Random truncations of Haar distributed matrices and bridges
aa r X i v : . [ m a t h . P R ] F e b RANDOM TRUNCATIONS OF HAAR DISTRIBUTEDMATRICES AND BRIDGES
C. DONATI-MARTIN AND A. ROUAULT
Abstract.
Let U be a Haar distributed matrix in U ( n ) or O ( n ). Ina previous paper, we proved that after centering, the two-parameterprocess T ( n ) ( s, t ) = X i ≤⌊ ns ⌋ ,j ≤⌊ nt ⌋ | U ij | converges in distribution to the bivariate tied-down Brownian bridge.In the present paper, we replace the deterministic truncation of U bya random one, where each row (resp. column) is chosen with prob-ability s (resp. t ) independently. We prove that the correspondingtwo-parameter process, after centering and normalization by n − / con-verges to a Gaussian process. On the way we meet other interestingconvergences. Introduction
Let us consider a unitary matrix U of size n × n . We fix two integers p < n and q < n and delete deterministically n − p rows and n − q columns.Let us call U p,q the (rectangular) matrix so obtained. It is well knownthat if U is Haar distributed in U ( n ), the random matrix U p,q ( U p,q ) ∗ hasa Jacobi matricial distribution and that if ( p/n, q/n ) → ( s, t ) ∈ (0 , , itsempirical spectral distribution converges to a limit D s,t (see for instance [5]),often called the generalized Kesten-McKay distribution. It is clear from theinvariance of the Haar distribution on U ( n ) that we can keep the first p rowsand the first q columns.In [10] we studied the trace of U p,q ( U p,q ) ∗ . It is the squared of the Frobe-nius (or Euclidean) norm of U p,q . Actually we set p = ⌊ ns ⌋ , q = ⌊ nt ⌋ andconsidered the process indexed by s, t ∈ [0 , n → ∞ to a bivariate tied-down Brownian bridge. Previously, Chapuy [4]proved a similar result for permutation matrices, with a n − / normalization.Besides, for purposes of random geometry analysis, in [13] (see also [12])B. Farrell deletes randomly and independently a proportion 1 − s of rows anda proportion 1 − t of columns. Let us call U s,t the matrix so obtained. Farrell Date : October 10, 2018.2010
Mathematics Subject Classification.
Key words and phrases.
Random Matrices, unitary ensemble, orthogonal ensemble,bivariate Brownian bridge, invariance principle. proved that (for fixed s, t ) the empirical spectral distribution of U s,t (cid:0) U s,t (cid:1) ∗ has an explicit limiting distribution which is precisely D s,t . Actually, Farrellconsidered first the (deterministic) DFT matrix F jk = 1 √ n e − π ( j − k − /n , j, k = 1 , . . . , n , and proved that a Haar unitary matrix has the same behaviour.It is then tempting to study the same statistic as above, when using thisrandom truncation. Of course, instead of the DFT, we can as well considerany (random or not) matrix whose all elements are of modulus n − / , forinstance a (normalized) complex Hadamard matrix.To define all random truncations simultaneously and get a two-parameterprocess, we define a double array of n auxiliary uniform variables, in orderto generate Bernoulli variables. We will prove below that after centering,we need a normalization to get a Gaussian limiting process.We use the Skorokhod space D ([0 , ). It consists of functions from[0 , to R which are at each point right continuous (with respect to thenatural partial order of [0 , ) and admit limits in all ”orthants”. Thespace D ([0 , ) is endowed with the topology of Skorokhod (see [3] for thedefinition).For the sake of completeness, we treat also the one-parameter process,i.e. truncation of the first column of the unitary matrix, and the case ofpermutation matrices.In Farell [13] is also mentioned another way of truncation, used in [18],consisting in drawing only one array of n Bernoulli variables to determinethe choice of rows and columns. We did not consider this model here, tomake the paper shorter.The rest of the paper is organized as follows. In Sec. 2 we introduce thebasic notations with the main statistics and the list of limiting processes.In Sec. 3 we present the statements of convergence in distribution for allmodels. In particuler Theorem 3.3 says that at the first order, the differencebetween DFT, Haar unitary or Haar orthogonal random matrices is notseen by the statistics T ( n ) at the limit since only the randomness of thetruncation contributes. Besides, Proposition 3.5 says that this difference isseen at the second order, by means of the process Z ( n ) . Sec. 4 is devotedto the proofs. The most delicate part is the proof of Prop. 3.5 (2). We firstprove the convergence of finite dimensional distributions after replacementof the indicator variables by Gaussian ones (Lindeberg’s strategy). Then weprove tightness of the process rescaled by n − / by application of Davydovand Zitikis’s criterion [8]. It could be noted that before that, we triedseveral tightness critera for the unscaled (two-parameter) process, such asBickel and Wichura [3] or Ivanoff [14], but they failed . In Sec. 6, we gather We hope to address a definite answer to the question of weak convergence of Z ( n ) ina forthcoming paper. ANDOM TRUNCATION AND BRIDGES 3 all the estimates of moments; i.e of polynomial integrals on U ( n ) and O ( n ).Finally in Sec. 7, we give the proof of the Lindeberg’s strategy.2. Notations
We introduce the random processes that we will consider in this paperand the various limiting processes involved.2.1.
The main statistics.
