Single-Equation Tests for Cointegration with GLS Detrended Data Appendix

Appendix Proof of Lemma 1: Assumption 1 along with the conditions on T 1=2 [T r] X t=1 for r 2 [0; 1], where 1=2 1=2 vt ) B(r) = 2 =4 11 ! 21 1=2 11 0 1=2 ! 2:1 (L) implies that W(r) 3 5; P r] ! 2:1 = ! 22 ! 21 111 ! 12 , and W = [W10 ; W2 ]0 . Using this notation, we have T 1=2 [T t=1 v1t ) P 0 1=2 1=2 1=2 1=2 1=2 [T r] 1=2 = ! 2:1 ! 21 11 so 11 W1 (r) and T t=1 v2t ) ! 21 11 W1 + ! 2:1 W2 . De…ne 0 1=2 1=2 0 Therefore ! 21 11 W1 = ! 2:1 W1 . Using Lemma 5.6 of Park = R2 =(1 R2 ). P that r] and Phillips (1988), T 1=2 [T t=1 v2t may be written in such way that it depends on only one 0 2 nuisance parameter, namely RP . Because = R2 =(1 R2 ), W12 = [R2 =(1 R2 )]1=2 W 1 (r) + m 1=2 W2 (r) where W 1 (r) = m i=1 Wi (r). The result follows since ut is a near-integrated process in terms of the errors v2t . Proof of Lemma 2: Consider …rst the cases py = px = 0 or py = px = 1. The results for ^ x follow from arguments as in ERS using the fact that the noise component of xt is a vector of random walks and the limit result (2). For ^ y , note that yt = dyt + 0 xt + ut and, hence, (My0 My ) 1 My0 Y = (My0 My ) 1 My0 (U + 0 X ) where Y = [y1 ; :::; yT ], U = [(dy1 + u1 ) ; :::; (dyT + uT ) ] and X = [x1 ; :::; xT ]. The results follows using arguments as in ERS for the …rst term using Lemma 1, and the results for ^ x for the second term. It remains to consider the case py = 0 and px = 1. Without loss of generality suppose xt is a scalar (m = 1), given by xt = 0x + 1x t + uxt = mx1t + mx2t + uxt : Then, ^ ( ^ 0x 0x = (my1 0 my1 ) 1 (my1 0 x ) ^ 0x = (my 0 my ) 1 my 0 f 0x mx + 1x mx + u g 1 1 2 1 1 x y 0 y y 0 x y 0 y y 0 1 1 0x ) = 1x (m1 m1 ) (m1 m2 ) + (m1 m1 ) (m1 ux ) From straightforward derivations, we have: (my1 0 my1 ) 1 (my1 0 mx2 ) ) (1 c + c2 =2) and (my1 0 my1 ) 1 (my1 0 ux ) ) ux1 A-1 Therefore: ( ^ 0x Hence, T 1=2 ( ^ 0x ) 1x (1 c + c2 =2) + ux1 ) 0. Now, 0x ) ( ^ 0y 0x ) ^ 0y = (my 0 my ) 1 (my 0 yt ) 1 1 1 y 0 y y 1 y 0 ^ 0y = (m1 m1 ) m1 f 0y m1 + 0y ) (my1 0 my1 = ) 1 (my1 0 0 x )+ 0 x +u g (my1 0 my1 ) 1 (my1 0 u ) From straightforward derivations, we have: (my1 0 my1 ) ) 1, (my1 0 0 x ) ) ux1 and (my1 0 u ) ) 0 u1 . Hence, ( b 0y ux1 + u1 and T 1=2 ( ^ 0y 0y ) ) 0y ) ) 0. Proof of Lemma 3: We …rst consider the limit of xdt . When py = px = 