S UPPLEMENT TO “T HE EFFECT OF RECURSIVE DETRENDING ON PANEL UNIT ROOT TESTS ”: A DDITIONAL DETAILS AND PROOFS Joakim Westerlund∗ Deakin University Australia

January 15, 2014

Abstract In this supplement, we (i) study the local asymptotic power of the tests statistic based on recursively detrending the level data directly, a possibility discussed in Remark 6 in Westerlund (2014), (ii) provide the proof of Lemmas A.1 and A.2 in Westerlund (2014), and (iii) provide the proof of Theorem 1 in the same paper.

1 Recursive detrending of the level data As an alternative to t–REC, in this section we consider a test statistic that is perhaps more natural, which is based on replacing Ri,t,p with Yi,t,p+1 . That is, rather than detrending and then accumulating the first-differenced data, we detrend the level data directly. The test statistic is given by t–LREC =

CNT,p √ , σˆ ϵ D NT,p

where CNT,p = D NT,p =

√

N

1 NT

1 NT 2

T

∑ ∑

Yi,t−1,p+1 yi,t,p ,

i =1 t = p +1 N

T

∑ ∑

i =1 t = p +1

2 Yi,t −1,p+1 .

∗ Deakin University, Faculty of Business and Law, School of Accounting, Economics and Finance, Melbourne Burwood Campus, 221 Burwood Highway, VIC 3125, Australia. Telephone: +61 3 924 46973. Fax: +61 3 924 46283. E-mail address: [email protected].

1

Hence, in this case, since yi,t,p = ∆Ui,t,p = (ρi − 1)Ui,t−1,p + ϵi,t,p , the numerator may be expanded as CNT,p =

√

NC1NT,p + C2NT,p ,

(1)

where 1 NT

C1NT,p =

√

C2NT,p =

N

T

∑ ∑

(ρi − 1)Ui,t−1,p Ui,t−1,p+1 ,

i =1 t = p +1

1 NT

N

T

∑ ∑

Ui,t−1,p+1 ϵi,t,p .

i =1 t = p +1

Lemma 2 provides the limits of

√

NC1NT,p and C2NT,p , which are enough to work out roughly

the local power properties of the test statistic. Lemma 2. Provided that Assumption 1 in Westerlund (2014) holds, as N, T → ∞ with κ ≥ 1/4 √ and NT −1 → 0,

√

√ NC1NT,p = σϵ2 N (α N µ1 c1,p + α2N µ2 c2,p ) + o p (1), √ C2NT,p = σϵ2 N (d0,p + α N µ1 d1,p + α2N µ2 d2,p ) + o p (1),

where 1 ( q + 1) ( ) p p +1 p p +1 bm,q−1 + bk,q−1 2q −1 − ∑ hk,p bk,q−1 − ∑ hk,p+1 bk,q−1 + ∑ ∑ hk,p hm,p+1 × , q (k + m + q) k =1 k =1 k =1 m =1 ( ) p 1 ( m − 1) ! ( k − 1 ) ! p +1 p = − ∑ hk,p + ∑ hk,p+1 hm,p (m + q)!(k + m + q) , ( q + 1) (k + q)! k∑ k =1 =1 m =1

cq,p =

dq,p

q

∑ (−1)n

bk,q =

n =0

(k − 1)!(1 + (−1)q ) 2q − n ( k − 1 ) ! + (−1)q+1 . (q + 1 − n)!(k + n)! 2( k + q ) ! ( k + q + 1)

Lemma 2 implies that CNT,p =

√

NC1NT,p + C2NT,p √ = σϵ2 N [d0,p + α N µ1 (c1,p + d1,p ) + α2N µ2 (c2,p + d2,p )] + o p (1)

(2)

√ as N, T → ∞, where the first term on the right-hand side, σϵ2 Nd0,p , is a bias that is there also under the unit root null. This is the standard situation in case of full-sample detrending (see Westerlund and Breitung, 2013, for a detailed discussion). The typical approach to ensure a 2

correctly centered asymptotic distribution under the null involves using a bias-adjusted test statistic of the type √ √ CNT,p − N σˆ ϵ2 d0,p σˆ ϵ Nd0,p √ t–LREC − √ = . D NT,p σˆ ϵ D NT,p

(3)

However, in this case d0,p = 0 for all p ≥ −1, suggesting that, as with t–REC, t–LREC does not require any correction for bias. Analogous to the analysis of t–REC (see Westerlund, 2014, Section 3), it is the values of

(c1,p + d1,p ) and (c2,p + d2,p ) that determine how κ should be set in order for the local power of t–LREC to be non-negligible and non-increasing. Some of these values are reported in Table 1. It is seen that, while (c1,p + d1,p ) > 0 for p < 1, (c1,p + d1,p ) = 0 for p ≥ 1. Hence, just as with t–REC, the local power of t–LREC when κ = 1/2 is non-negligible for p < 1 and negligible for p ≥ 1. We also see that when p ≥ 1, even if power is non-negligible for κ = 1/4, (c2,p + d2,p ) < 0 is still decreasing in p, which is again in agreement with the results for t–REC. Hence, while the appropriate value of κ is the same, power is still expected to decrease with increases in p ≥ 1. The values of (c2,p + d2,p ) are smaller in absolute value when compared to those reported in Table 1 in Westerlund (2014), suggesting that t–REC should be relatively more powerful than t–LREC. Table 1: Coefficients of the asymptotic distributions of t–LREC. p

−1 0 1 2 3 4

(c1,p + d1,p ) 0.5 0.25 0 0 0 0

(c2,p + d2,p ) 0.33333 0.16667 −0.016667 −0.003968 −0.001852 −0.001082

Notes: The reported numbers refer to the coefficients attached to the first-, and second–order bias terms in the asymptotic distribution of t–LREC. p refers to the order of the trend polynomial.

While informative about the mean of t–LREC, Lemma 2 is silent about the rest of the asymptotic distribution. The reason for not reporting the whole distribution is that t–REC stands out as the better test statistic; it more powerful and, as alluded in Westerlund (2014, Section 3), is more convenient in the presence serial and/or cross-section correlated error 3

terms. Proof of Lemma 2. The proof of Lemma 2 is similar to that of Lemma 1 in Westerlund (2014). The main difference is that in order to maintain the higher level of accuracy, when evaluating E(Ui,t Ui,s |ci ) and E(Ui,s ϵi,t |ci ) we need to account not only for the first-order term in α N , but also for higher-order terms. We begin by considering E(Ui,t Ui,s |ci ) for t ≥ s. Note that ρit = ∑rq=0 ( T −1 ci α N t)q /q! + O p (αrN+1 ). Making use of this we can show that 1 σϵ2 T

σϵ−2 T −1 E(Ui,t Ui,s |ci ) =

t

s

∑ ∑ ρit+s−m−l E(ϵi,m ϵi,l ) =

m =1 l =1

1 T

s

∑ ρit+s−2l

l =1

r

∑ (ci α N )q gT,q (s, t) + O(αrN+1 )

=

q =0 r

∑ (ci α N )q gq (v, w) + O(T −1 ) + O(αrN+1 ),

=

(4)

q =0

where gT,q (s, t) =

1 T

[ T −1 (t + s − 2l )]q , ∑ q! l =1 s

∫ v

gq (v, w) =

x =0

(w + v − 2x )q ( w + v ) q +1 − ( w − v ) q +1 dx = . q! 2( q + 1) !

Hence, making use of the fact that E(Ui,t,p Ui,s,p+1 |ci ) = E(Ui,t Ui,s |ci ) − t

+

s

t

n =2

k =2

∑ E(Ui,t Ui,n |ci )an,s,p − ∑ E(Ui,k Ui,s |ci )ak,t,p+1

s

∑ ∑ E(Ui,k Ui,n |ci )ak,t,p an,s,p+1 ,

k =2 n =2

we obtain σϵ−2 T −1 E(Ui,t,p Ui,t,p+1 |ci ) =

r

∑ (ci α N )q Iq,T,p (t) + O(αrN+1 ) + O(T −1 )

k =0 r

=

∑ (ci α N )q Iq,p (w) + O(αrN+1 ) + O(T −1 ),

q =0

4

(5)

where Iq,T,p (t) = gT,q (t, t) −

+

1 T2

t

=0 x =0 ∫uw ∫ w

+

∫ w u =0

k =2

k =2

1

k =2 n =2

∫ w ∫ u

u =0

u =0

t

1

k

+

∫w

t

∑ gT,q (k, t)Tak,t,p − T ∑ gT,q (k, t)Tak,t,p+1 t

t

∑ ∑ gT,q (n, k)T2 ak,t,p an,t,p+1 + T2 ∑ ∑ gT,q (k, n)T2 ak,t,p an,t,p+1 ,

Iq,p (w) = gq (w, w) −

Consider

1 T

x =u

∫ w

u =0

k =2 n = k

gq (u, w) a p (u, w)du −

∫ w

u =0

gq (u, w) a p+1 (u, w)du

gq ( x, u) a p (u, w) a p+1 ( x, w)dudx gq (u, x ) a p (u, w) a p+1 ( x, w)dudx.

gq (u, w) a p (u, w)du. By repeated integration by parts,

gq (u, w) a p (u, w)du p

∑ hk,p

=

k =1 p

∑

=

∑ hk,p

k =1

(

u =0

∫ w

hk,p

k =1 p

=

∫ w ((w + u)q+1 − (w − u)q+1 )uk−1

u =0

∫ w u =0

du 2( q + 1) w k ( ) ∫ 1 (2w)q+1 (q + 1) w ((w + u)q + (w − u)q )uk − du 2( q + 1) k k wk u =0 1 2( q + 1)

(2w)q+1 (q + 1)(2w)q w (q + 1)q − + k k ( k + 1) k ( k + 1)

×

∫ w ((w + u)q−1 − (w − u)q−1 )uk+1 u =0

wk

.. . p

∑ hk,p bk,q wq+1 ,

=

k =1

where

(k − 1)!2q−n

q

bk,q =

(k − 1)!(1 + (−1)q )

∑ (−1)n (q + 1 − n)!(k + n)! + (−1)q+1 2(k + q)!(k + q + 1) .

n =0

Moreover, ∫ u x =0

gq ( x, u) a p (u, w) a p+1 ( x, w)dx p

=

p +1

∑∑

hk,p hm,p+1

∑∑

hk,p hm,p+1

∑∑

hk,p hm,p+1

k =1 m =1 p p +1

=

k =1 m =1 p p +1

=

k =1 m =1

∫ u ((u + x )q+1 − (u − x )q+1 )uk−1 x m−1 x =0

2(q + 1)w2k

dx

∫ u ((u + x )q+1 − (u − x )q+1 ) x m−1 uk+m−1 x =0

2(q + 1)um w2k

bm,q uk+m+q , w2k 5

dx

) du

and hence ∫ w ∫ u u =0

p

x =0

gq ( x, u) a p (u, w) a p+1 ( x, w)dudx =

p +1

∑∑

hk,p hm,p+1

k =1 m =1

bm,q wm−k+q+1 . ( k + m + q + 1)

These results, together with the fact that gq ( x, u) a p (u, w) a p+1 ( x, w) is symmetric in ( x, u) and gq (w, w) = (2w)q+1 /2(q + 1), imply that Iq,p (w) can be rewritten as ( ) p p +1 p p +1 bk,q + bm,q 2q Iq,p (w) = − hk,p bk,q − ∑ hk,p+1 bk,q + ∑ ∑ hk,p hm,p+1 w q +1 , (q + 1) k∑ ( k + m + q + 1 ) =1 k =1 k =1 m =1 which in turn implies ∫ 1

Iq,p =

Iq,p (w)dw

w =0

1 ( q + 2) ( ) p p +1 p p +1 bm,q + bk,q 2q × − hk,p bk,q − ∑ hk,p+1 bk,q + ∑ ∑ hk,p hm,p+1 . (6) (q + 1) k∑ ( k + m + q + 1) =1 k =1 k =1 m =1

=

Hence, 1 2 σϵ T 2

T

∑

E(Ui,t−1,p Ui,t−1,p+1 |ci ) =

t = p +1

T

1 T

r

∑ ∑ (ci α N )q Iq,T,U (t) + O(αrN+1 ) + O p (T −1 )

t = p +1 q =0

r

=

∑ (ci α N )q Iq,p + O(αrN+1 ) + O p (T −1 ),

(7)

q =0

and therefore ( ) T 1 E (ρi − 1)Ui,t−1,p Ui,t−1,p+1 ci σϵ2 T t=∑ p +1

= =

1 σϵ2 T 2 1 T

T

T

∑

ci α N E(Ui,t−1,p Ui,t−1,p+1 |ci ) + O(α N T −1 )

t = p +1 r

∑ ∑ (ci α N )q+1 Iq,T,p (t) + O(αrN+2 ) + O(α N T −1 )

t = p +1 k =0

r

=

∑ (ci α N )q+1 Iq,p + O(αrN+2 ) + O(α N T −1 ).

