Supplementary Material For Testing Homogeneity of. High-dimensional Covariance Matrices

Transcript

1 Supplementary Material For Testing Homogeneity of High-dimensional Covariance Matrices Shurong Zheng, Ruitao Lin, Jianhua Guo, and Guosheng Yin 3 School of Mathematics & Statistics and KLAS, Northeast Normal University, China Department of Biostatistics, The University of Texas MD Anderson Cancer Center, Houston, Texas 77030, U.S.A. 3 Department of Statistics and Actuarial Science, University of Hong Kong, Hong Kong, China This Supplementary Material contains the proofs of the theorems and additional simulation results. In particular, Figure presents the simulation results for three samples with Gamma populations. For ease of exposition, the proof of Theorem is given first and that of Theorem is given later. We first provide two lemmas which are essential to the proofs of Theorems and. Lemmas Lemma Under Assumptions (A) (A), we have σ A {tr[(s S ) ] ˆµ µ A } N(0, )

2 where ˆµ = = (n n )n (n ) (trs ), and µ A = tr[(σ Σ ) ] + = n + (n ) tr(σ ) + = β n (n ) (e T l Σ e l ) σa = 4[n tr(σ )] + 4[n tr(σ )] + 8n +4[n tr(σ 4 ) + β n n [tr(σ Σ )] (e T l Σ e l ) ] +4{n tr[(σ Σ ) ] + β n 8[n tr(σ 3 Σ ) + β n +4n [tr(σ 4 ) + β (e T l Σ e l ) ] +4{n tr[(σ Σ ) ] + β n 8[n tr(σ 3 Σ ) + β n (e T l Σ / Σ Σ / e l ) } e T l Σ / Σ Σ / e l e T l Σ e l ] (e T l Σ / Σ Σ / e l ) } e T l Σ / Σ Σ / e l e T l Σ e l ]. Proof. Define r i = n / w i and w i = (w i,..., w pi ) T for i =,..., n and =,. We have (n ) n n tr(s ) = tr[( Σ / r ir T iσ / i= ) ] + n ( r T Σ r ) n r T Σ n r i r T iσ r, i= where r = n n i= r i for =,..., K. First, we will consider the bounded spectral norm case, that is, the maximum eigenvalue of Σ is bounded for =,. When tr(σ q ) = O(pq ) holds for q =,, 3, 4 and at least one integer in the index set {, }, we only need to consider the CLT of n /3 tr[(s S ) ], and its proof is almost the same as the proof for the CLT of tr[(s S ) ] when Σ and Σ have the bounded spectral norms. Specifically, for the bounded spectral norm case, the proof can be completed through the following steps.

3 Step. We will prove n r T Σ r = n n n i= rt i Σ r i. As E(n i<j rt i Σ r j ) = 0 and we have n Var(n n i= E(n i<j trσ +o p (). We have n r T Σ r = n r T iσ r j ) 4E(r T Σ r r T Σ r ) = 4n tr(σ ) 0, i<j rt i Σ r j = o p (). Moreover, as E(n n r T iσ r i ) = n E[(rT Σ r n i= rt i Σ r i ) = n trσ ) ] = n [tr(σ )+β i<j rt i Σ r j + trσ and (e T j Σ e j ) ] 0 from (.5) of Bai and Silverstein (004), we have n n i= rt i Σ r i n trσ = o p (), and thus Step. We have n r T Σ n i= Step.. We have n n i= r i r T iσ r = n n r T Σ r = n trσ + o p (). i j l +n r T iσ r i r T iσ r i = n n i j i= n +n (n i= r T iσ r j r T jσ r l + n r T iσ r i r T iσ r j + n (r T iσ r i n trσ ) j= r T iσ r j r T jσ r i i j n r T iσ r i r T iσ r i. i= trσ )(r T iσ r i n trσ ) + (n trσ ). As n n i= E[(rT i Σ r i n trσ ) ] = n [trσ + β p j= (et j Σ e j ) ] 0, we have n n i= (rt i Σ r i n Var[n n i= trσ ) = o p (). As n (r T iσ r i n n i= E(rT i Σ r i n trσ )] = n 3 [trσ + β we have n n i= (rt i Σ r i n trσ ) = o p (), and thus n n i= trσ ) = 0 and (e T j Σ e j ) ] 0, r T iσ r i r T iσ r i = (n trσ ) + o p (). (.) 3 j=

