No Eigenvalues Outside the Support of the Limiting Spectral Distribution of Large Dimensional Sample Covariance Matrices


 Mervyn Sims
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1 No igevalues Outside the Support of the Limitig Spectral Distributio of Large Dimesioal Sample Covariace Matrices By Z.D. Bai ad Jack W. Silverstei 2 Natioal Uiversity of Sigapore ad North Carolia State Uiversity Dedicated to the Sixtieth Birthday of MiTe Chao Summary Let B =/N )T /2 X XT /2 where X is N with i.i.d. complex stadardized etries havig fiite fourth momet, ad T /2 is a Hermitia square root of the oegative defiite Hermitia matrix T. It is kow that, as,if/n coverges to a positive umber, ad the empirical distributio of the eigevalues of T coverges to a proper probability distributio, the the empirical distributio of the eigevalues of B coverges a.s. to a oradom limit. I this paper we prove that, uder certai coditios o the eigevalues of T, for ay closed iterval outside the port of the limit, with probability there will be o eigevalues i this iterval for all sufficietly large. Research of this author coducted at the Departmet of Applied Mathematics, Natioal Su Yatse Uiversity, Kaohsiug, Taiwa 2 Supported by the Natioal Sciece Foudatio uder grat DMS AMS 99 subject classificatios. Primary 60F5; Secodary 62H99. Key Words ad Phrases. Radom matrix, empirical distributio fuctio of eigevalues, Stieltjes trasform.
2 . Itroductio For =, 2,... let X = X =X ij ), T = T,adT /2, deote, respectively, a N matrix cosistig of i.i.d. stadardized complex etries X =0, X 2 =),a oegative defiite matrix, ad ay square root of T. For ay square matrix A havig real eigevalues, let F A deote the empirical distributio fuctio e.d.f.) of its eigevalues. The matrix B =/N )T /2 XX T /2 ca be viewed as the sample covariace matrix of a broad class of radom vectors, T /2 X X j deotig the j th colum of X). Previous work o uderstadig the behavior of the eigevalues of B whe ad N are large but have the same order of magitude has bee o F B ad o the extreme eigevalues whe T = I, the idetity matrix. Assumig N = N) with /N c>0as ad F T D H, a proper p.d.f. it is kow that almost surely F B coverges weakly to a oradom p.d.f. F see Silverstei 995)). Provig this result, alog with describig F which ca be explicitly expressed i oly a few cases), is best achieved with the aid of the Stieltjes trasform, defied for ay p.d.f. G by m G z) λ z dgλ) z C+ {z C : Imz > 0}. Because of the iversio formula G[a, b]) = π lim η 0 b a Imm G ξ + iη)dξ a, b cotiuity poits of G), weak covergece of p.d.f. s ca be prove by showig covergece of Stieltjes trasforms. For each z C +, m = m F z) is a solutio to the equatio m = t c czm) z dht), which is uique i the set {m C : c)/z +cm C + }.LetB =/N )X TX. Sice the spectra of B ad B differ by N zero eigevalues, it follows that F B = ) I [0, ) + N N F B, from which we get /N) m F B z) = + z N m F B z) z C +, ad with F deotig the limit of F B we have F = c)i [0, ) + cf,
3 ad c) m F z) = + cm F z) z C +. z It follows that m F = z +tm F dht), for each z C +, m = m F z), is the uique solutio i C + to the equatio ) tdht) m = z c,.) +tm ad m F z) has a iverse, explicitly give by zm) = m + c tdht) +tm..2) Much of the aalytic behavior of F ca be iferred from these equatios see Silverstei ad Choi 995)). Ideed, cotiuous depedece of F o c ad H is readily apparet from.2), ad the iversio formula, ad it ca be show that F D H as c 0. Moreover, it is show i Silverstei ad Choi 995) that, away from zero, F has a cotiuous desity. As a example Figure a) is the graph of the desity whe c =. adh places mass.2,.4, ad.4 at, respectively,, 3, ad 0. The focus of this paper is o itervals of R + lyig outside the port of F. The iverse.2) ca be used to idetify these itervals, maily because, o ay such iterval, m F exists ad is icreasig. Cosequetly its iverse will also exist ad will be icreasig o the rage of this iterval. Silverstei ad Choi 995) cofirms each m i this rage is such that /m lies outside the port of H. Therefore, plottig.2) o R ad observig the rage of values where it is icreasig will yield the complemet of the port of F, ad, together with c to determie whether there is ay mass at zero), the complemet of the port of F. Figure b) provides a illustratio. It is the graph of.2) correspodig to the desity i a). 2
4 Fig. a) Graph of the limitig desity whe c =. ad H places mass.2,.4, ad.4 at, respectively,, 3, ad 0. b) The graph of x = m +c t+tm) dht) correspodig to a). The bold lies o the vertical axis idicate the port of the desity, the set i R + remaiig after removig itervals where the graph is icreasig. Usig the fact that the desity at x R + is equal to cπ) times the imagiary part of m F x) see Silverstei ad Choi 995)), the graph i a) was created by applyig Newto s method to.2) for values of z = x i the port. For large oe would ituitively expect o eigevalues to appear o a closed iterval outside the port of F. This of course caot be iferred from the limitig result o F B. The two importat cases whe T = I have bee settled. Here the port of F lies o [ c) 2, + c) 2 ], with the additio of zero whe c>. Whe the etries 3
