Integrals of Functions of Matrix Arguments and Their ... - Hikari Ltd

International Journal of Mathematical Analysis Vol. 11, 2017, no. 10, 479 - 492 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/ijma.2017.7462

Integrals of Functions of Matrix Arguments and Their Applications Daya K. Nagar, Sergio A. G´ omez-Noguera and Armando G´ omez Instituto de Matemáticas, Universidad de Antioquia Calle 67, No. 53–108, Medell´ın, Colombia (S.A.) c 2017 Daya K. Nagar, Sergio A. Gómez-Noguera and Armando Gómez. Copyright This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract In this article, we evaluate several integrals of scalar valued functions of matrix arguments. These results are useful in computing expected values of trace functions of random matrices.

Mathematics Subject Classification: 62H99, 60E05 Keywords: multivariate beta function; multivariate gamma function; random matrix; trace; zonal polynomials

1

Introduction

Several generalizations of the Euler’s gamma function are available in the scientific literature. The multivariate gamma function, which is frequently used in multivariate statistical analysis, is defined by Z Γm (a) = etr(−X) det(X)a−(m+1)/2 dX, X>0

where the integration is carried out over m × m symmetric positive definite matrices. By evaluating the above integral it is easy to see that m Y m−1 i−1 m(m−1)/4 , Re(a) > . Γm (a) = π Γ a− 2 2 i=1

480

Daya K. Nagar, Sergio A. Gómez-Noguera and Armando Gómez

The multivariate generalization of the beta function is given by Z Im det(X)a−(m+1)/2 det(Im − X)b−(m+1)/2 dX Bm (a, b) = 0

Γm (a)Γm (b) = = Bm (b, a), Γm (a + b) where Re(a) > (m − 1)/2 and Re(b) > (m − 1)/2. The multivariate gamma function defined above arises naturally in the probability density function (p.d.f.) of the Wishart matrix. If W ∼ Wm (n, Σ), n > m − 1, Σ > 0, then its p.d.f. is given by etr (−Σ−1 W /2) det(W )(n−m−1)/2 , 2nm/2 Γm (n/2) det(Σ)n/2

W > 0.

Further, if W1 ∼ Wm (n1 , Im ) and W2 ∼ Wm (n2 , Im ) are independent Wishart matrices, then the p.d.f.’s of X = (W1 + W2 )−1/2 W1 (W1 + W2 )−1/2 and Y = −1/2 −1/2 W2 W1 W2 are given by det(X)n1 /2−(m+1)/2 det(Im − X)n2 /2−(m+1)/2 , Bm (n1 /2, n2 /2)

0 < X < Im

and det(Y )n1 /2−(m+1)/2 det(Im + Y )−(n1 +n2 )/2 , Bm (n1 /2, n2 /2)

Y > 0,

respectively. Thus multivariate gamma and multivariate beta functions frequently occur in the densities of random matrices (e.g., see Gupta and Nagar [2]) and hence play pivotal role in multivariate statistical analysis. In this article we evaluate integrals which are closely connected to multivariate gamma and multivariate beta integrals. In multivariate statistical analysis these results are useful in computing expected values of trace functions of random matrices. Define f1a , f1b , f2a and f2b as Z etr(−B −1 X) det(X)a−(m+1)/2 f1a (g, h, i, j) = (tr X h )g (tr X j )i dX, det(B)a Γm (a) X>0 Z f1b (g, h, i, j) = X>0

etr(−B −1 X) det(X)a−(m+1)/2 (tr X −h )g (tr X −j )i dX, det(B)a Γm (a)

f2a (g, h, i, j) =

Im

det(X)a−(m+1)/2 det(Im − X)b−(m+1)/2 B(a, b) 0 h g × (tr(BX) ) (tr(BX)j )i dX Z

Integrals of functions of matrix arguments and their applications

481

and f2b (g, h, i, j) =

Im

det(X)a−(m+1)/2 det(Im − X)b−(m+1)/2 B(a, b) 0 −1 h g × (tr(BX ) ) (tr(BX −1 )j )i dX, Z

where g, h, i, j are non-negative integers. In sections 3,4,5, and 6, respectively, for specific values of g, h, i, j, we give explicit evaluations of f1a , f1b , f2a and f2b .