Let U ∈ U ( n ) be the unitary group of size n and let U i,j be the generic element of U . Let now R i , i = 1 , · · · , n and C j , j = 1 , · · · , n be two independent families of independent random vari-ables uniformly distributed on [0 , B ( n )0 = ⌊ ns ⌋ X (cid:0) | U i | − /n (cid:1) , s ∈ [0 , , (2.1) B ( n )0 = n X | U i | (1 R i ≤ s − s ) , s ∈ [0 , ! . (2.2)For the two parameters model, we introduce the same framework. Let T ( n ) s,t = X i ≤⌊ ns ⌋ ,j ≤⌊ nt ⌋ | U ij | , (2.3)and T ( n ) = (cid:16) T ( n ) s,t , s, t ∈ [0 , (cid:17) . Let now T ( n ) s,t := n X i,j =1 | U ij | R i ≤ s C j ≤ t . (2.4)We have clearly, since | U ij | and ( R i , C j ) are independent, T ( n ) s,t − E T ( n ) s,t = n X i,j =1 | U ij | (cid:2) R i ≤ s C j ≤ t − st (cid:3) . (2.5)Introducing the centering of Bernoulli variables, and using the fact that thematrix U is unitary, we may write T ( n ) st − E T ( n ) st = Z ( n ) st + n / W ( n ) s,t (2.6)where Z ( n ) s,t := n X i,j =1 | U ij | [ R i ≤ s − s ] (cid:2) C j ≤ t − t (cid:3) (2.7) W ( n ) s,t := sF ( n ) t + tG ( n ) s (2.8) C. DONATI-MARTIN AND A. ROUAULT with the empirical processes F ( n ) t := n − / n X j =1 [ C j ≤ t − t ] ; G ( n ) s := n − / n X i =1 [ R i ≤ s − s ] . (2.9)In the sequel, we denote, T ( n ) = (cid:16) T ( n ) s,t , s, t ∈ [0 , (cid:17) , Z ( n ) = (cid:16) Z ( n ) s,t , s, t ∈ [0 , (cid:17) ,F ( n ) = (cid:16) F ( n ) t , t ∈ [0 , (cid:17) , G ( n ) = (cid:16) G ( n ) t , t ∈ [0 , (cid:17) , and W ( n ) = (cid:16) W ( n ) s,t , s, t ∈ [0 , (cid:17) . Gaussian processes and bridges.
The classical Brownian bridgedenoted by B is a centered Gaussian process with continuous paths definedon [0 ,
1] of covariance E (cid:0) B ( s ) B ( s ′ ) (cid:1) = s ∧ s ′ − ss ′ . The bivariate Brownian bridge denoted by B , is a centered Gaussian pro-cess with continuous paths defined on [0 , of covariance E (cid:0) B , ( s, t ) B , ( s ′ , t ′ )) (cid:1) = ( s ∧ s ′ )( t ∧ t ′ ) − ss ′ tt ′ . The tied-down bivariate Brownian bridge denoted by W ( ∞ ) is a centeredGaussian process with continuous paths defined on [0 , of covariance E [ W ( ∞ ) ( s, t ) W ( ∞ ) ( s ′ , t ′ )] = ( s ∧ s ′ − ss ′ )( t ∧ t ′ − tt ′ ) . Let also W ( ∞ ) be the centered Gaussian process with continuous paths de-fined on [0 , of covariance E [ W ( ∞ ) ( s, t ) W ( ∞ ) ( s ′ , t ′ )] = ss ′ ( t ∧ t ′ ) + ( s ∧ s ′ ) tt ′ − ss ′ tt ′ . It can be defined also as W ( ∞ ) ( s, t ) = sB ( t ) + tB ′ ( s )where B and B ′ two independent one parameter Brownian bridges.At last we will meet the process denoted by B ⊗ B which is a centeredprocess with continuous paths defined on [0 , by B ⊗ B ( s, t ) = B ( s ) B ′ ( t )where B and B ′ are two independent Brownian bridges. This process isnot Gaussian, but it has the same covariance as W ( ∞ ) . ANDOM TRUNCATION AND BRIDGES 5 Convergence in distribution
We present unfied results in the cases of unitary and orthogonal groups.For this purpose we use the classical notation β ′ = β ( / , law −→ means conver-gence in distribution in D ([0 , Theorem 3.1.
We have : √ nB ( n )0 law −→ p β ′− B , (3.1) √ n B ( n )0 law −→ p β ′− B . (3.2)(3.1) is well known since at least Silverstein [17] (in the case β ′ = 1).Both results are a direct consequence of the fact that the vector ( | U i | , i =1 , . . . , n ) follows the Dirichlet ( β ′ , . . . , β ′ ) distribution on the simplex (see adetailed proof in Section 4).Let us continue with the two-parameters processes, where now law −→ meansconvergence in distribution in D ([0 , ), and fidi −→ means convergence in dis-tribution of finite dimensional marginals. The different normalizations areexplained in Remark 3.7.We begin with a recall of the convergence in the deterministic model oftruncation which was proved in Theorem 1.1 of [10]. Theorem 3.2.
Under the Haar measure on U ( n ) or O ( n ) , (cid:16) T ( n ) − E T ( n ) (cid:17) law −→ p β ′− W ( ∞ ) . In the sequel we will prove the following convergences for the second modelof truncation.
Theorem 3.3. (1)
Under the DFT model, or more generally when | U ij | =1 /n a.s. for every i, j , n − / (cid:16) T ( n ) − E T ( n ) (cid:17) law −→ W ( ∞ ) . (2) Under the Haar measure on U ( n ) or O ( n ) , n − / (cid:16) T ( n ) − E T ( n ) (cid:17) law −→ W ( ∞ ) . The proof uses the decomposition (2.6): T ( n ) − E T ( n ) = n / W ( n ) + Z ( n ) . The following trivial lemma (consequence of the convergence of empiricalprocesses F ( n ) and G ( n ) ) rules the behavior of W ( n ) and the following Propo-sition 3.5 provides a convergence in distribution of Z ( n ) hence a convergencein probability to 0 for n − / Z ( n ) . C. DONATI-MARTIN AND A. ROUAULT
Lemma 3.4. W ( n ) law −→ W ∞ . (3.3) Proposition 3.5. (1)
Under the DFT model, or more generally if | U ij | =1 /n a.s. for every i, j , Z ( n ) law −→ B ⊗ B . (2) Under the Haar model, Z ( n ) fidi −→ p β ′− W ( ∞ ) + B ⊗ B , (3.4) where B ⊗ B and W ( ∞ ) are independent, and n − / Z ( n ) P −→ . (3.5)Since our studies have for origin the article of Chapuy [4] which proved(3.6) on random permutations, we give now the complete behavior of theabove statistics in this case. Theorem 3.6.