5 of X come from the upper left portio of a doubly ifiite array of idepedet radom variables havig fiite fourth momet, Yi, Bai, ad Krishaiah 988), ad Bai ad Yi 993) show, respectively, the largest eigevalue of B coverges a.s. to + c) 2,ad the mi, N) th largest which is the smallest eigevalue whe c<) coverges a.s. to c) 2 we remark here that i Bai, Silverstei, ad Yi 988), it is prove that X 4 < is ecessary for the former to hold). xtesive computer simulatios, performed i order to show the importace of the spectral limitig results to the detectio problem i array sigal processig Silverstei ad Combettes 992)), resulted i o eigevalues appearig where there is o mass i the limit. Uder reasoably mild coditios, this paper will provide a proof of this pheomeo, agai i the form of a limit theorem as. It will be ecessary to impose stroger coditios o the eigevalues of T tha simply weak covergece of F T to H. For this, if we let F c,h deote F ad c = /N, the F c,h is the limitig oradom d.f. associated with the limitig ratio c ad d.f. H. As will be see, the coditios o H are reflected i F c,h. Theorem.. Assume a) X ij, i, j =, 2,... are i.i.d. radom variables i C with X =0, X 2 =,ad X 4 <. b) N = N) with c = /N c>0as. c) For each T = T is Hermitia oegative defiite satisfyig H F T D H, a p.d.f. d) T, the spectral orm of T is bouded i. e) B = /N )T /2 X XT /2, T /2 ay Hermitia square root of T, B = /N )XT X,whereX = X =X ij ), i =, 2,...,, j =, 2,...,N. f) Iterval [a, b] with a>0 lies outside the port of F c,h ad F c,h for all large The P o eigevalue of B appears i [a, b] for all large )=. Usig the results o the extreme eigevalues of /N )XX we see that the iterval ca also be ubouded. I particular we have Corollary. If T coverges to the largest umber i the port of H, the B coverges a.s. to the largest umber i the port of F. If the smallest eigevalue of T coverges to the smallest umber i the port of H, the c<c>) implies the smallest eigevalue of B B ) coverges to the smallest umber i the port of F F ). Theorem. is prove by showig the covergece of Stieltjes trasforms at a appropriate rate, uiform with respect to the real part of z over certai itervals, while the imagiary part of z coverges to zero. Besides relyig o stadard results o matrices, the proof requires wellkow bouds o momets of martigale differece sequeces, as 4
6 well as a extesio of Rosethal s iequality to radom quadratic forms. The proof of the latter will be give i the appedix. Statemets of most of the mathematical tools eeded will be give i the ext sectio. Sectio 3 establishes a rate of covergece of F B, eeded i provig the covergece of the Stieltjes trasforms. The latter will be broke dow ito two parts sectio 4 ad 5), while sectio 6 completes the proof. It is metioed here that Theorem. is actually oly part of the importat pheomea observed i simulatios. It ca be show that o ay iterval J H with edpoits outside the port of H, there correspods for c sufficietly small, a iterval J F,c with edpoits beig boudary poits of the port of F satisfyig F J F,c )=HJ H ). This should be viewed i the fiite but large dimesioal case as the eigevalues of B beig a smoothed deformatio of the eigevalues of T, cotiuous i the ratio of dimesio to sample size. Simulatios reveal that the umber of eigevalues of B appearig i J F,c is exactly the same as the umber of eigevalues of T i J H. The formulatio of the cojecture aturally arisig from this is simply F J F,c ) F c,h J F,c )) 0 a.s. Its truth is curretly beig ivestigated. 2. Mathematical tools We list i this sectio results eeded to prove Theorem.. Throughout the rest of the paper costats appearig i iequalities are represeted by K ad occasioally subscripted with variables they deped o. They are oradom ad may take o differet values from oe appearace to the ext. Refereced results below cocerig momets of sums of complex radom variables were origially prove for real variables. xtesio to the complex case is straightforward. Lemma 2. Burkholder 973)). Let {X k } be a complex martigale differece sequece with respect to the icreasig σfield {F k }.Theforp 2 p ) p/2 ) Xk K p Xk 2 F k ) + Xk p. Lemma 2.2 Burkholder 973)). With {X k } as above, we have, for p> p Xk Kp Xk 2) p/2. Lemma 2.3 Rosethal 970)) If {X k } are idepedet oegative, the for p ) p ) p ) Xk Kp Xk + X p k. 5
7 Lemma 2.4 Dilworth 993)). With {F k } as above, {X k } k a sequece of itegrable radom variables, ad q p< we have k= X k F k ) q ) p/q ) p/q p ) p/q X k q. q k= The followig lemma is foud i most probability textbooks. Lemma 2.5. Kolmogorov s iequality for submartigales). If X,...,X m is a submartigale, the for ay α>0 Pmax X k α) k m α X m ). The ext oe has a straightforward proof. Lemma 2.6. If for all t>0, P X >t)t p K for some positive p, the for ay positive q<p ) p X q K q/p. p q Lemma 2.7 proof i appedix). For X =X,...,X ) T etries, C matrix complex) we have for ay p 2 i.i.d. stadardized complex) X CX tr C p K p X 4 tr CC ) p/2 + X 2p tr CC ) p/2 ). Lemma 2.8 Corollary of Hor ad Johso 985)). For r s matrices A ad B with respective sigular values σ σ 2 σ q, τ τ 2 τ q,whereq = mir, s) we have σ k τ k B A for all k =, 2,...,q. Lemma ) of Hor ad Johso 99)). For Hermitia A =a ij ) with eigevalues λ,..., λ, ad covex f we have fa ii ) i= fλ i ). i= Lemma 2.0 Lemma 2.6 of Silverstei ad Bai 995)). Let z C + with v = Imz, A ad B with B Hermitia, ad r C. The tr B zi) B + rr zi) ) A = 6 r B zi) AB zi) r +r B zi) r A v.
