Linear Statistics of Matrix Ensembles in Classical Background
aa r X i v : . [ m a t h . C A ] J un Linear statistics of matrix ensembles in classical background
Yang Chen and Chao Min([email protected], [email protected])Department of Mathematics, University of Macau,Avenida da Universidade, Taipa, Macau, ChinaAugust 3, 2018
Abstract
Given a joint probability density function of N real random variables, { x j } Nj =1 , obtainedfrom the eigenvector-eigenvalue decomposition of N × N random matrices, one constructs arandom variable, the linear statistics, defined by the sum of smooth functions evaluated at theeigenvalues or singular values of the random matrix, namely, P Nj =1 F ( x j ) . For the jpdfs obtained from the Gaussian and Laguerre ensembles, we compute, in thispaper the moment generating function E β (exp( − λ P j F ( x j ))) , where E β denotes expectationvalue over the Orthogonal ( β = 1) and Symplectic ( β = 4) ensembles, in the form one plus aSchwartz function, none vanishing over R for the Gaussian ensembles and R + for the Laguerreensembles.These are ultimately expressed in the form of the determinants of identity plus a scalaroperator, from which we obtained the large N asymptotic of the linear statistics from suitablyscaled F ( · ) . The well-known joint probability density function for the eigenvalues { x j } Nj =1 of N × N Hermitianmatrices from an orthogonal ensemble ( β = 1), unitary ensemble ( β = 2) or symplectic ensemble1 β = 4) is given by [11] P ( β ) ( x , x , . . . , x N ) N Y j =1 dx j = C ( β ) N Y ≤ j 4. Here e − λ F ( x ) := 1 + f ( x ), and we do not indicate that f ( · ) also depends on λ. Throughout this paper, we assume f ( x ) lies in the Schwartz space, and 1 + f ( x ) = 0 over [ a, b ].The simplest, very well-studied, unitary case corresponds to β = 2, see [1, 4, 2, 3, 6, 15], andthe references therein.We state here, for later development, facts on orthogonal polynomials. Let ϕ j ( x ) := P j ( x ) p w ( x ), j = 0 , , , . . . , where P j ( x ) are the orthonormal polynomials of degree j with respect to the weight w ( x ) supported on [ a, b ] , Z ba P j ( x ) P k ( x ) w ( x ) dx = δ j,k , j, k = 0 , , , . . . . 2t is a well-known fact that, G (2) N ( f ) = det (cid:16) I + K (2) N f (cid:17) , where K (2) N f is an integral operator with kernel K (2) N ( x, y ) f ( y ). Here K (2) N ( x, y ) := P N − j =0 ϕ j ( x ) ϕ j ( y ),can be evaluated via the Christoffel-Darboux formula.With Gaussian background, where w ( x ) = e − x , x ∈ R , we have P j ( x ) := H j ( x ) c j , c j = π j p Γ( j + 1) , j = 0 , , , ..., where H j ( x ) are the Hermite polynomials of degree j .With Laguerre background, where w ( x ) = x α e − x , α > − , x ∈ R + , we have P j ( x ) := L ( α ) j ( x ) c ( α ) j , c ( α ) j = s Γ( j + α + 1)Γ( j + 1) , j = 0 , , , ..., where L ( α ) j ( x ) are the Laguerre polynomials of degree j . Properties of the Hermite and Laguerrepolynomials can be found in [13].We shall mainly deal with the β = 4 and β = 1 cases, which are more complicated than β = 2situation, especially so for β = 1.We always begin with the general case, and then apply the results obtained to the two classicalensembles, i.e., the Gaussian ensembles and Laguerre ensembles.Furthermore, in β = 1 situation, for convenience, N is taken to be even, and it is expeditiousto make use of the square root of the Gaussian weight, e − x / , x ∈ R and the square root of theLaguerre weight, x α/ e − x/ , α > − , x ∈ R + in later discussion.In this paper, we shall be concerned with the large N behavior of G ( β ) N ( f ) for the Gaussianensembles and Laguerre ensembles. For the Gaussian ensembles, we replace f ( x ) by f (cid:16) √ N x (cid:17) , in the orthogonal case, and f (cid:16) √ N x (cid:17) , in the symplectic case. For the Laguerre ensembles, wereplace f ( x ) by f (cid:16) √ N x (cid:17) , in the orthogonal case and f (cid:16) √ N x (cid:17) , in the symplectic case.We will ultimately give the mean and variance of the linear statistics, as N → ∞ , together withleading correction terms.For comparison purposes, we write down results in the GUE, where, w ( x ) = e − x , x ∈ R .Denote by µ ( GUE ) N and V ( GUE ) N the mean and variance of the linear statistics P Nj =1 F (cid:16) √ N x j (cid:17) . Itis shown in [3], that, µ ( GUE ) N → π Z ∞−∞ F ( x ) dx, N → ∞ , (1.2)3 ( GUE ) N → π Z ∞−∞ F ( x ) dx − Z ∞−∞ Z ∞−∞ (cid:20) sin( x − y ) π ( x − y ) (cid:21) F ( x ) F ( y ) dxdy, N → ∞ . (1.3)For the LUE, where w ( x ) = x α e − x , α > − , x ∈ R + . Denote by µ ( LUE, α ) N and V ( LUE, α ) N the meanand variance of the linear statistics P Nj =1 F (cid:0)p N x j (cid:1) . It is shown in[2], that, µ ( LUE, α ) N → Z ∞ B ( α ) ( x, x ) F ( x ) dx, N → ∞ , (1.4) V ( LUE, α ) N → Z ∞ B ( α ) ( x, x ) F ( x ) dx − Z ∞ Z ∞ B ( α ) ( x, y ) B ( α ) ( y, x ) F ( x ) F ( y ) dxdy, N → ∞ , (1.5)where B ( α ) ( x, y ) := J α ( x ) yJ ′ α ( y ) − J α ( y ) xJ ′ α ( x ) x − y x, (1.6) B ( α ) ( x, x ) := J α ( x ) − J α − ( x ) J α +1 ( x )2 x. (1.7)Here J α ( · ) is the Bessel function of order α .We want to point out, the motivation of this paper comes from [6], which provided results bothfor symplectic ensembles and orthogonal ensembles, and specialize to the Gaussian case, i.e., GSEand GOE. The [6] dealt with the situation where f ( . ) is the characteristic function of an interval(or the union of disjoint intervals), and focus on the distribution of the m th largest eigenvalue inthe GSE and GOE, while we are interested for ”smooth” f , and we also consider the Laguerre case,i.e., LSE and LOE.This paper is organized as follows. In Section 2, we recall a number of theorems, the operators D and ε, and end with two Lemmas relevant for later development. Section 3 begins with a generaldiscussion of the Symplectic ensembles in a general setting, followed by detailed discussions on theGSE and LSE cases and ends with the computation of the mean and variance of linear statistics forlarge N. Section 4 repeats the development in Section 3 but for the Orthogonal ensembles, whichis harder. We conclude in Section 5. For orientation purposes, we introduce here a number of results, which will be used throughout thispaper. 4 heorem 2.1. The Stirling’s formula [10] Γ( n ) = √ π n n − e − n (cid:2) O (cid:0) n − (cid:1)(cid:3) , n → ∞ . (2.1) Lemma 2.2. F (cid:18) − N, 1; 32 ; 2 (cid:19) = ( − N r π N + O (cid:0) N − (cid:1) , N → ∞ . Proof. From the integral representation of the hypergeometric function, F ( α, β ; γ ; z ) = Γ( γ )Γ( β )Γ( γ − β ) Z t β − (1 − t ) γ − β − (1 − tz ) − α dt, Re γ > Re β > , (2.2)we obtain, F (cid:18) − N, 1; 32 ; 2 (cid:19) = 12 Z (1 − t ) N √ − t dt. Let x = √ − t, then F (cid:18) − N, 1; 32 ; 2 (cid:19) = Z (2 x − N dx = Z √ (2 x − N dx + Z √ (2 x − N dx = ( − N Z √ (1 − x ) N dx + Z √ (2 x − N dx. (2.3)Consider the first integral in (2.3). Let x = √ 22 cos θ, θ ∈ h , π i , then Z √ (1 − x ) N dx = √ Z π (sin θ ) N +1 dθ = √ (cid:2) N Γ( N + 1) (cid:3) Γ(2 N + 2)= r π N (cid:2) O (cid:0) N − (cid:1)(cid:3) , N → ∞ , where use has been made of the Stirling’s formula (2.1), in the last equality.5ow consider the second integral in (2.3), Z √ (2 x − N dx ≤ Z √ (2 x − N dx = 1 − ( √ − N +1 N + 1) . Hence F (cid:18) − N, 1; 32 ; 2 (cid:19) = ( − N r π N + O (cid:0) N − (cid:1) , N → ∞ . Lemma 2.3. If α > , then n X m =0 ( − m m ! Q nl = m ( α + 2 l ) (cid:18) n + α n − m + 1 (cid:19) = 1(2 n + 1)! , n = 0 , , , . . . , where (cid:0) jk (cid:1) := Γ( j +1)Γ( k +1)Γ( j − k +1) .Proof. Firstly, n X m =0 ( − m m ! Q nl = m ( α + 2 l ) (cid:18) n + α n − m + 1 (cid:19) = 1(2 n + 1)! + Γ(2 n + α + 1)Γ (cid:0) α (cid:1) F (cid:0) − n − , α ; α ; 2 (cid:1) n +1 Γ(2 n + 2)Γ (cid:0) n + α + 1 (cid:1) Γ( α ) . (2.4)From (2.2), we see that, F (cid:16) − n − , α α ; 2 (cid:17) = Γ( α )Γ (cid:0) α (cid:1) Z t α − (1 − t ) α − (1 − t ) n +1 dt. Let t = cos (cid:18) θ (cid:19) , θ ∈ [0 , π ] , then Z t α − (1 − t ) α − (1 − t ) n +1 dt = − (cid:18) (cid:19) α − Z π (sin θ ) α − (cos θ ) n +1 dθ. We carry out mathematical induction in n to prove that for α > , R π (sin θ ) α − (cos θ ) n +1 dθ = 0 . Setting if n = 0, in the integral above, we have, Z π (sin θ ) α − cos θdθ = Z π (sin θ ) α − d sin θ = (sin θ ) α α (cid:12)(cid:12)(cid:12)(cid:12) π = 0 . R π (sin θ ) α − (cos θ ) k +1 dθ = 0, for α > 0, then Z π (sin θ ) α − (cos θ ) k +3 dθ = Z π (sin θ ) α − (cos θ ) k +1 cos θdθ = Z π (sin θ ) α − (cos θ ) k +1 (cid:0) − sin θ (cid:1) dθ = Z π (sin θ ) α − (cos θ ) k +1 dθ − Z π (sin θ ) α +1 (cos θ ) k +1 dθ = 0 . Hence, Z π (sin θ ) α − (cos θ ) n +1 dθ = 0 , α > . It follows that, for α > , F (cid:16) − n − , α α ; 2 (cid:17) = 0 , and from (2.4), the lemma follows. Theorem 2.4. Y ≤ j For any integrable functions p j ( x ) and q j ( x ) , j = 1 , , . . . , we have (cid:18)Z [ a,b ] N det ( p j ( x k ) , q j ( x k )) j =1 ,..., Nk =1 ,...,N dx · · · dx N (cid:19) = ( N !) det (cid:18)Z ba ( p j ( x ) q k ( x ) − p k ( x ) q j ( x )) dx (cid:19) Nj,k =1 , (cid:18)Z a ≤ x ≤···≤ x N ≤ b det ( p j ( x k )) Nj,k =1 dx · · · dx N (cid:19) = det (cid:18)Z ba Z ba sgn( y − x ) p j ( x ) p k ( y ) dxdy (cid:19) Nj,k =1 , where the determinant on the left of the first equality is a N × N determinant with alternatingcolumns { p j ( x ) } , { q j ( x ) } , { p j ( x ) } , { q j ( x ) } , . . . . In addition, N is even in the