2

Some Known Results

Let Cκ (X) be a zonal polynomial of an m × m symmetric matrix X corresponding to the ordered partition κ = (k1 , . . . , km ), k1 + · · · + km = k, k1 ≥ · · · ≥ km ≥ 0. For small values of k, explicit formulas for Cκ (X) are available as (James [4]), C(1) (X) = tr(X), 1 C(2) (X) = (tr X)2 + 2 tr(X 2 ) , 3 2 C(12 ) (X) = [(tr X)2 − tr(X 2 )], 3 1 C(3) (X) = [(tr X)3 + 6(tr X)(tr X 2 ) + 8 tr(X 3 )], 15 3 C(2,1) (X) = [(tr X)3 + (tr X)(tr X 2 ) − 2 tr(X 3 )], 5 1 C(13 ) (X) = [(tr X)3 − 3(tr X)(tr X 2 ) + 2 tr(X 3 )], 3 1 C(4) (X) = [(tr X)4 + 12(tr X)2 (tr X 2 ) + 12(tr X 2 )2 105 + 32(tr X)(tr X 3 ) + 48(tr X 4 )], 20 C(3,1) (X) = [(tr X)4 + 5(tr X)2 (tr X 2 ) − 2(tr X 2 )2 105 + 4(tr X)(tr X 3 ) − 8(tr X 4 )], 14 C(22 ) (X) = [(tr X)4 + 2(tr X)2 (tr X 2 ) + 7(tr X 2 )2 105 − 8(tr X)(tr X 3 ) − 2(tr X 4 )], 56 C(2,12 ) (X) = [(tr X)4 − (tr X)2 (tr X 2 ) − 2(tr X 2 )2 105 − 2(tr X)(tr X 3 ) + 4(tr X 4 )], 14 C(14 ) (X) = [(tr X)4 − 6(tr X)2 (tr X 2 ) + 3(tr X 2 )2 105

482


+ 8(tr X)(tr X 3 ) − 6(tr X 4 )]. 1 [(tr X)5 + 20(tr X)3 (tr X 2 ) + 60(tr X)(tr X 2 )2 C(5) (X) = 945 + 80(tr X)2 (tr X 3 ) + 160(tr X 2 )(tr X 3 ) + 240(tr X)(tr X 4 ) + 384(tr X 5 )], 35 C(4,1) (X) = [(tr X)5 + 11(tr X)3 (tr X 2 ) + 6(tr X)(tr X 2 )2 945 + 26(tr X)2 (tr X 3 ) − 20(tr X 2 )(tr X 3 ) + 24(tr X)(tr X 4 ) − 48(tr X 5 )], 90 C(3,2) (X) = [(tr X)5 + 6(tr X)3 (tr X 2 ) + 11(tr X)(tr X 2 )2 945 − 4(tr X)2 (tr X 3 ) + 20(tr X 2 )(tr X 3 ) − 26(tr X)(tr X 4 ) − 8(tr X 5 )], 225 C(3,12 ) (X) = [(tr X)5 + 3(tr X)3 (tr X 2 ) − 10(tr X)(tr X 2 )2 945 + 2(tr X)2 (tr X 3 ) − 4(tr X 2 )(tr X 3 ) − 8(tr X)(tr X 4 ) + 16(tr X 5 )], 252 C(2,2,1) (X) = [(tr X)5 + 5(tr X)(tr X 2 )2 − 10(tr X)2 (tr X 3 ) 945 − 10(tr X 2 )(tr X 3 ) + 10(tr X)(tr X 4 ) + 4(tr X 5 )], 300 C(2,13 ) (X) = [(tr X)5 − 4(tr X)3 (tr X 2 ) − 3(tr X)(tr X 2 )2 945 + 2(tr X)2 (tr X 3 ) + 10(tr X 2 )(tr X 3 ) + 6(tr X)(tr X 4 ) − 12(tr X 5 )], 42 C(15 ) (X) = [(tr X)5 − 10(tr X)3 (tr X 2 ) + 15(tr X)(tr X 2 )2 945 + 20(tr X)2 (tr X 3 ) − 20(tr X 2 )(tr X 3 ) − 30(tr X)(tr X 4 ) + 24(tr X 5 )], In an unpublished article, Kitchen [5] has tabulated zonal polynomials for k = 7 through k = 9 and has also included tables for k ≤ 6 given earlier by James [4]. From the above results, it is straightforward to show that (also see Mathai, Provost and Hayakawa [3]), 1 tr(X 2 ) = C(2) (X) − C(12 ) (X), 2 1 1 tr(X 3 ) = C(3) (X) − C(2,1) (X) + C(13 ) (X), 4 4 1 1 2 tr(X) tr(X ) = C(3) (X) + C(2,1) (X) − C(13 ) (X), 6 2 5 1 1 1 (trX)2 tr(X 2 ) = C(4) (X)+ C(3,1) (X)+ C(22 ) (X)− C(2,12 ) (X)− C(14 ) (X), 12 6 12 2