8 Lemma 2. Lemma 2.3 of Silverstei 995)). For z = x + iv C + let m z), m 2 z) be Stieltjes trasforms of ay two p.d.f. s, A ad B with A Hermitia oegative defiite, ad r C. The a) m z)a + I) max4 A /v, 2) b) tr Bm z)a+i) m 2 z)a+i) ) m 2 z) m z) B A max4 A /v, 2)) 2 c) r Bm z)a + I) r r Bm 2 z)a + I) r m 2 z) m z) r 2 B A max4 A /v, 2)) 2 r deotig uclidea orm o r). Lemma 2.2 Lemma 2.4 of Silverstei ad Bai 995)). For Hermitia A ad B F A F B raka B), here deotig orm o fuctios. Basic properties o matrices will be used throughout the paper, the two most commo beig: tr AB A tr B for Hermitia oegative defiite A ad B, ad for A ad r C, for which both A ad A + rr are ivertible r A + rr ) = + r A r) r A. At oe poit i Sectio 3 the twodimesioal Stieltjes trasform is eeded. Its defiitio ad relevat properties are give here. For a p.d.f. F x, y) defied o R 2 it is defied as mz,z 2 )= df x, y) x z )y z 2 ) for all z = x + iv, z 2 = x 2 + iv 2 v 0,v 2 0. Due to the iversio formula F [a, b] [c, d]) = π 2 lim mz,z 2 ) m z,z 2 ) mz, z 2 )+m z, z 2 )dx dx 2 v 0 [a,b] [c,d] v = v 2 = v, wheever F [a, b] [c, d])) = 0, weak covergece of p.d.f. s o R 2 is assured oce covergece of their Stieltjes trasforms is verified o a coutable collectio of poits z,z 2 ) dese i some ope set i C 2. 7
9 3. A rate o F B We begi by simplifyig our assumptios. Because of assumptio d) i Theorem. we ca assume T. For C > 0 let Y ij = X ij I [ Xij C] X ij I [ Xij C], Y = Y ij ) ad B = /N )T /2 Y Y T /2. Deote the eigevalues of B ad B by λ k ad λ k i decreasig order). Sice these are the squares of the k th largest sigular values of / N)T /2 ad / N)T /2 Y respectively), we fid usig Lemma 2.8 max k λ/2 k /2 λ k / N) X Y. Sice X ij Y ij = X ij I [ Xij >C] X ij I [ Xij >C], from Yi, Bai, ad Krishaiah 988) we have with probability oe lim max k λ/2 k /2 λ k + c) /2 X 2 I [ X >C]. Because of assumptio a) we ca make the above boud arbitrarily small by choosig C sufficietly large. Thus, i provig Theorem. it is eough to cosider the case where the uderlyig variables are uiformly bouded. I this case it is prove i Yi, Bai, ad Krishaiah 988) that there exists a sequece {k } satisfyig k / log such that for ay η> + c) 2 X /N )X X k η k for all sufficietly large. It follows the that λ max, the largest eigevalue of B, satisfies. P λ max K) =on l ), 3.) for ay K> + c) 2 ad ay positive l. Also, sice tr CC ) p/2 tr CC ) p/2, we get from Lemma 2.7 whe X is bouded X CX tr C p K p tr CC ) p/2 3.2) where K p also depeds o the distributio of X. From 3.2) we easily get X CX p K p tr CC ) p/2 + tr C p ). 3.3) Throughout the paper, variable z = x + iv will be the argumet of ay Stieltjes trasform. Let m = m F B ad m = m F B. For j =, 2,...,N,letq j =/ )X j X j deotig the j th colum of X), r j =/ N)T /2 X j,adb j) = Bj) = B r j rj. 8
10 I Silverstei 995) the formula m z) = N z + r j B j) zi) r j ) is derived. It is easy to verify Imr j /z)b j) I) r j 0. Therefore, for each j z + r j B j) zi) r j ) v, 3.4) It is also show i Silverstei 995) that tr zm z)t zi) m z) w z) = N where z + r j B j) zi) r j ) d j 3.5) d j = q j T /2 B j) zi) m z)t + I) T /2 q j tr m z)t + I) T B zi). The ext task is to prove for v = v N /7 ad for ay subsets S [0, ) cotaiig at most elemets the almost sure covergece of w z) max x S v 5 to zero. Let m j) z) = c ) z + c m F B j) z). From Lemma 2.0 we have max m z) m j) z) j N Nv. 3.6) Moreover,itiseasytoverifythatm j) z) is the Stieltjes trasform of a p.d.f., so that m j) z) v Write for each j Nd j = d j + d2 j + d3 j + d4 j where d j = qj T /2 B j) zi) m z)t +I) T /2 q j qj T /2 B j) zi) m j) z)t +I) T /2 d 2 j = qj T /2 B j) zi) m j) z)t + I) T /2 q j tr m j)z)t + I) T B j) zi) d 3 j = tr m j)z)t + I) T B j) zi) tr m j)z)t + I) T B zi). 9 q j