0, we have: ^ 0x = uxt xdt = xt T 1=2 d xt = T 1=2 ( ^ 0x T 1=2 ( ^ 0x uxt 0x ) 0x ) Using (2) and Lemma 2: 1=2 d x[T r] T 1=2 11 W1 (r) ) B1 (r) (A.1) When px = py = 1, we have: xdt = xt T 1=2 d xt = T ^ 0x 1=2 ^ 1x t = uxt T uxt 1=2 ( ^ 0x ( ^ 0x 0x ) T 0x ) 1=2 ^ ( ( ^ 1x 1x )t 1x )(t=T ) 1x Using Lemma 2: T 1=2 d x[T r] ) B1 (r) rHx Bd1 (r) = 1=2 d 11 W1 (r) When px = 1 and py = 0, ^ 0x = uxt + xdt = xt ( ^ 0x 1x t 0x ) and 1=2 d x[T r] T ) B1 (r) r 1x We now consider the limit of udt . From (7), we have: (^y udt = ut 0 y y ) mt y 0 ^0 x mt : + (A.2) Consider …rst the case with py = 0 and px = 0 or px = 1. Then: udt = ut T 1=2 d ut = T ( ^ 0y 1=2 ut 0 + 0 ^ 0x T 1=2 ( ^ 0y 0y ) + T 0y ) A-2 1=2 0 ^0 0x Using Lemmas 1 and 2: T 1=2 d u[T r] 1=2 ) ! 2:1 J12c (r) Consider now the case with py = 1, then (A.2) is ( ^ 0y udt = ut 0y ) ( ^ 1y 1y )t ( ^ 0y T 0y ) 0 0^ 1x (t=T ) 0 + 0 0 [ ^ 0x + ^ 1x t] and T 1=2 d ut = T 1=2 +T ut 1=2 1=2 T 0 ^0 0x + T 1=2 1=2 ( ^ 1y 1y )(t=T ) Since the estimates of the residuals and the test statistics are invariant to the value can without loss of generality set 01x = 0. Then, using Lemmas 1 and 2, we have: T 1=2 d u[T r] ) B2c (r) r(Hy 0 Hx ) + r 0 Hx = B2c (r) 0 1x , we 1=2 d rHy = ! 2:1 J12c (r). Lemma A.1. When the model is generated according to (1) with T ( T ! 1, we have: P 1=2 1=2 R 1 W1d W1d0 11 1. T 2 xdt xd0 11 t ) 0 P 1=2 1=2 R 1 d 2. T 2 xdt udt ) ! 2:1 11 0 W1d J12c R1 d 2 P ) 3. T 2 (udt )2 ) ! 2:1 0 (J12c P 1=2 R 1 d dW0 1=2 4. T 2 udt 1 0t ) 11 0 J12c P 1=2 R 1 5. T 2 xdt 1 0t ) 11 0 W1d dW0 1=2 1) = c, then, as d (1)2 (udT )2 ) ! 2:1 J12c 6. T 1 7. T 1 d d0 xT xT ) 1=2 d d 0 11 W1 (1)W1 (1) 1=2 11 Proof of Lemma A.1. The results follow straightfowardly from Lemmas 1 and 3, the continuous mapping theorem (CMT) and others results from Chang and Wei (1988) and Phillips (1987), with appropriate modi…cations for the fact that the variables are quasidi¤erenced. Proof of Theorem 1. The cointegration vector ^ is estimated by regressing ytd on xdt . In the …rst case (px = 0; py = 0) both variables are demeaned and for the second case (px = 1; py = 1) both variables are linearly detrended. In general, for both cases, # 1" # " T T X X (^ )= T 2 xdt xd0 T 2 xdt udt : t t=1 t=1 A-3 Using Lemma A.1 and the CMT, we have (^ ) ) 1=2 ! 2:1 1=2 11 1=2 ! 