Bootstrap Procedures for Detecting Multiple Persistence Shifts Chapter 2

Bootstrap Procedures for Detecting Multiple Persistence Shifts in Heteroskedastic Time Series: Supplement Supplement A: Proofs of the Theoretical Results P For a (d 1) vector v, kvk = ( di=1 vi2 )1=2 denotes the standard Euclidean norm while for a random variable v; kvkq = (E(jvjq )1=q denotes the Lq (q 1) norm. For a matrix p B, kBk denotes the Frobenius norm, i.e., kBk = tr(B 0 B) and MB = I PB ; PB = B(B 0 B) 1 B 0 . Let P denote the bootstrap probability measure and E the expectation with respect to P . De…ne the following quantities: (i) V (r) = diag(g 2 (r)Ip ; g(r)); (ii) DT = diag(aT 2 T 1 ; aT 1 T 1=2 ); (iii) For i = 1; :::; k + 1, Zi = (zTi 1 +1 ; :::; zTi )0 where zt = (yt 1 ; 1)0 for t = Ti 1 + 1; :::; Ti ; Z = (z1 ; :::; zT )0 , Y 1 = (y0 ; :::; yT 1 ), (T 1) = (1; :::; 1)0 ; (iv) zi = PT P i PTi 1 1 z , z = T (Ti Ti 1 ) 1 Tt=T z and z = (T T ) t 1 t i; 1 i i 1 t=1 zt , z 1 = t=T +1 +1 i 1 i 1 P T T 1 t=1 zt 1 . As a matter of notation, we will use C = C[0; 1] to denote the space of continuous functions on [0; 1] and D the space of right continuous with left limit processes on p w [0; 1], ‘!’to denote convergence in probability, ‘!’to denote weak convergence in the space w D endowed with the Skorohod metric, ‘!p ’to denote weak convergence in probability under the bootstrap measure (Giné and Zinn, 1990), and b:c to denote the integer part of its argument. Further, B1 (:) and B2 (:) denote standard independent Brownian motions on [0; 1] and B(:) = [B1 (:); B2 (:)]0 . For any stochastic process Z(:) de…ned over [0; 1], Z (i) (:) denotes R i 1 Z(:) demeaned over [ i 1 ; i ], i.e., Z (i) (r) = Z(r) ( i Z, r 2 [ i 1 ; i ]: Finally, i 1) i 1 Rb Rb for ease of presentation, all integrals of the form a f (r)dr are expressed as a f . We …rst state two lemmas that will be useful in developing the proofs of the main results. A-1 Lemma A.1 [Xu, 2008] Suppose fyt g is generated by the AR(p) model yt = + Pp j=1 j (yt j ) + et Pp j where all roots of (L) = 1 j=1 j L are outside the unit circle and fet g satis…es Assumptions A(3-4). Let yet j = yt j ; y p;t = (e yt 1 ; :::; yet p )0 and xt = (y 0 p;t ; 1)0 . Also, de…ne the [(p + 1) (p + 1)] matrix T = diag(T 1=2 ; :::; T 1=2 ; T 1=2 aT 1 ): Then P1 (a) y p;t = be with bj = ( j 1 ; :::; j p ) if j 1; j = 0 if j < 0; where P1 j=1 jj t j P1 1 (L) = j=0 j L ; 0 = 1; j < 1. j=0 j 1 0 R 2 g 0(p 1) P P p A and = 1 bj b0j . (b) aT 2 T 1 ( Tt=1 xt x0t ) T 1 ! Q where Q = @ j=1 0(1 p) 1 P w R (c) aT 2 T 1 Tt=1 xt et ! V dBp+1 ; where Bp+1 = (Bp0 ; B1 )0 with Bp is a p-vector Brownian motion with covariance matrix and B1 is a standard Brownian motion independent of Bp . Lemma A.2 Suppose fyt g is generated by the AR(p) model with yt = y t 1 + Pp 1 j=1 j yt j = 1: + et where f j g satis…es Assumption A(2) and fet g satis…es Assumptions A(3-4). Let e = (e1 ; :::; eT )0 , vt = yt , wt = ( yt 1 ; :::; yt p+1 )0 , W = (w1 ; :::; wT )0 , Wj = (wTj 1 +1 ; :::; wTj )0 (j = 1; :::; k + 1) and = ( 1 ; :::; p 1 )0 . Then P rc w R r ge(1)Bg;1 (r); (a) aT 1 T 1=2 bT t=1 et ! 0 gdB1 Rr P PbT rc w 1 1=2 (b) aP d(1)e g (1)Bg;1 (r), if d(1) 6= 0, with vt = 1 T T j=0 dj et j t=1 vt ! d(1) 0 gdB1 1 and j=0 j jdj j < 1; P rc w 2 (r) ge(r)2 ]; g (1)2 Bg;1 (c) aT 2 T 1 bT t=1 yt 1 et ! (1=2)d(1)[e (d) (aT 2 T 1 W 0W ) 1 = Op (1); 0 (e) kDT Z2i W2i k = Op (1); (f ) (g) aT 2 T 1=2 W 0 e = Op (1); h i P 0 0 1 0 aT 2 T 1 W 0 W aT 2 T 1 k=2 W Z (Z Z ) Z W 2i 2i 2i 2i 2i 2i i=1 1 = Op (1). Proof of Lemma A.2: (a) The result follows from Lemma 1 in Cavaliere and Taylor (2009). P P1 A2, yt = vt = 1 j=0 dj et j with j=0 j jdj j < 1, where d(L) = P1(b) Byj Assumption 1 j=0 dj L = (L) . Then, with the additional restriction d(1) 6= 0; the sequence fvt g satis…es A-2 P rc Assumption 1’in Cavaliere and Taylor (2009) and hence by their Theorem 3, aT 1 T 1=2 bT t=1 vt R w r ! d(1) 0 gdB1 . P rc (c) Note that from the Beveridge-Nelson (1981) decomposition, we have T 1 bT t=1 yt 1 et = PbT rc Pt 1 1 d(1)T t=2 f j=1 ej get + op (1). Next, using the fact that T 1 PbT rc Pt 1 t=2 f j=1 ej get = (1=2)[(T the result follows from (a) since aT 2 T 1 1=2 PbT rc t=1 PbT rc t=1 