11 ad d 4 j = tr m j)z)t + I) T B zi) tr m z)t + I) T B zi). I view of 3.4), it is sufficiet to show the a.s. covergece of d i j max j N,x S v 6 3.7) to zero for i =, 2, 3, 4. Usig A zi) /v for ay Hermitia matrix A we get from Lemma 2. c) ad 3.6) d j 6 X j 2 Nv 4. Usig 3.2) it follows that for ay ɛ>0, p 2, ad all sufficietly large d j P max ) j N,x S v 6 >ɛ P N K p Nv 0 ) p ɛ p p/2, so 3.7) a.s. 0 whe i = ad for ay v N /0, ]. Usig Lemma 2.0 ad Lemma 2. a) we fid max X j 2 ) j N 6 >ɛ/2 Nv0 v 6 d 3 j 4 v 8, so that 3.7) a.s. 0fori = 3 ad for ay v = N δ with δ [0, /8). We get from Lemma 2. b) ad 3.6) v 6 d 4 j 6 Nv 0, so that 3.7) a.s. 0fori =4,adforayδ [0, /0). Usig 3.2) we fid for ay p 2 v 6 d 2 j p /2 K p v 6p tr T p B j) zi) m j) z)t +I) T m j) z)t +I) B j) zi) T /2 ) p/2 = K p v 6p p tr m j)z)t + I) T m j) z)t + I) B j) zi) T B j) zi) ) p/2 usig Lemma 2. a)) K p v 6p p v 2 p/2 tr B j) zi) T B j) zi) ) p/2 0
12 = K p v 7 ) p tr T B j) zi) B j) zi) ) p/2 K p v 7 ) p /v2 ) p/2 = K p /2 v 8 ). p We have the for ay ɛ>0adp 2 P max v 6 d 2 N j N,x S j >ɛ) K p ɛ p /2 v 8 ). p Thus, max x S w z) v 5 a.s. 0 for ay oegative δ /7 sice we have show for ay positive l, wehaveforallp sufficietly large ad for all ɛ>0 Pmax w z) v 5 >ɛ) K p ɛ p l. x S Moreover, for the sequece {µ } with µ = N /68 we have for ay v = N δ with δ /7 Pµ max x S w z) v 5 >ɛ) K p ɛ p l. 3.8) We ow rewrite w totally i terms of m. With H F T, ad usig the idetity we have w = c z c m z) = c ) z dh t) m +tm c ) ) z = m z + c c z m + c m z) = m c c z m tdh t) +tm dh t) z c ) ) +tm m ). Let ω = z tdh t) + c. m +tm The ω = w zc /m. Returig ow to F c,h ad F c,h,letm 0 = m F c,h ad m 0 = m F c,h. The m 0 solves.), its iverse is give by.2), m 0 = z + c tdh t) +tm 0, 3.9) ad the iverse of m 0, deoted z 0,isgiveby z 0 m) = m + c tdh t) +tm. 3.0)
13 From 3.0) ad the iversio formula for Stieltjes trasforms it is obvious that F c,h D F c,h, as. Therefore, from assumptio f) a ɛ > 0 exists for which [a 2ɛ,b+2ɛ] also satisfies f). This iterval will stay uiformly bouded away from the boudary of the port of F c,h d for all large, so that for these both x [a 2ɛ,b+2ɛ] dx m0 x) is bouded ad /m 0 x) forx [a 2ɛ,b+2ɛ] stays uiformly away from the port of H. Therefore for all sufficietly large ) d x [a 2ɛ,b+2ɛ] dx m0 x) t 2 dh t) + tm 0 K. 3.) 2 x)) Let a = a ɛ, b = b+ɛ. O either,a ]or[b, ), each collectio of fuctios i λ, {λ x) : x [a, b]}, {λ x) 2 : x [a, b]}, form a uiformly bouded, equicotiuous family. It is straightforward, the, to show lim m 0 x) m 0 x) =0, 3.2) ad lim d dx m0 x) d dx m0 x) =0 3.3) see, for example, p. 7, problem 8 of Billigsley 968)). Sice for all x [a, b], λ [a,b ] c, ad positive v λ x + iv) λ x < v ɛ 2 we have for ay sequece of positive v covergig to 0 lim m 0 x + iv ) m 0 x) =0. 3.4) Similarly lim Imm 0 x + iv ) v d dx m0 x) =0. 3.5) 5. xpressios 3.), 3.2), 3.4), ad 3.5) will be eeded i the latter part of Sectio Let m 0 2 = Imm0.Wehavethefrom3.9) m 0 2 = v + m 0 t 2c 2 dh t) +tm 0 2 z + c tdh t) ) 2 +tm 0
