Sequential Panel Unit Root Tests for a Mixed Panel with ...Meg/MEG2004/Sul-Donggyu.pdfpanel unit root might not be meaningful when the power of a test under H1 ois poor. Hence to improve

Sequential Panel Unit Root Tests for a Mixed Panel with

Nonstationary and Stationary Time Series Data

Donggyu SulDepartment of Economics,University of Auckland

July 12, 2004

Abstract

The paper provides a new effective detrending method by using recursive mean adjust-ment. Based on recursive mean adjustment methods, univariate unit root tests as well aspanel unit root tests are proposed. More importantly, the maximum order tests under crosssectional dependence are developed to detect whether or not a panel contains nonstationaryunits. The cross sectional average of the second moment of the recursive time series meansgrows over time when a panel contains nonstationary units. To reduce the size distortionof tests, sequential panel unit root tests are proposed. The null hypothesis of panel unitroot (all units are nonstationary) is first tested. When the first null is rejected, the second

null hypothesis of mixed panel unit root (some units are nonstationary) is tested. MonteCarlo simulation shows that the sequential tests perform well with moderately large T andN.

Incomplete: Please do not quote.The final version of the paper will be available by the end of 2004 at

http://yoda.eco.auckland.ac.nz/~dsul013/working/working.htm

1

1 Introduction

Since Quah (1990, 94) opened the door of panel unit root testing literature, several important

theoretical development have been contributed by several researchers. Levin, Lin and Chu

(LLC, 2002) generalize Quah’s (1990, 94) panel unit root test under the alternative of a ho-

mogenous panel. Im, Pesaran and Shin (IPS, 2003), Choi (2001) and Maddala and Wu (MW,

1999) consider panel unit root tests under the alternative of a heterogeneous panel. More

recently Bai and Ng (BN, 2004a), Chang (2003), Moon and Perron (MP, 2004) and Phillips

and Sul (PS, 2003) propose panel unit root tests under cross sectional dependence. Along with

the theoretical development of panel unit root tests, their use in empirical research has grown

exponentially. The most important reason for their popularity is that panel unit root tests

reject the null hypothesis of unit root more often than univariate unit root tests. This is a

natural result because the goal of the panel unit root tests under the null hypothesis of unit

roots is to emplify the power of tests through the pooling of information across units.

However the unsolved and thorny problem of panel unit root tests is that the rejection of

the null hypothesis of panel unit roots does not imply that all units in the panel are stationary.

Consider the following two hypotheses of panel unit root.

H0o : ρi = 1 for all i H0

A : ρi < 1 for all i

H1o : ρi = 1 for all i H1

A : ρi < 1 for some i

Quah(1990) and LLC test the first null hypothesis H0o against H

0A. Some empirical researchers

think that the rejection ofH0o implies that all units in a panel are stationary. Unfortunately this

is a myth. Karlsson and Löthgren (2000) report the even when a half of panel are stationary,

the LLC test rejects H0o very often. The next null hypothesis H

1o has the same problem. IPS,

Choi, MW, BN, MP and PS tests are designed to reject H1o more often than LLC when there

is at least one stationary unit in a panel. Panel empirical researchers, however, want to test

whether or not a panel consists of pure stationary units rather than to test whether a panel

contains at least one stationary unit. For example, testing growth convergence or long run

purchasing power parity requires that all units in a panel are stationary. The rejection of H0o

or H1o does not provide any meaningful answer.

The null hypothesis of panel stationarity does not have such problems. The issue is rather

whether or not there is an efficient panel stationary test. Hadri (2000) proposes the panel

KPSS test under cross section independence. However as Bai and Ng (2004b) show, the finite

sample performance of the panel KPSS test is rather disappointing. Canor and Kilian (2001)

and Lee (1996) criticize KPSS test because the power of KPSS test is almost the same as the

size of the KPSS test based on Andrew (1992)’s prewhitening HAC estimator while the size

distortion of KPSS test based on Newey and West (1987)’s fixed bandwith HAC estimator is

serious. Sul, Phillips and Choi (2003) provide a reason that the KPSS test with Andrew’s

2

‘0.97 rule’1 is not consistent under the null hypothesis and propose a new boundary rule of

‘1− 1/√T ’ where T stands for the number time series observations. Under the new rule, thepower of the KPSS test becomes fairly good when compared with the KPSS test based on

NW’s HAC estimator, while the size distortion of the test becomes moderate under the new

rule. The problem is that the KPSS test does not have any power if the AR(1) coefficient

ρ is greather than 1 − 1/√T . For example, with T = 100, the KPSS test works only when

ρ ≤ 0.9. More importantly, Choi (2002) reports that even when the half of a panel consists ofnonstationary units, the size adjusted power of the KPSS test based on NW’s fixed bandwith

HAC estimator is very low. In other words, the KPSS test does not have enough power to

detect whether or not a panel is mixed.

Chang and Song (2002) consider more attractive null hypothesis of the mixed panel unit

root and propose a mixed panel unit root test.

H2o : ρi = 1 for some i H2

A : ρ < 1 for all i

The null hypothesis H2o is the most desirable hypothesis for empirical panel unit root tests.

The rejection of H2o implies that all units in a panel are stationary. Chang and Song (2002)’s

test, however, suffers from a lack of power. It is worthwhile noting that testing for a mixed

panel unit root might not be meaningful when the power of a test under H1o is poor. Hence

to improve the power of the mixed panel unit root test, one should improve the power of the

panel unit root test under the null H1o in the first place.

This paper proposes a new mixed panel unit root test for the cases of a constant and a

linear trend under the null hypothesis H2o . The new proposed test is based on the fact that the

second moment of the recursive mean is growing over time under the null hypothesis of H2o .

However, the proposed test suffers from a moderate size distortion especially when there are a

small number of nonstationary units in a panel. To reduce the size distortion while maintaining

good power of the tests, we introduce sequencial panel unit root tests. The proposed testing

procedure consists of two panel unit root tests. The first null hypothesis is H1o . If the null H

1o

is rejected, then the second null hypothesis of mixed panel unit, H2o is tested. To increase the

rejection rate of the null H1o , new panel unit root tests are proposed based on recursive mean

adjustment methods.

In the next section, I introduce Shin and So (2001)’s recursive mean adjustment for unit root

tests, propose a new efficient bias reduction method for the general autoregressive model with

a linear trend and construct a univariate unit root test by utilizing recursive mean adjustment.

In section 3, the new efficient and powerful panel unit root tests based on recursive mean

adjustment are proposed. Since the panel recursive mean adjusted estimator produces a smaller1The finite sample perforamcen of the prewhitening HAC estimator is heavily depending on the small sample

bias of the AR(1) coefficient in the prewhitening stage. Andrews (1992) suggests ‘the 0.97 rule’ that the estimate

of AR(1) coefficient is replaced by 0.97 when it is greather than 0.97.

3

variance than the least square dummy variables (LSDV) estimator, the pooled panel unit root

test (like the LLC type test) is more powerful than the meta test (like IPS, MW or Choi (2002)

). In section 4, a recursive mixed panel unit root test is proposed and its asymptotic properties

are studied. The cross sectional average of the second moment of the recursive mean with a

mixed panel grows very fast at order O¡T 3¢as the time series observations increase, while

that with a pure stationary panel grows less fast at order O(T ). I utilize this information to

construct a simple recursive mixed panel unit root test. Section 5 discuss the testing strategy

under cross section dependence and provides solutions. Section 6 shows the results of Monte

Carlo simulation. Section 7 provides a short application of purchasing power parity over 21

OECD countries and concludes.

