Supplementary Appendix to “Asymptotic Power of Sphericity Tests for High-dimensional Data” Alexei Onatski, Marcelo J. Moreira and Marc Hallin June 22, 2012

Abstract This note contains proofs of lemmas 4, 5, 6, 11, 12 and 13 in Onatski, Moreira and Hallin (2011), Asymptotic power of sphericity tests for highdimensional data, where we refer to for de…nitions and notation.

A

Proof of Lemma 4

The original contour K is such that the singularities z = integrand remain inside, whereas the singularity z =

1+h S h

1 ; :::; z

=

p

of the

remains outside the

domain encircled by K. Su¢ cient conditions for K to be similarly located with respect to the singularities of the integrand, and for f (z) and g(z) to be well-de…ned on K are min z0 (h) > max fbp ;

h2(0;h]

1g

(A1)

and h z0 (h) max < 1: h2(0;h] 1 + h S

(A2)

Alexei Onatski: University of Cambridge, [email protected]. Marcelo J. Moreira: FGV/EPGE, [email protected]. Marc Hallin: Université libre de Bruxelles and Princeton University, [email protected].

1

Hence, to establish Lemma 4 it is enough to show that (A1) and (A2) hold with probability approaching one as p; n ! 1 so that cp ! c: p 2 p c=h h . Consider the event E Let us …x a positive " such that " <

that holds if and only if the following four inequalities simultaneously hold:

(A3)

minh2(0;h] (z0 (h) bp ) > "; p 2 bp 1+ c < "=4; p 2 1+ c < "=4; 1 minh2(0;h]

1+h S h

z0 (h)

(A4) (A5) (A6)

> ":

Clearly, E implies (A1) and (A2). On the other hand, Pr (E) ! 1 as n; p ! 1 so that cp ! c. Indeed, by de…nition of z0 (h) and bp , z0 (h)

bp =

r

p

cp h

2

h

:

Therefore, as cp ! c, min (z0 (h)

h2(0;h]

bp ) ! min h2(0;h]

r

p

c h

2

h

=

r

c h

p h

2

;

which is larger than " by assumption. Hence, the probability of (A3) converges to p 2 p 2 one. Further, bp ! (1 + c) by de…nition, while 1 ! (1 + c) almost surely under our null hypothesis, as shown, for example, in Geman (1980). Thus, the probabilities of (A4) and (A5) converge to one too. Finally, by de…nition of z0 (h), h z (h) 1+h 0

= h + cp , so that

minh2(0;h]

1+h S h

z0 (h)

=

1+h S h

h

cp :

But under our null hypothesis S=p ! 1 in probability, as n; p ! 1 so that cp ! c. 2

This follows, for example, from Theorem 1.1 of Bai and Silverstein (2004). Hence, the probability of (A6) also converges to one. It remains to note that 1

Pr (E)

equals the probability of the union of the events complementary to (A3)-(A6).

B

Proof of Lemma 5

We have shown, in the proof of Lemma 4, that Pr (E) ! 1. Therefore, it is su¢ cient to prove Lemma 5 under the assumption that E holds. Event E implies that f (z) and g(z) are analytic at z0 (h) for any h 2 0; h . Furthermore, still under E,

f1

1 d2 f (z)jz=z0 (h) < 0: 2 dz 2

d f (z)jz=z0 (h) = 0 and f2 dz

Indeed, by de…nition, z0 (h) is a critical point of f (z) when h < p h < cp : Otherwise,

z0 (h)

at h =

p cp

r

bp

p h

cp h

p

cp : But E implies

2

=0<"

h, which contradicts (A3). Further, a direct computation based on

(3.3), (3.6), and (3.7)1 shows that

f2 =

1 4 (cp

h2 < 0: h2 ) (1 + h)2

First, let us focus on the analysis of H useful representation for the part of K1 e

H

K1

e

nf (z)

nf (z)

(A7)

g(z)dz: Olver (1997) derives a

g(z)dz that corresponds to a portion

of K1 close to its boundary point, which in our case is z0 (h): To make our exposition self-contained, we sketch Olver’s derivation; for details, we refer the reader to pages 1

Here and throughout this Supplement, numerical references are for equations in the main text.

3

121-124 of Olver’s book. Let us introduce new variables v and w by the equations

w2 = v = f (z)

(A8)

f0 ;

where the branch of w is determined by lim farg (w)g = 0 as z ! z0 (h) along K1 , and by continuity elsewhere. Consider w as a function of z: Since f1 = 0, there exists a small neighborhood of z0 (h), where the indicated branch of w(z) is an analytic function. Moreover, there exists a small number (h) > 0 such that w(z) maps the disk jz conformally on a domain

z0 (h)j < (h)

containing w = 0:

Let z1 (h) be a point of K1 chosen su¢ ciently close to z0 (h) to insure that the disk jwj

jf (z1 (h))

f0 j1=2 is contained in

: Then the portion [z0 ; z1 ]

[z0 (h); z1 (h)] of contour K1 can be deformed, without changing the value of the H integral [z0 ;z1 ] e nf (z) g(z)dz, to make its w(z) map a straight line. Transformation to the variable v gives I

e

nf (z)

nf0

g(z)dz = e

[z0 ;z1 ]

I

e

nv

'(v)dv;

(A9)

[0; (h)]

where (h) = f (z1 (h))

f0 ; '(v) =

g(z) ; f 0 (z)

(A10)

and the path for the integral on the right-hand side of (A9) is also a straight line. For small jvj = 6 0, '(v) has a convergent expansion of the form '(v) =

1 X

as v (s

1)=2

;

(A11)

s=0

in which the coe¢ cients as are related to fs and gs : The formulae for a0 ; a1 ; and