Assume that U is a random permutation of [ n ] . We have n − / W ( n ) law −→ W ( ∞ ) (3.6) n − / (cid:16) T ( n ) − E T ( n ) (cid:17) law −→ B , (3.7) n − / Z ( n ) law −→ W ( ∞ ) . (3.8) Remark 3.7.
Let M p,q = U p,q ( U p,q ) ∗ and M s,t = U s,t (cid:0) U s,t (cid:1) ∗ . For s, t fixed, the random variables T ( n ) s,t and T ( n ) s,t are linear functionals of the em-pirical spectral distribution of M ⌊ ns ⌋ , ⌊ nt ⌋ and M s,t respectively. For classicalmodels in Random Matrix Theory, the convergence of fluctuations of suchlinear functionals do not need a normalizing factor, since the variance isbounded (the eigenvalues are repelling each other). Here, this is indeed thecase for T ( n ) s,t (see [11] for the complete behavior for general tests functions).But, in the case of T ( n ) s,t , we have Var E [ T ( n ) s,t | R , . . . , R n , L , . . . , L n ] = O ( n ) ,which demands a normalization. Remark 3.8.
Going back to the decomposition (2.6), gathering the conver-gences in (3.3) (3.7) and (3.8) and the covariances, we recover the identity B , = W ∞ + W ∞ where the two processes in the RHS are independent, fact which was quotedin [9] section 2. Proofs
Proof of Theorem 3.1.
It is known that the vector ( | U i | , i = 1 , . . . , n )is distributed uniformly on the simplex. Actually, the result is a consequenceof the following proposition. ANDOM TRUNCATION AND BRIDGES 7
Proposition 4.1.
Let ( u i , i = 1 , . . . , n ) Dirichlet ( β ′ , . . . , β ′ ) distributed. (1) The process n / { P ⌊ ns ⌋ i =1 ( u i − /n ) , s ∈ [0 , } converges in distri-bution to p β ′− B . (2) The process n / { P n u i (1 R i ≤ s − s ) , s ∈ [0 , } converges in distri-bution to p β ′− B .Proof. It is easy to see that( u , · · · , u n ) law = (cid:18) g g + · · · + g n , · · · , g n g + · · · + g n (cid:19) (4.1)with g i independent and gamma distributed with parameter β ′ .(1) The above representation yields the following equality in law betweenprocesses n / { ⌊ ns ⌋ X i =1 ( u i − /n ) law = n S n (cid:18) S ⌊ ns ⌋ − β ′ ⌊ ns ⌋√ n − ⌊ ns ⌋ n S n − nβ ′ √ n (cid:19) where S k = P ki =1 g i . The WLLN and the Donsker’s theorem gives theconvergence to p β ′− ( B ( s ) − sB (1)) where B is the standard Brownianmotion.(2) The process n / { P n u i (1 R i ≤ s − s ) , s ∈ [0 , } is an example of aweighted empirical process. We may apply the Theorem 1.1 of Koul andOssiander [15] p.544. Set v ni = n u i be the weights. Under the followingconditions on the weights:(1) (cid:0) n − P i v ni (cid:1) / −→ γ (2) n − / max i v ni P −→ γ by an independent Brownianbridge. For the first one, we start from the representation (4.1), apply theSLLN twice and get n − X i v ni ! / −→ (cid:0) E g (cid:1) / E g = p β ′− . For the second one, we have: n − / max i v ni = n / max { g i , i = 1 , . . . , n } g + · · · + g n . Appealing again to the SLLN, it is enough to prove that max { g i , i = 1 , . . . , n } = o ( n / ) in probability, which is clear (this maximum is of order log n ). (cid:3) Proof of Theorem 3.6.
We notice that the double sums in the def-initions of T ( n ) or Z ( n ) are actually single sums where j = π ( i ) for π arandom permutation. Moreover, by exchangeability of the variables C j ,these single sums have (as processes in s, t ) the same distributions as if π was the identity permutation. In this case, we may apply the results of [16], C. DONATI-MARTIN AND A. ROUAULT since n − / [ T ( n ) − E T ( n ) ] and n − / Z ( n ) are precisely X Fn and Y n therein,respectively.4.3. Proof of Proposition 3.5.
Under the DFT model, we have Z ( n ) = F ( n ) ⊗ G ( n ) , so we have only to study the Haar model, in U ( n ) and in O ( n ).As long as possible, we keep the notation U ij for the generic element of U in both cases, but we use the notation O ij when we need to stress thatthe underlying unitary matrix is sampled according to the Haar measure in O ( n ).First, we prove (3.4), i.e. the convergence of finite dimensional distribu-tions, beginning with the 1-dimensional marginals since the notations areeasier to follow for the reader, and then using the Cramer-Wold device. Ina last part, we prove the tightness, using a criterion due to Davydov andZitikis [8].4.3.1. We first consider the convergence of the 1-marginals, i.e.for fixed s, t ∈ [0 , Z ( n ) ( s, t ) to p β ′− W ( ∞ ) ( s, t )+ B ⊗ B ( s, t ). By a scaling property, this is equivalent to prove the weakconvergence of Z ( n ) ( s, t ) p s (1 − s ) t (1 − t ) towards p β ′− N + N N where N i areindependent standard Gaussian variables. According to the next Proposi-tion, whose proof is given in the Appendix, we can replace the independentcentered random variables R i := R i ≤ s − s p s (1 − s ) , C j := C j ≤ t − t p t (1 − t ) , i, j = 1 , . . . , n by independent N (0 ,
1) random variables X i and Y j . Proposition 4.2.