483

7 1 1 1 (trX 2 )2 = C(4) (X)− C(3,1) (X)+ C(2,2) (X)− C(2,12 ) (X)+ C(14 ) (X), 6 12 6 4 1 1 1 1 3 (trX) tr(X ) = C(4) (X)+ C(3,1) (X)− C(22 ) (X)− C(2,12 ) (X) + C(14 ) (X), 8 4 16 4 1 1 1 1 4 tr(X ) = C(4) (X)− C(3,1) (X)− C(22 ) (X)+ C(2,12 ) (X)− C(14 ) (X), 6 24 12 8 11 3 3 3 2 (tr X) tr(X ) = C(5) (X) + C(4,1) (X) + C(3,2) (X) + C(3,12 ) (X) 20 10 20 1 1 − C(2,13 ) (X) − C(15 ) (X), 5 2 1 11 1 (tr X)(tr X 2 )2 = C(5) (X) + C(4,1) (X) + C(3,2) (X) − C(3,12 ) (X) 10 60 6 1 1 1 + C(22 ,1) (X) − C(2,13 ) (X) + C(15 ) (X), 12 20 4 13 1 1 2 3 (tr X) tr(X ) = C(5) (X) + C(4,1) (X) − C(3,2) (X) + C(3,12 ) (X) 40 20 40 1 1 1 − C(22 ,1) (X) + C(2,13 ) (X) + C(15 ) (X), 8 40 4 1 1 1 2 3 tr(X ) tr(X ) = C(5) (X) − C(4,1) (X) + C(3,2) (X) − C(3,12 ) (X) 8 8 40 1 1 1 − C(22 ,1) (X) + C(2,13 ) (X) − C(15 ) (X), 16 16 8 13 1 1 4 C(3,2) (X) − C(3,12 ) (X) (tr X) tr(X ) = C(5) (X) + C(4,1) (X) − 10 120 30 1 1 1 + C(22 ,1) (X) + C(2,13 ) (X) − C(15 ) (X), 24 40 8 1 1 1 5 tr(X ) = C(5) (X) − C(4,1) (X) − C(3,2) (X) + C(3,12 ) (X) 8 48 24 1 1 1 + C(22 ,1) (X) − C(2,13 ) (X) + C(15 ) (X). 96 32 16 The generalized hypergeometric coefficient (a)κ is defined as m Y i−1 (a)κ = a− , (1) 2 ki i=1 with (a)r = a(a + 1) · · · (a + r − 1) = (a)r−1 (a + r − 1) for r = 1, 2, . . ., and (a)0 = 1. Using (1), computation of (a)κ has been done for all the ordered partition of k. These coefficients are given in Table 1 for k ≤ 5. For an ordered partition ρ of r, ρ = (r1 , . . . , rm ), r1 ≥ · · · ≥ rm ≥ 0, r1 + · · · + rm = r, Γm (a, ρ) and Γm (a, −ρ), are defined by Γm (a, ρ) = (a)ρ Γm (a),

Γm (a, 0) = Γm (a)

(2)

and Γm (a, −ρ) =

(−1)r Γm (a) , (−a + (m + 1)/2)ρ

Re(a) > r1 +

m−1 , 2

(3)

484


(a)(1) (a)(2) (a)(12 ) (a)(3) (a)(2,1) (a)(13 ) (a)(4) (a)(3,1) (a)(22 ) (a)(2,12 ) (a)(14 ) (a)(5) (a)(4,1) (a)(3,2) (a)(3,12 ) (a)(22 ,1) (a)(2,13 ) (a)(15 )