4 Step.. We have n i j l E(rT i Σ r j r T j Σr l) = 0 and E(n i j l r T i Σr j r T j Σr l ) 0n tr(σ 4 ) + (n 3 + 4n 4 )[tr(σ 4 ) + β (e T j Σ e j ) + (trσ ) ] +4n 4 β (e T j Σ e j ) + 8n 4 β j= j= This leads to n i j l (rt i Σ r j r T j Σ r l ) = o p (). Step.3. We have n j= (e T j Σe l ) 4 0. i j ErT i Σ r j r T j Σ r i = (n )n tr(σ ) and n E( r T iσ r j r T jσ r i ) i j 6n 5 (n )β j= (e T j Σ e j ) + n 5 +n 5 (n )(n )(n 3)[tr(Σ )] +n 5 (n )(n )[tr(σ 4 ) + β (n )β As a result, we have Var(n i j rt i Σ r j r T j Σ r i ) 0; that is, Step.4. We have n i j j= (e T j Σ e l ) 4 (e T j Σ e j ) + (trσ ) ]. j= r T iσ r j r T jσ r i (n )n tr(σ ) = o p (). n = n r T iσ r i r T iσ r j i j i j (r T iσ r i n trσ )r T iσ r j + (n trσ ) i j r T iσ r j, and further, n (r T iσ r i n trσ )r T iσ r j i j n i (r T iσ r i n Then, we have n i j rt i Σ r i r T i Σ r j = o p (). trσ ) + n r T iσ r j r T jσ r i. 4 i j

5 Thus, we have tr(s ) = n n (n ) tr[( Σ / r ir T iσ / ) ] n + n (n ) (trσ ) n tr(σ ) + o p (). i= As trs = n (n ) ( n i= rt i Σ r i n r T Σ r ), we have n trs = n (n ) r T iσ r i (n ) trσ + o p (). i= As shown in Bai and Silverstein (004, pp ), n tr[( Σ / r ir T iσ / i= n )q ] tr[( i= where r i = n / w i, w i = ( w i,..., w pi ) T, Σ / r i r T iσ / )q ] = o p (), q =, w li = [Var(w li δ { wli n η n })] / [w li δ { wli n η n } E(w li δ { wli n η n })], w li c n η n, E w li = 0, E( w li ) = and E( w4 li ) < for l =,..., p and i =,..., n with η n 0, nη n and c being a positive constant. For simplicity, we rename the variable w li simply as w li and proceed by assuming that w li n η n, Ew li = 0, E(w li ) = and E(w4 li ) < with η n 0 and n η n. Let B = n i= Σ/ r ir T i Σ/, then trs = n (n ) trb (n ) trσ + o p () and tr(s ) = n (n ) tr(b ) n + n (n ) (trσ ) n tr(σ ) + o p (). (.) Step 3. In this step, we show that tr[(s S ) ] Etr[(S S ) ] = n (n ) [tr(b ) Etr(B )] + n (n ) [tr(b ) Etr(B )] n n (n ) (n ) [tr(b B ) Etr(Σ Σ )] + o p () is asymptotically normal. When Σ = Σ = Σ, we have Etr(S ) + Etr(Σ ) Etr(S S ) 5