Let ( R i ) and ( C j ) two independent sequences of iid cen-tered variables, with variance 1, and finite third moment. Consider also ( X i ) and ( Y j ) two independent sequences of iid standard Gaussian variables. Wedefine A n = P ni,j =1 | U ij | R i C j and B n = P ni,j =1 | U ij | X i Y j . Then, A n and B n have the same limiting distribution. We thus study the bilinear non symmetric formΣ n := n X i,j =1 X i | U ij | Y j built from the non symmetric matrix ˜ U = (cid:0) | U ij | (cid:1) i,j ≤ n . This matrix is aso-called ”doubly stochastic” matrix ([2], [7]). As a Markovian matrix, ithas 1 as dominant eigenvalue. Proposition 4.3.
The sequence of random variables Σ n := n X i,j =1 X i | U ij | Y j ANDOM TRUNCATION AND BRIDGES 9 converges in distribution towards p β ′− N + N N where N , N , N areindependent and N (0 , distributed.Proof. We know that E | U ij | = n − . We may remark that E [Σ n | X, Y ] = n − n X i,j =1 X i Y j = (cid:18) X + · · · + X n √ n (cid:19) (cid:18) Y + · · · + Y n √ n (cid:19) =: S ′ n and that S ′ n law = N N . Set S n := Σ n − S ′ n = n X i,j =1 X i V ij Y j with V ij = | U ij | − n − . We will use the characteristic function and conditioning. Set (
X, V ) :=( X i , i ≤ n, V ij , i, j ≤ n ). Conditionnally upon ( X, V ), the vector ( S n , S ′ n ) isGaussian andcov( S n , S ′ n | X, V ) = n − n X i,j,k,ℓ =1 X i X k V kℓ cov( Y j , Y ℓ )= n − n X i,j,k =1 X i X k V kj = n − n X i,k =1 X i X k n X j =1 V kj = 0since U is unitary. The variables S n and S ′ n are then independent, condi-tionnally upon ( X, V ). Moreover E [exp i θS ′ n | ( X, V )] = E [exp i θS ′ n | X ] = exp − θ P ni =1 X i ) n (4.2)and E [exp i θS n | ( X, V )] = exp − θ b S n (4.3)where b S n := n X j =1 n X i =1 V ij X i ! = n X i,j =1 X i X j H ij , with H ij = ( V V ∗ ) ij , i, j = 1 , . . . , n . This implies E exp i θ Σ n = E exp − θ n − ( n X i =1 X i ) + b S n ! (4.4) Since ( P X i ) /n is distributed as N (where N is N (0 , b S n := n X j =1 n X i =1 V ij X i ! −→ β ′− , (4.5)we will conclude that E exp i θ Σ n → E exp − θ N + β ′− )and this limit is the characteristic function of N N + p β ′− N .It is clear that (4.5) is implied bylim n E b S n = β ′− (4.6)lim n E ( b S n ) = β ′− . (4.7)To prove these assertions, we will need some joint moments of elementsof the matrix H , which are themselves affine functions of moments of thematrix U .The first limit in (4.6) is easy to obtain since E b S n = X i E H ii = n E H = n E ( V ) = n Var V = ( n Var | O | = n +3) n +2 in the orthogonal case ,n Var | U | = n − n +1 in the unitary case, (4.8)where we refer to (5.1).To prove (4.7) we expand b S n : b S n = n X i,j,k,ℓ =1 X i X j X k X ℓ H ij H kℓ . In the expectation, the only non vanishing terms are obtained according tothe decompostion E b S n = X i = j,k = ℓ = i + X i = k,j = ℓ = i + X i = ℓ,j = k = i + X i = j = k = ℓ (4.9)= (1) + (2) + (3) + (4) . (4.10)Since ( E X ) = 1 and E X = 3 we have successively(1) = n ( n − E ( H H ) (4.11)(2) = (3) = n ( n − E ( H ) (4.12)(4) = 3 n E H . (4.13)Now E H = n E V + n ( n − E (cid:0) V V (cid:1) (4.14) E ( H H ) = n E (cid:0) V V (cid:1) + n ( n − E (cid:0) V V (cid:1) (4.15) E ( H ) = n E (cid:0) V V (cid:1) + n ( n − E ( V V V V ) . (4.16) ANDOM TRUNCATION AND BRIDGES 11 this last equality coming from the symmetry of H .From Lemma 5.1 (see (5.7) with k = 4), E b S n = n ( n − E (cid:0) V V (cid:1) + 2 n ( n − E ( V V V V ) + o (1) . (4.17)and from Lemma 5.3, we conclude that (4.7) holds, which ends the proof of4.3. (cid:3) Finite-dimensional marginals.
We now consider the convergence ofthe finite dimensional distributions, following the same scheme of proof as inthe case of the 1-dimensional marginal. The Lindeberg’s strategy statementis now:
Proposition 4.4.