Table 1: Values of (a)κ a a(a + 1) a(a − 1/2) a(a + 1)(a + 2) a(a + 1)(a − 1/2) a(a − 1/2)(a − 1) a(a + 1)(a + 2)(a + 3) a(a + 1)(a + 2)(a − 1/2) a(a + 1/2)(a + 1)(a − 1/2) a(a + 1)(a − 1)(a − 1/2) a(a − 1/2)(a − 1)(a − 3/2) a(a + 1)(a + 2)(a + 3)(a + 4) a(a + 1)(a + 2)(a + 3)(a − 1/2) (a + 1/2)(a + 1)a(a − 1/2)(a + 2) (a − 1)(a + 1)a(a − 1/2)(a + 2) (a − 1)(a − 1/2)a(a + 1/2)(a + 1) (a − 3/2)(a − 1)(a − 1/2)a(a + 1) (a − 2)(a − 3/2)(a − 1)(a − 1/2)a

respectively. The coefficients (−a+(m+1)/2)ρ , for r ≤ 5, are given in Table 2. Lemma 2.1. Let Z be an m × m complex symmetric matrix whose real part is positive definite and let T be an m × m arbitrary complex symmetric matrix. Then Z etr(−ZR) det(R)a−(m+1)/2 Cκ (T R) dR R>0

= Γm (a, κ) det(Z)−a Cκ (T Z −1 ),

Re(a) >

m−1 . 2

(4)

Lemma 2.2. Let Z be an m × m complex symmetric matrix whose real part is positive definite and let T be an m × m arbitrary complex symmetric matrix. Then Z etr(−ZR) det(R)a−(m+1)/2 Cκ (T R−1 ) dR R>0

= Γm (a, κ) det(Z)−a Cκ (T Z) =

(−1)k Γm (a) det(Z)−a Cκ (T Z), (−a + (m + 1)/2)κ

Re(a) > k1 +

m−1 . 2

(5)

(−a + (m + 1)/2)(1) (−a + (m + 1)/2)(2) (−a + (m + 1)/2)(12 ) (−a + (m + 1)/2)(3) (−a + (m + 1)/2)(2,1) (−a + (m + 1)/2)(13 ) (−a + (m + 1)/2)(4) (−a + (m + 1)/2)(3,1) (−a + (m + 1)/2)(22 ) (−a + (m + 1)/2)(2,12 ) (−a + (m + 1)/2)(14 ) (−a + (m + 1)/2)(5) (−a + (m + 1)/2)(4,1) (−a + (m + 1)/2)(3,2) (−a + (m + 1)/2)(3,12 ) (−a + (m + 1)/2)(22 ,1) (−a + (m + 1)/2)(2,13 ) (−a + (m + 1)/2)(15 )

Table 2: Values of (−a + (m + 1)/2)κ −(a − (m + 1)/2) (a − (m + 1)/2)(a − (m + 3)/2) (a − (m + 1)/2)(a − m/2) −(a − (m + 1)/2)(a − (m + 3)/2)(a − (m + 5)/2) −(a − (m + 1)/2)(a − (m + 3)/2)(a − m/2) −(a − (m + 1)/2)(a − m/2)(a − (m − 1)/2) (a − (m + 1)/2)(a − (m + 3)/2)(a − (m + 5)/2)(a − (m + 7)/2) (a − (m + 1)/2)(a − (m + 3)/2)(a − (m + 5)/2)(a − m/2) (a − (m + 1)/2)(a − (m + 2)/2)(a − (m + 3)/2)(a − m/2) (a − (m + 1)/2)(a − (m + 3)/2)(a − (m − 1)/2)(a − m/2) (a − (m + 1)/2)(a − m/2)(a − (m − 1)/2)(a − (m − 2)/2) −(a − (m + 1)/2)(a − (m + 3)/2)(a − (m + 5)/2)(a − (m + 7)/2)(a − (m + 9)/2) −(a − (m + 1)/2)(a − (m + 3)/2)(a − (m + 5)/2)(a − (m + 7)/2)(a − m/2) −(a − (m + 2)/2)(a − (m + 3)/2)(a − (m + 1)/2)(a − m/2)(a − (m + 5)/2) −(a − (m − 1)/2)(a − (m + 3)/2)(a − (m + 1)/2)(a − m/2)(a − (m + 5)/2) −(a − (m + 3)/2)(a − m/2)(a − (m + 1)/2)(a − (m + 2)/2)(a − (m + 3)/2) −(a − (m − 2)/2)(a + (m − 1)/2)(a − m/2)(a − (m + 1)/2)(a − (m + 3)/2) −(a − (m − 3)/2)(a − (m − 2)/2)(a − (m − 1)/2)(a − m/2)(a − (m + 1)/2)