w e2t ! et )2 Rr 1 T t=1 g(s)2 0 PbT rc e2t ] + op (1) ge(r)2 . PT 1 yt j yt j 0 , j; j 0 2 (d) The entries in the matrix aT 2 T 1 W 0 W are of the form T t=1 f1; :::; p 1g. When = 1, f yt g isPan AR(p 1) process with all roots outside the unit circle. Then by Lemma A.1(b), T 1 Tt=1 yt j yt j 0 = Op (1) and the result follows. (e) We have aT 1 T aT 1 T 1=2 1=2 ybT rc = Op (1) uniformly in r 2 [0; 1]. For a …xed j 2 f1; :::; p PT2i t=T2i yt 1 +1 j w = aT 1 T ! d(1)e g (1)fBg;1 ( 1=2 yT2i 2i ) Bg;1 ( PT2i 1 +1 aT 1 T j 2i 1 )g 1=2 yT2i 1 +1 j 1=2 yt j ) 1g, = Op (1): Further, aT 2 T 1 PT2i t=T2i ( 2i 1 +1 w yt 1 yt j = t=T2i 2 ! (1=2)fd(1)2 ge(1)2 [Bg;1 ( 2i 1 ) 0 g where j = E( yt yt j ) = ge(1) Op (1) and the result follows. + ( 2i P1 2 s=0 2i 1 ) (aT 1 T 2 Bg;1 ( 2i ) 0 + 1=2 1 yt 1 )(aT 1 T 2i 1 )] + ::: + j 1 = Op (1) 0 ds ds+j . Hence, all entries in the matrix DT Z2i W2i are (f) The result follows by applying Lemma A.1(c) to the sequence f yt g. (g) First, observe that aT 2 T 1 W 0 W = Op (1) by Lemma A.1(b). Next, we have 2 3 R 2i R 2i 2 2 2 d(1) ge(1) 2i 1 Bg;1 d(1)e g (1) 2i 1 Bg;1 w 0 5 W1;i a2T DT Z2i Z2i DT ! 4 R 2i d(1)e g (1) 2i 1 Bg;1 2i 2i 1 0 Denote the limit of DT Z2i W2i by W2;i . Thus combining the results in (e), P 0 2 0 1 0 T 1 k=2 [DT Z2i W2i ] i=1 [W2i Z2i DT ] (aT DT Z2i Z2i DT ) P w 1 0 1 ! T 1 k=2 )Op (1) = op (1) i=1 W2;i W1;i W2;i = Op (T and the result follows. N Proof of Theorem 1: We prove the result for Model 1a and k even. The proofs for the e and E^ be the vector of residuals in the i-th other tests are very similar and omitted. Let E i i A-3 (1) (1) regime under H0 and Ha;k , respectively, for i = 1; :::; k + 1. Denote ^ 2i = (^ 2i 1; c^2i )0 , i = 1; :::; k=2; where ^ 2i and c^2i are the OLS estimates obtained from regime 2i. Then we have ei = E E^2i = = (W 0 W ) for i = 1; :::; k + 1 Wi ; W2i ^ Y2i E^2i+1 = where Yi W2i+1 ^ ; Y2i+1 (A.1) for i = 1; :::; k=2 Z2i ^ 2i ; for i = 0; :::; k=2 (1) W 0 e under H0 . Further, ^ and ^ 2i satisfy the …rst order conditions 1 0 ^ Z2i E2i = 0; for i = 1; :::; k=2 ; (A.2) Pk=2 Pk=2 ^ ^ (A.3) i=1 W2i E2i + i=0 W2i+1 E2i+1 = 0: Pk=2 0 (1) Under H0 , from (A:3), we have ^ = (W 0 W ) 1 (W 0 e i=1 W2i Z2i ^ 2i ). Next, from (A:2), h i 0 0 ^ ) + DT Z 0 E2i aT 2 DT 1 ^ 2i = (a2T DT Z2i Z2i DT ) 1 DT Z2i W2i ( (A.4) 2i for i = 1; :::; k=2, where E2i = (eT2i ^ = [W 0 W Pk=2 1 +1 ; :::; eT2i )0 . Solving for ( ^ 0 0 W2i0 Z2i (Z2i Z2i ) 1 Z2i W2i ] 1 [W 0 e ) we obtain Pk=2 0 0 W2i0 Z2i (Z2i Z2i ) 1 Z2i E2i ] (A.5) 0 2 0 so that using Lemma A.2, and noting that the limits of W2i Z2i DT , aT DT Z2i Z2i DT and 0 DT Z2i E2i are Op (1) uniformly in i, jj ^ i=1 [W 0 W jj h Pk=2 i=1 kW 0 ek + = [Op (aT 2 T Pk=2 i=1 1 i=1 0 0 W2i0 Z2i (Z2i Z2i ) 1 Z2i W2i ] 1 0 kW2i0 Z2i DT k (DT Z2i Z2i DT ) 1 0 kDT Z2i E2i k )][Op (a2T T 1=2 ) + (k=2)fOp (1)Op (a2T )Op (1)g] = Op (T i 1=2 ) Also, ^ )jj 0 0 jj(DT Z2i Z2i DT ) 1 DT Z2i W2i ( 0 0 jj(DT Z2i Z2i DT ) 1 jjjjDT Z2i W2i jjjj( = Op (a2T )Op (1)Op (T 1=2 ) = Op (a2T T ^ )jj 1=2 ) (A.6) Using (A:6) in (A:4), we have 0 0 aT 2 DT 1 ^ 2i = (a2T DT Z2i Z2i DT ) 1 DT Z2i E2i + op (1): Next, ^ = (W 0 W ) jj ^ jj 1 Pk=2 i=1 (A.7) fW2i0 Z2i ^ 2i g so that (W 0 W ) 1 = Op (aT 2 T 1 Pk=2 i=1 kW2i0 Z2i DT k DT 1 ^ 2i )(k=2)fOp (1)Op (a2T )g = Op (T A-4 1 ): e2i = E^2i + Z2i ^ 2i + W2i ( ^ ) and for We can write, from (A:1), for i = 1; :::; k=2, E ^ ^ e ). Thus the numerator of the F statistic can be i = 0; :::; k=2, E2i+1 = E2i+1 + W2i+1 ( written as Pk=2 e 0 e P (1) (1) 0 e0 e SSR0 SSR1a;k = E^2i0 E^2i g + k=2 E^2i+1 E^2i+1 g (A.8) i=1 fE2i E2i i=0 fE2i+1 E2i+1 Pk=2 1 1 0 0 = i=1 (DT ^ 2i ) (DT Z2i Z2i DT )DT ^ 2i P +( ^ )0 k=2 (W 0 Z2i DT )(D 1 ^ 2i )+( ^ )0 (W 0 W )( ^ ) 2i i=1 T where jj( ^ )0 Pk=2 jj ^ 1 0 i=1 (W2i Z2i DT )(DT ^ 2i )jj 1 = Op (T jj Pk=2 i=1 k(W2i0 Z2i DT )k (DT 1 ^ 2i ) )(k=2)fOp (1)Op (a2T )g = Op (a2T T 1 ) and jj( ^ )0 (W 0 W )( ^ jj ^ )jj 1 = Op (T jj kW 0 W k jj ^ )Op (a2T T )Op (T 1 jj ) = Op (a2T T 1 ) Then, using (A:7) in (A:8), we have (1) aT 2 (SSR0 (1) Pk=2 0 0 0 E2i Z2i DT (a2T DT Z2i Z2i DT ) 1 DT Z2i E2i + op (1) P T2i 2 k=2 (yt 1 y2i; 1 )et g2 P faT T 1 t=T 2i 1 +1 (A.9) = [ P 2i aT 2 T 2 Tt=T (yt 1 y2i; 1 )2 i=1 2i 1 +1 