2 Models and Unit Root Tests with Recursive Mean Adjust-ment

2.1 Models

The autoregression models considered in the paper fall into the following two categories:

M1: (Unknown Constant)

(yit = ai (1− ρi) + ρiyit−1 + εit

yit = ai + xit, xit = ρixit−1 + εit

M2: (Linear Trends)

(yit = ai (1− ρ) + biρi + bi (1− ρi) t+ ρiyit−1 + εit

yit = ai + bit+ xit, xit = ρxit−1 + εit

where t (t = 1, .., T ) indexes the time series observations while the index i (i = 1, ..., N) stands

for the ith cross sectional unit. The regression error εit is a stationary process. In the unit

root cases, the initialization of xit is taken to be xi0 = Op (1) and is uncorrelated with {εit}t≥1.

2.2 Univariate Unit Root Tests with Recursive Mean Adjustement

We consider the univariate unit root test (N = 1). The limiting distribution is obtained by

letting T → ∞. Shin and So (1999) originally introduce recursive mean adjustment in au-

toregressions originally to reduce the small sample bias of least square estimators. Later Shin

and So (2001, 2002) extend their recursive mean adjustment to unit root test for the case of

unknown mean. Phillips, Park and Chang (2002) study more extensively the properties of

recursive mean adjustment, while Sul, Phillips and Choi (2003) provide the reason why recur-

sive mean adjustment reduces the small sample bias and show that the recursive detrending

proposed by Shin and So (2002) does not work properly. To fix ideas, let consider M1 with iid

4

εit and define recursive mean

yt−1 =1

t− 1t−1Xs=1

ys

and rewrite the regression of M1 in deviation from mean form:

yt − yt−1 = a (1− ρ) + ρ (yt−1 − yt−1)− (1− ρ) yt−1 + εt (1)

From M1,

(1− ρ) yt−1 = a (1− ρ) + (1− ρ) xt−1. (2)

Plugging (2) to (1) yields

yt − yt−1 = ρ (yt−1 − yt−1) + t (3)

where t = − (1− ρ) xt−1 + εt. When ρ = 1, t = εt. Since the regression error t does not

contain overall mean of yt−1, the small sample bias of ρ is signficantly reduced.This principle, however, cannot directly apply to the case of linear trend. Phillip, Park and

Chang (2002) propose a recursive detrending method to make the regression error become a

martingale difference sequence. Their detrending estimator suffers from serious upward bias

when ρ < 1. Here we provide a new detrending method to reduce the small sample bias. For

M2 with N = 1, observe this:

2yt−1 =2

t− 1t−1Xs=1

ys = 2a+ b (t− 1) + 2xt−1,

yt−1 − 2yt−1 = −a+ (xt−1 − 2xt−1)

yt − 2yt−1 = a (1− ρ) + bρ+ b (1− ρ) t+ ρ (yt−1 − 2yt−1)− 2 (1− ρ) yt−1 + εt

= −a (1− ρ) + bρ+ ρ (yt−1 − 2yt−1)− 2 (1− ρ) xt−1 + εt

The trend is eliminated but the constant is still present. Taking an overall mean adjustment

yields

yt − y − 2 (yt−1 − µ) = ρ [yt−1 − y−1 − 2 (yt−1 − µ)] + (ut − u)

where µ = T−1P

yt−1, y−1 = T−1P

yt−1 and ut = −2 (1− ρ) xt−1 + εt. This procedure

reduces the small sample bias significantly. Table 1 shows the dramatic bias reduction from

the recursive mean adjustment. It is worthywhile noting that the variance of the recursive

mean adjusted (RMA) estimator is far less than that of the OLS estimator, especially for the

case of the linear trend. We also investigated whether or not the recursive mean adjustment

works with a general AR(p) specification by means of Monte Carlo simulation and found that

the proposed new estimator works very well.2 In the next section, we provide explicit bias

formulae for RMA estimators.2The simulation results are not reported here to save the space but will be available upon request of the

author.

5

As Table 1 revealed, the relative variance of the RMA estimator compared to that of the

OLS estimator decreases as T increases. This useful fact can be used for testing unit roots.

Shin and So (2001) already proposed a univariate unit root test based on recursive mean

adjustment for the case of an unknown constant. Here we complete their task by adding the

case of a linear trend. Let

wct = yt − yt−1, wτ

t = yt − 2yt−1 − y + 2µ

wct−1 = yt−1 − yt−1, wτ

t−1 = yt−1 − 2yt−1 − y−1 + 2µ

and consider the RMA estimator of ρ in the following general autorregressions

wct = ρwc

t−1 +pX

j=1

φj∆yt−j + ut for M1

wτt = ρwτ

t−1 +pX

j=1

φj (∆yt−j −∆y−j) + ut for M2

where ∆y−j is the sample mean of ∆yt−j . The RMA ADF tests are given by

tr =ρ− 1se (ρ)

where se (ρ) is the standard error of ρ. The limiting distribution of the test statistics under

the local to unity alternative ρ = 1 + c/T follows as

Proposition 1 Assume ut is a sequence of i.i.d. errors with E(u1) = 0 and σ2 = E¡u21¢<∞.

Then For M2,

tr =⇒·Z 1

0

hJ − J

idW

¸ ·Z 1

0

³J − J

´2dr

¸−1/2+ cΦ

(·Z 1

0

³J − J

´2dr

¸1/2+

·Z 1

0

³J − J

´µJ −

Z 1

0Jdr

¶dr

¸ ·Z 1

0

³J − J

´2dr

¸−1/2)where

Φ =

Ã1−

p−1Xi=1

φi

!−1, J = J (r) =

Z r

0ec(r−s)dW (s)

J = J (r) = r−1Z r

0J (s) ds, J = 2J −

Z 1

0J (s) ds+ 2

Z 1

0Jdr

Here W (r) is a standard Brownian motion on [0, 1] while J (r) is the Ornstein-Uhlenbeck

process. The proof of Proposition 1 is straightforward, hence it is omitted.3 The critical values3For M1, Shin and So (2001) provided the limiting distribution of the test statistics given by t =⇒hR 10

£J − J

¤dW

i hR 10

¡J − J

¢2dri−1/2

+ cΦ

½hR 10

¡J − J

¢2dri1/2

+hR 10

¡J − J

¢Jdr

i hR 10

¡J − J

¢2dri−1/2¾

6

of tr with c = 0 (for testing the unit root hypothesis) are given in Table 2.4

Table 1: Bias, variance and mean square error (MSE) of RMA and OLS estimators.

Constant Linear Trend

(A) (B) (C) (D) (A) (B) (C) (D)

ρ=0.92,T=30 -0.046 -0.144 0.97 0.51 -0.012 -0.254 0.68 0.28

ρ=0.94,T=30 -0.050 -0.149 0.95 0.49 -0.023 -0.265 0.63 0.26

ρ=0.96,T=30 -0.054 -0.155 0.93 0.47 -0.036 -0.277 0.60 0.25

ρ=0.98,T=30 -0.060 -0.161 0.93 0.46 -0.051 -0.292 0.56 0.24

ρ=1.00,T=30 -0.065 -0.167 0.92 0.44 -0.068 -0.309 0.52 0.24

ρ=1.00,T=50 -0.040 -0.104 0.85 0.43 -0.042 -0.193 0.45 0.22

ρ=1.00,T=70 -0.029 -0.075 0.82 0.42 -0.030 -0.141 0.40 0.20

ρ=1.00,T=100 -0.020 -0.053 0.81 0.41 -0.022 -0.100 0.39 0.19

ρ=1.00,T=200 -0.010 -0.027 0.77 0.41 -0.011 -0.051 0.36 0.18

Note: (A) =Bias of recursive mean adjusted estimator; (B) = Bias of OLS estimator; (C) =

Variance ratio of recursive mean adjusted estimator to OLS estimator; (D) = MSE ratio of

recursive mean adjusted estimator to OLS estimator.