485

486


Lemma 2.3. Let T be an m × m symmetric matrix, then Z Im det(R)a−(m+1)/2 det(Im − R)b−(m+1)/2 Cκ (T R) dR 0

=

Γm (a, κ)Γm (b) Cκ (T ) Γm (a + b, κ)

(6)

Lemma 2.4. Let T be an m×m arbitrary complex symmetric matrix. Then Z Im det(R)a−(m+1)/2 det(Im − R)b−(m+1)/2 Cκ (R−1 T ) dR 0

Γm (a, −κ)Γm (b) Cκ (T ) Γm (a + b, −κ) Γm (a)Γm (b)(−a − b + (m + 1)/2)κ = Cκ (T ), (−a + (m + 1)/2)κ Γm (a + b) =

Re(a) > k1 +

m−1 . (7) 2

Lemma 2.1 and Lemma 2.3 are given in Constantine [1]. Results given in Lemma 2.2 and Lemma 2.4 were derived by Khatri [6].

3

Type 1(a) Integrals

Consider the integral Z f1a (1, 2, 0, −) = X>0 2

etr(−B −1 X) det(X)a−(m+1)/2 (tr X 2 ) dX. det(B)a Γm (a)

Writing (tr X ) in terms of zonal polynomials and integrating the resulting expression by using (4), we obtain Z etr(−B −1 X) det(X)a−(m+1)/2 C(2) (X) dX f1a (1, 2, 0, −) = det(B)a Γm (a) X>0 Z 1 etr(−B −1 X) det(X)a−(m+1)/2 − C(12 ) (X) dX 2 X>0 det(B)a Γm (a) 1 = (a)(2) C(2) (B) − (a)(12 ) C(12 ) (B). 2 Now, substituting for (a)(2) , (a)(12 ) , C(2) (B) and C(12 ) (B) above, we have a f1a (1, 2, 0, −) = [(tr B)2 + (2a + 1) tr(B 2 )]. 2 3 Writing (tr X ) in terms of zonal polynomials and integrating the resulting expression by using (4), we get Z etr(−B −1 X) det(X)a−(m+1)/2 (tr X 3 ) dX f1a (1, 3, 0, −) = a det(B) Γm (a) X>0 1 1 = (a)(3) C(3) (B) − (a)(2,1) C(2,1) (B) + (a)(13 ) C(13 ) (B). 4 4


487

Now, substituting for (a)(3) , (a)(2,1) (a)(13 ) , C(3) (B), C(2,1) (B) and C(13 ) (B) above, we have a f1a (1, 3, 0, −) = (tr B)3 + 3(2a + 1)(tr B) tr(B 2 ) + 2(2a2 + 3a + 2) tr(B 3 ) . 4 Following the procedure described above, we evaluate f1a (g, h, i, j) explicitly for selected values of g, h, i, j given by a f1a (1, 2, 1, 1) = a(tr B)3 + (2a2 + a + 2)(tr B) tr(B 2 ) + 2(1 + 2a) tr(B 3 ) . 2 a f1a (2, 1, 1, 2) = a2 (trB)4 + a(5 + a + 2a2 )(trB)2 (trB 2 ) + (2 + a + 2a2 )(trB 2 )2 2 + 4(1 + a + 2a2 )(tr B)(tr B 3 ) + 6(1 + 2a)(tr B 4 ) , f1a (2, 2, 0, −) =

f1a (1, 3, 1, 1) =

f1a (1, 4, 0,−) =

f1a (3, 1, 1, 2) =

a a(tr B)4 + 2(2 + a + 2a2 )(tr B)2 (tr B 2 ) 4 + (1 + 2a)(2 + a + 2a2 )(tr B 2 )2 + 8(1 + 2a)(tr B)(tr B 3 ) +2(5 + 2a(5 + 4a))(tr B 4 ) ,

