LOCAL ASYMPTOTIC POWER OF BREITUNG’S TEST Mehdi Hosseinkouchack Goethe University Frankfurt y December 2011

Abstract In this paper we derive the local asymptotic power function of the unit root test proposed by Breitung (2002, Journal of Econometrics, 108, pp. 343-363). Breitung’s test is a nonparametric test and is free of nuisance parameters. We compare the local power curve of the Breitungs’test with that of the DickeyFuller test. This comparison is in fact a quanti…cation of the loss of power that one has to accept when applying a nonparametric test.

1

Introduction

Breitung (2002) proposes a nonparametric unit root test (BR) that is invariant to short-run dynamics. He carefully picks di¤erent processes in his simulation study to shed light on this property. Breitung (2002) provides the limiting distribution of his test statistic under a local alternative and using a simulation study illustrates its behavior. Davidson, Magnus, and Wiegerinck (2008) show that the test statistic proposed by Breitung (2002) always lies in the interval [0; 1= 2 ]. In this paper we derive the characteristic function of this test statistic under a local alternative. This in turn leads to a rigorous analysis of the local asymptotic power for BR. We compare the power curves of the BR for a no intercept and an intercept case to those of a Dickey-Fuller type test. This comparison is in fact a quanti…cation of the loss of power that one has to accept when applying a robust nonparametric test. Let fyt gT1 be an observed time series for which we are interested in testing the null of yt being I (1) against an I (0) alternative. Consider the following decomposition for yt yt = t + t , (1) where t = 0 dt is the deterministic component of yt and t is its stochastic component. dt may be zero, a constant or a time trend. In this model, no speci…c The author is thankful to Jörg Breitung for helpful comments. Department of Economics, Frankfurt University, Grueneburgplatz 1, 60322 Frankfurt am Main, Germany (e-mail: [email protected]) y

1

assumption is made regarding the short-run dynamics of yt . To test whether yt is I (1), Breitung (2002) proposes to use the following statistic

1

b %T = T

1

T

PT

l=1

PT

Pl

t=1 bt

2

2 t=1 bt

(2)

;

where bt , t = 1; 2; :::; T , are the OLS residuals from regressing yt on dt . Under the null of yt being I (1), Breitung (2002) shows that w

w

b %T ! Rj =

R1 Rr 0

R01 0

w ej (s) ds

2

dr

w ej2 (s) ds

, j = 1; 2; 3,

(3)

where ! denotes weak convergence, w e1 (s) = w (s) e2 (s) R= w (s) R R for dt = 0, w w (s) ds for dt = 1, and w e3 (s) = w (s) (4 6s) w (s) ds (12s 6) sw (s) ds for dt = [1; t]0 . w (s) represents a standard Wiener process. Using the results developed in Phillips (1987), he further shows that under a sequence of local alternatives w

(c)

b %T ! R j =

R1 Rr 0

0

R1 0

Jej (s) ds

2

dr

Jej2 (s) ds

, j = 1; 2; 3,

(4)

R1 with Je1 (s) = J (s) for dt = 0, Je2 (s) = J (s) J (s) ds for dt = 1, and Je3 (s) = 0 R1 R1 J (s) (4 6s) 0 J (s) ds (12s 6) 0 sJ (s) ds for dt = [1; t]0 , where J (s) is an Ornstein-Uhlenbeck (OU) process whose dynamics are given by dJ (s) = cJ (s) ds + dw (s).

2

Limiting characteristic functions

In this section we derive the characteristic functions of the limiting distribution of Breitung’s test statistic for the no intercept and intercept cases.1 We do so for the expressions given in (4) leaving the null case be recovered when c ! 0. We write (c)

Rj =

Nj ; j = 1; 2; Dj

where Nj and Dj are the numerator and denominator of equation (4), respectively. For j = 1; 2, we may write (c)

Pr Rj

1

z

Nj z ; Dj = Pr (zDj Nj 0) ; Z h 1 1 11 (c) = + Im j ( 2 0

= Pr

The trend case is tedious, hence not pursued here.