2:1 1=2 11 Z 1 1 W1d W1d0 0 d c: Z 1 d W1d J12c 0 with all terms as de…ned in the text. In the case where px = 1; py = 0 we follow Hansen (1992). First, note that xt = 0x + 1x t + uxt and we can deduce that xdt = xt ^ 0x , with ^ 0x ) 1x (1 c + c2 =2) + ux1 . Note that a similar result holds for yt . Let 1x be the m (m 1) matrix which spans the null space of 1x and let h i C = C1 C2 i h 0 0 1 1=2 : = ) ) ( ( 1x 1x 1x 1x 11 1x 1x Given that ( 1x0 11 1x ) is positive de…nite (because 11 is positive de…nite), C is wellde…ned. Therefore, we have: 3 3 2 2 0 1=2 0 d 0 d ( ( ) ) x C x 1x 1x 11 1x 1 t t 5 5=4 C0 xdt = 4 0 1 0 d 0 d ( 1x ( 1x 1x ) ) xt C2 xt 2 3 ( 1x ( 1x0 11 1x ) 1=2 )0 f 0x + 1x t + uxt ^ 0x g 5 = 4 0 1 0 ^ ( 1x ( 1x 1x ) ) f 0x + 1x t + uxt 0x g 2 3 0 1=2 0 ^ ( 1x ( 1x 11 1x ) ) f( 0x 0x ) + 1x t + uxt g 5 = 4 0 1 0 ^ ( 1x ( 1x 1x ) ) f( 0x 0x ) + 1x t + uxt g De…ning the weight matrix DT = diag(Im 1 T 1=2 ; T ), we have using Lemma 2 and the fact that 1x and 1x are orthogonals: 2 3 0 1=2 1=2 0 T ( 1x ( 1x 11 1x ) ) fux[T r] + op (1)g 5 DT 1 C0 xd[T r] = 4 T 1 ( 1x ( 01x 1x ) 1 )0 fux[T r] + 1x t + op (1)g 2 3 d W1(m 1) d 5 W1R ) 4 : r In the last step, we used the fact that W1(m 1) = C10 B1 (r) = 1x ( 0 1x 11 1x ) 1=20 see Hansen (1992, p. 93, eq. (12)). For udt , we have: udt = ( ^ 0y 0y ) A-4 + 0^ 0x + ut B1 (r); and T 1=2 d u[T r] ) J12c (r). We also have 1 1 T 0 DT C T X 1 xdt xd0 t CDT t=1 and 3=2 T 1 0 DT C T X xdt udt ) t=1 T 1=2 DT C 1 ( ^ " ) = T 1 DT 1 C0 1=2 ) ! 2:1 T X ) 1=2 ! 2:1 Z d d0 W1R W1R Z d J12c W1R 1 xdt xd0 t CDT t=1 Z 1 d d0 W1R W1R Z # 1 " T 3=2 DT 1 C0 T X t=1 xdt udt # 1=2 ! 2:1 e dc d W1R J12c 0 0 d = [W1(m where W1R 1) ; r] . Before proving Theorem 2, we introduce some auxiliary results. Lemma A.2. Let the data be generated according to (1) with = 1+c=T and Assumptions 1-2 holding. Let u^t be the residuals from the cointegration regression (9) estimated using GLS detrended variables with a non-centrality parameter = 1+c=T , and s2 be the estimate of the long-run variance. Then, as T ! 1: PT 1. T 2 2. T 1 2 u^T 3. T 1 t=1 ) ! 2:1 PT d0 d c Ac u^2t ) ! 2:1 d d0 d c Ac (1) c ^t u^t ) ! 2:1 c t=1 u 4. s2 ) ! 