − , n + 1], then B δ,M ( x, h ) = q − h d − ν Z A ( Ms ) A ( δs ) ψ ( x/A − ( t ))( A − ( t )) d d t + o δ,M,h (1) X n ∈ J x,δ,M a − dn . As s → ∞ , by change of variable t = A ( s/u ), the integral on the RHS is Z /δ /M ψ ( uω )( s/u ) d sA ′ ( s/u ) u d u = [1 + o δ,M,h (1)] α Z /δ /M ψ ( uω )( s/u ) d A ( s/u ) u d u = [1 + o δ,M,h (1)] αA ( s ) s d Z /δ /M ψ ( uω ) u d − α − d u. Then (5.2) follows by noting that X n ∈ J x,δ a − dn = [1 + o δ,M (1)] Z A ( Ms ) A ( δs ) d t ( A − ( t )) d = [1 + o δ,M (1)] αA ( s ) s d Z /δ /M u d − − α d u. For the general case, the following corollary will be used. Corollary 5.2. Under the condition (5.1) , given δ > and h > , as s → ∞ sup ω ∈ S d − , η ≥ δ | r η,h ( sω ) − h d − ν ̺ η ( ω ) | = o δ,h (1) . (5.8) Proof. From the preceding proof, it suffices to show that given M > δ , as s → ∞ ,sup ω ∈ S d − , δ ≤ η ≤ M | r η,h ( sω ) − h d − ν ̺ η ( ω ) | = o δ,h (1) . (5.9)For δ ≤ η < θ ≤ M , 0 ≤ ̺ η ( ω ) − ̺ θ ( ω ) ≪ R /η /θ u d − α − d u ≪ η α − d − θ α − d . By F ∗ n ( x + ∆ h ) ≪ a − dn and the bound at the end of the preceding proof, for s ≫ δ ≤ r η,h ( sω ) − r θ,h ( sω ) ≪ s d A ( s ) X A ( ηs ) ≤ n ≤ A ( θs ) a − dn ≪ δ η α − d − θ α − d . Thus (5.9) holds if sup ω ∈ S d − ,η ∈ E | r η,h ( sω ) − h d − ν ̺ η ( ω ) | = o δ,h (1), where E is a finite set in [ δ, M ]with its adjacent elements being arbitrarily close. Then Theorem 3.1 under the additional condition(5.1) can be invoked to finish the proof. 17 roof of Theorem 3.1, general case. Suppose T X = ( L, W, Z ). Put ˜ X = T X . Then S n ( ˜ X ) /a n weakly converges to a stable law with density ˜ ψ . For x = 0, put s = | T x | and ω = x/s . In general ω S d − and | ω | is a variable in x . However, T ω ∈ S d − and since T is nonsingular, | ω | ≥ η forsome constant η > 0. Then T − x = ( T ω ) s and by S n ( X ) = T − S n ( ˜ X ), r δ,h ( x ) = | ω | d s d A ( | ω | s ) X n ≥ A ( δ | ω | s ) P { S n ( ˜ X ) ∈ ( T ω ) s + ∆ h } . Applying Corollary 5.2 to S n ( ˜ X ), as s → ∞ , the RHS is | ω | d s d A ( | ω | s ) A ( s ) s d " h d − ν αq − Z / ( δ | ω | )0 ˜ ψ ( uT ω ) u d − α − d u + o δ,h (1) . By change of variable u = v/ | ω | and ˜ ψ ( x ) = | det T | − ψ ( T − x ), the above quantity is equal to[1 + o (1)] h d − ν ̺ δ ( ω/ | ω | ) + o δ,h (1). The proof is complete by noting ω/ | ω | = x/ | x | . n contribution This section proves Theorem 3.2. First some notation. For n ≥ k ≥ 1, denote by x n :1 , . . . , x n : n a permutation of x , . . . , x n such that | x n : i | are sorted in decreasing order and S n : k ( x ) = x n :1 + · · · + x n : k . Define | x n :0 | = ∞ , S n :0 ( x ) = 0, and x n : k = 0 for k > n . Recall that according to(2.7), κ = ⌊ d/α ⌋ .The proof follows from four lemmas. For the first two, fix γ ∈ ( dα − ( κ + 1) − , κ + 1 > d/α . Define ζ n,s = a − γn s γ , n ≥ , s > . Lemma 6.1. Let k = κ + 1 and b = [ d/ ( kγ ) + α ] / . Note that kγb > d and b ∈ (0 , α ) . Given δ ∈ (0 , , for s ≫ δ , X n ≤ A ( δs ) sup | x |≥ s P { S n ( X ) ∈ x + hI d , | X n : k | > ζ n, | x | } ≪ h δ kγb + α − d A ( s ) s d . Lemma 6.2. Fix k ≥ and ǫ, δ ∈ (0 , . For n ≥ and x ∈ R d , denote E n,x = { S n ( X ) ∈ x + hI d , | X n : k | > ζ n, | x | ≥ | X n : k +1 |} . For s ≫ ǫ,δ , X n ≤ A ( δs ) sup | x |≥ s P { E n,x , | S n : k ( X ) | ≤ (1 − ǫ ) | x |} ≪ h,k A ( δs ) s d Z ∞ / (2 δ ) u d − e − ǫu − γ d u. In particular, if k ≥ , X n ≤ A ( δs ) sup | x |≥ s P { E n,x , | X n :1 | ≤ (1 − ǫ ) | x | /k } ≪ h,k A ( δs ) s d Z ∞ / (2 δ ) u d − e − ǫu − γ d u. For the next two lemmas, define S ′ n ( X ) = n X i =1 X i {| X i | > a n } . emma 6.3. Given < δ < ǫ < , for s ≫ ǫ,δ , X n ≤ A ( δs ) sup | x |≥ s P { S n ( X ) ∈ x + hI d , | S n ( X ) − S ′ n ( X ) | ≥ ǫ | x |} ≪ h A ( δs ) s d Z ∞ / (2 δ ) u d − e − ǫu d u. Lemma 6.4. Fix < θ < ν < and < η < ǫ < ν − θ . For s ≫ θ,ν,η,ǫ,h , P { S n ( X ) ∈ sω + hI d , | X n :1 | > νs, | S n ( X ) − S ′ n ( X ) | < ηs } ≪ h na − dn sup | t | >θs K ( t, a n , ǫ/θ, h ) . Proof of Theorem 3.2. Since K ( t, a, η, h ) is increasing in η , if (3.7) holds for some η > 0, then itholds for all larger η with everything else unchanged. Therefore, to prove (3.7), η > < θκ < 1, let η < / ( θκ ) − 1. Fix ν = ν ( η, θ ) ∈ ((1 + η ) θ, /κ ). Thensup ω F ∗ n ( sω + hI d ) ≤ P i =1 C n,i ( s ), where the supremum is taken over ω ∈ S d − and C n, ( s ) = sup ω P { S n ( X ) ∈ sω + hI d , | X n : κ +1 | > ζ n,s } ,C n, ( s ) = sup ω P { S n ( X ) ∈ sω + hI d , | X n :1 | ≤ ζ n,s } ,C n, ( s ) = sup ω P { S n ( X ) ∈ sω + hI d , | X n : κ +1 | ≤ ζ n,s < | X n :1 | ≤ νs } ,C n, ( s ) = sup ω P { S n ( X ) ∈ sω + hI d , | S n ( X ) − S ′ n ( X ) | ≥ . θηs } ,C n, ( s ) = sup ω P { S n ( X ) ∈ sω + hI d , | X n :1 | > νs, | S n ( X ) − S ′ n ( X ) | < . θηs } . Put ǫ n ( s ) = P i =1 C n,i ( s ). Apply Lemma 6.1 to bound P n ≤ A ( δs ) sup r ≥ s C n, ( r ) and Lemma 6.2with k = 0 to bound P n ≤ A ( δs ) sup r ≥ s C n, ( r ). Fixing ǫ ′ > ν ≤ (1 − ǫ ′ ) /κ , C n, ( s ) ≤ κ X k =1 sup ω P { S n ( X ) ∈ sω + hI d , | X n : k +1 | ≤ ζ n,s < | X n : k | , | X n :1 | ≤ (1 − ǫ ′ ) s/k } . Then apply Lemma 6.2 with k ≥ P n ≤ A ( δs ) sup r ≥ s C n, ( r ). Letting 0 < δ < . θη ,apply Lemma 6.3 to bound P n ≤ A ( δs ) sup r ≥ s C n, ( r ). Together, these bounds yield (3.8). Finally,let ˜ η = 0 . θη and ˜ ǫ = θη . Then 0 < ˜ η < ˜ ǫ < ν − θ , so by Lemma 6.4, for s ≫ θ,η,ǫ,h C n, ( s ) = sup ω P { S n ( X ) ∈ sω + hI d , | X n :1 | > νs, | S n ( X ) − S ′ n ( X ) | < ˜ ηs }≪ h na − dn sup | t | >θs K ( t, a n , ˜ ǫ/θ, h ) , yielding the first term on the RHS of (3.7). Proof of Lemma 6.1. Denote f n ( x ) = P { S n ( X ) ∈ x + hI d , | X n : k | > ζ n, | x | } . Clearly, if n < k , then f n ( sω ) = 0. For n ≥ k , s > ω ∈ S d − , f n ( sω ) ≤ n k P { S n ( X ) ∈ sω + hI d , | X k : k | > ζ n,s } . For any z i ∈ R d , P { S n ( X ) ∈ sω + hI d | X i = z i , , i ≤ k } = P { S n − k ( X ) ∈ sω − S k ( z ) + hI d } , whichby Proposition 5.1 is O h ( a − dn − k ). Then X n ≤ A ( δs ) sup | x |≥ s f n ( x ) ≪ h X n ≤ A ( δs ) a − dn − k n k P {| X k : k | > ζ n,s } ≪ X n ≤ A ( δs ) a − dn n k q X ( ζ n,s ) k . (6.1)19or s ≫ δ 1, since q X ( s ) ≪ /A ( s ), X n ≤ A ( δs ) a − dn n k q X ( ζ n,s ) k ≪ X n ≤ A ( δs ) n k a dn A ( ζ n,s ) k ≍ Z A ( δs )1 t k d t ( A − ( t )) d A ( A − ( t ) − γ s γ ) k = Z δsa A ( u ) k A ′ ( u ) d uu d A ( u − γ s γ ) k , where the last line is due to change of variable t = A ( u ). By A ′ ( u ) ≍ A ( u ) /u , for s ≫ δ X n ≤ A ( δs ) a − dn n k q X ( ζ n,s ) k ≪ Z δsa A ( u ) k +1 d uu d +1 A ( u − γ s γ ) k ≪ A ( δs ) Z δsa u d +1 (cid:20) A ( u ) A ( u − γ s γ ) (cid:21) k d u. For a ≤ u ≤ δs , since u < u − γ s γ and b ∈ (0 , α ), by Potter’s Theorem ([1], Th. 1.5.6), A ( u ) /A ( u − γ s γ ) ≪ [ u/ ( u − γ s γ )] b = ( u/s ) bγ . Then by kγb > d , Z δsa u d +1 (cid:20) A ( u ) A ( u − γ s γ ) (cid:21) k d u ≪ Z δs ( u/s ) kγb u d +1 d u ≪ s d δ kγb − d kγb − d . By A ( δs ) ≍ δ α A ( s ) for s ≫ δ 1, the above two displays together imply X n ≤ A ( δs ) a − dn n k q X ( ζ n,s ) k ≪ A ( s ) s d δ kγb + α − d kγb − d . This combined with (6.1) finishes the proof. Proof of Lemma 6.2. Fixing k ≥ ǫ, δ ∈ (0 , f n ( x ) = P { E n,x , | S n : k ( X ) | ≤ (1 − ǫ ) | x |} .If n < k , then E n,x = ∅ and so f n ( x ) = 0. Let n ≥ k . Define g n ( x ) = P (cid:8) S n ( X ) ∈ x + hI d , | S k ( X ) | ≤ (1 − ǫ ) | x | , | X i | > ζ n, | x | ≥ | X j | , i ≤ k < j (cid:9) . Then f n ( x ) ≤ n k g n ( x ). Let Y j = X k + j . By S n ( X ) = S k ( X ) + S n − k ( Y ), for s > ω ∈ S d − , if S n ( X ) ∈ sω + hI d and | S k ( X ) | ≤ (1 − ǫ ) s , then | S n − k ( Y ) | ≥ | S n ( X ) | − | S k ( X ) | ≥ ǫs − h √ d , so g n ( sω ) ≤ P (cid:26) S n − k ( Y ) ∈ sω − S k ( X ) + hI d , | S n − k ( Y ) | ≥ ǫs − h √ d, and | X k : k | > ζ n,s ≥ | Y n − k :1 | (cid:27) . By conditioning on X i = z i , i ≤ k , with | z i | > ζ n,s , g n ( sω ) ≤ P {| X k : k | > ζ n,s } M n,s = q X ( ζ n,s ) k M n,s , where M n,s = sup | y |≥ ǫs − h √ d P { S n − k ( Y ) ∈ y + hI d , | Y n − k :1 | ≤ ζ n,s } . By Proposition 4.2, there is C > { F, A } , such