A remark on the maximum eigenvalue for circulant matrices

Transcription

1 IMS Collections High Dimensional Probability V: The Luminy Volume Vol 5 ( c Institute of Mathematical Statistics, 009 DOI: 04/09-IMSCOLL5 A remark on the imum eigenvalue for circulant matrices W lodek Bryc and Sunder Sethuraman University of Cincinnati and Iowa State University Abstract: We point out that the method of Davis-Mikosch [Ann Probab 7 ( ] gives for a symmetric circulant n n matrix composed of iid entries with mean 0 and finite ( + δ-moments in the first half-row that the imum eigenvalue is on the order n log n, and the fluctuations are Gumbel Let {X 0,X,} be iid mean-zero, variance, random variables For m, consider the (m + (m + palindromic circulant matrix X 0 X X X m X m X m X ( X m X m X 0 X X X m X X X 3 X m X m X m X 0 In this note, we observe, for circulant matrices (, that an argument of [] for the imum of periodograms easily applies to deduce that the imum eigenvalue is on the order m log m, and the fluctuations are Gumbel (Theorem In particular, a sort of universality with respect to the entries {X i }, much discussed in other contexts in the random matrix literature, is established for the asymptotic imum eigenvalue distribution We refer to [3] for more discussion of random circulant matrices, and note the result for Gaussian entries is as well given in [3, Corollary 5] Theorem Suppose X,X, are iid with E(X = 0, E(X =,and E( X s < for some s> Denotebyλ m the imum eigenvalue of (, and let a m = logm log(4π log m/ ( logm Then lim m P (( λm a m logm x = G(x where G(x =exp( e x The proof follows closely the method used to prove [, Theorem ] which is based on Einmahl s multivariate extension of the Komlos-Major-Tusnady theorem Department of Mathematical Sciences, University of Cincinnati, 855 Campus Way, PO Box 005, Cincinnati, OH , USA, WlodzimierzBryc@UCedu Department of Mathematics, 396 Carver Hall, Iowa State University, Ames, IA 500, USA, sethuram@iastateedu Keywords and phrases: imum, eigenvalue, circulant, Gumbel, random, matrix AMS 000 subject classifications: Primary 5A5 79

2 80 W Bryc and S Sethuraman (cf Lemma 3 Indeed, Lemmas 4, 5 are similar to [, Lemmas 33, 34] with analogous proofs The well known Bonferroni inequalities (Lemma and Lemma 3 are stated as [, Lemmas 3, 3] Lemma Let A,,A n be measurable events Then for every k n/, k d= ( d S d P (A A n where S d = j < j d n P (A j A jd ( d S d, k The next statement is Einmahl s Corollary (b, page 3, in combination with the Remark on page 3 [] Lemma 3 Let ξ,,ξ n be independent random vectors in R d Assume that the moment generating function of {ξ i } exists in a neighborhood of the origin, and that d= cov(ξ + + ξ n =B n I d, where B n > 0 and I d is the d dimensional identity matrix Let η k be independent N(0,σ cov(ξ k random vectors for k n independent of {ξ i },and0 <σ Let ξk = ξ k + η k for k n, andwritep n as the density of Bn / n ξ k Choose 0 <α</ such that ( α Let n E ξ k 3 exp(α ξ k B n (3 β n = β n (α = B 3/ n If n E ξ k 3 exp(α ξ k (4 x c αb / n, σ c β n log β n and B n c 3 α, where c,c,c 3 are constants depending only on d, then (5 p n(x = φ (+σ I d (xexp( T n (x with T n (x c 4 β n ( x 3 +, where φ C is the density of the d-dimensional centered Gaussian vector with covariance matrix C and c 4 is a constant depending only on d Let now {X j } j 0 be as in Theorem For j, m 0, define X j = X j E(X Xj m /s /s X m Lemma 4 We have as that m ( πjk X k m + m + m ( πjk X (m k = O(m / m + X (m j =

3 A remark on the imum eigenvalue for circulant matrices 8 Proof First, we can add and subtract (m + / X0 on the left-side Since we can replace with + m X 0 + ( πjk m + m X 0 X0 m /s + m = 0, ( πjk X k m + ( πjk X k m + Xk m /s Now, by Borel-Cantelli, as t P ( X t >t /s <, wehave X t t /s for all t N(ω as Then, m m X t X t Xt m = X /s t Xt >m /s N(ω N(ω X t Xt >m /s + m t=n(ω+ X t Xt >m /s + t>n(ω for m {N(ω, X s,, X N(ω s } Hence, the sums m ( πjk X k m + and agree for all large m as We finish by noting the extra term m X t Xt >t /s X t Xt >t /s = 0 ( πjk X k m + Xk m /s [ X 0 X 0 X0 m /s] = E[X 0 X0 m /s] = O(m / For d, define v d (t = ( πjt m+,,( πj dt m+ with respect to distinct integers j,,j d m Let also {N j } be a sequence of iid N(0, random variables independent of {X j } Lemma 5 For d, let p m be the density of E[ X ](m +[ ( X0 + σ m N 0 vd (0 + m ( Xk + σ m N k vd (k] where σ m = E[ X ]s mifm c5 log m s m for c 5 =/ ( δ/s > 0 and some 0 <δ<, then, uniformly for x 3 = o(m / /s, p m (x =φ (+smi d (x( + o(