6 = = n + (n ) tr(σ ) + = n n n (n ) (trσ) + = β n (n ) (e T l Σ e l ), where p Etr(S ) = n (n ) + tr(σ p(n ) ) + n n pn (n ) (trσ ) + β wn p(n ) (e T l Σ e l ) = p tr(σ ) + n + p(n ) tr(σ ) + n n pn (n ) (trσ ) + β wn p(n ) (e T l Σ e l ). Since = (n n )n (n ) (trσ), we need to establish the CLT of tr(s S ) ˆµ, where ˆµ = = (n n )n (n ) (trs ). Let E l be the conditional expectation given {x,..., x l } and E l,s be the conditional expectation given {x,..., x l, S }. Based on the martingale difference central limit theory, it can be derived that conditional on S, σ / A [tr(s ) Etr(S ) tr(s S ) + tr(σ S ) n n n (n ) {(trs ) (trσ ) d }] N(0, ), where (.3) σ A = σ 0A + 4σ 0A 4σ 0A + 4n (trσ ) σ 330A 4n (trσ )σ 30A + 8n (trσ )σ 30A, with σ 0A = σ 0A = σ 330A = σ 0A = n n [(E l E l )tr(b )], [(E l,b E l,b )tr(b B )], n [(E l,b E l,b )tr(b )], n [(E l E l )tr(b )][(E l,b E l,b )tr(b B )], 6

7 σ 30A = σ 30A = n n [(E l E l )tr(b )][(E l,b E l,b )trb ], [(E l,b E l,b )(trb )][(E l,b E l,b )tr(b B )]. Moreover, we have σ / A [tr(s ) Etr(S ) [tr(σ S ) tr(σ Σ )] n n n (n ) {(trs ) (trσ ) d }] N(0, ), where (.4) σ A = σ 440A + 4σ 550A + 4n (trσ ) σ 660A 4σ 450A 4n (trσ )σ 460A + 8n (trσ )σ 560A, with σ 440A = σ 550A = σ 660A = σ 450A = σ 460A = σ 560A = n n [(E l E l )tr(b )], [(E l E l )tr(b Σ )], n [(E l E l )trb ], n [(E l E l )tr(b )][(E l E l )tr(b Σ )], n [(E l E l )tr(b )][(E l E l )trb ], n [(E l E l )tr(b Σ )][(E l E l )trb ]. Step 4. Next, we show that µ A = Etr(S ) + Etr(S ) tr(σ Σ ) = n n n (n ) (trσ ) = tr[(σ Σ ) ] + n + (n ) tr(σ ) + n + (n ) tr(σ ) + β n (n ) = (e T Σ e ) + β n (n ) = (e T Σ e ). 7

8 Moreover, we have σ 0A = σ 0A = n [(E l E l )tr(b )] = 4n [tr(σ 4 ) + β (e T l Σ e l ) ] + (n trσ ) n [tr(σ ) + β +4[n tr(σ )] + 8(n trσ )n [tr(σ 3 ) + β +(n trσ ) n [tr(σ ) + β n [(E l,b E l,b )tr(b B )] = n tr[(σ B ) ] + β n = σ 330A = = σ 0A = σ 30A = (e T l Σ e l ) ], (e T l Σ / B Σ / e l ) [tr(σ Σ )] + [n tr[(σ Σ ) ] + β n n n n n n [(E l,b E l,b )trb ] E(r T l Σ r l ) = n tr(σ ) + β n e T l Σ e l e T l Σ e l ] (e T l Σ e l ), [(E l E l )tr(b )][(E l,b E l,b )tr(b B )] = n [tr(σ 3 B ) + β [ +(n trσ )n tr(σ B ) + β = [n tr(σ 3 Σ ) + β n = e T l Σ / B Σ / e l e T l Σ e l )] [ +(n trσ )n tr(σ Σ ) + β n n (e T l Σ e l ) ] (e T l Σ / Σ Σ / e l ) ] + o p (), ] e T l Σ e l e T l Σ / B Σ / e l e T l Σ / Σ Σ / e l e T l Σ e l ] [(E l E l )tr(b )][(E l,b E l,b )tr(trb )] [(E l E l )tr(b )][(E l E l )trb ] 8 e T l Σ e l e T l Σ / Σ Σ / e l ] + o p (), l