14 For ay real x, by Lemma 2. a), t m 0 2 ) dh t) 2c +tm 0 2 = c tdh t) Im +tm 0 It follows that c T I + Tm 0 ) 4c /v. t m0 2c 2 dh t) +tm 0 2 v + m 0 2 c t 2 dh t) +tm 0 2 /2 < Kv 2, 3.7) for some positive costat K. Let m = m + im 2,wherem = Re m, m 2 = Imm.Wehavem satisfyig m = z + c tdh t) +tm ω, 3.8) ad m 2 = v + m 2 c t 2 dh t) +tm + Imω 2 z + c tdh t) ω ) +tm From 3.9) ad 3.8) we get m m 0 = m m 0 t )c 2 dh t) ) z + c tdh t) +tm ω +tm )+tm 0 ) z + c tdh t) +tm 0 ) +m m 0 ω. 3.20) From CauchySchwarz, 3.6), 3.7) ad 3.9) we get, whe Imω/v < t 2 dh t) +tm c )+tm 0 ) ) ) z + c tdh t) +tm ω z + c tdh t) +tm 0 t 2 dh t) +tm 2 c z + c tdh t) ω = +tm m 2 c t 2 dh t) +tm 2 v + m 2 c t 2 dh t) +tm 2 2 /2 + Imω 3 t 2 dh t) +tm c 0 2 z + c tdh t) /2 +tm 2 0 t m0 2c 2 dh t) +tm 0 2 v + m 0 2 c t 2 dh t) +tm 0 2 /2 /2
15 t m0 2c 2 dh t) +tm 0 2 v + m 0 2 c t 2 dh t) +tm 0 2 /2 Kv ) We claim that o the set {λ max K },wherek > + c) 2, for all sufficietly large, m 2 µ v wheever x µ v. Ideed, whe x v or x λ max + v m Rem K + µ v K + µ v ) 2 + v 2 2µ v for large. Whe v <x<λ max + v m Imm v K + v ) 2 + v 2 µ v for large. Thus the claim is prove. Therefore, whe x µ v, o the set { w v} {λ 4 max K } we have for large z 2µ v ad Therefore, by 3.20) ad 3.2), we have Imω) c zw /m Kµ 2 v 2 w <v. m m K 0 v 2 m m 0 ω = K v 2 c zm 0 w K v 4 µ w. It is easy to verify that for large, whe either x >µ v, w >v 4,orλ max >K Therefore, for large, we have m m 0 3µ v +2v I [ w >v 4 ] + I [λmax >K ]). max v x S +3µ m z) m 0 K µ max w v 5 x S +2v 2 ) max I[ w >v x S 4 ] + I [λmax >K ]. Therefore, from 3.) ad 3.8) we fid for ay positive ɛ ad l Pv for all p sufficietly large, wheever δ /7. max m z) m 0 x S >ɛ) K p ɛ p l 3.22) 4
16 We ow assume the elemets of S to be equally spaced betwee ad. Sice for x x 2 2 /2, ad whe x,for large m x + iv ) m x 2 + iv ) 2 /2 v 2 m 0 x + iv ) m 0 x 2 + iv ) 2 /2 v 2, m x + iv ) 2 /2 + v I [λmax >K ] m 0 x + iv ) 2 /2 we coclude from 3.22) ad 3.), that for ay positive ɛ ad l Pv m x + iv ) m 0 x + iv ) >ɛ) K p ɛ p l 3.23) x R for all sufficietly large p, wheever δ /7. Let 0 ) deote expectatio ad k ) deote coditioal expectatio with respect to the σfield geerated by r,,r k. Let l, l > 0 be arbitrary. Choose l >l,letp be suitably large so that 3.23) holds with l replaced by l, adr lp/ll ) is greater tha. Sice k v l x R m x + iv ) m 0 x + iv ) l ), k =0,...,N forms a martigale, it follows from Jese s iequality, Lemmas 2.5, 2.6 ad 3.23) that for ay positive ɛ ɛ r v rl Pmax kv l m x + iv ) m 0 x + iv ) l ) >ɛ) k N x R x R m x + iv ) m 0 x + iv ) rl ) ɛ r K l/l p wheever δ /7. I particular, we have for δ /7 lim max k x R m x + iv ) mx 0 + iv ) 2 ) k N v 2 Let λ λ 2 λ N be the eigevalues of B ad write m j = m out j + m i j j =, 2 l l l l = 0 with probability. 3.24). where m out 2 x + iv) = N λ j [a,b ] v x λ j ) 2 + v 2, 5
17 m out x + iv) = N λ j [a,b ] Defie the sequece {G m } m= of fuctios o R 2 by x λ j x λ j ) 2 + v 2. G Nj)+)+kx,x 2 )= k F B x )F B x2 ), for k =0,,...,N). Clearly each G m is a probability distributio fuctio o R 2,ad whe m = Nj)+)+k, the twodimesioal Stieltjes trasform, mg) m x +iv,x 2 + iv 2 )ofg m is k m x + iv )m x 2 + iv 2 ). Obviously, whe δ = 0, 3.24) implies that, with probability oe x,x 2 R m m G) x + iv,x 2 + iv 2 ) m 0 x + iv )m 0 x 2 + iv 2 ) 0 as m for coutably may v,v 2 ) formig a dese subset of a ope set i the first quadrat bouded uiformly away from the two axes). We coclude that with probability oe, G