Table 2: Critical values for RMA ADF test

1% 2.5% 5% 10% 20%

T Constant

25 -2.664 -2.257 -1.925 -1.556 -1.126

50 -2.593 -2.223 -1.906 -1.549 -1.132

100 -2.561 -2.200 -1.898 -1.549 -1.135

250 -2.541 -2.189 -1.887 -1.545 -1.135

500 -2.531 -2.185 -1.887 -1.545 -1.135

Linear Trend

25 -2.700 -2.278 -1.930 -1.545 -1.098

50 -2.603 -2.212 -1.889 -1.523 -1.092

100 -2.553 -2.184 -1.873 -1.518 -1.093

250 -2.529 -2.170 -1.862 -1.513 -1.092

500 -2.518 -2.159 -1.857 -1.509 -1.092

4The local asymptotic power comparison with the ADF tests can easily be done. To save space as well as to

maintain the focus of the paper, we don’t report the results here but note that the local asymptotic power with

RMA is much better than that with ADF tests.

7

3 Panel Unit Root Test

In this section, the panel unit root tests based on the RMA estimator are proposed under cross

sectional independence. Two types of the RMA panel unit root tests are considered. The first

test is similar to Choi (2001) and Maddala and Wu (1999)’s Meta analysis. The Fisher test

statistics can be deriven directly from the univariate RMA unit root tests developed in the

previous section. Let pi denote the probability value of tr statistics of the following regressions:

wcit = ρiw

cit−1 +

pXj=1

φij∆yit−j + uit for M1 (4)

wτit = ρiw

τit−1 +

pXj=1

φij (∆yit−j −∆yi−j) + uit for M2 (5)

The null hypothesis of the panel unit roots is given by

H1o : ρi = 1 for all i

while the alternative is

H1A : ρi < 1 for some i

The RMA Fisher test statistic distribution is given by:

R = −2NXi=1

ln pi ∼ χ22N (6)

Here we consider only that the recursive mean adjusted Fisher test because Choi (2001) and

Phillips and Sul (2003) report that the Fisher test performs best among other Meta statistics

performs.

The second test is the pooled RMA test. Even when ρi 6= ρ, by pooling cross sectional

information, the pooled RMA unit root test provides a greater rejection rate than the RMA

Fisher test. Further detail will be provided in Section 6. Here we study the N−asymptoticproperties of the pooled RMA estimator for the case of ρi = 1 for the case of AR(1). Consider

panel AR(1) regressions are given by

wcit = ρwc

it−1 + uit for M1,

wτit = ρwτ

it−1 + uit for M2.

Denote

ρpr =

PNi=1

PTt=1w

citw

cit−1PN

i=1

PTt=1

¡wcit−1

¢2 for M1

ρpr =

PNi=1

PTt=1w

τitw

τit−1PN

i=1

PTt=1

¡wτit−1

¢2 for M2

8

and their t−statistics aretpr =

ρpr − 1se¡ρpr¢ .

Following Harris and Tzavalis (1999), as N →∞ for fixed T, we have

Proposition 2 The probability limit of the pooled recursive mean adjusted estimator underthe null hypothesis of panel unit root is given by

plimN→∞¡ρpr − 1

¢= 0, for M1 and M2

andρpr − 1qV ar

¡ρpr¢ →L N (0, 1)

where V ar¡ρpr¢= σ2u

³PNi=1

PTt=1

¡wkit−1

¢2´−1for k = c and τ for constant and linear trend,

respectively.

Appendix A provides the proof of Propostion 2 but it is instructive to sketch its ouline here.

For fixed effects case, it is easy to see why the pooled RMA estimator is consistent. Under the

null, we have

plimN→∞1

N

NXi=1

TXt=1

yit−1uit = plimN→∞1

N

NXi=1

TXt=1

"Ã1

t− 1t−1Xs=1

yis

!uit

#= 0

since Eyit−juit = 0 for j > 0 as long as uit is not serially correlated. For the linear trend case,note that

plimN→∞1

N

NXi=1

ÃTXt=1

yit−1

!ÃTXt=1

uit

!

= 2× plimN→∞1

N

NXi=1

ÃTXt=1

Ã1

t− 1t−1Xs=1

yis

!!ÃTXt=1

uit

!

as long as uit is not serially correlated. For panel AR(p) regressions, the above results hold

as long as the enough augumented terms are included in the regressions. Choi, Mark and Sul

(2004) find that when ρ < 1, the pooled RMA estimator is upward biased and its inconsistency

is given by ρ¡T−1 lnT

¢with moderately large T.5 For the linear trend case, also the pooled

RMA estimator is slightly upward biased when ρ < 1 but its bias vanishes rapidly as T

increases.5The expression of the approximation is O(T−1 lnT ) which is greater than O

¡T−1

¢. However, the actual

bias in absolute value is far less than that of LSDV with fixed effects. See Choi, Mark and Sul (2004) for exact

biases formulae.

9

It is worthywhile noting that the pooled RMA estimator is more efficient than the pooled

mean unbiased estimator proposed by Harris and Tzavalis (1999) and Phillips and Sul (2004).

Let ρc be the pooled mean unbiased estimator. That is,

ρc = ρ+ bias (1, T )

where ρ is the pooled OLS estimator and bias (1, T ) is the mean bias function provided by

Harris and Tzavalis (1999) and Phillips and Sul (2004). For the case of unknown constant,

bias (1, T ) = 3/T while for the case of the linear trend, it becomes 7.5/T with moderately

large T. The variance of the mean unbiased estimator ρc is larger than the variance of ρprasymptotically. Harris and Tzavalis (1999) provide the asymptotic variance of ρc under the

assumption of normal error of εit which is given by

plimN→∞V (ρc) =

3(17T 2+14T+14)

5T (T+2)3for M1

15(193T 2−728T+1147)112(T+2)3(T−2) for M2

As T →∞, the variance ratio is given by

limT→∞

"plimN→∞

V (ρc)

V¡ρpr¢# ' ( 3.12 for M1

2.87 for M2

For small T, the exact asymptotic variance ratios for panel AR(1) model are plotted in Figure

1. Even in small T (lets say T > 15), the pooled RMA estimator is more efficient than the

pooled mean unbiased estimator for both the fixed effects and the incidental linear trend.

Maddala and Wu (1999) and Karlsson and Löthgren (2000) report that the IPS and the

Fisher tests reject the null of panel unit roots than the LLC test under the alternative more

often. One of the reasons is that the variance of the pooled mean unbiased estimator is

hampered by the biases. In other words, the variance of the pooled mean unbiased estimator

is always larger than the variance of the LSDV estimator which results in the power loss. For

the pooled RMA estimator, this is no longer true. Even when ρi 6= ρ under the alternative,

the power of the pooled RMA estimator is much better than that of the Fisher test considered

in (6). Figure 2 shows the results of a Monte Carlo simulation with T = 25 of which ρi is

generated from U (0.9, 1.0) where U stands for the uniform distribution. Note that the Fisher

test based on the univariate RMA-ADF test is superior to the Fisher test based on ADF test

in terms of power as well as size. Moreover, the pooled RMA test is more powerful than the

Fisher test based on RMA-ADF test. See other Monte Carlo results later in the paper for a

more detail comparison.

10

0

1

2

3

4

0 50 100 150 200T

Varia

nce

ratio

Fixed Effects

Incidental Trend

Figure 1: Asymptotic variance ratio of the pooled mean unbiased estimator to the pooled

recursive mean adjusted estimator for panel AR(1) case.