second equality. heorem 2.6. If A, B are Hilbert-Schmidt operators on a Hilbert space H , then [7] det( I + AB ) = det( I + BA ) . Following [14, 15, 16], we introduce here operators ε and D which will be crucial for laterdevelopment. Let ε be the integral operator with kernel ε ( x, y ) := 12 sgn( x − y ) , then for any integrable function g defined on [ a, b ] ,εg ( x ) = Z ba ε ( x, t ) g ( t ) dt = 12 (cid:18)Z xa g ( t ) dt − Z bx g ( t ) dt (cid:19) , x ∈ [ a, b ] . It is clear that ε ( y, x ) = − ε ( x, y ), i.e., ε t = − ε , where t denotes transpose.Let D be the operator that acts by differentiation, thus for any differentiable function g definedon [ a, b ] , Dg ( x ) = dg ( x ) dx = g ′ ( x ) . With further conditions on g ( x ), we prove an easy lemma on the commutator [ D, ε ] . Lemma 2.7. For any function g ∈ C [ a, b ] and g ( a ) = g ( b ) = 0 , Dεg ( x ) = εDg ( x ) = g ( x ) , i.e., Dε = εD = I .Proof. For any function g ∈ C [ a, b ] and g ( a ) = g ( b ) = 0, we have( Dε ) g ( x ) = ddx Z ba ε ( x, t ) g ( t ) dt = 12 ddx (cid:20)Z xa g ( t ) dt − Z bx g ( t ) dt (cid:21) = 12 g ( x ) + 12 g ( x ) = g ( x ) , and ( εD ) g ( x ) = Z ba ε ( x, t ) g ′ ( t ) dt = 12 Z xa g ′ ( t ) dt − Z bx g ′ ( t ) dt = 12 [ g ( x ) − g ( a )] − 12 [ g ( b ) − g ( x )]= g ( x ) − 12 [ g ( a ) + g ( b )]= g ( x ) . u ⊗ v the integral operator with kernel ( u ⊗ v )( x, y ) := u ( x ) v ( y ). We have the followinglemma. Lemma 2.8. If A is an integral operator with kernel A ( x, y ) , then A ( u ⊗ v ) = ( Au ) ⊗ v, ( u ⊗ v ) A = u ⊗ ( A t v ) . Taking β = 4 in (1.1) gives the symplectic ensembles and generating function becomes, G (4) N ( f ) = C (4) N Z [ a,b ] N Y ≤ j 2, which ultimately expresses K (4) N in terms of K (2 , N, . Theorem 3.1. K (2 , N, = K (2 , N, ε, K (1 , N, = DK (2 , N, ε, K (1 , N, = DK (2 , N, , K (2 , N, εD = K (2 , N, . Proof. First of all, K (2 , N, ε = − N − X j,k =0 ( µ jk ψ j ⊗ ψ ′ k ) ε = N − X j,k =0 µ jk ψ j ⊗ ( εψ ′ k )= N − X j,k =0 µ jk ψ j ⊗ ψ k = K (2 , N, . Secondly, for any integrable function g ( x ) defined on [ a, b ], we have DK (2 , N, g ( x ) = ddx Z ba K (2 , N, ( x, y ) g ( y ) dy = Z ba ∂∂x K (2 , N, ( x, y ) g ( y ) dy = Z ba K (1 , N, ( x, y ) g ( y ) dy = K (1 , N, g ( x ) , DK (2 , N, = K (1 , N, . It follows that K (1 , N, = DK (2 , N, ε. Similarly, DK (2 , N, g ( x ) = ddx Z ba K (2 , N, ( x, y ) g ( y ) dy = Z ba ∂∂x K (2 , N, ( x, y ) g ( y ) dy = Z ba K (1 , N, ( x, y ) g ( y ) dy = K (1 , N, g ( x ) , i.e., K (1 , N, = DK (2 , N, . Finally, we find K (2 , N, εDg ( x ) = K (2 , N, Dg ( x ) = Z ba K (2 , N, ( x, y ) g ′ ( y ) dy = K (2 , N, ( x, y ) g ( y ) | ba − Z ba g ( y ) ∂∂y K (2 , N, ( x, y ) dy. Note that K (2 , N, ( x, y ) : = N − X j,k =0 ψ j ( x ) µ jk ψ k ( y )= N − X j,k =0 ψ j ( x ) µ jk π k ( y ) p w ( y )and w ( a ) = w ( b ) = 0 , hence K (2 , N, ( x, a ) = K (2 , N, ( x, b ) = 0 . Continuing, K (2 , N, εDg ( x ) = − Z ba g ( y ) ∂∂y K (2 , N, ( x, y ) dy = Z ba K (2 , N, ( x, y ) g ( y ) dy = K (2 , N, g ( x ) , K (2 , N, εD = K (2 , N, . The proof is complete.According to Theorem 3.1, K (4) N can be written as K (4) N = DK (2 , N, ε DK (2 , N, K (2 , N, ε K (2 , N, = D I K (2 , N, ε K (2 , N, K (2 , N, ε K (2 , N, = : e A e B, where e A = D I , e B = K (2 , N, ε K (2 , N, K (2 , N, ε K (2 , N, . From (3.2) and using Theorem 2.6, we have h G (4) N ( f ) i = det (cid:18) I + (cid:16) e A e B (cid:17) f (cid:19) = det (cid:18) I + e A (cid:16) e Bf (cid:17)(cid:19) = det (cid:18) I + (cid:16) e Bf (cid:17) e A (cid:19) = det (cid:18) I + e B (cid:16) f e A (cid:17)(cid:19) . Since e B (cid:16) f e A (cid:17) = K (2 , N, K (2 , N, K (2 , N, K (2 , N, f D I = K (2 , N, ε K (2 , N, K (2 , N, ε K (2 , N, f D f = K (2 , N, εf D K (2 , N, fK (2 , N, εf D K (2 , N, f , then h G (4) N ( f ) i = det I + K (2 , N, εf D K (2 , N, fK (2 , N, εf D I + K (2 , N, f . The computation below reduces the above into a determinant of scalar operators. We subtractrow 1 from row 2, h G (4) N ( f ) i = det I + K (2 , N, εf D K (2 , N, f − I I . h G (4) N ( f ) i = det I + K (2 , N, εf D + K (2 , N, f K (2 , N, f I = det (cid:16) I + K (2 , N, εf D + K (2 , N, f (cid:17) . Remark . The above result agrees with [6] for the GUE case if we take f = − µ χ J , where χ J is thecharacteristic function of the interval J .Now we use the commutator [ D, f ] := Df − f D to obtain a better suited result for our purpose.For a given function smooth g ( x ), we have[ D, f ] g ( x ) = Df g ( x ) − f Dg ( x )= ( f ( x ) g ( x )) ′ − f ( x ) g ′ ( x )= f ′ ( x ) g ( x ) , this is, [ D, f ] = f ′ . It follows that f D = Df − [ D, f ] = Df − f ′ . (3.3)Taking this into account, we have the following theorem. Theorem 3.2. h G (4) N ( f ) i = det (cid:16) I + 2 K (2 , N, f − K (2 , N, εf ′ (cid:17) , where the kernel of K (2 , N, reads, K (2 , N, ( x, y ) = − N − X j,k =0 ψ j ( x ) µ jk ψ ′ k ( y ) . In the case of the Gaussian weight w ( x ) = e − x , x ∈ R , we again follow the discussions [6, 15], andchoose a special ψ j to simplify M (4) as much as possible. To proceed, let ψ j +1 ( x ) := 1 √ ϕ j +1 ( x ) , ψ j ( x ) := − √ εϕ j +1 ( x ) , j = 0 , , , . . . , (3.4)13here ϕ j ( x ) is given by ϕ j ( x ) = H j ( x ) c j e − x , c j = π j p Γ( j + 1) , (3.5)and H j ( x ) , j = 0 , , . . . are the usual Hermite polynomials with the orthogonality condition Z ∞−∞ H m ( x ) H n ( x )e − x dx = c n δ mn . We show in the next lemma that, this definition satisfies (3.1), i.e., ψ j ( x ) = π j ( x )e − x , j = 0 , , , . . . ,where π j ( x ) is a polynomial of degree j. Lemma 3.3. ψ j ( x )e x , j = 0 , , , . . . is a polynomial of degree j .Proof. If the index is odd, then it is clear that ψ j +1 ( x )e x is a polynomial of degree 2 j + 1. Foreven index, ψ j ( x )e x = − √ εϕ j +1 ( x )) e x = − √ x ε (cid:18) c j +1 H j +1 ( x )e − x (cid:19) = − √ c j +1 e x (cid:20)Z x −∞ H j +1 ( y )e − y dy − Z ∞ x H j +1 ( y )e − y dy (cid:21) , j = 0 , , , · · · . Since H j +1 ( y )e − y , j = 0 , , . . . is an odd function, we have0 = Z ∞−∞ H j +1 ( y )e − y dy = Z x −∞ H j +1 ( y )e − y dy + Z ∞ x H j +1 ( y )e − y dy. Hence Z ∞ x H j +1 ( y )e − y dy = − Z x −∞ H j +1 ( y )e − y dy, and we find, ψ j ( x )e x = − √ c j +1 e x Z x −∞ H j +1 ( y )e − y dy. From mathematical induction, it follows that Z x −∞ y k e − y dy = − e − x (cid:2) x k − + ( k − x k − + ( k − k − x k − + · · · + ( k − k − · · · , k = 1 , , , · · · . Since H j +1 ( y ) is a linear combination of y, y , . . . , y j +1 , we see that R x −∞ H j +1 ( y )e − y dy is equalto e − x multiplying a polynomial of degree 2 j . It follows that ψ j ( x )e x is a polynomial of degree2 j . The proof is complete. 14sing (3.4) to compute M (4) := (cid:18)Z ∞−∞ (cid:0) ψ j ( x ) ψ ′ k ( x ) − ψ ′ j ( x ) ψ k ( x ) (cid:1) dx (cid:19) N − j,k =0 , we obtain the following lemma. Lemma 3.4. ( Dieng , Tracy − Widom ) M (4) = · · · − · · · · · · − · · · ... ... ... ... ... ... · · · · · · − N × N . It is clear that (cid:0) M (4) (cid:1) − = − M (4) , so µ j, j +1 = − , µ j +1 , j = 1, and µ jk = 0 for other cases.Hence K (2 , N, ( x, y ) = − N − X j,k =0 ψ j ( x ) µ jk ψ ′ k ( y )= N − X j =0 ψ j ( x ) ψ ′ j +1 ( y ) − N − X j =0 ψ j +1 ( x ) ψ ′ j ( y )= 12 " N − X j =0 ϕ j +1 ( x ) ϕ j +1 ( y ) − N − X j =0 εϕ j +1 ( x ) ϕ ′ j +1 ( y ) . (3.6)Recall that the Hermite polynomials H j satisfy the differentiation formulas [10] H ′ j ( x ) = 2 xH j ( x ) − H j +1 ( x ) , j = 0 , , , . . . , (3.7) H ′ j ( x ) = 2 jH j − ( x ) , j = 0 , , , . . . . (3.8)Using the fact that H j ( x ) = c j ϕ j ( x ) e x , (3.7) becomes ϕ ′ j ( x ) = xϕ j ( x ) − p j + 1) ϕ j +1 ( x ) , j = 0 , , , . . . , (3.9)and similarly, (3.8) becomes, ϕ ′ j ( x ) = − xϕ j ( x ) + p jϕ j − ( x ) , j = 0 , , , . . . . (3.10)15ombining (3.9) and (3.10), to eliminate ϕ j ( x ) , we obtain ϕ ′ j ( x ) = r j ϕ j − ( x ) − r j ϕ j +1 ( x ) , j = 0 , , , . . . . (3.11)Using (3.11) to replace ϕ ′ j +1 ( y ), we find, N − X j =0 εϕ j +1 ( x ) ϕ ′ j +1 ( y )= N − X j =0 εϕ j +1 ( x ) "r j + 12 ϕ j ( y ) − p j + 1 ϕ j +2 ( y ) = N − X j =0 r j + 12 εϕ j +1 ( x ) ϕ j ( y ) − N − X j =0 p j + 1 εϕ j +1 ( x ) ϕ j +2 ( y )= N X j =0 r j + 12 εϕ j +1 ( x ) ϕ j ( y ) − N X j =0 p jεϕ j − ( x ) ϕ j ( y ) − r N + 12 εϕ N +1 ( x ) ϕ N ( y )= − N X j =0 "p jεϕ j − ( x ) − r j + 12 εϕ j +1 ( x ) ϕ j ( y ) − r N + 12 εϕ N +1 ( x ) ϕ N ( y ) . (3.12)To proceed further, using Lemma 2.7, together with (3.11) we find ϕ j ( x ) = εDϕ j ( x ) = ε ϕ ′ j ( x )= p j εϕ j − ( x ) − r j + 12 εϕ j +1 ( x ) . (3.13)Substituting (3.13) into (3.12), it follows that N − X j =0 εϕ j +1 ( x ) ϕ ′ j +1 ( y ) = − N X j =0 ϕ j ( x ) ϕ j ( y ) − r N + 12 εϕ N +1 ( x ) ϕ N ( y ) . Hence, (3.6) becomes, K (2 , N, ( x, y ) = 12 " N − X j =0 ϕ j +1 ( x ) ϕ j +1 ( y ) + N X j =0 ϕ j ( x ) ϕ j ( y ) + r N + 12 εϕ N +1 ( x ) ϕ N ( y ) = 12 " N X j =0 ϕ j ( x ) ϕ j ( y ) + r N + 12 εϕ N +1 ( x ) ϕ N ( y ) = 12 " S N ( x, y ) + r N + 12 εϕ N +1 ( x ) ϕ N ( y ) , where S N ( x, y ) := N X j =0 ϕ j ( x ) ϕ j ( y ) = r N + 12 ϕ N +1 ( x ) ϕ N ( y ) − ϕ N +1 ( y ) ϕ N ( x ) x − y , and here the last equality comes from the Christoffel-Darboux formula.By Theorem 3.2, we have the following theorem.16 heorem 3.5. h G (4) N ( f ) i = det I + S N f − S N εf ′ + r N + 12 ( εϕ N +1 ⊗ ϕ N f ) + 12 r N + 12 ( εϕ N +1 ⊗ εϕ N ) f ′ ! . N behavior of the GSE moment