m x,x 2 ) coverges weakly to F c,h x )F c,h x 2 ). Sice the itegrads of [a,b ] c [a, b ] c d k F Bx )F Bx 2 ) x x ) 2 + v 2 )x x 2 ) 2 + v 2 ) ad d k F Bx ) [a,b ] x x c ) 2 + v 2 o their respective domais are uiformly bouded ad equicotiuous for x [a, b], it follows as i 3.3) that max k m i 2 x + iv) k d v dx m0 x) for ay v = v 0. Therefore, from 3.24) ad 3.25) we have max k N 2 a.s. 0, 3.25) v 2 k m out 2 x + iv ) ) 2 a.s ) From 3.26) we ca ifer a boud o the umber of eigevalues i [a, b]. Notice NF B A) is the umber of eigevalues of B i the set A. Lete deote the left side of 3.26). For ay x [a, b] e N 2 max k k N max k N λ j [a,b] [x v,x+v ] x λ j ) 2 + v 2 N 2 k F B {[a, b] [x v,x+ v ]} ) 2 4vN 4 2, 6 2
18 ad sice the umber of itervals of legth 2v eeded to cover [a, b] is b a)/2v,we fid k F B {[a, b]} ) 2 b a) 2 ve 2. Therefore max ) k F B {[a, 2 b]} = oa.s. v)=o 2 a.s. N 2/7 ), k N which implies max kf B {[a, b]}) =oa.s. v )=o a.s. N /7 ). k N The above argumets apply to [a,b ] as well, so we also have ad max k F B {[a,b ]} ) 2 = oa.s. v)=o 2 a.s. N 2/7 ) 3.27) k N max k N kf B {[a,b ]}) =o a.s. v )=o a.s. N /7 ). 3.28) 4. Covergece of m m We ow restrict δ =/68, that is, v = v = N /68. Our goal is to show that Nv m m 0 a.s. as 4.) Write D = B zi, D j = D r j rj,add jj = D r j rj + r jrj ) j j. The m = tr D ). Let us also deote α j = rj D 2 j r j N tr D 2 j T ), a j = N tr D 2 j T ), β j = γ j = rj D j r j N tr D, b +rj = D j r j +N tr T D ), j T )), ˆγ j = rj D j r j N tr D j T ). We first derive bouds o momets of γ j ad ˆγ j. Usig 3.2) we fid for all p 2 ˆγ j p K p N p tr T /2 D j T D Usig Lemmas 2.2 ad 2.0 we have for p 2 γ j ˆγ j p = γ ˆγ p = N j T /2 ) p/2 K p N p/2 v p. 4.2) j tr T D j tr T D j=2 7 p
19 = N = N j=2 j j ) r j D j T D +rj D j r j j=2 p j tr T D D j ) j tr T D D j ) j r j p N K p N p j j ) r j D j T D j=2 j r j +rj D j r j 2) p/2 K p N p/2 v p. Therefore γ j p K p N p/2 v p. 4.3) We ext prove that b is bouded for all. We have b ad β both bouded i absolute value by z /v see 3.4)). From the equatio relatig m to the β j s above 3.4)) we have β = zm. Usig 3.24) we get m z)) m 0 z) = ov ). Sice m 0 is bouded for all, x [a, b] adv we have β K. Sice b = β + β b γ we get b = β + β b γ K + K /2 2 v 3 N /2 K. Sice m x + iv ) m x 2 + iv ) x x 2 v 2, we see that 4.) will follow from max x S Nv m m 0 a.s. where S ow cotais 2 elemets, equally spaced i [a, b]. We write m m = j tr D j tr D + = = r j D 2 ) j r j [ j j ] +rj D j r j j j ) j j ) r j D 2 j rj D 2 j r j +N tr T D j r j N tr T D j + N tr T D j ) 2 8 r j D j r j )
20 + j j ) r j D 2 j r j N tr T D j + N tr T D j rj D j r j ) 2 ) 2 + rj D j r j ) = b j α j b2 j a j ˆγ j b2 j j )α j γ j rj D 2 j r j β j γj 2 ) W W 2 W 3. Let F j be the spectral distributio of the matrix k j r krk. From Lemma 2.2 ad 3.27) we get max j F j [a,b ])) 2 = on 2/7 )=ov) 8 a.s. 4.4) j Defie B j = I [j F j [a,b ]) v 4 ] [ j F j [a,b ])) 2 v 8 ]. The B j = I [j F j [a,b ]) v 4 ] [ j F j [a,b ])) 2 v 8 ] a.s. ad we have Therefore we have for ay ε>0, [ P max v x S N ) P [B j =0], i.o. =0. ) P max Nv W > ε, i.o. x S ] N j α j ) >ε P max v x S ) N [B j =] ) j α j )B j >ε, i.o., ) ) [B j =0], i.o. where ε = if ε/nb ) > 0 sice b is bouded. Note that for each x R { j α j )B j } forms a martigale differece sequece. By Lemma 2. ad 3.2), we have for each x [a, b] adp 2 v p j α j )B j N K p j v j α j )B j 2 ) p/2 + 9 ) v j α j )B j p