4

a2 are given, for example, on p.86 of Olver (1997). We use them in the statement of Lemma 5. Finally, de…ne 'k (v), k = 0; 1; 2; ::: by the relations 'k (0) = ak and

'(v) =

k 1 X

as v (s

1)=2

+ v (k

1)=2

s=0

'k (v) for v 6= 0:

(A12)

Then the integral on the right-hand side of (A9) can be rearranged in the form I

e

nv

'(v)dv =

[0; (h)]

k 1 X

as

s+1 2

s=0

"k;1 (h) + "k;2 (h) ;

n(s+1)=2

(A13)

where

"k;1 (h) = "k;2 (h) =

k 1 X

Is=0

s+1 ; (h)n 2 nv (k 1)=2

e

v

as ; (s+1)=2 n

'k (v) dv;

(A14) (A15)

[0; (h)]

and x

( ; x) = e x

Z

1

e

xt

(1 + t)

1

dt

0

is the incomplete Gamma function. This completes our sketch of Olver’s derivation. The remaining part of the proof of Lemma 5 is mostly concerned with two auxiliary lemmas establishing uniform asymptotic properties of "k;1 (h) and "k;2 (h) : The …rst of these two lemmas provides explicit forms for (h), z1 (h), and (h) allowing further analysis of their dependence on h. Lemma A1. Let B ( ; R) and B ( ; R) denote, respectively, the open and closed balls in the complex plane with center at r(h) = min z0 (h) i r(h), 9 26

max fbp ;

and (h) = f (z1 (h))

1g ;

1+h S h

and radius R: Further, let

z0 (h) , (h) =

1 r(h), 3 24

f0 : If event E holds, then, 5

z1 (h) = z0 (h) +

(i) For any

1;

2

from B (z0 (h); (h)), we have jw ( 2 )

w ( 1 )j >

1 2

1=2

f2

j

2

1 j;

(ii) The function w(z) is a one-to-one mapping of B (z0 (h); (h)) on an open set : The inverse function z(w) is analytic in (iii) There exist positive constants 2

1

and

;

such that Re (h) >

2

1

and Im (h) <

for all h 2 0; h ;

(iv) B 0; 2 j (h)j1=2 is contained in

:

Proof. Throughout this proof, we simplify the notation and write z0 , z1 , r, , and

instead of z0 (h), z1 (h), r(h), (h), and (h), respectively. First, we show 1=2

that w(z) is analytic in B (z0 ; ) and that w0 (z0 ) = f2 : Let f (j) (z) denote the j-th order derivative of f (z): Consider the Taylor expansion of f (j) (z) at z0 :

f (j) (z) =

k X 1 (j+s) f (z0 ) (z s! s=0

z0 )s + Rj;k+1 :

In general, for any z 2 B (z0 ; R), the remainder Rj;k+1 satis…es jz z0 jk+1 max f (j+k+1) (t) : (k + 1)! jt z0 j R

jRj;k+1 j

(A16)

From de…nition (3.3) of f (z), we have

f

(s)

cp (t) = ( 1)s 2

If t 2 B z0 ; 12 r , then jt

1

(s

1)!

j > 12 (z0 jt

js+1 >

Z

(t

) for any 1 2s+1

6

(z0

)

s

dFp ( ) for s

2:

(A17)

in the support of Fp : Therefore, )s r;

and using (A17) we get

f (s+1) (t) <

s2s+1 (s) f (z0 ) for s r

Combining this with (A16), we obtain for k + j jz

jRj;k+1 j

(A18)

2:

2 2 and z 2 B z0 ; k+1 k+j

k j 2

z0 jk (k+j) f (z0 ) : 2k!

r ,

(A19)

Further, since Rj;k =

1 (k+j) f (z0 ) (z k!

2 and z 2 B z0 ; k+1 2 k+j

(A19) implies that, for k + j 1 f (k+j) (z0 ) jz 2k!

z0 )k + Rj;k+1 ;

z0 jk < jRj;k j <

k j 2

r ,

3 f (k+j) (z0 ) jz 2k!

z0 jk :

(A20)

Next, since f (1) (z0 ) = 0, inequalities (A20) imply that

jf (z)

f (z0 )j = jR0;2 j >

1 (2) f (z0 ) jz 4

z0 j2

1 jf2 j jz 2

z0 j2

(A21)

for any z 2 B z0 ; 235 r : Since f2 6= 0, inequality (A21) implies that f (z)

f (z0 )

does not have zeros in B z0 ; 235 r except a zero of the second order at z = z0 : Therefore,

s

f (z) (z

f (z0 ) w (z) = 2 (z z0 ) z0 ) 1=2

is analytic inside B z0 ; 235 r , which includes B (z0 ; ), and converges to f2

as

1=2

z ! z0 : This implies that w (z) is analytic in B (z0 ; ) and w0 (z0 ) = f2 . Now, let us show that, for any z 2 B (z0 ; ), jw0 (z)

w0 (z0 )j <

7

1 0 jw (z0 )j : 2

(A22)

Indeed, since w0 (z) = 1=2

and w0 (z0 ) = f2

f 0 (z) 1 = (f (z) 2w (z) 2

f0 )

1=2

f 0 (z)

6= 0,

w0 (z) = w0 (z0 )

R0;3 1+ f2 (z z0 )2

1 2

1+

R1;2 2f2 (z z0 )

:

(A23)

Note that for any y1 and y2 such that jy2 j < 1, 1 + y1 p 1 + y2

jy1 j + jy2 j ; 1 jy2 j

1

(A24)

where the principal branch of the square root is used. This follows from the facts p p that, for jy2 j < 1, 1 + y2 1 jy2 j and 1 + y1 1 + y2 jy1 j + jy2 j : Setting y1 =

R0;3 R1;2 and y2 = 2f2 (z z0 ) f2 (z z0 )2

and using (A23), (A20) and the fact that, for any z 2 B (z0 ; ), f (3) (z0 ) jz f (2) (z0 )

1 z0 j < ; 3

which follows from (A18), we get w0 (z) w0 (z0 )

1 1 < : 2

Hence, (A22) holds. Finally, let

1

and

2

be any two points in B (z0 ; ), and let (t) = (1

t 2 , where t 2 [0; 1] : We have Z

1

(w0 ( (t))

w0 (z0 )) dt =

0

w ( 2) 2

8

w ( 1) 1

w0 (z0 ) :

t)

1

+

Therefore, using (A22), we obtain w ( 2)

w ( 1)

2

w0 (z0 ) <

1 1=2

This inequality and the fact that w0 (z0 ) = f2

1 0 jw (z0 )j : 2

imply part (i) of the lemma.