(8)

k =0

Consequently, letting ( ) p p +1 p p +1 bm,q−1 + bk,q−1 1 2q −1 cq,p = − ∑ hk,p bk,q−1 − ∑ hk,p+1 bk,q−1 + ∑ ∑ hk,p hm,p+1 , ( q + 1) q (k + m + q) k =1 k =1 k =1 m =1 by the law of large numbers given in Corollary 1 of Phillips and Moon (1999) (the conditions of which can be verified using the same steps as in the proof of Lemma 1 in Westerlund, 6

2014), 1 NT

C1NT,p =

= σϵ2

N

T

∑ ∑

r

∑ µ q +1 α N

q +1

q =0

= σϵ2

(ρi − 1)Ui,t−1,p Ui,t−1,p+1

i =1 t = p +1

r +1

Iq,p + O p (αrN+2 ) + O p (α N T −1 ) + o p (α N )

∑ µq α N cq,p + O p (αrN+2 ) + O(α N T −1 ) + o p (α N ), q

(9)

q =1

which holds jointly as N, T → ∞. Moreover, while c1,p and c2,p cancel out against the correq

1 sponding coefficients in C2NT,p , c3,p does not. This means that ∑rq+ =1 µq α N cq,p can be truncated

at r = 1, with an error of order O(α2N ). Next, consider C2NT,p . In the proof of Lemma 1 in Westerlund (2014) it is shown that if t > s, then E(Ui,s ϵi,t |ci ) = 0. A first-order approximation for E(Ui,s ϵi,t |ci ) for s ≥ t is also given, which can be extended in the following obvious fashion to account also for higherorder terms: σϵ−2 E(Ui,s ϵi,t |ci ) =

1 σϵ2

s

∑

m =1

ρis−m E(ϵi,m ϵi,t ) = ρis−t =

( c i α N ) q −1 [ T (s − t)]q + O(αrN+1 ). q! q =0 r

∑

This result implies that by using the results given in Proof of Lemma 1, for t > s, σϵ−2 E(Ui,s,p ϵi,t,p |ci )

= − = −

1 σϵ2 T 1 σϵ2 T

s

∑

k =2

+

1 2 σϵ T 2

s

s

s

∑ ∑ E(Ui,k ϵi,n |ci )T2 ak,s,p+1 an,t,p

k =2 n =2 s

1

k −1

∑ E(Ui,s ϵi,k |ci )Tak,t,p + σ2 T2 ∑ ∑ E(Ui,k ϵi,n |ci )T2 ak,s,p+1 an,t,p + O(T −1 )

k =2

(c α = ∑ i N q! q =0 r

E(Ui,s ϵi,k |ci ) Tak,t,p +

)q

(

ϵ

1 − T

k =2 n =2

s

1 ∑ [T −1 (s − k)]q , Tak,t,p + T2 k =2

s

k −1

∑ ∑ [T −1 (k − n)]q T2 ak,s,p+1 an,t,p

k =2 n =2

O(αrN+1 ) + O( T −1 ),

where 1 T

[ T −1 (s − k)]q Tak,t,p = ∑ q! k =2 s

∫ v u =0

(v − u)q a p (u, w)du + O( T −1 ) q! ∫ v

p

=

∑

k =1

hk,p

u =0

7

( v − u ) q u k −1 du + O( T −1 ). q!wk

)

By again using repeated integration by parts, ∫ v u =0

( v − u ) q u k −1 du = q!wk

∫ v

( v − u ) q −1 u k du wk u =0 ∫ v 1 ( v − u ) q −2 u k +1 du (q − 2)!k(k + 1) u=0 wk 1 (q − 1)!k

= .. .

∫

v u k + q −1 ( k − 1) ! du ( k + q − 1 ) ! u =0 w k (k − 1)!vk+q , (k + q)!wk

= = and therefore 1 T

p [ T −1 (s − k)]q (k − 1)!vk+q + O ( T −1 ). Ta = hk,p k,t,p ∑ ∑ k q! ( k + q ) !w k =2 k =1 s

A similar calculation reveals that 1 T2

q

k −1

[ T −1 (k − n)]q 2 T ak,s,p+1 an,t,p ∑∑ q! k =2 n =2 s

=

∫ v ∫ u (u − x )q u =0 p +1

=

x =0

q!

p

∑ ∑

hk,p+1 hm,p

k =1 m =1

a p+1 (u, v) a p ( x, w)dxdu + O( T −1 )

(m − 1)!vm+q + O ( T −1 ) , (m + q)!(k + m + q)wm

and so, by insertion of s = t − 1, 1 T

t −1

[ T −1 (t − 1 − k)]q Tak,t,p = ∑ q! k =2

p

∑ hk,p

k =1

(k − 1)!wq + O ( T −1 ), (k + q)!

and 1 T2

t −1 k −1

[ T −1 (k − n)]q 2 T ak,t−1,p+1 an,t,p q! k =2 n =2

∑∑ p +1

=

p

∑ ∑

k =1 m =1

hk,p+1 hm,p

(m − 1)!wq + O ( T −1 ). (m + q)!(k + m + q)

8

This means that 1 σϵ2 T

T

∑

E(Ui,t−1,p+1 ϵi,t,p |ci )

t = p +1 r

∑ ( ci α N )q

=

q =0

1 T

×

(

T

∑

t = p +1

1 − T

[ T −1 (s − k)]q 1 Tak,t,p + 2 ∑ q! T k =2 s

k −1

[ T −1 (k − n)]q 2 T ak,s,p+1 an,t,p ∑∑ q! k =2 n =2 s

)

+ O(αrN+1 ) + O( T −1 ) r

∑ (ci α N )q dq,p + O(αrN+1 ) + O(T −1 ),

=

(10)

q =0

where dq,p

1 = ( q + 1)

(

( m − 1) ! ( k − 1 ) ! p +1 p − ∑ hk,p + ∑ ∑ hk,p+1 hm,p ( k + q ) ! k =1 m =1 (m + q)!(k + m + q) k =1 p

) .

Another application of Corollary 1 of Phillips and Moon (1999) now yields 1 √ C2NT,p = N

1 NT

N

T

∑ ∑ Ui,t−1,p+1 ϵi,t,p

i =1 t =2 r q µq α N c2q,p σϵ2 q =0

∑

=

+ O p (αrN+1 ) + O p ( T −1 ) + o p (1)

(11)

In order to maintain the same order of accuracy as for C1NT,p , the above sum may be trun-



cated at r = 2.

2 Proof of Lemmas A.1 and A.2 Proof of Lemma A.1. q

q

q

0 ). Consider E (U We begin with E( IiT,p i,t,p Ui,s,p ). By the definition of Ui,t,p , )( )] [( q

q

E(Ui,t,p Ui,s,p ) = E

q

Ui,t −

t

∑ Ui,k ak,t,p

q

q

Ui,s −

k =2 q

q

= E(Ui,t Ui,s ) − t

+

s

s

q

n =2

t

∑ E(Ui,t Ui,n )an,s,p − ∑ E(Ui,k Ui,s )ak,t,p q

q

n =2

k =2

∑ ∑ E(Ui,k Ui,n )ak,t,p an,s,p , q

s

∑ Ui,n an,s,p

q

k =2 n =2

9

q

q

where ak,t,p = d′k,p (∑tn=2 dn,p d′n,p )−1 dt,p for p ≥ 1 and ak,t,−1 = ak,t,0 = 0. Here, t

∑ E(Ui,k Ui,s ) q

q

q

= E[(Ui,s )2 ] +

k =2 t

s

t

∑ E(Ui,k Ui,s ) + ∑ q

q

k =2 s

∑ ∑ E(Ui,k Ui,n ) q

s −1

q

k = s +1

s

k −1

q

q

E(Ui,k Ui,s ),

∑ E[(Ui,k )2 ] + ∑ ∑ E(Ui,k Ui,n )

=

k =2 n =2

q

q

k =2 n =2

k =2 s

s

∑ ∑

+

q

k =2 n = k +1

t

s

∑ ∑ E(Ui,k Ui,n ),

q

q

E(Ui,k Ui,n ) +

q

q

k = s +1 n =2

which in turn implies T −1 E(Ui,t,p Ui,s,p ) q

q

= T −1 E(Ui,t Ui,s ) − q

− + +

1 T2 1 T3 1 T3

q

s −1

s

1 T2

∑ E(Ui,t Ui,n )Tan,s,p − T −2 E[(Ui,s )2 ]Tas,t,p q

q

q

n =2

t

1

∑ E(Ui,k Ui,s )Tak,t,p − T2 ∑ q

q

k =2

q

k = s +1

s

1

q

E(Ui,k Ui,s ) Tak,t,p k −1

s

∑ E[(Ui,k )2 ]T2 ak,t,p ak,s,p + T3 ∑ ∑ E(Ui,k Ui,n )T2 ak,t,p an,s,p q

k =2 s

q

q

k =2 n =2

s

∑ ∑

q

k =2 n = k +1

q

E(Ui,k Ui,n ) T 2 ak,t,p an,s,p +

1 T3

t

s

∑ ∑ E(Ui,k Ui,n )T2 ak,t,p an,s,p . q

q

k = s +1 n =2

When t = ⌊wT ⌋, where ⌊ x ⌋ is the integer part of x, we have |(tT −1 )n − wn | = O( T −1 ), which holds for any integer n ≥ 1 and all w ∈ [0, 1]. This, together with the fact that Ui,t = ∑ts=1 [ T −1 (t − s)]q ϵi,s , suggests that for s = ⌊vT ⌋, w ≥ v and q = 0, q

0 0 σϵ−2 T −1 E(Ui,t Ui,s )=

Define a p (v, w) = d p (v)

′

(∫

1 σϵ2 T

t

∑

s

∑

(12)

m = p +1 n = p +1

w

′

u =0

E(ϵi,m ϵi,n ) = σϵ2 T −1 s = σϵ2 v + O( T −1 ). ) −1

d p (u)d p (u) du

d p ( w ),

with d p (v) = (1, v, ..., v p−1 )′ for p ≥ 1 and a−1 (v, w) = a0 (v, w) = 0. Hence, letting JT = diag( T −1/2 , T −3/2 , ..., T −( p+1/2) ), we have ( ) −1 t √ √ ′ ′ T JT dt,p = a p (v, w) + O( T −1 ), Tak,t,p = Tdk,p JT JT ∑ dn,p dn,p JT

(13)

n =2

which holds uniformly in (v, w) ∈ [0, 1] × (0, 1]. If w → 0, then, by the mean value theorem, a p (v, w) → d p (v)′ (wd p (w)d p (w)′ )−1 d p (w). Define the p × p matrices    1 1/2 1 x . . . x p −1   ∫ 1  2 p 1/3 x ... x   1/2  x  dx =  .  . Hp = .. . . .    .. .. .. .. x =0  .   .. x p −1 x p . . . x 2( p −1)

... ... .. .

1/p 1/( p + 1) .. .

1/p 1/( p + 1) . . . 1/(2p − 1)

10

     

and D p (w) = diag(1, w, ..., w p−1 ). By variable substitution of x = u/w yields   1 u . . . u p −1  ∫ w ∫ w  u2 . . . up  u  ′   du d p (u)d p (u) du = . . .  .  .. . . . .. u =0 u =0  .  p − 1 p 2 ( p − 1 ) u u ... u   1 xw . . . ( xw) p−1  ∫ 1  xw ( xw)2 . . . ( xw) p     dx = w .. .. .. ..   x =0  . . . . 

( xw) p−1 ( xw) p . . . ( xw)2( p−1)

= wD p (w) H p D (w) p . H p is the so-called Hilbert matrix, whose inverse has typical element hmk,p = [ H p−1 ]mk , where (

hmk,p = (−1)

m+k

p+m−1 ( m + k − 1) p−k

)(

p+k−1 p−m

)(

m+k−2 m−1

)2

(see Choi, 1983). Hence, since D p (w) and H p are invertible with D p (w)−1 d p (w) = (1, ..., 1)′ = p

ι p (a p-vector of ones), letting hk,p = ∑m=1 hkm,p , we have (∫ w ) −1 ′ ′ a p (v, w) = d p (v) d p (u)d p (u) du d p (w) u =0

=

1 1 d p (v)′ D p (w)−1 H p−1 D p (w)−1 d p (w) = d p (v)′ D p (w)−1 H p−1 ι p w w

=

v k −1 ∑ wk k =1 p

p

p +1

∑

hkm,p =

m =1

∑

k =1

v k −1 hk,p+1 . wk

(14)

Making use of these results, because T −2 E[(Ui,s )2 ] and T −3 ∑sk=2 E[(Ui,k )2 ] are both O( T −1 ), q

q

0 0 σϵ−2 T −1 E(Ui,t,p Ui,s,p ) 0 0 = T −1 E(Ui,t Ui,s )−

− + =

1 T2 1 T3

t

1 T2

s

∑

0 0 E(Ui,t Ui,n ) Tan,s,p −

n =2

∑

0 0 E(Ui,k Ui,s ) Tak,t,p +

s

s

k = s +1

∑ ∑

1 T3

s

=

s −1

∑ E(Ui,k0 Ui,s0 )Tak,t,p

k =2

k −1

0 ) T 2 ak,t,p an,s,p ∑ ∑ E(Ui,k0 Ui,n

k =2 n =2

0 0 E(Ui,k Ui,n ) T 2 ak,t,p an,s,p +

k =2 n = k +1 0 IT,p (s, t) + O( T −1 )

1 T2

1 T3

I p0 (v, w) + O( T −1 ),

11

t

s

0 ) T 2 ak,t,p an,s,p + O( T −1 ) ∑ ∑ E(Ui,k0 Ui,n

k = s +1 n =2

(15)

where 0 IT,p (s, t) =

+ + I p0 (v, w)

s 1 − 2 T T 1 T3 1 T3

s

1

s −1

n =2

k =2

k −1

s

t

1

∑ nTan,s,p − T2 ∑ kTak,t,p − T2 ∑ s

1

s

∑ ∑ nT2 ak,t,p an,s,p + T3 ∑ ∑

k =2 n =2 t

sTak,t,p

k = s +1

kT 2 ak,t,p an,s,p

k =2 n = k +1

s

∑ ∑ nT2 ak,t,p an,s,p ,

k = s +1 n =2 ∫ v

= v−

∫ v ∫x=u0

xa p ( x, v)dx −

+ =0 ∫uw

0 ∫x = v

u=v

x =0

+

∫ v u =0

ua p (u, w)du −

∫ w

∫ v ∫ v u=v

xa p (u, w) a p ( x, v)dudx +

u =0

va p (u, w)du

ua p (u, w) a p ( x, v)dudx

x =u

xa p (u, w) a p ( x, v)dudx.

