Technical Note: Speci…cation Sensitivities in Right-Tailed Unit Root Testing for Financial Bubbles Shu-Ping Shi The Australian National University Peter C. B. Phillips Yale University, University of Auckland, University of Southampton & Singapore Management University Jun Yu Singapore Management University February 8, 2011

Before proving Proposition 4.1, we …rst introduce a few lemmas. 1 j=0

(L) "t =

j "t j ;

where

1 j j=0

< 1 and f"t g is an i:i:d P[T r] sequence with mean zero,P variance and …nite fourth moment. De…ne MT (r) = 1=T s=1 us t with r 2 0 ; 1] and t = P s=1 us . We have: P1 P1 P[r (1) Pts=1 us = (1) ts=1 "s + t 0 , where t = j=0 j " j and j=0 j "t j , 0 = 1 , which is absolutely summable. = j i=1 j+i 1 P[T r] 2 p (2) T t=1 "t ! 2 r: P[T r] L (3) T 1=2 t=1 "t ! W (r) : h i P[T r] P L (4) T 1 t=1 ts=11 "s "t ! 12 2 W (r)2 r : Rr P[T r] L (5) T 3=2 t=1 "t t ! rW (r) 0 W (s) ds : P p [T r] (6) T 1 t=1 t 1 0 "t ! 0: Lemma 0.1 Let ut =

2

p

(7) T 1=2 [T r] 0 ! 0: p L (8) T MT (r) ! (1) W (r) : h i P L [T r] (9) T 1 t=1 t 1 "t ! 12 (1) 2 W (r)2 r : Rr P[T r] L (10) T 3=2 t=1 t 1 ! (1) 0 W (s) ds: Rr P[T r] p (11) T 5=2 t=1 t 1 t ! (1) 0 W (s) sds: 1

j

(12) T (13) T

R L 2 2 (1)2 r t=1 t 1 ! 0 P[T r] p 3=2 u ! 0; 8j t=1 t 1 t j 2

P[T r]

W (s)2 ds: 0:

Pt

1 Lemma 0.2 De…ne yt = ~ t + s=1 us and ut = (L) "t = 1 j <1 j=0 j "t j ; where j=0 j 2 and f"t g is an i:i:d sequence with mean zero, variance and …nite fourth moment. Then, we have Z r [T r] X L 3=2 (a) T yt 1 "t ! ~ rW (r) W (s) ds : 0

t=1

(b) T

2

[T r] X

~ 2 r : 2

p

yt

1

!

yt2

1

!

yt

1 ut j

t=1

[T r]

(c) T

3

X t=1

(d) T

2

[T r] X

~2 3 r : 3

p

t=1

p

! 0; 8j

0:

The proof of Lemma 0.1 and Lemma 0.2 can be found in Phillips (1987), Phillips and Perron (1988), Phillips and Solo (1991) and Hamilton (1994). P Lemma 0.3 De…ne yt = ~ T t + ts=1 us ; ~ T = (1) T and ut = (L) "t = 1 j=0 j "t j ; 1 where j=0 j j < 1 and f"t g is an i:i:d sequence with mean zero, variance 2 and …nite fourth moment. Then, if > 1=2, we have (a1) T

1

[T r] X

yt

1 "t

t=1

(b1) T

3=2

[T r] X

2

[T r] X

yt

yt2

1

L

1

t=1

(d1) T

3=2

[T r] X

yt

1 2

L

t=1

(c1) T

L

! !

!

= 1=2, we have (a2) T

1

[T r] X t=1

yt

1 "t

L

!

(1)

Z

(1)

i r :

r

W (s) ds:

0

2

(1)2

Z

r

W (s)2 ds:

0

1 ut j

t=1

and if

h (1) W (r)2

2

rW (r)

p

! 0; ; j = 0; 1; Z

0

2

:

r

W (s) ds +

1 h W (r)2 2

r

i

:

[T r] X

3=2

(b2) T

L

yt

1

t=1

!