P 2i T faT 1 T 1=2 Tt=T e g2 ] + 2i 1 +1 t T2i T2i 1 SSR1a;k ) = i=1 Using Lemma A.2(a,c) in (A:9), we have 2 n (1) aT 2 (SSR0 (1) w SSR1a;k ) ! ge(1)2 Finally, noting that aT 2 (T k=2 P6 6 4 i=1 (2i) Bg;1 ( + (1) p k) 1 SSR1a;k ! Proof of Theorem 2: We can write (0) SSR0 2i ) R1 0 (0) o2 n (2i) Bg;1 ( 4 1 2i g2 2i 1 R 2i 2i 1 o2 2 g e(1) 2 fg e( h i2 (2i) Bg;1 (r) dr 2i 1 ) fBg;1 ( 2i ) Bg;1 ( 2i ) 2 g e( 2i 1 ) 2i 1 )g 2 2 g 3 7 7 5 ge(1)2 , the result follows.N SSR1;k = DR (1; k + 1) Pk+1 i=1 DU (i; i) where DU (i; j) [DU (i; j), resp.,] is the sum of squared residuals from the unrestricted (restricted, resp.,) regression using data from segments i to j (inclusively). Let Y( 1)1;i , Z1;i , W1;i , and E1;i denote the vectors or matrices containing elements of Y 1 , Z, W and e, respectively, A-5 belonging to the partition from segment 1 to i (inclusively), for i = 1; :::; k + 1. Further, 0 0 0 0 0 E1;i for W1;i and Mj = W1;i W1;i , Li = W1;i Z1;i , Ki = Z1;i E1;i , Hi = Z1;i de…ne Si = Z1;i 0 1 0 0 1 0 i = 1; :::; k+1. Finally, let AT = (W MZ W ) W MZ e and AT = (W MZ W ) W MZ e, where Z = diag(Z1 ; :::; Zk+1 ). Then, from Bai and Perron (1998, eq. (39) and (41), pg. 73-74), P (0) (0) SSR0 SSR1;k = ki=1 FT;i + DR (1; 1) DU (1; 1) where 1 0 Si+1 Hi+1 Si+1 + Si0 Hi 1 Si + (Si+1 FT;i = + + 1 0 Hi+1 2Si+1 Ki+1 AT 2(Mi+1 Mi )0 (AT Si )[Hi+1 2Si0 Hi 1 Ki AT 2(Si+1 0 AT ) + (AT AT ) (Li+1 Hi ] 1 (Si+1 0 Si ) [Hi+1 Li )(AT Si ) Hi ] 1 (Ki+1 Ki )AT AT ) (A.10) = T1 + T2 + T3 We now analyze each of the terms T 1-T 3 in (A:10). For T 1: 1 0 Si+1 Hi+1 Si+1 + Si0 Hi 1 Si + (Si+1 Si )0 [Hi+1 Hi ] 1 (Si+1 Si ) PTi+1 2 P i+1 PTi+1 1 2 1 1=2 et 1 g 1 fT 1=2 Tt=1 yet 1 et g2 ] = i+1 [fT t=1 y t=1 et g + fT P P P i i i + i 1 [fT 1=2 Tt=1 et g2 + fT 1 Tt=1 yet2 1 g 1 fT 1=2 Tt=1 yet 1 et g2 ] PTi+1 PTi+1 PTi+1 1 1=2 2 1 +( i+1 et2 1 g 1 fT 1=2 t=T ye e g2 ] i ) [fT t=Ti +1 et g + fT t=Ti +1 y i +1 t 1 t T1 = +op (a2T ) using Lemma A.1(b) where yet aT 2 T 1 w ! ge(1)2 2 1 2 i+1 Bg;1 ( i+1 ) For T 2: ge(1)2 f i Bg;1 ( i 1=2 T 2 = 2(T i+1 ) i+1 ( Si+1 )0 (T 1=2 2[T 1 + . Then, from Lemma A.1, we have j 1 2 Bg;1 ( i) + ( i+1 i+1 ) Bg;1 ( i )]2 2 2 fe g ( i )2 Bg;2 ( i+1 ) ge( i+1 )2 Bg;2 ( i )g i+1 Bg;1 ( i )g + ge( i )2 ge( i+1 )2 fe g ( i+1 )2 ge( i )2 g i+1 i) Hi+1 ) 1 T Si )]0 [T (Si+1 i i) 1 [Bg;1 ( 3 1 2 2 2 1 2 fe g ( i+1 )g Bg;2 ( i+1 ) + fe g ( i )g Bg;2 ( i )+ 5 1 fe g 2 ( i+1 ) ge2 ( i )g [Bg;2 ( i+1 ) Bg;2 ( i )]2 +e g (1)2 4 " = yt j 1 1 Ki+1 T 1=2 AT Hi )] 1 T (Hi+1 2(T 1 1=2 (Ki+1 Si )0 (T 1 Hi ) 1 T 1 # : Ki T 1=2 AT Ki )T 1=2 AT : 0 De…ne e p 1 = ( 11 12 ; 12 13 ; :::; 1(p 1) 1p ) , where ij is the (i; j) element of de…ned in Lemma A.1. Then, using Lemma A.1(a)-(c), we have aT 2 (T 1=2 Si+1 )0 (T aT 2 (T 1=2 Si )0 (T 1 aT 2 [T 1=2 (Si+1 Si )]0 [T w ! (1= 1 Hi+1 ) 1 T Hi ) 1 T 1=2 [Bg;2 ( i+1 ) 11 ) 1 w Ki+1 ! (1= w 1=2 Bg;2 ( i+1 ) e 0p 1 ; 11 ) 1=2 Bg;2 ( i ) e 0p 1 ; 11 ) 1 (Hi+1 Hi )] 1 T 1 (Ki+1 Ki ) Bg;2 ( i )] e 0p 1 : 1 Ki ! (1= A-6 Using Lemma A.1, it can further be shown that p T 1 T 1=2 1e e 0 e2 (1); 11 p 1 p 1 g W 0 PZ W ! w 1=2 e 11 ) p 1 Bg;2 (1); W 0 PZ e ! (1= p aT 2 T 3 = aT 2 [2(Mi+1 aT 2 FT;i 1 p W 0 PZ W ! 1=2 T W 0 PZ e AT ! 0. Hence, aT 2 T 2 = op (1). For T 3: so that AT since AT T Mi )0 (AT 1e e 0 e2 (1); 11 p 1 p 1 g w ! (1= 11 )1=2 e p 1 Bg;2 (1); AT )0 (Li+1 AT ) + (AT Li )(AT p AT )] ! 0 p AT ! 0. From (A:10), we then obtain # " 2 2 2 2 fe g ( ) B ( ) g e ( ) B ( )g f B ( ) B ( )g w i g;2 i+1 i+1 g;2 i i g;1 i+1 i+1 g;1 i + ! ge(1)2 ge( i )2 ge( i+1 )2 fe g ( i+1 )2 ge( i )2 g i i+1 ( i+1 i) The result follows noting that [T (0) p 2(k + 1)] 1 aT 2 SSR1;k ! ge(1)2 .N Proof of Theorem 3(a): We will prove the theorem for the bootstrap test based on F1a ( ; k) for k even. The bootstrap statistic is given by F1a ( ; k) = (T k)(SSR0 ;(1) ;(1) ;(1) SSR1a;k )=[kSSR1a;k ] where SSR0 ;(1) ;(1) SSR1a;k = = (1) PT (1) (1) yt 1 )2 t=1 (yt Pk=2 PT2i (1) t=T2i 1 +1 (yt i=1 P PT2i+1 (1) + k=2 t=T2i +1 (yt i=0 (A.11) (1) y2i (1) (1) ^ 2i (yt 1 (1) y2i; 1 ))2 (1) yt 1 )2 (1) In (A:12), ^ 2i denotes the slope estimate from an OLS regression of yt (1) 1 yt (t = T2i (1) 1 + 1; :::; T2i ; i = 1; :::; k=2). Since yt ;(1) (A.12) (1) 1 = yt (1) + et for t on a constant and T; we have ;(1) aT 2 (SSR0 SSR1a;k ) " # PT2i (1) (1) (1) 2 k=2 2 1 X [a T f(y y )e g] t t 1 2i; 1 T t=T2i 1 +1 (1) (A.13) = (T2i T2i 1 )[aT 1 e2i ]2 + PT2i (1) (1) 2 2 2 a T (y y ) 2i; 1 i=1 T t=T2i 1 +1 t 1 (1) Next, we establish an invariance principle for the sequence faT 1 et ; t = 1; :::; T g. To this end, (1) (1) let Ft be the -…eld generated by fvs ; s tg. Since et = et vt , faT 1 et ; Ft g is a martingale di¤erence array. Further, uniformly over r 2 [0; 1], aT 2 T 1 PbT rc t=1 (1) [et ]2 aT 2 T A-7 1 PbT rc t=1 p e2t ! 