generating function To proceed with the large N investigation, write, h G (4) N ( f ) i , as h G (4) N ( f ) i =: det( I + T ) , where T := S N f − S N εf ′ + r N + 12 ( εϕ N +1 ⊗ ϕ N f ) + 12 r N + 12 ( εϕ N +1 ⊗ εϕ N ) f ′ . (3.14)We find, log det( I + T ) = Tr log( I + T ) = Tr T − 12 Tr T + 13 Tr T − · · · . This is obtained by the trace-log expansion of log det( I + ν T ), where 0 < ν < , follow by acontinuation to ν = 1. We assume that similar continuation also holds in other cases. Theorem 3.6. lim N →∞ √ N S N (cid:18) x √ N , y √ N (cid:19) = sin( x − y ) π ( x − y ) , lim N →∞ √ N S N (cid:18) x √ N , x √ N (cid:19) = 1 π . The next theorem characterizes the large N asymptotic of various “scaled” quantities. Theorem 3.7. ϕ N (cid:18) x √ N (cid:19) = ( − N π − N − cos x + O (cid:16) N − (cid:17) , N → ∞ ,εϕ N (cid:18) x √ N (cid:19) = 2 − ( − N π − N − sin x + O (cid:16) N − (cid:17) , N → ∞ ,εϕ N +1 (cid:18) x √ N (cid:19) = − − N − + O (cid:16) N − (cid:17) , N → ∞ . roof. From the asymptotic expansion of Hermite polynomials [13], H n ( x ) e − x = λ n h cos (cid:16) √ n + 1 x − nπ (cid:17) + O (cid:16) n − (cid:17)i , n → ∞ , (3.15)where λ n = Γ(2 n + 1)Γ ( n + 1) and λ n +1 = Γ(2 n + 3)Γ( n + 2) (4 n + 3) − . We have ϕ N (cid:18) x √ N (cid:19) = 1 π N p Γ(2 N + 1) H N (cid:18) x √ N (cid:19) e − x N = 1 π N p Γ(2 N + 1) · Γ(2 N + 1)Γ( N + 1) " cos r N + 14 N x − N π ! + O (cid:16) N − (cid:17) = 1 π N · p Γ(2 N + 1)Γ( N + 1) h cos( x − N π ) + O (cid:16) N − (cid:17)i = ( − N π − N − cos x + O (cid:16) N − (cid:17) , N → ∞ , where we have used the Stirling’s formula (2.1).A straightforward computation gives, εϕ N ( x ) = 12 (cid:18)Z x −∞ ϕ N ( y ) dy − Z ∞ x ϕ N ( y ) dy (cid:19) = 12 (cid:18)Z −∞ ϕ N ( y ) dy + Z x ϕ N ( y ) dy − Z x ϕ N ( y ) dy − Z ∞ ϕ N ( y ) dy (cid:19) = 12 (cid:18) Z x ϕ N ( y ) dy + Z −∞ ϕ N ( y ) dy − Z ∞ ϕ N ( y ) dy (cid:19) = 1 π N +1 p Γ(2 N + 1) (cid:18) Z x H N ( y )e − y dy + Z −∞ H N ( y )e − y dy − Z ∞ H N ( y )e − y dy (cid:19) = 1 π N p Γ(2 N + 1) Z x H N ( y )e − y dy. So we find, for large N , εϕ N (cid:18) x √ N (cid:19) = 1 π N p Γ(2 N + 1) Z x √ N H N ( y )e − y dy = 1 π N p Γ(2 N + 1) · √ N Z x H N (cid:18) y √ N (cid:19) e − y N dy = 1 π N p Γ(2 N + 1) · √ N · Γ(2 N + 1)Γ( N + 1) Z x h ( − N cos y + O (cid:16) N − (cid:17)i dy = p Γ(2 N + 1) π N √ N Γ( N + 1) h ( − N sin x + O (cid:16) N − (cid:17)i = 2 − ( − N π − N − sin x + O (cid:16) N − (cid:17) , N → ∞ , εϕ N +1 ( x ) = 12 (cid:20)Z x −∞ ϕ N +1 ( y ) dy − Z ∞ x ϕ N +1 ( y ) dy (cid:21) = 1 π N + p Γ(2 N + 2) (cid:20)Z x −∞ H N +1 ( y )e − y dy − Z ∞ x H N +1 ( y )e − y dy (cid:21) = 1 π N + p Γ(2 N + 2) (cid:20)Z −∞ H N +1 ( y )e − y dy + Z x H N +1 ( y )e − y dy − Z x H N +1 ( y )e − y dy − Z ∞ H N +1 ( y )e − y dy (cid:21) = 1 π N + p Γ(2 N + 2) (cid:20) Z x H N +1 ( y )e − y dy − Z ∞ H N +1 ( y )e − y dy (cid:21) = 1 π N + p Γ(2 N + 2) (cid:20)Z x H N +1 ( y )e − y dy − Z ∞ H N +1 ( y )e − y dy (cid:21) , continuing, we see that, εϕ N +1 (cid:18) x √ N (cid:19) = 1 π N + p Γ(2 N + 2) Z x √ N H N +1 ( y )e − y dy − π N + p Γ(2 N + 2) Z ∞ H N +1 ( y )e − y dy = 1 π N + p Γ(2 N + 2) · √ N Z x H N +1 (cid:18) y √ N (cid:19) e − y N dy − π N + p Γ(2 N + 2) Z ∞ H N +1 ( y )e − y dy. By (3.15), we have H N +1 (cid:18) y √ N (cid:19) e − y N = Γ(2 N + 3)Γ( N + 2) (4 N + 3) − h ( − N sin y + O (cid:16) N − (cid:17)i , N → ∞ . It follows that Z x H N +1 (cid:18) y √ N (cid:19) e − y N dy = Γ(2 N + 3)Γ( N + 2) (4 N + 3) − h ( − N (1 − cos x ) + O (cid:16) N − (cid:17)i , N → ∞ . On the other hand, from [8], Z ∞ H N +1 ( y )e − y dy = 2 ( − N Γ(2 N + 2)Γ( N + 1) F (cid:18) − N, 1; 32 ; 2 (cid:19) . It follows from Lemma 2.2, that, Z ∞ H N +1 ( y )e − y dy = 2 ( − N Γ(2 N + 2)Γ( N + 1) (cid:20) ( − N r π N + O (cid:0) N − (cid:1)(cid:21) , N → ∞ . 19y the Stirling’s formula (2.1), we obtain εϕ N +1 (cid:18) x √ N (cid:19) = h − ( − N π − N − (1 − cos x ) + O (cid:16) N − (cid:17)i − h − N − + O (cid:16) N − (cid:17)i = − − N − + O (cid:16) N − (cid:17) , N → ∞ . We are now in a position to compute Tr T and Tr T as N → ∞ , using Theorem 3.6 and Theorem3.7. The estimates provided by Theorem 3.7 are instrumental in the large N computations thatfollows.In what follows, we replace f ( x ) by f (cid:16) √ N x (cid:17) , and note that f ′ ( √ N x ) = 1 √ N ddx f ( √ N x ) . The f ′ that appears in the trace will be accordingly interpreted.We first consider Tr T , which reads,Tr T = Tr S N f − Tr 12 S N εf ′ + Tr r N + 12 ( εϕ N +1 ⊗ ϕ N f ) + Tr 12 r N + 12 ( εϕ N +1 ⊗ εϕ N ) f ′ . Tr T has four parts.First of all, we find Tr S N f = Z ∞−∞ S N ( x, x ) f (cid:16) √ N x (cid:17) dx = Z ∞−∞ √ N S N (cid:18) x √ N , x √ N (cid:19) f ( x ) dx → π Z ∞−∞ f ( x ) dx, N → ∞ . The second term reads,Tr 12 S N εf ′ = 12 Z ∞−∞ Z ∞−∞ S N ( x, y ) ε ( y, x ) f ′ (cid:16) √ N x (cid:17) dxdy = 14 Z ∞−∞ (cid:20)Z ∞ x S N ( x, y ) dy − Z x −∞ S N ( x, y ) dy (cid:21) f ′ (cid:16) √ N x (cid:17) dx. To proceed further, let u = √ N x, v = √ N y, 20t follows that,Tr 12 S N εf ′ = 14 √ N Z ∞−∞ (cid:20)Z ∞ u √ N S N (cid:18) u √ N , v √ N (cid:19) dv − Z u −∞ √ N S N (cid:18) u √ N , v √ N (cid:19) dv (cid:21) f ′ ( u ) du → √ N π Z ∞−∞ (cid:20)Z ∞ u sin( u − v ) u − v dv − Z u −∞ sin( u − v ) u − v dv (cid:21) f ′ ( u ) du, N → ∞ . Now let t = u − v , it follows that,Tr 12 S N εf ′ → √ N π Z ∞−∞ (cid:20)Z −∞ sin tt dt − Z ∞ sin tt dt (cid:21) f ′ ( u ) du = 14 √ N π Z ∞−∞ (cid:16) π − π (cid:17) f ′ ( u ) du = 0 , N → ∞ . For the third term, we have,Tr r N + 12 ( εϕ N +1 ⊗ ϕ N f ) = r N + 12 Z ∞−∞ εϕ N +1 ( x ) ϕ N ( x ) f (cid:16) √ N x (cid:17) dx = r N + 12 · √ N Z ∞−∞ εϕ N +1 (cid:18) x √ N (cid:19) ϕ N (cid:18) x √ N (cid:19) f ( x ) dx = − ( − N √ πN Z ∞−∞ cos x f ( x ) dx + O (cid:0) N − (cid:1) , N → ∞ , where use has been made of Theorem 3.7.Finally to the fourth term, and take note of Theorem 3.7,Tr 12 r N + 12 ( εϕ N +1 ⊗ εϕ N ) f ′ = 12 r N + 12 Z ∞−∞ εϕ N +1 ( x ) εϕ N ( x ) f ′ (cid:16) √ N x (cid:17) dx = 12 r N + 12 · √ N Z ∞−∞ εϕ N +1 (cid:18) x √ N (cid:19) εϕ N (cid:18) x √ N (cid:19) f ′ ( x ) dx = O (cid:0) N − (cid:1) , N → ∞ . Therefore, the large N expansion of Tr T , reads,Tr T = 1 π Z ∞−∞ f ( x ) dx − ( − N √ πN Z ∞−∞ cos x f ( x ) dx + O (cid:0) N − (cid:1) , N → ∞ . T , with T given by, (3.14), there are 10 traces:Tr T = Tr S N f S N f − Tr S N f S N εf ′ + Tr √ N + 2 S N f ( εϕ N +1 ⊗ ϕ N f )+ Tr r N + 12 S N f ( εϕ N +1 ⊗ εϕ N ) f ′ + Tr 14 S N εf ′ S N εf ′ − Tr r N + 12 S N εf ′ ( εϕ N +1 ⊗ ϕ N f ) − Tr 12 r N + 12 S N εf ′ ( εϕ N +1 ⊗ εϕ N ) f ′ + Tr (cid:18) N + 12 (cid:19) ( εϕ N +1 ⊗ ϕ N f )( εϕ N +1 ⊗ ϕ N f )+ Tr (cid:18) N + 12 (cid:19) ( εϕ N +1 ⊗ ϕ N f )( εϕ N +1 ⊗ εϕ N ) f ′ + Tr 14 (cid:18) N + 12 (cid:19) ( εϕ N +1 ⊗ εϕ N ) f ′ ( εϕ N +1 ⊗ εϕ N ) f ′ . (3.16)In the following, we calculate the trace on the right side of (3.16), term by term.The first term:Tr S N f S N f = Z ∞−∞ Z ∞−∞ S N ( x, y ) f (cid:16) √ N y (cid:17) S N ( y, x ) f (cid:16) √ N x (cid:17) dydx = Z ∞−∞ Z ∞−∞ (cid:20) √ N S N (cid:18) x √ N , y √ N (cid:19)(cid:21) f ( x ) f ( y ) dxdy → π Z ∞−∞ Z ∞−∞ (cid:20) sin( x − y ) x − y (cid:21) f ( x ) f ( y ) dxdy, N → ∞ . The second term reads,Tr S N f S N εf ′ = Z ∞−∞ Z ∞−∞ Z ∞−∞ S N ( x, y ) f (cid:16) √ N y (cid:17) S N ( y, z ) ε ( z, x ) f ′ (cid:16) √ N x (cid:17) dxdydz = 12 Z ∞−∞ Z ∞−∞ S N ( x, y ) (cid:20)Z ∞ x S N ( y, z ) dz − Z x −∞ S N ( y, z ) dz (cid:21) f ′ (cid:16) √ N x (cid:17) f (cid:16) √ N y (cid:17) dxdy. A change of variables, u = √ N x, v = √ N y, w = √ N z, giveTr S N f S N εf ′ = 12 √ N Z ∞−∞ Z ∞−∞ √ N S N (cid:18) u √ N , v √ N (cid:19)(cid:20)Z ∞ u √ N S N (cid:18) v √ N , w √ N (cid:19) dw − Z u −∞ √ N S N (cid:18) v √ N , w √ N (cid:19) dw (cid:21) f ′ ( u ) f ( v ) dudv → √ N π Z ∞−∞ Z ∞−∞ sin( u − v ) u − v (cid:20)Z ∞ u sin( v − w ) v − w dw − Z u −∞ sin( v − w ) v − w dw (cid:21) f ′ ( u ) f ( v ) dudv, N → ∞ . 22o proceed further, let t = v − w , we see that, Z ∞ u sin( v − w ) v − w dw − Z u −∞ sin( v − w ) v − w dw = Z v − u −∞ sin tt dt − Z ∞ v − u sin tt dt = 2 Z v − u sin tt dt = 2 Si( v − u ) , where Si( x ) is the sine integral Si( x ) := Z x sin tt dt. Since Si( − x ) = − Si( x ) , it follows thatTr S N f S N εf ′ → √ N π Z ∞−∞ Z ∞−∞ sin( u − v ) u − v Si( v − u ) f ′ ( u ) f ( v ) dudv = − π √ N Z ∞−∞ Z ∞−∞ sin( u − v ) u − v Si( u − v ) f ′ ( u ) f ( v ) dudv, N → ∞ . The third term:Tr √ N + 2 S N f ( εϕ N +1 ⊗ ϕ N f )= √ N + 2 Z ∞−∞ Z ∞−∞ S N ( x, y ) f (cid:16) √ N y (cid:17) εϕ N +1 ( y ) ϕ N ( x ) f (cid:16) √ N x (cid:17) dxdy = √ N + 2 · √ N Z ∞−∞ Z ∞−∞ √ N S N (cid:18) x √ N , y √ N (cid:19) f ( y ) εϕ N +1 (cid:18) y √ N (cid:19) ϕ N (cid:18) x √ N (cid:19) f ( x ) dxdy = − ( − N π √ πN Z ∞−∞ Z ∞−∞ sin( x − y ) x − y cos x f ( x ) f ( y ) dxdy + O (cid:0) N − (cid:1) , N → ∞ . The fourth term:Tr r N + 12 S N f ( εϕ N +1 ⊗ εϕ N ) f ′ = r N + 12 Z ∞−∞ Z ∞−∞ S N ( x, y ) f (cid:16) √ N y (cid:17) εϕ N +1 ( y ) εϕ N ( x ) f ′ (cid:16) √ N x (cid:17) dxdy = r N + 12 · √ N Z ∞−∞ Z ∞−∞ √ N S N (cid:18) x √ N , y √ N (cid:19) f ( y ) εϕ N +1 (cid:18) y √ N (cid:19) εϕ N (cid:18) x √ N (cid:19) f ′ ( x ) dxdy = O (cid:0) N − (cid:1) , N → ∞ . The fifth term, with the change of variables, u = √ N x, v = √ N y, w = √ N z, τ = √ N t, 23e see that,Tr 14 S N εf ′ S N εf ′ = 164 N Z ∞−∞ Z ∞−∞ (cid:20)Z ∞ w √ N S N (cid:18) u √ N , v √ N (cid:19) dv − Z w −∞ √ N S N (cid:18) u √ N , v √ N (cid:19) dv (cid:21)(cid:20)Z ∞ u √ N S N (cid:18) w √ N , τ √ N (cid:19) dτ − Z u −∞ √ N S N (cid:18) w √ N , τ √ N (cid:19) dτ (cid:21) f ′ ( u ) f ′ ( w ) dudw = O (cid:0) N − (cid:1) , N → ∞ . The sixth term,Tr r N + 12 S N εf ′ ( εϕ N +1 ⊗ ϕ N f )= 18 N r N + 12 Z ∞−∞ Z ∞−∞ (cid:20)Z ∞ w √ N S N (cid:18) u √ N , v √ N (cid:19) dv − Z w −∞ √ N S N (cid:18) u √ N , v √ N (cid:19) dv (cid:21) ϕ N (cid:18) u √ N (cid:19) εϕ N +1 (cid:18) w √ N (cid:19) f ( u ) f ′ ( w ) dudw = O (cid:0) N − (cid:1) , N → ∞ . The seventh term, becomes,Tr 12 r N + 12 S N εf ′ ( εϕ N +1 ⊗ εϕ N ) f ′ = 116 N r N + 12 Z ∞−∞ Z ∞−∞ (cid:20)Z ∞ w √ N S N (cid:18) u √ N , v √ N (cid:19) dv − Z w −∞ √ N S N (cid:18) u √ N , v √ N (cid:19) dv (cid:21) εϕ N (cid:18) u √ N (cid:19) εϕ N +1 (cid:18) w √ N (cid:19) f ′ ( u ) f ′ ( w ) dudw = O (cid:16) N − (cid:17) , N → ∞ . The eighth term can be computed in a similar manner, and we have,Tr (cid:18) N + 12 (cid:19) ( εϕ N +1 ⊗ ϕ N f )( εϕ N +1 ⊗ ϕ N f )= N + N Z ∞−∞ Z ∞−∞ εϕ N +1 (cid:18) u √ N (cid:19) εϕ N +1 (cid:18) v √ N (cid:19) ϕ N (cid:18) u √ N (cid:19) ϕ N (cid:18) v √ N (cid:19) f ( u ) f ( v ) dudv = O (cid:0) N − (cid:1) , N → ∞ . Proceeding in a similar manner with the ninth term, we have,Tr (cid:18) N + 12 (cid:19) ( εϕ N +1 ⊗ ϕ N f )( εϕ N +1 ⊗ εϕ N ) f ′ = N + N Z ∞−∞ Z ∞−∞ εϕ N +1 (cid:18) u √ N (cid:19) εϕ N +1 (cid:18) v √ N (cid:19) εϕ N (cid:18) u √ N (cid:19) ϕ N (cid:18) v √ N (cid:19) f ′ ( u ) f ( v ) dudv = O (cid:16) N − (cid:17) , N → ∞ . T , becomesTr 14 (cid:18) N + 12 (cid:19) ( εϕ N +1 ⊗ εϕ N ) f ′ ( εϕ N +1 ⊗ εϕ N ) f ′ = N + N Z ∞−∞ Z ∞−∞ εϕ N +1 (cid:18) u √ N (cid:19) εϕ N +1 (cid:18) v √ N (cid:19) εϕ N (cid:18) u √ N (cid:19) εϕ N (cid:18) v √ N (cid:19) f ′ ( u ) f ′ ( v ) dudv = O (cid:0) N − (cid:1) , N → ∞ . Hence, the large N behavior of (3.16), reads,Tr T = 1 π Z ∞−∞ Z ∞−∞ (cid:20) sin( x − y ) x − y (cid:21) f ( x ) f ( y ) dxdy + 12 π √ N Z ∞−∞ Z ∞−∞ sin( x − y ) x − y Si( x − y ) f ′ ( x ) f ( y ) dxdy − ( − N π √ πN Z ∞−∞ Z ∞−∞ sin( x − y ) x − y cos x f ( x ) f ( y ) dxdy + O (cid:0) N − (cid:1) , N → ∞ . We are now in a position to compute the mean and variance of the (scaled) linear statistics P Nj =1 F (cid:16) √ N x j (cid:17) , which are obtained as the coefficients of λ and λ , of log det( I + T ) . Since f (cid:16) √ N x (cid:17) ≈ − λF (cid:16) √ N x (cid:17) + λ F (cid:16) √ N x (cid:17) , we replace f with − λF + λ F in the expression of Tr T and Tr T . A minor rearrangement gives,log det ( I + T )= − λ ( π Z ∞−∞ F ( x ) dx − ( − N √ πN Z ∞−∞ cos x F ( x ) dx + O (cid:0) N − (cid:1) ) + λ ( π Z ∞−∞ F ( x ) dx − π Z ∞−∞ Z ∞−∞ (cid:20) sin( x − y ) x − y (cid:21) F ( x ) F ( y ) dxdy − π √ N Z ∞−∞ Z ∞−∞ sin( x − y ) x − y Si( x − y ) F ′ ( x ) F ( y ) dxdy − ( − N √ πN (cid:20)Z ∞−∞ cos x F ( x ) dx − π Z ∞−∞ Z ∞−∞ sin( x − y ) x − y cos x F ( x ) F ( y ) dxdy (cid:21) + O (cid:0) N − (cid:1) ) , N → ∞ . Denote by µ ( GSE ) N and V ( GSE ) N the mean and variance of the linear statistics P Nj =1 F (cid:16) √ N x j (cid:17) , thenwe have obtained, the large N corrections of these quantities. Theorem 3.8. As N → ∞ , µ ( GSE ) N = 12 µ ( GUE ) N − ( − N √ πN Z ∞−∞ cos x F ( x ) dx + O (cid:0) N − (cid:1) , ( GSE ) N = 12 V ( GUE ) N − π √ N Z ∞−∞ Z ∞−∞ sin( x − y ) x − y Si( x − y ) F ′ ( x ) F ( y ) dxdy − ( − N √ πN (cid:20)Z ∞−∞ cos x F ( x ) dx − π Z ∞−∞ Z ∞−∞ sin( x − y ) x − y cos x F ( x ) F ( y ) dxdy (cid:21) + O (cid:0) N − (cid:1) , where µ ( GUE ) N and V ( GUE ) N for N → ∞ are given in (1.2) and (1.3) respectively. We study the case with the Laguerre background, namely, the weight, w ( x ) = x α e − x , α > , x ∈ R + . The idea is to choose special ψ j so that M (4) takes on the simplestpossible form. To this end, let ψ j +1 ( x ) := 1 √ ϕ ( α − j +1 ( x ) , j = 0 , , , . . . , (3.17) ψ j ( x ) := − √ ε e ϕ ( α − j +1 ( x ) , j = 0 , , , . . . , (3.18)where ϕ ( α − j ( x ) and e ϕ ( α − j ( x ) are given by ϕ ( α − j ( x ) = L ( α − j ( x ) c ( α − j x α e − x , j = 0 , , , . . . , (3.19) e ϕ ( α − j ( x ) = L ( α − j ( x ) c ( α − j x α − e − x , j = 0 , , , . . . . Here L ( α ) j ( x ) , j = 0 , , , . . . , are the Laguerre polynomials, with the orthogonality condition, Z ∞ L ( α ) j ( x ) L ( α ) k ( x ) x α e − x dx = (cid:16) c ( α ) j (cid:17) δ jk , c ( α ) j = s Γ( j + α + 1)Γ( j + 1) . It is easy to see that Z ∞ ϕ ( α − j ( x ) e ϕ ( α − k ( x ) dx = δ jk , j, k = 0 , , , . . . . We now prove that (3.17) and (3.18) satisfy (3.1), i.e., ψ j ( x ) = π j ( x ) x α e − x , j = 0 , , , . . . , where π j ( x ) is a polynomial of degree j . Theorem 3.9. ψ j ( x ) = π j ( x ) x α e − x , j = 0 , , , . . . , where π j ( x ) is a polynomial of degree j . roof. We prove this by considering two cases, j odd, and j even. If j = 2 n + 1, then by (3.17), π n +1 ( x ) = 1 √ c ( α − n +1 L ( α − n +1 ( x ) . Let j = 2 n , then ψ n ( x ) = − √ ε e ϕ n +1 ( x ) = − √ c ( α − n +1 Z x L ( α − n +1 ( y ) y α − e − y dy, and we have used the fact R ∞ L ( α − n +1 ( y ) y α − e − y dy = 0 , n = 0 , , , . . . . Let us rewrite the above as Z x L ( α − n +1 ( y ) y α − e − y dy = b π n ( x ) x α e − x , n = 0 , , , . . . , (3.20)where b π n ( x ) := −√ c ( α − n +1 π n ( x ) . Take a derivative on both sides, L ( α − n +1 ( x ) x α − e − x = (cid:18) x b π ′ n ( x ) + α b π n ( x ) − x b π n ( x ) (cid:19) x α − e − x , which becomes, L ( α − n +1 ( x ) = x b π ′ n ( x ) + α b π n ( x ) − x b π n ( x ) . (3.21)Note that equation (3.20) is equivalent to (3.21). Now we seek to solve (3.21).Suppose b π n ( x ) := a ( n ) + a ( n ) x + a ( n ) x + · · · + a n − ( n ) x n − + a n ( n ) x n , we see that the right side of (3.21) is equal to x b π ′ n ( x ) + α b π n ( x ) − x b π n ( x )= α a ( n ) + (cid:20)(cid:16) α (cid:17) a ( n ) − a ( n ) (cid:21) x + (cid:20)(cid:16) α (cid:17) a ( n ) − a ( n ) (cid:21) x + · · · + (cid:20)(cid:16) n − α (cid:17) a n − ( n ) − a n − ( n ) (cid:21) x n − + (cid:20)(cid:16) n + α (cid:17) a n ( n ) − a n − ( n ) (cid:21) x n − a n ( n ) x n +1 . (3.22)27n the other hand, the left side of (3.21) is equal to L ( α − n +1 ( x ) = (cid:18) n + α n + 1 (cid:19) − (cid:18) n + α n (cid:19) x + 12! (cid:18) n + α n − (cid:19) x − (cid:18) n + α n − (cid:19) x + · · · + 1(2 n )! (cid:18) n + α (cid:19) x n − n + 1)! (cid:18) n + α (cid:19) x n +1 . (3.23)Compare the coefficients of (3.22) and (3.23), we have the equations α a ( n ) = (cid:0) n + α n +1 (cid:1)(cid:0) α (cid:1) a ( n ) − a ( n ) = − (cid:0) n + α n (cid:1)(cid:0) α (cid:1) a ( n ) − a ( n ) = (cid:0) n + α n − (cid:1)(cid:0) α (cid:1) a ( n ) − a ( n ) = − (cid:0) n + α n − (cid:1) ... (cid:0) n − α (cid:1) a n − ( n ) − a n − ( n ) = − n − (cid:0) n + α (cid:1)(cid:0) n + α (cid:1) a n ( n ) − a n − ( n ) = − n )! (cid:0) n + α (cid:1) − a n ( n ) = − n +1)! (cid:0) n + α (cid:1) . (3.24)By solving the first 2 n + 1 equations in (3.24), we find, a ( n ) = α (cid:0) n + α n +1 (cid:1) a ( n ) = α ( α +2) (cid:0) n + α n +1 (cid:1) − α +2 (cid:0) n + α n (cid:1) a ( n ) = α ( α +2)( α +4) (cid:0) n + α n +1 (cid:1) − α +2)( α +4) (cid:0) n + α n (cid:1) + 12! 1 α +4 (cid:0) n + α n − (cid:1) a ( n ) = α ( α +2)( α +4)( α +6) (cid:0) n + α n +1 (cid:1) − α +2)( α +4)( α +6) (cid:0) n + α n (cid:1) + 12! 1( α +4)( α +6) (cid:0) n + α n − (cid:1) − 13! 1 α +6 (cid:0) n + α n − (cid:1) ... a n ( n ) = α ( α +2)( α +4) ··· ( α +4 n ) (cid:0) n + α n +1 (cid:1) − α +2)( α +4) ··· ( α +4 n ) (cid:0) n + α n (cid:1) + 12! 1( α +4)( α +6) ··· ( α +4 n ) (cid:0) n + α n − (cid:1) − 13! 1( α +6) ··· ( α +4 n ) (cid:0) n + α n − (cid:1) + · · · − n − α +4 n − α +4 n ) (cid:0) n + α (cid:1) + n )! 1 α +4 n (cid:0) n + α (cid:1) . The last equation of (3.24), simplifies to, a n ( n ) = 2(2 n + 1)! . By Lemma 2.3, we see that linear system (3.24) is solvable. Hence b π n ( x ) is a polynomial of degree2 n. The proof is complete. 28ith (3.17) and (3.18), we compute M (4) := (cid:0)R ∞ (cid:0) ψ j ( x ) ψ ′ k ( x ) − ψ ′ j ( x ) ψ k ( x ) (cid:1) dx (cid:1) N − j,k =0 , resultingin the following theorem. Theorem 3.10. M (4) = · · · − · · · · · · − · · · ... ... ... ... ... ... · · · · · · − N × N . Proof. Let m jk be the ( j, k )-entry of M (4) , i.e., m jk := Z ∞ (cid:0) ψ j ( x ) ψ ′ k ( x ) − ψ ′ j ( x ) ψ k ( x ) (cid:1) dx, j, k = 0 , , . . . , N − . We compute m j,k by considering four cases: ( j, k ) = (even , odd) , (odd , even) , (even , even) , (odd , odd) . For the (even , odd) case, m j, k +1 = Z ∞ (cid:0) ψ j ( x ) ψ ′ k +1 ( x ) − ψ ′ j ( x ) ψ k +1 ( x ) (cid:1) dx = Z ∞ ψ j ( x ) ψ ′ k +1 ( x ) dx − Z ∞ ψ ′ j ( x ) ψ k +1 ( x ) dx = (cid:18) ψ j ( x ) ψ k +1 ( x ) | ∞ − Z ∞ ψ ′ j ( x ) ψ k +1 ( x ) dx (cid:19) − Z ∞ ψ ′ j ( x ) ψ k +1 ( x ) dx = − Z ∞ ψ ′ j ( x ) ψ k +1 ( x ) dx = − Z ∞ (cid:18) − √ e ϕ ( α − j +1 ( x ) (cid:19) (cid:18) √ ϕ ( α − k +1 ( x ) (cid:19) dx = Z ∞ e ϕ ( α − j +1 ( x ) ϕ ( α − k +1 ( x ) dx = δ jk . For the (odd , even) case, since M (4) is antisymmetric, we have m j +1 , k = − m k, j +1 = − δ j,k . , even) case and j > k , m j, k = Z ∞ (cid:0) ψ j ( x ) ψ ′ k ( x ) − ψ ′ j ( x ) ψ k ( x ) (cid:1) dx = Z ∞ ψ j ( x ) ψ ′ k ( x ) dx − Z ∞ ψ ′ j ( x ) ψ k ( x ) dx = ψ j ( x ) ψ k ( x ) | ∞ − Z ∞ ψ ′ j ( x ) ψ k ( x ) dx = − Z ∞ ψ ′ j ( x ) ψ k ( x ) dx = − Z ∞ − √ c ( α − j +1 L ( α − j +1 ( x ) x α − e − x ! (cid:0) π k ( x ) x α e − x (cid:1) dx = √ c ( α − j +1 Z ∞ L ( α − j +1 ( x ) π k ( x ) x α − e − x dx = 0 , since π k ( x ) is a polynomial of degree 2 k which is less than 2 j + 1.For the (odd , odd) case and j > k , m j +1 , k +1 = Z ∞ (cid:0) ψ j +1 ( x ) ψ ′ k +1 ( x ) − ψ ′ j +1 ( x ) ψ k +1 ( x ) (cid:1) dx = Z ∞ ψ j +1 ( x ) ψ ′ k +1 ( x ) dx − Z ∞ ψ ′ j +1 ( x ) ψ k +1 ( x ) dx = Z ∞ ψ j +1 ( x ) ψ ′ k +1 ( x ) dx − (cid:18) ψ j +1 ( x ) ψ k +1 ( x ) | ∞ − Z ∞ ψ j +1 ( x ) ψ ′ k +1 ( x ) dx (cid:19) = 2 Z ∞ ψ j +1 ( x ) ψ ′ k +1 ( x ) dx = 1 c ( α − j +1 c ( α − k +1 Z ∞ L ( α − j +1 ( x ) x α e − x (cid:18) x (cid:16) L ( α − k +1 ( x ) (cid:17) ′ + α L ( α − k +1 ( x ) − xL ( α − k +1 ( x ) (cid:19) x α − e − x dx = 1 c ( α − j +1 c ( α − k +1 Z ∞ L ( α − j +1 ( x ) (cid:18) x (cid:16) L ( α − k +1 ( x ) (cid:17) ′ + α L ( α − k +1 ( x ) − xL ( α − k +1 ( x ) (cid:19) x α − e − x dx = 0 , since x (cid:16) L ( α − k +1 ( x ) (cid:17) ′ + α L ( α − k +1 ( x ) − xL ( α − k +1 ( x ) is a polynomial of degree 2 k + 2 which is less than2 j + 1.If j < k , due to the fact that M (4) is antisymmetric, m j, k = − m k, j = 0 , j +1 , k +1 = − m k +1 , j +1 = 0 . Thus m j, k = m j +1 , k +1 = 0 , j, k = 0 , , . . . , N − . This is just the desired form of M (4) .It’s clear that (cid:0) M (4) (cid:1) − = − M (4) , so µ j, j +1 = − , µ j +1 , j = 1, and µ jk = 0 for other cases.The rest of this subsection is devoted to the determination of K (2 , N, ( x, y ), K (2 , N, ( x, y ) = − N − X j,k =0 ψ j ( x ) µ jk ψ ′ k ( y )= N − X j =0 ψ j ( x ) ψ ′ j +1 ( y ) − N − X j =0 ψ j +1 ( x ) ψ ′ j ( y )= 12 N − X j =0 ϕ ( α − j +1 ( x ) e ϕ ( α − j +1 ( y ) − N − X j =0 ε e ϕ ( α − j +1 ( x ) h ϕ ( α − j +1 ( y ) i ′ . From (3.19), we see that, h ϕ ( α − j ( x ) i ′ = 1 c ( α − j x α − e − x (cid:20) x (cid:16) L ( α − j ( x ) (cid:17) ′ + α − x L ( α − j ( x ) (cid:21) . (3.25)Recall that the Laguerre polynomials L ( α − j ( x ) , satisfy the differentiation formulas [8], x (cid:16) L ( α − j ( x ) (cid:17) ′ = jL ( α − j ( x ) − ( j + α − L ( α − j − ( x ) , j = 0 , , , . . . , (3.26) x (cid:16) L ( α − j ( x ) (cid:17) ′ = ( j + 1) L ( α − j +1 ( x ) + ( x − j − α ) L ( α − j ( x ) , j = 0 , , , . . . . (3.27)Summing (3.26) and (3.27), and divide by 2, gives, x (cid:16) L ( α − j ( x ) (cid:17) ′ = j + 12 L ( α − j +1 ( x ) − j + α − L ( α − j − ( x ) + x − α L ( α − j ( x ) , j = 0 , , , . . . , or x (cid:16) L ( α − j ( x ) (cid:17) ′ + α − x L ( α − j ( x ) = j + 12 L ( α − j +1 ( x ) − j + α − L ( α − j − ( x ) , j = 0 , , , . . . . Hence (3.25) becomes, h ϕ ( α − j ( x ) i ′ = 1 c ( α − j x α − e − x (cid:20) j + 12 L ( α − j +1 ( x ) − j + α − L ( α − j − ( x ) (cid:21) = 12 p ( j + 1)( j + α ) e ϕ ( α − j +1 ( x ) − p j ( j + α − e ϕ ( α − j − ( x ) . (3.28)31e see that h ϕ ( α − j i ′ is a linear combination of e ϕ ( α − j +1 and e ϕ ( α − j − , just like the GSE case studiedin the last section. Replacing j by 2 j + 1 in (3.28), to find, h ϕ ( α − j +1 ( y ) i ′ = s ( j + 1) (cid:18) j + α + 12 (cid:19) e ϕ ( α − j +2 ( y ) − s(cid:18) j + 12 (cid:19) (cid:16) j + α (cid:17) e ϕ ( α − j ( y ) . A straightforward computation shows that, N − X j =0 ε e ϕ ( α − j +1 ( x ) h ϕ ( α − j +1 ( y ) i ′ = N − X j =0 ε e ϕ ( α − j +1 ( x ) s ( j + 1) (cid:18) j + α + 12 (cid:19) e ϕ ( α − j +2 ( y ) − N − X j =0 ε e ϕ ( α − j +1 ( x ) s(cid:18) j + 12 (cid:19) (cid:16) j + α (cid:17) e ϕ ( α − j ( y )= N X j =0 s j (cid:18) j + α − (cid:19) ε e ϕ ( α − j − ( x ) e ϕ ( α − j ( y ) − N X j =0 s(cid:18) j + 12 (cid:19) (cid:16) j + α (cid:17) ε e ϕ ( α − j +1 ( x ) e ϕ ( α − j ( y )+ s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y )= N X j =0 "s j (cid:18) j + α − (cid:19) ε e ϕ ( α − j − ( x ) − s(cid:18) j + 12 (cid:19) (cid:16) j + α (cid:17) ε e ϕ ( α − j +1 ( x ) ϕ ( α − j ( y )+ s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y ) . From (3.28) and Theorem 2.7, ϕ ( α − j ( x ) = ε D ϕ ( α − j ( x )= ε h ϕ ( α − j ( x ) i ′ = ε (cid:20) p (2 j + 1)(2 j + α ) e ϕ ( α − j +1 ( x ) − p j (2 j + α − e ϕ ( α − j − ( x ) (cid:21) = s(cid:18) j + 12 (cid:19) (cid:16) j + α (cid:17) ε e ϕ ( α − j +1 ( x ) − s j (cid:18) j + α − (cid:19) ε e ϕ ( α − j − ( x ) . Hence, the sum, P N − j =0 ε e ϕ ( α − j +1 ( x ) h ϕ ( α − j +1 ( y ) i ′ simplifies immediately, and leads to, N − X j =0 ε e ϕ ( α − j +1 ( x ) h ϕ ( α − j +1 ( y ) i ′ = − N X j =0 ϕ ( α − j ( x ) e ϕ ( α − j ( y ) + s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y ) . 32t follows that, K (2 , N, ( x, y ) = 12 N − X j =0 ϕ ( α − j +1 ( x ) e ϕ ( α − j +1 ( y ) + 12 N X j =0 ϕ ( α − j ( x ) e ϕ ( α − j ( y ) − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y )= 12 N X j =0 ϕ ( α − j ( x ) e ϕ ( α − j ( y ) − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y )= 12 S N ( x, y ) − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y ) , where S N ( x, y ) := N X j =0 ϕ ( α − j ( x ) e ϕ ( α − j ( y ) = − p (2 N + 1)(2 N + α ) ϕ ( α − N +1 ( x ) e ϕ ( α − N ( y ) − e ϕ ( α − N +1 ( y ) ϕ ( α − N ( x ) x − y . Here we used the Christoffel-Darboux formula in the last equality.By Theorem 3.2, we have the following theorem. Theorem 3.11. h G (4) N ( f ) i = det I + S N f − S N εf ′ − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ ! . N behavior of the LSE moment generating function Now consider the scaling limit of h G (4) N ( f ) i , write h G (4) N ( f ) i =: det( I + T ) , where T : = S N f − S N εf ′ − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) − s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ . Theorem 3.12. lim N →∞ y N S N (cid:18) x N , y N (cid:19) = B ( α − ( x, y ) , N →∞ x N S N (cid:18) x N , x N (cid:19) = B ( α − ( x, x ) , where B ( α − ( x, y ) and B ( α − ( x, x ) are given by (1.6) and (1.7) with α replaced by α − . Theorem 3.13. e ϕ ( α − N (cid:18) x N (cid:19) = 2 (2 N ) J α − ( x ) x + O (cid:16) N − (cid:17) , N → ∞ ,ε e ϕ ( α − N (cid:18) x N (cid:19) = (2 N ) − (cid:20)Z x J α − ( y ) dy − (cid:21) + O (cid:16) N − (cid:17) , N → ∞ ,ε e ϕ ( α − N +1 (cid:18) x N (cid:19) = (2 N ) − Z x J α − ( y ) dy + O (cid:16) N − (cid:17) , N → ∞ . Proof. Recall the asymptotic formula of the Laguerre polynomials [13], L ( α ) N ( x ) x α e − x = Γ( N + α + 1)Γ( N + 1) (cid:18) N + α + 12 (cid:19) − α J α (cid:16)p (4 N + 2 α + 2) x (cid:17) + x O (cid:16) N α − (cid:17) , N → ∞ . We find, e ϕ ( α − N (cid:18) x N (cid:19) = s Γ(2 N + 1)Γ(2 N + α ) L ( α − N (cid:18) x N (cid:19) (cid:18) x N (cid:19) α − e − x N = s Γ(2 N + 1)Γ(2 N + α ) · (cid:18) x N (cid:19) − (cid:20) Γ(2 N + α )Γ(2 N + 1) (cid:16) N + α (cid:17) − α − J α − ( x ) + O (cid:16) N α − (cid:17)(cid:21) = 2 (2 N ) J α − ( x ) x + O (cid:16) N − (cid:17) , N → ∞ , where we have used the formula,Γ( n + a )Γ( n + b ) = n a − b (cid:2) O (cid:0) n − (cid:1)(cid:3) , n → ∞ . Proceeding to ε e ϕ ( α − N ( x ), we have, ε e ϕ ( α − N ( x )= 12 (cid:20)Z x e ϕ ( α − N ( y ) dy − Z ∞ x e ϕ ( α − N ( y ) dy (cid:21) = 12 s Γ(2 N + 1)Γ(2 N + α ) (cid:20)Z x L ( α − N ( y ) y α − e − y dy − Z ∞ x L ( α − N ( y ) y α − e − y dy (cid:21) = 12 s Γ(2 N + 1)Γ(2 N + α ) (cid:20) Z x L ( α − N ( y ) y α − e − y dy − Z ∞ L ( α − N ( y ) y α − e − y dy (cid:21) = 12 s Γ(2 N + 1)Γ(2 N + α ) " Z x L ( α − N ( y ) y α − e − y dy − α Γ (cid:0) N + α (cid:1) Γ( N + 1) , Z ∞ L ( α − N ( y ) y α − e − y dy = 2 α Γ (cid:0) N + α (cid:1) Γ( N + 1) . Continuing, ε e ϕ ( α − N (cid:18) x N (cid:19) = 12 s Γ(2 N + 1)Γ(2 N + α ) " N Z x L ( α − N (cid:18) y N (cid:19) (cid:18) y N (cid:19) α − e − y N ydy − α Γ (cid:0) N + α (cid:1) Γ( N + 1) = (2 N ) − (cid:16) N + α (cid:17) − α − s Γ(2 N + α )Γ(2 N + 1) Z x J α − ( y ) dy − α − s Γ(2 N + 1)Γ(2 N + α ) · Γ (cid:0) N + α (cid:1) Γ( N + 1) + O (cid:16) N − (cid:17) = (2 N ) − (cid:20)Z x J α − ( y ) dy − (cid:21) + O (cid:16) N − (cid:17) , N → ∞ . Similarly, ε e ϕ ( α − N +1 ( x )= 12 (cid:20)Z x e ϕ ( α − N +1 ( y ) dy − Z ∞ x e ϕ ( α − N +1 ( y ) dy (cid:21) = 12 s Γ(2 N + 2)Γ(2 N + α + 1) (cid:20)Z x L ( α − N +1 ( y ) y α − e − y dy − Z ∞ x L ( α − N +1 ( y ) y α − e − y dy (cid:21) = 12 s Γ(2 N + 2)Γ(2 N + α + 1) (cid:20) Z x L ( α − N +1 ( y ) y α − e − y dy − Z ∞ L ( α − N +1 ( y ) y α − e − y dy (cid:21) = s Γ(2 N + 2)Γ(2 N + α + 1) Z x L ( α − N +1 ( y ) y α − e − y dy, where we have used the fact that Z ∞ L ( α − N +1 ( y ) y α − e − y dy = 0 . It follows that ε e ϕ ( α − N +1 (cid:18) x N (cid:19) = s Γ(2 N + 2)Γ(2 N + α + 1) Z x N L ( α − N +1 ( y ) y α − e − y dy = 14 N s Γ(2 N + 2)Γ(2 N + α + 1) Z x L ( α − N +1 (cid:18) y N (cid:19) (cid:18) y N (cid:19) α − e − y N ydy = (2 N ) − (cid:16) N + 1 + α (cid:17) − α − s Γ(2 N + α + 1)Γ(2 N + 2) Z x J α − ( y ) dy + O (cid:16) N − (cid:17) = (2 N ) − Z x J α − ( y ) dy + O (cid:16) N − (cid:17) , N → ∞ . 