[T r]

2

(c2) T

X

L

yt2

1

t=1

[T r] X

3=2

(d2) T

(1)2

!

yt

1 r+ 2

Z

1 3 r + 3

2

(1)

1 ut j

t=1

r

W (s) ds :

0

Z

r

Z

W (s)2 ds + 2

0

r

W (s) sds :

0

p

! 0; ; j = 0; 1;

:

Proof: (a) From (1) of Lemma 0.1, T

1

[T r] X

yt

1 "t

1

=T

t=1

[T r] X

(1)

t T

t=1

=

1

+

t 1 X

(1)

X

1

t"t

1

(1) T

t=1

+ L

!

1

(1) T

[T r] t 1 X X

:

i r Rr 0

(b) From Lemma 0.1(10), T

3=2

[T r] X

yt =

1

"s "t + T

h (1) W (r)2 n (1) rW (r)

1 2 2

(1) T

(1) (1)

L

!

[T r] X

[T r] X

(1)2 T

yt2 =

t=1

L

!

(

2(1+ )

[T r] X

T

[T r] X t=1

yt

1 ut j

t 1

t2 + T

2

= T

[T r] X

[T r] X

(~ T t +

3

2

W (r)

[T r] X

r

io

if

> 1=2

if

= 1=2

t

(s) R r ds 0 W (s) ds 2 t

if if

+ 2 (1) T

2

> 1=2 = 1=2 [T r] X

t

t

t=1

+2

1=2;

t=1

1 2

h

t=1

t=1 t=1 R 2 (1)2 r W (s)2 ds h 0 Rr 2 1 3 (1) 3 r + 2 0 W (s)2 ds 3=2

"t

0

3=2

t+T

t=1 Rr 0 W 1 r 2 +

(d)From (5) and (13) of Lemma 0.1, when 3=2

"t

W (s) ds +

(c) From (11) and (12) of Lemma 0.1, 2

[T r] X

"t

t=1

[T r] X

3=2

t=1

T

0

t=1

t=1 s=1

8 <

t 1

s=1

[T r]

(1) T

"s +

!

t ) ut j

Rr 0

i if W (s) sds if

> 1=2 = 1=2

=

[T r] X

3=2

(1) T

tut

3=2

+T

j

t=1

=

(1)

3=2

(L) T

[T r] X

t ut j

t=1

[T r]

X

t"t

j

+T

t=1

3=2

[T r] X

t ut j

t=1

p

! 0:

We now derive the asymptotic distributions of SADF. Proof of Proposition 4.1. First, consider Case 1. The regression model is yt =

p 1 X

k r

yt

k

+

r yt 1

+ "t :

k=1

Let r be the window size (fractional) with r 2 [r0 ; 1], r0 the smallest fraction considered and t = [T r]. Assume the initial value y0 = 0. Under the null hypothesis that r = 0; we have P 1 1 2 2 p 1 p 1 yt = ts=1 us ; where ut = r (L) "t with r (L) = 1 : rL rL r L The deviation of the OLS estimate ^r from the true value r is given by 2 3 12 3 [T r] [T r] X X ^r Xt Xt0 5 4 X t "t 5 (1) r =4 t=1

t=1

where Xt = [ut 1 ut 2 : : : ut p+1 yt 1 ]0 ; r = [ 1r 2r : : : pr 1 r ]0 . P[T r] From (13) of Lemma 0.1, we know that the probability limit of t=1 Xt Xt0 is block diagonal. P[T r] P[T r] Therefore, we only need to obtain the last row of t=1 Xt Xt0 and t=1 Xt "t to calculate the ADF statistics. Then, the deviation of the OLS estimate ^ r from the true value is characterized by T 1 yt 1 " t T ^r = ; (2) T 2 yt2 1 where denotes summation over t = 1; 2; ; [T r] : Since Equation (2) is a continuous function of T 1 yt 1 "t and T 2 yt2 1 , based on (9) and (12) of Lemma 0.1, we have L

T ^r !