9 σ 30A = = (n tr(σ 3 ) + β n = = σ 440A = σ 550A = σ 660A = σ 450A = e T l Σ e l e T l Σ e l ) +(n trσ )[n tr(σ ) + β n n n (e T l Σ e l ) ], [(E l,b E l,b )(trb )][(E l,b E l,b )tr(b B )] [(E l E l )trb ][(E l,b E l,b )tr(b B )] n E l,b (r T l Σ r l n trσ )[r T l Σ / B Σ / r l n tr(σ B )] = [n tr(σ Σ ) + β n n [(E l E l )tr(b )] = 4n [tr(σ 4 ) + β (e T l Σ e l )(e T l Σ / Σ Σ / e l )], (e T l Σ e l ) ] + (n trσ ) n [tr(σ ) + β +4[n tr(σ )] + 8(n trσ )n [tr(σ 3 ) + β +(n trσ ) n [tr(σ ) + β n [(E l E l )tr(b Σ )] = [n tr[(σ Σ ) ] + β n n n (e T l Σ e l ) ], e T l Σ e l e T l Σ e l ] (e T l Σ / Σ Σ / e l ) ], [(E l E l )trb ] = [n tr(σ ) + β n [(E l E l )tr(b )][(E l E l )trb Σ ] = n [tr(σ 3 Σ ) + β [ +(n trσ )n tr(σ Σ ) + β (e T l Σ e l ) ], e T l Σ / Σ Σ / e l e T l Σ e l ] e T l Σ e l e T l Σ / Σ Σ / e l ], (e T l Σ e l ) ] l 9

10 σ 460A = σ 560A = n [(E l E l )tr(b )][(E l E l )trb ] = [n tr(σ 3 ) + β n e T l Σ e l e T l Σ e l ] [ +(n trσ ) n tr(σ ) + β n n Thus, we have (e T l Σ e l ) ], [(E l E l )trb Σ ][(E l E l )trb ] = [n tr[(σ Σ )] + β n (e T l Σ e l )(e T l Σ / Σ Σ / e l )]. σ A = σ 0A + 4σ 0A + 4n (trσ ) σ 330A 4σ 0A 4n (trσ )σ 30A + 8n (trσ )σ 30A = 4[n tr(σ 4 ) + β n (e T l Σ e l ) ] + 4[n tr(σ )] + 8n (n trσ Σ ) n and +4[n tr[(σ Σ ) ] + β n 8[n tr(σ 3 Σ ) + β n (e T l Σ / Σ Σ / e l ) ] e T l Σ / Σ Σ / e l e T l Σ e l ], σ A = σ 440A + 4σ 550A + 4n (trσ ) σ 660A 4σ 450A 4n (trσ )σ 460A + 8n (trσ )σ 560A = 4n [tr(σ 4 ) + β (e T l Σ e l ) ] + 4[n tr(σ )] +4[n tr[(σ Σ ) ] + β n 8(n tr(σ 3 Σ ) + β n Step 5. According to (.3) and (.4), we have (e T l Σ / Σ Σ / e l ) ] e T l Σ / Σ Σ / e l e T l Σ e l )]. σ A {tr[(s S ) ] ˆµ µ A } N(0, ), where µ A = tr[(σ Σ ) ] + = n + (n ) tr(σ ) + = β n (n ) (e T l Σ e l ) 0

11 and σ A = σ A + σ A = 4[n tr(σ )] + 4[n tr(σ )] + 8n n [tr(σ Σ )] [ ] +4 n tr(σ 4 ) + β n (e T l Σ e l ) +4{n tr[(σ Σ ) ] + β n 8[n tr(σ 3 Σ ) + β n +4n [ tr(σ 4 ) + β ] (e T l Σ e l ) +4{n tr[(σ Σ ) ] + β n 8[n tr(σ 3 Σ ) + β n As a result, the proof of Lemma is completed. (e T l Σ / Σ Σ / e l ) } e T l Σ / Σ Σ / e l e T l Σ e l ] (e T l Σ / Σ Σ / e l ) } e T l Σ / Σ Σ / e l e T l Σ e l )]. Lemma Under Assumptions (A) (A), we have σ A σ A 3 σ A 3 σa 3 j / tr(s S ) ˆµ µ A tr(s S 3 ) ˆµ 3 µ A 3 where 0 = (0, 0) T and I is the identity matrix with N(0, I ), ˆµ ij = (n n )n (n ) (trs ), =i,j µ Aij = tr[(σ i Σ j ) ] + =i,j σ Aij = 4[n i tr(σ i )] + 4[n j +4[n i tr(σ 4 i ) + β i n i n + (n ) tr(σ ) + β n (n ) =i,j tr(σ j)] + 8n i n j [tr(σ i Σ j )] (e T l Σ i e l ) ] +4{n i tr[(σ i Σ j ) ] + β i n i (e T l Σ / i Σ j Σ / i e l ) } (e T l Σ e l ),