0 since E faT 2 T 1 PbT rc t=1 (1) ([et ]2 P rc 2 2 1 g2 e2t )g2 = E faT 2 T 1 bT t=1 et (vt P rc 1 2 4 CT 2 bT t=1 (aT et ) = op (1); (1) (where C is a positive constant), using the fact that under H0 , aT 1 et = aT 1 et + aT 1 wt0 ( P rc 2 p R r 2 ) = aT 1 et + Op (T 1=2 ). Also, aT 2 T 1 bT t=1t et ! 0 g uniformly over r 2 [0; 1] (by Lemma 2 in CavaliereP and Taylor, 2008(a)). Then, applying Theorem 2.1 in Hansen (1992) with ST (:) = T 1=2 [T:] t=1 vt , we obtain R P rc 1 (1) w Rr 1 r T 1=2 bT a e = a e dS (s) ! e(1)Bg;1 (r): (A.14) T p 0 g(s)dB1 (s) = g bT sc t T T t=1 0 Using (A:14), we have 2 aT T 2 T2i X t=T2i aT 2 T 1 1 +1 T2i X t=T2i (1) (yt 1 f(yt ( 2i 1 ) 2i ge(1) 2 Z 2i (2i) [Bg;1 (s)]2 2i 1 w 2 g (1)2 4 y2i; 1 )et g !p (1=2)e (1) (1) (1) 1 1 +1 (1) w aT 1 T 1=2 e2i !p w (1) y2i; 1 )2 !p 1 ge(1)[Bg;1 ( 2i ) Bg;1 ( (2i) fBg;1 ( 2i )g2 2 ge(1) fe g( (2i) fBg;1 ( 2i 1 )g2 2 2i ) 2i 1 )] ;(1) p ge( 2i 2 1) g 3 5 (A.15) w Substituting (A:15) in (A:13) and noting that (T k) 1 SSR1a;k ! ge(1)2 , F1a ( ; k) !p 0 0 ( ; k) is the weak limit of F1a ( ; k) as stated in Theorem 1. The rest of ( ; k), where F1a F1a the proof follows from the proof of Theorem 5 in Hansen (2000). The bootstrap BP test for k breaks is given by G1 (k) = [T 2(k + 1)](SSR0 ;(0) ;(0) ;(0) SSR1;k )=[kSSR1;k ] where SSR0 ;(0) ;(0) SSR1;k = = PT (0) (0) e(0) e (0) (et 1 t=1 (et Pk+1 PTi (0) (0) ei i=1 t=Ti 1 +1 (et (0) (0) e 1 ))2 (0) (0) ^ i (et 1 (A.16) (0) ei; 1 ))2 (A.17) In (A:16) (resp., A:17), e (0) (resp., ^ i ) denotes the slope estimate from an OLS regression (0) (0) (0) of et on a constant and et 1 (t = 1; :::; T ) (resp., et 1 (t = Ti 1 + 1; :::; Ti )). After some algebra, we have P (0) (0) (0) [aT 2 T 1=2 Tt=1 f(et 1 ei; 1 )et g]2 ;(0) ;(0) 1 (0) 2 2 aT (SSR0 SSR1;k ) = T aT e P (0) (0) aT 2 T 1 Tt=1 (et 1 ei; 1 )2 " # PTi (0) (0) (0) 2 2 k+1 1=2 X [a T f(e e )e g] t 1 i; 1 t T t=Ti 1 +1 (0) + (Ti Ti 1 )[aT 1 ei ]2 + : (A.18) PTi (0) (0) 2 2 1 a T (e e ) i=1 i; 1 T t=Ti 1 +1 t 1 A-8 (0) Next, we establish an invariance principle for the sequence faT 1 et ; t = 1; :::; T g. In particular, we will show that for r 2 [0; 1], P rc 1 (0) w R r T 1=2 bT (A.19) t=1 aT et !p 0 g(s)dB1 (s): (0) (0) To this end, let Ft be the -…eld generated by fvs ; s tg. Since et = eet vt , f aT 1 et ; Ft g is a martingale di¤erence array. Further, uniformly over r 2 [0; 1], aT 2 T 1 since E faT 2 T 1 PbT rc t=1 PbT rc t=1 (0) ([et ]2 (0) [et ]2 aT 2 T 1 PbT rc t=1 p ee2t ! 0 P rc 2 2 ee2t )g2 = E faT 2 T 1 bT et (vt 1 g2 t=1 e P rc 1 4 et ) = op (1) CT 2 bT t=1 (aT e P rc 2 p using the fact that aT 1 eet = aT 1 et +Op (T 1=2 ) [eq. (A.7) in Xu, 2008]. Also, aT 2 T 1 bT et ! t=1 e Rr 2 P[T:] 1=2 g . Then, again applying Theorem 2.1 in Hansen (1992) with ST (:) = T t=1 vt , we 0 have R P rc 1 (0) w Rr 1 r T 1=2 bT e e dS (s) ! T p 0 g(s)dB1 (s): bT sc t=1 aT et = aT 0 (0) (0) Noting that faT 2 et et 1 ; Ft g is a martingale di¤erence array, we can show, using similar arguments as above, that T 1=2 bT rc X t=1 (0) (0) w aT 2 et et 1 !p Z r g 2 (s)dB2 (s) (A.20) 0 ;(0) p for r 2 [0; 1], where B2 (:) is independent of B1 (:). Finally, since aT 2 T 1 SSR1;k ! ge(1)2 , w G1 ( ; k) !p G01 ( ; k) using (A:19) and (A:20) in (A:18), where G01 ( ; k) is the weak limit of G1 ( ; k) as de…ned in Theorem 2. Hence, following the proof of Theorem 5 in Hansen w w (2000), pk;G1 ! U [0; 1], pU Dmax ! U [0; 1].N Proof of Theorem 3(b): The proof of this result follows directly from part (a) and is hence omitted. p p (1) Proof of Theorem 3(c): We will prove pm;W1 ! 0 and pm;G1 ! 0 under Ha;m with m even. p p (1) Consequently, pW max ! 0 and pU Dmax ! 