Let ( β i ( s )) and ( γ i ( s )) be two independent sequences ofindependent Brownian bridges. Define G n ( s, t ) = P ni,j =1 | U ij | β i ( s ) γ j ( t ) .Then, the processes Z n and G n have the same finite dimensional limitingdistributions. We then consider the convergence of the finite-dimensional distributionsof the process G n ( s, t ) = X i,j | U ij | β i ( s ) γ j ( t ) = S n ( s, t ) + S ′ n ( s, t )where S ′ n ( s, t ) = n P i,j β i ( s ) γ j ( t ), S n ( s, t ) = P i,j V ij β i ( s ) γ j ( t ) and ( β i ),( γ j ) are two independent sequences of independent bridges.Let ( s , t ) , . . . , ( s K , t K ) ∈ [0 , , α , . . . , α K ∈ R and defineΣ n = K X l =1 α l G n ( s l , t l ) := S n + S ′ n according to the decomposition of G n . Conditionnally to ( V, β i , i ≤ n ),( S n , S ′ n ) is a Gaussian vector with covariancecov( S, S ′ | ( V, β )) = 1 n K X l,l ′ =1 n X i,j,i ′ ,j ′ =1 α l α l ′ V ij β i ( s l ) β i ′ ( s l ′ ) cov( γ j ( t l ) , γ j ′ ( t l ′ ))= 1 n K X l,l ′ =1 n X i,i ′ =1 α l α l ′ β i ( s l ) β i ′ ( s l ′ ) g ( t l , t l ′ ) n X j =1 V ij = 0where g ( t, t ′ ) denotes the covariance of the Brownian bridge and we use that P j V ij = 0. Thus S n and S ′ n are conditionnally independent given ( V, β ).Thus, E [exp i θ Σ n | ( V, β )] = E (cid:0) E (cid:2) exp i θS ′ n | ( V, β ) (cid:3) E [exp i θS n | ( V, β )] (cid:1) E (cid:2) exp i θS ′ n | ( V, β ) (cid:3) = exp − θ n K X l,l ′ =1 α l α l ′ g ( t l , t l ′ )( n X i =1 β i ( s l ))( n X i =1 β i ( s l ′ ) law = exp − θ K X l,l ′ =1 α l α l ′ g ( t l , t l ′ ) β ( s l ) β ( s l ′ ) , and E [exp i θS n | ( V, β )] = exp (cid:18) − θ b S n (cid:19) where b S n = K X l,l ′ =1 α l α l ′ g ( t l , t l ′ ) n X i,k =1 H i,k β i ( s l ) β k ( s l ′ ) . We now prove the convergence in probability of b S n to L := β ′− K X l,l ′ =1 α l α l ′ g ( s l , s l ′ ) g ( t l , t l ′ ) . We have first E ( b S n ) = K X l,l ′ =1 α l α l ′ g ( t l , t l ′ ) g ( s l , s l ′ ) n X i =1 E ( H ii )and P i E ( H ii ) = n E ( H ) −→ β ′− . Now, b S n = K X l,l ′ ,p,p ′ =1 n X i,j,k,q =1 α l α l ′ α p α p ′ g ( t l , t l ′ ) g ( t p , t p ′ ) H ik H jq β i ( s l ) β k ( s l ′ ) β j ( s p ) β q ( s p ′ ) . Taking the expectation, non zero terms are obtained when the indexes i, j, k, q are equal 2 by 2. Moreover, the only non null contribution at thelimit is given for the combination i = k, j = q , i = j from Lemma 5.3. Wethus obtain that E ( b S n ) converges to β ′− K X l,l ′ ,p,p ′ =1 α l α l ′ α p α p ′ g ( t l , t l ′ ) g ( t p , t p ′ ) g ( s l , s l ′ ) g ( s p , s p ′ ) = L proving the convergence in probability of b S n as desired. Thus, we haveobtained E (exp i θ Σ n ) −→ E exp − θ X l,l ′ α l α l ′ g ( t l , t l ′ ) β ( s l ) β ( s l ′ ) + L . ANDOM TRUNCATION AND BRIDGES 13
Since L = E X l α l p β ′− W ( ∞ ) ( s l , t l ) ! and E exp − θ X l,l ′ α l α l ′ g ( t l , t l ′ ) β ( s l ) β ( s l ′ ) = E exp iθ X l α l B ⊗ B ( s l , t l ) ! , we have proved (3.4) in the sense of the finite dimensional distributions.4.5. Tightness of n − / Z ( n ) . For the tightness, owing to the structure ofthe process Z n we apply a criterion due to Davydov and Zitikis [8] whichcan be reformulated as follows. Theorem 4.5 (Davydov-Zitikis) . Let ξ ( n ) be a sequence of stochastic pro-cesses in D ([0 , such that ξ ( n ) fidi −→ ξ and ξ ( n ) (0) = 0 . Assume that (1) there are constants α ≥ β > , c ∈ (0 , ∞ ) and a n ↓ such that forall n ≥ , we have sup s ′ ,t ′ E (cid:16) | ξ ( n ) ( s + s ′ , t + t ′ ) − ξ ( n ) ( s ′ , t ′ ) | α | (cid:17) ≤ c || ( s, t ) || β , (4.18) whenever || ( s, t ) || ≥ a n , where || ( s, t ) || = max( | s | , | t | ) . (2) the process ξ ( n ) can be written as the difference of two coordinate-wise non-decreasing processes ξ ( n )1 and ξ ( n )2 such that ξ ( n )2 satisfies sup s,t ≤ max { ξ ( n )2 ( s, t + a n ) − ξ ( n )2 ( s, t ) , ξ ( n )2 ( s + a n , t ) − ξ ( n )2 ( s, t ) } = o P (1) , (4.19) then the process ξ ( n ) converges weakly to ξ . We will prove the two following lemmas.
Lemma 4.6. Z ( n ) satisfies (4.18). Lemma 4.7.