12 . 8[n i tr(σ 3 i Σ j ) + β i n i +4[n j tr(σ 4 j) + β j n j e T l Σ / i Σ j Σ / i e l e T l Σ i e l ] (e T l Σ je l ) ] +4{n j tr[(σ j Σ i ) ] + β j n j 8(n j tr(σ 3 jσ i ) + β j n j σ A 3 = 4[n tr(σ )] + 4[n tr(σ 4 ) + β n 4[n tr(σ 3 Σ 3 ) + β n 4[n tr(σ 3 Σ ) + β n (e T l Σ / j Σ i Σ / j e l ) } e T l Σ / j Σ i Σ / j e l e T l Σ je l )], +4[n tr(σ Σ Σ 3 Σ ) + β n (e T l Σ e l ) ] e T l Σ / Σ 3 Σ / e l e T l Σ e l ] e T l Σ / Σ Σ / e l e T l Σ e l ] e T l Σ / Σ Σ / e l e T l Σ / Σ 3 Σ / e l ]. Proof. Similar to Lemma, we first consider Σ with the bounded spectral norm for all =,, 3. When tr(σ q ) = O(pq ) for q =,, 3, 4 and at least one in the index set {,, 3 }, the proof mimics that in the bounded spectral norm case. Similar to the proof of Lemma, it can be shown that tr(s S ) ˆµ µ A and tr(s S 3 ) ˆµ 3 µ A 3 are asymptotically distributed as a bivariate normal distribution with the asymptotic variances σa, σa 3 and asymptotic covariance σ A 3. For simplicity of presentation, we consider the computation of σ A3, which is given by where σ A3 = σ 0A σ 30A σ 0A + 4 σ 30A + 4n (trσ ) σ 660A (n trσ )σ 460A +4(n trσ )σ 560A (n trσ )σ 460A + 4(n trσ )σ 760A, σ 760A = n [(E l E l )tr(b B 3 )][(E l E l )tr(b )],

13 σ 0A = σ 0A = σ 30A = σ 30A = n n [(E l E l )tr(b )], [(E l E l )tr(b )][(E l E l )tr(b B )], n [(E l E l )tr(b )][(E l E l )tr(b B 3 )], n [(E l E l )tr(b B )][(E l,b3 E l,b3 )tr(b B 3 )]. Then, we have σ 760A = σ 0A = σ 0A = σ 30A = n [(E l E l )tr(b B 3 )][(E l E l )tr(b )] = [n tr(σ 3 Σ ) + β n n [(E l E l )tr(b )] = 4n [tr(σ 4 ) + β (e T l Σ e l )(e T l Σ / Σ 3 Σ / e l )], (e T l Σ e l ) ] + 4(n trσ ) n [tr(σ ) + β +4[n tr(σ )] + 8(n trσ )n [tr(σ 3 ) + β n [(E l E l )tr(b )][(E l E l )tr(b B )] = n [tr(σ 3 B ) + β +(n trσ )n [tr(σ B ) + β = n [tr(σ 3 Σ ) + β +(n trσ )n [tr(σ Σ ) + β n (e T l Σ e l ) ] e T l Σ e l e T l Σ e l ], e T l Σ / B Σ / e l e T l Σ e l ] e T l Σ e l e T l Σ / B Σ / e l ] e T l Σ / Σ Σ / e l e T l Σ e l ] [(E l E l )tr(b )][(E l E l )tr(b B 3 )] = n [tr(σ 3 B 3 ) + β e T l Σ e l e T l Σ / Σ Σ / e l ] + o p (), e T l Σ / B 3 Σ / e l e T l Σ e l ] 3