W (r)2 r : Rr 2 r (1) 0 W (s)2 ds

To calculate the t-statistic of ^ r ; we need to …nd the standard error of ^ r : Since the variance ^ of r is 82 3 12 39 2 3 1 [T r] [T r] [T r] = < X X X V ar ^r = V ar 4 Xt Xt0 5 4 Xt "t 5 = 2r 4 Xt Xt0 5 : : ; t=1

t=1

4

t=1

Therefore, variance of ^ r is

2 r

multiplied by the last element of 2 r ; 2 yt 1

V ar ^ r =

hP

[T r] 0 t=1 Xt Xt

i

1

; which is

so the variances of T ^ r can be calculated such that V ar T ^ r =

2 r

T

2

1 : W (s)2 ds

L

yt2 1

2Rr r (1) 0

!

Hence, the t-statistic of ^ r is ^ ADFr =

r

=

se ^ r

T ^r se T ^

L

!

r

2

hR

W (r)2 r 0

r

W (s)2 ds

i1=2 :

By CML, the asymptotic distribution of the SADF statistic is 8 9 > > < = W (r)2 r L SADF (r0 ) ! sup : h i 1=2 > r2[r0 ;1] > : 2 R r W (s)2 ds ; 0 Second, consider Case 3 where the regression model is yt =

r

+

r yt 1

+

p 1 X

k r

yt

k

+ "t :

k=1

P Under the null hypothesis that r = T and r = 0; we have yt = ~ T t + ts=1 us ; where 1 2 2 p 1 p 1 1 ~ T = r (1) T and ut = r (L) "t with r (L) = 1 . rL rL r L The deviation of the OLS estimate ^r from the true value r is given by ^r

r

2 3 [T r] X =4 Xt Xt0 5 t=1

12

3 [T r] X 4 X t "t 5 ;

(3)

t=1

where Xt = [~ T + ut 1 ~ T + ut 2 : : : ~ T + ut p+1 1 yt 1 ]0 ; = [ 1r 2r : : : pr 1 r r ]0 . As we can observe, the limiting distributions of (a1) (d1) of Lemma 0.3 are exactly the same as (9); (10); (12); (13) of Lemma 0.1 respectively. Therefore, if > 1=2, the limiting distribution of the SADF statistic under Case 2 is identical with that of the case when the regression includes a constant and the true process is a random walk without drift. From (d1) 5

P[T r] of Lemma 0.3, we know that the probability limit of t=1 Xt Xt0 is block diagonal. Therefore, P[T r] we only need to obtain the last 2 2 components of t=1 Xt Xt0 and the last 2 1 component P[T r] of t=1 Xt "t , which are 1 yt 1

yt yt2

1

"t

and

yt

1

respectively, where denotes summation over t = 1; 2; and (a1) of Lemma 0.3, the scaling matrix should be equation (3) by T , results in br b

T

r

r

=

r

and the matrix

T

(1)

hP

T

T 1=2 0 0 T

Rrr

W (s) ds

[T r] t=1 Xt "t

1

yt

r)

L

!

1 ( 2) ( 2)

T

;

> ;

=

1 "t

r

; [T r] : Based on (3) of Lemma 0.3 = diag T 1=2 ; T : Pre-multiplying 18

1

( 2) 1

"t

Under the null hypothesis that T 1=2 (^ r T ^r

i

i

9 > =

T 1=2 0 = 0 T 1 Rr r (1) r R 0 W (s) ds 2 2 r 2 r r (1) 0 W (s) ds

yt yt2

1

hP

1 T

( 2) ( 2)

[T r] 0 t=1 Xt Xt

1

r 0 1

2 3 [T r] X 14 Xt Xt0 5 t=1

yt

L

r

T 1

T 1=2 0 0 T

> :

1

Consider the matrix

!