a a(tr B)4 + 3(1 + a + 2a2 )(tr B)2 (tr B 2 ) + 3(1 + 2a)(tr B 2 )2 4 + 2(2a3 + 3a2 + 8a + 3)(tr B)(tr B 3 ) + 6(2a2 + 3a + 2)(tr B 4 ) ,

a (tr B)4 + 6(1 + 2a)(tr B)2 (tr B 2 ) + (5 + 10a + 8a2 )(tr B 2 )2 8 + 8(2a2 + 3a + 2)(trB)(trB 3 ) + 2(4a3 + 12a2 + 21a + 10)(trB 4 ) , a 3 a (tr B)5 + a2 (2a2 + a + 9)(tr B)3 (tr B 2 ) 2 + 3a(2a2 + a + 4)(trB)(trB 2 )2 + 2a(6a2 + 3a + 7)(trB)2 (trB 3 ) + 4(4a2 + 2a + 3)(tr B 2 )(tr B 3 ) + 6(6a2 + 3a + 2)(tr B)(tr B 4 ) + 24(2a + 1)(tr B 5 ) ,

f1a (2, 2, 1, 1) =

a 2 a (tr B)5 + 2a(2a2 + a + 4)(tr B)3 (tr B 2 ) 4 + (2a2 + a + 4)(2a2 + a + 2)(tr B)(tr B 2 )2 + 4(6a2 + 3a + 2)(tr B)2 (tr B 3 ) + 4(2a + 1)(2a2 + a + 4)(tr B 2 )(tr B 3 ) + (16a3 + 20a2 + 58a + 24)(tr B)(tr B 4 ) + 8(8a2 + 10a + 5)(tr B 5 ) ,

488


a 2 a (tr B)5 + a(6a2 + 3a + 7)(tr B)3 (tr B 2 ) 4 + 3(6a2 + 3a + 2)(tr B)(tr B 2 )2 + 2(2a4 + 3a3 + 14a2 + 6a + 3)(tr B)2 (tr B 3 ) + 2(2a3 + 3a2 + 20a + 9)(tr B 2 )(tr B 3 ) + 6(4a3 + 6a2 + 10a + 3)(tr B)(tr B 4 ) + 24(2a2 + 3a + 2)(tr B 5 ) , a f1a (1, 3, 1, 2) = a(tr B)5 + 2(4a2 + 2a + 3)(tr B)3 (tr B 2 ) 8 + 3(2a + 1)(2a2 + a + 4)(tr B)(tr B 2 )2 + 2(2a3 + 3a2 + 20a + 9)(tr B)2 (tr B 3 ) + 2(4a4 + 8a3 + 31a2 + 32a + 15)(tr B 2 )(tr B 3 ) + 6(12a2 + 16a + 9)(tr B)(tr B 4 ) + 24(2a3 + 5a2 + 7a + 3)(tr B 5 ) , a f1a (1, 4, 1, 1) = a(tr B)5 + 2(6a2 + 3a + 2)(tr B)3 (tr B 2 ) 8 + (8a3 + 10a2 + 29a + 12)(tr B)(tr B 2 )2 + 4(4a3 + 6a2 + 10a + 3)(tr B)2 (tr B 3 ) + 4(12a2 + 16a + 9)(tr B 2 )(tr B 3 ) + 2(4a5 + 12a4 + 45a2 + 46a + 24)(tr B)(tr B 4 ) + 8(4a3 + 12a2 + 21a + 10)(tr B 5 ) , a (tr B)5 + 10(2a + 1)(tr B)3 (tr B 2 ) f1a (1, 5, 0, −) = 16 + 5(8a2 + 10a + 5)(tr B)(tr B 2 )2 + 20(2a2 + 3a + 2)(tr B)2 (tr B 3 ) + 20(2a3 + 5a2 + 7a + 3)(tr B 2 )(tr B 3 ) + 10(4a3 + 12a2 + 21a + 10)(tr B)(tr B 4 ) + 4(4a4 + 20a3 + 65a2 + 80a + 37)(tr B 5 ) . f1a (1, 3, 2, 1) =