For any sequence ( b n ) with b n → + ∞ , b − n Z ( n ) satisfies (4.19). Proof of Lemma 4.6 . Let us compute the moment of order 6 of the in-crements. Let s, s ′ , t, t ′ ∈ [0 ,
1] such that s + s ′ ≤ , t + t ′ ≤ B =] s ′ , s + s ′ ] × ] t ′ , t + t ′ ]. Recall that the increment of Z n over B is given by Z n ( B ) = Z n ( s + s ′ , t + t ′ ) − Z n ( s + s ′ , t ′ ) − Z n ( s ′ , t + t ′ ) + Z n ( s ′ , t ′ ) . It is easy to see that E ( Z n ( B ) ) = E ( Z n ( s, t ) ) . We have, using the notation B i = (1 ( L i ≤ s ) − s ) and B ′ j = (1 ( C j ≤ t ) − t ) : E ( Z n ( B ) ) = X i k ,j k ,k =1 ,... E ( Y k =1 | U i k j k | ) E ( Y k =1 B i k ) E ( Y k =1 B ′ j k ) . (4.20) Since the B i and B ′ j are independent and centered, in the RHS of the aboveequation, the non null term in the sum are obtained when the i k (resp. the j k ) are equal at least 2 by 2. We now use the following properties: for some C > • | E ( B ki ) | ≤ Cs for 2 ≤ k ≤ • | E (( B ′ j ) k ) | ≤ Ct for 2 ≤ k ≤ • E ( Q k =1 | U i k j k | ) = O ( n ) (see (5.7))It follows that ( C may change from line to line) E ( Z n ( B ) ) ≤ C (cid:18) stn + st n + st n + s tn + s t n + s t n + s t (cid:19) . (4.21)Note that if s ≥ n and t ≥ n , E ( Z n ( B ) ) ≤ Cs t . (4.22)Define || ( s, t ) || = sup( | s | , | t | ). If || ( s, t ) || ≥ n , we can see from (4.21) that E ( Z n ( B ) ) ≤ C || ( s, t ) || . (4.23)We now consider the increment Z n ( s + s ′ , t + t ′ ) − Z n ( s ′ , t ′ ). It is easy tosee that Z n ( s + s ′ , t + t ′ ) − Z n ( s ′ , t ′ ) = Z n ( B ) + Z n ( B ) + Z n ( B )where B = [0 , s ′ ] × [ t ′ , t ′ + t ] and B = [ s ′ , s ′ + s ] × [0 , t ′ ]. Therefore, E ( |Z n ( s + s ′ , t + t ′ ) −Z n ( s ′ , t ′ ) | ) ≤ C ( E ( Z n ( B ) )+ E ( Z n ( B ) )+ E ( Z n ( B ) )) . From (4.21), we can see that in any case, for i = 1 ,
2, if || ( s, t ) || ≥ n : E ( Z n ( B i ) ) ≤ C || ( s, t ) || . For example, if s ≤ n ≤ t , E ( Z n ( B ) )) ≤ C sn ≤ Cst ≤ Ct . We have thus proved the following: if || ( s, t ) || ≥ n , E ( |Z n ( s + s ′ , t + t ′ ) − Z n ( s ′ , t ′ ) | ) ≤ C || ( s, t ) || , which is exactly (4.18) with a n = n − , α = 6 and β = 3. (cid:3) Proof of Lemma 4.7 . Indeed, we can write: Z n ( s, t ) = Ξ ( s, t ) − Ξ ( s, t )withΞ ( s, t ) = X i,j | U ij | L i ≤ s C j ≤ t + st , Ξ ( s, t ) = s X j C j ≤ t + t X i L i ≤ s ANDOM TRUNCATION AND BRIDGES 15 and both these processes are coordinate-wise non-decreasing. Let us nowcheck (4.19) with ξ ( n )1 = Ξ /b n and ξ ( n )2 = Ξ /b n . We have to prove that for δ = 1 /n ,sup s,t ≤ max { Ξ ( s, t + δ ) − Ξ ( s, t ) , Ξ ( s + δ, t ) − Ξ ( s, t ) } = o P ( b n ) . (4.24)Owing to the symmetry of the roles played by C j and L i , we will focus onthe first term in the above maximum. We haveΞ ( s, t + δ ) − Ξ ( s, t ) = s X j t ≤ C j ≤ t + δ + δ X i L i ≤ s ≤ X j t ≤ C j ≤ t + δ + nδ (4.25):= N n ( t, /n ) + 1 . (4.26)We have to prove that, for every ε > P (sup t N n ( t, /n ) > εb n ) → . (4.27)or, equivalently, P (sup t N n ( t, /n ) > ⌊ εb n ⌋ ) → . (4.28)The event { sup t N n ( t, /n ) > ⌊ εb n ⌋} means that there is a subintervalof [0 ,
1] of length 1 /n which contains at least ⌊ εb n ⌋ points of the sample( C , . . . , C n ). Denote by (cid:16) C ( n )(1) ≤ C ( n )(2) , . . . , ≤ C ( n )( n ) (cid:17) the reordered sample.We have (denoting C = 0 by convention) { sup t N n ( t, /n ) > ⌊ εb n ⌋} ⊂ {∃ ≤ k ≤ n : C ( n )( k + ⌊ b n ε ⌋ ) − C ( n )( k ) ≤ /n } (4.29)hence, by the union bound P (sup t N n ( t, /n ) > ⌊ εb n ⌋ ) ≤ X k P (cid:16) C ( n )( k + ⌊ b n ε ⌋ ) − C ( n )( k ) ≤ /n (cid:17) . (4.30)It is well known that the spacings follow the Dirichlet distribution of pa-rameter (1 , , . . . ,
1) so that C ( k + r ) − C ( k ) law = g + · · · + g r g + · · · + g n , (4.31)where g i are i.i.d. and exponential. From (4.30), we get P (sup t N n ( t, /n ) > ⌊ εb n ⌋ ) ≤ n P (cid:18) g + · · · + g ⌊ b n ε ⌋ g + · · · + g n ≤ n (cid:19) ≤ (4.32) ≤ n P ( g + · · · + g n > n ) + n P ( g + · · · + g ⌊ b n ε ⌋ ≤ . But we know (Chernov bound) that P ( g + · · · + g n > n ) ≤ exp − nh (2) (4.33) P ( g + · · · + g ⌊ b n ε ⌋ ≤ ≤ exp −⌊ b n ε ⌋ h (cid:18) ⌊ b n ε ⌋ (cid:19) (4.34)where h ( x ) = x − − log x . We conclude that (4.19) is fulfilled. (cid:3) Moments If U is Haar distributed matrix on U ( n ) or O Haar distributed on O ( n ),let us denote by u the generic element | U ij | or O ij . We know that u followsthe beta distribution on [0 ,
1] with parameter ( β ′ , ( n − β ′ ) so that E | U i,j | = 1 n , E | U i,j | = 2 n ( n + 1) , Var | U i,j | = n − n ( n + 1) (5.1) E | O i,j | = 1 n , E | O i,j | = 3 n ( n + 2) , Var | O i,j | = 2( n + 3) n ( n + 2) , (5.2)and more generally E | U ij | k = ( n − k !( n − k )! (5.3) E | O ij | k = (2 k )!! n ( n + 2) . . . ( n + k − . (5.4) Lemma 5.1.