14 σ 30A = +(n trσ )n [tr(σ B 3 ) + β = n [tr(σ 3 Σ 3 ) + β +(n trσ )n [tr(σ Σ 3 ) + β n Thus, we obtain e T l Σ e l e T l Σ / B 3 Σ / e l ] e T l Σ / Σ 3 Σ / e l e T l Σ e l ] [(E l E l )tr(b B )][(E l,b3 E l,b3 )tr(b B 3 )] = n [tr(b Σ B 3 Σ ) + β = [n tr(σ Σ Σ 3 Σ ) + β n e T l Σ e l e T l Σ / Σ 3 Σ / e l ] + o p (), e T l Σ / B Σ / e l e T l Σ / B 3 Σ / e l ] σ A3 = 4[n tr(σ )] + 4n [tr(σ 4 ) + β +4(n trσ ) n [tr(σ ) + β +8(n trσ )n [tr(σ 3 ) + β 4[n tr(σ 3 Σ 3 ) + β n e T l Σ / Σ Σ / e l e T l Σ / Σ 3 Σ / e l ] + o p (). (e T l Σ e l ) ] (e T l Σ e l ) ] l e T l Σ e l e T l Σ e l ] 4(n trσ )[n tr(σ Σ 3 ) + β n 4[n tr(σ 3 Σ ) + β n 4(n trσ )[n tr(σ Σ ) + β n +4[n tr(σ Σ Σ 3 Σ ) + β n e T l Σ / Σ 3 Σ / e l e T l Σ e l ] e T l Σ e l e T l Σ / Σ 3 Σ / e l ] e T l Σ / Σ Σ / e l e T l Σ e l ] +4(n trσ ) [n tr(σ ) + β n 4(n trσ )[n tr(σ 3 ) + β n 4 e T l Σ e l e T l Σ / Σ Σ / e l ] e T l Σ / Σ Σ / e l e T l Σ / Σ 3 Σ / e l ] (e T l Σ e l ) ] e T l Σ e l e T l Σ e l ]

15 4(n trσ )(n trσ )[n tr(σ ) + β n +4(n trσ )[n tr(σ Σ ) + β n 4(n trσ )[n tr(σ 3 ) + β n (e T l Σ e l ) ] (e T l Σ e l )(e T l Σ / Σ Σ / e l )] e T l Σ e l e T l Σ e l ] 4(n trσ )(n trσ )[n tr(σ ) + β n +4(n trσ )[n tr(σ 3 Σ ) + β n = 4[n tr(σ )] + 4[n tr(σ 4 ) + β n 4[n tr(σ 3 Σ 3 ) + β n 4[n tr(σ 3 Σ ) + β n +4[n tr(σ Σ Σ 3 Σ ) + β n (e T l Σ e l ) ] (e T l Σ e l )(e T l Σ / Σ 3 Σ / e l )] (e T l Σ e l ) ] e T l Σ / Σ 3 Σ / e l e T l Σ e l ] e T l Σ / Σ Σ / e l e T l Σ e l ] e T l Σ / Σ Σ / e l e T l Σ / Σ 3 Σ / e l ]. Generally, σ A 3 is obtained by replacing n i, Σ i and β i by n i, Σ i and β i in σ A3. That is, σ A 3 = 4[n tr(σ )] + 4[n tr(σ 4 ) + β n 4[n tr(σ 3 Σ 3 ) + β n 4[n tr(σ 3 Σ ) + β n +4[n tr(σ Σ Σ 3 Σ ) + β n The proof of Lemma is completed. (e T l Σ e l ) ] e T l Σ / Σ 3 Σ / e l e T l Σ e l ] e T l Σ / Σ Σ / e l e T l Σ e l ] e T l Σ / Σ Σ / e l e T l Σ / Σ 3 Σ / e l ]. 5