that M n,s ≪ h ( ζ − dn,s + a − dn − k ) e − ǫs/ζ n,s + Cn/A ( ζ n,s ) s ≫ ǫ n ≥ k . For n ≤ A ( δs ), as ζ n,s = a − γn s γ ≥ a n , M n,s ≪ h,k a − dn e − ǫ ( s/a n ) − γ . As aresult, g n ( sω ) ≪ h,k a − dn q X ( ζ n,s ) k e − ǫ ( s/a n ) − γ , and hence f n ( sω ) ≤ n k g n ( sω ) ≪ h,k a − dn n k q X ( ζ n,s ) k e − ǫ ( s/a n ) − γ ≪ h,k a − dn e − ǫ ( s/a n ) − γ , (6.2)where the last bound is due to nq X ( ζ n,s ) ≪ A ( a n ) /A ( ζ n,s ) ≤ 1. Take sum over n ≤ A ( δs ). Sincefor n ≥ t ∈ [ n, n + 1], a n ≤ A − ( t ) ≪ a n , then for s ≫ δ X n ≤ A ( δs ) sup | x |≥ s f n ( x ) ≪ h,k Z A (2 δs )1 A − ( t ) d e − ǫ ( s/A − ( t )) − γ d t. By change of variable t = A ( s/u ), or u = s/A − ( t ), and use A ′ ( x ) ≍ A ( x ) /x for x > 0, the lastintegral is no greater than Z ∞ / (2 δ ) s/u ) d e − ǫu − γ A ′ ( s/u ) su − d u ≪ s − d Z ∞ / (2 δ ) u d − e − ǫu − γ A ( s/u ) d u ≤ s − d A (2 δs ) Z ∞ / (2 δ ) u d − e − ǫu − γ d u. Combining the above two displays then finishes the proof.To prove Lemmas 6.3 and 6.4, define τ n = n X i =1 {| X i | > a n } . Then P { τ n = m } = (cid:0) nm (cid:1) q X ( a n ) m [1 − q X ( a n )] n − m and by q X ( a n ) ≪ /A ( a n ) = 1 /n , P { τ n = m } ≤ n ! O (1 /n ) m m !( n − m )! ≪ O (1) m m ! . (6.3)Conditioning on τ n = m ,( S n ( X ) − S ′ n ( X ) , S ′ n ( X )) ∼ ( S n − m ( b ( n ) ) , S m ( u ( n ) )) , where b ( n )1 , b ( n )2 , . . . , u ( n )1 , u ( n )2 , . . . are independent, with P { b ( n ) i ∈ d x } = ( P { X ∈ d x | | X | ≤ a n } if q X ( a n ) < δ (d x ) else , and P { u ( n ) i ∈ d x } = ( P { X ∈ d x | | X | > a n } if q X ( a n ) > δ (d x ) else , where δ is the unit measure concentrated at 0. 21 roof of Lemma 6.3. Denote f n ( x ) = P { S n ( X ) ∈ x + hI d , | S n ( X ) − S ′ n ( X ) | ≥ ǫ | x |} . For n ≤ A ( δs )and ω ∈ S d − , f n ( sω ) = n X m =0 P { S n − m ( b ( n ) ) + S m ( u ( n ) ) ∈ sω + hI d , | S n − m ( b ( n ) ) | ≥ ǫs } P { τ n = m } . (6.4)For each m ≤ n , P { S n − m ( b ( n ) ) + S m ( u ( n ) ) ∈ sω + hI d , | S n − m ( b ( n ) ) | ≥ ǫs } = Z P { S n − m ( b ( n ) ) ∈ sω − z + hI d , | S n − m ( b ( n ) ) | ≥ ǫs } P { S m ( u ( n ) ) ∈ d z }≤ sup | x |≥ ǫs − h √ d P { S n − m ( b ( n ) ) ∈ x + hI d } . (6.5)For n with q X ( a n ) < P { S n − m ( b ( n ) ) ∈ x + hI d } = P { S n − m ( X ) ∈ x + hI d , | X n − m :1 | ≤ a n } P {| X n − m :1 | ≤ a n } . Since q X ( a n ) ≪ /n , for all n ≥ q X ( a n ) < ≤ m ≤ n , P {| X n − m :1 | ≤ a n } ≥ [1 − q X ( a n )] n ≫ 1. Then P { S n − m ( b ( n ) ) ∈ x + hI d } ≪ P { S n − m ( X ) ∈ x + hI d , | X n − m :1 | ≤ a n } . Let s > n be the largest n with a n < s . For n ≤ n , if | x | > na n + h √ d , then the RHS is 0. Otherwise, as | x | ≪ 1, the RHS is O ( a − dn − m e −| x | /a n ). On theother hand, for n > n , since a n ≥ s , by applying Proposition 4.2 to the RHS P { S n − m ( b ( n ) ) ∈ x + hI d } ≪ h ( a − dn − m + a − dn ) e −| x | /a n + C ( n − m ) /A ( a n ) ≪ h a − dn − m e −| x | /a n . (6.6)From the discussion for n ≤ n , it is seen the bound holds for all n with q X ( a n ) < 1. For n with q X ( a n ) = 1, as b ( n ) i ≡ P { S n − m ( b ( n ) ) ∈ x + hI d } = { ∈ x + hI d } ≤ {| x | ≤ h √ d } . Since thereare only a finite number of n with q X ( a n ) = 1, the bound in (6.6) still holds. Combining the boundwith (6.4)–(6.5), for | x | ≥ ǫs − h √ d , f n ( sω ) ≪ h e − ǫs/a n n X m =0 a − dn − m P { τ n = m } Since by (6.3), n X m =0 a − dn − m P { τ n = m } ≪ a − dn X m ≤ n/ P { τ n = m } + a − d X n/ Put f n ( sω ) = P { S n ( X ) ∈ sω + hI d , | X n :1 | > νs, | S n ( X ) − S ′ n ( X ) | < ηs } .For s ≫ η,h 1, if S n ( X ) ∈ sω + hI d and | S n ( X ) − S ′ n ( X ) | < ηs , then S ′ n ( X ) = 0, yielding τ n ≥ | X n :1 | > a n . Therefore, if q X ( a n ) = 0, then f n ( sω ) = 0 and the bound in Lemma 6.4 triviallyholds. In the rest of the proof, let q X ( a n ) > 0. Then f n ( sω ) ≤ n X m =1 P m ( sω ) P { τ n = m } , (6.7)where, with n being fixed, for each m = 1 , . . . , n , P m ( sω ) = P { S n − m ( b ( n ) ) + S m ( u ( n ) ) ∈ sω + hI d , | u ( n ) m :1 | > νs, | S n − m ( b ( n ) ) | < ηs } = Z | y | <ηs P { S m ( u ( n ) ) ∈ sω − y + hI d , | u ( n ) m :1 | > νs } P { S n − m ( b ( n ) ) ∈ d y }≤ m Z | y | <ηs P { S m ( u ( n ) ) ∈ sω − y + hI d , | u ( n )1 | > νs } P { S n − m ( b ( n ) ) ∈ d y } . Denote T = sω − ( u ( n )2 + · · · + u ( n ) m ). By independence of T and u ( n )1 , with the latter following thedistribution of X conditioned on | X | > a n , P { S m ( u ( n ) ) ∈ sω − y + hI d , | u ( n )1 | > νs } = P { X ∈ T − y + hI d , | X | > νs | | X | > a n }≤ P { X ∈ T − y + hI d , | X | > νs } q X ( a n ) . For y with | y | < ηs , if X ∈ T − y + hI d and | X | > νs , then | T | ≥ | X | − | y | − h √ d > ( ν − η ) s − h √ d .For s ≫ η,ν,θ,h 1, ( ν − η ) s − h √ d > θs and hence P m ( sω ) ≤ mq X ( a n ) Z | y | <ηs P { X ∈ T − y + hI d , | X | > νs } P { S n − m ( b ( n ) ) ∈ d y }≤ mq X ( a n ) Z | y | <ηs P { X ∈ T − y + hI d , | T | > θs } P { S n − m ( b ( n ) ) ∈ d y } Let G n,m ( t, s ) = Z | y | <ηs F ( t − y + hI d ) P { S n − m ( b ( n ) ) ∈ d y } . Then by Fubini’s theorem, the last inequality yields P m ( sω ) ≤ mq X ( a n ) Z | t | >θs G n,m ( t, s ) P { T ∈ d t } ≤ mq X ( a n ) sup | t | >θs G n,m ( t, s ) . (6.8)Given v ∈ hI d , ηsB d is covered by disjoint cubes z + hI d with z ∈ ( v + h Z d ) ∩ ( ηs + h √ d ) B d .If y ∈ z + hI d , then − y + hI d ⊂ − z + hJ d , where J d = ( − , d . As a result, for any t ∈ R d , G n,m ( t, s ) ≤ X z ∈ ( v + h Z d ) ∩ ( ηs + h √ d ) B d Z z + hI d F ( t − y + hI d ) P { S n − m ( b ( n ) ) ∈ d y }≤ X z ∈ ( v + h Z d ) ∩ ( ηs + h √ d ) B d F ( t − z + hJ d ) P { S n − m ( b ( n ) ) ∈ z + hI d } . P { S n − m ( b ( n ) ) ∈ z + hI d } , G n,m ( t, s ) ≪ h a − dn − m X z ∈ ( v + h Z d ) ∩ ( ηs + h √ d ) B d F ( t − z + hJ d ) e −| z | /a n . Let u = z − v . Then z ∈ ( v + h Z d ) ∩ ( ηs + h √ d ) B d implies u ∈ ( h Z d ) ∩ ( ηs + 2 h √ d ) B d , yielding G n,m ( t, s ) ≪ h a − dn − m X u ∈ ( h Z d ) ∩ ( ηs +2 h √ d ) B d F ( t − u − v + hJ d ) e −| u + v | /a n . Take average over v ∈ hI d . By z = u + v and Fubini’s theorem, G n,m ( t, s ) ≪ h a − dn − m X u ∈ ( h Z d ) ∩ ( ηs +2 h √ d ) B d Z z ∈ u + hI d F ( t − z + hJ d ) e −| z | /a n d z ≤ a − dn − m Z ( ηs +3 h √ d ) B d F ( t − z + hJ d ) e −| z | /a n d z ≪ h a − dn − m Z ( ηs +4 h √ d ) B d F ( t − z + hI d ) e −| z | /a n d z. Now for s ≫ η,ǫ,h 1, if | t | > θs and z ∈ ( ηs + 4 h √ d ) B d , then | z | < ( ǫ/θ ) | t | and so the last integralis no greater than K ( t, a n , ǫ/θ, h ). Combining the bound with (6.8) and then with (6.7), f n ( sω ) ≤ q X ( a n ) sup | t | >θs K ( t, a n , ǫ/θ, h ) n X m =1 ma − dn − m P { τ n = m } . Similar to the argument at the end of the proof of Lemma 6.3, n X m =1 ma − dn − m P { τ n = m } ≪ a − dn X ≤ m ≤ n/ m P { τ n = m } + X n/ Proof of Theorem 2.4. Let X ∼ F . For t ∈ R d with | t | ≫ h a > K ( t, a, / , h ) ≤ Z | z |≤| t | / d z Z { x ∈ t − z + hI d } F (d x )= Z | z |≤| t | / d z Z | x | > | t | / { x ∈ t − z + hI d } F (d x ) ≤ Z | x | > | t | / F (d x ) Z { x − t + z ∈ hI d } d z = h d q X ( | t | / . θ > η ≤ / 3, by K ( t, a, η, h ) ≤ K ( t, a, / , h ), X n ≤ A ( δs ) na − dn sup | t | >θs K ( t, a n , η, h ) ≪ q X ( s/ 3) ˜ A d ( δs ) , (7.1)where ˜ A d is defined in (3.11). Since α ∈ (0 , O ( ˜ A d ( δs ) /A ( s )) ≍ δ α − d A ( s ) /s d , andhence the proof follows from Theorem 2.3. Proof of Proposition 2.5. The inequality in (7.1) still holds but now q X ( s ) = o (1 /A ( s )) as s → ∞ (cf. (3.2)). By Theorem 2.3, it suffices to verify q X ( s ) ˜ A d ( s ) = o ( A ( s ) /s d ) in each case. The valueof δ is irrelevant. If d = 3, then by α = 2 > d/ q X ( s ) ˜ A ( s ) ≍ q X ( s ) A ( s ) /s = o ( A ( s ) /s ).If d = 4, then the proof directly follows from ˜ A ( s ) ≪ R s u − A ( u ) d u (cf. (3.11)). Finally, if d ≥ 5, then ˜ A d ( δs ) ≤ ˜ A d ( ∞ ) < ∞ and A ( s ) ≍ s for s ≫ 1. So q X ( s ) = o ( s − d ) implies q X ( s ) ˜ A d ( s ) = o ( A ( s ) /s d ). Proof of Theorem 2.7. Since the integral in (2.11) is increasing in η , assume