A direct calculation reveals that ∫ v u =0

ua p (u, v)du =

∫ v u =0

ua p (u, w)du =

∫ v

k =1 p

∫ v

∑ hk,p ∑

hk,p

∑

hk,p

k =1 p

∫ w u=v

p

va p (u, w)du =

k =1

p

u =0

v uk du = ∑ hk,p , ( k + 1) vk k =1 p

u =0

∫ w u=v

v k +1 uk du = h ∑ k,p (k + 1)wk , wk k =1 ( ) p vuk−1 1 v k +1 du = ∑ hk,p v− k . k wk w k =1

Similarly, ∫ v ∫ u u =0

x =0

p

xa p (u, w) a p ( x, v)dudx =

p

∑∑

hk,p hm,p

∑∑

hk,p hm,p

k =1 m =1 p p

=

u =0

∫ v

k =1 m =1 p p

=

∫ v ∫ u u k −1 x m

u =0

x =0

wk vm

dudx

uk+m du ( m + 1) w k v m v k +1

∑ ∑ hk,p hm,p (m + 1)(k + m + 1)wk ,

k =1 k =1

∫ v ∫ v u =0

x =u

p

ua p (u, w) a p ( x, v)dudx =

∫ v ∫ v

p

∑∑

hk,p hm,p

∑∑

hk,p hm,p

k =1 m =1 p p

=

k =1 m =1 p p

=

u =0

∫ v u =0

u k x m −1 dudx m x =u wk v ( ) 1 uk+m k u − m du v mwk v k +1

∑ ∑ hk,p hm,p (k + 1)(k + m + 1)wk ,

k =1 k =1

12

and ∫ w ∫ v u=v

p

x =0

xa p (u, w) a p ( x, v)dudx =

∫ w ∫ v

p

∑∑

hk,p hm,p

k =1 m =1 p p

u=v

x =0

u k −1 x m dudx wk vm

∫ w

vuk−1 dv u = v ( m + 1) w k k =1 m =1 ) ( p p 1 v k +1 = ∑ ∑ hk,p hm,p v− k , k ( m + 1) w k =1 m =1

=

from which we deduce

hk,p hm,p

(

) (2k + 1)v v k +1 = v − ∑ hk,p − k ( k + 1) k ( k + 1) w k k =1 ) ( p p v k +1 v − , + ∑ ∑ hk,p hm,p k(m + 1) k(k + 1)(k + m + 1)wk k =1 m =1 p

I p0 (v, w)

∑∑

(16)

Let 0 IT,p =

I p0 =

1 T2 ∫ 1

t −1

T

∑ ∑

0 IT,p (s, t),

t = p +1 s = p +1

∫ w

v =0

w =0

I p (v, w)0 dvdw.

From |(tT −1 )n − wn | = O( T −1 ), it is not difficult to show that with t = ⌊wT ⌋ and s = ⌊vT ⌋, 0 ( s, t ) − I 0 ( v, w )| = O ( T −1 ), which in turn implies | I 0 − I 0 | = O ( T −1 ). Therefore, | IT,p p p T,p 0 E( IiT,p ) =

=

1 T2 I p0

T

t −1

∑ ∑

t = p +1 s = p +1

0 0 0 −1 σϵ−2 T −1 E(Ui,s −1,p Ui,t−1,p ) = IT,p + O ( T )

+ O ( T −1 ).

(17)

As for I p0 , by substitution of ∫ 1

∫ w

vdvdw =

∫ 1 w =0

w =0 v =0 ∫ w k +1 v v =0 w k

dvdw =

1 2

∫ 1 w =0

1 ( k + 2)

w2 dw = ∫ 1 w =0

1 , 6

w2 dw =

13

1 , 3( k + 2)

we obtain I p0

∫ 1

∫ w

w =0 ∫ 1

v =0 ∫ w

w =0

v =0

I1,p (v, w)dvdw

=

p

vdvdw −

= p

∑∑

+

k =1

∫ w (

∫ 1

p

hk,p hm,p

∑ hk,p

w =0

k =1 m =1 p

∫ w ( (2k + 1)v

∫ 1

v =0

w =0

v k +1 − k ( k + 1) k ( k + 1) w k

v =0

v v k +1 − k(m + 1) k(k + 1)(k + m + 1)wk

) dvdw ) dvdw

2k + 5 1 − ∑ hk,p 6 k =1 6(k + 1)(k + 2)

=

p

p

∑∑

+

hk,p hm,p

k =1 m =1

(k + 1)(k + m + 3) + 2(m + 1) , 6(k + 1)(k + 2)(m + 1)(k + m + 1)

(18)

as was to be shown. q

q

q

Next, consider E( JiT,p ). In particular, consider E(Ui,t,p ϵi,s,p ). By the definitions of Ui,t,p and ϵi,s,p , q

E(Ui,t,p ϵi,s,p ) [( q Ui,t

= E

−

)(

t

∑

k =2 q

= E(Ui,t ϵi,s ) − q

= E(Ui,t ϵi,s ) − 1 T

−

t

∑∑

k =2 n =2

∑ ϵi,n an,s,p

n =2 t

t

s

∑ E(Ui,t ϵi,n )an,s,p − ∑ E(Ui,k ϵi,s )ak,t,p + ∑ ∑ E(Ui,k ϵi,n )ak,t,p an,s,p q

n =2

1 T

q

q

k =2 n =2

k =2

s

1

s −1

∑ E(Ui,t ϵi,n )Tan,s,p − T ∑ E(Ui,k ϵi,s )Tak,t,p q

n =2

q

k =2

E(Ui,k ϵi,s ) Tak,t,p +

k −1

s

s

)]

s

ϵi,s −

q

k = s +1

1 T2

+

∑

q Ui,k ak,t,p

1 T2

t

s

∑ ∑ E(Ui,k ϵi,n )T2 ak,t,p an,s,p q

k = s +1 n =2

q

E(Ui,k ϵi,n ) T 2 ak,t,p an,s,p +

1 T2

s

s

∑ ∑

E(Ui,k ϵi,n ) T 2 ak,t,p an,s,p + O( T −1 ), q

k =2 n = k +1

where the last equality holds because E(Ui,s ϵi,s ) as,t,p = O( T −1 ), suggesting that q

t

∑ E(Ui,k ϵi,s )ak,t,p q

=

k =2 t

s

∑ ∑ E(Ui,k ϵi,n )ak,t,p an,s,p q

1 T

s −1

1

t

∑ E(Ui,k ϵi,s )Tak,t,p + T ∑ q

k =2

k = s +1

t

s

1 T2

k = s +1 n =2

+

1 T2

∑ ∑ E(Ui,k ϵi,n )T2 ak,t,p an,s,p

+

1 T2

=

k =2 n =2

E(Ui,k ϵi,s ) Tak,t,p + O( T −1 ), q

∑ ∑ E(Ui,k ϵi,n )T2 ak,t,p an,s,p

s

q

k −1

q

k =2 n =2 s s

∑ ∑

E(Ui,k ϵi,n ) T 2 ak,t,p an,s,p + O( T −1 ). q

k =2 n = k +1

14

Consider σϵ−2 E(Ui,t ϵi,s ). If t ≥ s, then q

σϵ−2 E(Ui,t ϵi,s ) = q

1 σϵ2

t

∑ [T −1 (t − m)]q E(ϵi,m ϵi,s ) = σϵ−2 [T −1 (t − s)]q E(ϵi,s2 ) = [T −1 (t − s)]q ,

m =1

whereas if t ≤ s, then σϵ−2 E(Ui,t ϵi,s ) = 0. Insertion into the expression for E(Ui,t,p ϵi,s,p ) leads q

q

to the following for t ≥ s: σϵ−2 E(Ui,t,p ϵi,s,p ) = JT,p (s, t) + O( T −1 ), q

q

(19)

where JT,p (s, t) = [ T −1 (t − s)]q − q

+

1 T2

t

∑

1 T

s

∑ [T −1 (t − k)]q Tak,s,p −

k =2

s

∑ [T −1 (k − n)]q T2 ak,t,p an,s,p +

k = s +1 n =2

1 T2

t

1 T

∑

[ T −1 (k − s)]q Tak,t,p

k = s +1 s

k −1

∑ ∑ [T −1 (k − n)]q T2 ak,t,p an,s,p .

k =2 n =2

Let ∫ 1

q

Jp =

w =0

∫ w

q

v =0

J p (v, w)dvdw,

where q J p (v, w)

= (w − v) − q

∫ w ∫ v

+ ∫uv=v ∫ xu=0

+ u =0

x =0

∫ v u =0

(w − u) a p (u, v)du − q

∫ w u=v

(u − v)q a p (u, w)du

(u − x )q a p (u, w) a p ( x, v)dxdu (u − x )q a p (u, w) a p ( x, v)dxdu,

with (w, v) ∈ [0, 1] × [0, 1]. Since | JT,p (s, t) − J p (v, w)| = O( T −1 ) uniformly in (w, v), and q

q

t = ⌊wT ⌋ and s = ⌊vT ⌋, we have, by the continuous mapping theorem, E( JiT,p ) = J p + O( T −1 ). q

q

(20)

15

Consider J p0 . From ∫ v u =0

x =0

k =1 p

∫ w

∑ hk,p

k =1 p

∫ w ∫ v

hk,p hm,p

k =1 m =1 p k =1 p

m =1 u = v

p

1 = ∑ ∑ hk,p hm,p mk k =1 m =1

∫ v ∫ u

p

p

x =0

a p (u, w) a p ( x, v)dxdu =

∑∑

hk,p hm,p

∑∑

hk,p hm,p

∑∑

hk,p hm,p

k =1 m =1 p p

= =

k =1 m =1

we obtain p

1 = 1 − ∑ hk,p k k =1

(

vk 2− k w

)

p

+

(

x =0

u =0

u =0

du

vk 1− k w x =0

∫ 1

∫ w vk

w =0

w =0

∫ w

v =0 w k

v =0

∫ 1

dvdw =

dvdw =

w =0

( hk,p hm,p

k =1 m =1

w 1 dw = , ( k + 1) 2( k + 1)

1 , 2

∫ w (

) vk 2 − k dvdw = w w =0 v =0 ) ∫ 1 ∫ w ( k 1 v − dvdw = k (m + k )wk w=0 v=0 mk

16

wk

vm

vk 1 − mk k(m + k)wk

we get ∫ 1

, dxdu

u m + k −1 du mwk vm

Here, by using ∫ 1

)

vk , m(m + k )wk

p

∑∑

mwk

u k −1 x m −1 dxdu wk vm

∫ v ∫ u u k −1 x m −1

∫ v

k =1 m =1 p p

J p0 (v, w)

u=v

p ∫ w u k −1

∑ hk,p hm,p ∑

=

u =0

u=v

p

∑∑

a p (u, w) a p ( x, v)dxdu =

u k −1 1 du = ∑ hk,p , k k v k =1 ( ) p u k −1 vk 1 1 − du = h , ∑ k,p k wk wk k =1 p

u =0

∑ hk,p

a p (u, w)du =

∫ w ∫ v u=v

∫ v

a p (u, v)du =

∫ w u=v

p

2k + 1 , 2( k + 1) k+m+1 , 2m(m + k )(k + 1)

) .

(21)

which in turn implies J p0

∫ 1

∫ w

w =0 ∫ 1

v =0 ∫ w

=

J p0 (v, w)dvdw

) ∫ ∫ w ( vk 1 1 2 − k dvdw = dvdw − ∑ hk,p k w =0 v =0 w w =0 v =0 k =1 ( ) ∫ 1 ∫ w p p 1 vk − + ∑ ∑ hk,p hm,p dvdw k (m + k )wk w=0 v=0 mk k =1 m =1 p

p

p

p

k+m+1 2k + 1 1 − ∑ hk,p + ∑ ∑ hk,p hm,p . 2 k =1 2k(k + 1) k=1 m=1 2m(m + k)(k + 1)

=

(22)

J p1 can be evaluated in the same way. In particular, since ∫ w u=v

p

ua p (u, w)du =

∑ hk,p

k =1

∫ w uk u=v

p

1 du = ∑ hk,p k ( k + 1) w k =1

(

v k +1 w− k w

) ,

0 ), we have that, via the results provided in the proof of E( IiT,p

∫ v u =0

∫ w u=v

p

(w − u) a p (u, v)du =

∑ hk,p

k =1 p

(u − v) a p (u, w)du =

∑ hk,p

k =1

( (

v w − k ( k + 1)

) ,

w v v k +1 − + ( k + 1) k k ( k + 1) w k

) .

Similarly, since ∫ w ∫ v u=v

x =0

u =0

x =0

ua p (u, w) a p ( x, v)dxdu =

∫ v ∫ u

ua p (u, w) a p ( x, v)dxdu =

(

v k +1 w− k w 1 v k +1 , m ( k + m + 1) w k 1 m ( k + 1)

) ,

we can show that ∫ w ∫ v

(u − x ) a p (u, w) a p ( x, v)dxdu ( ) w v ( k − m ) v k +1 = ∑ ∑ hk,p hm,p − − , m(k + 1) (m + 1)k mk (k + 1)(m + 1)wk k =1 m =1

u=v

x =0 p p

and ∫ v ∫ u u =0

x =0

p

(u − x ) a p (u, w) a p ( x, v)dxdu =

p

∑∑

k =1 m =1

17

hk,p hm,p

v k +1 . m(m + 1)(k + m + 1)wk

It follows that J p1 (v, w)

= (w − v) −

∫ v

∫ w ∫ v

u =0

∫ w

(w − u) a p (u, v)du −

u=v

(u − v) a p (u, w)du ∫ v ∫ u

(u − x ) a p (u, w) a p ( x, v)dxdu + (u − x ) a p (u, w) a p ( x, v)dxdu u = v x =0 u =0 x =0 ( ) p 1 v k +1 (2k + 1)w − (2k + 1)v + k = (w − v) − ∑ hk,p k ( k + 1) w k =1 ( ) p p w v v k +1 + ∑ ∑ hk,p hm,p − + , (23) m(k + 1) (m + 1)k k(k + 1)(k + m + 1)wk k =1 m =1 +

which we can use to show that ∫ 1

J p1 =

∫ w

w =0

∫ 1

v =0

∫ w

= w =0 p

J p1 (v, w)dvdw

(w − v)dvdw

v =0 ∫ 1

) ( v k +1 1 (2k + 1)w − (2k + 1)v + k dvdw − ∑ hk,p w w =0 v =0 k ( k + 1 ) k =1 ( ) ∫ 1 ∫ w p p w v v k +1 + ∑ ∑ hk,p hm,p − + dvdw (m + 1)k k(k + 1)(k + m + 1)wk w =0 v =0 m ( k + 1 ) k =1 m =1 ∫ w

p

1 (2k + 1)(k + 2) + 2 − ∑ hk,p 6 k =1 6k(k + 1)(k + 2)

=

p

p

∑∑

+

hk,p hm,p

k =1 m =1

(m(k − 1) + 2k)(k + 2)(k + m + 1) + 2m(m + 1) . 6mk(k + 1)(k + 2)(m + 1)(k + m + 1)

q

Finally, consider E(KiT,p ). For t > s, 1 T2

s

s

∑ ∑ E(Ui,k ϵi,n )T2 ak,s,p an,t,p q

k =2 n =2

=

1 T2

+

1 T2

=

1 T2

s

s

1

k −1

∑ E(Ui,k ϵi,k )T2 ak,s,p ak,t,p + T2 ∑ ∑ E(Ui,k ϵi,n )T2 ak,s,p an,t,p

k =2 s

q

k =2 n =2

s

∑ ∑

k =2 n = k +1 s

k −1

q

E(Ui,k ϵi,n ) T 2 ak,s,p an,t,p

∑ ∑ E(Ui,k ϵi,n )T2 ak,s,p an,t,p + O(T −1 ), q

k =2 n =2

18

q

(24)

q

leading to the following expression for E(Ui,s,p ϵi,t,p ): q

E(Ui,s,p ϵi,t,p ) t

s

q

k =2

= − = − =

1 T 1 T

s

s

t

∑ E(Ui,s ϵi,k )ak,t,p − ∑ E(Ui,k ϵi,t )ak,s,p + ∑ ∑ E(Ui,k ϵi,n )ak,s,p an,t,p

q

= E(Ui,s ϵi,t ) −

1

s

q

q

k =2 n =2

k =2 s

∑ E(Ui,s ϵi,k )Tak,t,p + T2 ∑ ∑ E(Ui,k ϵi,n )T2 ak,s,p an,t,p q

k =2 n =2

k =2 s

∑

q

q

E(Ui,s ϵi,k ) Tak,t,p +

k =2 2 q σϵ KT,p (s, t) + O( T −1 )

=

1 T2

s

k −1

∑ ∑ E(Ui,k ϵi,n )T2 ak,s,p an,t,p + O(T −1 ) q

k =2 n =2 2 q σϵ K p (v, w) + O( T −1 ),

(25)

where q

KT,p (s, t) = − q K p (v, w)

= −

1 T

s

∑ [T −1 (s − n)]q Tan,t,p +

n =2

∫ v

u =0

(v − u)q a p (u, w)du +

1 T2

s

k −1

∑ ∑ [T −1 (k − n)]q T2 ak,s,p an,t,p ,

k =2 n =2

∫ v ∫ u u =0

x =0

(u − x )q a p (u, v) a p ( x, w)dxdu.