2

i ;z ) d ;

where the inversion formula is due to Imhof (1961). (c) To derive expressions for j ( 1 ; 2 ), j = 1; 2, we use two theorems which are stated below. The …rst theorem (see Tanaka, 1996, pp. 110) is due to Girsanov (1960) (see Liptser et al., 1977, for a more general discussion). Theorem 1. Let X = fX (r) : 0 r 1g and Y = fY (r) : 0 r 1g be OU processes on C = C[0; 1] de…ned by dX (r) = dY (r) =

X (r) dr + dw (r) ; Y (r) dr + dw (r) ; X (0) = Y (0) = 0:

Let X and Y be probability measures on (C; B (C)) induced by X and Y , respectively, by the relation X

(A) = Pr (! : X 2 A) ,

Then measures d d

X Y

(A) = Pr (! : Y 2 A) , A 2 B (C) .

are equivalent and Z 1 (h) = exp ( ) h (r) dh (r) X

and

Y

Y

2

2

2

0

Z

1

h2 (r) dr ;

0

where the left side is the Radon-Nikodym derivative evaluated at h 2 C with h (0) = X (0). Remark 1 Theorem 1 is a special case of the Girsanov (1960)’s theorem as it is applied to OU processes. Remark 2 Theorem 1 does not restrict the choicehof p ing = i , with i = 1; leads to dd X (h) = exp (i

(or ). In particular, i choosR1 1) 0 h (r) dh (r) .

Y

R1 Rr 2 Theorem 1 is not directly applicable to functions like 0 0 w (t) ds dr which appear in the numerator of (3) and (4). Tanaka (1996, pp. 117) provides the following theorem which proves Rpractical in our derivations. r Theorem 2. Let Fg (r) = 0 Fg 1 (s) ds with F0 (r) = w (r) and dYg (t) = Yg (r) dr + dFg (r) with Yg (0) = 0; where g 2 N. Then probability measures Fg and Yg are equivalent and d d

Fg

(f ) = exp

Yg

"

Z

0

1

dg f (r) d dtg

dg f (r) dtg

2

+

2

Z

0

1

dg f (r) dtg

2

dr ;

where f (0) = 0 and f (r) is g times continuously di¤erentiable on [0; 1]. Using these theorems we are now able to state our main results. Theorem 3. Joint asymptotic cf of (Nj ; Dj ) for j = 1; 2 are as follows (a) for j = 1 (c) 1

( 1;

2)

=

"

ec 1

2 1

2 2 1

#

1=2

3

#

2 1

1

+

2c

1

3 1 2 1

2 1 3

+

+

4 1

2 1

+

(b) for j = 2 (c) 2

( 1;

2)

=

"

c

e 3

2 2 2

2 2

5 5 2 2

6 9 2 2 2+ 2 9 2 +3e 2 2 2 + 4 2 (1 2 +24ic 2 2 1 22 2

3e

+2 +4

3 2 3 2

2

2 2 3 2 3 2

h

5 4 2 2

6c3

3 2c 3 2c 2 2 2 2 2 2 2 2 2 2

3

3 2 2 2

5 2

2 2

#

+

2) 2 2 2 2+

+

1

2

+ (2i

1

2 1

+

2

2

cosh

1

sinh

1

2

cos

1

cosh

1

+2

+2 1 1

2

sinh 2 1

+

+

1 2 1

c2 )2 =4 and

2

4(

1 2 1+

6 2

2i

1

cosh

2

2 2 2 2

+ 2i

1

2 2

2 2

4 2

2 2

2 1

+ 2 1

2

+ +2

cosh

1

sinh =i

1

1

p

sinh 2 1

1

;

c2 + 2i 2 ;