8 > <

T

;

1 "t

T

> < > :

T

T

1

2 3 [T r] X 14 X t "t 5 t=1

1 yt

3=2

T T

1

3=2 2

A0;r = C0;r =

r

yt 1 yt2 1

;

T 1=2 "t T 1 yt 1 " t

L

!

"

1 2 2 r

rW h

1

C0;r D0;r

=

(r)

(1) W (r)2

r

1 B0;r r + A20;r

r

2 r

6

2

Z

r

r

i

#

:

B0;r C0;r + A0;r D0;r A0;r C0;r rD0;r

W (s) ds; B0;r = W (s)2 ds; r (1) 0 0 h i 1 2 2 (1) W (r) r : r W (r) ; D0;r = 2 r r r (1)

:

(4)

with Z

> ;

;

= 0;

r A0;r A0;r B0;r

( 2) 1

9 > =

Therefore, we have A0;r C0;r rD0;r : B0;r r + A20;r

L

T ^r !

To calculate the t-statistic of ^ r ; we need to …nd the standard error of ^ r . We know that ^r ^

var

1

2 r

=

r

yt

1

so the variances of T ^ r can be calculated as follows: ( 1 T 1=2 (^ r T 1=2 0 r) 2 var = r 0 T T ^r 2 r

L

!

1

yt yt2

1 1

1 yt

yt yt2

1

B0;r A0;r A0;r r

B0;r r + A20;r

1 1

T 1=2 0 0 T

1

)

1

:

Hence, the t-statistic of ^ r is T ^r ADFr = se T ^

1 2r

L

r

!

r1=2

By CMT, we have L

SADF (r0 ) ! sup

8 > <

1 2r

r2[r0 ;1] > : r1=2

h

W (r)2

nR r 0

h

0

i

W (s)2 dsr

W (r)2

nR r

r

r

i

W (s)2 dsr

Rr

W (s) dsW (r) o : Rr 2 1=2 0 W (s) ds

0

9 > = W (s) dsW (r) 0 : o Rr 2 1=2 > ; W (s) ds

Rr

0

P[T r] If = 1=2, we know that the probability limit of t=1 Xt Xt0 is block diagonal from (d2) P[T r] of Lemma 0.3. Similarly, we only need to obtain the last 2 2 components of t=1 Xt Xt0 and P[T r] the last 2 1 component of t=1 Xt "t to calculate the ADF statistics. Based on (3) of Lemma p T ; T . Consider the 0.1 and (a2) of Lemma 0.3, the scaling matrix should be T = diag hP i [T r] 1 0 matrix T 1 t=1 Xt Xt T ; ( 2) ( 2)

p

L

!

T 0 "

0 T

1

1 yt

1

yt yt2

1 1

p

T 0

r (1)

1 2r

+

Rr r 0

W (s) ds

(1)2

0 T h

1

Rr 1 2 r + r 0 W (s) ds R Rr 2 r W (s)2 ds + 2 r 0 r 0

(1) 1 3 3r

+ 7

i W (s) sds

#

=

1 0

0 (1)

1 2r

p

!

T 0 "

r

( 2) 1

(1)

r

0

0

r

r

n

T (^ r T ^r

+

;

(s) ds Rr 2 r 0 W (s) sds

1 "t

Rr

rW (r) "

(1)

r)

0

rW

r

=T

and

r

L

!