The following holds: ∫ v x =0

p

a p ( x, w)dx =

k =1 p

∫ v ∫ u u =0

x =0

∑ hk,p

a p (u, v) a p ( x, w)dxdu =

∫ v x =0

x k −1 vk dx = h , k,p ∑ wk kwk k =1 p

p

∑∑

hk,p hm,p

∑∑

hk,p hm,p

∑∑

hk,p hm,p

k =1 m =1 p p

=

∫ v x =0

∫ v ∫ u u =0

x =0

(v − x ) a p ( x, w)dx =

k =1 m =1 ∫ v p

∑ hk,p

k =1 p

(u − x ) a p (u, v) a p ( x, w)dxdu =

x =0

∑∑

hk,p hm,p

∑∑

hk,p hm,p

19

vk wm

dxdu

u m + k −1 du mvk wm

vm , m(m + k )wm p

∫ v

p

k =1 m =1

u =0

x =0

vx k−1 − x k v k +1 dx = h , k,p ∑ wk k ( k + 1) w k k =1

k =1 m =1 p p

=

u =0

∫ v

k =1 m =1 p p

=

∫ v ∫ u u k −1 x m −1

u =0

uk+m du m ( m + 1) w m v k

v m +1 , m(m + 1)(m + k + 1)wm

from which we obtain

= −

K0p (v, w)

∫ v

∫ v ∫ u

a p ( x, w)dx +

x =0 p

p

K1p (v, w) = −

x =0 p

x =0

a p (u, v) a p ( x, w)dxdu

vk vm + h h ∑ ∑ k,p m,p m(m + k)wm , kwk k=1 m=1

= − ∑ hk,p k =1 ∫ v

u =0 p

(v − x ) a p ( x, w)dx +

∫ v ∫ u u =0 p

p

x =0

(26)

(u − x ) a p (u, v) a p ( x, w)dxdu

v k +1 v m +1 h h . + m,p k,p ∑ ∑ m(m + 1)(m + k + 1)wm k ( k + 1 ) w k k =1 m =1

= − ∑ hk,p k =1

(27)

Hence, since |KT,p (s, t) − K p (v, w)| = O( T −1 ) for all v ∈ [ T −1 (s − 1), T −1 s] and w ∈ [ T −1 (t − q

q

1), T −1 t], we can show that σϵ−2 E(Ui,s,p ϵi,t,p ) = KT,p (s, t) + O( T −1 ) = K p (v, w) + O( T −1 ), q

q

q

(28)

which in turn implies q

E(KiT,p ) =

t −1

T

1 T2

∑ ∑

KT,p (s, t) + O( T −1 ) = K p + O( T −1 ), q

t = p +1 s = p +1

q

(29)

where q Kp

∫ w

∫ 1

q

= v =0

w =0

K p (v, w)dvdw.

Here, K0p =

∫ 1

∫ w

w =0 p

v =0

= − ∑ hk,p k =1 p

= − ∑ hk,p k =1

K0p (v, w)dvdw ∫ 1

∫ w

w =0

v =0

p

p

vk dvdw + ∑ ∑ hk,p hm,p kwk k =1 m =1 p

∫ 1 w =0

∫ w v =0

vm dvdw m(m + k )wm

p

1 1 + ∑ hk,p hm,p 2m(m + 1)(m + k) , 2k(k + 1) k∑ =1 m =1

(30)

and K1p =

∫ 1

∫ w

w =0 p

v =0

= − ∑ hk,p p

+

k =1 p

∑∑

k =1 m =1 p

∫ 1

∫ w

w =0

v =0

v k +1 dvdw k ( k + 1) w k

∫ 1

hk,p hm,p

= − ∑ hk,p k =1

K1p (v, w)dvdw

w =0

∫ w v =0

v m +1 dvdw m(m + 1)(m + k + 1)wm p

p

1 1 + ∑ hk,p hm,p 3m(m + 1)(m + k + 1)(m + 2) . 3k(k + 1)(k + 2) k∑ =1 m =1 20

(31)

0 ) and E ( K 1 ). The proof of the lemma is This establishes the required results for E(KiT,p iT,p



therefore complete. Proof of Lemma A.2. Consider E( L0iT,p ). Because of symmetry, 1 σϵ2 T 4

E( L0iT,p ) =

1 2 σϵ T 4

=

t −1

∑ ∑

∑

t = p +1 s = p +1 k = p +1 t −1

T

∑ ∑

t = p +1 s = p +1

s −1

∑ ∑

∑

σϵ2 T 4 t= p+1 s= p+1 k= p+1 T

2 T3

=

s −1

∑

2 2 σϵ T 4

T

t −1

∑ ∑

s −1

∑

t = p +1 s = p +1 k = p +1

0 0 E(Ui,s −1,p Ui,k−1,p )

0 0 −1 E(Ui,s −1,p Ui,k−1,p ) + O ( T )

0 IT,p (k, s) + O( T −1 )

t = p +1 s = p +1 k = p +1

∫ 1

= 2

t −1

∑ ∑

0 0 E(Ui,s −1,p Ui,k−1,p )

0 2 E[(Ui,s −1,p ) ] +

t −1

T

2

=

t −1

T

∫ w ∫ v

w =0

v =0

u =0

I p0 (u, v)dudvdw + O( T −1 ),

(32)

0 ( k, s ) and I 0 ( u, v ) are as in the proof of Lemma A.1. The integral can be evaluated where IT,p p

in the same way as before. Specifically, since ∫ 1

∫ w ∫ v

w =0

∫ 1 ∫ 1

v =0

u =0

∫ w ∫ v

w =0

v =0

∫ w ∫ v

w =0

v =0

ududvdw =

uk dudvdw = vk

u =0 u k +1

u =0

vk

dudvdw =

1 , 24 ∫ 1 ∫ w 1 1 vdvdw = , ( k + 1 ) w =0 v =0 6( k + 1) 1 ( k + 2)

∫ 1

w =0

∫ w

v =0

v2 dvdw =

1 , 12(k + 2)

we have ∫ 1

∫ w ∫ v

w =0

v =0

∫ 1

u =0 ∫ w

= w =0 p

I p0 (u, v)dudvdw ∫ v

ududvdw ∫ v (

v =0 u =0 ∫ 1 ∫ w

) u k +1 (2k + 1)u − − ∑ hk,p dudvdw k ( k + 1) k ( k + 1) v k w =0 v =0 u =0 k =1 ) ∫ 1 ∫ w ∫ v ( p p u k +1 u 1 − + ∑ ∑ hk,p hm,p dudvdw = I p0 , k 4 k(k + 1)(m + k + 1)v w =0 v =0 u =0 k ( m + 1 ) k =1 m =1 and therefore E( L0iT,p ) =

1 0 I + O ( T −1 ) . 2 p

(33)

21

0 ) and E ( M1 ). By using the same steps as in the proof of Lemma A.1, Consider E( MiT,p iT,p

we can show that q

E( MiT,p )

∑ ∑

t −1

T

t −1

T

s −1

∑

t −1

t −1

∑ ∑ ∑ s −1

∑

t = p +1 s = p +1 k = p +1

∫ w ∫ v

= v =0

w =0 q

u =0

q

∑ ∑

s −1

∑

t = p +1 s = p +1 k = p +1

q

E(Ui,s−1,p ϵi,k,p )

E(Ui,s−1,p ϵi,k,p ) E(Ui,s−1,p ϵi,k,p ) E(Ui,s−1,p ϵi,k,p ) + O( T −1 ) q

t = p +1 s = p +1 k = s +1 t −1

t −1

T

q

t = p +1 s = p +1 k = p +1 T

1 2 σϵ T 3

q

q

t = p +1 s = p +1 k = s +1

∑ ∑

∫ 1

t −1

∑ ∑

q

E(Ui,s−1,p ϵi,s,p ) +

∑ ∑ ∑

T

1 T3

=

q

E(Ui,s−1,p ϵi,k,p )

q

t = p +1 s = p +1

1 2 σϵ T 3

+

t −1

T

1 2 σϵ T 3

=

∑

t = p +1 s = p +1 k = p +1

1 σϵ2 T 3

+

t −1

∑ ∑

1 2 σϵ T 3

=

t −1

T

1 2 σϵ T 3

=

q

JT,p (k, s) +

1 T3

q

t −1

t −1

∑ ∑ ∑

t = p +1 s = p +1 k = s +1

∫ 1

J p (u, v)dudvdw + q

T

w =0

∫ w ∫ w v =0

u=v

KT,p (s, k ) + O( T −1 ) q

K p (v, u)dudvdw + O( T −1 ), q

(34)

q

where JT,p (k, s), KT,p (s, k), J p (u, v) and K p (v, u) are as in the proof of Lemma A.1. A straightforward calculation reveals that ∫ w ∫ v

∫ 1 w =0

∫ 1 ∫

v =0

u =0

∫ w ∫ v

w =0 v =0 u =0 1 ∫ w ∫ w

w =0 ∫ 1

v =0 ∫ w

u=v ∫ w

w =0

v =0

u=v

J p0 (u, v)dudvdw = J p1 (u, v)dudvdw =

K0p (v, u)dudvdw = K1p (v, u)dudvdw =

1 0 J , 3 p 1 1 J , 4 p 1 0 K , 3 p 1 1 K . 4 p

0 ) and E ( M1 ) are implied by this. The required results for E( MiT,p iT,p



3 Proof of Theorem 1 We begin with some notation. Under the unit root null, by the properties of OLS, yi,t,p = ϵi,t,p . Hence, letting ϕi = (ϕ1,i , ..., ϕq,i )′ and xi,t = (yi,t−1,p , ..., yi,t−q,p )′ , we have yi,t,p = ϕi′ xi,t + λi f t,p + ei,t,p .

(35)

22

This equation can be written in matrix notation as yi,p = xi ϕi + f p λi + ei,p ,

(36)

where yi = (yi,h , ..., yi,T )′ , f = ( f h , ..., f T )′ and ei = (ei,h , ..., ei,T )′ are ( T − h + 1) × 1, and xi = ( xi,h , ..., xi,T )′ is ( T − h + 1) × q. The matrix notation yd = xϕ′ + f p λ′ + ed

(37)

will also be used, where y = (y1 , ..., y N ) and e = (e1 , ..., e N ) are ( T − h + 1) × N, x =

( x1 , ..., x N ) is ( T − h + 1) × Nq and ϕ = diag(ϕ1′ , ..., ϕ′N ) is N × Nq. In what follows the representations in (35)–(37) will be used interchangeably. It is going to be convenient to let PA = A( A′ A)−1 A′ , such that M A = IT −h+1 − PA . In this notation, we have that, under the null, Mxi yi,p = Mxi ( f p λi + ei,p ) = f p λi + ei,p − Pxi f p λi = f p λi + vi ,

(38)

where the first equality holds because Mxi xi = 0 and vi = Mxi ei,p − Pxi f p λi = ei,p − Pxi ( f p λi + ei,p ) is ( T − h + 1) × 1. It follows that, letting λ = N −1 ∑iN=1 λi and v = N −1 ∑iN=1 vi , 1 fˆ = N

N

∑ Mx yi,p = i

i =1

fp

1 N

N

1

N

∑ λi + N ∑ vi =

i =1

f p λ + v.

(39)

i =1

Let us further define g = fˆ − f p λ. Direct substitution from (39) yields g = fˆ − f p λ = v. The following decomposition of f p λi is very useful: f p λi = f p λλ

−1

−1 −1 −1 −1 λi = fˆλ λi − ( fˆ − f p λ)λ λi = fˆλ λi − gλ λi ,

(40)

such that under the null, Mxi yi,p can be written as −1 −1 Mxi yi,p = f p λi + vi = fˆλ λi − gλ λi + vi .

