A Two Sample Test for Mean Vectors with Unequal Covariance Matrices

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1 A Two Sample Test for Mean Vectors wth Unequal Covarance Matrces Tamae Kawasak 1 and Takash Seo 2 1 Department of Mathematcal Informaton Scence Graduate School of Scence, Tokyo Unversty of Scence, Tokyo, Japan 2 Department of Mathematcal Informaton Scence Faculty of Scence, Tokyo Unversty of Scence, Tokyo, Japan Abstract In ths paper, we consder testng the equalty of two mean vectors wth unequal covarance matrces. In the case of equal covarance matrces, we can use Hotellng s T 2 statstc, whch follows the F dstrbuton under the null hypothess. Meanwhle, n the case of unequal covarance matrces, the T 2 type test statstc does not follow the F dstrbuton, and t s also dffcult to derve the exact dstrbuton. In ths study, we propose an approxmate soluton to the problem by adjustng the degrees of freedom of the F dstrbuton. That s, we derve an extenson of the results derved by Yanaghara and Yuan (2005). Asymptotc expansons up to the term of order N 2 for the frst and second moments of the test statstc are gven, where N s the total sample sze mnus two, and a new result of the approxmate degrees of freedom s obtaned. Fnally, numercal comparson s presented by a Monte Carlo smulaton. Keywords Approxmate degrees of freedom; F approxmaton; Hotellng s T 2 statstc; Multvarate Behrens-Fsher problem; Two sample problem. 1 Introducton Let x 1,..., x j,..., x n be p-dmensonal random vectors from N p (µ, Σ ), = 1, 2, j = 1, 2,..., n. We consder the followng hypothess test problem: H 0 : µ 1 = µ 2 vs. H 1 : µ 1 µ 2, (1.1) where Σ 1 Σ 2. A natural statstc for testng (1.1) s ( T = (x 1 x 2 ) S1 + S ) 1 2 (x 1 x 2 ), n 1 n 2 where x = 1 n n j=1 x j, S = 1 n 1 n j=1 (x j x )(x j x ). 1

2 When n 1 = n 2 and Σ 1 = Σ 2, the T statstc s reduced to the two sample Hotellng s T 2 statstc. Then, under the null hypothess n (1.1), (n p 1)T/{p(n 2)} follows the F dstrbuton wth p and n p 1 degrees of freedom, where n = n 1 + n 2. To consder the tests for equalty of two mean vectors s a fundamental problem. Mean comparson wth unequal varances s ntrnscally dffcult, and s well known as the Behrens-Fsher problem. Welch (1938) and Scheffé (1943) proposed approxmate solutons for the unvarate case. One of the earlest methods for solvng the multvarate Behrens-Fsher problem was derved by Bennett (1951) based on an extenson of Scheffé s (1943) unvarate soluton. Some approxmate solutons were consdered by James (1954), Yao (1965), Johansen (1980), Nel et al. (1990), and Km (1992). Nel et al. (1986) obtaned the exact null dstrbuton of T, and Krshnamoorthy and Yu (2004) proposed a modfcaton to the soluton. Recently, Krshnamoorthy and Yu (2012) proposed a soluton extendng the modfed Nel and Van der Merwe s test procedure n ther earler study to the case of ncomplete data wth a monotone pattern. The problem concerns the dfference between the mean vectors of two normal populatons wth the case of a monotone mssng pattern when Σ 1 = Σ 2, as suggested by Seko, Kawasak and Seo (2011). Grón and del Castllo (2010) studed the multvarate Behrens-Fsher dstrbuton, whch s defned as the convoluton of two ndependent multvarate Student t dstrbutons. Yanaghara and Yuan (2005) provded three approxmate solutons to the multvarate Behrens-Fsher problem that are two F approxmatons wth approxmate degrees of freedom and modfed Bartlett corrected statstc. However, these solutons are not good approxmatons when the dfference between the covarance matrces s large. Our goal s to gve a new approxmate soluton by an extenson of Yanaghara and Yuan (2005). The followng secton presents a dervaton of the man result by approxmate degrees of freedom and presents the proof. In Secton 3, we compare four approxmate procedures by Monte Carlo smulaton and evaluate the advantages of the proposed procedures. In the Appendx, we present certan formulas used to derve the man result. 2