1=2

4i

+

4 2

+3

2 1

q 2

2i 2 ) =2

2 1

+

1

2 1+ 2 1+

c

1

2 1

+

2c

2

c

(c2

1c

+

c

r

=

1

4 1

2 1 1

c + where

c2

2 1

2 3

+

2 5 2+

2 2 1

1) 4i 2

i

3+

sinh

sinh

1

cosh 2 2

2

cosh

2

i

2

1

2

cosh

cosh

2 2 2 +4 4 6 ( 2 1) 2 2 i 2 cosh 2 cosh 2 2 +2 2 3c2 52 42 + 3 72 42 12 22 21 cosh 2 sinh 2 2 2 +2 2 4i 22 1 3 42 + 3 + 22 i 1 cosh 2 sinh 2 2 2 +2 2 3c 42 62 + 4 22 42 i 1 4 22 2 21 cosh 2 sinh 2 2 2 +4 52 6 22 i 1 + i ( 1 3 2 ) 42 cosh 2 sinh 2 2 2 6 4 +6 62 2 2 + 2i 2 2 cosh 2 sinh 2 2 12 62 52 c3 12 32 32 22 22 + 2i 1 + 12c 62 52 22 + 22 4c 32 32 2 3 22 + 3 + 22 22 i 1 + 4 21 6 ( 2 1) 22 22 i 2 2 2 2 24 2 c 2 i 1 22 2 2 2 + 2i 1 cosh 2 6 4 4 2 2 6 2 c 22 4 22 2 21 cosh 2 sinh 2 2 2 2 + 4 2 2i 1 2 6 2 c2 52 42 22 2 cosh 2 sinh 2 2 4 2 2 2 2 4 2i 1 3 2 + i 1 2 42 42 i ( 1 3 2 ) cosh 2 sinh 2 2 2 22 6 52 42 i 2 + 2 22 i 1 62 2 3 + 22 i 1 3 62 i 2 cosh 2 sinh 2 2 4 2 2 6 2 52 82 + 2 62 i 2 cosh 2 sinh 2 + 24 2 i 1 22 2 2 2+2 c+ 2 i 1 4 2 2 2 24 2 i 1 22 6c3 52 42 42 + 22 sinh 2 2 2 2 + 2 c + 2 i 1 sinh 2 2 2 2 2 6 22 62 62 + 4 22 42 22 + 22 i 1 4 42 2+ 2+2 2 1 sinh 2

2c 8c +12

h

2 1

2

2 1

3

5 4 2 2

3

4 2

+

4 4 2 2c ( 2

2 2

+

2 2

2 2 2

6+ 1)

4 2

+

2 2 2 2

2 2 2

2 2 4 2

6

4 2

+ 3+

sinh

+ 2 2 1=2

2

4

4 2 2 2 4 2

+ 6+

sinh

2

2 2

4 2

i

1

i

sinh

2

sinh

2

where 2 = 1 and 2 = Proof. See Appendix.

1.

(0)

Remark 3 For R1 , we have (0) 1

; x ) = 4i + 2x2

(

q

0:5

p

2i + 2 i + x2

x2 2

with 1 = ix i 2i + and distribution for j = 1 is recovered.

2

=

cos

q

1

cosh

ix

i

+ ix

2

1 2

sin

1

sinh

0:5 2

p

2i + x2 2 . Hence the null

(0)

Remark 4 For R2 , we have (0) 2

with

3

1

and

2

(

sin

;x ) =

1

sinh

1=2 2

;

1 2

given under Remark 3.

Comparison study

To shed light on the relative power of BR with respect to other tests we take, as a benchmark, the OLS estimate of , b, in yt = with t =

0

dt + t ; t = 1; :::; T , t 1 + ut , and x0 = 0.

Further, let ut take an MA(1) structure as P1 P1 P ut = 1 l=0 l=0 j l j < 1, and l=0 l "t l ,

l

6= 0.

Under a local alternative of the form = 1 c=T , Nabeya and Tanaka (1990, theorem4) provide the asymptotic characteristic fucntion of b for dt = 0 and dt = 1.2 This type of structure on the innovation is due to Phillips (1987). As a result, one in fact allows for a temporal dependence and heteroskedasticity in the innovations. This P1 in2 turn P1leads 2to asymptotics of the OLS estimator to involve the ratio R = = ( l=0 l l=0 l ) . Choosing this benchmark one would be able to see that how a nonparametric test involves a power loss when a parametrization of the above form is in fact true. We choose R 2 f0:5; 1; 2g to see the e¤ect of R as well. Figure (1) shows the power curves of the benchmark test and that of BR for di¤erent values of the parameter c for the no intercept case. As this …gure shows, R has little e¤ect on the power properties of the OLS. Further, we see that power loss due to application of a nonparametric test, namely BR, is realtively high with respect to the OLS. As the parameter c increases the power of BR increases only slightly while the power curve of OLS gets close to one very fast. 2

They also give a treatment for the trend case.