4

Type 1(b) Integrals

Consider the integral Z f1b (1, 2, 0, −) = X>0

etr(−B −1 X) det(X)a−(m+1)/2 (tr X −2 ) dX. det(B)a Γm (a)

Writing (tr X −2 ) in terms of zonal polynomials and integrating the resulting expression by using (5), we get obtains


f1b (1, 2, 0, −) =

489

C(12 ) (B −1 ) C(2) (B −1 ) 1 − . (−a + (m + 1)/2)(2) 2 (−a + (m + 1)/2)(12 )

Now, substituting for (−a + (m + 1)/2)(2) , (−a + (m + 1)/2)(12 ) , C(2) (B −1 ) and C(12 ) (B −1 ) above, we have f1b (1, 2, 0, −) =

(tr B −1 )2 + (2a − m − 1) tr(B −2 ) . 2 (a − m/2)(a − (m + 1)/2)(a − (m + 3)/2)

Writing (tr X −3 ) in terms of zonal polynomials and integrating the resulting expression using (5), we obtain Z etr(−B −1 X) det(X)a−(m+1)/2 (tr X −3 ) dX f1b (1, 3, 0, −) = a Γ (a) det(B) m X>0 (−1)3 C(3) (B −1 ) 1 (−1)3 C(2,1) (B −1 ) = − (−a + (m + 1)/2)(3) 4 (−a + (m + 1)/2)(2,1) 1 (−1)3 C(13 ) (B −1 ) + . 4 (−a + (m + 1)/2)(13 ) Now, substituting for (−a + (m + 1)/2)(3) , (−a + (m + 1)/2)(2,1) , (−a + (m + 1)/2)(13 ) , C(3) (B −1 ), C(2,1) (B −1 ) and C(13 ) (B −1 ) above, we have f1b (1, 3, 0, −) =

8 2(tr B −1 )3 (2a − m − 5)(2a − m − 3)(2a − m − 1)3 + 3(2a − m − 1) tr(B −1 ) tr(B −2 ) + (2a − m − 1)2 tr(B −3 ) .

Similarly, following the procedure described above, we get f1b (1, 2, 1, 1) =

5

8 (2a − m − 3)(tr B −1 )3 (2a − m − 5)(2a − m − 3)(2a − m − 1)3 + {(2a − m − 2)2 + 3}(tr B −1 ) tr(B −2 ) + 4(2a − m − 1) tr(B −3 ) .

Type 2(a) Integrals

Consider the integral Z Im det(X)a−(m+1)/2 det(Im − X)b−(m+1)/2 f2a (1, 2, 0, −) = (tr(BX)2 ) dX. B (a, b) m 0 Writing (tr(BX)2 ) in terms of zonal polynomials and integrating resulting expression by using (6), one obtains f2a (1, 2, 0, −) =

(a)(2) 1 (a)(12 ) C(2) (B) − C(12 ) (B). (a + b)(2) 2 (a + b)(12 )

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Now, substituting for (a)(2) , (a + b)(2) , (a)(12 ) , (a + b)(12 ) , C(2) (B) and C(12 ) (B) above, we have f2a (1, 2, 0, −) =

a[b(tr B)2 + {(2a + 1)(a + b) − 1} tr(B 2 )] . 2(a + b − 1/2)(a + b)(a + b + 1)

Similarly, we get f2a (1, 3, 0, −) =

a b(b − a)(tr B)3 4(a + b − 1)4 (a + b − 1/2) + 3b{(2a + 1)(a + b) − 2}(tr B) tr(B 2 )

+ {4 + 4(b2 − a2 ) + 2(a + b)(3 + 2a)(a2 + ab − 1)} tr(B 3 ) , a f2a (1, 2, 1, 1) = b {a(1 + a + b) − 1} (tr B)3 2(a + b − 1)4 (a + b − 1/2) + (2a2 + a + 2)(a + b)2 + (2a2 − 8a − 1)(a + b) + 2(a2 − a + 1) (tr B) tr(B 2 ) + 2b {(2a + 1)(a + b) − 2}tr(B 3 ) .