3 2 Approxmate Degrees of Freedom Assumng the standard regularty condton n /n = O(1), = 1, 2, then, as n Yanaghara and Yuan (2005), we can wrte where z = T = z W 1 z = z z U, (2.1) n1 n ( 2 n Σ 1/2 (x 1 x 2 ), W = Σ 1/2 n 2 n S 1 + n ) 1 n S 2 Σ 1/2, By approxmatng the dstrbuton of U as Σ = n 2 n Σ 1 + n 1 n Σ 2, U = z z z W 1 z. we have U χ2 ν ϕ, (2.2) ν pϕ T χ2 p/p χ 2 ν/ν F p,ν. Note that when Σ 1 = Σ 2 and n 1 = n 2, U s exactly dstrbuted as χ 2 ν/ϕ, where ν = n p 1 and ϕ = n 2. In general, the constants ν and ϕ can be gven usng the followng theorems for the frst and second moments of U. Theorem 2.1 Let U = z z/z W 1 z be defned by (2.1). Then, an asymptotc expanson up to the term of order N 2 for E[U] can be expanded as where E[U] = 1 θ 1 N + 1 N 2 (θ 2 θ 3 ) + O(N 3 ), (2.3) N =n 2, θ 1 = 1 p(p + 2) =1 c {p(a (1) ) 2 + (p 2)a (2) }, 3

4 1 θ 2 = p(p + 2)(p + 4) =1 { 1 θ 3 = p(p + 2)(p + 4)(p + 6) d {4p 2 a (3) + (p 2)(3p + 4)a (1) a (2) + p(p + 2)(a (1) ) 3 }, =1 c 2 { p 2 (5p + 14)a (4) + 4(p + 3)(p + 2)(p 2)a (1) a (3) +p(p + 3)(p 2)(a (2) ) 2 + 2(p 3 + 5p 2 + 7p + 6)a (2) (a (1) ) 2 p(p + 4)(a (1) ) 4} + 4(p + 3)(p + 2)(p 2)ψ 1 + 4p(p + 2)(p 2)ψ 2 + 4p(p + 4)(p + 2)ψ 3 2p(p 2)ψ 4 2(p + 3)(p 2)ψ 5 2p(p + 4)(p 2)ψ 6 2p(p + 4)ψ 7 + 2p(p + 4)(3p + 2)ψ 8 }, wth ψ k, k = 1, 2,..., 7 gven by ψ 1 = c 1 c 2 {a (1) 1 b(1, 2, 1) + a (1) 2 b(2, 1, 1) }, ψ 2 = c 1 c 2 b (2, 2, 1), ψ 3 = c 1 c 2 a (1) 1 a(1) 2 b(1, 1, 1), ψ 4 = c 1 c 2 a (2) 1 a(2) 2, ψ 5 = c 1 c 2 {a (2) 1 (a(1) 2 )2 + (a (1) 1 )2 a (2) 2 }, ψ 6 = c 1 c 2 (b (1, 1, 1) ) 2, ψ 7 = c 1 c 2 (a (1) 1 )2 (a (1) 2 )2, ψ 8 = c 1 c 2 b (1,1,2), and c = (n n ) 2 (n 2) n 2, d = (n n ) 3 (n 2) 2 (n 1) n 3 (n 1) 2, a (l) = tr(σ Σ 1 ) l, = 1, 2, l = 1, 2, 3, 4, b (q, r, s) = tr{(σ 1 Σ 1 ) q (Σ 2 Σ 1 ) r } s, (q, r, s) = (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 2), (2, 2, 1). Proof. Let ρ = n 1 n n 2 ( = 1, 2), Ω n = n 1 Σ 2 Σ 1 2 and V = n 1(Σ 1/2 S Σ 1/2 I p ), = 1, 2. Then, W 1 can be expanded as where W 1 = I p 1 N V + 1 N V 2 1 N N V N 2 V 4 + O p (N 5/2 ), (2.4) V = =1 ρ 1 Ω V Ω. 4