5

;

r= 0.5

r= 1

r= 2

4

5 c

6

BR

0.8 0.7 0.6

power

0.5 0.4 0.3 0.2 0.1 0 0

1

2

3

7

8

9

10

Figure 1: Power curves for the no intercept case for a 5% level test. Power is calculated via inverting the corresponding characteristic function of each test statistic. Figure (2) shows the power curve of the benchmark test and the BR for di¤erent values of the parameter c for the intercept case. In this case we observe more distortions in the power curve of OLS when R is varied. The power of BR is comparebale in value to those of OLS for di¤erent values of R. We also observe that he power of BR is slightly higher compared to the no intercept case while, and as expected, the power of the OLS is lower in this case. r= 0.5

r= 1

4

5 c

r= 2

BR

0.5

0.4

power

0.3

0.2

0.1

0 0

1

2

3

6

7

8

9

10

Figure 2: Power curves for the intercept case for a 5% level test. Power is calculated via inverting the corresponding characteristic function of each test statistic.

4

Conclusion

In this paper we derive the asymptotic local power of the non parametric unit root test by Breitung (2002). The analysis done here is compliment to the simulation study of Breitung (2002). We observe that an OLS estimator based test exhibits a higher local power than Breitung’s test for a no intercept case where the variance parameters have little e¤ect on the local power of OLS. For an intercept case, the local power of Breitung’s test becomes at least comparable to the local power of OLS when the local-to-unity parameter is fairly small. Breitung’s test exhibits higher power for the intercept case compared to the no intercept case. When an intercept 6

is present local power features of OLS are heavily a¤ected by variance parameters, a fact which might lead to wrong decisions. Hence in such cases where a relatively good estimate of the variance parameters is not possible, Breitung’s test could be a good alternative to parametric tests like DF.

Appendix A Unless stated, integrations are taken from 0 to 1. In the proofs provided below we repeatedly make use of the following two expressions for dY (r) = Y (r) dr + w (r) Z

0

and

Z

r

dY (r) dr

2

dY (s) ds

dY (s) d ds

2 2

dr =

1 dY (r) 1 = r; 2 dr 2 1 1 = [ Y (r) + w (r)]2 r; 2 2

Z

2

2

Y (r) dr + Y (1) +

Proof of Theorem 3. (a) We have ( Z Z (c) 1

2

r

( 1;

2)

= E exp i

J (s) ds

1

dr + i

0

2

Z

Z

w2 (r) dr:

)

J 2 (r) dr ;

(A.1)

where dJ (r) = cJ (r) dr + dw (r). Using Theorems 1 …srt and then Theorem 2 we derive an expression for (A.1), thus " Z Z # Z Z 2 r c2 (c) w (s) ds dt + i 2 w2 (r) dr + c w (r) dw (r) ; 1 ( 1 ; 2 ) = E exp i 1 2 0 " Z Z 2 2 c2 + 2i 2 dY1 (r) 1 2 = E exp i 1 Y1 (r) dr + dr 2 dr Z dY1 (r) dY1 (r) ( 1 c) d ; dr dr r q where dY1 (r) = we can write (c) 1

( 1;

2)

1 Y1

(r) dr+w (r). Choosing 1c

= E exp 1

(

2

+

2 1c

+2

2 1

c) Y1 (1) w (1)

Introducing X (r) with dX (r) =

1X

1i 2

1

=

c2 +2i 2

2 1

2

2i

c2 + 2i 2 (1) + 2 c 2 c 1 w (1) + 1 : 2 2 Y12

Z

1

+

c2 +2i 2

w2 (r) dr

(r)+dw (r) with X (0) = 0 and using Theorem

7

2

2

1 we get (c) 1

( 1;