4 Mixed Panel Unit Root Test

Consider the following two null hypotheses.

H1o : ρ = 1 for all i H1

A : ρi < 1 for some i

H2o : ρi = 1 for some i H2

A : ρi < 1 for all i

If the first null is not rejected, then there is no need to consider the second hypothesis as long

as the test under the first null hypothesis does not suffer from a size distortion. When the

first null hypothesis of the panel unit root is rejected, the mixed panel becomes an important

issue. Since the first alternative H1A implies that there are some stationary units in the panel,

the rejection of the first null naturally leads to test the second null hypothesis.

This section provides the asymptotic behavior of the pooled OLS estimator where a panel

consists of nonstationary and stationary units and proposes a convienient test for mixed panel

unit root test under the second null hypothesis.

4.1 Properties of a Mixed Panel

The main feature of a mixed panel is that the dominant root converges to unity as T increases.

Let α be the percentage of stationary units in the panel. We further assume that the fraction

11

0

0.2

0.4

0.6

0.8

1

0 12 24 36 48 60 72 84 96N

Pow

er

Pooled Test for M1

Meta Test for M1

Pooled Test or M2

Meta Test for M2

Figure 2: Power Comparison: pooled v.s. Meta test under heterogeneity of ρi with ρi −U (0.9, 1.0) and T = 25.

of stationary units does not depend on the size N. In other words, the number of stationary

units, αN or nonstationary units, (1− α)N goes to infinity as N → ∞. For simplicity, we

assume the cross sectional homogeneity: That is ρi = ρ for all i.

The probability limit of the pooled OLS estimator with the mixed panel approaches unity

as T increases. To fix ideas, consider the case of a known constant and let σ2 = 1 for notational

convinience. Then the probability limit of the pooled OLS estimator for fixed T is given by

plimN→∞ρNT =α ρ1−ρ2 + (1− α) T+12

α 11−ρ2 + (1− α) T+12

= 1−(1− ρ)α 1

1−ρ2α 11−ρ2 + (1− α) T+12

(7)

= 1− c0Tfor moderately large T

where

c0 =(1− α) (1 + ρ)

2α.

Here we assume the variance of the innovation error is identical for both stationary and non-

stationary cases. Otherwise, the definition of α should be changed to account for the variance

ratio of the innovation error of the stationary units to that of nonstationary units. Nonethless,

the probability limit of the pooled estimator has the features of a local to unity. As T increases,

the nonstationary units dominate the stationary units, which results in the pooled OLS es-

12

timator approaching unity. Moon and Phillips (2002, 2004) provide the estimation method

for the local unity parameter c in the case of a homogeneous panel by utilizing generalized

method of moments. If the panel is known to be mixed, then the local unity parameter c can

be estimated.

However it is hard to detect whether or not a panel is mixed, especially under fixed effects

or incidental linear trend case. As Nickell (1987) and Phillips and Sul (2004) show, the pooled

OLS etimator suffers from a serious downward bias but its bias is decreasing over time. Let

ρNt be the pooled OLS estimator by using panel data upto the time t. For the case of fixed

effects, with all stationary units, the probability limit of ρNt is given by

plimN→∞ρNt = ρ− 1 + ρ

t+O

¡t−2¢

while with a mixed panel, it becomes

plimN→∞ρNt = 1−c1t+O

¡t−2¢

where

c1 = 3α− 2µ1− α

α

¶µ1− 5ρ+ ρ2

1 + ρ+ ρ2 (1− ρ)

¶> 0

In practice, the true parameters ρ and c1 are unknown. Moreover, as more time series obser-

vations are available, the probability limit of the pooled OLS estimate with a pure stationary

panel is increasing over time and that with a mixed panel as well.

The pooled recursive mean adjusted estimator provides asymptotically clear distinct be-

tween two panels. For the case of pure stationary panel, the pooled recursive mean adjusted

estimator is upward biased but its bias is decreasing over t.With moderate large t, the asymp-

totic bias can be approximed as

plimN→∞ρNt = ρ+ ρt−1 ln t+O¡t−2¢> 0 for ρ > 0

In contrast, for the case of a mixed panel, the asymptotic bias of the pooled recursive mean

adjusted estimator is given by

plimN→∞ρNt = 1−µ

6

1 + ρ

¶µ1− α

α

¶1

t+O

¡t−2 ln t

¢This opposite relationship holds even for the linear trend case. With large T and N, it is

possible to distinguish whether or not a panel consists of pure stationary units by utilizing this

information. However, in finite N, it is rather ambiguous since the variance of ρNt is too large

to distinguish the mixed panel from the pure stationary panel.

13

4.2 Maximum Order Test

The clear difference between nonstationary and stationary panel is the different rate of growing

variance. To fix ideas, consider the case of known constant as follow.

yit = ρyit−1 + εit

From the direct calculation, we have

plimN→∞1

TN

NXi=1

TXt=1

y2it =

(σ2

1−ρ2 for stationary yitσ2

2 (T + 1) for nonstationary yit(8)

That is, the second moment of stationary yit is constant over time while that of nonstation-

ary yit is growing over time. Note that when ρi 6= ρ for stationary yit, the limit becomes

σ2³N−1PN

i=1

¡1− ρ2i

¢−1´ ≤ σ2¡1− ρ2

¢−1. Hence the order does not change at all. Define

ηit =1t

Pts=1 y

2is. The cross sectional mean of ηit will be varying over time when yit is not

stationary. However this information is not of much use when for testing panel unit roots espe-

cially with small T. With given N, the approximated confidence interval of the cross sectional

mean of ηit even with stationary yit is given by σ2/¡1− ρ2

¢ ± 2N−1/2. With fixed N, this

implies the cross sectional mean of ηit can move around σ2/¡1− ρ2

¢over time. Hence in order

to figure it out whether or not yit is stationary, one needs a very large T dimension.

Alternatively, the second moment of recursive mean provides much more helpful informa-

tion. From the direct calculation, we have

plimN→∞1

N

NXi=1

Ã1

t

tXs=1

yis

!2=

(σ2

(1−ρ)2 t−1 +O

¡t−2¢

for stationary yitσ2

3 t+O (1) for nonstationary yit(9)

Define κit =¡1t

Pts=1 yis

¢2. It is worthywhile noting that the maximum order of κit does not

change even under the ARMA(p,q) case. This is because yit is Op

¡t1/2

¢and κit is Op

¡t−1¢

when yit is stationary, while yit is Op

¡t3/2

¢and κit is Op (t) when yit is nonstationary. Hence

the cross sectional mean of κit at time t is decreasing over t when κit is stationary while it is

increasing when κit is nonstationary. We utilize this finding to develop a maximum order test

to detect whether or not a panel consists of nonstationary units. First consider the unknown

mean case.

Fixed Effects Take another recursive mean of κit over t and denote it as zit. Take cross

sectional average and denote it as zNt. As N →∞, the following proposition holds.

14

Proposition 3 Assume that plimN→∞ 1N

PNi=1 a

2i = µ2a < ∞ where ai is unknown constants.

For every t, as N →∞, the probability limits of zNt is given by

plimN→∞zNt = plimN→∞1

N

NXi=1

tXs=1

Ã1

s

sXk=1

yikyi1

!2=

(ln t+O (1) for all I(0)

16

¡t2 + 2t+ ln t

¢+O (1) for all I(1)

,

Appendix B provides detail proof of Proposition 3. From (9), for ARMA(p,q) cases, it is

obvious that the recursive sum of κit is bounded by Op (ln t) as long as yit is stationary while

that of κit is bounded by Op

¡t2¢as long as yit is nonstationary. Here zNt is normalized by

the initial value of yi1 to make it be independent of variance of εit. Note that the maximum

order does not change even when ρi 6= ρ with stationary κit. The issue here is rather the time

varying behavior of zNt with finite T, especially when ρi is near unity. Since zNt is increasing

over time either when yit is I (0) or I (1) , it is necessary to divide it by tβ to distinguish

between stationary and nonstationary units when β ≤ 1.6 Let SNt = zNt × t−β and considerits probability limit.