5 Note that U = z z/z W 1 z. It follows from (2.4) that we can expand U as U =1 + 1 N Q N (Q2 1 Q 2 ) + 1 N N (Q 3 2Q 1 Q 2 + Q 3 1) + 1 N 2 (Q4 1 Q 4 + 2Q 1 Q 3 + Q 2 2 3Q 2 1Q 2 ) + O p (N 5/2 ), where Q = z V z/z z, = 1, 2, 3, 4. Note that V and z are ndependent, and so are z V z/z z and z z, as well as z V 2 z/z z and z z (see Fang et al., 1990, p.30). In the same way as n Yanaghara and Yuan (2005), the followng results can be obtaned after a good deal of calculaton: () E[Q 1 ] = 0, () E[Q 2 ] = 1 p E[Q 2 1] = =1 2 p(p + 2) () NE[Q 3 ] = 1 p NE[Q1 Q 2 ] = NE[Q 3 1 ] = c {(a (1) ) 2 + a (2) }, =1 =1 c {(a (1) ) 2 + 2a (2) }, d {4a (3) + 3a (1) a (2) + (a (1) ) 3 }, 2 p(p + 2) =1 8 p(p + 2)(p + 4) d {6a (3) + 5a (1) a (2) + (a (1) ) 3 }, =1 d {8a (3) + 6a (1) a (2) + (a (1) ) 3 }, (v) E[Q 4 ] = 1 { c 2 {5a (4) + 4a (1) a (3) + (a (2) ) 2 + 2a (2) (a (1) ) 2 } p =1 } +2(2ψ 1 + 2ψ 2 + 2ψ 3 + ψ 6 + 3ψ 8 ) + O(N 1 ), { E[Q ] = c 2 {48a (4) +32a (1) a (3) +12(a (2) ) 2 +12a (2) (a (1) ) 2 +(a (1) ) 4 } p(p + 2)(p + 4)(p + 6) =1 } +2(16ψ ψ 2 + 8ψ 3 + 4ψ 4 + 2ψ 5 + 8ψ 6 + ψ ψ 8 ) + O(N 1 ), E[Q 1 Q 3 ] = { 2 c 2 {8a (4) + 7a (1) a (3) + (a (2) ) 2 + 2a (2) (a (1) ) 2 } p(p + 2) =1 } +(7ψ ψ 2 + 4ψ 3 + 2ψ 6 + 6ψ 8 ) + O(N 1 ), 5

6 { E[Q 2 1 2] = c 2 {14a (4) + 8a (1) a (3) + 7(a (2) ) 2 + (a (1) ) 4 + 6a (2) (a (1) ) 2 } p(p + 2) =1 } +2(4ψ 1 + 4ψ 2 + 4ψ 3 + ψ 4 + ψ 5 + 6ψ 6 + ψ ψ 8 ) + O(N 1 ), E[Q 2 1Q 2 ] = 2 p(p + 2)(p + 4) { =1 c 2 {40a (4) + 28a (1) a (3) + 10(a (2) ) a (2) (a (1) ) 2 + (a (1) ) 4 } +28ψ ψ ψ 3 + 4ψ 4 + 3ψ ψ 6 + 2ψ ψ 8 } + O(N 1 ). Usng the above results, we can show (2.3). Ths completes the proof of Theorem 2.1. In addton, we have the followng result. Corollary 2.1 If Σ 1 = Σ 2, n 1 = n 2, then E[U] = (p 1) + N N 2 p 2 (p 2) (p + 2)(p + 6) + O(N 3 ). Smlarly, as the result of asymptotc expanson for E[U 2 ],we have the followng theorem. Theorem 2.2 Let U = z z/z W 1 z be defned by (2.1). Then, an asymptotc expanson up to the term of order N 2 for E[U 2 ] can be expanded as where 1 θ 4 = p(p + 2) E[U 2 ] = 1 2 N (θ 1 θ 4 ) + 1 N 2 (2θ 5 θ 6 ) + O(N 3 ), (2.5) =1 1 θ 5 = p(p + 2)(p + 4) { c (a (1) ) 2 } + 2a (2), =1 d { 4(p 2 3p + 4)a (3) + 3p(p 4)a (1) a (2) + p 2 (a (1) ) 3}, 6