(41)

The following cumulative sum notation will be used extensively: t

Ft,p =

∑

f s,p ,

s=h

with analogous definitions of Gt , Xi,t and Ei,t,p in terms of gt , xi,t and ei,t,p , respectively. The

( T − h + 1)-rowed vectors of stacked observations on the lags of these cumulative sums 23

will be written in an obvious notation as F−1,p = ( Fh,p , ..., FT −1,p )′ , G−1 = ( Gh , ..., GT −1 )′ , Xi,−1 = ( Xi,h , ..., Xi,T −1 )′ and Ei,−1,p = ( Ei,h,p , ..., Ei,T −1,p )′ . The matrix notations of Xi,t−1 and Ei,t−1,p are given by X−1 = ( X1,−1 , ..., X N,−1 ) and E−1,p = ( E1,−1,p , ..., EN,−1,p ), respectively. Throughout Ft is going to denote a generic sigma-field. Unless otherwise stated, Ft is assumed to be generated by { f s }ts=1 and {es }ts=1 . As in Westerlund (2014), M < ∞, ⌊ x ⌋, → p and →d are going to denote a generic positive number, the integer part of x, convergence in probability, and convergence in distribution, respectively. For a matrix A, tr( A) and || A|| = √ tr( A′ A) are going to denote its trace and Frobenius (Euclidean) norm, respectively. Before we come to the proof of Theorem 1 we state the following very useful lemma. Lemma A.3. Provided that Assumptions 3 and 4 in Westerlund (2014) hold,

| T −1/2 f p′ g| = O p ( N −1/2 ) + O p ( T −1/2 ), NT −1 g′ g = O p ( NT −1 ) + O(1), 2 | T [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ]| = O p (( NT )−1/2 ) + O p ( T −1 ) + O p ( N −1 ), √ | T −1/2 g′ ei,p | = O p ( N −1 T ) + O p ( N −1/2 ) + O p ( T −1/2 ),

| T −1 g′ Ei,p | = O p ( N −1/2 ) + O p ( T −1/2 ), | T −1 g′ Fp | = O p ( N −1/2 ) + O p ( T −1/2 ), ′ −1/2 | T −1 G− ) + O( T −1/2 ), 1 ei,p | = O ( N ′ −1/2 | T −1 G− ) + O( T −1/2 ), 1 f p | = O( N ′ −1 −1/2 | T −1 G− ) + O p ( T −1 ), 1 g | = O p ( N ) + O p (( NT ) ′ −1/2 | T −2 G− ) + O p ( T −1/2 ), 1 Ei,−1,p | = O p ( N ′ −1/2 | T −2 G− ) + O p ( T −1/2 ). 1 F−1,p | = O p ( N

Proof of Lemma A.3. We start with T −1/2 f p′ g, which, via the definitions of g and v, can be written as T −1/2 f p′ g = T −1/2 f p′ v =

N

1 √

N

f p′ [ei,p − Px ( f p λi + ei,p )], ∑ T i

i =1

where N

1 √

N

∑ T

i =1

f p′ Pxi ( f p λi + ei,p ) =

N

1 √

N

f p′ xi ( xi′ xi )−1 xi′ ( f p λi + ei,p ). ∑ T i =1

24

Since xi is stationary and ergodic with E(ei,t xi,t ) = E[ E(ei,t |Ft−1 ) xi,t ] = 0, and similarly E( xi,t f t′ ) = 0, it is clear that || T −1/2 f p′ xi ||, || T −1 xi′ xi || and || T −1/2 xi′ ei,p || are all O p (1). Hence, 1 √ N T

≤

N

∑

f p′ Pxi ( f p λi

i =1 N

1 1 √ TN

= Op (T

+ ei,p )

∑ ||T −1/2 f p′ xi ||||(T −1 xi′ xi )−1 ||||T −1/2 xi′ ( f p λi + ei,p )||

i =1 −1/2

).

Consider ( NT )−1/2 ∑iN=1 f p′ ei,p . Because f t and ei,t are independent by Assumption 4 in Westerlund (2014), the mean is clearly zero. As for the variance, by using the results provided in the proof of Lemma 1 in Westerlund (2014), ′ 2 2 E(ei,p ei,p ) = σe,i diag[(1 − ah,h,p ), ..., (1 − a T,T,p )] = σe,i ( IT −h+1 − A T −h+1,p ),

where A T −h+1,p = diag( ah,h,p , ..., a T,T,p ) and ak,t,p = d′k,p (∑tn=1 dn,p d′n,p )−1 dt,p . Also, by crosssection independence, E(ei,p e′j,p ) = 0 for all i ̸= j. Thus, since the autocovariance structure 2 replaced by σ2 , letting γ −1 N σ2 , we can of f t,p is the same as that of ei,t,p with σe,i ∑i=1 e,i e,N = N f

show that [(

√

E

= =

)(

N

1 NT

∑ f p′ ei,p

√

i =1 N

N

1 NT

N

)]

∑ f p′ ej,p

j =1

i =1 j =1

1 NT

′ )(1 − at,t ) ∑ σe,i2 E[ f p′ ( IT−h+1 − AT−h+1,p ) f p ] = γe,N T ∑ E( f t,p f t,p

∑

∑ E[ f p′ E(ei,p e′j,p | f ) f p ] =

1 NT

N

1 NT

N

i =1

1

i =1

= γe,N σ2f

′ | f ) fp] ∑ E[ f p′ E(ei,p ei,p

1 T

T

t=h

T

∑ (1 − at,t,p )2 ≤ M < ∞,

t=h

where we have made use of the fact that T −1 ∑tT=h (1 − at,t,p )2 = 1 + O( T −1 ). In order to see that this is so, note that by Assumption 2, letting k = ⌊vT ⌋ and t = ⌊wT ⌋, ) −1 ( 1 t ′ ′ JT dt,p Tak,t,p = dk,p JT JT ∑ dn,p dn,p JT T n =1 (∫ w ) −1 ′ ′ → d p (v) d p (u)d p (u) du d p (w) = a p (v, w) u =0

as T → ∞, which holds uniformly in (v, w) ∈ [0, 1] × (0, 1]. If w → 0, then, by the mean value theorem, a p (v, w) → d p (v)′ (wd p (w)d p (w)′ )−1 d p (w). Hence, | ak,t,p | = O( T −1 ), suggesting

25

that T −1 ∑tT=h (1 − at,t,p )2 = 1 + O( T −1 ). Hence, 1 √ NT



N

f p′ ei,p

∑

i =1

It follows that

= O p (1).

N 1 ′ = √ ∑ f p [ei,p − Pxi ( f p λi + ei,p )] N T i =1 N N 1 1 1 ′ ′ ≤ √ √ f p ei,p + √ ∑ f p Pxi ( f p λi + ei,p ) ∑ N NT i=1 N T i =1

| T −1/2 f p′ g|

= O p ( N −1/2 ) + O p ( T −1/2 ).

(42)

This establishes the first result. Next, consider NT −1 g′ g. By the definitions of g and v, NT −1 g′ g = NT −1 v′ v 1 NT

=

1 NT

=

N

N

∑ ∑ [ei,p − Px ( f p λi + ei,p )]′ [ej,p − Px ( f p λ j + e j,p )] i

j

i =1 j =1 N

N

′ ′ e j,p − ei,p Px ( f p λ j + e j,p ) − ( f p λi + ei,p )′ Px e j,p ∑ ∑ [ei,p j

i

i =1 j =1

+ ( f p λi + ei,p )′ Pxi Px j ( f p λ j + e j,p )] = I1 + I2 + I3 ,

(43)

where I1 =

1 NT

I2 = − I3 =

N

′ e j,p , ∑ ∑ ei,p

i =1 j =1

2 NT

1 NT

N

N

N

′ Px ( f p λ j + e j,p ), ∑ ∑ ei,p j

i =1 j =1

N

N

∑ ∑ ( f p λi + ei,p )′ Px Px ( f p λ j + ej,p ). i

j

i =1 j =1

Consider I1 , which we write as I1 =

1 NT

N

1

N

N

′ ei,p + e′ e j,p . ∑ ei,p NT ∑ ∑ i,p

i =1

i =1 j ̸ = i

26

The mean of the second term on the right is clearly zero. The variance is given by ( )2  N N N N N N 1 1 ′ ′ ′ ei,p e j,p  = e j,p ek,p en,p ) E ∑ ∑ ∑ E(ei,p ∑ ∑ NT i=1 j̸= j ( NT )2 i=1 ∑ j ̸ =i k =1 n ̸ = k

=

1 ( NT )2

N

N

′ e j,p e′j,p en,p ) ∑ ∑ E(en,p

j =1 n ̸ = j N

N

=

1 1 T N2

=

2 T −1 γe,N T −1 tr[( IT −h+1

2 2 −1 σe,j T tr[( IT −h+1 − A T −h+1,p )2 ] ∑ ∑ σe,n

j =1 n ̸ = j

− AT −h+1,p )2 ] + O( N −1 )

= O ( T −1 ) + O ( N −1 ) , where the third equality follows from noting that, by cross-section independence and the cyclical property of the trace, ′ 2 ′ E[en,p E(e j,p e′j,p )en,p ] = σe,j E[en,p ( IT −h+1 − AT −h+1,p )en,p ] 2 ′ = σe,j tr[ E(en,p en,p )( IT −h+1 − AT −h+1,p )] 2 2 = σe,j σe,n tr[( IT −h+1 − A T −h+1,p )2 ],

where T −1 tr[( IT −h+1 − A T −h+1,p )2 ] = T −1 ∑tT=h (1 − at,t,p )2 = 1 + O( T −1 ). Thus, since the variance is O( T −1 ) + O( N −1 ), the second term in I1 is O p ( T −1/2 ) + O p ( N −1/2 ). Hence, since the first term is clearly O p (1), we can show that

| I1 | = O p (1) + O p ( T −1/2 ) + O p ( N −1/2 ) = O p (1), The corresponding result for I2 is given by 2 N N ′ | I2 | ≤ ei,p Px j ( f p λ j + e j,p ) ∑ ∑ NT i=1 j=1 √ N N 2 N 1 ′ √ ei,p x j ||( T −1 x ′j x j )−1 |||| T −1/2 x ′j ( f p λ j + e j,p )|| ≤ ∑ ∑ T N j =1 NT i=1 √ = O p ( NT −1 ). Also, from

|( f p λi + ei,p )′ Pxi Px j ( f p λ j + e j,p )| = || T −1/2 ( f p λi + ei,p )′ xi ||||( T −1 xi′ xi )−1 |||| T −1 xi x j ||||( T −1 x ′j x j )−1 || × || T −1/2 x ′j ( f p λ j + e j,p )|| = O p (1), 27

(44)

we get

| I3 | ≤

N

N 1 T N2

N

∑ ∑ |( f p λi + ei,p )′ Px Px ( f p λ j + ej,p )| = O p ( NT −1 ). i

j

i =1 j =1

√ which in turn implies, with O p ( NT −1 ) < O p ( NT −1 ), | NT −1 g′ g| ≤ | I1 | + | I2 | + | I3 | = O p (1) + O p ( NT −1 ).

(45)

2 Consider [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ]. By adding and subtracting terms, 2 fˆ′ fˆ − λ f p′ f p = ( fˆ − f p λ)′ fˆ + λ f p′ ( fˆ − f p λ) = g′ fˆ + λ f p′ g,

suggesting 2

2

2

( fˆ′ fˆ)−1 − (λ f p′ f p )−1 = ( fˆ′ fˆ)−1 [ fˆ′ fˆ − λ f p′ f p H ](λ f p′ f p )−1 = ( fˆ′ fˆ)−1 [ g′ fˆ + λ f p′ g](λ f p′ f p )−1 , whose order can be deduced from (42) and (45). In fact,

| T −1/2 g′ fˆ| ≤ | T −1/2 g′ f p | +

√

TN −1 | NT −1 g′ g| √ = [O p ( N −1/2 ) + O p ( T −1/2 )] + TN −1 [O p (1) + O p ( NT −1 )] √ = O p ( N −1/2 ) + O p ( T −1/2 ) + O p ( TN −1 ).

2 Therefore, since T −1 λ f p′ f p and T −1 fˆ′ fˆ are O p (1), we can show that 2 T |( fˆ′ fˆ)−1 − (λ f p′ f p )−1 | 2 ≤ T −1/2 |( T −1 fˆ′ fˆ)−1 || T −1/2 g′ fˆ + T −1/2 λ f p′ g||( T −1 λ f p′ f p )−1 | √ = T −1/2 [O p ( N −1/2 ) + O p ( T −1/2 ) + O p ( TN −1 )]

= O p (( NT )−1/2 ) + O p ( T −1 ) + O p ( N −1 ),

(46)

as required. Next, consider T −1/2 g′ ei,p =

1 √

N

N T

∑ [ej,p − Px ( f p λ j + e j,p )]′ ei,p . j

j =1

We begin by noting that,

|( f p λ j + e j,p )′ Px j ei,p | ≤ || T −1/2 ( f p λ j + e j,p )′ x j ||||( T −1 x ′j x j )−1 |||| T −1/2 x ′j ei,p || = O p (1). Moreover, by cross-section independence,

N

1 √

N

∑ e′j,p ei,p = N −1 T j =1

√

1 1 ′ TT −1 ei,p ei,p + √ √ N NT 28

N

∑ e′j,p ei,p = O p ( N −1 j ̸ =i

√

T ) + O p ( N −1/2 ),

which we can use to show that N 1 −1/2 ′ ′ |T g ei,p | = √ ∑ [e j,p − Px j ( f p λ j + e j,p )] ei,p N T j =1 N 1 1 1 N ′ ≤ √ ∑ e j,p ei,p + √ ∑ |( f p λ j + ej,p )′ Pxj ei,p | N T j =1 T N j =1 √ = O p ( N −1 T ) + O p ( N −1/2 ) + O p ( T −1/2 ).