21 N K p N K p v N p p j v 2 N 2 B j tr T /2 ) p/2 B j j tr D 2 j D 2 j ) + Nv p ) p/2 ) D 2 j T D 2 j T /2 ) + Nv α p p ) N p tr T /2 D 2 T D 2 T /2 )) p/2 N ) p/2 K p v N p p B j j tr D 2 j D 2 j ) + v p N ). p/2 Let λ kj deote the k th smallest eigevalue of k j r kr k.wehave B j j tr D 2 j D 2 j = [ B j j λ kj / [a b ] λ kj x) 2 + v 2 ) 2 + λ kj [a b ] ] λ kj x) 2 + v) 2 2 ɛ 4 + B j v 4 j F j [a,b ])) KN 2. Therefore P max v x S ) j α j )B j >ε 2 K p,ε N p/68 which is summable whe p>204. Therefore, max x S W = o/n v ) a.s. Provig max W 2 = o/n v ) a.s. 4.5) x S is hadled the same way. We get usig Lemma 2., 3.2), ad the fact that a j N v 2 v p j a j ˆγ j )B j N K p v N p p N K p j v j a j ˆγ j )B j 2 ) p/2 + ) p/2 B j j a j 2 tr D j D j ) + N p v p ) v j a j ˆγ j )B j p ) tr D j D j ) p/2 N ) p/2 K p v N p p B j j a j 2 tr D j D j ) + v 2p N ). p/2 20
22 This time B j j a j 2 tr D j D j ) B j j N 2 k λ kj x) 2 + v) 2 2 k λ kj x) 2 + v 2 B j N 2 j ɛ 4 + v 4 F j [a, b]))ɛ 2 + v 2 F j [a, b])) KN 2, so that 4.5) also holds. Usig Lemmas 2.2, 2.0, 3.2), ad 4.3) we get v j j )α j γ j r j D 2 j r j β j γj 2 ) p K p v p N p/2 α γ p + v p γ 2p ) K p vn p p/2 N p tr D 2 D 2 ) p ) /2 N p/2 v p + v 3p N p ) K p vn p p/2 N p N p/2 v 2p N p/2 v p + v 3p N p )=2K p N p/2 v 2p. Thus we get max x S W 3 = o/n v ) a.s. ad, cosequetly, 4.). 2
23 5. Covergece of expected value Our ext goal is to show that, for v = N /68. m m 0 = O/N ). 5.) We begi by derivig a idetity similar to 3.5). Write B zi zm z)t zi) = N r jr j zm z))t. Takig first iverses ad the expected value we get = z N zm T zi) B zi) [ N ] = zm T zi) r j rj zm z))t B zi) β j [ m z)t +I) r j r j B j) zi) N m z)t +I) T B zi) ] = z Nβ [ m z)t + I) r r D N m z)t + I) T D ]. Takig the trace o both sides ad dividig by N/z, weget dh t) c + zc m z)) +tm We first show = β [ r D m T + I) r N tr m T + I) T D ]. 5.2) N tr m T + I) T D tr m T + I) T D From 4.4) we get = ON ). 5.3) tr D D ) 2 ɛ 2 + v 2 F [a,b ])) 2 KN 2 5.4) ad tr D 2 D 2 ɛ 4 + v 4 F [a,b ])) KN. 5.5) Also, because of 3.24) ad the fact that /m 0 z) stays uiformly away from the eigevalues of T for all x [a, b], we must have m z)t + I) K 5.6) 22
24 Therefore, from 3.3), 4.3), 5.4) 5.6), ad the fact that b is bouded, we get left side of 5.3) = N β rd m T + I) T D r N b rd m T + I) T D r + β b γ rd m T + I) T D r ) KN KN N tr T /2 N tr D D D m T + I) T D T /2 + v γ 2 ) /2 rd m T + I) T D r 2 ) /2 ) + v Thus 5.3) holds. From 3.2), 5.4), ad 5.6) we get N /2 v N tr D 2 D 2 + tr D D ) 2 ) /2 ) KN. rd m T + I) r N tr D m T + I) T 2 Next we show KN 2 tr D D KN 5.7) N 2 tr m T + I) T D tr m T + I) T D 2 KN. 5.8) Let β j = +r j D j r j, b = +N tr T D2 ), ad γ j = rj D j r j N tr D j T )). As i the previous sectio, both β j ad b are bouded i absolute value by z /v ad γ j satisfies the same boud as i 4.3). Moreover, if we let X ) deote X without its first colum, the oe ca easily derive N tr N X ) T X ) N N z)i) = N tr N X ) T X ) zi) = zn ) ad coclude that β j ad cosequetly b are bouded. 23 β j, j=2