7 { 1 θ 6 = c 2 {2(p + 1)(5p 2 14p + 24)a (4) + 4(p 4)(2p 2 + 5p + 6)a (1) a (3) p(p + 2)(p + 4)(p + 6) =1 + (p 2)(p 4)(2p + 3)(a (2) ) 2 + 2(p + 2)(2p 2 p + 12)a (2) (a (1) ) 2 3(p 2 + 2p 4)(a (1) ) 4 } + 4(p 4)(2p 2 + 5p + 6)ψ 1 + 8p(p 2)(p 4)ψ 2 + 8p(p 2 + 4p + 2)ψ 3 6(p 2)(p 4)ψ 4 6(p 4)(p + 2)ψ 5 +4(p + 3)(p 2)(p 4)ψ 6 6(p 2 + 2p 4)ψ 7 +12(p 3 +p 2 2p + 8)ψ 8 }. Proof. In the same way, we can expand U 2 as U 2 =1 + 2 N Q N (3Q2 1 2Q 2 ) + 2 N N (Q 3 3Q 1 Q 2 + 2Q 3 1) + 1 N 2 (5Q4 1 2Q 4 + 6Q 1 Q 3 + 3Q Q 2 1Q 2 ) + O p (N 5/2 ). By calculatng the expectatons of the above results, we can show (2.5). Ths completes the proof of Theorem 2.2. In addton, we have the followng result. Corollary 2.2 If Σ 1 = Σ 2, n 1 = n 2, then E[U 2 ] = (p 2) + N N 2 p5 + 6p 4 p 3 92p 2 60p O(N 3 ). (p + 2)(p + 4)(p + 6) It follows from (2.2) that E[U] ν ϕ, E[U 2 ] ν(ν + 2) ϕ 2. Therefore, usng the asymptotc expansons of E[U] and E[U 2 ] n Theorems 2.1 and 2.2, the new approxmaton to the values of ν and ϕ are gven by ν KS = ϕ KS = 2(N 2 Nθ 1 + θ 2 θ 3 ) 2 N 2 (N 2 2Nθ 1 + 2Nθ 4 + 2θ 5 θ 6 ) (N 2 Nθ 1 + θ 2 θ 3 ) 2, (2.6) N 2 ν N 2 Nθ 1 + θ 2 θ 3, (2.7) respectvely. We can propose a new procedure as follows. 7

8 (I) Hgh Order Procedure T KS = ν KS pϕ KS T a F p,νks where ν KS and ϕ KS are gven by (2.6) and (2.7), respectvely, and where a means approxmately followng. If θ 2 = θ 3 = θ 5 = θ 6 = 0, then ν KS = (N θ 1) 2 Nθ 4 θ 2 1 /2(= ν Y ), ϕ KS = ν Y N N θ 1 (= ϕ Y ), (2.8) and these values are the same as the results of Yanaghara and Yuan (2005). In addton, they slghtly adjust the coeffcent ν Y to whch wll be used to obtan the F approxmaton. 3 Numercal Studes ν M = (N θ 1) 2 Nθ 4 θ 1, (2.9) In ths secton, we perform a Monte Carlo smulaton n order to nvestgate the accuracy of our procedure (I) and to compare t wth the followng three procedures: (II) Yanaghara and Yuan s (2005) Procedure where ν Y and ϕ Y are gven by (2.8). T Y = ν Y pϕ Y T a F p,νy (III) Modfed Yanaghara and Yuan s (2005) Procedure where T M = ν M pϕ M T a F p,νm ϕ M = ν MN N θ 1 and ν M s gven by (2.9). 8