St = plimN→∞zNt × t−β

Figure 3 shows time varying properties of St with β = 0.9. In finite T, the time varying pattern

of St can be summarized as follows. Firstly, for fixed positive value of β ≤ 1, St has a humpshape. Let t∗ be the time t when St reaches at its maximum value. As ρ → 1, t∗ → ∞.

Secondly, for fixed value of ρ, as β increases, t∗ approaches unity.Consider the following null hypothesis of mixed panel unit root

H2o : ρi = 1 for some i

H2A : ρi < 1 for all i

Under the null hypothesis of a mixed panel unit root, the maximum order of St is O¡t2−β

¢as

long as the fraction of nonstationary units (1− α) > 0. Testing the above null hypothesis is

straightforward. Consider the following simple linear regression

SNt = d1 + d2t+ et, t = t0, ..., T (10)

where SNt is the cross sectional average of zNt× t−β. Note that t starts from t0 rather than the

initial observation due to hump shape of SNt. Under the null hypothesis of H2o , as N → ∞,

the sign of d2 must be positive, while under the alternative, they must not be significantly

positive. Hence the null hypothesis of H2o can be rewritten as

H2o : d2 ≥ 0

6Here β is bounded by unity rather than two. The regression in (10) provides the reason clearly. As T →∞,

if β > 1, then the probability limit of d2 goes to zero.

15

0.2

0.4

0.6

0.8

1

1.2

1.4

0 25 50 75 100T

0.8

0.9

0.95

0.98

0.99

Figure 3: Probability limit of St with β = 0.9 for AR(1) case.

Note that as T → ∞, under the null hypothesis of H2o , the t−statistic of d2 converges to a

normal distribution when β = 1 while plimT→∞d2 = +∞ when β < 1. To reduce the size

distortion of the test, we set β = 0.9 but use the normal table to test the null hypothesis of

d2 ≥ 0. It is necessary since the size distortion could be huge with a large fraction of stationaryunits α.

Incidental Linear Trends For the case of linear trend, define qNt as

qNt =1

N

NXi=1

tXs=1

yisyi1− 2

s

sXq=1

yiqyi1

2For fixed t, as N →∞, the probability limit of qNt is given by

Proposition 4 Assume that plimN→∞ 1N

PNi=1 b

2i = µ2b <∞ where bi is unknown trend coeif-

ficient. For every t, as N →∞, the probability limits of qNt is given by

plimN→∞qNt = plimN→∞1

N

NXi=1

tXs=1

yisyi1− 2

s

sXq=1

yiqyi1

2

=

(5

1−ρ2 tσ2 +O (1) for all I(0)¡

118

¢σ2t3 +O (1) for all I(1)

. (11)

16

0.5

0.75

1

1.25

1.5

1.75

2

2.25

0 25 50 75 100T

Prob

abilit

y lim

it of

Qt

0.8

0.9

0.95

0.98

0.99

Figure 4: Probability limit of Qt with γ = 1.8 for AR(1) case

See Appendix B for the detail proof of Proposition 4. Comparing to the fixed effects case,

qNt is O(t) when yit is stationary.7 Let QNt = qNt × t−γ and consider its probability limit.

Qt = plimN→∞qNt × t−γ

Note that γ must be greater than unit but less than two. Figure 4 shows time varying properties

of Qt with γ = 1.8. Compared to Figrue 3, the maximum time t∗ for the linear trend case ismuch greater than that for the constant case. This is because qNt is Op (t) rather than Op (ln t)

when yit is stationary. Of course, an increasing value of γ results in smaller t∗ but it also affectsthe size of the test.

The following regression is considered to test the null hypothesis of mixed panel unit root.

QNt = f1 + f2t+ vt (12)

Similar to (10), as T → ∞, plimT→∞f2 = +∞ with γ < 2. If γ = 2, then the t−statistics off2 follows a normal distribution asymptotically. To secure the size of the test, we set γ = 1.8

which is equal to 2×β.7Readers may wonder why we didn’t take additional recursive mean over qNt. Note that (t)

−2 qNt ≈ 5×κNt

with ρ < 1 while (t)−2 qNt ≈ κNt/6 with ρ = 1. Taking additional recursive mean of qNt yields 5×zNt for ρ < 1

but it becomes zNt/12. Hence in the finite T, the performance of the recursive mean of qNt becomes worse that

that of qNt.

17

5 Sequential Panel Unit Root Tests under Cross Sectional De-pendence

Bai and Ng (2002), Forni, Hallin, Lippi and Reichlin (2000), Moon and Perron (2002), and

Phillips and Sul (2003) study parametric structures of cross section dependence. Phillips and

Sul (2004) distinguish macro panel data from micro panel data in the terms of the number

of common factors. Usually macro panel data is cross sectionally aggregated data so that the

number of common factor is limited whereas the micro panel data might have a number of

common factors.

There are two types of factor models in nonstationary panel literature. Bai and Ng (2002)

consider a direct factor structure for the data of the form

yit =KXs=1

λisFst +mit. (13)

where Fst is the s − th common factor while λis is the factor loading coefficient for the s −th common factor. Usually the idiosyncratic term mit is assumed to be cross sectionally

independent. Bai and Ng (2002) proposed how to estimate the number of factors in panel

data, while more recently Bai and Ng (2004) have suggest how to estimate Fst from yit. Once

the idiosyncratic term mit is subtracted from yit, the mixed panel unit root as well as the panel

unit root tests proposed in the previous section can be directly applied.

The second model is studied by Forni, Lippi and Reichlin (1999), Moon and Perron (2002),

and Phillips and Sul (2003). In all these studies, common factors enter into the regression

errors.

yit = ai + ρiyit−1 + uit, uit =KXs=1

δisθst + εit, (14)

where the errors uit depend on K factors {θst : s = 1, ...,K} with factor loadings {δis : s =1, ...,K}, and εit is assumed to be iid(0, σ2i ). Note that the first factor model is more general

than the second factor model in the sense that Fst can have a different serial correlation

structure to mit.

However, these two models can be complimentary to each other when mit follows the com-

mon factor structure like (14). For example, all panel empirical studies for growth convergence

are based on the following simple linear specification

yit = µt + yoit (15)

where yit is an individual per capital income of the i−th economy, µt is the common growthcomponent and yoit is the deviation between yit and µt. The deviation part, y

oit usually has

cross sectional dependence which can be modelled by (14). This specification is rather like a

combination between (13) and (14). That is, mit in (13) has the common factor structure like

18

(14). Nonethless when the cross sectional dependence is best modelled as (13), the mixed panel

unit root tests developed in the previous section can be directly applied to the estimated mit

as Bai and Ng (2004) suggest. Otherwise, the orthogonalization method proposed by Phillips

and Sul (2003) is required to eliminate factor loading coefficients δis. With orthogonalized

panel data, the mixed panel unit root test can be carried on. The latter procedure is rather

complicated. Here I provide step by step procedure how to perform sequential panel unit root

tests.

Step 1 Obtain an individual RMA estimator ρri from (4) for M1 and (5) for M2. Denote the

residual εit

εit = yit − ρriyit−1 −pX

j=1

φij∆yit−j for M1

εit = yit − ρriyit−1 −pX

j=1

φij (∆yit−j −∆yi−j) for M2

If ρri ≥ 1, then set ρri = 1. Note that εit has non-zero mean for M1 and has non-zero

mean and linear trend for M2 when ρ < 1.