Part (ii) of the lemma is a simple consequence of part (i) and of the analyticity of w(z) in B(z0 ; ), established above. Indeed, by the open mapping theorem, an open set. Next, by (i), w(z) is one-to-one mapping of B(z0 ; ) on non-zero derivative in B(z0 ; ): Further, let Fix w~ 2 : Then

(w) be de…ned on

(w) ~ = z~ for a unique z~ in B(z0 ; ): If w 2

is

and has a

by

(w (z)) = z:

and

(w) = z, we

have (w) w

(w) ~ w~

=

z w (z)

z~ : w (~ z)

By (i), w ! w~ as z ! z~, and the latter equality implies z(w)

(w) is an analytic inverse of w(z) on

0

(w) ~ =

1 : w0 (~ z)

Therefore,

.

To see that part (iii) holds, note that cp Re = 2 and for any

such that 0

Z

ln

z1 z0

dFp ( ) ;

(A25)

< z0 , we have z1 z0

1+

i r : 9 26 z0

When E holds, the latter expression is bounded from below by a …xed constant that is strictly larger than one for all h 2 0; h : Therefore, when E holds, (A25) implies that Re >

1

> 0, for all h 2 0; h , where

9

1

is …xed:

Next, by de…nition of , we have

Im =

1 2

h r 1 + h 9 26

cp

Z

arg

z1 z0

dFp ( ) :

But h h r< z0 1+h 1+h

cp + h;

which is smaller than a …xed positive number for all h 2 0; h when E holds: Here the boundedness of h is obvious whereas the boundedness of cp follows from (A4). Further, arg for all h 2 0; h because Re zz10

z1 z0

<

2

1: Hence, there exists

2

such that jIm j <

2

for all h 2 0; h : Finally, part (iv) of the lemma can be established as follows. Note that by part (i), w z0 + ei for any

w (z0 ) >

2

jw0 (z0 )j

2 [0; 2 ] : Therefore, for any w1 such that jw1

w(z0 )j

4

jw0 (z0 )j, we

have min w1

w z0 + ei

>

4

jw0 (z0 )j :

By a corollary to the maximum modulus theorem (see Rudin (1987), p.212), the latter inequality implies that the function w (z) region

w1 has a zero in B(z0 ; ): Thus,

includes B(0; 4 jw0 (z0 )j). On the other hand, 2 j j1=2 <

4

10

jw0 (z0 )j :

Indeed, consider the identity

= f (1) (z0 ) (z1

z0 ) + R0;2 :

Since f (1) (z0 ) = 0, (A20) together with (A7) imply

j j< 1=2

Since w0 (z0 ) = f2

and jz1

z0 j =

3 jf2 j jz1 2 1 r, 9 26

2 j j1=2 < Therefore,

z0 j2 :

the latter inequality implies that

4

jw0 (z0 )j :

includes B(0; 2 j j1=2 ):

Before proceeding with the proof of Lemma 5, we still need one more auxiliary lemma. Lemma A2. Under the null hypothesis, supz2 so that cp ! c, where

1

1

jg(z)j = Op (1) as n; p ! 1

z0 (h)j < 21 r(h) and Op (1) is uniform

= z : jRe(z)

over h 2 0; h : Proof. First, consider the case when g(z) = exp p X

p (z)

ln (z

j)

p

j=1 p

=

X

ln 1

j=1

This statistic

p (z)

j

Z

ln (z

p

z

Z

ln 1

1 2

p (z)

, where

) dFp ( )

z

dFp ( ) :

is a special form of a linear spectral statistic

p (')

p X

' ( j)

j=1

11

p

Z

' ( ) dFp ( )

studied by Bai and Silverstein (2004). According to their Theorem 1.1, if ' ( ) h p 2i is analytic on an open set containing interval Ic 0; (1 + c) , then the sequence f Pr (j

p

p

(')j

(')g is tight. That is, for any for every

B) > 1

p

> 0 there exists a bound B such that

(') from the sequence:

A close inspection of Bai and Silverstein’s (2004, pp.562-563) proof of tightness reveals that the bound B can be chosen so that it depends on ' ( ) only through its supremum over an open area A that includes Ic and where ' ( ) is analytic. In particular, if we denote by

a family of functions ' (x), each of

which is analytic in the area A = x : sup sup'2 supx2A j' (x)j < 1, then sup'2 j Let

= '(x)

x z

ln 1

2

:z2

2

2Ic p

jx

j < " , and if

is such that

(')j is tight.