without loss of gen-erality that 0 < η < θ > 0. For s ≫ n ≥ 1, and t , z ∈ R d with | t | ≥ θs and | z | ≤ η | t | , | t − z | d A ( | t − z | ) ≍ | t | d A ( | t | ). Then K ( t, a, η, h ) ≪ | t | d A ( | t | ) Z | z | <η | t | φ ( t − z ) e −| z | /a d z ≪ | t | d A ( | t | ) "Z | z | <η | t | [ φ ( t − z ) − T ] + d z + T Z e −| z | /a d z ≤ o (1) A ( | t | ) | t | d + O ( a d ) | t | d A ( | t | ) , which combined with (3.11) yields (2.12). Proof of Proposition 2.10. Let Y ∈ R be independent of ξ with ϕ Y ( θ ) = e − Cf α ( θ ) . As noted beforeProposition 2.10, Y is strictly stable. It actually has L´evy density c { x > } x − − α for someconstant c > 0. Let X = Y ξ . Then q X ( s ) = q Y ( s ) ≍ s − α for s ≫ 1. For Γ ⊂ S d − , P {| X | >s, X/ | X | ∈ Γ } = P { Y > s } P { ξ ∈ Γ } + P { Y < − s } P { ξ ∈ − Γ } , so by P { Y < − s } ≤ E e − s − Y ≪ e − s ([29], Th. 25.17) P {| X | > s, X/ | X | ∈ Γ } /q X ( s ) → P { ξ ∈ Γ } , s → ∞ . Then X is in the domain of attraction of a stable law with the same L´evy measure as G , with a n = n /α being norming constant ([28], Th. 4.2). By Y ∈ D ( α ) and (3.5), ( n/a n ) c X ( a n ) =( n/a n ) c Y ( a n ) E ξ converges, so by (3.5) again, S n ( X ) /a n weakly converges to a strictly stable law.Since G is strictly stable, if α = 1, then the limiting law is G . However, if α = 1, then the limitinglaw is G ( x − x ), where x need not be 0. Let g be the density of Y and λ be the density of ξ withrespect to the spherical measure σ on S d − . Then for E ⊂ R d , P { X ∈ E } = Z { yu ∈ E } g ( y ) λ ( u )d y σ (d u )= Z r> { ru ∈ E } [ g ( r ) λ ( u ) + g ( − r ) λ ( − u )]d r σ (d u ) . For x = ru = 0 with r = | x | , letting h ( x ) = c [ g ( r ) λ ( u ) + g ( − r ) λ ( − u )] /r d − with c > R { x ∈ E } h ( x ) d x , showing that X has density h .25ince sup u [ r d h ( ru ) /q X ( r )] ≪ (sup λ ) r α [ g ( r ) + g ( − r )] ≪ 1, it is seen that the function φ ( x ) inTheorem 2.7 is bounded and hence (2.10) holds for the limiting law of S n ( X ) /a n . For α = 1, thiscompletes the proof. If α = 1, one can only conclude that (2.10) holds for G ( x − x ). However,consider X + x , whose corresponding limiting law is G . Since | x | d P { X + x ∈ x + hI d } A ( | x | ) ≪| x | d A ( | x | ) + | x − x | d P { X ∈ x − x + hI d } A ( | x − x | ) is bounded, a repeat of argument showsthat (2.10) holds for G . Proof of Proposition 2.11. It suffices to show that X ∼ F satisfies (3.9). Part of the argument issimilar to that in [5], so only parts that are different will be shown in detail. First, the support of ν is unbounded, otherwise E e | tX | < ∞ for all t and F ∈ D (2) ([29], Th. 25.17). Let ν ( · ) = ν ( · \ B d )and λ = ν − ν . Let µ = ν ( R d ). Then S n ( X ) ∼ S S n ( N ) ( Z ) + S n ( W ) + nv , where N i , Z j and W k are independent with N i i.i.d. ∼ Poisson( µ ), Z j i.i.d. ∼ ν /µ , and W k i.i.d., ID with L´evy measure λ and mean 0, and v ∈ R d is a constant. Fix M > ∨ (4 µ ) /α and ǫ > 0. Let Y = Z + v/µ and G the distribution of Y . Let V = W + v − N v/µ . Then F ∗ n ( sω + hI d ) = P { S S n ( N ) ( Y ) + S n ( V ) ∈ sω + hI d }≤ sup | t | <ǫs P { S S n ( N ) ( Y ) ∈ sω − t + hI d } + P {| S n ( V ) | ≥ ǫs }≤ X k ≤ A ( Mδs ) P { S n ( N ) = k } sup | y | > (1 − ǫ ) s G ∗ k ( y + hI d ) + R n ( s ) + R ′ n ( s ) , where R n ( s ) = P { S n ( N ) > A ( M δs ) } , R ′ n ( s ) = P {| S n ( V ) | > ǫs } . It can be shown that X n ≤ A ( δs ) F ∗ n ( sω + hI d ) ≪ X k ≤ A ( Mδs ) sup | y | > (1 − ǫ ) s G ∗ k ( y + hI d ) + X n ≤ A ( δs ) [ R n ( s ) + R ′ n ( s )] . As in [5], P n ≤ A ( δs ) R n ( s ) = o ( A ( s ) /s d ) as s → ∞ . Let V = ( ξ , . . . , ξ d ). Then R ′ n ( s ) ≤ P dj =1 R ′ nj ( s ), where R ′ nj ( s ) = P {| S n ( ξ j ) | > ǫs/d } . Each ξ j ∈ R has mean zero and E e tξ j < ∞ for all t . Fix b ∈ (0 , α ∧ s ≫ b /δ and n ≤ A ( δs ), if 1 ≤ n ≤ s b , then R ′ nj ( s ) = P { S n ( ξ j ) > ǫs/d } + P {− S n ( ξ j ) > ǫs/d }≤ [( E e ξ j ) n + ( E e − ξ j ) n ] e − ǫs/d = O (1) n e − ǫs/d ≪ e − ǫs/ d . If s b < n < A ( δs ), then, letting σ j = E [ ξ j ] and σ = max σ j , ǫs/ ( dσ √ n ) ≥ ( ǫ/δ ) A − ( n ) / ( dσ √ n ) ≥ ηn c for some constants η > < c < / 6. By Cram´er’s large deviation ([26], Th. 5.23), R ′ nj ( s ) ≤ P {| S n ( ξ j ) | / ( σ j √ n ) > ǫs/ ( dσ √ n ) } ≪ − Φ( ηn c ) ≤ − Φ( ηs bc ), where Φ is the distribution functionof N (0 , P n ≤ A ( δs ) R ′ n ( s ) = o ( A ( s ) /s d ).Since N n /n D → µ and S n ( V ) /a n D → 0, by S n ( X ) /a n ∼ S N n ( Y ) /a n + S n ( V ) /a n , it can be seenthat µ /α Y is in the domain of attraction without centering of the same stable law as X . By theassumption on φ ν ( x ), and Theorems 2.7 and 3.2, X k ≤ A ( Mδs ) sup | y | > (1 − ǫ ) s G ∗ k ( y + hI d ) ≪ h δA ν ( s ) /s d . By following almost line by line the argument in [16], p. 572–573, q X ( s ) ∼ /A ν ( s ). 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Theory Probab.Appl. , 2, 335–349. ppendixA Proofs for the lattice-nonlattice composition For a set E in a Euclidean space, denote by span( E ) the linear subspace spanned by elements of E . If M is a matrix, denote by csp( M ) the linear subspace spanned by the column vectors of M .If the rank of M is equal to its number of columns, then M is said to be of full column rank. Proof of Proposition 2.1. Let Γ = Γ X , whereΓ X = { v ∈ R d : there is a ∈ R such that h v, X i ∈ a + Z a.s. } = { v ∈ R d : | ϕ X (2 πv ) | = 1 } . (A.1)As in the proof of [31], T6.1, Γ plays an important role. The first step is to show that it is alattice. The first line in (A.1) implies that Γ is an additive subgroup of R d , so it suffices to showthat 0 is not a cluster point of Γ. Let u n ∈ Γ such that u n → 0. Let V n = span( u i , i ≥ n ). Since R d ⊃ V ⊃ V ⊃ . . . , there is k , such that V k = V k +1 = · · · . Let X ∗ be i.i.d. ∼ X and ξ = X − X ∗ .Then almost surely, h u n , ξ i ∈ Z for all n . Since h u n , ξ i → 0, this implies h u n , ξ i = 0 for n ≥ k largeenough. But then ξ ∈ V ⊥ n = V ⊥ k . By assumption, ξ is not concentrated in any linear subspace ofdimension d − 1. Then V k = { } , giving u k = u k +1 = · · · = 0, so 0 is not a cluster point of Γ.Let V = span(Γ) and r = dim( V ). Suppose r ≥ 1. By a fundamental theorem on lattices(cf. [34], Lemma 3.4), Γ = M Z r for some M ∈ R d × r of rank r . Let v , . . . , v r be the column vectorsof M and a = ( a , . . . , a r ) such that h v i , X i ∈ a i + Z . Then X ∈ Λ = { x ∈ R d : M ′ x ∈ a + Z r } .By π V ( x ) = HM ′ x , where H = M ( M ′ M ) − , π V ( X ) = HM ′ X ∈ H ( a + Z r ), so π V ( X ) is lattice.If v ∈ R d \ V , then v Γ, so | ϕ X (2 πv ) | < 1. Thus V has the property stated in (1). To continue,assume the following result is true for now. Lemma A.1. There is K ∈ Z r × r with det K = ± such that Ka = (0 , . . . , , z ν , z ν +1 , . . . , z r ) ,where z ν ∈ [0 , ∞ ) ∩ Q and z ν +1 , . . . , z r ∈ (0 , ∞ ) \ Q are rationally independent. Let Q ∈ R d × ( d − r ) be of full column rank such that Q ′ M = O . Define T = (cid:18) KM ′ Q ′ (cid:19) , Y = KM ′ X, Z = Q ′ X, β i = z i − ⌊ z i ⌋ , i = ν, . . . , r. (A.2)Then T X = ( Y, Z ), β ν ∈ Q ∩ [0 , β ν +1 , . . . , β r ∈ (0 , \ Q are rationally independent. Put β = (0 , . . . , , β ν , β ν +1 , . . . , β r ). Since Y ∈ K ( a + Z r ) = β + Z r , T has the property stated in (3).To show T has the property stated in (2), if u = ( k, ∈ Z r × { } , then by h u, T X i = h k, Y i ∈h k, β i + Z , | ϕ T X (2 πu ) | = 1. Conversely, if | ϕ T X (2 πu ) | = 1, then | ϕ X (2 πT ′ u ) | = 1, so T ′ u = M k ∈ Γfor some k ∈ Z r . Write u = ( w, v ) with w ∈ R r . By (A.2), H ′ T ′ u = H ′ ( M K ′ w + Qv ) = K ′ w .As the LHS is also H ′ M k = k , K ′ w = k , giving w = ( K ′ ) − k ∈ Z r . On the other hand,(Id d − M H ′ ) T ′ u = (Id d − M H ′ )( M K ′ w + Qv ) = Qv and the LHS is also (Id d − M H ′ ) M k = 0.Thus Qv = 0. Since Q is of full column rank, v = 0 and hence u = ( w, ∈ Z r × { } .So far it has been assumed that r = dim( V ) > 0. If r = 0, then Γ = V = { } . Consequently, | ϕ X (2 πv ) | < v = 0 and T = Id d has the property stated in (2)–(3).To show that V is unique, let W be a linear subspace such that π W ( X ) is lattice and | ϕ X (2 πv ) | < v W . By definition, Γ ⊂ W , so V = span(Γ) ⊂ W . If V = W , then W ∩ V ⊥ = ∅ and π W ∩ V ⊥ ( X ) = π W ∩ V ⊥ ( π W ( X )) is lattice. It follows that there is 0 = u ∈ W ∩ V ⊥ , such that h u, X i = h u, π W ∩ V ⊥ ( X ) i ∈ c + Z for some c . But then u ∈ Γ ⊂ V . The contradiction shows V = W and hence the uniqueness of V . 