0. The proofs for the alternatives Hb;m and (0) H1;m can be established using similar arguments. The proof proceeds in two steps: (i) we …rst show that the bootstrap counterparts F1a (m), F1b (m) and G1 (m) of F1a (m), F1b (m) and (1) G1 (m), respectively, are each Op (1) under Ha;m ; (ii) F1a (m) (hence W1 (m)) and G1 (m) both diverge with T . For (i), …rst note that for s 2 [0; 1], P aT 1 ebT sc = aT 1 ebT sc + aT 1 m=2 1)hbT sc 1 I(bT sc 2 [T2i0 1 + 1; T2i0 ]) + Op (T 1=2 ) i=1 ( 2i A-9 so that aT 2 T 1 PbT rc t=1 e2t = aT 2 T 1 PbT rc e2t +aT 2 T t=1 PbT rc Pm=2 t=1 i=1 ( 1)2 h2t 1 I(t 2 [T2i0 1 +1; T2i0 ])+op (1) 2i P (2i) R r g 2 + m=2 1)2 ( 11 ) 0 g(s)2 I(s 2 [ 02i 1 ; 02i ])ds i=1 ( 2i PbT sc p (2i) R s g 2 if s 2 [ 02i 1 ; 02i ] and where aT 2 T 1 t=T 0 +1 h2t 1 ! 11 0 p ! Rr 1 2i 1 2i 1 of (2i) with (2i) de…ned analogously to Therefore, we have for r 2 [0; 1], aT 1 T 1=2 PbT rc (1) et = aT 1 t=1 Rr 0 (A.21) V (r) 0 (2i) 11 is the (1; 1) element in Lemma A.1 but now speci…c to regime 2i. Rr w ebT sc dST (s) !p 0 g1 (s)dB1 (s) (A.22) Bg;1 (r) Rr P (2i) 0 0 1=2 2 ])ds] . Note that g (s)2 = ; where g1 (s) = g(s)[1 + m=2 ( 1) ( )I(s 2 [ 2i 11 2i 2i 1 i=1 0 1 V (r). Then, the results stated in (A:15) all hold with Bg;1 (:) replaced by Bg;1 (:). Further, R1 ;(1) w (T k) 1 SSR1a;k !p 0 g1 (s)2 = V (1). Thus, F1a (m) = Op (1). Entirely analogous arguments can be used to establish F1b (m) = Op (1). Next, we show that G1 (m) is stochastically (1) bounded under Ha;m . First, note that we can write where T (e eet = yt e (yt y (1) y 1) 1 (wt w)0 e 1) = Op (1) since Ha;m involves a mix of I(1) and I(0) regimes. Further, ey y 1 = y y (e 1 1=2 = Op (aT T 1)y ) 1 1 =T Op (aT T 1=2 (yT y0 ) (e 1=2 ) = Op (aT T 1)y 1 ): Thus, in an I(1) regime, i.e., t 2 [T2i + 1; :::; T2i+1 ], i = 0; :::; m=2, we have aT 1 eet = aT 1 et + aT 1 (1 1 = aT et + Op (T e )yt 1=2 1 + Op (T 1=2 ) = aT 1 et + Op (T 1 )Op (T 1=2 ) + Op (T 1 + 1; :::; T2i ], i = 1; :::; m=2, we have aT 1 eet = aT 1 et + ( 2i 1)aT 1 ht 1 1=2 + Op (T (A.24) ): Combining (A:23) and (A:24), we have for t = 1; :::; T , aT 1 eet = aT 1 et + so that for r 2 [0; 1], Pm=2 i=1 aT 2 T +aT 2 T 1 PbT rc Pm=2 t=1 i=1 ( ) (A.23) ): In an I(0) regime, i.e., t 2 [T2i 1=2 1 2i ( 2i 1)aT 1 ht 1 I(t 2 [T2i0 PbT rc 1 + 1; :::; T2i0 ]) P rc 2 e2t = aT 2 T 1 bT t=1 e t=1 et 1)2 h2t 1 I(t 2 [T2i0 1 + 1; T2i0 ]) A-10 p + op (1) ! V (r) R P rc (0) w 1 r ee dST (s) !p Bg;1 (r) where V (r) is de…ned in (A:21). Hence, aT 1 T 1=2 bT t=1 et = aT 0 bT sc ;(0) and the limits in (A:19) and (A:20) now hold with g(:) replaced by g1 (:). Also, aT 2 T 1 SSR1;k p ! V (1). Thus, G1 (m) = Op (1). To show step (ii), note that since 0 2 m and F1a (m) = sup 2 m F1a ( ; m), it is su¢ cient to show that F1a ( 0 ; m) = Op (T ): De…ne P P = ( Tt=1 wt wt0 ) 1 Tt=1 wt yt Then, ht 1 = yt e2i = 2i e2i ; t 2 [T2i0 1 2i 1 ; 1 i = 1; :::; m=2: + 1; T2i0 ]. We have 0 P PT2i+1 0 ) + et g2 yt wt0 )2 = m=2 0 +1 fwt ( i=0 t=T2i P PT2i0 f( 2i 1)ht 1 + wt0 ( + m=2 ) + et g2 0 i=1 t=T2i 1 +1 PT2i0 PT 2 Pm=2 2 ( 1) ( e + h2t 1 ) = 0 2i t i=1 t=1 t=T2i 1 +1 PT2i0 P 1)( t=T ht 1 wt ) +2( )0 m=2 0 i=1 ( 2i 2i 1 +1 P +( )0 (W 0 W )( ) + 2( )0 W 0 e + 2 m=2 i=1 ( 2i (1) SSR0 PT 1 + yT2i0 = t=1 ( Let Z (1) = diag(Ze1 ; :::; Zem+1 ), where Zei (1) (1) (1) the [(m + 1) 1] vector 1 = (0; PT2i0 P 1) t=T (W 0 W ) 1 m=2 0 i=1 ( 2i 2i 1 (1) SSR0 = PT 2 t=1 et + 1; 0; 4 h w, +1 t 1 t 2 1; ::; 0)0 . Noting that PT 2 t=1 et + aT 2 01 Z (1)0 MW Z (1) Next, we have (with ^ denoting the estimate of 0 (1) SSR1a;m = m=2 T2i+1 X X 0 +1 i=0 t=T2i ^ ) + et g2 + fwt0 ( m=2 X P = (W 0 W ) 1 [ m=2 i=1 ( = [Op (aT 2 T 1 2i 1 +1 ht 1 et ): ^ 2i ) = (W 0 W ) 1 W 0 e + 1 PT2i0 t 1 +1 0 t=T2i 1 +1 wt (yt (A.25) W 0 e] = a2T Op (1) = + Op (T 1=2 ): 2i 1 )][Op (a2T T 1=2 ) + Op (a2T T 1=2 )] = Op (T A-11 1=2 under the alternative), 8 0 T2i < ( X ^ 2i )(yt 2i : +(w w )0 ( 0 i=1 t=T2i Then, noting that ^ 0 t=T2i e0 W (W 0 W ) 1 W 0 e + 2 01 Z (1)0 e: Using (i) e0 W (W 0 W ) 1 W 0 e = a2T [aT 2 T 1=2 e0 W (aT 2 T 1 W 0 W ) 1 aT 2 T Op (a2T ); (ii) 01 Z (1)0 e = Op (a2T T 1=2 ), we have from (A:25) (1) PT2i0 is the …rst column of Zei = (hT2i0 1 ; :::; hT2i0 1 ) and 0 (1)0 MW Z (1) 1 1Z aT 2 SSR0 = aT 2 1)( 1 92 y2i; 1 ) = : ^) + e ; y2i; 1 ) + W 0 e] 1=2 ) t we can show, after some simpli…cations, (1) aT 2 SSR1a;m = aT 2 Combining (A:25) and (A:26), (1) PT 2 t=1 et (1) aT 2 (SSR0 (A.26) + Op (1): SSR1a;m ) = aT 2 01 Z (1)0 MW Z (1) 1 + Op (T 1=2 ): (A.27) Now, since regime 2i (i = 1; :::; m=2) is I(0), aT 2 01 Z (1)0 MW Z (1) 1 = aT 2 Op (a2T T ) = Op (T ). Since this term is positive and dominant in (A:27), F1a ( 0 ; m) diverges to positive in…nity at rate T . Entirely analogous arguments can be used to show the divergence of G1 (m) at rate T . The details are omitted.N Proof of Theorem 4: To prove this result, it is su¢ cient to show that lim P ( min fpi g < T !1 1 i l+1 (A.28) l+1 ) as the rest of the proof follows the same arguments as in the proof of Theorem 2 in Kejriwal (2019). First, note that P ( min fpi g < 1 i l+1 l+1 ) =1 P ( min fpi g l+1 ) 1 i l+1 l+1 i=1 = 1 P (pi l+1 ) h ;(i) l+1 P (fp1;W1 < i=1 1 = 1 ;(i) l+1 g \ fp1;G1 < i g) l+1 (A.29) where the second equality follows from the independence of the test statistics across segments ;(i) ;(i) and the third from the fact that pi = maxfp1;W1 ; p1;G1 g. Next, from Theorem 3(a,b), it follows that under the null hypothesis of l breaks, we have for any segment i 2 f1; :::; l + 1g, ;(i) P (fp1;W1 < ;(i) P (fp1;W1 < ;(i) ;(i) l+1 g \ fp1;G1 < l+1 g) P (fp1;W1 < l+1 g \ fp1;G1 < l+1 g) P (fp1;G1 < ;(i) ;(i) l+1 g) ! l+1 if i is I(1); l+1 g) ! l+1 if i is I(0): (A.30) Thus, using (A:30) in (A:29), we have lim P ( min fpi g < T !1 1 i l+1 l+1 ) 1 1 l+1 l+1 = which proves (A:28).N Proof of Theorem 5: The proof uses arguments very similar to those used in proving Theorems 3-4 and is hence omitted. A-12 Supplement B: Detailed Simulation Results Supplement B presents detailed simulation results to assess the …nite sample performance our procedures and to provide a comparison with existing approaches. Following Cavaliere and Taylor (2008b, CT henceforth), we consider the following three speci…cations for the volatility process: Model 1 (Single Volatility Break): t = 0 + ( 1 0:5T ); Model 0 )I(t 2 (Trending Volatility): t = 0p +( 1 0 )(t 1=(T 1)); Model 3 (Near-Integrated Stochastic Volatility): t = 0 exp(0:5 bt = T ), bt = (1 c=T )bt 1 + kt , kt i:i:d: N (0; 1), b0 = 0. We set 0 = 1 in all cases, := 0 = 1 2 f1; 1=3; 3g for Models 1 and 2, = 5 and c 2 f0; 10g for Model 3. Next, we generate an ARMA(1,1) process fzt gTt=1 as: zt = zt 1 +et et 1 , z0 = 0, et = t "t with "t i:i:d: N (0; 1). While our theory does not formally allow for moving average processes, we nevertheless include this case in our simulations as a robustness check. The wild bootstrap is implemented using a two point distribution, i.e., vt 2 f 1; 1g with equal probability. We also experimented with the standard normal distribution for vt but found that our tests perform noticeably better when using the two-point distribution relative to the normal. The level of trimming is set at = :15, T 2 f200; 400g and 1000 replications are used. We report results for the non-trending case only (those for the trending case are qualitatively similar). The lag length in the KPZ and BP procedures is selected using BIC with maximal value set to …ve.1 We report the performance of the tests H (k; ); k = 1; 2 and Hmax1 ( ) = maxfH (1; ); H (2; )g as well as their non-robust (homoskedasticity-based) asymptotic counterparts H(k; ); k = 1; 2 and Hmax1 ( ) = maxfH(1; ); H(2; )g. The ratio-based bootstrap tests of CT are designed to test the I(0) null hypothesis while our tests allow the process to be either I(1) or I(0) under the null. Further, while our tests are based on a …nite order autoregressive model, the CT tests are non-parametric and based on a mixing-type assumption for the innovations. Given that conducting a full and fair comparison of tests with di¤erent underlying models and null hypotheses is not possible, we did not include the CT tests in our analysis. Finite Sample Size. With no persistence change, fyt g is generated by DGP-0: yt = yt 1 + zt , y0 = 0. Table B-1 reports the empirical size of 5% asymptotic and bootstrap tests. The asymptotic tests are considerably oversized indicating their lack of robustness to nonstationary