= E exp

2)

"

2 1

+

1

p

1c

2

2 1c

+

+2

Z

1i 2 2

e

2

2 t

e

Z c2 + 2i 2 + 21 X 2 (r) dr 1( 2 c+ 1 c+ 1 2 X (1) + 1 : 2 2

X (r) dr c) e X (1) 1

1

1

=i

c+ 2

1

Now as

R

e

r

e

1r

X (r) dr

2 1

c2 + 2i 2 simpli…es this expression to " Z 2 2 (c) 1c + 1c + 2 1i 2 2 e e 1 ( 1 ; 2 ) = E exp 2 Z 1 c) e X (1) e 1 r X (r) dr 1( 1

Choosing

Z

1

c+ 2

1

X 2 (1) +

1

2 r

X (r) dr

:

X (r) dr and X (1) are jointly normally distributed we can write (c) 1

( 1;

2)

= jI2

1 2

1j

2A1

1

e2(

1

c+

1)

;

where I2 is an identity matrix of size 2, 1 ( 1 2 1 ( 1 2 1

A1 = and the element of can be shown to be 11

=

12

=

22

=

1

e2

1 2 1

c) e 1 ; e2 1 2 R which is the covariance matrix of e r X (r) dr and X (1)

1

(

1

2 1 c+2 1 i 2

, e

1 (

1 2 1 2 1c +

c + 1) c) e 1

)2

1

"

e2 )

(1 2

1

2

1

+

2e

1

e

1

1

e 2

e

,

+ 1

e 1

1

+

e

+

e

1

(e2 2

1)

#

:

A bit of calculus leads to the desirable result. Proof of Theorem 3. (b) The derivation is exactly the same for part (a) in

8

the sense that we start with ( Z (c) 2

( 1;

2 ) = E exp i

+

i

Z

1

3

Z

1

2

r

J (s) ds

dr

2i

1

0

Z

1

J (s) ds

0

2

J (s) ds

+i

2

Z

J 2 (s) ds

i

Z Z

r

rJ (s) dsdr

0

2

Z

2

J (s) ds

)

:

and use Theorem 1 to obtain an expression in w (r) rather than J (r) as ( Z Z Z 1 Z Z r 2 r (c) w (s) ds dr 2i 1 w (s) ds rw (s) dsdr 2 ( 1 ; 2 ) = E exp i 1 0

+

i

1

3

+ i

2

i

2

Z

c2 2

0

Z

0

2

w (s) ds 2

w (s) ds + c

Z

w (r) dw (r) :

Using Theorem 2 with dY2 (r) = 2 Y2 (r) dr + w (r) we obtain Z Z 1 Z Z r dY2 (s) dY2 (s) (c) 2 Y2 (r) dr 2i 1 ds r dsdr 2 ( 1 ; 2 ) = E exp i 1 ds ds 0 0 Z Z 2 2 2 i 1 dY2 (r) c2 dY2 (r) 2 + i 2 dr + i 2 + dr 3 dr 2 2 dt Z dY2 (r) dY2 (r) d + (c : 2) dr dr r q

Choosing (c) 2

( 1;

2

c2 +2i 2

=

2 ) = E exp

2i

2i

2

1

Z

0

+

i

+ i + (c

1

3 2

i

2

1

c2 +2i 2

+

2

we can write Z Z

1

(

2 Y2 (s) + w (s)) ds

Z

2X

r

r(

2 Y2

(s) + w (s)) dsdr

0

2

[

2 Y2

(r) + w (r)] dr Z 2 Y2 (1) + w2 (r) dr

2 c2 + 2 2 2 1 ) [ Y2 (1) + w (1)]2 2

Introducing X (t) with dX (t) =

2

1 2

:

(t) + dw (t) with X (0) = 0 and using Theorem

9

1 we get (c) 2

( 1;

2 ) = E exp Z Z

2i

1

+

r

r

2

+

i

3

2)