0

(r)

W (s) ds +

rW (r)

Under the null hypothesis that p

Rr 1 r + r 2 0 W R 2 r W (s)2 ds + r 0

1 0

0 (1)

"t yt

r

=

i

1

0 T

1 3 3r

W (s) ds

r 0

[T r] t=1 Xt "t

T

L

+

hP

1

and the matrix

Rrr

Rr

W (r)2

W (r)

0

W (s) ds +

r

= 0;

0 (1)

r

1 2 r

h

r 1

Ar;

r

h

1 2 r

r

io

W (r)2

1

Ar; Br;

#

r r

r

i

Cr; Dr;

#

:

r r

where Ar; Br; Cr;

r

r

r

1 = r+ 2 1 = r3 + 3

r

Z

r

W (s) ds;

0

2 r

Z

r

2

W (s) ds + 2

0

= W (r) ; Dr;

r

r

Z

= rW (r)

Z

r

W (s) sds

0

r

W (s) ds +

0

1 2

r

h

W (r)2

i r :

We can see that ^ r converges at rate T to the following distribution L

T ^r !

0

r

(1)

r

1

Ar;

Ar; Br;

r

1

Cr; Dr;

r r

r

r

=

(1)

r

rDr; r Ar; r Cr; Br; r r A2r; r

r

:

We know that var

p

T (^ r T ^r

r)

= L

!

2 r

( p

2 r

T 0

1 0

0 T 0 (1)

8

1

1 yt

1

1

r Ar;

r

yt yt2 Ar; Br;

p

T 0

1 1 1

r r

1 0

0 T 0 (1)

1

) 1

:

1

;

Hence, the t-statistic of ^ r is: ADFr =

T ^r se T ^

rDr;

L

!

r

r1=2

Ar; r Cr;

r

Br; r r

r

1=2

A2r;

:

r

By CMT, we have L

SADF (r0 ) ! sup

r2[r0 ;1]

(

rDr;

r

Ar; r Cr;

r1=2 Br; r r

A2r;

r

1=2 r

)

:

Third, consider Case 4. The regression model is the same as Case 2. P Under the null hypothesis that r = 0 and r = r , we have yt = ~ r + ut = ~ r t + ts=1 us ; where 1 1 2 2 p 1 p 1 . ~ r = (1) r and ut = (L) "t with (L) = 1 r L rL rL The deviation of the OLS estimate ^r from the true value r is given by 2 3 12 3 [T r] [T r] X X ^r Xt Xt0 5 4 X t "t 5 (5) r =4 t=1

t=1

where Xt = [~ r + ut 1 ~ r + ut 2 : : : ~ r + ut k 1 yt 1 ]0 , r = [ 1r 2r : : : pr 1 r r ]0 . P[T r] From (d) of Lemma 0.2, we know that the probability limit of t=1 Xt Xt0 is block diagonal. P[T r] Therefore, we only need to obtain the last 2 2 components of t=1 Xt Xt0 and the last 2 1 P[T r] component of of Lemma 0.2, t=1 Xt "t . Based on (3) of Lemma 0.1 and (a) hP i the scaling matrix 3 [T r] 1 1 1=2 0 should be T = diag T ; T 2 . Consider the matrix T t=1 Xt Xt T ; ( 2) ( 2)

1

T 1=2 0 1 yt 1 0 T 3=2 hP i [T r] and the matrix T 1 X " t t t=1

yt yt2

( 2) 1

T 1=2 0 0 T 3=2

1

Under the null hypothesis that T 1=2 (^ r T 3=2 ^ r 2

=

2r r 6r 3 ~ r 1

1

r

=

rW

L

1 "t r

and

!

~r

r

= 0,

1 2 r 2 ~rr ! 1 1 2 3 2 2 ~rr 3 ~rr Rr rW (r) + 3 R0 W (s) ds r 2 0 W (s) ds r rW (r)

r)

r

L

!

1 2 2 ~rr

1 2 2 ~rr 1 2 3 3 ~rr

;

"t yt

T 1=2 0 0 T 3=2

1

L

9

r

rW (r)

1

~r :

r

(r) Rr 0

W (s) ds

:

r W (r) Rr rW (r) 0 W (s) ds

;

(6)

We can see that L T ^ r ! 6r

3

~r

1

r

rW (r)

2

p

r 1 2 ~ 2 rr

2 r

r

W (s) ds :

0

The variances of T ^ r can be calculated as follows: ( 1 T 1=2 (^ r T 1=2 0 r) 2 var = r 0 T 3=2 T 3=2 ^ r !