, where

n p = z : Re(z) > 1 + c

2

o + 2" :

This family of functions satis…es the above requirements. Indeed,

sup x2A;z2

so that each of '( ) 2

2

p 2 (1 + c) + " x = <1 p 2 z (1 + c) + 2"

is analytic in A: Moreover, since by de…nition

ln 1

x = ln 1 z

x + i arg 1 z

x ; z

we have sup sup j' (x)j < ln j1 '2

Rj +

x2A

2

;

where R

sup x2A;z2

Therefore, sup'2 j

p

2

x < 1: z

(')j is tight and supz2 12

2

jg(z)j = Op (1), where Op (1) does

not depend on h: It remains to note that, as p; n ! 1 so that cp ! c, z0 (h) inf h2(0;h]

1 r(h) 2

!

p 1 h+1 c+h 1 1+ c + 2 2 h

almost surely. Therefore, for a su¢ ciently small ", Pr ( supz2

1

2

> 1+

2)

1

p

c

2

! 1, and thus,

jg(z)j = Op (1), where Op (1) is uniform over h 2 0; h :

Now, consider the case when np

g(z) = exp

Since, as has just been shown, supz2 prove that supz2

1

h z 1+hS

p+2 ln 1 2

1 2

exp

1

n hz 21+h p (z)

(z) 2

p

:

= Op (1), we only need to

g~ (z) = Op (1), where

g~ (z) = exp

np

p+2 Re ln 1 2

h z 1+hS

n h Re z 2 1+h

:

We have

Re ln 1

h z 1+hS

h z > ln 1 1+hS

= ln 1

Note that (A6) and the de…nition of

1

h Re z 1+h S

imply that

h Re z <1 1+h S for any z 2

1:

In general, for any real x such that 0 < x < 1, we have

ln (1

x) >

13

x 1

x

:

:

Therefore, for any z 2

1,

h Re z 1+h S

ln 1

>

h Re z 1+h

S

1

h Re z ; 1+h

and we can write "

p h Re z ln g~ (z) < 2cp 1 + h From the de…nition of

p

1

h Re z 1+h

S

1

#

1 :

(A26)

1,

h h Re z < 1+h 1+h Further, S

p

2 cp + n

1 r(h) + z0 (h) 2

+ ::: +

p

<

3 hz0 (h) 3 = (h + cp ) : 2 1+h 2

p = Op (1) by Theorem 1.1 of Bai and Silver-

stein (2004). Combining these facts with (A26), we get supz2

1

g~ (z) = Op (1)

uniformly over h 2 0; h : Let us return to the proof of Lemma 5. Consider '(v)w as a function of w: According to (A8) and (A11), '(v)w has a convergent series representation

'(v)w =

1 X

as w s

(A27)

s=0

for su¢ ciently small jwj : Let us show that the series in (A27) converges for all w 2 : Indeed, from (A10), we see that '(v)w = (2w0 (z))

1

g (z) :

(A28)

By Lemma A1 (ii), z, viewed as the inverse of w(z), is analytic in : Further, g (z)

14

and w0 (z) are analytic in z ( )

B (z0 (h) ; (h)) : Finally, jw0 (z)j >

1 1=2 f 2 2

(A29)

1=2

for z 2 B (z0 (h); (h)) by Lemma A1 (i), and f2

6= 0 for h 2 0; h : Therefore,

'(v)w must be analytic in

and the series (A27) must converge there: n o 1=2 Now, formula (A7) implies that inf h2(0;h] f2 =h > 0. Therefore, from

Lemma A2 and (A29), we have

sup j'(v)wj

sup

w2

z2B(z0 (h); (h))

g (z) = h 1 Op (1); 2w0 (z)

(A30)

where Op (1) is uniform in h 2 0; h : By Lemma A1 (iii) and (iv), j (h)j > jRe (h)j >

1

and B 0; j 1 j1=2 is con-

tained in , where '(v)w is analytic. Using Cauchy’s estimates for the derivatives of an analytic function (see Theorem 10.26 in Rudin (1987)), (A27) and (A30), we get jas j

j 1j

s=2

sup j'(v)wj = h 1 Op (1): w2B (0;j 1 j1=2 )

Next, Olver (1997, ch. 4, pp.109-110) shows that j j ! 1; uniformly in the sector jarg ( )j us take

=

s+1 2

and

2

(A31)

( ; ) = O e

1n

!1

and jarg ( (h) n)j = arctan

Im (h) < arctan Re (h)

15

as

for an arbitrary positive : Let

= (h) n: Lemma A1 (iii) shows that

j (h) nj >

1

2 1

<

2

;

uniformly over h 2 0; h : Therefore, s+1 ; (h) n 2

=O e

(h)n

( (h) n)

s 1 2

= Op e

1 n 2 1

(A32)

for any integer s, uniformly over h 2 0; h : Equality (A32), the de…nition (A14) of "k;1 (h), and inequality (A31) imply that "k;1 (h) = h 1 Op (e

1 n 2 1

);

(A33)

where Op ( ) is uniform over h 2 0; h : Next, consider wk 'k (v) as a function of w: Since, by de…nition,

wk 'k (v) = ' (v) w

k 1 X

as w s ;

s=0

it can be interpreted as a remainder in the Taylor expansion of ' (v) w: As explained above, such an expansion is valid in

, which includes the ball B 0; 2 j (h)j1=2

by Lemma A1 (iv). By a general formula for remainders in Taylor expansions, for any w 2 B 0; j (h)j1=2 , k

w 'k (v)

jwjk k!

dk max (w' (v)) : k w2B (0;j (h)j1=2 ) dw

(A34)

Further, for any w 2 B 0; j (h)j1=2 , a ball with radius j 1 j1=2 centered in w is contained in the ball B 0; 2 j (h)j1=2

. Therefore, using (A30) and

Cauchy’s estimates for the derivatives of an analytic function (see Theorem 10.26 in Rudin (1987)), we get dk max (w' (v)) k w2B (0;j (h)j1=2 ) dw

k! j 1 j

16

k=2

sup jw' (v)j = h 1 Op (1):

w2

(A35)

Combining (A34) and (A35), we have

sup v2(0; (h)]

j'k (v)j = h 1 Op (1):