2 2

2

Z

Z

2

Z

s u)

2 (s

e

X (u) du + X (s) ds

0

s

u)

2 (s

e Z

1 2

2

X (u) ds + X (s) dsdr

Z

2

2

r 2 (r

e

0

2 c2 + 2 2 2

2

+ (c

2

0

1

+ i

1

0

0

i

Z

2

Z

u)

X (u) du + X (s) dr 2

1

e

2 (1

u)

X (u) du

+

0

Z

2

1

e

u)

2 (1

1 2

X (u) du + X (1)

0 2

X 2 (t) dt

2

X (1)2 +

2

Z

X 2 (r) dr !

!

;

2

p 2 which can be simpli…ed by choosing 2 = i c2 + 2i 2 as 2 Z 1 Z 1Z s (c) (s u) X (s) ds 2 1 e X (u) duds + 2 ( 1 ; 2 ) = E exp 0 0 0 Z Z r Z Z rZ s (s u) rX (s) dsdr re X (u) dsdsdr + 0

0

1

+

2

3

+

2

+ (c

0

Z Z

r

(r u)

e

0

c2 c 2 2 + e 2 2 Z 1 ) e X (1) e

Z

X (u) dudr + 1

e

u

Z

2

X (r) dr 2

X (u) du

0

u

X (u) du +

c

X 2 (1) +

2

0

+

c 2

The last expression can be written as (c) 2

( 1;

2)

= E exp f 2i 1 ( 2 n2 + n1 ) ( 2 n3 + n4 ) c2 2 c 22 2 2 2 i 1 i 2 ( 2 n2 + n1 )2 + i 2 2 + e n5 + 3 2 2 2 + (c 2 ) 2 e n6 n5 + 0:5 (c 2 2 ) n6 (1) + 0:5 ( 2 + 2

or compactly (c) 2

( 1;

2)

= E exp fn0 A2 ng

10

exp

2

+

2

2

c

;

c) ;

:

with 2

6 6 6 6 6 A2 = 6 6 6 6 4

1i 1 3

i 2

1 3 1

2

i 2

i 1 2 i 1 0

1 3 1 1 3 1

2

i i

i 1 2

i 1

0

i 2 2

i 1 2 2 0 0 0

i 1 2 0 0 0

0

0

0

and n = [n1 ; :::; n6 ]0

2 2 1 2 1 2

i 2

0

N (0; (c) 2

( 1;

0

2 ). 2)

0

1 2

2 2 c +c 2 2e 2 ) e 2 2 2

2i 2

0

(c

7 7 7 7 7 0 7; 7 0 7 2 7 5 2) 2e 2 2) 0

0 0 1 2

1 (c 2 1 (c 2

3

We can now simply write

= jI6

2A2

2j

1 2

1

e2(

2+

c)

;

where I6 is an identity matrix of size 6. We do not report the elements of 2 as to save space, the derivation of which only requires some primitive calculus. Simpli…cation of this last expression leads to the desirable results.

References Breitung, J. (2002). Nonparametric tests for unit roots and cointegration. Journal of econometrics 108, 343–363. Davidson, J., J. Magnus, and J. Wiegerinck (2008). Notes and problems: A general bound for the limiting distribution of breitung’s statistic. Econometric Theory 24, 1443–1455. Girsanov, I. (1960). On transforming a certain class of stochastic processes by absolutely continuous substitution of measures. Theory of Probability and its Applications 5, 285–301. Imhof, J. P. (1961). Computing the distribution of quadratic forms in normal variables. Biometrika 4, 419–426. Liptser, R., A.N., and A. Shiryaev (1977). Statistics of random processes: General Theory. Springer Verlag. Nabeya, S. and K. Tanaka (1990). A general approach to the limiting distribution for estimators in time series regression with nonstable autoregressive errors. Econometrica 58, 145–163. Phillips, P. C. B. (1987). Towards a uni…ed asymptotic theory for autoregressions. Biometrika 74, 535–547. Tanaka, K. (1996). Time Series Analysis: Nonstationary and Noninvertible Distribution Theory. Wiley.

11