Spurious logarithms and the KPSS statistic

Economics Letters 76 (2002) 383–391 www.elsevier.com / locate / econbase

Spurious logarithms and the KPSS statistic Robert M. de Jong*, Peter Schmidt Michigan State University, Department of Economics, 217 Marshall Hall, East Lansing, MI 48824, USA Received 18 December 2001; accepted 25 February 2002

Abstract This paper analyzes the asymptotic behavior of two types of so-called KPSS tests when a logarithm transformation has been applied spuriously to data that are themselves an integrated time series. Although a different limit distribution is obtained, the asymptotic convergence behavior of the KPSS statistic is reminiscent of that of integrated time series, and it is shown that the KPSS test cannot distinguish consistently between an integrated time series and the logarithm of an integrated time series.  2002 Elsevier Science B.V. All rights reserved. Keywords: Unit roots; Unit root test; Weak convergence; Non-linearity JEL classification: C22; C32

1. Introduction Let w t be an observable series. In this paper we ask whether we can distinguish the case that w t is I(1) from the case that log(w t ), or more precisely loguw t u, is I(1). This is a matter of some importance in practice because the asymptotic theory for cointegrating regressions assumes that the variables in question are I(1), and it is not clear whether standard unit root tests can be trusted to distinguish whether w t or loguw t u is I(1). Empirically, regressions are often run in logarithms and it is not clear whether standard theory applies. That is, if we regress loguy t u on loguw t u, standard theory applies if these logarithmic variables are I(1) and cointegrated, but a different and as yet undeveloped theory would apply if y t and w t are I(1) in levels (but the regression is still run in logarithms). In this paper we ask whether the KPSS tests of Kwiatkowski et al. (1992) can be used to make this distinction. The KPSS tests are most commonly used to test the null hypothesis of stationarity, but they are also standard tests of the unit root hypothesis (e.g. Shin and Schmidt, 1992; Harvey, 2001). * Corresponding author. Fax: 11-517-432-1068. E-mail address: [email protected] (R.M. de Jong). 0165-1765 / 02 / $ – see front matter PII: S0165-1765( 02 )00067-8

 2002 Elsevier Science B.V. All rights reserved.

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For a sample w 1 , . . . ,w n , define w¯ as the sample mean, and define the demeaned data as w t 2 w¯ (t 5 1, . . . ,n). There are three types of KPSS tests; the first one is based on the actual data w t , the second is based on the demeaned data, and the third type considers demeaned and detrended data. The asymptotic distribution of all three statistics is a functional of Brownian motion. In this paper, we investigate the consequences for the first two types of KPSS statistics if the data w t are generated as the logarithm of an integrated process. We find that the KPSS statistic cannot distinguish consistently between the case that w t is I(1) and the case that loguw t u is I(1).

2. The KPSS statistic, assumptions, and main results The KPSS statistic ‘in levels’ is defined as

O O n

21

2

n St t51 KPSS 1 5 ]]] n 2 wt

(1)

t51 tw

where St 5 o j51 w j , and the w t are observed. It is well-known that for I(1) w t , under regularity conditions, 1

a

E 1E

O O n

2

2

24 2 W(r) dr da St n d t 51 0 0 n 21 KPSS 1 5 ]]]] → ]]]]]] n 1 22 2 n wt W(r)2 dr t51

(2)

E 0

d

where ‘ →’ denotes convergence in distribution, while for mean zero I(0) w t , again under regularity conditions,

O O n

22

2

1 n St d t 51 KPSS 1 5 ]]]] → W(r)2 dr. n 21 2 0 n wt

E

(3)

t51

In the above equations, the KPSS 1 statistic is written in such a way as to make clear what the weighting with respect to n needs to be for the numerator and the denominator of the KPSS statistic. The KPSS statistic that considers data in deviations from mean is

O n

21

S¯ t n t 51 KPSS 2 5 ]]]] n 2 (w t 2 w¯ )

O

t51

2

(4)

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t where S¯t 5 o j51 (w t 2 w¯ ). For KPSS 2 we have, for I(1) w t , under regularity conditions, 1

d

a

E 1E

2

¯ dr W(r)

2

da

n KPSS 2 → ]]]]]], 21

0

0 1

(5)