35e now use Theorem 3.12 and Theorem 3.13 to compute Tr T and Tr T as N → ∞ . In thecomputations below, we replace f ( x ) by f (cid:16) √ N x (cid:17) and f ′ ( x ) by f ′ (cid:16) √ N x (cid:17) = d √ N d √ x f (cid:16) √ N x (cid:17) . Consider Tr T , which reads,Tr T = Tr S N f − Tr 12 S N εf ′ − Tr s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) − Tr 12 s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ . So we compute Tr T by calculating the four terms in the right side. The first term,Tr S N f = Z ∞ S N ( x, x ) f (cid:16) √ N x (cid:17) dx = Z ∞ x N S N (cid:18) x N , x N (cid:19) f ( x ) dx → Z ∞ B ( α − ( x, x ) f ( x ) dx, N → ∞ . The second term,Tr 12 S N εf ′ = 12 Z ∞ Z ∞ S N ( x, y ) ε ( y, x ) f ′ (cid:16) √ N x (cid:17) dxdy = 14 Z ∞ (cid:20)Z ∞ x S N ( x, y ) dy − Z x S N ( x, y ) dy (cid:21) f ′ (cid:16) √ N x (cid:17) dx. Let u = √ N x, v = p N y, thenTr 12 S N εf ′ = 116 N Z ∞ (cid:20)Z ∞ u N S N (cid:18) u N , v N (cid:19) vdv − Z u N S N (cid:18) u N , v N (cid:19) vdv (cid:21) u f ′ ( u ) du → N Z ∞ (cid:20)Z ∞ u B ( α − ( u, v ) dv − Z u B ( α − ( u, v ) dv (cid:21) u f ′ ( u ) du, N → ∞ . s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) = s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ ε e ϕ ( α − N +1 ( x ) e ϕ ( α − N ( x ) f (cid:16) √ N x (cid:17) dx = q(cid:0) N + (cid:1) (cid:0) N + α (cid:1) N Z ∞ ε e ϕ ( α − N +1 (cid:18) x N (cid:19) e ϕ ( α − N (cid:18) x N (cid:19) xf ( x ) dx = 12 Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) J α − ( x ) f ( x ) dx + O (cid:0) N − (cid:1) , N → ∞ . The fourth term,Tr 12 s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ = 12 s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ ε e ϕ ( α − N +1 ( x ) ε e ϕ ( α − N ( x ) f ′ (cid:16) √ N x (cid:17) dx = q(cid:0) N + (cid:1) (cid:0) N + α (cid:1) N Z ∞ ε e ϕ ( α − N +1 (cid:18) x N (cid:19) ε e ϕ ( α − N (cid:18) x N (cid:19) x f ′ ( x ) dx = 116 N Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) (cid:20)Z x J α − ( y ) dy − (cid:21) x f ′ ( x ) dx + O (cid:0) N − (cid:1) , N → ∞ . Therefore,Tr T = Z ∞ B ( α − ( x, x ) f ( x ) dx − Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) J α − ( x ) f ( x ) dx − N Z ∞ (cid:20)Z ∞ x B ( α − ( x, y ) dy − Z x B ( α − ( x, y ) dy (cid:21) x f ′ ( x ) dx − N Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) (cid:20)Z x J α − ( y ) dy − (cid:21) x f ′ ( x ) dx + O (cid:0) N − (cid:1) , N → ∞ . T , where there are 10 traces,Tr T = Tr S N f S N f − Tr S N f S N εf ′ − Tr 2 s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N f (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) − Tr s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N f (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ + Tr 14 S N εf ′ S N εf ′ + Tr s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N εf ′ (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) + Tr 12 s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N εf ′ (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ + Tr (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) + Tr (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ + Tr 14 (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ . In the following, we need to calculate the traces on the right side term by term. The first term,Tr S N f S N f = Z ∞ Z ∞ S N ( x, y ) f (cid:16)p N y (cid:17) S N ( y, x ) f (cid:16) √ N x (cid:17) dxdy = Z ∞ Z ∞ y N S N (cid:18) x N , y N (cid:19) x N S N (cid:18) y N , x N (cid:19) f ( x ) f ( y ) dxdy → Z ∞ Z ∞ B ( α − ( x, y ) B ( α − ( y, x ) f ( x ) f ( y ) dxdy, N → ∞ . The second term,Tr S N f S N εf ′ = 18 N Z ∞ Z ∞ v N S N (cid:18) u N , v N (cid:19) (cid:20)Z ∞ u w N S N (cid:18) v N , w N (cid:19) dw − Z u w N S N (cid:18) v N , w N (cid:19) dw (cid:21) u f ′ ( u ) f ( v ) dudv → N Z ∞ Z ∞ B ( α − ( u, v ) (cid:20)Z ∞ u B ( α − ( v, w ) dw − Z u B ( α − ( v, w ) dw (cid:21) u f ′ ( u ) f ( v ) dudv, N → ∞ . The third term,Tr 2 s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N f (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) = 12 N s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ Z ∞ N S N (cid:18) u N , v N (cid:19) v e ϕ ( α − N (cid:18) u N (cid:19) ε e ϕ ( α − N +1 (cid:18) v N (cid:19) uf ( u ) f ( v ) dudv = Z ∞ Z ∞ B ( α − ( u, v ) J α − ( u ) (cid:20)Z v J α − ( t ) dt (cid:21) f ( u ) f ( v ) dudv + O (cid:0) N − (cid:1) , N → ∞ . s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N f (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ = 14 N s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ Z ∞ N S N (cid:18) u N , v N (cid:19) v f ( v ) ε e ϕ ( α − N +1 (cid:18) v N (cid:19) ε e ϕ ( α − N (cid:18) u N (cid:19) u f ′ ( u ) dudv = 18 N Z ∞ Z ∞ B ( α − ( u, v ) (cid:20)Z u J α − ( t ) dt − (cid:21) (cid:20)Z v J α − ( t ) dt (cid:21) u f ′ ( u ) f ( v ) dudv + O (cid:0) N − (cid:1) , N → ∞ . The fifth term,Tr 14 S N εf ′ S N εf ′ = 14 Z ∞ Z ∞ Z ∞ Z ∞ S N ( x, y ) ε ( y, z ) f ′ (cid:16) √ N z (cid:17) S N ( z, t ) ε ( t, x ) f ′ (cid:16) √ N x (cid:17) dxdydzdt = 116 Z ∞ Z ∞ (cid:20)Z ∞ z S N ( x, y ) dy − Z z S N ( x, y ) dy (cid:21) (cid:20)Z ∞ x S N ( z, t ) dt − Z x S N ( z, t ) dt (cid:21) f ′ (cid:16) √ N x (cid:17) f ′ (cid:16) √ N z (cid:17) dxdz = 1256 N Z ∞ Z ∞ (cid:20)Z ∞ w N S N (cid:18) u N , v N (cid:19) vdv − Z w N S N (cid:18) u N , v N (cid:19) vdv (cid:21)(cid:20)Z ∞ u N S N (cid:18) w N , τ N (cid:19) τ dτ − Z u N S N (cid:18) w N , τ N (cid:19) τ dτ (cid:21) u w f ′ ( u ) f ′ ( w ) dudw = O (cid:18) N (cid:19) , N → ∞ . The sixth term,Tr s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N εf ′ (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) = 132 N s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ Z ∞ (cid:20)Z ∞ w v N S N (cid:18) u N , v N (cid:19) dv − Z w v N S N (cid:18) u N , v N (cid:19) dv (cid:21) f ′ ( w ) ε e ϕ ( α − N +1 (cid:18) w N (cid:19) e ϕ ( α − N (cid:18) u N (cid:19) u w f ( u ) dudw = 116 N Z ∞ Z ∞ (cid:20)Z ∞ w B ( α − ( u, v ) dv − Z w B ( α − ( u, v ) dv (cid:21) (cid:20)Z w J α − ( t ) dt (cid:21) J α − ( u ) w f ′ ( w ) f ( u ) dudw + O (cid:0) N − (cid:1) , N → ∞ . s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) S N εf ′ (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ = 164 N s(cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ Z ∞ (cid:20)Z ∞ w v N S N (cid:18) u N , v N (cid:19) dv − Z w v N S N (cid:18) u N , v N (cid:19) dv (cid:21) f ′ ( w ) ε e ϕ ( α − N +1 (cid:18) w N (cid:19) ε e ϕ ( α − N (cid:18) u N (cid:19) f ′ ( u ) uwdudw = O (cid:18) N (cid:19) , N → ∞ . The eighth term,Tr (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) = (cid:0) N + (cid:1) (cid:0) N + α (cid:1) N Z ∞ Z ∞ ε e ϕ ( α − N +1 (cid:18) u N (cid:19) ε e ϕ ( α − N +1 (cid:18) v N (cid:19) e ϕ ( α − N (cid:18) u N (cid:19) e ϕ ( α − N (cid:18) v N (cid:19) uvf ( u ) f ( v ) dudv = 14 Z ∞ Z ∞ (cid:20)Z u J α − ( t ) dt (cid:21) (cid:20)Z v J α − ( t ) dt (cid:21) J α − ( u ) J α − ( v ) f ( u ) f ( v ) dudv + O (cid:0) N − (cid:1) , N → ∞ . The ninth term,Tr (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ e ϕ ( α − N f (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ = (cid:0) N + (cid:1) (cid:0) N + α (cid:1) N Z ∞ Z ∞ ε e ϕ ( α − N +1 (cid:18) u N (cid:19) ε e ϕ ( α − N +1 (cid:18) v N (cid:19) ε e ϕ ( α − N (cid:18) u N (cid:19) e ϕ ( α − N (cid:18) v N (cid:19) uv f ′ ( u ) f ( v ) dudv = 116 N Z ∞ Z ∞ (cid:20)Z u J α − ( t ) dt (cid:21) (cid:20)Z v J α − ( t ) dt (cid:21) (cid:20)Z u J α − ( t ) dt − (cid:21) J α − ( v ) u f ′ ( u ) f ( v ) dudv + O (cid:0) N − (cid:1) , N → ∞ . The tenth term,Tr 14 (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ (cid:16) ε e ϕ ( α − N +1 ⊗ ε e ϕ ( α − N (cid:17) f ′ = 164 N (cid:18) N + 12 (cid:19) (cid:16) N + α (cid:17) Z ∞ Z ∞ ε e ϕ ( α − N +1 (cid:18) u N (cid:19) ε e ϕ ( α − N +1 (cid:18) v N (cid:19) ε e ϕ ( α − N (cid:18) u N (cid:19) ε e ϕ ( α − N (cid:18) v N (cid:19) uv f ′ ( u ) f ′ ( v ) dudv = O (cid:18) N (cid:19) , N → ∞ . T = Z ∞ Z ∞ B ( α − ( x, y ) B ( α − ( y, x ) f ( x ) f ( y ) dxdy − Z ∞ Z ∞ B ( α − ( x, y ) J α − ( x ) (cid:20)Z y J α − ( z ) dz (cid:21) f ( x ) f ( y ) dxdy + 14 Z ∞ Z ∞ (cid:20)Z x J α − ( z ) dz (cid:21) (cid:20)Z y J α − ( z ) dz (cid:21) J α − ( x ) J α − ( y ) f ( x ) f ( y ) dxdy − N Z ∞ Z ∞ B ( α − ( x, y ) (cid:20)Z ∞ x B ( α − ( y, z ) dz − Z x B ( α − ( y, z ) dz (cid:21) x f ′ ( x ) f ( y ) dxdy − N Z ∞ Z ∞ B ( α − ( x, y ) (cid:20)Z x J α − ( z ) dz − (cid:21) (cid:20)Z y J α − ( z ) dz (cid:21) x f ′ ( x ) f ( y ) dxdy + 116 N Z ∞ Z ∞ (cid:20)Z ∞ x B ( α − ( y, z ) dz − Z x B ( α − ( y, z ) dz (cid:21) (cid:20)Z x J α − ( z ) dz (cid:21) J α − ( y ) x f ′ ( x ) f ( y ) dxdy + 116 N Z ∞ Z ∞ (cid:20)Z x J α − ( z ) dz (cid:21) (cid:20)Z y J α − ( z ) dz (cid:21) (cid:20)Z x J α − ( z ) dz − (cid:21) J α − ( y ) x f ′ ( x ) f ( y ) dxdy + O (cid:0) N − (cid:1) , N → ∞ . Now we want to see the mean and variance of the linear statistics P Nj =1 F (cid:0)p N x j (cid:1) , so we needto obtain the coefficients of λ and λ , firstly we know f (cid:16) √ N x (cid:17) ≈ − λF (cid:16) √ N x (cid:17) + λ F (cid:16) √ N x (cid:17) , then we replace f with − λF + λ F in the expression of Tr T and Tr T , similar to previousdiscussions, denote by µ ( LSE, α ) N and V ( LSE, α ) N the mean and variance of the linear statistics P Nj =1 F (cid:0)p N x j (cid:1) ,we have the following theorem. Theorem 3.14. As N → ∞ , µ ( LSE, α ) N = 12 µ ( LUE, α − N − Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) J α − ( x ) F ( x ) dx − N Z ∞ (cid:20)Z ∞ x B ( α − ( x, y ) dy − Z x B ( α − ( x, y ) dy (cid:21) x F ′ ( x ) dx − N Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) (cid:20)Z x J α − ( y ) dy − (cid:21) x F ′ ( x ) dx + O (cid:0) N − (cid:1) , ( LSE, α ) N = 12 V ( LUE, α − N − Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) J α − ( x ) F ( x ) dx + 12 Z ∞ Z ∞ B ( α − ( x, y ) J α − ( x ) (cid:20)Z y J α − ( z ) dz (cid:21) F ( x ) F ( y ) dxdy − Z ∞ Z ∞ (cid:20)Z x J α − ( z ) dz (cid:21) (cid:20)Z y J α − ( z ) dz (cid:21) J α − ( x ) J α − ( y ) F ( x ) F ( y ) dxdy − N Z ∞ (cid:20)Z ∞ x B ( α − ( x, y ) dy − Z x B ( α − ( x, y ) dy (cid:21) x F ( x ) F ′ ( x ) dx − N Z ∞ (cid:20)Z x J α − ( y ) dy (cid:21) (cid:20)Z x J α − ( y ) dy − (cid:21) x F ( x ) F ′ ( x ) dx + 116 N Z ∞ Z ∞ B ( α − ( x, y ) (cid:20)Z ∞ x B ( α − ( y, z ) dz − Z x B ( α − ( y, z ) dz (cid:21) x F ′ ( x ) F ( y ) dxdy + 116 N Z ∞ Z ∞ B ( α − ( x, y ) (cid:20)Z x J α − ( z ) dz − (cid:21) (cid:20)Z y J α − ( z ) dz (cid:21) x F ′ ( x ) F ( y ) dxdy − N Z ∞ Z ∞ (cid:20)Z ∞ x B ( α − ( y, z ) dz − Z x B ( α − ( y, z ) dz (cid:21) (cid:20)Z x J α − ( z ) dz (cid:21) J α − ( y ) x F ′ ( x ) F ( y ) dxdy − N Z ∞ Z ∞ (cid:20)Z x J α − ( z ) dz (cid:21) (cid:20)Z y J α − ( z ) dz (cid:21) (cid:20)Z x J α − ( z ) dz − (cid:21) J α − ( y ) x F ′ ( x ) F ( y ) dxdy + O (cid:0) N − (cid:1) , where µ ( LUE, α − N and V ( LUE, α − N for N → ∞ are given in (1.4) and (1.5) respectively with α replaced by α − . For the orthogonal ensembles, β = 1, the equation (1.1), becomes, G (1) N ( f ) = C (1) N Z [ a,b ] n Y ≤ j 48n the computations below, we replace f ( x ) by f (cid:16) √ N x (cid:17) and f ′ ( x ) by f ′ (cid:16) √ N x (cid:17) = d √ N dx f (cid:16) √ N x (cid:17) . Similarly to the GSE case, we have the following theorems. Theorem 4.5. lim N →∞ √ N S N (cid:18) x √ N , y √ N (cid:19) = sin( x − y ) π ( x − y ) , lim N →∞ √ N S N (cid:18) x √ N , x √ N (cid:19) = 1 π . Theorem 4.6. ϕ N − (cid:18) x √ N (cid:19) = − ( − N π − N − sin x + O (cid:16) N − (cid:17) , N → ∞ ,εϕ N (cid:18) x √ N (cid:19) = ( − N − π − N − sin x + O (cid:16) N − (cid:17) , N → ∞ ,εϕ N − (cid:18) x √ N (cid:19) = − − N − + O (cid:16) N − (cid:17) , N → ∞ ,ε ( ϕ N − f ) (cid:18) x √ N (cid:19) = − ( − N − π − N − (cid:20)Z x −∞ sin y f ( y ) dy − Z ∞ x sin y f ( y ) dy (cid:21) + O (cid:16) N − (cid:17) , N → ∞ . Using Theorem 4.5 and Theorem 4.6 to compute Tr T and Tr T as N → ∞ . we haveTr T = 1 π Z ∞−∞ (cid:2) f ( x ) + 2 f ( x ) (cid:3) dx − π √ N Z ∞−∞ (cid:20)Z ∞ x sin( x − y ) x − y f ( y ) dy − Z x −∞ sin( x − y ) x − y f ( y ) dy (cid:21) f ′ ( x ) dx − πN Z ∞−∞ sin x (cid:2) f ( x ) + 2 f ( x ) (cid:3) dx − ( − N √ πN Z ∞−∞ sin x f ′ ( x ) dx + O (cid:16) N − (cid:17) , N → ∞ , and Tr T = 1 π Z ∞−∞ Z ∞−∞ (cid:20) sin( x − y ) x − y (cid:21) (cid:2) f ( x ) + 2 f ( x ) (cid:3) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + 2 π √ N Z ∞−∞ Z ∞−∞ sin( x − y ) x − y Si( x − y ) f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy − π √ N Z ∞−∞ Z ∞−∞ (cid:20)Z ∞ x sin( y − z ) y − z f ( z ) dz − Z x −∞ sin( y − z ) y − z f ( z ) dz (cid:21) sin( x − y ) x − yf ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + O (cid:0) N − (cid:1) , N → ∞ . P Nj =1 F (cid:16) √ N x j (cid:17) , since f (cid:16) √ N x (cid:17) ≈ − λF (cid:16) √ N x (cid:17) + λ F (cid:16) √ N x (cid:17) , we replace f with − λF + λ F in the expression of Tr T and Tr T , we havelog det ( I + T )= − λ ( π Z ∞−∞ F ( x ) dx − πN Z ∞−∞ sin x F ( x ) dx − ( − N √ πN Z ∞−∞ sin x F ′ ( x ) dx + O (cid:16) N − (cid:17) ) + λ ( π Z ∞−∞ F ( x ) dx − π Z ∞−∞ Z ∞−∞ (cid:20) sin( x − y ) x − y (cid:21) F ( x ) F ( y ) dxdy − π √ N Z ∞−∞ (cid:20)Z ∞ x sin( x − y ) x − y F ( y ) dy − Z x −∞ sin( x − y ) x − y F ( y ) dy (cid:21) F ′ ( x ) dx − π √ N Z ∞−∞ Z ∞−∞ sin( x − y ) x − y Si( x − y ) F ′ ( x ) F ( y ) dxdy + O (cid:0) N − (cid:1) ) , N → ∞ . Denote by µ ( GOE ) N and V ( GOE ) N the mean and variance of the linear statistics P Nj =1 F (cid:16) √ N x j (cid:17) , thenwe have the following theorem. Theorem 4.7. As N → ∞ , µ ( GOE ) N = µ ( GUE ) N − πN Z ∞−∞ sin x F ( x ) dx − ( − N √ πN Z ∞−∞ sin x F ′ ( x ) dx + O (cid:16) N − (cid:17) , V ( GOE ) N = 2 V ( GUE ) N − π √ N Z ∞−∞ (cid:20)Z ∞ x sin( x − y ) x − y F ( y ) dy − Z x −∞ sin( x − y ) x − y F ( y ) dy (cid:21) F ′ ( x ) dx − π √ N Z ∞−∞ Z ∞−∞ sin( x − y ) x − y Si( x − y ) F ′ ( x ) F ( y ) dxdy + O (cid:0) N − (cid:1) , where µ ( GUE ) N and V ( GUE ) N for N → ∞ are given in (1.2) and (1.3) respectively. We now specialize results obtained in Section 4.1 to the situation where w is taken to be the squareroot of the Laguerre weight, namely, w ( x ) = x α e − x , α > − , x ∈ R + . Again we choose a special ψ j so that M (1) is as simple as possible. Let ψ n +1 ( x ) := ddx ϕ ( α +1)2 n ( x ) , ψ n ( x ) := e ϕ ( α +1)2 n ( x ) , n = 0 , , , . . . , (4.4)50here ϕ ( α +1) j ( x ) and e ϕ ( α +1) j ( x ) are given by ϕ ( α +1) j ( x ) := L ( α +1) j ( x ) c ( α +1) j x α +1 e − x , j = 0 , , , . . . , (4.5) e ϕ ( α +1) j ( x ) := L ( α +1) j ( x ) c ( α +1) j x α e − x , j = 0 , , , . . . . It’s easy to see that Z ∞ ϕ ( α +1) j ( x ) e ϕ ( α +1) k ( x ) dx = δ jk , j, k = 0 , , , . . . . The next theorem shows that (4.4) satisfies (4.1), namely, ψ j ( x ) x − α e x is a polynomial of degree j . Theorem 4.8. ψ j ( x ) x − α e x , j = 0 , , , . . . is a polynomial of degree j .Proof. We prove this by considering two cases, j even and j odd. It is clear for even j , that ψ j x − α/ e x/ is up to a constant multiple of L ( α +1)2 j ( x ) . If j is odd, then we find, ψ n +1 ( x ) x − α e x = h ϕ ( α +1)2 n ( x ) i ′ x − α e x = c ( α +1)2 n L ( α +1)2 n ( x ) x α +1 e − x ! ′ x − α e x = 1 c ( α +1)2 n (cid:20) x (cid:16) L ( α +1)2 n ( x ) (cid:17) ′ + (cid:16) α (cid:17) L ( α +1)2 n ( x ) − xL ( α +1)2 n ( x ) (cid:21) , n = 0 , , , . . . . Thus ψ n +1 ( x ) x − α/ e x is a polynomial of degree 2 n + 1. The proof is complete.We use (4.4) to compute M (1) := (cid:0)R ∞ ψ j ( x ) εψ k ( x ) dx (cid:1) N − j,k =0 , resulting in the following theorem. Theorem 4.9. M (1) = · · · − · · · · · · − · · · ... ... ... ... ... ... · · · · · · − N × N . roof. Let m jk be the ( j, k )-entry of M (1) , again separating into four cases: ( j, k ) = (even , odd),(odd , even), (even , even) and (odd , odd). We find, for the (even , odd) case, m j, k +1 = Z ∞ ψ j ( x ) εψ k +1 ( x ) dx = Z ∞ e ϕ ( α +1)2 j ( x ) (cid:16) εDϕ ( α +1)2 k ( x ) (cid:17) dx = Z ∞ e ϕ ( α +1)2 j ( x ) ϕ ( α +1)2 k ( x ) dx = δ jk . For the (odd , even) case, note that M (1) is antisymmetric, then m j +1 , k = − m k, j +1 = − δ jk . The (even , even) case, m j, k = Z ∞ ψ j ( x ) εψ k ( x ) dx = Z ∞ e ϕ ( α +1)2 j ( x ) ε e ϕ ( α +1)2 k ( x ) dx = 12 c ( α +1)2 j c ( α +1)2 k Z ∞ L ( α +1)2 j ( x ) x α e − x (cid:18)Z x L ( α +1)2 k ( y ) y α e − y dy − Z ∞ x L ( α +1)2 k ( y ) y α e − y dy (cid:19) dx = 12 c ( α +1)2 j c ( α +1)2 k Z ∞ L ( α +1)2 j ( x ) x α e − x (cid:18) Z x L ( α +1)2 k ( y ) y α e − y dy − Z ∞ L ( α +1)2 k ( y ) y α e − y dy (cid:19) dx = 12 c ( α +1)2 j c ( α +1)2 k (cid:18) Z ∞ L ( α +1)2 j ( x ) x α e − x dx Z x L ( α +1)2 k ( y ) y α e − y dy − Z ∞ L ( α +1)2 j ( x ) x α e − x dx Z ∞ L ( α +1)2 k ( y ) y α e − y dy (cid:19) = 12 c ( α +1)2 j c ( α +1)2 k " α +2 Γ (cid:0) j + 1 + α (cid:1) Γ (cid:0) k + 1 + α (cid:1) j ! k ! − α +2 Γ (cid:0) j + 1 + α (cid:1) Γ (cid:0) k + 1 + α (cid:1) j ! k ! = 0 , where we have used the fact Z ∞ L ( α +1)2 j ( x ) x α e − x dx = 2 α +1 Γ (cid:0) j + 1 + α (cid:1) j ! , j = 0 , , , . . . . , odd) case. If j < k , m j +1 , k +1 = Z ∞ ψ j +1 ( x ) εψ k +1 ( x ) dx = Z ∞ h ϕ ( α +1)2 j ( x ) i ′ εDϕ ( α +1)2 k ( x ) dx = Z ∞ h ϕ ( α +1)2 j ( x ) i ′ ϕ ( α +1)2 k ( x ) dx = Z ∞ c ( α +1)2 j (cid:20) x (cid:16) L ( α +1)2 j ( x ) (cid:17) ′ + (cid:16) α (cid:17) L ( α +1)2 j ( x ) − xL ( α +1)2 j ( x ) (cid:21) x α e − x c ( α +1)2 k L ( α +1)2 k ( x ) x α +1 e − x dx = 1 c ( α +1)2 j c ( α +1)2 k Z ∞ L ( α +1)2 k ( x ) (cid:20) x (cid:16) L ( α +1)2 j ( x ) (cid:17) ′ + (cid:16) α (cid:17) L ( α +1)2 j ( x ) − xL ( α +1)2 j ( x ) (cid:21) x α +1 e − x dx = 0 , since, 2 j + 1, the degree of the polynomials x (cid:16) L ( α +1)2 j ( x ) (cid:17) ′ + (cid:0) α + 1 (cid:1) L ( α +1)2 j ( x ) − xL ( α +1)2 j ( x ), is lessthan 2 k .If j > k , due to the fact that M (1) is antisymmetric, m j +1 , k +1 = − m k +1 , j +1 = 0 , hence m j +1 , k +1 = 0 , j, k = 0 , , . . . . It is the desired result for M (1) . The proof is complete.We begin here a series of computations analogues to those in derivation of the LSE problem,ending up with an expression for h G (1) N ( f ) i as a scalar Fredholm determinant.It’s obvious that (cid:0) M (1) (cid:1) − = − M (1) , so µ j, j +1 = − , µ j +1 , j = 1, and µ jk = 0 for other cases,hence K (2 , N, ( x, y ) = N − X j,k =0 µ jk εψ j ( x ) ψ k ( y )= − N − X j =0 εψ j ( x ) ψ j +1 ( y ) + N − X j =0 εψ j +1 ( x ) ψ j ( y )= N − X j =0 ϕ ( α +1)2 j ( x ) e ϕ ( α +1)2 j ( y ) − N − X j =0 ε e ϕ ( α +1)2 j ( x ) h ϕ ( α +1)2 j ( y ) i ′ . x , we find, h ϕ ( α +1) j ( x ) i ′ = 1 c ( α +1) j (cid:20) x (cid:16) L ( α +1) j ( x ) (cid:17) ′ + α + 2 − x L ( α +1) j ( x ) (cid:21) x α e − x . Recall that the Laguerre polynomials L ( α +1) j satisfy the differentiation formulas [8] x (cid:16) L ( α +1) j ( x ) (cid:17) ′ = jL ( α +1) j ( x ) − ( j + α + 1) L ( α +1) j − ( x ) , j = 0 , , , . . . , (4.6) x (cid:16) L ( α +1) j ( x ) (cid:17) ′ = ( j + 1) L ( α +1) j +1 ( x ) − ( j + α + 2 − x ) L ( α +1) j ( x ) , j = 0 , , , . . . . (4.7)The sum of (4.6) and (4.7) divided by 2, gives, x (cid:16) L ( α +1) j ( x ) (cid:17) ′ = j + 12 L ( α +1) j +1 ( x ) − j + α + 12 L ( α +1) j − ( x ) − α + 2 − x L ( α +1) j ( x ) , j = 0 , , , . . . , which is the same as, x (cid:16) L ( α +1) j ( x ) (cid:17) ′ + α + 2 − x L ( α +1) j ( x ) = j + 12 L ( α +1) j +1 ( x ) − j + α + 12 L ( α +1) j − ( x ) . Hence the derivative of ϕ ( α +1) j ( x ) becomes, h ϕ ( α +1) j ( x ) i ′ = 1 c ( α +1) j (cid:20) j + 12 L ( α +1) j +1 ( x ) − j + α + 12 L ( α +1) j − ( x ) (cid:21) x α e − x = 12 p ( j + 1)( j + α + 2) e ϕ ( α +1) j +1 ( x ) − p j ( j + α + 1) e ϕ ( α +1) j − ( x ) . (4.8)So replacing j by 2 j , we see that, h ϕ ( α +1)2 j ( y ) i ′ = s(cid:18) j + 12 (cid:19) (cid:18) j + α + 22 (cid:19) e ϕ ( α +1)2 j +1 ( y ) − s j (cid:18) j + α + 12 (cid:19) e ϕ ( α +1)2 j − ( y ) . Continuing, we compute the summation to give, N − X j =0 ε e ϕ ( α +1)2 j ( x ) h ϕ ( α +1)2 j ( y ) i ′ = N − X j =0 s(cid:18) j + 12 (cid:19) (cid:18) j + α + 22 (cid:19) ε e ϕ ( α +1)2 j ( x ) e ϕ ( α +1)2 j +1 ( y ) − N − X j =1 s j (cid:18) j + α + 12 (cid:19) ε e ϕ ( α +1)2 j ( x ) e ϕ ( α +1)2 j − ( y )= N X j =1 "s(cid:18) j − (cid:19) (cid:16) j + α (cid:17) ε e ϕ ( α +1)2 j − ( x ) − s j (cid:18) j + α + 12 (cid:19) ε e ϕ ( α +1)2 j ( x ) ϕ ( α +1)2 j − ( y )+ 12 p N ( N + α + 1) ε e ϕ ( α +1) N ( x ) e ϕ ( α +1) N − ( y ) . ϕ ( α +1)2 j − ( x ) = ε h ϕ ( α +1)2 j − ( x ) i ′ = s j (cid:18) j + α + 12 (cid:19) ε e ϕ ( α +1)2 j ( x ) − s(cid:18) j − (cid:19) (cid:16) j + α (cid:17) ε e ϕ ( α +1)2 j − ( x ) . The sum simplifies to N − X j =0 ε e ϕ ( α +1)2 j ( x ) h ϕ ( α +1)2 j ( y ) i ′ = − N X j =1 ϕ ( α +1)2 j − ( x ) e ϕ ( α +1)2 j − ( y ) + 12 p N ( N + α + 1) ε e ϕ ( α +1) N ( x ) e ϕ ( α +1) N − ( y ) . The computations above gives a compact form for K (2 , N, ( x, y ) : K (2 , N, ( x, y ) = N − X j =0 ϕ ( α +1)2 j ( x ) e ϕ ( α +1)2 j ( y ) + N X j =1 ϕ ( α +1)2 j − ( x ) e ϕ ( α +1)2 j − ( y ) − p N ( N + α + 1) ε e ϕ ( α +1) N ( x ) e ϕ ( α +1) N − ( y )= N − X j =0 ϕ ( α +1) j ( x ) e ϕ ( α +1) j ( y ) − p N ( N + α + 1) ε e ϕ ( α +1) N ( x ) e ϕ ( α +1) N − ( y )= S N ( x, y ) − p N ( N + α + 1) ε e ϕ ( α +1) N ( x ) e ϕ ( α +1) N − ( y ) , where S N ( x, y ) := N − X j =0 ϕ ( α +1) j ( x ) e ϕ ( α +1) j ( y ) = − p N ( N + α + 1) ϕ ( α +1) N ( x ) e ϕ ( α +1) N − ( y ) − e ϕ ( α +1) N ( y ) ϕ ( α +1) N − ( x ) x − y . Here we have used the Christoffel-Darboux formula in the last equality.By Theorem 4.2, we have the following theorem. Theorem 4.10. h G (1) N ( f ) i = det (cid:18) I + S N ( f + 2 f ) − S N εf ′ − S N f εf ′ − p N ( N + α + 1) h ε e ϕ ( α +1) N ⊗ e ϕ ( α +1) N − (cid:0) f + 2 f (cid:1) i − p N ( N + α + 1) (cid:16) ε e ϕ ( α +1) N ⊗ ε e ϕ ( α +1) N − (cid:17) f ′ − p N ( N + α + 1) h ε e ϕ ( α +1) N ⊗ ε (cid:16) e ϕ ( α +1) N − f (cid:17)i f ′ (cid:19) . N behavior of the LOE moment generating function Now we consider the scaling limit of h G (1) N ( f ) i . We write h G (1) N ( f ) i =: det( I + T ) , T : = S N ( f + 2 f ) − S N εf ′ − S N f εf ′ − p N ( N + α + 1) h ε e ϕ ( α +1) N ⊗ e ϕ ( α +1) N − (cid:0) f + 2 f (cid:1) i − p N ( N + α + 1) (cid:16) ε e ϕ ( α +1) N ⊗ ε e ϕ ( α +1) N − (cid:17) f ′ − p N ( N + α + 1) h ε e ϕ ( α +1) N ⊗ ε (cid:16) e ϕ ( α +1) N − f (cid:17)i f ′ . In the computations below, we replace f ( x ) by f (cid:16) √ N x (cid:17) and f ′ ( x ) by f ′ (cid:16) √ N x (cid:17) = d √ N d √ x f (cid:16) √ N x (cid:17) . Similarly to the LSE case, we have the following two theorems. Theorem 4.11. lim N →∞ y N S N (cid:18) x N , y N (cid:19) = B ( α +1) ( x, y ) , lim N →∞ x N S N (cid:18) x N , x N (cid:19) = B ( α +1) ( x, x ) , where B ( α +1) ( x, y ) and B ( α +1) ( x, x ) are given by (1.6) and (1.7) with α replaced by α + 1 . Theorem 4.12. e ϕ ( α +1) N − (cid:18) x N (cid:19) = 2 N J α +1 ( x ) x + O (cid:16) N − (cid:17) , N → ∞ ,ε e ϕ ( α +1) N (cid:18) x N (cid:19) = N − (cid:20)Z x J α +1 ( y ) dy − (cid:21) + O (cid:16) N − (cid:17) , N → ∞ ,ε e ϕ ( α +1) N − (cid:18) x N (cid:19) = N − Z x J α +1 ( y ) dy + O (cid:16) N − (cid:17) , N → ∞ ,ε (cid:16) e ϕ ( α +1) N − f (cid:17) (cid:18) x N (cid:19) = 12 N − (cid:20)Z x J α +1 ( y ) f ( y ) dy − Z ∞ x J α +1 ( y ) f ( y ) dy (cid:21) + O (cid:16) N − (cid:17) , N → ∞ . T and Tr T as N → ∞ . We obtainTr T = Z ∞ B ( α +1) ( x, x ) (cid:2) f ( x ) + 2 f ( x ) (cid:3) dx − Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) J α +1 ( x ) (cid:2) f ( x ) + 2 f ( x ) (cid:3) dx − N Z ∞ (cid:20)Z ∞ x B ( α +1) ( x, y ) dy − Z x B ( α +1) ( x, y ) dy (cid:21) x f ′ ( x ) dx − N Z ∞ (cid:20)Z ∞ x B ( α +1) ( x, y ) f ( y ) dy − Z x B ( α +1) ( x, y ) f ( y ) dy (cid:21) x f ′ ( x ) dx − N Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) (cid:20)Z x J α +1 ( y ) dy (cid:21) x f ′ ( x ) dx − N Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) (cid:20)Z x J α +1 ( y ) f ( y ) dy − Z ∞ x J α +1 ( y ) f ( y ) dy (cid:21) x f ′ ( x ) dx + O (cid:0) N − (cid:1) , N → ∞ , T = Z ∞ Z ∞ B ( α +1) ( x, y ) B ( α +1) ( y, x ) (cid:2) f ( x ) + 2 f ( x ) (cid:3) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy − Z ∞ Z ∞ B ( α +1) ( x, y ) J α +1 ( x ) (cid:20)Z y J α +1 ( z ) dz − (cid:21) (cid:2) f ( x ) + 2 f ( x ) (cid:3) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + 14 Z ∞ Z ∞ (cid:20)Z x J α +1 ( z ) dz − (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) J α +1 ( x ) J α +1 ( y ) (cid:2) f ( x ) + 2 f ( x ) (cid:3) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy − N Z ∞ Z ∞ B ( α +1) ( x, y ) (cid:20)Z ∞ x B ( α +1) ( y, z ) dz − Z x B ( α +1) ( y, z ) dz (cid:21) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy − N Z ∞ Z ∞ B ( α +1) ( x, y ) (cid:20)Z ∞ x B ( α +1) ( y, z ) f ( z ) dz − Z x B ( α +1) ( y, z ) f ( z ) dz (cid:21) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy − N Z ∞ Z ∞ B ( α +1) ( x, y ) (cid:20)Z x J α +1 ( z ) dz (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy − N Z ∞ Z ∞ B ( α +1) ( x, y ) (cid:20)Z x J α +1 ( z ) f ( z ) dz − Z ∞ x J α +1 ( z ) f ( z ) dz (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + 14 N Z ∞ Z ∞ (cid:20)Z ∞ x B ( α +1) ( y, z ) dz − Z x B ( α +1) ( y, z ) dz (cid:21) (cid:20)Z x J α +1 ( z ) dz − (cid:21) J α +1 ( y ) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + 14 N Z ∞ Z ∞ (cid:20)Z ∞ x B ( α +1) ( y, z ) f ( z ) dz − Z x B ( α +1) ( y, z ) f ( z ) dz (cid:21) (cid:20)Z x J α +1 ( z ) dz − (cid:21) J α +1 ( y ) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + 14 N Z ∞ Z ∞ (cid:20)Z x J α +1 ( z ) dz − (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) (cid:20)Z x J α +1 ( z ) dz (cid:21) J α +1 ( y ) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + 18 N Z ∞ Z ∞ (cid:20)Z x J α +1 ( z ) dz − (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21)(cid:20)Z x J α +1 ( z ) f ( z ) dz − Z ∞ x J α +1 ( z ) f ( z ) dz (cid:21) J α +1 ( y ) x f ′ ( x ) (cid:2) f ( y ) + 2 f ( y ) (cid:3) dxdy + O (cid:0) N − (cid:1) , N → ∞ . Now we want to see the mean and variance of the linear statistics P Nj =1 F (cid:0)p N x j (cid:1) , so we needto obtain the coefficients of λ and λ , firstly we know f (cid:16) √ N x (cid:17) ≈ − λF (cid:16) √ N x (cid:17) + λ F (cid:16) √ N x (cid:17) , then we replace f with − λF + λ F in the expression of Tr T and Tr T , similar to previous58iscussions, denote by µ ( LOE, α ) N and V ( LOE, α ) N the mean and variance of the linear statistics P Nj =1 F (cid:0)p N x j (cid:1) ,then we have the following theorem. Theorem 4.13. As N → ∞ , µ ( LOE, α ) N = µ ( LUE, α +1) N − Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) J α +1 ( x ) F ( x ) dx − N Z ∞ (cid:20)Z ∞ x B ( α +1) ( x, y ) dy − Z x B ( α +1) ( x, y ) dy (cid:21) x F ′ ( x ) dx − N Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) (cid:20)Z x J α +1 ( y ) dy (cid:21) x F ′ ( x ) dx + O (cid:0) N − (cid:1) , V ( LOE, α ) N = 2 V ( LUE, α +1) N − Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) J α +1 ( x ) F ( x ) dx + 2 Z ∞ Z ∞ B ( α +1) ( x, y ) J α +1 ( x ) (cid:20)Z y J α +1 ( z ) dz − (cid:21) F ( x ) F ( y ) dxdy − Z ∞ Z ∞ (cid:20)Z x J α +1 ( z ) dz − (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) J α +1 ( x ) J α +1 ( y ) F ( x ) F ( y ) dxdy − N Z ∞ (cid:20)Z ∞ x B ( α +1) ( x, y ) dy − Z x B ( α +1) ( x, y ) dy (cid:21) x F ( x ) F ′ ( x ) dx − N Z ∞ (cid:20)Z ∞ x B ( α +1) ( x, y ) F ( y ) dy − Z x B ( α +1) ( x, y ) F ( y ) dy (cid:21) x F ′ ( x ) dx − N Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) (cid:20)Z x J α +1 ( y ) dy (cid:21) x F ( x ) F ′ ( x ) dx − N Z ∞ (cid:20)Z x J α +1 ( y ) dy − (cid:21) (cid:20)Z x J α +1 ( y ) F ( y ) dy − Z ∞ x J α +1 ( y ) F ( y ) dy (cid:21) x F ′ ( x ) dx + 12 N Z ∞ Z ∞ B ( α +1) ( x, y ) (cid:20)Z ∞ x B ( α +1) ( y, z ) dz − Z x B ( α +1) ( y, z ) dz (cid:21) x F ′ ( x ) F ( y ) dxdy + 12 N Z ∞ Z ∞ B ( α +1) ( x, y ) (cid:20)Z x J α +1 ( z ) dz (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) x F ′ ( x ) F ( y ) dxdy − N Z ∞ Z ∞ (cid:20)Z ∞ x B ( α +1) ( y, z ) dz − Z x B ( α +1) ( y, z ) dz (cid:21) (cid:20)Z x J α +1 ( z ) dz − (cid:21) J α +1 ( y ) x F ′ ( x ) F ( y ) dxdy − N Z ∞ Z ∞ (cid:20)Z x J α +1 ( z ) dz − (cid:21) (cid:20)Z y J α +1 ( z ) dz − (cid:21) (cid:20)Z x J α +1 ( z ) dz (cid:21) J α +1 ( y ) x F ′ ( x ) F ( y ) dxdy + O (cid:0) N − (cid:1) , where µ ( LUE, α +1) N and V ( LUE, α +1) N for N → ∞ are given in (1.4) and (1.5) respectively with α replaced by α + 1 . 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