(47)

The same arguments used for establishing (47) can be used to show that

| T −1 ( f p λ j + e j,p )′ Px j Ei,p | = O p ( T −1/2 ), and therefore 1 NT

T −1 g′ Ei,p =

1 NT

=

N

1

N

∑ e′j,p Ei,p + N ∑ T −1 ( f p λ j + ej,p )′ Px Ei,p j

j =1

j =1

N

∑ e′j,p Ei,p + O p (T −1/2 ),

(48)

j =1

where the first term on the right is O p ( N −1/2 ). Similarly, T −1 g′ Fp =

N

1 NT

∑ e′j,p Fp +

j =1

1 N

N

∑ T −1 ( f p λ j + ej,p )′ Px Fp = O p ( N −1/2 ) + O p (T −1/2 ). j

(49)

j =1

′ e , where Consider T −1 G− 1 i,p t

Gt =

∑

1 N

∑

1 N

1 N

N

s=h t

=

s=h

=

N

∑ vi,s

i =1 N

∑ [ei,s,p − xi,s′ (xi′ xi )−1 xi′ ( f p λi + ei,p )]

i =1

∑ [Ei,t,p − Xi,t′ (xi′ xi )−1 xi′ ( f p λi + ei,p )],

i =1

and therefore G−1 =

1 N

N

∑ [Ei,−1,p − Xi,−1 (xi′ xi )−1 xi′ ( f p λi + ei,p )].

i =1

It follows that ′ T −1 G− 1 ei,p =

=

N

1 NT

j =1

1 NT

∑ E′j,−1,p ei,p − NT ∑ ( f p λ j + ej,p )′ x j (x′j x j )−1 Xj,′ −1 ei,p ,

∑ [Ej,−1,p − Xj,−1 (x′j x j )−1 x′j ( f p λ j + ej,p )]′ ei,p N

j =1

1

N

j =1

29

(50)

where, by Lemma 2.1 of Park and Phillips (1989), || T −1 Xi,′ −1 ei,p || = O p (1), suggesting

| T −1 ( f p λ j + e j,p )′ x j ( x ′j x j )−1 X j,′ −1 ei,p | ≤ T −1/2 || T −1/2 ( f p λ j + e j,p )′ xi ||||( T −1 xi′ xi )−1 |||| T −1 Xi,′ −1 ei,p || = O p ( T −1/2 ). As for the remaining term, by the definition of Ei,−1,p , ( )2  N 1 E E′j,−1,p ei,p  NT j∑ =1 1 ( NT )2

= = = + = =

N

j =1 k =1 N

N

T t −1 s −1

T

1 ( NT )2

j =1 k =1 t = h s = h l = h m = h

1 ( NT )2

∑∑∑∑ ∑

j =1 k =1 t = h l = h m = h

2 ( NT )2

∑∑∑∑∑ ∑

∑∑∑∑∑ ∑ N

N

N

N

T t −1 t −1

E[e j,l,p E(ei,t,p |Ft−1 )en,s,p ek,m,p ]

j =1 k =1 t = h s = h l = h m = h N

N

T t −1 t −1

j =1 k =1 t = h l = h m = h

1 ( NT )2

∑∑ ∑

= O( N

E[e j,l,p ei,t,p en,s,p ek,m,p ]

E[e j,l,p E(ei,t,p en,t,p |Ft−1 )ek,m,p ]

T t −1 t −1 s −1

1 ( NT )2

1 1 N T2

=

N

′ Ek,−1,p ] ∑ ∑ E[E′j,−1,p ei,p en,p

∑∑∑∑ ∑ N

T

t −1

2 σe,i (1 − at,t,p ) E[ E(e j,l,p ek,m,p |Ft−1 )]

2 2 σe,i σe,j (1 − at,t,p )(1 − am,m,p )

j =1 t = h m = h t −1

T

∑∑

2 σe,i γe,N (1 − at,t,p )(1 − am,m,p )

t=h m=h −1

),

where the last result is due to the fact that 1 T2

T

∑

t −1

∑ (1 − at,t,p )(1 − am,m,p ) =

t=h m=h

1 T2

T

∑ t + O ( T −1 ) →

t=h

∫ 1 v =0

vdv =

1 2

as T → ∞. Hence, 1 N ′ −1/2 ), NT ∑ Ej,−1,p ei,p = O( N j =1 and so we obtain

|T

−1

′ G− 1 ei,p |

1 ≤ NT

= O( N

N

∑

j =1



1 N

E′j,−1,p ei,p +

−1/2

) + O( T

−1/2

).

30

N

∑ |T −1 ( f p λ j + ej,p )′ x j (x′j x j )−1 Xj,′ −1 ei,p |

j =1

(51)

The proof of ′ −1/2 | T −1 G− ) + O( T −1/2 ) 1 f p | = O( N

(52)

is almost identical to the one just given, and is therefore omitted. ′ g. Write Consider T −1 G− 1

1 N2 T

′ T −1 G− 1g =

N

N

∑ ∑ [Ej,−1,p − Xj,−1 (x′j x j )−1 x′j ( f p λ j + e j,p )]′ [ei,p − Px ( f p λi + ei,p )] i

j =1 j =1

= J1 + J2 + J3 + J4 , where J1 =

1 N2 T

J2 = − J3 = − J4 =

N

j =1 i =1

1 N2 T 1 N2 T

1 N2 T

N

∑ ∑ E′j,−1,p ei,p , N

N

∑ ∑ E′j,−1,p Px ( f p λi + ei,p ), i

j =1 i =1 N

N

∑ ∑ ( f p λ j + ej,p )′ x j (x′j x j )−1 Xj,′ −1 ei,p ,

j =1 i =1

N

N

∑ ∑ ( f p λ j + ej,p )′ x j (x′j x j )−1 Xj,−1 Px ( f p λi + ei,p ). i

j =1 i =1

′ e ||, We start with J1 . By using the same steps as in the proof for || T −1 G− 1 i,p ( )2  1 N N ′ E ∑ Ei,−1,p e j,p  NT i∑ =1 j =1

= = = + = = =

N

N

N

N

1 ( NT )2

i =1 j =1 m =1 l =1

1 ( NT )2

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ E(ei,k,p e j,t,p el,s,p em,n,p )

1 ( NT )2

′ Em,−1,p ) ∑ ∑ ∑ ∑ E(Ei,′ −1,p e j,p el,p N

N

N

N

T

T t −1 s −1

i =1 j =1 m =1 l =1 t = h s = h k = h n = h N

N

N

N

T t −1 t −1

∑ ∑ ∑ ∑ ∑ ∑ ∑ E[ei,k,p E(ej,t,p el,t,p |Ft−1 )em,n,p ]

i =1 j =1 m =1 l =1 t = h k = h n = h N

N

N

N

T t −1 t −1 s −1

2 ( NT )2

i =1 j =1 m =1 l =1 t = h s = h k = h n = h

1 ( NT )2

∑ ∑ ∑ ∑ ∑ ∑ σe,j2 (1 − at,t,p )E(ei,k,p em,n,p )

1 T2 1 T2

N

∑ ∑ ∑ ∑ ∑ ∑ ∑ ∑ E[ei,k,p E(e j,t,p |Ft−1 )el,s,p em,n,p ] N

N

N

T t −1 t −1

i =1 j =1 m =1 t = h k = h n = h N

T t −1

∑ ∑ ∑ ∑ σe,j2 σe,i2 (1 − at,t,p )(1 − ak,k,p )

i =1 j =1 t = h k = h T t −1

2 ≤ M, ∑ ∑ (1 − at,t,p )(1 − ak,k,p )γe,N

t=h k=h

31

and so we obtain that 1 1 N N ′ | J1 | = Ej,−1,p ei,p = O p ( N −1 ). ∑ ∑ N NT i=1 j=1 Similar calculations reveal that 1 1 N 1 | J2 | = − √ ∑ √ NT N j=1 NT

N

∑

i =1



E′j,−1,p xi ||( T −1 xi′ xi )−1 |||| T −1/2 xi′ ( f p λi

+ ei,p )||

= O p (( NT )−1/2 ), with | J3 | being of the same order. As for J4 , we have

| T −1 ( f p λ j + e j,p )′ x j ( x ′j x j )−1 X j,−1 Pxi ( f p λi + ei,p )| ≤ T −1 || T −1/2 ( f p λ j + e j,p )′ x j ||||( T −1 x ′j x j )−1 |||| T −1 X j,′ −1 xi || × ||( T −1 xi′ xi )−1 |||| T −1/2 xi′ ( f p λi + ei,p )|| = O p ( T −1 ), with which we obtain

| J4 | ≤

1 N2

N

N

∑ ∑ ||T −1 ( f p λ j + ej,p )′ x j (x′j x j )−1 Xj,−1 Px ( f p λi + ei,p )|| = O p (T −1 ). i

j =1 i =1

Hence, by adding the terms, ′ −1 −1/2 | T −1 G− ) + O p ( T −1 ). 1 g | ≤ | J1 | + | J2 | + | J3 | + | J4 | = O p ( N ) + O p (( NT )

(53)

′ E Next, consider T −2 G− 1 i,−1,p . By using calculations similar to those used for evaluating

J1 before, ( 1 E √ NT 2 suggesting 1 NT 2

N

∑

j =1

)2  ∑ E′j,−1,p Ei,−1,p  ≤ M, N

j =1



E′j,−1,p Ei,−1,p

1 1 = √ √ N NT 2

N

∑

j =1



E′j,−1,p Ei,−1,p

= O p ( N −1/2 ).

But we also have 1 N ′ ′ −1 ′ λ ( f λ + e ) x ( x x ) X E p j j j,p j j i, − 1,p ∑ j j, − 1 NT 2 j =1

≤

1 1 √ TN

= Op (T

N

∑ |λ j |||T −1/2 ( f p λ j + e j,p )′ x j ||||(T −1 x′j x j )−1 ||||T −2 Xj,′ −1 Ei,−1,p ||

j =1

−1/2

), 32

and therefore ′ | T −2 G− 1 Ei,−1,p | 1 N ′ ′ ′ − 1 ′ = [ Ej,−1,p − ( f p λ j + e j,p ) x j ( x j x j ) X j,−1 ] Ei,−1,p ∑ 2 NT j=1 1 N ′ 1 N ′ ′ −1 ′ ≤ E E + ( f λ + e ) x ( x x ) X E p i, − 1,p j j,p j j i, − 1,p ∑ ∑ j, − 1,p j j, − 1 NT 2 NT 2 j=1 j =1

= O p ( N −1/2 ) + O p ( T −1/2 ).

(54)

The proof of ′ −1/2 | T −2 G− ) + O p ( T −1/2 ) 1 F−1,p | = O p ( N

(55)

is almost identical to the previous one. It is therefore omitted. This completes the proof of



the lemma. Proof of Theorem 1.

2 is known. We then show that the asymptotic results are unafWe start by assuming that σe,i 2 is replaced by σ 2 . ˆ e,i fected if σe,i

−2 ′ Consider N −1/2 T −1 ∑iN=1 σe,i Ri,−1,p ri,p , the numerator of the infeasible version of t–RREC.

Making use of (41) to substitute for Mxi yi,p in ri,p = M fˆ Mxi yi,p ,

√

N

1

−2 ′ σe,i Ri,−1,p ri,p ∑ NT i =1

=

√

N

1 NT 1

∑ σe,i−2 Ri,′ −1,p M fˆ Mx yi,p i

i =1 N

∑ σe,i−2 Ri,′ −1,p M fˆ[ fˆH −1 λi − gH −1 λi + ei,p − Pxi ( f p λi + ei,p )] NT i=1 = K1 + K2 + K3 ,

=

√

where K1 =

√

N

1

−2 ′ σe,i Ri,−1,p M fˆ ei,p , ∑ NT

K2 = − √ K3 = − √

1

i =1 N

−2 ′ σe,i Ri,−1,p M fˆ gλ ∑ NT

1 NT

−1

λi ,

i =1 N

∑ σe,i−2 Ri,′ −1,p M fˆ Px ( f p λi + ei,p ). i

i =1

33

(56)

We begin by considering K2 . In particular, let us consider T −1 Ri,′ −1,p g. Note how

( M f p λ − M fˆ ) = g( fˆ′ fˆ)−1 g′ + 2g( fˆ′ fˆ)−1 f p′ + f p [( fˆ′ fˆ)−1 − ( f p′ f p )−1 ] f p′ .

(57)

2

This, together with Pf p λ = f p λ(λ f p′ f p )−1 λ f p′ = f p ( f p′ f p )−1 f p′ = Pf p and M f p Mxi ( f p λi + ei,p )

= M f p ( IT −h+1 − Pxi )( f p λi + ei,p ) = M f p [ei,p + Pxi ( f p λi + ei,p )] = ei,p + xi ( xi′ xi )−1 xi′ ( f p λi + ei,p ) − f p ( f p′ f p )−1 f p′ [ei,p + Pxi ( f p λi + ei,p )], suggests that ri,p can be expanded in the following fashion: ri,p = M fˆ Mxi ( f p λi + ei,p )

= M f p Mxi ( f p λi + ei,p ) − ( M f p λ − M fˆ ) Mxi ( f p λi + ei,p ) = ei,p + xi ( xi′ xi )−1 xi′ ( f p λi + ei,p ) − f p ( f p′ f p )−1 f p′ [ei,p + Pxi ( f p λi + ei,p )] 2 2 − [ g( fˆ′ fˆ)−1 g′ + λg( fˆ′ fˆ)−1 f p′ + λ f p ( fˆ′ fˆ)−1 g′ + λ f p (( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ) f p′ ]

× Mxi ( f p λi + ei,p ),

(58)

and therefore Ri,−1,p

= Ei,−1,p + Xi,−1 ( xi′ xi )−1 xi′ ( f p λi + ei,p ) − F−1,p ( f p′ f p )−1 f p′ [ei,p + Pxi ( f p λi + ei,p )] − [ G−1 ( fˆ′ fˆ)−1 g′ + λG−1 ( fˆ′ fˆ)−1 f p′ + λF−1,p ( fˆ′ fˆ)−1 g′ 2 2 + λ F−1,p (( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ) f p′ ] Mxi ( f p λi + ei,p ).