Step 2 If ρri < 1, demean εit for M1 or detrend εit for M2. If ρri ≥ 1, then demean εit for

M2. Construct sample variance and covariance matrix Σε.

Step 3 Construct the orthogonal complement matrix Fδ based on Σε. Transform the data yitby premultiplying Fδ and denote y

+it = Fδyit.Note that y

+it is cross sectionally independent

as T →∞.

Step 4 Perform either the Fisher test with RMA-ADF or the pooled test where the null

hypothesis is ρi = 1 for all i. If the null cannot be rejected, then all units in the panel

are I(1). If the null can be rejected, then next perform the maximum order test. The

null hypothesis is ρi = 1 for some i.

The above 4-step procedure provides a consistent test either under the null hypothesis or

under alternative as T → ∞. This is because the sample (or intitial) convariance matrix isestimated under the assumption of heterogeneity of ρi Note that the regression error εit 6= uit

under the alternative. Hence to construct consistent estimator of Σε, the ordinary demeaning

or detrending is needed after obtaining εit in Step 1. When the null hypothesis of panel unit

roots H1o is rejected, the maximum order test under cross sectional dependence can be carried

out using y+it .

19

6 Simulation Experiments

We considered a number of Monte Carlo experiments with various data generating processes.

Since the results of Monte Carlo studies are similar over different data generating processes,

only results for two sets of Monte Carlo experiments are reported here.8

6.1 Design of Data Generating Process

The data generating process is given by

yit = ai + bit+ xit, xit = ρixit−1 + uit, (16)

uit = δiθt + εit (17)

or equivalently

yit = a+i + βit+ ρiyit−1 + uit

where a+i = ai (1− ρi) + biρi, βi = bi (1− ρi) while εit ∼ iid N (0, 1) over i and t, θt ∼ iid

N (0, 1) over t, and for (ρi, δi) parameter selections that are detailed below. The primary

distinction is between the homogeneous case where ρi = ρ for all i and the heterogeneous

case where ρi differs across individuals i. For cross section independence case, δi = 0 while

for cross section dependence case, δi ∼ U [1, 3] where U is uniform distribution. Here the lag

length is assumed to be known. Panel data are generated under four specifications which differ

according to their degree of the cross sectional dependence and whether or not the homogeneity

restriction is imposed on ρ. These specifications are as follows:

Case I: (AR(1) under No Cross-sectional Dependence) For the case of homogeneity, we setρi = 0.9. For the case of heterogeneity, we set ρi ∼ U [0.85, 0.95] for M1 while ρi ∼U [0.80, 0.90] for M2.

Case II: (AR(1) under High Cross-sectional Dependence) We set δi ∼ U [1, 3].

Each experiment involves 5,000 replications of panel samples of (N, T ) observations. Recall

that α is the fraction of stationary units. I consider five different values of α = (1, 3/4, 1/2, 1/4, 0) .

We set T = 30, 50 and 100 while N = 20, 40 and 80 so that when α = 3/4, the number of non-

stationary units become 5, 10 and 20 forN = 20, 40 and 80 respectively. It is worthy noting that

asymptotically the true value of the pooled estimators with mixed panels depend on the size

of T. From the formulae in (7), these values with T = 30, 50 and 100 are (0.990,0.994,0.997),

(0.975,0.983,0.991) and (0.975,0.983,0.991) for α = 3/4, 1/2 and 1/4.

The initial starting point t0 must be selected for testing the null hypothesis of mixed

panel unit root. Theoretically with large T, the initial starting point t0 could be the initial8Other Monte Carlo results are available upon request of the author.

20

observation. With finite T, the optimal starting point t0 must be the maximum values of Stand Qt to increase the power of the mixed panel unit root test and to reduce the size distortion

as well. The problem is that the maximum values of St and Qt are dependent of ρ. As it is

shown in Figure 3 and 4, the maximum value t∗ becomes the last observation T when ρ is nearunity with small T . Note that the power of the panel unit root test with small T under H1

o is

also very poor when ρ is near unity. Also it is meaningless to test the null hypothesis of mixed

panel unit root, H2o when the power of the test under H

1o is poor. With extensive numerical

simulations based on the exact asymptotic formulae of St and Qt, we set t0 = 25.

6.2 Simulation Results

Table 3 consists shows the finite sample performances of three panel unit roots tests. They are

the Fisher test based on the DF-GLS, the Fisher test based on RMA-DF and the pooled RMA

test under no cross section dependence. The rejection rate is evaluated based on 5% test. All

rejection rates are not size adjusted. As it is shown in Figure 2, the power of the pooled RMA

test is better than those of the two Fisher tests for both homogeneity and heterogeneity of ρi.

More interestingly, the rejection rates increase as N and T increase even when α = 1/4.

Table 4 reports the rejection rates of the maximum order test and the sequential mixed

panel unit root tests under cross section dependence. For the case of the maximum order test,

the power increases as either T or N increases. However, the size of the test decreases as N

increases for fixed T . Also the size of the test increases as the propotion of stationary units

(α) increases. In contrast, the sequential test combining with the pooled RMA and maximum

order tests provides reduces the size distortion significantly without much sacrificing the power

of the test for the moderately large T.

7 Application and Concluding Remarks

We apply the sequential mixed panel unit roots test for long run purchasing power parity. The

data used in the paper covers the period 1948-1998 for 21 OECD countries was taken from the

International Financial Statistics. The series involved annual price indices for each country

and real exchange rates calculated from the individual national price indices and the end of

the period spot exchange rates. We use BIC to determine the appropriate lag and end up with

AR(1) specification.

Table 5 reports the results. The Fisher test based on RMA-ADF rejects the null hypothesis

of panel unit roots for all numeraire currencies, while the pooled RMA test rejects 19 out of

21 numeraire currencies. Hence the first null H1o or H

0o is rejected fairly strongly. Next, we

test the null hypothesis of mixed panel unit roots. The maximum order test cannot reject

the null hypothesis for all numeraire currencies that there are some nonstationary units in the

21

panel. Therefore if real exchange rates don’t have any nonlinear adjustment, they are not all

stationary.

This paper provides the sequential mixed panel unit roots test and applies the test to test

long run purchasing power parity. We found that long run purchasing power parity does not

hold for all bilateral currencies. Since the proposed mixed panel unit roots test is constructed

based only on linear enviroment, threshold type mixed panel unit roots tests will be promising.

22

Table 3: Rejection Rates of the Fisher Test based on DF-GLS, RMA-ADF and the pooledRMA test.