volatility, consistent with the large sample results in Section 4. In contrast, the proposed bootstrap tests are robust to I(1) or I(0) processes maintaining empirical size close to the nominal 5% level across the di¤erent volatility speci…cations. The same is generally true for the di¤erent error structures considered. The H tests are more accurately sized when 2 f:5; 1g but less so when = :7, since the tests are a hybrid of the KPZ and BP tests which each have size close to 5% when = 1 and = :5; respectively. When 1 We also experimented with larger values and found that they yielded comparable size but lower power, especially for the multiple break and sequential tests. The BIC was computed under the null model for each test. No qualitative di¤erences were observed if the lag selection was implemented under the alternative model. The modi…ed information proposed by Cavaliere et al. (2015) yielded no improvement in our setup. B-1 = :7, the BP tests are mildly over-sized while the KPZ tests diverge at rate T , hence the mild size distortions. Similar reasoning explains the slightly higher sizes for the hybrid tests when T increases, especially when = :7 and with MA(1) errors. Finite Sample Power. We consider DGPs with one and two breaks. The results are reported only for the case = = 0 and are brie‡y summarized for the other cases. The DGPs in the one break case are DGP-1: yt = y t 1 +z t if t [T 01 ], yt = y t 1 +z t otherwise; DGP-2: yt = y t 1 +z t if t [T 01 ], yt y [T 01 ] = (y t 1 y [T 01 ] ) + z t , otherwise; DGP-3: yt = 1 yt 1 +z t if t [T 01 ], yt = 2 yt 1 +z t , otherwise. For DGP-1 and DGP-2, 2 f:5; :7g and for DGP0 3, 1 ; 2 2 f:2; :9g. We de…ne = 2 1 . The break fraction is 1 = :5. Table B-2 reports the size-adjusted power of the tests. Several features are worth noting. First, the bootstrap tests are broadly comparable to their asymptotic counterparts, with neither class of tests uniformly dominating the other. Second, the proposed tests are generally powerful against the di¤erent persistence change alternatives; an exception being the case of an I(1) regime with high volatility (e.g., DGP-1 with = 1=3 and DGP-2 with = 3). This occurs since the process is dominated by the I(1) regime and the tests behave as with a stable I(1) process. In Table B-3, we show that power improves considerably if the I(0) regime is longer and/or the volatility shift is less prominent. Third, the proposed tests have substantial power against I(0)-preserving breaks (DGP-3), a feature that distinguishes these tests from most existing persistence change tests (e.g., the ratio-based tests) that only have trivial power against such breaks given that the latter tests are designed to detect changes between I(1) and I(0) regimes. Hence, using the KPZ tests to control size in the I(1) case causes little power loss relative to using the BP tests in isolation. Fourth, the hybrid tests are generally more powerful with deterministic (Models 1 and 2) rather than stochastic volatility (Model 3). With serially correlated errors, the results (Tables B-4 and B-5) are qualitatively similar except that power is lower for all the tests relative to = = 0. With two breaks, the DGPs are speci…ed as follows: For t [T 0 1] For [T 0 1] 0 1] +1 [T 0 2] y [T 0 1] [T yt = y t 1 +z t yt y [T DGP-5 yt = y t 1 +z t yt = y t 1 +z t yt y [T DGP-6 yt = yt = yt = 1 2 yt 1 +z t ) + zt For t DGP-4 1 yt 1 +z t = (y t t 0 2] +1 yt = y t 1 +z t 0 2] = (y t 1 y [T 0 2] ) + zt 1 yt 1 +z t The true break fractions are ( 01 ; 02 ) = (:3; :8). The results reported in Table B-6 are broadly similar to the one break case. The e¤ect of underspecifying the number of breaks can be seen by comparing the power of H (1) and H (2) for DGPs with two breaks, where the former is generally less powerful than the latter, though not in all cases. Interestingly, when the …rst regime has lower persistence relative to the second, H (1) has higher power than H (2) even though the former is based on a misspeci…ed model. However, Hmax1 has adequate power in most cases, often close to that of the most powerful test amongst H (1) and B-2 H (2). This highlights the practical advantage of using Hmax1 to detect the presence