Z

1 yt

1

1

1 2 2 ~rr 1 2 3 3 ~rr

1 1

4r 1 6r 2 ~ r 1

2 r

=

yt yt2

T 1=2 0 0 T 3=2

1

)

1

6r 2 ~ r 1 12r 3 ~ r 2

Hence, the t-statistic of ^ r is T 3=2 ^ r ADFr = se T 3=2 ^

L

r

!

Rr 0

By CMT, we have L

SADF (r0 ) ! sup

r2[r0 ;1]

(R r 0

Rr sdW (s) 0 W (s) ds : Rr 1=2 2 0 s ds

) Rr sdW (s) W (s) ds 0 : Rr 2 ds 1=2 s 0

Finally, consider Case 5 where the regression model is yt =

r+

r yt

1+

rt +

p 1 X

k r

p 1 X

k r ut k

yt

k

+ "t

k=1

This equation can equivalently be written as yt =

r

+

r

t+

rt +

+ "t

k=1

where ut k = yt k (1) r r , and r r (1) ; r = r (1) r (1 r) ; r = r; r = r + 1 1 2 2 p 1 p 1 = y (1) (t 1) with (L) = 1 L L L : Under the null t 1 r t r r r r hypotheses Pthat r = 0 and r = 0 and with the assumption that y0 = 0; we have ut = r (L) "t and t = ts=1 us : Consider a hypothetical regression of yt on ut 1 ; ; ut p+1 ; a constant, t and a time trend with the sample from 1 to [T r] ; producing the OLS estimates 2 3 2 3 12 3 ^r 1 t yt t 1 2 4 ^ 5=4 5 4 5 t 1 yt t 1t t 1 t 1 r 2 t yt t t ^r t 1t 10

The maintained hypothesis is that = r (1) r ; = 0 and = 0; we have = (L) 0 ; = 0 and = 0: The deviation of the OLS estimation from these true values are given by 2 3 2 3 12 3 ^r 1 t "t r (1) r t 1 2 4 5=4 5 4 5 ^ t 1 "t t 1 t 1t t 1 r 2 t"t t t ^r t 1t p T ; T; T 3=2 and based on Based on Lemma 0.1, the scaling matrix should be T = diag Lemma 0.2, we have 2 p 3 12 32 p 3 1 1 t T 0 0 T 0 0 t 1 2 4 0 T 54 0 T 0 5 4 0 5 t 1 t 1t t 1 2 3=2 t t 0 0 T 0 0 T 3=2 t 1t 2 3 T 1 1 T 3=2 t 1 T 2 t = 4 T 3=2 t 1 T 2 2t 1 T 5=2 t t 1 5 T 3 t2 Rr 3 1 2 r (1) r R 0 W (s) ds r 2 R R L r 2 (1)2 r W (s)2 ds ! 4 (1) r 0r W (s) ds (1) r 0 W (s) sds 5 r 0 R r 1 2 1 3 (1) r 0 W (s) sds 2r 3r Rr 2 32 32 1 2 r W (s) ds r 1 0 0 1 2 0 R R R (1) r 0 5 4 0r W (s) ds R0r W (s)2 ds 0r W (s) sds 5 4 0 =4 0 r 1 2 1 3 0 0 1 0 2r 3r 0 W (s) sds 2

T

2

t

T

5=2

t

t 1

and

2 p

T 4 0 0 2

0 T 0

0 0 T 3=2 0

r

=4 0 0

2 r

r

0

3 5

(1)