This equality together with (A31) and the fact that, by de…nition, 'k (0) = ak imply that max j'k (v)j = h 1 Op (1);

(A36)

v2[0; (h)]

where Op (1) is uniform in h 2 0; h : For "k;2 (h), the substitution of variable v = (h) nx in the integral (A15) yields "k;2 (h) = n

(k+1)=2

Z

n

e

(h)x

x

k 1 2

(h)

k+1 2

'k (v) dx:

0

Therefore,

"k;2 (h) n

(k+1)=2

< <

max j'k (v)j

v2[0; (h)]

max j'k (v)j

v2[0; (h)]

Z

n

e

Re (h)x

x

k 1 2

0

Z

1

e

Re (h) y j (h)j

y

k 1 2

j (h)j

k+1 2

dx (A37)

dy:

0

But by Lemma A1 (iii), Re (h) Re (h) > > j (h)j jRe (h)j + jIm (h)j

1 1

+

2

for all h 2 0; h : Therefore, the integral in (A37) is bounded uniformly over h 2 0; h : Using (A36), we conclude that "k;2 (h) = h 1 Op n

17

(k+1)=2

:

(A38)

Combining (A9), (A13), (A33), and (A38), we get I

e

nf (z)

nf0

g(z)dz = e

k 1 X

s+1 2

s=0

[z0 ;z1 ]

as n(s+1)=2

Op (1) + (k+1)=2 hn

!

(A39)

;

where Op (1) is uniform in h 2 0; h : Let us now consider the contribution of K+ n[z0 ; z1 ], that is, the part of contour H K+ excluding the segment [z0 ; z1 ], to the contour integral K+ e nf (z) g(z)dz: On K1 , cp f0 ) = 2

Re (f (z)

Z

ln 1 + i

Im z z0 (h)

dFp ( )

is an increasing function of Im (z) : Hence, on K1 n[z0 ; z1 ], Re (f (z)

f0 ) > Re

1:

Therefore, I

e

nf (z)

e

g(z)dz

K1 n[z0 ;z1 ]

nf0

e

n

1

I

K1 n[z0 ;z1 ]

= e

nf0

e

n

1

= e

nf0

e

n

1

jg(z)dzj

j3z0 (h)j Op (1)

h 1 Op (1):

(A40)

For the horizontal part K2 of K+ , consider …rst the case when g(z) = exp I

e

nf (z)

1 2

p

g(z)dz

(z) : We have

=

K2

I

e

n h z 2 1+h

K2

=

n h 21+h

p Y

(z

j)

1 2

dz

e

p 2

ln(3z0 (h))

j=1

I

n

h

e 2 1+h z dz

K2

1

e

n 2

(cp ln(3z0 (h))

18

h z (h) 1+h 0

):

(A41)

But

h z (h) 1+h 0

h + cp , so that

cp ln (3z0 (h))

h z0 (h) > cp ln (z0 (h)) 1+h

h > 2f0 + cp :

Combining such a lower bound with (A41), we get I

e

nf (z)

g(z)dz = e

nf0

n c 2 p

h 1O e

nf0

=e

h 1 Op e

n c 4

(A42)

;

K2

where Op e

n c 4

does not depend on h:

For the case when

g(z) = exp

np

p+2 ln 1 2

h z 1+hS

n hz 21+h

(z) 2

p

;

we have I

e

nf (z)

g(z)dz

=

K2

I

1

np p+2 2

h z 1+hS

e

ln(3z0 (h))

I

(z

1 2

j)

dz

j=1

K2 p 2

p Y

1

h z 1+hS

1

h x 1+hS

np p+2 2

dz :

K2

Further, I

K2

1

h z 1+hS

np p+2 2

dz

Z

z0 (h) 1

2S 1 + h = np p h

19

1

np p+2 2

dx

h z0 (h) 1+h S

np + p2 2

:

Hence, we can write I

nf (z)

e

2S 1 + h e np p h

g(z)dz

np p 2

ln 1

h z0 (h) 1+h S

p 2

ln(3z0 (h))

(A43)

:

K2

Now, for any real x such that 0 < x < 1, we have ln (1 np

p 2

ln 1

h z0 (h) 1+h S

< (p

cp ) S

x : 1 x

x) >

1

hz0 (h) 1+h

Hence,

n hz0 (h) : 2 1+h

But (p

cp ) S

1

hz0 (h) 1+h

= 1 + Op n

1

:

The Op (n 1 ) quantity here is uniform over h 2 0; h in view of the facts that S

p = Op (1) by Theorem 1.1 of Bai and Silverstein (2004), hz0 (h) = jh + cp j 1+h

h + cp

for all h 2 0; h , and n and p diverge to in…nity at the same rate. Therefore, (A43) implies I

e

nf (z)

g(z)dz =

n h 21+h

1

e

n 2

(cp ln(3z0 (h))

h z (h) 1+h 0

) O (1) ; p

(A44)

K2

which, similarly to (A41), implies (A42). Combining (A39), (A40), and (A42), we get I

K+

e

nf (z)

g(z)dz = e

nf0

k 1 X

s+1

as n(s+1)=2

s=0

20

Op (1) + (k+1)=2 hn

!

:

(A45)

Finally, note that I

e

nf (z)

g(z)dz =

K

I

e

nf (z)

I

g(z)dz

e

nf (z)

g(z)dz;

~ K

K+

~ is a contour that coincides with K but has the opposite orientation: where K As explained in Olver (1997, pp.121-122), as with odd s in the asymptotic expanH sion for K~ e nf (z) g(z)dz coincides with the corresponding as in the asymptotic H expansion for K+ e nf (z) g(z)dz: However, as with even s in the two expansions

di¤er by the sign. Therefore, coe¢ cients as with odd s cancel out, but those with even s double in the di¤erence of the two expansions. Setting k = 2m, we have I

e

nf (z)

g(z)dz = 2e

m X1

nf0

K

a2s

1 s+ 2

s=0

ns+1=2

Op (1) + m+1=2 hn

!