AR(1) with no cross section dependence

— Part A: Cases of Homogeneity ρ

Sample Constant Linear Trend

α=1 α=34 α=1

2 α=14 α=0 α=1 α=3

4 α=12 α=1

4 α=0

Fisher Test based on ADF-GLS

T=30,N=20 0.76 0.55 0.32 0.15 0.05 0.17 0.13 0.10 0.07 0.05

T=30,N=40 0.96 0.83 0.52 0.22 0.05 0.28 0.20 0.13 0.08 0.05

T=30,N=80 1.00 0.98 0.80 0.33 0.05 0.44 0.30 0.18 0.10 0.05

T=50,N=20 1.00 0.95 0.71 0.28 0.04 0.57 0.39 0.22 0.11 0.04

T=50,N=40 1.00 1.00 0.94 0.45 0.05 0.85 0.64 0.37 0.15 0.04

T=50,N=80 1.00 1.00 1.00 0.72 0.04 0.99 0.88 0.60 0.23 0.04

Fisher Test based on RMA-ADF

T=30,N=20 0.77 0.55 0.32 0.14 0.05 0.19 0.14 0.10 0.07 0.04

T=30,N=40 0.97 0.83 0.53 0.22 0.05 0.31 0.23 0.14 0.08 0.05

T=30,N=80 1.00 0.98 0.80 0.32 0.04 0.51 0.35 0.21 0.10 0.05

T=50,N=20 1.00 0.95 0.69 0.27 0.05 0.58 0.40 0.23 0.11 0.05

T=50,N=40 1.00 1.00 0.93 0.45 0.05 0.85 0.64 0.38 0.16 0.04

T=50,N=80 1.00 1.00 1.00 0.71 0.04 0.99 0.88 0.60 0.23 0.04

Pooled RMA Test

T=30,N=20 0.96 0.66 0.32 0.12 0.03 0.24 0.15 0.09 0.05 0.03

T=30,N=40 1.00 0.92 0.55 0.17 0.03 0.45 0.27 0.15 0.08 0.04

T=30,N=80 1.00 1.00 0.82 0.27 0.03 0.72 0.46 0.23 0.09 0.03

T=50,N=20 1.00 0.91 0.54 0.17 0.03 0.75 0.46 0.23 0.09 0.03

T=50,N=40 1.00 1.00 0.79 0.27 0.03 0.97 0.75 0.39 0.13 0.03

T=50,N=80 1.00 1.00 0.97 0.44 0.03 1.00 0.96 0.66 0.21 0.03

23

Table 3: -Continue

— Part B: Cases of Heterogeneity ρ

Sample Constant Linear Trend

α=1 α=34 α=1

2 α=14 α=0 α=1 α=3

4 α=12 α=1

4 α=0

Fisher Test based on ADF-GLS

T=30,N=20 0.81 0.61 0.37 0.15 0.05 0.44 0.31 0.20 0.10 0.04

T=30,N=40 0.98 0.88 0.58 0.25 0.05 0.65 0.43 0.27 0.13 0.05

T=30,N=80 1.00 0.99 0.83 0.38 0.05 0.92 0.73 0.47 0.20 0.04

T=50,N=20 1.00 0.98 0.78 0.27 0.04 0.97 0.87 0.61 0.25 0.04

T=50,N=40 1.00 1.00 0.96 0.52 0.04 1.00 0.97 0.78 0.35 0.04

T=50,N=80 1.00 1.00 1.00 0.78 0.04 1.00 1.00 0.98 0.60 0.04

Fisher Test based on RMA-ADF

T=30,N=20 0.82 0.62 0.37 0.15 0.05 0.48 0.35 0.21 0.11 0.05

T=30,N=40 0.98 0.89 0.59 0.25 0.05 0.70 0.46 0.29 0.15 0.05

T=30,N=80 1.00 0.99 0.83 0.36 0.04 0.95 0.78 0.50 0.21 0.05

T=50,N=20 1.00 0.97 0.77 0.27 0.05 0.96 0.85 0.58 0.25 0.04

T=50,N=40 1.00 1.00 0.96 0.52 0.05 1.00 0.96 0.77 0.35 0.05

T=50,N=80 1.00 1.00 1.00 0.77 0.04 1.00 1.00 0.97 0.59 0.04

Pooled RMA Test

T=30,N=20 0.96 0.68 0.34 0.11 0.03 0.62 0.41 0.21 0.09 0.03

T=30,N=40 1.00 0.93 0.56 0.18 0.03 0.87 0.59 0.33 0.13 0.04

T=30,N=80 1.00 1.00 0.82 0.28 0.03 1.00 0.90 0.56 0.20 0.03

T=50,N=20 1.00 0.91 0.55 0.16 0.03 1.00 0.86 0.49 0.17 0.03

T=50,N=40 1.00 1.00 0.79 0.28 0.03 1.00 0.97 0.70 0.24 0.03

T=50,N=80 1.00 1.00 0.97 0.45 0.03 1.00 1.00 0.95 0.42 0.03

24

Table 4: Rejection Rates of Sequential Mixed Panel Unit Root TestAR(1) with high cross section dependence