12

"t

3

2

(r)

i

3 0 0 5: 1

r

3

7 (1) W (r)2 r 5 Rr t"t rW (r) W (s) ds 3 0 32 0 h W (r) i 7 6 2 1 0 54 r 5: 2 W (r) Rr r rW (r) 0 W (s) ds 4

t 1 "t

L 6 5! 4

rW h

0 (1) 0

1 2 2 r

r

Under the null hypothesis that r = 0 and r = 0; 2 p 3 T (^ r r (1) r ) 4 5 T ^r 3=2 T ^r Rr 2 3 12 3 1 2 r W (s) ds r 1 0 0 2 0 R Rr Rr L 2 5 5 4 r W (s) ds ! r4 0 r (1) r 0 0 0 W (s) ds 0 W (s) sds R r 1 1 2 3 0 0 1 2r 3r 0 W (s) sds 11

1

2

1 4 0 0 2 L

where

0 (1) r 0 r

!4 0 0

r

0 r (1) 0

Ar =

Z

3 0 0 5 1 1

3 32 0 h W (r) i 6 7 2 1 r 5 r r (1) 0 5 4 2 W (r) Rr 0 1 rW (r) 0 W (s) ds 3 2 3 1 1 0 r Ar 2 r2 Dr 5 4 5 4 Ar Br Cr Er 5 0 1 2 1 3 Cr 3 r Fr r 2r 12

1 4 0 0 32

0

r

W (s) ds; Br =

0

Z

r

W (s) ds; Cr =

0

Dr = W (r) ; Er =

2

1h W (r)2 2

Z

r

W (s) sds; Z r i r ; Fr = rW (r) W (s) ds: 0

0

Therefore, we can see that ^ converges at rate T to the following distribution T ^r

L

^ r (1) = T r !

Er r 3

6Fr (2Cr Ar r) + 2Dr r (3Cr 2Ar r) : 12Ar Cr r 12Cr2 + Br r3 4A2r r2

The variances of T ^ can be calculated as follows: 02 p 31 T ^r V ar @4 T ^ r 5A T 3=2 ^ r 82 p 3 12 32 p 3 > 1 t T 0 0 T 0 0 < t 1 2 54 0 T = 2r 4 0 T 0 5 4 0 5 t 1 t 1t t 1 > 2 : t t 0 0 T 3=2 0 0 T 3=2 t 1t 2 3 1 T 1 1 T 3=2 t 1 T 2 t = 2r 4 T 3=2 t 1 T 2 2t 1 T 5=2 t 1 t 5 2 5=2 T t T t t 1 T 3 t2 3 12 3 2 32 0 0 0 0 r Ar 12 r2 r r L 1 1 !4 0 0 5 4 Ar Br Cr 5 4 0 0 5: r (1) r (1) 1 3 1 2 Cr 3 r 0 0 0 0 r r 2r

Therefore,

V ar T ^ r

L

r (1) !

12Ar Cr r

r3 12Cr2 + Br r3

4A2r r2

:

Hence, the t-statistic of ^ r is ^ ADFr =

r

se ^ r

L

!

Er r 3

6Fr (2Cr

r3=2 (12Ar Cr r 12

Ar r) + 2Dr r (3Cr 12Cr2

+ Br

r3

2Ar r) : 2 2 4Ar r )1=2

1

9 > = > ;

1

By CMT, the asymptotic SADF distribution is

L

SADF (r0 ) !

sup r22[r0 ;1]

(

Er r 3

6Fr (2Cr

r3=2 (12Ar Cr r

Ar r) + 2Dr r (3Cr 12Cr2 + Br r3

2Ar r)

4A2r r2 )1=2

)

:

REFERENCES Hamilton, J., 1994, 1994, Time Series Analysis. Princeton University Press. Phillips, P.C.B., 1987, Time series regression with a unit root, Econometrica 55, 277-301. Phillips, P.C.B., and Perron, P., 1988, Testing for a unit root in time series regression. Biometrika, 75(2):335–346. Phillips, P.C.B., and Solo, V., 1992, Asymptotics for linear processes. The Annals of Statistics, 20:971–1001.

13