¯ E W(r) dr 2

0

while for I(0) w t with arbitrary constant mean, 1

d

E

¯ 2 dr, KPSS 2 → W(r)

(6)

0

¯ 5 W(r) 2 e0 W(s) ds. Below, it is assumed that instead of being a stationary or an where W(r) integrated process, the w t that we observe are instead generated as w t 5 logux t u, where x t satisfies assumptions implying that x t is I(1). This situation would occur if the original data were I(1), but a logarithm transformation has been (incorrectly) applied to these data. Define z t 5 logun 21 / 2 x t u, ¯ It will be assumed that implying that w t 5 z t 2 (1 / 2) log(n) and w t 2 w¯ 5 z t 2 z. 1

x t 5 x t21 1 vt ,

(7)

where vt is generated according to

Of´ `

vt 5

k t 2k

,

(8)

k50

where ´t is assumed to be a sequence of i.i.d. random variables with mean zero, and it is assumed that o `k 50 fk ± 0. In addition, we will assume that x 0 is an arbitrary random variable that is independent of all vt . The following regularity conditions will be assumed to hold for ´t : Assumption 1. (a) o `k50 k ufk u , ` and Eu´t u p , ` for some p.2. (b) The distribution of ´t is absolutely continuous with respect to the Lebesgue measure and has characteristic function c (s) for which lim s→` sh c (s) 5 0 for some h . 0. The following theorem is the key to determining the behavior of the KPSS statistic if a logarithm transformation has been applied to an I(1) process. In the theorem below, ‘ ⇒ ’ denotes weak convergence; the theorem below provides an extension of results in de Jong (2001), where a result similar to the one below is shown, but only pointwise for a 5 1. Theorem 1. Under Assumption 1

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386

a

[an]

n

21

O logun

21 / 2

xt u ⇒

t 51

E logusW(r)u dr

(9)

0

and a

[an]

n

21

O (logun

21 / 2

x t u) ⇒ 2

t 51

E (logusW(r)u) dr, 2

(10)

0

where

s 2 5lim E(n 21 / 2 x n )2 [ (0, `). n →`

(11)

Note that the results of Theorem 1 are certainly not as straightforward as they may look at first sight. Because the logarithm function has a pole at 0, we cannot simply apply the continuous mapping theorem in order to arrive at even the pointwise result for a 5 1, and the assertion that logun 21 / 2 x [nr] u ⇒ logus W(r)u is not correct. To prove the above result, a separate proof is required, using results of de Jong (2001). In de Jong (2001), it is shown that

O T(n

1

n

n

21

21 / 2

d

E

x t ) → T(s W(r)) dr

t51

(12)

0 K

will typically hold as long as for all K . 0, e2K uT(x)u dx , `, in spite of possible poles in the function 21 / 2 T(.) and in spite of the fact that in the presence of poles in T(.), the assertion T(n x [nr] ) ⇒ T(s W(r)) is not true in general. In addition, it can be shown that for functions with poles that are non-integrable (such as T(x) 5 uxu 2 f for f . 1), the result of Eq. (12) is incorrect in general. Using Theorem 1, first, the denominators of both KPSS statistics are analyzed: Lemma 1. Under Assumption 1

O w →1 / 4, n

21

n ((log(n))

22

2 p t

(13)

t51

and n

1

1

0

0

2

O (w 2 w¯ ) →E 1loguW(r)u 2E loguW(s)u ds2 dr. n

21

2 d

t

t51

(14)

For the KPSS 1 and KPSS 2 statistics, the following results can now be shown: Theorem 2. Under Assumption 1 1

d

log(n)(n KPSS 1 2 (1 / 3)) → 2 (4 / 3) logus u 1 2 21

E (r 2 1) loguW(r)u dr 2

0

(15)

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387

and 1

a

1

2

E 1E (loguW(r)u 2E loguW(s)u ds2 dr da

d

n 21 KPSS 2 → ]]]]]]]]]]]]] . 1 1 2 0

0

0

(16)

E 1loguW(r)u 2E loguW(s)u ds2 dr 0

0

The above result implies that when a logarithm transformation has been applied incorrectly to I(1) data, the asymptotic convergence behavior of both KPSS tests will be identical to the situation where w t is I(1), in the sense that the scaling factors are the same, although a different limit distribution results. That implies that the KPSS test cannot be a suitable means for distinguishing integrated processes from integrated processes to which a logarithm transformation was applied. Our results do not settle the question of whether some other statistic could be used to distinguish consistently between an integrated process and the logarithm of an integrated process. Work on this question continues.