of at least one break. Tables B-7 and B-8 provide results with serial correlation showing the results to be qualitatively similar. Number of Breaks. We assess the adequacy of the sequential algorithm to estimate the number of breaks with data generated by DGP 0-6. We set A = 2 and = :10. The results are in Table B-9, with Pc and Po denoting the probability of correct selection and over-estimation, respectively. The procedure is generally reliable in the stable case (DGP-0) or with a single break (DGP 1-3). Its performance deteriorates in the two breaks case when the probability of underestimation can be non-negligible. For instance, in DGP-5 with an abrupt increase in volatility (model 1, = 1=3), the breakpoint estimate used to partition the sample is typically close to the second true breakpoint, so that the …rst segment includes an I(0) to I(1) break while the second is I(0). Whether a second break is selected depends on the power of the single break test in the …rst segment, which is relatively low (Table B-2). Similarly, with decreasing volatility, the breakpoint is estimated near the …rst true date so that selecting an additional break depends on the power of the single break test in the I(1)-I(0) case. Results reported in Table B-10 show a notable improvement as the magnitude of the volatility shift decreases and/or the volatility shift occurs near the second persistence break in the increasing volatility case and near the …rst break otherwise. Further re…nement of the algorithm is a potentially interesting topic for future research. Disentangling Trend and Persistence Shifts. Table B-11 reports the probabilities of selecting the true model based on the procedure proposed in Section 6 for disentangling mean shifts and persistence breaks. In addition to DGPs 1-3, a DGP with a pure mean shift, denoted DGP-0 ; is considered where the data are generated as: yt = ( )I(t > [T 01 ]) + yt 1 + et ; where et = t "t ; "t i:i:d: N (0; 1) and the same three speci…cations for t (Models 1-3) are used. The …ndings indicate that performance is generally satisfactory and improves as T increases. B-3 Notes to Tables 1. Table B-1 reports the empirical size of asymptotic and bootstrap tests with nominal size 5%. The tests H1 ; H2 ; Hmax are the tests of Kejriwal et al. (2013) and the tests H1 ; H2 ; Hmax are their bootstrap counterparts. 2. Table B-2 reports the size-adjusted power of 5% bootstrap tests in the single break case with breakpoint 01 = :5 and serially uncorrelated errors ( = = 0). 3. Table B-3 reports the size-adjusted power of 5% bootstrap tests under di¤erent change points and volatility intensity in the single break case with breakpoint 01 = :5 and serially uncorrelated errors ( = = 0) with abrupt volatility change (Model 1). 4. Table B-4 reports the size-adjusted power of 5% bootstrap tests in the single break case with breakpoint 01 = :5 and AR(1) errors ( = :5; = 0). 5. Table B-5 reports the size-adjusted power of 5% bootstrap tests in the single break case with breakpoint 01 = :5 and MA(1) errors ( = 0; = :5). 6. Table B-6 reports the size-adjusted power of 5% bootstrap tests in the two breaks case with breakpoint vector ( 01 ; 02 ) = (:3; :8) and serially uncorrelated errors ( = = 0). 7. Table B-7 reports the size-adjusted power of 5% bootstrap tests in the two breaks case with breakpoint vector ( 01 ; 02 ) = (:3; :8) and AR(1) errors ( = :5; = 0). 8. Table B-8 reports the size-adjusted power of 5% bootstrap tests in the two breaks case with breakpoint vector ( 01 ; 02 ) = (:3; :8) and MA(1) errors ( = 0; = :5). 9. Table B-9 reports the probabilities of selecting the true number of breaks from the sequential procedure with serially uncorrelated errors ( = = 0) and level = :10. 10. Table B-10 reports the probabilities of selecting the true number of breaks from the sequential procedure under di¤erent abrupt volatility break points and intensities in the two breaks case with breakpoint vector ( 01 ; 02 ) = (:3; :8); serially uncorrelated errors ( = = 0) and level = :10. 11. Table B-11 reports the probabilities of selecting the true model based on the procedure proposed in Section 6 for disentangling mean shifts and persistence breaks. B-4