6

Type 2(b) Integrals

Consider the integral Z Im det(X)a−(m+1)/2 det(Im − X)b−(m+1)/2 f2b (1, 2, 0, −) = tr(X −1 B)2 dX. B (a, b) m 0 Writing tr(X −1 B)2 in terms of zonal polynomials and integrating the resulting expression by using (7), one obtains f2b (1,2,0,−) =

(−a + (m + 1)/2)(12 ) (−a + (m + 1)/2)(2) C(2)(B)− C(12 )(B). (−a − b + (m + 1)/2)(2) 2(−a − b + (m + 1)/2)(12 )

Now, substituting for (−a + (m + 1)/2)(2) , (−a − b + (m + 1)/2)(2) , (−a + (m + 1)/2)(12 ) , (−a − b + (m + 1)/2)(12 ) , C(2) (B) and C(12 ) (B) above, we have f2b (1,2,0, −) =

(2a − m − 1)[2{(2a − m)(2a + 2b − m − 3) − 4b}tr B 2 − b(trB)2 ] . 2(2a + 2b − m − 1)(2a + 2b − m − 3)(2a + 2b − m)

Further, f2b (1, 3, 0, −) =

1 (2a + 2b − m − 5)(2a + 2b − m − 3)(2a + 2b − m − 1)3 h × 2b(2a − m − 1)(2b − 2a + m + 1)(tr B)3 − 6b(2a − m − 1){(2a − m − 2)(2a + 2b − m − 1) − 4}(trB)(trB 2 ) + (2a − m)(2a + 2b − m − 5)(2a + 2b − m + 1)(2a − m − 3) i 3 − 4b{(2a + 2b − m + 3)(2a − m − 4) + 8} (tr B ) .


7

491

Applications

As an illustration, we will compute the expected value of tr W 5 , where the random matrix W follows a Wishart distribution with parameters n and Σ. By using the density of W given in Section 1, we have Z etr (−Σ−1 W /2) det(W )(n−m−1)/2 5 E(tr W ) = (tr W 5 ) dW. nm/2 Γ (n/2) det(Σ)n/2 2 m W >0 Now, comparing the right hand side of the above expression with that of f1a (g, h, i, j) given in Section 1, it is straightforward to see that E(tr W 5 ) = f1a (1, 5, 0, −) with a = n/2 and B = 2Σ. Finally, substituting appropriately in the expression for f1a (1, 5, 0, −) given in Section 3, we obtain E(tr W 5 ) = n (tr Σ)5 + 10(n + 1)(tr Σ)3 (tr Σ2 ) + 5(2n2 + 5n + 5)(tr Σ)(tr Σ2 )2 + 10(n2 + 3n + 4)(tr Σ)2 (tr Σ3 ) + 5(n3 + 5n2 + 14n + 12)(tr Σ2 )(tr Σ3 ) + 5(n3 + 6n2 + 21n + 20)(tr Σ)(tr Σ4 ) + (n4 + 10n3 + 65n2 + 160n + 148)(tr Σ5 ) . Acknowledgements. The research work of DKN and AG was supported by SUI, Universidad de Antioquia [project no. 2015-5323].

References [1] A. G. Constantine, Some non-central distribution problems in multivariate analysis, Annals of Mathematical Statistics, 34 (1963), 1270–1285. https://doi.org/10.1214/aoms/1177703863 [2] A. K. Gupta and D. K. Nagar, Matrix Variate Distributions, Chapman & Hall/CRC, Florida, 2000. [3] A. M. Mathai, Serge B. Provost and Takesi Hayakawa, Bilinear Forms and Zonal Polynomials, Springer-Verlag, New York, 1995. https://doi.org/10.1007/978-1-4612-4242-0 [4] A. T. James, Distributions of matrix variates and latent roots derived from normal samples, Annals of Mathematical Statistics, 35 (1964), 475– 501. https://doi.org/10.1214/aoms/1177703550 [5] James O. Kitchen, Extended tables for zonal polynomials, North Carolina State University Institute of Statistics mimeo series no. 565, 1968. https://repository.lib.ncsu.edu/handle/1840.4/2754

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[6] C. G. Khatri, On certain distribution problems based on positive definite quadratic functions in normal vectors, Annals of Mathematical Statistics, 37 (1966), no. 2, 468–479. https://doi.org/10.1214/aoms/1177699530 Received: May 3, 2017; Published: May 17, 2017