9 (IV) Modfed Bartlett Procedure (see Yanaghara and Yuan (2005)) ( T MB = (N β 1 + β 2 ) log 1 + T ) a N β χ 2 p, 1 where 2 β 1 =, β2 = (p + 2) γ 2 2(p + 4) γ 1, γ 2 2 γ 1 2( γ 2 2 γ 1 ) γ 1 = 1 c {(â (1) ) â (2) }, γ 2 = c {2(p + 3)(â (1) ) 2 + 2(p + 4)â (2) }, p p(p + 2) =1 â (l) = tr(s S 1 ) l, S = n 2 n S 1 + n 1 n S 2. For each of parameter, the smulaton was carred out for 1,000,000 trals based on normal random vectors. Wthout loss of generalty, we can assume that µ 1 = µ 2 = 0. We compare the followng type I errors for four procedures: (I) α 1 = P(T KS > F α;p,νks ), (II) α 2 = P(T Y > F α;p,νy ), (III) α 3 = P(T M > F α;p,νm ), (IV) α 4 = P(T MB > c α;p ), where F α;m,n s the upper 100α percentle of the F dstrbuton wth m and n degrees of freedom and c α;p s the upper 100α percentle of the ch-square dstrbuton wth p degrees of freedom. We choose α = 0.05, 0.01, p = 4, 8, and the sample szes (n 1, n 2 ) = (10, 10), (10, 20), (20, 10), (20, 20), (50, 50), (50, 80), (80, 50), (80, 80) for (I) (IV). We note that the second degree of freedom of F dstrbuton for the test statstc (I) (III) changes wth each smulaton of 1,000,000 trals. In practcal use, we must estmate a (l) ths paper, we use the consstent estmators of a (l) procedure (IV). =1 and b (q,r,s) for (I) (III) snce Σ and Σ are unknown. In and b (q,r,s), whch are the same as that of Table 1 presents the emprcal szes α j, j = 1, 2, 3, 4 n the case of Σ 1 = dag(η, η 2,, η p ) and Σ 2 = I, where η = 1, 5(5)20. We note that Σ 1 1 and the dfference between Σ 1 and Σ 2 s large when η s large. Tables 2 and 3 present the emprcal szes α j, j = 1, 2, 3, 4 n the case of Σ 1 = σ 2 I and Σ 2 = I, where σ 2 = 0.1(0.2)0.9,1 n Table 2, and σ 2 = 2, 5(5)30 n Table 3. We note that the emprcal szes for the case of Σ 1 1 and Σ 1 > 1 are gven n Tables 2 and 3, respectvely. The last row of each of these tables ndcates the average absolute dscrepancy (AAD). In ths context, see Yanaghara and Yuan (2005). 9

10 Table 1 Emprcal szes ( α 1 α 4 ) when p = 4, 8, and η = 1, 5(5)20 α = 0.05 α = 0.01 p n 1 n 2 η α 1 α 2 α 3 α 4 α 1 α 2 α 3 α AAD AAD Note : AAD = 100 α 100α /20 10

11 Table 2 Emprcal szes ( α 1 α 4 ) when p = 4, 8 and σ 2 = 0.1(0.2)0.9, 1 α = 0.05 α = 0.01 p n 1 n 2 σ 2 α 1 α 2 α 3 α 4 α 1 α 2 α 3 α AAD Note : AAD = 100 α 100α /24 11

12 Table 2 (Contnued) α = 0.05 α = 0.01 p n 1 n 2 σ 2 α 1 α 2 α 3 α 4 α 1 α 2 α 3 α AAD Note : AAD = 100 α 100α /24 12

13 Table 3 Emprcal szes ( α 1 α 4 ) when p = 4, 8 and σ 2 = 2, 5(5)30 α = 0.05 α = 0.01 p n 1 n 2 σ 2 α 1 α 2 α 3 α 4 α 1 α 2 α 3 α AAD Note : AAD = 100 α 100α /28 13