Appendix A Proof of Theorem 1. Below, only the first result of Theorem 1 will be proven, since the second can be shown analogously. From the proof of de Jong (2001), it follows that for all a[[0, 1], pointwise in a, a

[an]

Zn (a) 5 n

O logun

21

21 / 2

d

E

x t u → logus W(r)u dr.

t 51

(A.1)

0

This is because by Theorem 1 of de Jong (2001,

O logucn

1

n

n

21

21 / 2

d

E

x t u → logucW(r)u dr,

t51

(A.2)

0

and therefore for each a [ [0, 1], [an]

n

21

O logun

x t u 5 a(an)

t 51

and because a a

1

[an ] 21 / 2

21

O logua

t 51

1/2

(an)

21 / 2

d

x t u →a

E logua

1/2

W(r)u dr,

(A.3)

0

1/2

W(r) is distributed identically to W(ar), the last expression can be rewritten as

1

a

0

0

E loguW(ar)u dr 5E loguW(r)u dr.

(A.4)

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Also, note that from de Jong (2001), it follows that for some large N not depending on ´,

O n

lim lim sup En

´ →0

21

n →`

ulogun

21 / 2

x t uuI(un

21 / 2

x t u , ´) 5 0.

(A.5)

t 5N 11

Since we have pointwise convergence by the result of Eq. (A.1), it suffices to show stochastic equicontinuity. To show this, note that, for all ´ . 0 and a $ b and a 2 b , ´, defining ‘empty summations’ as zero,

U O

[an]

sup

uZn (a) 2 Zn (b)u 5

a,b :ua 2b u, ´

sup

n

a,b :ua 2b u, ´

# o P (1) 1 n

21

logun

21 / 2

xt u

t5[bn ]11 n

O

21

ulogun

21 / 2

x t uuI(un

U

21 / 2

x t u # ´)

t5N 11

(A.6)

[an ]

1

sup

n

O O

21

ulogun

21 / 2

x t uuI(´ # un

ulogun

21 / 2

x t uuI(un

21 / 2

x t u # 1)

t 5[bn]11

a,b :ua2b u, ´

[an ]

1

sup

n

21

21 / 2

x t u . 1),

t 5[bn]11

a,b :ua2b u, ´

where the first term is uniform in a and b. By the result of Eq. (A.5), the second term satisfies

O n

lim lim sup En

´ →0

21

n →`

ulogun

21 / 2

x t uuI(un

21 / 2

x t u # ´) 5 0.

(A.7)

t 5N 11

The third term can be bounded almost surely by [an]

sup

n

21

a,b :ua 2b u, ´

O

ulog(´)u # ´ ulog(´)u

(A.8)

t5[bn ]11

which also goes to 0 as ´ → 0. For the last term, we have the upper bound of

´ sup ulogun 21 / 2 x t uuI(un 21 / 2 x t u . 1),

(A.9)

1#t #n

and since sup ulogun 21 / 2 x t uuI(un 21 / 2 x t u . 1) 5 OP (1)

(A.10)

1#t #n

by the continuous mapping theorem, the result now follows. h Proof of Lemma 1. Note that, since w t 5 z t 2 (1 / 2) log(n),

O w 5n n

n 21 (log (n))22

2 t

t 51

O z 1 (1 / 4) 2 n n

21

(log(n))22

2 t

t51

Oz. n

21

((log(n))22 log(n)

t

(A.11)

t51

Obviously the lemma is therefore complete if we can show that the first and last term in the above equation converge to 0 in probability. To show the first result, note that by Theorem 1,

R.M. de Jong, P. Schmidt / Economics Letters 76 (2002) 383 – 391 1

O z → E (logusW(r)u) dr n

n

389

21

2 d t

2

t51

(A.12)

0 22

implying that this term converges to 0 in probability at rate (log(n)) . Also, 1

O z →E logusW(r)u dr, n

n

21

d

(A.13)

t

t51

0 21

which implies that the last term in Eq. (A.11) converges to 0 at rate (log(n)) . For the second part of the lemma, simply note that by Theorem 1, n