25 It is also clear that the bouds i 4.4), 5.4), ad 5.5) hold whe two colums of X are removed. Moreover, with F 2 deotig the e.d.f of j,2 r jrj we get tr D2 D 2 ) 4 ɛ 2 + v 2 F 2 [a,b ])) 4 KN 4 ɛ 8 + v 8 F 2 [a,b ])) 2 ) KN 4 ad tr D2 2 D 2 2 ) 2 ɛ 4 + v 4 F 2 [a,b ])) 2 KN 2. With these facts ad 3.3) ad 5.6) we have left side of 5.8) = 2N 2 N 2 j=2 N j=2 j j )tr m T + I) T D 2 β j r j D j m T + I) T D j r j 2 =2N b β 2 b γ 2 )r2d 2 m T + I) T D2 r 2 2 KN r2d 2 m T + I) T D2 r 2 2 +v 2 γ 2 4 r2d 2 m T + I) T D2 r 2 4 ) /2 ) KN 3 tr D2 2 D 2 2 ) +tr D2 D 2 ) 2 + v 2 N v 2 tr D2 2 D 2 2 ) 2 + tr D2 D 2 ) 4 ) /2 ) KN 3 N 2 + Nv 4 ) KN. Thus we get 5.8) Notice we get the same result if m T + I) is removed from all the expressios, that is, we have just show Moreover, from 3.2) ad 5.4), whe p =2 γ ˆγ 2 KN. ˆγ 2 KN 2 tr D D KN. 24
26 Therefore γ 2 KN. 5.9) From 4.3), 5.2), 5.3), 5.7) 5.9) we get KN + c dh t) + zc m z)) +tm [ β rd m T + I) r ] N tr m T + I) T D [ = KN + b 2 γ β γ) 2 rd m T +I) r ] N tr m T +I) T D KN + γ 2 + v 2 γ 4 ) /2 N /2 ) KN +N + v 2 N 2 v 4 ) /2 N /2 ) KN. As i Sectio 3 we let w = z dh t) +tm z) m z)) ad ω = z + c m tdh t) +tm. The w KN, ω = w zc /m, ad equatio 3.20) together with the steps leadig to 3.2) hold with m replaced with its expected value. From 3.0) it is clear that m 0 must be uiformly bouded away from 0 for all x [a, b] ad all. From 3.24) we see that m must also satisfy this same property. Therefore ω KN. ad Usig 3.), 3.2), 3.4), ad 3.5) it follows that m 0 is bouded i m 0 2c t 2 dh t) v + m 0 2 c +tm 0 2 t 2 dh t) +tm
27 is bouded away from for all. Therefore, we get for all sufficietly large which is 5.). m m Kc 0 zm 0 w KN, 6. Completig the proof From the last two sectios we get m z) mz) 0 = o/n v ) a.s. 6.) whe v = N /68. It is clear from the argumets used i Sectios 3 5 that 6.) is true whe the imagiary part of z is replaced by a costat multiple of v.ifactwehave max k {,2...,34} m x + i kv ) m 0 x + i kv ) = o/n v )=ov 67 ) a.s. We take the imagiary part ad get max k {,2...,34} df B λ) F c,h λ)) x λ) 2 + kv 2 = ov66 ) a.s. Upo takig differeces we fid max k k 2 v 2 df B λ) F c,h λ)) x λ) 2 + k v)x 2 λ) 2 + k 2 v) 2 = ov66 ) a.s. max k,k 2,k 3 distict v) 2 2 df B λ) F c,h λ)) x λ) 2 + k v)x 2 λ) 2 + k 2 v)x 2 λ) 2 + k 3 v) 2 = ov66 ) a.s.. v) 2 33 df B λ) F c,h λ)) x λ) 2 + v)x 2 λ) 2 +2v) 2 x λ) 2 +34v) 2 = ov66 ) a.s. Thus df B λ) F c,h λ)) x λ) 2 + v)x 2 λ) 2 +2v) 2 x λ) 2 +34v) 2 = o) a.s. 26
28 We split up the itegral ad get I [a,b ] c df B λ) F c,h λ)) x λ) 2 + v)x 2 λ) 2 +2v) 2 x λ) 2 +34v) 2 + v 68 x λ j ) 2 + v)x 2 λ j ) 2 +2v) 2 x λ j ) 2 +34v) 2 = o) a.s. 6.2) λ j [a,b ] Now if, for each term i a subsequece satisfyig 6.2), there is at least oe eigevalue cotaied i [a, b], the the sum i 6.2), with x evaluated at these eigevalues, will be uiformly bouded away from 0. Thus, at these same x values, the itegral i 6.2) must also stay uiformly bouded away from 0. But the itegral coverges to zero a.s. sice the itegrad is bouded ad with probability oe, both F B ad F c,h coverge weakly to the same limit havig o mass o {a,b }. Thus, with probability oe, o eigevalues of B will appear i [a, b] for all sufficietly large. This completes the proof of Theorem.. APPNDIX We give here a proof Lemma 2.7. We first prove Lemma A. For X =X,...,X ) T i.i.d. stadardized complex) etries, B Hermitia oegative defiite matrix we have for ay p X BX p K p tr B) p + X 2p tr B ). p A.) Proof. Notice the result is trivially true for p =. Forp>wehave p p p) X BX p K p X i 2 B ii + X i X j B ij + X j X i B ij i= i= Usig Lemmas 2.3 ad 2.9 p = K p X i 2 B ii +2 X i X j B ij p). A.2) p X i 2 B ii K p tr B) p + i= j=2 ) X 2p B ii ) p i= K p tr B) p + X 2p tr B p ). i<j 27
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