14 Table 3 (Contnued) α = 0.05 α = 0.01 p n 1 n 2 σ 2 α 1 α 2 α 3 α 4 α 1 α 2 α 3 α AAD Note : AAD = 100 α 100α /28 14

15 From Table 1, t s seen that the proposed approxmatons α 1 are very good for cases when η s large. In contrast, t seems that other α are farther from α as η becomes large. It may also be noted that α 1 s stable and a good approxmaton to α when n 1 and n 2 are large. From Table 2, we can see that α 3 s AAD and α 4 s AAD are lower than the others, and ther approxmatons are good for the case of p = 4, an that α 1 s a good approxmaton for the case of p = 8. On the other hand, t seems from Table 2 that the emprcal szes are almost unchanged except for α 2. It can be seen from Table 3 that α 3 are closer to α when p = 4. Meanwhle, the behavor of α 1 resembles the behavor of α 4. In addton, t s seen from Table 3 that α 1 and α 3 are good approxmatons when p = 8. In concluson, when the dfference between covarance matrces s large, the approxmate upper percentle of the null dstrbuton of T by the method (hgh order procedure) proposed n ths paper s better than those of other procedures. Appendx In ths Appendx, we present some results of expectaton: A.1 Let u N p (0, I) wth A and B are p p symmetrc matrces, then (1) E[u Au] = tra, (2) E[u u(u Au)] = (p + 2)(trA), (3) E[(u Au) 2 ] = 2(trA 2 ) + (tra) 2, (4) E[(u Au)(u Bu)] = 2(trAB) + (tra)(trb), (5) E[(u Au) 3 ] = 8(trA 3 ) + 6(trA 2 )(tra) + (tra) 3, (6) E[(u Au) 2 (u Bu)] = 8(trA 2 B) + 4(trAB)(trA) + 2(trA 2 )(trb) + (tra) 2 (trb), (7) E[(u Au) 4 ] = 48(trA 4 ) + 32(trA 3 )(tra) + 12(trA 2 )(tra) (trA 2 ) 2 + (tra) 4. 15

16 A.2 Let S W p (n, Σ) and V = n(s Σ) wth A, B and C are p p symmetrc matrces, then (1) E[(trAV ) 2 ] = 2 tr(aσ) 2, (2) E[tr(AV ) 2 ] = tr(aσ) 2 + (traσ) 2, (3) E[(trAV BV )] = (trav BV ) + (traσ)(trbσ), (4) E[{tr(AV )} 3 ] = 8 n tr(aσ) 3, (5) E[tr(AV ) 3 ] = 1 n [4 tr(aσ) 3 + 3{tr(AΣ) 2 }(traσ) + (traσ) 3 ], (6) E[(trAV ){tr(av ) 2 }] = 4 n [tr(aσ) 3 + (traσ){tr(aσ) 2 }], (7) E[(trAV ) 2 (trbv )] = 8 n {tr(aσ) 2 BΣ}, (8) E[(trAV ) 4 ] = 12{tr(AΣ) 2 } 2 + O(N 1 ) (9) E[tr(AV ) 4 ] = 5 tr(aσ) 4 +4{tr(AΣ) 3 }(traσ)+2{tr(aσ) 2 }(traσ) 2 +{tr(aσ) 2 } 2 +O(N 1 ), (10) E[(trAV ){tr(av ) 3 }] = 6[ tr(aσ) 4 + (traσ){tr(aσ) 3 }] + O(N 1 ), (11) E[{tr(AV ) 2 } 2 ] = 5{tr(AΣ) 2 } tr(aσ) 4 + 2{tr(AΣ) 2 }(traσ) 2 + (traσ) 4 + O(N 1 ), (12) E[(trAV ) 2 {tr(av ) 2 }] = 8 tr(aσ) 4 + 2{tr(AΣ) 2 } 2 + 2{tr(AΣ) 2 }(traσ) 2 + O(N 1 ), (13) E[(trAV ) 3 (trbv )] = 12{tr(AΣ) 2 }(traσbσ) + O(N 1 ), (14) E[(trAV ) 2 (trbv ) 2 ] = 8(trAΣBΣ) 2 + 4{tr(AΣ) 2 }{tr(bσ) 2 } + O(N 1 ), (15) E[(trAV ) 2 (trbv )(trcv )] = 8(trAΣBΣ)(trAΣCΣ) + 4{tr(AΣ) 2 }(trbσcσ) + O(N 1 ). 16