1

1

2

O (w 2 w¯ ) 5 n O (z 2 z¯ ) →E 1loguW(r)u 2E loguW(s)u ds2 dr. n

21

n

2 d

21

2

t

t

t51

t51

0

h

(A.14)

0

Proof of Theorem 2. First note that given the result of Lemma 1,

O O n

23

22

2

n (log(n)) St t51 log(n)(n 21 KPSS 1 2 1 / 3) 5 log(n) ]]]]]] 2 1/3 n 21 22 2 n (log(n)) wt

1

t51

2

(A.15)

will have a limit distribution identical to that of

F

O S 2 (1 / 12)G n

23

4 log(n) n (log(n))

22

2 t

(A.16)

t51

if we can find a limit distribution for the last statistic. Next, note that

O z 2 (1 / 2)t log(n) ; S 2 (1 / 2)t log(n) t

St 5

Z t

j

(A.17)

j51

and therefore

O S 5O (S ) 1 (1 / 4)(log(n)) O t 2 log(n) O tS . n

n

2 t

n

Z 2 t

t51

2

t51

n

2

Z t

(A.18)

t 51

t51

Noting that

O t 5 (1 / 12)n (log(n)) 1 O(n (log(n)) ), n

(1 / 4)(log(n))

2

2

3

2

2

2

(A.19)

t 51

it follows that

F

O S 2 (1 / 12)G 5 o (1) 1 n n

23

log(n) n (log(n))

22

2 t

t51

P

O (S ) 2 n O tS . n

23

(log(n))

21

n

Z 2 t

t51

23

Z t

t51

(A.20)

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390

The second term on the left-hand side in the last equation converges to 0 in probability because by Z a Theorem 1, n 21 S [an ] ⇒ e0 logus W(r)u dr, implying that by the continuous mapping theorem,

O (S ) 5 n O (n n

n

23

1

n

21

Z 2 t

t51

21

Z 2 d t

S ) →

t51

a

2

E 1E logusW(r)u dr2 da. 0

(A.21)

0

For the last term, again by the continuous mapping theorem, 1

2n

a

1

O tS → 2E a1E logusW(r)u dr2 da 5E (1 / 2)(r 2 1) logusW(r)u dr, n

23

Z d t

2

t51

0

0

(A.22)

0

and the last expression equals 1

E

2

2 (1 / 3) logus u 1 (1 / 2)(r 2 1) loguW(r)u dr.

(A.23)

0

After multiplication by 4 as suggested by Eq. (A.16), we now obtain the result for KPSS 1 . To show the result for KPSS 2 , note that

O SO O n

D

t

2

O O n

23 Z 2 (w j 2 w¯ ) n n 23 (S¯ t ) t51 j51 t 51 21 n KPSS 2 5 ]]]]]]]] 5 ]]]]] , n n 22 2 22 2 (w t 2 w¯ ) n (z t 2 z¯ ) n t 51

(A.24)

t 51

21 Z a 1 and because n S¯ [an] ⇒ e0 (logus W(r)u 2 e0 logus W(s)u ds) dr by Theorem 1, using Lemma 1 and the continuous mapping theorem, it now follows that 1

d

a

1

2

E 1E (logusW(r)u 2E logusW(s)u ds2 dr da

n KPSS 2 → ]]]]]]]]]]]]]] , 1 1 2 21

0

0

0

(A.25)

E 1logusW(r)u 2E logusW(s)u ds2 dr 0

0

and noting that the s cancel out, we obtain the result of the theorem for KPSS 2 . h

References de Jong, R.M., 2001. A continuous mapping theorem-type result without continuity, mimeo. Michigan State University, available at http: / / www.msu.edu / user / dejongr Harvey, A., 2001. A unified approach to testing for stationarity and unit roots, mimeo. University of Cambridge.

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Kwiatkowski, D., Phillips, P.C.B., Schmidt, P., Shin, Y., 1992. Testing the null hypothesis of stationarity against the alternative of a unit root: How sure are we that economic time series have a unit root? Journal of Econometrics 54, 159–178. Shin, Y., Schmidt, P., 1992. The KPSS stationarity test as a unit root test. Economics Letters 38, 387–392.