;

which establishes Lemma 5.

C

Proof of Lemma 6

Fix 0 < " <

q ~ c=h

(A4) and (A5) hold,

p ~ h

2

, and consider the event E1 that holds if and only if

~ z0 (h)

bp > "

and minh2[h;1 ~ )

1+h S h

~ z0 (h)

> ":

h ~ 1 , the integrals in The fact that, with probability approaching 1, for all h 2 h;

~ can be established along (2.9) and (2.10) do not change as K is deformed into K(h) the same lines as in the proof of Lemma 4 by replacing event E with event E1 . Similarly, an equivalent, for h same steps. Hence, since Re f (z)

~ of Lemma 2A, is easily proved along the h, ~ f z0 h 21

is an increasing function of Im z

~ , on K1 h I

nf (z)

e

g(z)dz

~ )) nf (z0 (h

e

I

~) K1 (h

~) K1 (h ~ )) nf (z0 (h

= e

jg(z)dzj (A46)

Op (1) :

Further, as in (A41) and (A44), we have I

e

nf (z)

g(z)dz

1

n h 21+h

=

~) K2 (h

~ )) nf (z0 (h

= e

n 2

e

(cp ln(3z0 (h~ ))

Combining (A46) and (A47), we get

e

nf (z)

g(z)dz = e

I

e

nf (z)

g(z)dz = e

~ )) nf (z0 (h

~) K+ (h

Similarly,

K

~ )) nf (z0 (h

(h~ )

Op (1):

Op (1):

Lemma 6 follows from the latter two equalities.

D

Proof of Lemma 11

Consider I (h)

Z

bp

ln (z0 (h)

ap

22

)

(h~ )) O (1) p (A47)

Op (1):

I

h z 1+h 0

p

( )d ;

where

p

( ) is de…ned in (3.2). Making the substitution

p 2 cp cos

= 1 + cp

and replacing z0 (h) by the right-hand side of (3.7), we get

I (h) =

2

Z

0

=

1

Z

2

0

p ln h + h 1 cp + 2 cp cos sin2 d p 1 + cp 2 cp cos 2 p p ln cp =h + hei sin2 d : p 1 + cp 2 cp cos

Further, changing the variable of integration from

1 I (h) = 2 i

I

jzj=1

to z = ei , we get

h p p p p ln cp =h + hz cp =h + hz p p 2 cp z z cp 1

1

i

(z

2

z 1)

dz:

(A48)

Representing the logarithm of a product as a sum of logarithms, splitting the integral into two parts corresponding to the summands, and changing the variable of integration in the second integral from z to z 1 , we get

I(h) =

1 2 i

I

p p 2 cp =h + hz (z z 1 ) dz: p p cp z z cp 1

ln

jzj=1

If h <

p

(A49)

p p cp =h + hz is analytic inside the ball jzj

cp , then function ln

1:

p Therefore, if cp < 1, the integrand in (A49) has singularities only at zero and cp : p If cp > 1, the singularities are at zero and 1=cp : If cp = 1, the only singularity

is at zero. Computing the residues of the integrand at the singularity points and using Cauchy’s theorem, we get

I (h) =

8 > < > :

cp 1 cp 1 cp cp

ln (1 + h) +

ln 1 +

h cp

+

h cp

+ ln chp

h cp

1 cp

+

ln

23

cp h

if h <

p

cp and cp < 1

if h <

p

1

cp and cp

:

(A50)

If h >

p

cp , then represent the logarithm in (A48) in the form

ln

z

q p cp =h + h

z

1

q

cp =h +

p

h

;

and proceed as above to get

I (h) =

8 > < > :

cp 1 cp

ln (h + cp ) +

1 cp cp

1 h

ln (1 + h) +

+

1 h

1 cp

p cp and cp < 1 . p if h > cp and cp 1

ln h if h >

+ ln h

(A51)

Now, it is straightforward to verify that Lemma 11 follows from (A50), (A51), and from the facts that

f0 =

that

h z (h) 1+h 0

max 0; 1

E

1 2

h z0 (h) 1+h

cp

Z

ln (z0 (h)

) dFp ( ) ;

= h + cp , and that the Marchenko-Pastur distribution has mass

cp 1 at zero.

Proof of Lemma 12

Let z0j = lim z0 (hj ) as n; p ! 1: As follows from Bai and Silverstein (2004, p. 563), p

(z0 (hj )) =

I

ln (z0 (hj )

I

ln (z0j

z) Mp (z) dz

C

and p

(z0j ) =

C

24

z) Mp (z) dz;

where C is a …xed contour of integration encircling the support of the MarchenkoPastur distribution, but not z0 (hj ) and z0j , and

Mp (z) =

p X

(

j

z)

1

p

j=1

Z

(x

z)

1

dFp (x) :

Therefore,

p

(z0 (hj ))

p

(z0j ) =

I

ln

z0 (hj ) z z0j z

Mp (z) dz:

C

Further, as can be shown using arguments similar to those given on p.563 of Bai and Silverstein (2004), I

ln

z0 (hj ) z z0j z

Mp (z) dz =

C

where

I

ln

z0 (hj ) z z0j z

^ p (z) dz + op (1); M

C

n o ^ p (z) ; p = 1; 2; ::: is a tight sequence of random continuous functions M

on C. On the other hand, as n; p ! 1; ln

z0 (hj ) z z0j z

!0

uniformly over C. Hence, I

ln

z0 (hj ) z z0j z

^ p (z) dz = op (1); M

(z0 (hj ))

p

C

and thus p

(z0j ) = op (1):