17 A.3 Let S W p (n, Σ) and V = n(s Σ) wth A, B, C and D are p p symmetrc matrces where = 1, 2, then (1) E[tr(AV 1 ) 2 (BV 2 ) 2 ] = tr(aσ) 2 (BΣ) 2 + (traσ){traσ(bσ) 2 } + {tr(aσ) 2 BΣ}(trBΣ) +(traσ)(trbσ)(traσbσ), (2) E[(trAV 1 BV 2 ) 2 ] = 2{tr(AΣ) 2 (BΣ) 2 } + {tr(aσ) 2 }{tr(bσ) 2 } + (traσbσ) 2, (3) E[{tr(AV 1 ) 2 }{tr(bv 2 ) 2 }] = {tr(aσ) 2 }{tr(bσ) 2 } + {tr(aσ) 2 }(trbσ) 2 +(traσ) 2 {tr(bσ) 2 } + (traσ) 2 (trbσ) 2, (4) E[(trAV 1 ) 2 (trbv 2 ) 2 ] = 4{tr(AΣ) 2 }{tr(bσ) 2 }, (5) E[{tr(AV 1 ) 2 }(trbv 2 ) 2 ] = 2{tr(AΣ) 2 }{tr(bσ) 2 } + 2(trAΣ) 2 {tr(bσ) 2 }, (6) E[(trAV 1 ){trav 1 (BV 2 ) 2 }] = 2{tr(AΣ) 2 (BΣ) 2 } + 2{tr(AΣ) 2 BΣ}(trBΣ), (7) E[(trAV 1 )(trbv 2 )(trav 1 BV 2 )] = 4{tr(AΣ) 2 (BΣ) 2 }, (8) E[trAV 1 BV 1 CV 2 DV 2 ] = (traσbσcσdσ) + (traσbσcσ)(trdσ) +(traσcσdσ)(trb) + (traσcσ)(trbσ)(trdσ). A.4 The followng results are presented as supplementary expectatons: (1) E[tr(Ω 2 Ω 1V 1 Ω 1 Ω 2V 2 ) 2 ] = 3tr(Ω 1 Ω 1 Ω 2Ω 2 )2 + (trω 1 Ω 1 Ω 2Ω 2 )2, (2) E[(trΩ 1 Ω 1V 1 )(trω 2 Ω 1V 1 Ω 1 Ω 2V 2 Ω 2 Ω 2V 2 )] = 2{tr(Ω 1 Ω 1) 2 Ω 2 Ω 2}(trΩ 2 Ω 2) +2{tr(Ω 1 Ω 1) 2 (Ω 2 Ω 2) 2 }, (3) E[(trΩ 2 Ω 1V 1 Ω 1 Ω 2V 2 ) 2 ] = 2tr(Ω 1 Ω 1 Ω 2Ω 2 )2 + 2(trΩ 1 Ω 1 Ω 2Ω 2 )2, (4) E[(trΩ 1 Ω 1V 1 )(trω 2 Ω 2V 2 )(trω 2 Ω 1V 1 Ω 1 Ω 2V 2 )] = 4{tr(Ω 1 Ω 1) 2 (Ω 2 Ω 2) 2 }, where the notatons are defned by Secton 2. 17

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