2:1 d c d0 d c Ac d c d0 c + Z 1 0 d d0 c D c f0 Wcd dW d c 0 d where Wcd = W1d0 ; J12c , d0 c Z = 1 d W1d J12c 0 Adc = Z 1 1 W1d W1d0 = 1 0 Z 0 1 2 Z 1 W1d W1d0 6 0 Wcd Wcd0 = 6 4 Z 1 d0 W1d J12c 0 A-5 Z h 1 Z0 0 d0 W1d J12c 1 d 2 (J12c ) d0 c 3 7 7 5 1 i ; 2 Adc (1) = 4 W1d (1)Wd1 (1)0 d (1) W1d (1)J12c d d d J12c (1)W1d (1)0 J12c (1)J12c (1) 2 3 I 5 D=4 0 0 1+ 3 5 f 0 = [W0 ; W12 ]0 , W12 = [R2 =(1 R2 )]1=2 W 1 (r) + W2 , J12c (r) is an Ornstein-Uhlenbeck W 1 Rs process such that J12c (r) = W12 (r) + c 0 e(s r)c W12 (r)dr. Proof of Lemma A.2. Note …rst that u^t = udt ( ^ )0 xdt . For part (1) and the cases with px = py = 0 and px = py = 0: 32 3 2 T h i T 2 P xd xd0 T 2 P xd ud ^ X ( ) t t t t 54 5 u^2t = T 2 (^ )0 1 4 P P 2 d d0 2 d d0 1 u T u x T u t=1 t t t t i h 1=2 d0 1=2 ) 1 11 c ! 2:1 3 2 Z 1 Z 1 2 3 1=2 1=2 1=2 d d0 d d0 1=2 1=2 1=2 d W J W W ! 11 2:1 11 1 12c 7 1 1 6 11 ! 2:1 11 c 74 0 6 5 Z 10 5 4 1=2 1=2 Z 1 d d0 d 2 1 J12c W1 ! 2:1 (J12c ) ! 2:1 11 0 ! 2:1 d0 d c Ac 0 d c d0 where d0 1 ], and the other terms are as de…ned in the text and in Lemma A.2. c = [ c For the case where px = 1 and py = 0, we have T h 2 T X u^2t t=1 i 0 ^ = T DT C ( ) 1 2 32 PT PT 1 1 0 1 0 1 d d0 3=2 d d T DT C t=1 xt xt CDT T DT C t=1 xt ut T 4 54 P P T 3=2 DT 1 C0 Tt=1 udt xd0 T 2 Tt=1 (udt )2 t 2 32 3 d h i R Wd Wd0 R Wd J d e 1 1 1 12c c d0 d d 54 5 ! 1=2 ) e d0 1 4 R R 2:1 c Ac c c d d0 d 2 J12c W1 (J12c ) 1 1=2 1 A-6 1=2 DT C ( ^ 1 1 ) 3 5 0 d d 0 = [W1(m where, here, W1d = W1R 1) ; r] and J12c = J12c . For part (2), 2 32 3 h i T 1 xd xd0 T 1 xd ud ^ ( ) T T T T 54 5 T 1 u^2T = (^ )0 1 4 1 d d0 1 d d0 T uT xT T uT uT 1 3 2 1=2 1=2 1=2 d0 1=2 d d d0 h i W (1)W (1) ! W (1)J (1) 11 2:1 11 1 1 12c 1 11 1=2 d0 1=2 5 ) 1 4 1=2 1=2 11 c ! 2:1 2 d d0 d ! 2:1 (J12c (1)) ! 2:1 11 J12c (1)W1 (1) 2 3 1=2 1=2 d ! 2:1 11 c d d 4 5 ! 2:1 d0 c Ac (1) c 1 d0 where d0 c = [ c ; 1], and the other terms as de…ned in the text and in Lemma A.2. For the case where px = 1 and py = 0, the result follows using similar modi…cations as in part (a) 0 d 0 d = [W1(m with the limit result de…ned with W1d = W1R 1) ; r] and J12c = J12c . For part (3), the proof follows using arguments in Phillips and Ouliaris (1990), see also Corollary A.1 of Pesavento (2004). For part (4), following the arguments of Phillips and Ouliaris (1990), s2 is a zero frequency )0 ; 1]0 vt . Since spectral density estimate based on u^t = [ ( ^ )0 ; 1]0 zdt = [ ( ^ 1=2 1=2 1=2 0 0 d0 1=2 [ (^ )0 ; 1]0 ) [ d0 11 ; 1] . 11 ; 1] , we can condition on the value of [ c ! 2:1 c ! 