(59)

It follows that T −1 Ri,′ −1,p g

= T −1 Ei,′ −1,p g + T −1 ( f p λi + ei,p )′ xi ( xi′ xi )−1 Xi,′ −1 g ′ − T −1 [ei,p + ( f p λi + ei,p )′ Pxi ] f p ( f p′ f p )−1 F−′ 1,p g ′ ˆ′ ˆ −1 ′ ˆ′ ˆ −1 ′ − T −1 ( f p λi + ei,p )′ Mxi [ g( fˆ′ fˆ)−1 G− 1 + λ f p ( f f ) G−1 + λg ( f f ) F−1,p 2 2 + λ f p (( fˆ′ fˆ)−1 − f p (λ f p′ f p )−1 ) F−′ 1,p ] g,

34

(60)

where, by Lemma A.3, T −1 ( f p λi + ei,p )′ xi ( xi′ xi )−1 Xi,′ −1 g

= T −1/2 T −1/2 ( f p λi + ei,p )′ xi ( T −1 xi′ xi )−1 T −1 Xi,′ −1 g = T −1/2 [O p ( N −1/2 ) + O p ( T −1/2 )] = O p (( NT )−1/2 ) + O p ( T −1 ), ′ ′ T −1 ei,p f p ( f p′ f p )−1 F−′ 1,p g = T −1/2 T −1/2 ei,p f p ( T −1 f p′ f p )−1 T −1 F−′ 1,p g

= T −1/2 [O p ( N −1/2 ) + O p ( T −1/2 )] = O p (( NT )−1/2 ) + O p ( T −1 ), and T −1 ( f p λi + ei,p )′ Pxi f p ( f p′ f p )−1 F−′ 1,p g

= T −1 T −1/2 ( f p λi + ei,p )′ xi ( T −1 xi′ xi )−1 T −1/2 xi′ f p ( T −1 f p′ f p )−1 T −1 F−′ 1,p g = T −1 [O p ( N −1/2 ) + O p ( T −1/2 )] = O p ( N −1/2 T −1 ) + O p ( T −3/2 ). Also, ′ ˆ′ ˆ −1 ′ ˆ′ ˆ −1 ′ | T −1 ( f p λi + ei,p )′ Mxi [ g( fˆ′ fˆ)−1 G− 1 + λ f p ( f f ) G−1 + λg ( f f ) F−1,p 2 2 + λ f p (( fˆ′ fˆ)−1 − f p (λ f p′ f p )−1 ) F−′ 1,p ] g|

′ ˆ′ ˆ −1 ′ ˆ′ ˆ −1 ′ ≤ | T −1 ( f p λi + ei,p )′ [ g( fˆ′ fˆ)−1 G− 1 + λ f p ( f f ) G−1 + λg ( f f ) F−1,p 2

2

+ λ f p (( fˆ′ fˆ)−1 − f p (λ f p′ f p )−1 ) F−′ 1,p ] g|, where ′ −1/2 −1/2 ′ T −1 ( f p λi + ei,p )′ g( fˆ′ fˆ)−1 G− T ( f p λi + ei,p )′ g( T −1 fˆ′ fˆ)−1 T −1 G− 1g = T 1g √ = T −1/2 [O p ( N −1 T ) + O p ( N −1/2 ) + O p ( T −1/2 )]

× [O p ( N −1 ) + O p (( NT )−1/2 ) + O p ( T −1 )] = O p ( N −2 ) + O p (( NT )−1 ) + O p ( T −2 ), ′ −1 ′ −1 ˆ′ ˆ −1 −1 ′ T −1 ( f p λi + ei,p )′ f p ( fˆ′ fˆ)−1 G− f f ) T G−1 g 1 g = T ( f p λi + ei,p ) f p ( T

= O p ( N −1 ) + O p (( NT )−1/2 ) + O p ( T −1 ), and 2 T −1 ( f p λi + ei,p )′ f p (( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ) F−′ 1,p g 2 = T −1 ( f p λi + ei,p )′ f p T [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ] T −1 F−′ 1,p g

= [O p ( N −1 ) + O p (( NT )−1/2 ) + O p ( T −1 )][O p ( N −1/2 ) + O p ( T −1/2 )] = O p ( N −3/2 ) + O p ( N −1 T −1/2 ) + O p ( N −1/2 T −1 ) + O p ( T −3/2 ). 35

Thus, by adding the results, T −1 Ri,′ −1,p g = T −1 Ei,′ −1,p g + O p ( N −1 ) + O p (( NT )−1/2 ) + O p ( T −1 ).

(61)

By using this and the proof of Lemma A.3, where it is shown that 1 NT

T −1 Ei,′ −1,p g = we obtain

| K2 | = ≤ = = =

N

∑ Ei,′ −1,p ej,p + O p (T −1/2 ),

j =1

N 1 −1 −2 ′ √ NT ∑ σe,i Ri,−1,p M fˆ gλ λi i =1 1 N −2 −1 ′ −1 √ ∑ σ T R λ gλ i i,−1,p e,i N i =1 √ 1 N −2 −1 ′ − 1 −1/2 √ ∑ σ T E ) + O p ( T −1/2 ) + O p ( NT −1 ) i,−1,p gλ λi + O p ( N e,i N i =1 √ 1 1 N N −2 ′ −1 √ σe,i Ei,−1,p e j,p λ λi + O p ( N −1/2 ) + O p ( T −1/2 ) + O p ( NT −1 ) ∑ ∑ N NT i=1 j=1 √ O p ( N −1/2 ) + O p ( T −1/2 ) + O p ( NT −1 ).

As for K3 , since

| T −1 Ri,′ −1,p M fˆ Pxi ( f p λi + ei,p )| ≤ | T −1 Ri,′ −1,p Pxi ( f p λi + ei,p )| ≤ T −1/2 || T −1 Ri,′ −1,p xi ||||( T −1 xi′ xi )−1 |||| T −1/2 xi′ ( f p λi + ei,p )|| = O p ( T −1/2 ), we obtain

| K3 | ≤

√

N

1 N

N

∑ σe,i−2 |T −1 Ri,′ −1,p M fˆ Px ( f p λi + ei,p )||λi | = O p (

√

i

NT −1/2 ).

(62)

i =1

K1 requires more work. We begin by expanding it in the following way: K1 =

= = −

√ √ √ √

1 NT 1

N

∑ σe,i−2 Ri,′ −1,p M fˆ Mx ei,p i

i =1 N

∑ σe,i−2 Ri,′ −1,p M f NT 1

i =1 N

∑ σe,i−2 Ri,′ −1,p M f NT 1

Mxi ei,p − √ p

N

1

∑ σe,i−2 Ri,′ −1,p ( M f λ − M fˆ) Mx ei,p NT

p

ei,p − √

i =1 N

i

N

1

∑ σe,i−2 Ri,′ −1,p M f NT

p

Pxi ei,p

i =1

−2 ′ σe,i Ri,−1,p ( M f λ − M fˆ ) Mx ei,p . ∑ NT i =1

p

i =1

i

p

36

(63)

The summand of the last term can be evaluated using the expansion in (57);

| T −1 Ri,′ −1,p ( M f p λ − M fˆ ) Mxi ei,p | ≤ | T −1 Ri,′ −1,p ( M f p λ − M fˆ )ei,p | = | T −1 Ri,′ −1,p [ g( fˆ′ fˆ)−1 g′ + λg( fˆ′ fˆ)−1 f p′ 2 2 + λ f p ( fˆ′ fˆ)−1 g′ + λ f p (( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ) f p′ ]ei,p |.

The only term here whose order is unknown is Ri,′ −1,p f p . By using the same arguments leading up to (61) we can show that T −1 Ri,′ −1,p f p

= T −1 Ei,′ −1,p f p + T −1 ( f p λi + ei,p )′ xi ( xi′ xi )−1 Xi,′ −1 f p ′ − T −1 [ei,p + ( f p λi + ei,p )′ Pxi ] f p ( f p′ f p )−1 F−′ 1,p f p

− T −1 ( f p λi + ei,p )′ Mxi 2 2 × [ g( fˆ′ fˆ)−1 g′ + λg( fˆ′ fˆ)−1 f p′ + λ f p ( fˆ′ fˆ)−1 g′ + λ f p (( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ) f p′ ] f p , (64),

where the first term on the right dominates the others. Hence, since this term is O p (1), we have that T −1 Ri,′ −1,p f p = O p (1).

(65)

By using this and Lemma A.3,

| T −1 Ri,′ −1,p ( M f p λ − M fˆ ) Mxi ei,p | ≤ T −1/2 | T −1 Ri,′ −1,p g||( T −1 fˆ′ fˆ)−1 || T −1/2 g′ ei,p | + T −1/2 |λ|| T −1 Ri,′ −1,p g||( T −1 fˆ′ fˆ)−1 || T −1/2 f p′ ei,p | + T −1/2 |λ|| T −1 Ri,′ −1,p f p ||( T −1 fˆ′ fˆ)−1 || T −1/2 g′ ei,p | 2 2 + T −1/2 λ | T −1 Ri,′ −1,p f p || T [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ]|| T −1/2 f p′ ei,p | √ = T −1/2 [O p ( N −1/2 ) + O p ( T −1/2 )][O p ( N −1 T ) + O p ( N −1/2 ) + O p ( T −1/2 )]

+ T −1/2 [O p ( N −1/2 ) + O p ( T −1/2 )] √ + T −1/2 [O p ( N −1 T ) + O p ( N −1/2 ) + O p ( T −1/2 )] + T −1/2 [O p (( NT )−1/2 ) + O p ( T −1 ) + O p ( N −1 )] = O p ( N −1 ) + O p (( NT )−1/2 ) + O p ( T −1 ),

37

suggesting that N 1 − 2 ′ √ NT ∑ σe,i Ri,−1,p ( M f p λ − M fˆ ) Mxi ei,p i =1 √ 1 N −2 −1 ′ ≤ N ∑ σe,i | T Ri,−1,p ( M f p λ − M fˆ ) Mxi ei,p | N i =1 √ = N [O p ( N −1 ) + O p (( NT )−1/2 ) + O p ( T −1 )] √ = O p ( N −1/2 ) + O p ( T −1/2 ) + O p ( NT −1 ).

(66)

Moreover, from

| T −1 Ri,′ −1,p M f p Pxi ei,p | ≤ | T −1 Ri,′ −1,p Pxi ei,p | ≤ T −1/2 || T −1 Ri,′ −1,p xi ||||( T −1 xi′ xi )−1 |||| T −1/2 xi′ ei,p || = O p ( T −1/2 ), we obtain 1 √ NT

N

∑

i =1



−2 ′ σe,i Ri,−1,p M f p Pxi ei,p

√

1 N −2 −1 ′ σe,i | T Ri,−1,p M f p Pxi ei,p | N i∑ =1 √ = O p ( NT −1/2 ).

≤

N

(67)

It remains to consider the first term on the right-hand side of (63). It is given by

√

1

N

∑ σe,i−2 Ri,′ −1,p M f NT

p

ei,p

i =1

= =

√ √

1

N

N

1

∑ σe,i−2 Ri,′ −1,p ei,p − √ NT ∑ σe,i−2 Ri,′ −1,p f p ( f p′ f p )−1 f p′ ei,p NT 1

i =1 N

−2 ′ σe,i Ri,−1,p ei,p + O p ( ∑ NT

i =1

√

NT −1/2 ) + O p ( N −1/2 ),

i =1

where the last equality holds because N 1 −2 ′ ′ −1 ′ √ NT ∑ σe,i Ri,−1,p f p ( f p f p ) f p ei,p i =1 √ N 1 N −2 −1 ′ ≤ √ ∑ σe,i |T Ri,−1,p f p ||(T −1 f p′ f p )−1 ||T −1/2 f p′ ei,p | T N i =1 √ √ √ = NT −1/2 [O p (1) + O p ( N −1 T )] = O p ( NT −1/2 ) + O p ( N −1/2 ).

38

(68)

The summand in the first term on the right is, via the definition of Ri,−1,p , T −1 Ri,′ −1,p ei,p = T −1 Ei,′ −1,p ei,p + T −1 ( f p λi + ei,p )′ xi ( xi′ xi )−1 Xi,′ −1 ei,p ′ − T −1 [ei,p + ( f p λi + ei,p )′ Pxi ] f p ( f p′ f p )−1 F−′ 1,p ei,p ′ ˆ′ ˆ −1 ′ − T −1 ( f p λi + ei,p )′ Mxi [ g( fˆ′ fˆ)−1 G− 1 + λ f p ( f f ) G−1 2 2 + λg( fˆ′ fˆ)−1 F−′ 1,p + λ f p (( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ) F−′ 1,p ]ei,p ,

where, by using the same arguments as when evaluating T −1 Ri,′ −1,p g, T −1 ( f p λi + ei,p )′ xi ( xi′ xi )−1 Xi,′ −1 ei,p = O p ( T −1/2 ), ′ T −1 ei,p f p ( f p′ f p )−1 F−′ 1,p g = O p ( T −1 ),

T −1 ( f p λi + ei,p )′ Pxi f p ( f p′ f p )−1 F−′ 1,p ei,p = O p ( T −1 ). ′ e , whose effect on the crossThe last term is dominated by T −1 ( f p λi + ei,p )′ f p ( fˆ′ fˆ)−1 G− 1 i,p

sectional sum of T −1 Ri,′ −1,p ei,p is given by

√

1 NT

N

∑ σe,i−2 T −1 ( f p λi + ei,p )′ f p ( fˆ′ fˆ)−1 G−′ 1 ei,p

i =1

1 N −2 ′ √ = T −1 f p′ f p ( T −1 fˆ′ fˆ)−1 T −1 G− ∑ σe,i λi ei,p 1 N i =1 √ N 1 N −2 −1/2 + √ ∑ σe,i T (ei,p )′ f p (T −1 fˆ′ fˆ)−1 T −1 G−′ 1 ei,p T N i =1 √ = O p ( T −1/2 ) + O p ( N −1/2 ) + NT −1/2 [O p ( T −1/2 ) + O p ( N −1/2 )] √ = O p ( T −1/2 ) + O p ( N −1/2 ) + O p ( NT −1 ). √ √ √ It follows that, with O p ( T −1/2 ) < O p ( NT −1/2 ) and O p ( NT −1/2 ) > O p ( NT −1 ),

√

N

1

1 −2 ′ σe,i Ri,−1,p ei,p = √ ∑ NT i=1 N

Consider

T −1 Ei,′ −1,p ei,p

=

T −1

N

∑ σe,i−2 T −1 Ei,′ −1,p ei,p + O p ( N −1/2 ) + O p (

√

NT −1/2 ). (69)

i =1

1 ∑tT=h ∑ts− =h ei,s,p ei,t,p . We begin by considering the mean.

Since E(ei,t ei,s ) = 0 for all t > s = h, ..., T, t t −1

t −1

∑ ∑ E(ei,k ej,i )ak,t,p

=

∑ ∑ ∑ E(ei,n ei,k )ak,t,p as,t−1,p

=

k =2 j = h t

t −1

k=h

t −1 t −1

s

k =2 s = h n =2

By using this and

∑sn=2 an,t,p an,s,p

E( T −1 Ei,′ −1,p ei,p )

1 = T

T t −1

∑∑

∑ ak,t,p σe,i2 , t −1

s

s

∑ ∑ ∑ E(ei,n ei,k )ak,t,p an,s,p = ∑ ∑ an,t,p an,s,p σe,i2 .

k =2 s = h n =2

s = h n =2

= as,t,p , we can show that

−2 σe,i E(ei,k,p ei,t,p )

t=h k=h

39

1 = T

T t −1

(

s

∑ ∑ ∑ an,t,p an,s,p − as,t,p

t=h s=h

n =2

)

= 0.