4 8 W Bryc and S Sethuraman Proof Apply Lemma 3 to the centered vectors X 0 v d (0, X v d (, X m v d (m where, after some calculation, cov( X0 v d (0 + X v d ( + + X m v d (m = B m I d and [ B m = E X m ( ] πk +4 m + Choose for a fixed constant c 6 > 0, = (m +E X Note for each 0 t m, that Then, for large m, v d (t = α = c 6 m /s d / d l= αe X 0 v d (0 3 exp{ α X 0 v d (0} + α ( πjl t m + d m E X t v d (t 3 exp { α X t v d (t } 8d 3/ α(m +E X 3 exp { α X d /} 0dc 6 m /s E X 3 exp { } c 6 } 0dc 6 exp{4c 6 m δ/s E X +δ where 0 <δ<ischosensothate X +δ < Then, ( holds with α = α for sufficiently small c 6 Now choose β m = Bm 3/ E X 0 v d (0 3 exp{ α X 0 v d (0} +B 3/ m m E X t v d (t 3 exp { α X t v d (t } 8d 3/ B 3/ m (m +E X 3 exp { α X d /} Then, β m const(bm 3/ m +( δ/s E X +δ const(m c5 where c 5 =/ ( δ/s > 0 Next, we consider (4 We can choose x so that Then, we can choose σ = s m so that x c αb / m const(m / /s s m const(m c5 log m and note Noting (5, we have B m m c 3 α m /s p m = φ (+s m I d (xexp ( Tm (x with T m (x c 4 βm ( x 3 +

5 A remark on the imum eigenvalue for circulant matrices 83 However, uniformly over x 3 = o(m / /s, T m (x c 4 βm ( x 3 + const(m / /s c5 = const(m δ/s = o( Proof of Theorem From properties of circulants, we have that the eigenvalues of ( are λ j = X 0 + m ( πkj m+ Xk for 0 j m, andalsoλ j = λ m+ j for j m Sincem / logm 0, the variable X 0 in the expression for λ j can be replaced by X 0 We will also be able to omit the contribution of λ 0 to the imum By Lemma 4, it will be enough to prove [ (6 X 0 m ( πjk ] logm + X k a m G m + m + m + To this end, let σ m = E[ X ]s m = E[ X ]m c5 log m Wefirstshow [ ( X 0 + σ m N 0 logm E[ X ](m + (7 + E[ X ](m + For j m, let m λ X+N j = ( X0 + σ m N 0 + E[ X ](m +[ Since e e u statement P ( πjk ( m + X ] k + σ m N k a m m G ( πjk ( m + X ] k + σ m N k = d= ( d (e du /d!, by Lemma, (7 will follow from the ( λ X+N j >a m + u logm,,λ X+N j d >a m + u logm (8 = m d exp( du( + o( uniformly over the d-tuples j < <j d m for each d asm Let A d m denote the event in the probability on the left-side Then, noting s m = m c5 log m, φ (+s m I d (xdx = m d exp{ du}( + o( A d m as m Note that we can neglect the parts in (8 when there is l d such that λ X+N j l 3 > m / /s ɛ for a small ɛ>0 Indeed, given d ands>choose0<ε</ /s and γ>such that γ(/ /s ε >d+ Note also / E[ X ] formlarge enough Then, by Rosenthal s inequality there is a constant C(γ such that P m X0 + ( πjl k m + C(γ m γ( /s ɛ ( C(γ 3 X k >m /s ɛ ( m 3γ/ E X k + me X 3γ k=0 m γ(/ /s ε + m = o(m d m γ( 4/s ε

6 84 W Bryc and S Sethuraman On the other hand, also P N0 m ( πjl k 3 + σ m N k m + m + m + >m / /s ε = o(m d We conclude by Lemma 5 (which does not depend on the choice of j,,j d that (8 holds To deduce (6, note E[ X ] E[X ] =, and (9 σm N 0 m ( X0 + σ m N 0 + m + X0 m + σm N 0 + ( πjk ( σ m N k m + m ( πjk m + m ( πjk X k m + m + m ( πjk σ m N k m + m + ( X k + σ m N k Let (m + / λ N j = σ m N 0 + m (πjk/m +σ mn k for j m One can calculate that that {λ N j }m j= are iid N(0,σ m variables Hence, to finish, the bounds (9 correspond to the imum of m iid N(0,σm random variables, well known to be on order σ m logm m c 5 log m 0in probability Acknowledgement The authors benefited from a discussion with Magda Peligrad References [] Davis, R A and Mikosch, T (999 The imum of the periodogram of a non-gaussian sequence Ann Probab [] Einmahl, U (989 Extensions of results of Komlos, Major and Tusnady to the multivariate case J Multivariate Anal [3] Meckes, M W (009 Some results on random circulant matrices IMS Coll 5 3-3