29o show that ν , r , and q are unique, suppose 0 ≤ µ ≤ s ≤ d , q ∗ ∈ N , and B ∈ R d × d isnonsingular, such that | ϕ BX (2 πu ) | = 1 ⇐⇒ u ∈ Z s × { } , (A.3)and BX = ( Y ∗ , Z ∗ ) with Y ∗ ∈ γ + Z s , Z ∗ ∈ R d − s , and γ = (0 , . . . , , γ µ , γ µ +1 , . . . , γ s ), where γ µ = p ∗ /q ∗ with 0 ≤ p ∗ < q ∗ being coprime, and γ ν +1 , . . . , γ s ∈ (0 , \ Q are rationally independent.By B ′ u ∈ Γ ⇐⇒ | ϕ BX (2 πu ) | = 1 and (A.3), Γ = B ′ ( Z s ×{ } ). Since B is nonsingular, a comparisonof dimensions yields s = dim( V ) = r .Let r ≥ 1, otherwise nothing remains to be shown. Then by Γ = T ′ ( Z r × { } ) = B ′ ( Z r × { } ), T ′ Z r = B ′ Z r , where T , B ∈ R r × d consist of the first r rows of T and B , respectively. Thenthere are J, J ∗ ∈ Z r × r such that T = J B and J ∗ T = B , giving J ∗ J B = J ∗ T = B . Since therows of B are linearly independent, J ∗ J = Id r . Thus J − = J ∗ . On the other hand, B X = Y ∗ .Then T X = J B X = J Y ∗ ∈ J ( γ + Z r ) = J γ + J Z r = J γ + Z r . Since T X = Y ∈ β + Z r ,then β − J γ = ( b , . . . , b r ) ∈ Z r . Let J = ( g ij ). Each β i = c i + g i,µ +1 γ µ +1 + . . . + g ir γ r with c i = b i + g i γ + . . . + g iµ γ µ = b i + g iµ γ µ ∈ Z . Since β i , i > ν , are rationally independent, this leadsto µ ≤ ν . Likewise, ν ≤ µ . Thus µ = ν . For i ≤ ν , g i,ν +1 γ ν +1 + · · · + g ir γ r = β i − c i ∈ Z . By rationalindependence of γ ν +1 , . . . , γ r , β i = c i . In particular, β ν = kγ ν − l with k = g νν and l = − b ν .Likewise, γ ν = k ∗ β ν − l ∗ with k ∗ , l ∗ ∈ Z . As a result ( kk ∗ − β ν = k ( γ ν + l ∗ ) − β ν = l + kl ∗ ∈ Z .Since β ν = p/q with 0 ≤ p < q being coprime, q | kk ∗ − 1, so k ∗ and q are coprime. Then γ ν = p ∗ /q with p ∗ = k ∗ p − l ∗ q being coprime with q . Thus q ∗ = q , completing the proof. Proof of Proposition 2.2. (1) Let Kξ = ( ζ , . . . , ζ ν − , p + qζ ν ), where K ∈ Z ν × ν with det K = ± ≤ p < q are coprime, and ζ = ( ζ , . . . , ζ ν ) ∈ Z ν is strongly aperiodic. By ξ ∈ Z ν , h t, ξ i ∈ Z for t ∈ Z ν . Conversely, if h t, ξ i ∈ Z , then letting s = ( K ′ ) − t , h s, Kξ i = s ζ + · · · + s ν − ζ ν − + qs ν ζ ν + ps ν ∈ Z . The strong aperiodicity of ζ implies s , . . . , s ν − , qs ν , ps ν ∈ Z . Since p and q are coprime,then s ν ∈ Z . Thus s ∈ Z ν and t = K ′ s ∈ Z ν . This shows ξ is aperiodic.Conversely, let ξ be aperiodic. Define Γ = Γ ξ as in (A.1). Then Γ is a lattice. Since Z ν ⊂ Γ,by Smith normal form ([34], Th. 3.7), there are linearly independent u , . . . , u ν ∈ Γ and integers1 ≤ n ≤ · · · ≤ n ν with n i | n i +1 , such that, letting M = ( u , . . . , u ν ) and D = diag( n , . . . , n ν ),Γ = M Z ν , Z ν = M D Z ν . (A.4)By u i ∈ Γ, h u i , ξ i ∈ s i + Z for some s i . In matrix form, M ′ ξ ∈ s + Z ν , where s = ( s , . . . , s ν ).Define K = DM ′ , Z = Kξ , and b = Ds . From the second identity in (A.4), Z ν = K ′ Z ν , giving K, K − ∈ Z ν × ν . Then Z ∈ Z ν and is aperiodic. Meanwhile, Z = DM ′ ξ ∈ D ( s + Z ν ) = b + D Z ν .Then from Z ∈ ( b + D Z ν ) ∩ Z ν and D Z ν ⊂ Z ν , b ∈ Z ν . Let Z , Z , . . . , Z ′ , Z ′ , . . . be i.i.d. ∼ Z .For m, n ≥ S m ( Z ) − S n ( Z ′ ) ∈ Z b + D Z ν ⊂ Z ν . By aperiodicity of Z , for every standard basevector e i of R ν , there are m and n such that P { S m ( Z ) − S n ( Z ′ ) = e i } > e i ∈ Z b + D Z ν . As a result, Z b + D Z ν = Z ν . Let s i ∈ Z and v i = ( v i , . . . , v iν ) ∈ Z ν , such that bs i + Dv i = e i . Write b = ( b , . . . , b ν ). By comparing the coordinates, b i s i + n i v ii = 1 , b j s i + n j v ij = 0 , j = i. Thus, each pair of b i and n i are coprime. For j > i , as n j = 0, n j | b