2:1 1=2 1=2 0 Now, given the assumptions on vt , [ d0 11 ; 1] vt is also a linear process satisfying c ! 2:1 the conditions of Berk (1974). Hence, s2 is a consistent estimate of (2 times) the spectral 1=2 1=2 0 density function at frequency zero of [ d0 11 ; 1] vt given by: c ! 2:1 i i0 h h 1=2 1=2 d0 1=2 d0 1=2 1 1 11 11 c ! 2:1 c ! 2:1 2 32 3 1=2 1=2 d i h ! ! 11 12 2:1 11 c 1=2 d0 1=2 54 5 = 1 4 11 c ! 2:1 ! 21 ! 22 1 d0 1=2 11 + ! 21 )( c 0 d0 d0 d d c c c c 1=2 = ( ! 2:1 = ! 2:1 f 1=2 d 11 c) 1 ! 2:1 ! 22 g 1=2 ! 2:1 + 1=2 + ( ! 2:1 d0 c 1=2 11 ! 12 + ! 22 ) after some algebra. Note that ! 2:11 ! 22 = ! 22 [! 22 = ! 22 f! 22 [1 ! 21 R2 ]g Therefore, collecting terms, we have: 2 h i Im d0 s2 ) ! 2:1 1 4 0 c 1 ! 21 1 11 ! 12 ]g 1 ! 22 1 1 = ! 22 [ ]= =1+ 2 ! 22 (1 R ) 1 R2 1 1 11 ! 12 ] 1+ = ! 22 f! 22 [1 0 A-7 32 54 d c 1 3 5 ! 2:1 d d0 c D c: 0 For the case where px = 1 and py = 0, the arguments are similar with appropriate modi…cations. Proof of Theorem 2. Most of proofs of the limiting distributions of the tests follow from Lemma A.2 and results in Phillips and Ouliaris (1990) and Pesavento (2004). We provide a brief outline. We have P c2 T 2 Tt=1 u^2t 1 cT 1 u^2T GLS M PT; = : s2 P 2 d d d d 1 2 From Lemma A.2, c2 T 2 Tt=1 u^2t 1 ) c2 ! 2:1 d0 u^T ) c! 2:1 d0 c Ac c , cT c Ac (1) c and s ) d GLS . The limit dis! 2:1 d0 c D c , from which we readily obtain the limit distribution of M PT; GLS GLS de…ned by tribution of M PT; is obtained similarly. For the test M Zb M ZbGLS = T 2T 1 2 u^T s2 ; PT 2 2 u ^ t t=1 P d d the limit distribution is obtained using the facts that 2T 2 Tt=1 u^2t 1 ) 2! 2:1 d0 c Ac c , d 1 2 d0 d d 2 d0 GLS T u^P limit distribution of M SB = T ) ! 2:1 c Ac (1) c and s ) ! 2:1 c D c . Similarly, PT the T 2 d0 d d 2 2 2 2 2 ^t 1 ) ! 2:1 c Ac c and s ) ^t =s ], follows using the facts that T [T t=1 u t=1 u d d0 GLS GLS GLS ! 2:1 c D c . The result for M Zt^ = M SB M Zb then follows automatically. In the GLS GLS GLS and Ztb , the results follow using arguments similar cases of the statistics ADF , Zb to those of Phillips and Ouliaris (1990) and Corollary A.1 of Pesavento (2004). Note that for the case px = 1 and py = 0, similar results are obtained using the appropriate modi…cations. Proof of Theorem 3. The proofs are extensions of the OLS-based versions analyzed by Pesavento (2004) using the results of Lemma A.2. A-8