The latter equality implies that the vectors (S (S

p;

p

(z01 ) ; :::;

p

p;

p

(z0 (h1 )) ; :::;

p

(z0 (hr ))) and

(z0r )) simultaneously diverge, or converge, in distribution,

25

to the same limit. Now, according to Theorem 1.1 of Bai and Silverstein (2004), (S p

(z0r )) converges in distribution to a Gaussian vector ( ;

1 ; :::; r )

p;

p

(z01 ) ; :::;

with means E =

0,

E

j

=

1 2 i

I

ln (z0j

z)

cm3 (z) dz; (1 + m (z))3 cm2 (z) (1 + m (z))

(A52)

covariances

Cov

Cov

j; k

j;

1

I I

ln (z0j z1 ) ln (z0k z2 ) dm (z1 ) dm (z2 ) dz1 dz2 ; 2 dz1 dz2 (m (z1 ) m (z2 ))2 (A53) I I 1 z2 ln (z0j z1 ) dm (z1 ) dm (z2 ) dz1 dz2 ; (A54) 2 2 dz2 (m (z1 ) m (z2 ))2 dz1

=

2

=

and variance

Var ( ) =

1 2

2

I I

z1 z2 dm (z1 ) dm (z2 ) dz1 dz2 ; 2 dz2 (m (z1 ) m (z2 )) dz1

(A55)

where m (z) =

(1

c) z

1

+ cm(z)

with m (z) given by (3.6) where cp is replaced by c: That is,

m (z) =

z+c

1+

q (z 2z

c

1)2

4c

;

(A56)

where the branch of the square root is chosen so that the real and the imaginary q parts of (z c 1)2 4c have the same signs as the real and the imaginary parts of z

c

1, respectively. The contours of integration in (A52)-(A55) are closed,

oriented counterclockwise, enclose zero and the support of the Marchenko-Pastur distribution with parameter c, and do not enclose z0j and z0k . 26

The expressions for E j , Cov

j; k

, Cov

and Var ( ) can be simpli…ed

j;

along the same steps as in Bai and Silverstein (2004, pp.596-599). Exactly following the derivation of their formula 5.13, we get

E

j

=

ln ((z0j

a) (z0j 4

b))

1 2

Z

b

a

q

ln (z0j 4c

(x

x)

(A57)

dx; 2

c

1)

p 2 p 2 c) and b = (1 + c) : p Making substitution x = 1 + c 2 c cos as in the above proof of Lemma 11,

where a = (1

and using similar steps to those used in that proof, we obtain Z q p 1 q ln c=hj + hj ei dx = 2 0 a 4c (x c 1)2 Z q q p 1 1 z ln = c=hj + hj z dz = ln c=hj : 2 i jzj=1 1 2

Z

b

ln (z0j

x)

2

d

Using this in (A57), we get

E

j

1 = ln 4 =

1 ln 1 2

q p c=hj + hj

2

q c=hj

c 1 h2j :

For the covariance Cov

j; k

p hj

2

!

ln

q

c=hj

we use formula 1.16 of Bai and Silverstein

(2004), to get

Cov

j; k

=

1 2

2

I I

ln (z0j

z (m1 )) ln (z0k (m1 m2 )2

z (m2 ))

dm1 dm2 ;

(A58)

where z (m) =

1 c + : m 1+m

(A59)

Note that substituting m (z) as de…ned in (A56) in the right-hand side of (A59),

27

we get z, so (A59) describes a function inverse to m (z) : Let us split the double integral in (A58) into three parts according to the decomposition

Cov

j; k

=

1 Var( j ) + Var ( k ) 2

Var

j

k

;

where 1

Var( j ) =

2

2 1

Var( k ) =

2

2

I I I I

ln (z0j

z (m1 )) ln (z0j (m1 m2 )2

z (m2 ))

ln (z0k

z (m1 )) ln (z0k (m1 m2 )2

z (m2 ))

I I ln

z0j z(m1 ) z0k z(m1 )

dm1 dm2 ;

(A60)

dm1 dm2 ;

(A61)

dm1 dm2 :

(A62)

and

Var

j

k

=

1 2

2

(m1

ln

z0j z(m2 ) z0k z(m2 )

m2 )2

The contours of integration over m1 and m2 in (A60-A62) are obtained from the contours of integration over z1 and z2 in (A53) by transformation m (z) : Recall that by assumption the contours over z1 and z2 intersect the real line to the left of zero and in between the upper boundary of the support of the Marchenko-Pastur p 2 distribution, (1 + c) , and min fz0j ; z0k g : Therefore, as can be shown using the de…nition (A56) of m (z), the m1 -contour and m2 -contour are clockwise oriented p 1 and intersect the real line in between (1 + c) and min fm (z0j ) ; m (z0k )g = max hj (hj + c) contours enclose 0, and

(1 + hk )

1

1

; hk (hk + c) hj (hj + c)

1 1

and to the right of zero. In particular, both and

hk (hk + c) 1 , but not

1,

(1 + hj )

1

:

Without loss of generality, assume that the m2 -contour encloses the m1 -contour.