j s i , so n j | s i . Then by b i ( s i /n j ) n j + n i v ii = 1, n j and n i are coprime. By n i | n j , this gives n i = 1. As a result, n = · · · = n ν − = 1. Put q = n ν and let 0 ≤ p < q such that q | ( b ν − p ). Let ζ = D − ( Z − pe ν ).By Z ∈ b + D Z ν and b − pe ν ∈ D Z ν , ζ ∈ Z ν . If h t, ζ i ∈ s + Z , where t ∈ R ν and s ∈ R , then h M t, ξ i = h t, M ′ ξ i = h t, D − Z i ∈ c + Z with c = s + p h t, D − e ν i . Then M t ∈ Γ = M Z ν , so t ∈ Z ν .Thus ζ is strongly aperiodic. By Kξ = Z = pe ν + Dζ , the proof is complete.302) Let ζ = L − β ν e ν . Then ζ ∈ Z ν . Since for u ∈ R d , | ϕ T X (2 πu ) | = 1 ⇐⇒ u ∈ Z ν × { } ,then for v ∈ R ν , | ϕ ζ (2 πv ) | = | ϕ L (2 πv ) | = 1 ⇐⇒ v ∈ Z ν , so ζ is strongly aperiodic. Then by (1), DL = ( ζ , . . . , ζ ν − , p + qζ ν ) ∈ Z ν is aperiodic. Proof of Lemma A.1. Recall Q , R , and their quotient R / Q are vector spaces over the field Q , Let¯ a = (¯ a , . . . , ¯ a r ) with ¯ a i = a i + Q ∈ R / Q . First, if ¯ a = 0, then there are linearly independent¯ u , . . . , ¯ u s ∈ R / Q , 1 ≤ s ≤ r , such that ¯ a = A ¯ u , where A ∈ Q r × s is of full column rank and¯ u = (¯ u , . . . , ¯ u s ). Equivalently, a − Au ∈ Q r . Note that u i are rationally independent. Bymultiplying A by a large m ∈ N and dividing u by m , A can be assumed to be in Z r × s . It isknown that there are P ∈ Z r × r and R ∈ Z s × s with | det P | = | det R | = 1, such that P A = (cid:0) DO (cid:1) R , where D = diag( d , . . . , d s ) with d i ∈ N and d i | d i +1 (cf. [23], Th. III.5). Let DRu = v .Then P ( a − Au ) = P a − ( v, ∈ Q r , so P a = (˜ v, w ), where ˜ v = v + y for some y ∈ Q s and w ∈ Q r − s . The coordinates of ˜ v are rationally independent. On the other hand, similar to A ,there is M ∈ Z ( r − s ) × ( r − s ) with det M = ± 1, such that M w = ( q, , . . . , ∈ Q r − s with q ≥ K = (cid:0) Id s M (cid:1) P gives K a = (˜ v, q, , . . . , a = 0, then a ∈ Q r . Following the treatment of the above w , the result follows. B Proofs regarding distributions in the domain of attraction Proof of (3.1) and (3.2) . Let X ∼ F ∈ D ( α ). If α = 2, then (3.2) is part of [28], Th. 4.1. For any c > V X ( s ) ≤ V X ( cs ) ≤ V X ( s ) + c s q X ( s ), which by (3.2) gives V X ( cs ) /V X ( s ) → s → ∞ .Then (3.1) follows. If α ∈ (0 , q X ∈ R − α , which leads to both(3.1) and (3.2) (cf. [1], Th. 1.6.4). Proof of (3.4) . For the univariate case, see [1], p. 347. For the multivariate case, first, let α ∈ (0 , a n is the infimum of all s such that P {| X | > s, X/ | X | ∈ E } ≤ γ ( E ) /n ≤ P {| X | ≥ s, X/ | X | ∈ E } , where γ is a nonzero measure on S d − and E is any fixed subset of S d − with γ ( E ) > 0. ByTh. 14.10 of [29], γ is finite. Letting E = S d − and c = γ ( S d − ), it follows that a n can be any s satisfying q X ( s ) ≤ c/n ≤ q X ( s − ). Then by (3.2) and (3.3), a n can (also) be taken to be anysequence such that A ( a n ) ∼ cn .Let α = 2 and b ( u ) = h u, Σ u i , where Σ is the covariance matrix of the limiting normal distri-bution. If E | X | < ∞ , then (3.4) follows from the Central Limit Theorem. Suppose E | X | = ∞ .By Th. 2.4 of [28], a n can be any sequence such that for any ǫ > 0, (i) nq X ( ǫa n ) → n/a n )[ m V ( ǫa n , u ) − h c V ( ǫa n ) , u i ] → b ( u ) for any u ∈ S d − . Since | c V ( s ) | = o ( V X ( s )) as s → ∞ ([28], (4.5)), by (3.1) and (4.6), (ii) is equivalent to n/A ( a n ) → P i b ( e i ). Once (ii) is satisfied, by(3.2), (i) is satisfied. Then the claim on a n follows. Proof of (3.5) . For the univariate case, see [1], p. 347. For the multivariate case, according tothe last comment on p. 190 in [28], b n can be taken to be ( n/a n ) c X ( ta n ) + γ , where γ is anyconstant vector, and t > {| x | = t } has measure 0 under the L´evymeasure of the limiting stable law. From the characterization of the L´evy measure (cf. [28], (3.4)–(3.5)), t can be any positive number. It follows that any b n satisfying (3.4) must be of the form( n/a n ) c X ( a n ) + γ + ǫ n for some constant vector γ , where ǫ n → n → ∞→ ∞