16 Proofs of Theorem and By Lemma and the delta method, under the conditions of Lemma, we have σ AK (T K ˆµ K µ AK ) d N(0, ), where ˆµ K = ω ˆµ = < K < K and µ AK = i<j K ω ijµ Aij and [ ω (n n )n (n ) (trs ) ], =, σ AK = ωijσ Aij + ω ij ω j σ Aijj i<j K + i<j< ω ij ω j σ Aijj + i<<j j<i< ω ij ω j σ Aijj, with the weights {ω ij, i, j K} and ω ij = ω ji. By (3) and (33) in Cai, Liu and Xia (03), we have a.s. a.s. max l l p δ l l s(n, n, p) 0.5 max i j p 0.5 max i j p (σ ij σ ij) ˆθ ij/n + ˆθ ij/n (σ ij σ ij) θ ij/n + θ ij/n s(n, n, p) 4 log p log log p s(n, n, p) 4 log p log log p Then, there exists a pair of, K satisfying (σ ij σ ij) 0.5 max s(n, n, p) + 4 log p. i j p θ ij/n + θ ij/n Then we have and max δ l l s(n, n, p) a.s. 0.5 log log p l l p max δ l l s(n, n, p) a.s. > 0 l l p 6

17 Therefore, we have That is, T K a.s. = K 0. Then, we have max {I{ max δ ijl l > s(n i, n j, p)}} a.s. =. i<j K l l p σ AK (T K K 0 ˆµ K µ AK ) d N(0, ). We first focus on the proof of the case with K =, and that of K 3 can be shown in a similar way. When K =, the power function is g (Σ, Σ ) = P HA (T ˆµ > ˆµ + z αˆσ ) ( T ˆµ µ A = P HA > ˆµ ) µ A + z αˆσ σ A > P HA ( T ˆµ µ A σ A > z αˆσ σ A Step. When p tra 0 and tra > c 0 for some positive constant c 0, σ A /ˆσ. That is, (z αˆσ )/σ A z α = o p (). Then, when n, n are large enough, we have g (Σ, Σ ) > α. Step. We have g (Σ, Σ ) > P HA ((T ˆµ µ A )σ A > (ˆµ µ A + z αˆσ )σ A ). When tr(a ), we have (ˆµ µ A )σ A in probability. Then, the power function satisfies g (Σ, Σ ). Step 3. We have σ A ). g (Σ, Σ ) = P HA (T ˆµ > ˆµ + z αˆσ ) ( T ˆµ µ A = P HA > ˆµ ) µ A + z αˆσ σ A > P HA ( T ˆµ µ A σ A > z αˆσ σ A > P HA ( T ˆµ µ A σ A σ A ) + T K σ A > z αˆσ σ A ). Because T ˆµ µ A σ A is asymptotically normal under H and T K a.s. = K 0, if (σ ij σ ij) 0.5 max s(n, n, p) + 4 log p, i j p θ ij/n + θ ij/n 7

18 then the power function g (Σ, Σ ) tends to one. The proof of Theorem is completed. Moreover, Theorem is a special case of Theorem. 8

19 References Bai, Z. D. and Silverstein, J. W. (004). CLT for linear spectral statistics of large dimensional sample covariance matrices. Ann. Probab., 3, Schott, J. R. (007). A test for the equality of covariance matrices when the dimension is large relative to the sample sizes. Comput. Stat. Data Anal., 5, Srivastava, M. S. and Yanagihara, H. (00). Testing the equality of several covariance matrices with fewer observations than the dimension. J. Multivar. Anal., 0,

20 (a) Size, n =n =n 3 =00 Percentage T 3 T 3 SC SY p (b) Power, n =n =n 3 =00 Percentage p (c) Size, n =60, n =00, n 3 =50 Percentage p (d) Power, n =60, n =00, n 3 =50 Percentage p Figure : Simulation results for testing the equality of three covariance matrices with Gamma populations under Scenarios 4 in comparison with two existing tests of Schott (007) (SC) and Srivastava and Yanagihara (00) (SY). 0