28

For …xed m2 , we have I

I d z (m1 ) z (m1 )) dm1 dm = dm1 1 2 (z0j z (m1 )) (m1 m2 ) m2 ) I 1=m21 c= (m1 + 1)2 dm1 ; = (z0j + 1=m1 c= (m1 + 1)) (m1 m2 )

ln (z0j (m1

(A63)

where the …rst equality follows from integration by parts and the fact that z (m1 )) is a single-valued function along the m1 -contour. To see this,

ln (z0j

note that

ln (z0j

z0j m1 + (1 + hj ) z (m1 )) = ln m1 + 1

1

+ ln m1 +

hj hj + c

ln m1 :

The …rst of the latter three terms is a single-valued function along the m1 -contour because it does not have singularities inside the contour. The second and the third terms are not single-valued, but their changes after passing once along the contour cancel each other. Now, the integrand in (A63) has …rst-order poles at 0, and at

(1 + hj )

1

hj (hj + c) 1 , m2 ,

1

and no other singularities: As explained above, only the …rst

two of the above poles are enclosed by the m1 -contour. Using Cauchy’s residue theorem, we get I

ln (z0j (m1

z (m1 )) dm1 = 2 i m2 )2

1 1 + m2 m2 + hj (hj + c)

29

1

!

:

(A64)

Let us denote

hj (hj + c)

1

as

j:

Using (A64) and (A60), we get

I 2 i 1 Var( j ) = ln (z z (m )) 0j 2 2 2 m2 I 2 i 1 = ln 1 z0j1 z (m2 ) 2 2 m2 ! I 1 2 i m2 + (1 + hj ) = ln 2 2 m2 + 1 I 1 2 i m2 j ln 2 2 m2 m2

1

dm2

m2

j

1

dm2

m2 1 m2

j

1

dm2

m2 1

m2

j

dm2 : j

By Cauchy’s residue theorem, the …rst term in the latter expression is equal to c 1 h2j : The second term equals zero because the integrand has antih i2 m2 j 1 derivative 2 ln m2 which is a single-valued function along the contour. 2 ln 1

Similarly, we can show that

Var( k ) =

2 ln 1

c 1 h2k

and that 2

Var(

j

k ) = 2 ln

(1 c 1 hj hk ) : 1 c 1 h2j (1 c 1 h2k )

Combining these results, we get

Cov

j; k

=

c 1 h2j

ln 1

ln 1

c 1 h2k

2

(1 c 1 hj hk ) ln 1 c 1 h2j (1 c 1 h2k ) =

For Cov

j;

2 ln 1

c 1 hj hk :

and Var ( ), an analysis similar to but simpler than that leading

to the above formula for Cov

j; k

shows that Cov

2c:

30

j;

=

2hj and Var ( ) =

F

Proof of Lemma 13

First, note that CLR =

p X

q( j )

p

j=1

where q(x) = x

Z

q(x)dFp (x) ;

1: Also, recall that, as shown in the proof of Lemma 12,

ln x

p

(z0 (h)) =

p

(z0 ) + op (1);

where z0 = lim z0 (h) and

p

(z0 ) =

p X

s( j )

j=1

with s(x) = ln (z0

p

Z

s(x)dFp (x) ;

x) : Therefore, in view of Theorem 1.1 of Bai and Silver-

stein (2004), CLR and

p

(z0 (h)) jointly converge in distribution to a Gaussian

vector with covariance 1

R=

2

2

I I

s(z1 )q(z2 ) dm (z1 ) dm (z2 ) dz1 dz2 : 2 dz2 (m (z1 ) m (z2 )) dz1

(A65)

Here m (z) is as de…ned in (A56), and the contours of integration are closed, oriented counterclockwise, enclose the support of the Marchenko-Pastur distribution with parameter c < 1, and do not enclose z0 : Further, we will choose such contours so that the z1 -contour encloses 0, but the z2 -contour does not. Using Formula 1.16 of Bai and Silverstein (2004) we can simplify (A65) to get

R=

1 2

2

I I

ln (z0

z (m1 )) (z (m2 ) ln z (m2 ) (m1 m2 )2

where z (m) =

1 c + m 1+m 31

1)

dm1 dm2 ;

and the contours of integration over m1 and m2 are obtained from the contours of integration over z1 and z2 in (A65) by the transformation m (z) : In particular, m1 -contour is oriented clockwise and encloses

h h+c

and 0 but not

whereas m2 -contour is oriented counterclockwise and encloses h h+c

1 c 1

1 and

and

1 , 1+h

1 but not

and 0:

Using (A64), we can write R = R1 + R2 + R3 , where i

R1 = R2 = R3 =

Since

1 m2

+

i i

I

I

1 1 + m2 m2 + hj (hj + c)

I

1 m2 +hj (hj +c)

1

1 1 + m2 m2 + hj (hj + c)

1

1 1 + m2 m2 + hj (hj + c)

1

1

!

! !

z (m2 ) dm2 ;

ln z (m2 ) dm2 ; and dm2 :

is analytic in the area enclosed by the m2 -contour,

R3 = 0: Further, using Cauchy’s theorem and the fact that

z (m2 ) =

we get R1 =

c 1 + ; m2 1 + m2

2h: Finally, integrating R2 by parts, and using the fact that ln z (m2 )

is a single-valued function on the m2 -contour, we get

R2 =

i

I

c (1+m2 )2 1 + m2c+1 m2

1 m22

ln m2 + ln m2 + hj (hj + c)

1

dm2 :

The integrand in the above integral has only two singularities in the area enclosed by the m2 -contour: a pole at

1 c 1

and a pole at

1: Therefore, by Cauchy’s residue

theorem, we get R2 = 2 ln (1 + h) : To summarize, R = R1 + R2 + R3 = 2 ln (1 + h), which establishes Lemma 13.

32

2h +

References [1] Bai, Z.D. and J.W. Silverstein (2004) “CLT for Linear Spectral Statistics of Large-Dimensional Sample Covariance Matrices”, Annals of Probability 32, 553-605. [2] Geman, S. (1980) “A limit theorem for the norm of random matrices”, Annals of Probability 8, 252–261. [3] Olver, F.W.J. (1997) Asymptotics and Special Functions, A K Peters, Natick, Massachusetts. [4] Rudin, W. (1987) Real and Complex Analysis, 3rd edition, McGraw-Hill, New York.

33