The variance can be evaluated in the following fashion: E[( T −1 Ei,′ −1,p ei,p )2 ] =

= + =

1 T2

T t −1 s −1

T

∑ ∑ ∑ ∑ σe,i−4 E(ei,k,p ei,n,p ei,s,p ei,t,p )

t=h s=h k=h n=h 1 T t −1 t −1 −4 σ E[ei,k,p ei,n,p E((ei,t,p )2 |Ft−1 )] T 2 t=h k=h n=h e,i 2 T t −1 t −1 s −1 −4 σ E[ei,k,p ei,n,p ei,s,p E(ei,t,p |Ft−1 )] T 2 t=h s=h k=h n=h e,i 1 T t −1 t −1 −2 (1 − at,t,p )σe,i E(ei,k,p ei,n,p ), T 2 t=h k=h n=h

∑∑∑

∑∑∑∑ ∑∑∑

where the first equality holds because of symmetry, while the second uses E[(ei,t,p )2 ] = (1 − 2 . Consider t at,t,p )σe,i ∑k=h ∑lj=h E(ei,k,p ei,j,p ), which for t ≥ l = h, ..., T can be written as t

l

∑ ∑ E(ei,k,p ei,j,p )

k=h j=h

t

l

l

t

k

t

k

l

∑ ∑ E(ei,k ei,j ) − ∑ ∑ ∑ E(ei,s ei,j )a j,k,p − ∑ ∑ ∑ E(ei,s ei,j )as,k,p

=

t

l

k = h s =2 j = h

k = h s = h j =2

k=h j=h

k

s

∑ ∑ ∑ ∑ E(ei,n ei,l )al,k,p an,s,p ,

+

k = h s = h l =2 n =2

2 ) = σ2 , where, since ei,t is serially uncorrelated with E(ei,t e,i t

l

∑ ∑ E(ei,k ei,j )

2 = [l − ( p + 1)]σe,i ,

k=h j=h l

t

∑∑

k

∑ E(ei,s ei,j )a j,k,p =

k = h s = h j =2 t

k

l

∑ ∑ ∑ E(ei,s ei,j )as,k,p

l

k

∑ ∑ a j,k,p σe,i2 ,

k=h j=h l

=

k = h s =2 j = h

k

l

t

k = h s =2 j = h

(

=

k

l

∑ ∑ ∑ E(ei,s ei,j )as,k,p + ∑ ∑ ∑ E(ei,s ei,j )as,k,p l

k = l +1 s =2 j = h

k

t

)

l

∑ ∑ as,k,p + ∑ ∑ as,k,p

k=h s=h

2 σe,i .

k = l +1 s = h

Also, t

l

k

s

∑ ∑ ∑ ∑ E(ei,n ei,m )am,k,p an,s,p

l

=

l

s

k

∑ ∑ ∑ ∑ E(ei,n ei,m )am,k,p an,s,p

k = h s = h m =2 n =2

k = h s = h m =2 n =2

t

+

l

k

s

∑ ∑ ∑ ∑ E(ei,n ei,m )am,k,p an,s,p ,

k = l +1 s = h m =2 n =2

40

where l

l

k

s

∑ ∑ ∑ ∑ E(ei,n ei,m )am,k,p an,s,p

k = h s = h m =2 n =2 l

k

k

s

∑∑∑ ∑

=

l

k = h s = h n =2 m =2

(

l

k

l

k

k=h l

(

s

l

s=h n=h

)

k

k=h

2 σe,i

s = k +1 n = h

k

)

l

∑ ∑ as,k,p + ∑

=

E(ei,n ei,m ) am,k,p an,s,p

k = h s = k +1 n =2 m =2

∑ ∑ ∑ an,k,p an,s,p + ∑ ∑ an,k,p an,s,p

=

s

∑ ∑ ∑ ∑

E(ei,n ei,m ) am,k,p an,s,p +

s=h

2 σe,i ,

ak,s,p

s = k +1

and t

l

k

s

∑ ∑∑ ∑

t

k = l +1 s = h m =2 n =2

l

s

∑ ∑∑

E(ei,n ei,m ) am,k,p an,s,p =

t

l

∑ ∑ as,k,p σe,i2 ,

2 an,k,p an,s,p σe,i =

k = l +1 s = h n =2

k = l +1 s = h

suggesting that t

l

∑ ∑ E(ei,k,p ei,j,p )

k=h j=h

= [l − ( p

k

∑∑

k

s=h

(

( 2 a j,k,p σe,i

s = k +1

[l − ( p + 1)] −

l

k

t

t

l

∑ ∑ as,k,p σe,i2

l

)

l

∑ ∑ as,k,p + ∑ ∑

k=h s=h

2 σe,i

k = l +1 s = h

k = l +1 s = h k

)

l

∑ ∑ as,k,p + ∑ ∑ as,k,p

k=h s=h 2 σe,i +

ak,s,p

l

−

)

l

∑ ∑ as,k,p + ∑

k=h

=

−

l

k=h j=h

(

l

+

2 + 1)]σe,i

2 2 σe,i = bl,p σe,i ,

ak,s,p

k = h s = k +1

with an implicit definition of bl,p . It follows that E[( T −1 Ei,′ −1,p ei,p )2 ] =

1 T2

T t −1 t −1

∑∑

∑ (1 − at,t,p )σe,i−2 E(ei,k,p ei,n,p ) =

t=h k=h n=h

1 T2

T

∑ (1 − at,t,p )bt−1,p

t=h

= ΣT , whose limit as T → ∞ can be deduced by noting that by the Cauchy–Schwarz inequality,

| T −2 ∑tT=h at,t,p bt−1,p | = O( T −1 ). Also, T

−1

b⌊vT ⌋,p → v −

∫ v w =0

∫ w u =0

∫ v

a p (u, w)dudw +

w =0

∫ v u=w

a p (w, u)dudw = b p (v),

from which it follows that ΣT =

1 T2

T

∑ (1 − at,t,p )bt−1,p =

t=h

1 T2

T

∑ bt−1,p + O(T −1 ) →

t=h

41

∫ 1 v =0

b p (v)dv = Σ

as T → ∞. Hence, 1 √ N

N

∑ σe,i−2 T −1 Ei,′ −1,p ei,p →d N (0, Σ)

(70)

i =1

as N, T → ∞, which in turn implies K1

= = − =

√ √ √ √

1

N

−2 ′ σe,i Ri,−1,p M fˆ Mx ei,p ∑ NT i

1 NT 1 NT 1

i =1 N

∑ σe,i−2 Ri,′ −1,p M f p ei,p − √

i =1 N

N

1

−2 ′ σe,i Ri,−1,p M f ∑ NT

p

Pxi ei,p

i =1

∑ σe,i−2 Ri,′ −1,p ( M f λ − M fˆ) Mx ei,p i

p

i =1 N

′ ei,p + O p ( N −1/2 ) + O p ( T −1/2 ) + O p ( ∑ σe,i−2 Ei,p NT

√

NT −1/2 )

i =1

→d N (0, Σ)

(71)

as N, T → ∞ with N/T = o (1). Thus, by combining (62), (62) and (71), (56) becomes

√

1

N

−2 ′ σe,i Ri,−1,p ri,p ∑ NT

=

K1 + K2 + K3

=

√

i =1

1

N

′ ei,p + O p ( N −1/2 ) + O p ( ∑ σe,i−2 Ei,p NT

√

NT −1/2 )

i =1

→d N (0, Σ).

(72)

as N, T → ∞ with N/T → 0. −2 ′ Consider N −1 T −2 ∑iN=1 σe,i Ri,−1,p Ri,−1,p , the denominator of the infeasible version of t–

RREC. The summand can be expanded in the following fashion: T −2 Ri,′ −1,p Ri,−1,p

= T −2 Ei,′ −1,p Ri,−1,p + T −2 ( f p λi + ei,p )′ xi ( xi′ xi )−1 Xi,′ −1 Ri,−1,p ′ − T −2 [ei,p + ( f p λi + ei,p )′ Pxi ] f p ( f p′ f p )−1 F−′ 1,p Ri,−1,p ′ ˆ′ ˆ −1 ′ ˆ′ ˆ −1 ′ − T −2 ( f p λi + ei,p )′ Mxi [ g( fˆ′ fˆ)−1 G− 1 + λ f p ( f f ) G−1 + λg ( f f ) F−1,p 2 2 + λ f p [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ] F−′ 1,p ] Ri,−1,p .

(73)

The first term here is also the leading term. The evaluation of this term can be done in the

42

same way as before, giving T −2 Ei,′ −1,p Ri,−1,p

= T −2 Ei,′ −1,p Ei,−1,p + T −2 Ei,′ −1,p Xi,−1 ( xi′ xi )−1 xi′ ( f p λi + ei,p ) − T −2 Ei,′ −1,p F−1,p ( f p′ f p )−1 f p′ [ei,p + Pxi ( f p λi + ei,p )] − T −2 Ei,′ −1,p [ G−1 ( fˆ′ fˆ)−1 g′ + λG−1 ( fˆ′ fˆ)−1 f p′ + λF−1,p ( fˆ′ fˆ)−1 g′ 2

2

+ λ F−1,p [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ] f p′ ] Mxi ( f p λi + ei,p ) = T −2 Ei,′ −1,p Ei,−1,p + O p ( N −1/2 ) + O p ( T −1/2 ), which in turn implies that the denominator simplifies to 1 N

N

∑ σe,i−2 T −2 Ri,′ −1,p Ri,−1,p =

i =1

1 N

N

∑ σe,i−2 T −2 Ei,′ −1,p Ei,−1,p + O p ( N −1/2 ) + O p (T −1/2 ).

(74)

i =1

The expected value of the remaining term is given by ( ) 1 N −2 −2 ′ 1 N T t −1 t −1 −2 E σ T E E = i, − 1,p ∑ ∑ ∑ σe,i E(ei,k,p ei,n,p ) i,−1,p e,i N i∑ NT 2 i∑ =1 =1 t = h k = h n = h

=

1 T2

T

∑ bt−1,p →

t=h

∫ 1

v =0

b p (v)dv = Σ

which we can use to show that 1 N

N

∑ σe,i−2 T −2 Ri,′ −1,p Ri,−1,p

=

i =1

1 N

N

∑ σe,i−2 T −2 Ei,′ −1,p Ei,−1,p + O p ( N −1/2 ) + O p (T −1/2 )

i =1

→ p Σ.

(75)

The results in (72) and (75) prove the asymptotic distribution of t–RREC in the case when 2 is known. In what remains we show how the effect of replacing σ2 by σ 2 is negligible. ˆ e,i σe,i e,i

In analogy to (73) we have 2 ′ σˆ e,i = T −1 ri,p ri,p

′ = T −1 ei,p ri,p + T −1 ( f p λi + ei,p )′ xi ( xi′ xi )−1 xi′ ri,p ′ − T −1 [ei,p + ( f p λi + ei,p )′ Pxi ] f p ( f p′ f p )−1 f p′ ri,p

− T −1 ( f p λi + ei,p )′ Mxi [ g( fˆ′ fˆ)−1 g′ + λ f p ( fˆ′ fˆ)−1 g′ + λg( fˆ′ fˆ)−1 f p′ 2

2

+ λ f p [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ] f p′ ]ri,p ,

43

(76)

where the leading term is again given by the first term on the right, which is ′ ′ ′ T −1 ei,p ri,p = T −1 ei,p ei,p + T −1 ei,p Xi,−1 ( xi′ xi )−1 xi′ ( f p λi + ei,p ) ′ − T −1 ei,p F−1,p ( f p′ f p )−1 f p′ [ei,p + Pxi ( f p λi + ei,p )] ′ − T −1 ei,p [ g( fˆ′ fˆ)−1 g′ + λ f p ( fˆ′ fˆ)−1 g′ + λg( fˆ′ fˆ)−1 f p′ 2 2 + λ f p [( fˆ′ fˆ)−1 − (λ f p′ f p )−1 ] f p′ ] Mxi ( f p λi + ei,p )

′ = T −1 ei,p ei,p + O p ( N −1/2 ) + O p ( T −1/2 ),

where ′ 2 −1 ′ 2 T −1 ei,p ei,p = σe,i T tr( IT −h+1 ) + T −1/2 T −1/2 tr[ei,p ei,p − σe,i IT − h + 1 ] 2 = σe,i + O p ( T −1/2 ),

′ e ′ where the third equality follows from writing ei,p i,p = tr( ei,p ei,p ) to which we then add and 2 I ′ subtract σe,i T − h+1 . As for the fourth and last equality, making use of the fact that E ( ei,p ei,p ) = T 2 (I −1 T Ta σe,i ∑t=h t,t,p = O(1) and E[(ei,t,p )4 ] = T − h+1 − A T − h+1,p ), tr( A T − h+1,p ) = ∑t=h at,t,p = T 4 ) + o (1) = O (1) (which can be shown using the same steps as in Westerlund, 2014, Proof E(ei,t

of Lemma 1), it is possible to show that, by a central limit theorem, ′ 2 T −1/2 tr[ei,p ei,p − σe,i IT − h + 1 ] ′ 2 2 −1/2 = T −1/2 tr[ei,p ei,p − σe,i ( IT −h+1 − AT −h+1,p )] − σe,i T tr( A T −h+1,p ) ′ ′ 2 −1/2 = T −1/2 tr[ei,p ei,p − E(ei,p ei,p )] − σe,i T tr( A T −h+1,p )

= O p (1) + O( T −1/2 ). 2 therefore simplifies to The expression for σˆ e,i 2 2 σˆ e,i = σe,i + O p ( N −1/2 ) + O p ( T −1/2 ).

(77)



The completes the proof of the theorem.

44

References Choi, M.-D. (1983). Tricks or Treats with the Hilbert Matrix. The American Mathematical Monthly 90, 301–312. Park, J. Y., and P. C. B. Phillips (1989). Statistical inference in regressions with integrated processes: part 2. Econometric Theory 5, 95–131. Phillips, P. C. B., and H. R. Moon (1999). Linear Regression Limit Theory of Nonstationary Panel Data. Econometrica 67, 1057–1111. Westerlund, J., and J. Breitung (2013). Lessons from a Decade of IPS and LLC. Econometric Reviews 32, 547–591. Westerlund, J. (2014). The Effect of Recursive Detrending on Panel Unit Root Tests. Unpublished manuscript.

45