Let k ∈ N , (1) For every choice of indices i = ( i , . . . , i k ) and j = ( j , . . . , j k ) , E (cid:0) | U i j | . . . | U i k j k | (cid:1) = O ( 1 n k ) (5.5)(2) For every choice of indices i = ( i , . . . , i k ) and j = ( j , . . . , j k ) , E (cid:0) | O i j | . . . | O i k j k | (cid:1) = O ( 1 n k ) (5.6)(3) For every choice of indices i = ( i , . . . , i k ) and j = ( j , . . . , j k ) , forthe unitary and orthogonal cases, E ( V i j . . . V i k j k ) = O ( 1 n k ) (5.7) Proof. (1) follows from the Weingarten formula giving the moments of Haarunitary matrix coefficients (see [6], Corollary 2.4). E (cid:0) U i j . . . U i k ,j k ¯ U i ¯1 j ¯1 . . . ¯ U i ¯ k ,j ¯ k (cid:1) = X α,β ∈S k ˜ δ α i ˜ δ β j Wg( n, βα − ) (5.8) ANDOM TRUNCATION AND BRIDGES 17 where ˜ δ α i = 1 if i ( s ) = i ( α ( s )) for every s ≤ k and 0 otherwise, and theasymptotics of the Weingarten function (see [6], Proposition 2.6):Wg( n, σ ) = O ( n − k −| σ | ) (5.9)where | σ | = k − σ ). Note that the maximal order is obtained for theidentity permutation Id for which | Id | = 0 and in this case, Wg( n, Id ) = n − k (1 + o (1)).(2) follows first from the Weingarten formula giving the moments of Haarorthogonal matrix coefficients (see [6], Corollary 3.4). For every choice ofindices i = ( i , . . . , i k , i ¯1 , . . . , i ¯ k ) and j = ( j , . . . , j k , j ¯1 , . . . , j ¯ k ), E (cid:0) O i j . . . O i k j k O i ¯1 j ¯1 . . . O i ¯ k j ¯ k (cid:1) = X p ,p ∈M k δ p i δ p j Wg O ( n ) ( p , p ) (5.10)where M k denotes the set of pairings of [2 k ], Wg O ( n ) is the orthogonalWeingarten matrix and δ p i (resp. δ p j ) is equal to 1 or 0 if i (resp. j ) isconstant on each pair of p (resp. p ) or not.We then use asymptotics for the orthogonal Weingarten matrix (see [6],Theorem 3.13): Wg O ( n ) ( p , p ) = O ( n − k − l ( p ,p ) )for some metric l on M k .(3) follows from (1), resp. (2), and the definition of V in terms of | U | . (cid:3) .We now need to have precise asymptotics for some moments of U up toorder 8. Proposition 5.2. (1) E ( | U | | U | ) = 4 n + O ( 1 n ) (5.11) E ( | U | | U | | U | | U | ) = 1 n + O ( 1 n ) (5.12) E ( | U | | U | ) = 2 n + O ( 1 n ) (5.13) E ( | U | | U | | U | ) = 1 n + O ( 1 n ) (5.14)(2) E ( | O | | O | ) = 9 n + O ( 1 n ) (5.15) E ( | O | | O | | O | | O | ) = 1 n + O ( 1 n ) (5.16) E ( | O | | O | ) = 3 n + O ( 1 n ) (5.17) E ( | O | | O | | O | ) = 1 n + O ( 1 n ) (5.18) Proof.
Let us first prove (1). We shall give the details for the first formula(5.11), the other ones are similar. From (5.8) and (5.9), the main contribu-tion in (5.8) is given by the pairs ( α, β ) for which βα − = Id . We now giveall the permutations α and β giving a non null contribution in (5.8) for thecomputation of E ( | U | | U | ). Tha admissible permutations α are givenby their cycle decomposition: α = Id = (1) , (2) , (3) , (4) α = (12) , (3) , (4) α = (1) , (2) , (34) α = (12) , (34)and the same for the β .There are 4 pairs of permutations giving βα − = Id , those corresponding to α = β . Therefore, we obtain (5.11).For (5.12), the admissible α are the same and the corresponding β are givenby β = Id = (1) , (2) , (3) , (4) β = (13) , (2) , (4) β = (1) , (3) , (24) β = (13) , (24) . Thus, there is only one pair ( α , β ) giving the main contribution.The proof for the moments of order 6 is similar.Let us now prove (2). We refer to the paper [1]. In particular, theseauthors define I n (cid:18) a cb d (cid:19) = E ( O a O b O c O d ) , I n − ( a, b ) = E ( O a O b ) . (5.19)In the sequel we denote m !! = ( m − m − . . . We need I n (cid:18) (cid:19) , I n (cid:18) (cid:19) , I n (cid:18) (cid:19) , I n (cid:18) (cid:19) The first and the third are ruled by Theorem C therein, so that I n (cid:18) (cid:19) = n !2!!4!!( n + 4)!! n !!( n + 2)!!( n + 5)!! = 3( n + 3) n ( n − n + 2)( n + 4) = 3 n + O ( 1 n )and I n (cid:18) (cid:19) = n !4!!4!!( n + 6)!!( n + 2)!!( n + 2)!!( n + 7)!!= 9( n + 3)( n + 5)( n + 6)( n + 4)( n + 2)( n + 1) n ( n −
1) = 9 n + O ( 1 n )Now, from Definition 3.1 therein I n (cid:18) (cid:19) = [ I n − (2 , Φ (cid:18) (cid:19) ANDOM TRUNCATION AND BRIDGES 19 and from Theorem 6.2Φ (cid:18) (cid:19) = n − n − u + u + u + u )with u = ( n + 6)!!( n + 7)!! , u = u = − n + 4)!!( n + 7)!! , u = 4 ( n + 2)!!( n + 7)!!Now, from Theorem 1.2 therein I n − (2 ,
2) = ( n − n + 2)!!which gives I n (cid:18) (cid:19) = ( n + 4 n + 7)( n + 6)( n + 4)( n + 2)( n + 1) n ( n −
1) = 1 n + O ( 1 n )The last one is managed with the same scheme: I n (cid:18) (cid:19) = I n − (2 , I n − (2)Φ (cid:18) (cid:19) and from Theorem 4.3 thereinΦ (cid:18) (cid:19) = ( n − n + 2)!!] ( n − n !!( n + 5)!!and since I n − (2) = E O = n we have got I (cid:18) (cid:19) = ( n − n + 2)!! n ( n + 2)!! n !!( n + 5)!! = ( n + 1) n ( n + 4)( n + 2) = 1 n + O ( 1 n ) (cid:3) Lemma 5.3. (1)
In the unitary case lim n →∞ n ( n − E (cid:0) V V (cid:1) = 1 (5.20) and lim n →∞ n ( n − E ( V V V V ) = 0 (5.21)(2) In the orthogonal case lim n →∞ n ( n − E (cid:0) V V (cid:1) = 4 (5.22) and lim n →∞ n ( n − E ( V V V V ) = 0 (5.23) Proof.
We develop V V = ( | U | − n | U | + 1 n )( | U | − n | U | + 1 n ) to obtain E ( V V ) = E ( | U | | U | ) − n E ( | U | | U | ) + 2 n E ( | U | )+ 4 n E ( | U | | U | ) − n E ( | U | ) + 1 n , (5.24)in the unitary case and the same expression with U ij replaced by O ij in theorthogonal case. It remains to make the substitutions from Proposition 5.2.The proof of the second limit is similar. (cid:3) Appendix : Proofs of Theorem 4.2 and Theorem 4.4
We define D n = P ij | U ij | R i Y j and we prove that A n and D n have thesame limit in law. We can write A n := n X j =1 Λ j ( n ) C j := S n − + Λ n ( n ) C n , D n := n X j =1 Λ j ( n ) Y j where Λ j ( n ) = P ni =1 | U ij | R i are independent of C j and Y j . Let F be asmooth function with a bounded third derivative. The first step consists inreplacing C n by Y n and to compare E ( F ( S n − + Λ n ( n ) C n )) and E ( F ( S n − +Λ n ( n ) Y n )). Using a Taylor expansion, E ( F ( S n − + Λ n ( n ) C n )) = E ( F ( S n − )) + E ( F ′ ( S n − )) E (Λ n ( n )) E ( C n )+ 12 E ( F ′′ ( S n − )) E (Λ n ( n ) ) E ( C n ) + O ( E (Λ n ( n ) )) E ( C n )and a similar expression for E ( F ( S n − + Λ n ( n ) Y n )). All the terms in the twoexpressions, but the last, are equal. We thus need to estimate E (Λ n ( n ) )By centering, E (Λ n ( n ) ) = n X i =1 E ( | U in | ) E ( R i ) = O ( 1 n ) (6.1)from (5.3) or (5.4). We repeat the operation of replacement of C j by Y j from j = n − E ( F ( A n )) − E ( F ( D n )) = O ( 1 n ) . In the same way D n and B n have the same limit in law, by exchanging therole of i and j .We can extend this proof for the finite dimensional distributions. Theproof is the analogue as above, using a Taylor expansion of the two expres-sions, involving a smooth function F of k variables, F ( S n − ( s , t )+Λ n ( s )(1 ( C n ≤ t ) − t ) , . . . , S n − ( s k , t k )+Λ n ( s k )(1 ( C n ≤ t k ) − t k ))and F ( S n − ( s , t ) + Λ n ( s ) γ n ( t ) , . . . , S n − ( s k , t k ) + Λ n ( s k ) γ n ( t k )) . We then use that the process (1 ( C i ≤ t ) − t ) has the same covariance as theBrownian bridge and the estimate (6.1). (cid:3) ANDOM TRUNCATION AND BRIDGES 21
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Universit´e Versailles-Saint Quentin, LMV UMR 8100, Bˆatiment Fermat, 45avenue des Etats-Unis, F-78035 Versailles Cedex
E-mail address : [email protected] URL : http://lmv.math.cnrs.fr/annuaire/catherine-donati/ Universit´e Versailles-Saint Quentin, LMV UMR 8100, Bˆatiment Fermat, 45avenue des Etats-Unis, F-78035 Versailles Cedex
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