—Part A: Homogeneity ρ

Sample Constant Linear Trend

Maximum Order Test

α=1 α=34 α=1

2 α=14 α=0 α=1 α=3

4 α=12 α=1

4 α=0

T=30,N=20 0.80 0.24 0.18 0.12 0.02 0.59 0.18 0.12 0.07 0.01

T=30,N=40 0.90 0.23 0.17 0.09 0.00 0.57 0.16 0.10 0.04 0.00

T=30,N=80 0.96 0.20 0.15 0.08 0.00 0.55 0.13 0.09 0.04 0.00

T=50,N=20 0.87 0.27 0.20 0.14 0.02 0.67 0.23 0.17 0.11 0.01

T=50,N=40 0.97 0.25 0.19 0.12 0.00 0.68 0.20 0.14 0.07 0.00

T=50,N=80 1.00 0.23 0.18 0.11 0.00 0.72 0.19 0.14 0.07 0.00

T=100,N=20 0.95 0.26 0.20 0.15 0.02 0.82 0.25 0.19 0.12 0.01

T=100,N=40 1.00 0.25 0.20 0.12 0.00 0.87 0.24 0.18 0.10 0.00

T=100,N=80 1.00 0.24 0.19 0.11 0.00 0.93 0.21 0.16 0.09 0.00

Sequential Panel Unit Roots Tests: The pooled RMA & Maximum Order Tests

T=30,N=20 0.74 0.04 0.01 0.00 0.00 0.12 0.01 0.00 0.00 0.00

T=30,N=40 0.90 0.06 0.01 0.00 0.00 0.21 0.01 0.00 0.00 0.00

T=30,N=80 0.96 0.06 0.01 0.00 0.00 0.35 0.01 0.00 0.00 0.00

T=50,N=20 0.87 0.06 0.01 0.00 0.00 0.38 0.01 0.00 0.00 0.00

T=50,N=40 0.97 0.07 0.01 0.00 0.00 0.60 0.01 0.00 0.00 0.00

T=50,N=80 1.00 0.08 0.02 0.00 0.00 0.71 0.01 0.00 0.00 0.00

T=100,N=20 0.95 0.07 0.01 0.00 0.00 0.82 0.03 0.00 0.00 0.00

T=100,N=40 1.00 0.08 0.01 0.00 0.00 0.87 0.03 0.00 0.00 0.00

T=100,N=80 1.00 0.09 0.03 0.00 0.00 0.93 0.04 0.00 0.00 0.00

25

— Table 4: Continue

—Part A: Heterogeneity ρ

Sample Constant Linear Trend

Maximum Order Test

α=1 α=34 α=1

2 α=14 α=0 α=1 α=3

4 α=12 α=1

4 α=0

T=30,N=20 0.72 0.27 0.21 0.12 0.02 0.62 0.25 0.19 0.10 0.01

T=30,N=40 0.76 0.23 0.16 0.10 0.00 0.65 0.20 0.14 0.08 0.00

T=30,N=80 0.78 0.20 0.15 0.09 0.00 0.67 0.17 0.13 0.07 0.00

T=50,N=20 0.80 0.27 0.21 0.12 0.02 0.68 0.26 0.20 0.11 0.01

T=50,N=40 0.83 0.25 0.19 0.11 0.00 0.76 0.24 0.18 0.10 0.00

T=50,N=80 0.84 0.23 0.17 0.11 0.00 0.78 0.21 0.17 0.10 0.00

T=100,N=20 0.88 0.28 0.22 0.13 0.02 0.78 0.29 0.22 0.13 0.01

T=100,N=40 0.91 0.25 0.20 0.12 0.00 0.88 0.25 0.19 0.11 0.00

T=100,N=80 0.91 0.23 0.18 0.11 0.00 0.92 0.24 0.19 0.12 0.00

Sequential Panel Unit Roots Tests: The pooled RMA & Maximum Order Tests

T=30,N=20 0.56 0.04 0.01 0.00 0.00 0.21 0.01 0.00 0.00 0.00

T=30,N=40 0.71 0.05 0.01 0.00 0.00 0.39 0.01 0.00 0.00 0.00

T=30,N=80 0.76 0.06 0.01 0.00 0.00 0.58 0.02 0.00 0.00 0.00

T=50,N=20 0.78 0.05 0.01 0.00 0.00 0.57 0.02 0.00 0.00 0.00

T=50,N=40 0.82 0.06 0.01 0.00 0.00 0.74 0.02 0.00 0.00 0.00

T=50,N=80 0.84 0.08 0.02 0.00 0.00 0.78 0.04 0.00 0.00 0.00

T=100,N=20 0.87 0.05 0.01 0.00 0.00 0.78 0.03 0.01 0.00 0.00

T=100,N=40 0.91 0.08 0.02 0.00 0.00 0.88 0.05 0.00 0.00 0.00

T=100,N=80 0.91 0.10 0.02 0.00 0.00 0.92 0.07 0.01 0.00 0.00

26

Table 5: Are Real Exchange Rates All Stationary?

Numeraire country Pooled RMA Test Fisher Test Maximum Order Test

Australia -2.27(0.01) 64.47(0.00) 43.67(1.00)

Austria -1.77(0.04) 61.27(0.01) 55.01(1.00)

Belgium -2.09(0.02) 85.95(0.00) 56.39(1.00)

Canada -2.16(0.02) 64.54(0.00) 51.90(1.00)

Denmark -0.84(0.20) 64.15(0.01) 36.95(1.00)

Finland -3.25(0.00) 61.26(0.01) 66.01(1.00)

France -2.20(0.01) 57.07(0.02) 55.58(1.00)

Germany -2.18(0.01) 55.81(0.03) 43.02(1.00)

Greece -2.25(0.01) 64.18(0.00) 55.05(1.00)

Ireland -2.79(0.00) 57.67(0.02) 56.91(1.00)

Italy -2.51(0.01) 55.40(0.03) 71.23(1.00)

Japan -2.24(0.01) 54.06(0.04) 64.39(1.00)

Netherlands -0.75(0.23) 68.16(0.00) 47.46(1.00)

New Zealand -2.65(0.00) 69.31(0.00) 67.97(1.00)

Norway -2.36(0.01) 58.97(0.02) 53.51(1.00)

Portugal -2.19(0.01) 54.84(0.04) 60.78(1.00)

Spain -2.62(0.00) 56.44(0.03) 63.56(1.00)

Sweden -2.51(0.01) 59.07(0.02) 49.49(1.00)

Switzerland -1.82(0.03) 72.19(0.00) 63.66(1.00)

U.K. -1.99(0.02) 67.92(0.00) 56.53(1.00)

U.S. -2.50(0.01) 64.92(0.00) 76.98(1.00)

The number in parenthesis is the probability value.

27

8 Appendix

We restate models here first and take the following assumptions.

M1: (Unknown Costant)

(yit = ai (1− ρi) + ρiyit−1 + εit

yit = ai + xit, xit = ρixit−1 + εit

M2: (Linear Trends)

(yit = ai (1− ρ) + biρi + bi (1− ρi) t+ ρiyit−1 + εit

yit = ai + bit+ xit, xit = ρxit−1 + εit

Assumption 1 The εit have zero mean, finite 2+2ν moments for some ν > 0, are independent

over i and t with E(ε2it) = σ2i for all t, and limN→∞ 1N

PNi=1 σ

2i = σ2.

Assumption 2 limN→∞ 1N

PNi=1 ai = µa <∞, limN→∞ 1

N

PNi=1 bi = µb <∞, limN→∞ 1

N

PNi=1 a

2i =

µ2a <∞ and limN→∞ 1N

PNi=1 b

2i = µ2b <∞.

The folllowing lemmas are useful to prove Propostion 2 and 3. Denote

Ei =plimN→∞ 1N

PNi=1, then we have

Lemma 1 (Stationary xt)

1-1 EiPT

t=2 xitxit−1 = (T − 1)ρσ2x

1-2 Ei³PT−1

t=1 xit

´2= σ2x

³T − 1 + 2ρ

1−ρPT−2

k=1 (1− ρk)´

1-3 EiPT

t=2

³1

t−1Pt−1

s=1 xis

´2= σ2x

PTt=2

³1

t−1´2

1−ρ2(1−ρ)2

³t− 1− 2ρ1−ρt−1

1−ρ2´

1-4 Ei³PT

t=21

t−1Pt−1

s=1 xis

´2= σ2x

PT−1j=1

³PT−1s=j

1s

´2+2σ2x

PT−1k=2 ρ

k−1PT−kj=1

³PT−1s=j

1s

´³PT−1s=j+(k−1)

1s

´1-5 Ei

PTt=2

³xit−1 1

t−1Pt−1

s=1 xis

´= σ2x

PTt=2

³1

t−1´³

1−ρt−11−ρ

´1-6 Ei

PTt=2

³xit

1t−1

Pt−1s=1 xis

´= σ2xρ

PTt=2

³1

t−1´³

1−ρt−11−ρ

´1-7 Ei

³PTt=2 xit−1

´³PTt=2

1t−1

Pt−1s=1 xis

´= σ2x (T − 1) + σ2x

³PT−2k=1

hPT−k−1j=1 ρj

PT−1s=k+j

1s

i´+ σ2x

³PT−2k=1

hPT−k−1j=1 ρT−k−j

PT−1s=j

1s

i´1-8 Ei

³PTt=2 xit

´³PTt=2

1t−1

Pt−1s=1 xis

´= σ2x (T − 1)+ σ2x

³PT−2k=1

hPT−k−1j=1 ρj

PT−1s=k+j

1s

i´+

σ2x

³PT−2k=1

hPT−k−1j=1 ρT−k−j

PT−1s=j

1s

i´− σ2x

³PT−1j=1 ρ

j−1PT−1s=j

1s

´+ σ2x

³PT−1j=1 ρ

T−j−1PT−1s=j

1s

´

28

Lemma 2 ( Unit Root xt ):

2-1 EiPT

t=2 xitxit−1 = σ2PT−1

t=1 t

2-2 Ei³PT−1

t=1 xit

´2= σ2

nPT−1t=1

Ptj=1 j +

PT−2t=1 t

PT−1j=t+1 1

o2-3 Ei

PTt=2

³1

t−1Pt−1

s=1 xis

´2= σ2

PTt=2

³1

t−1´2 ¡

16 t+

13 t3 − 1

2t2¢

2-4 Ei³PT

t=21

t−1Pt−1

s=1 xis

´2= σ2

PT−1j=1 j

³PT−1s=j

1s

´2+ 2

PT−1k=2

PT−kj=1 j

³PT−1s=j

1s

´³PT−1s=j+(k−1)

1s

´2-5 Ei

PTt=2

³xit−1 1

t−1Pt−1

s=1 xis

´= σ2

PTt=2

³1

t−1´Pt

s=2 (s− 1)

2-6 EiPT

t=2

³xit

1t−1

Pt−1s=1 xis

´= σ2

PTt=2

³1

t−1´Pt

s=2 (s− 1)

2-7 Ei³PT

t=2 xit−1´³PT

t=21

t−1Pt−1

s=1 xis

´= σ2

2 (T − 1)2 T − σ2PT−2

k=1

PT−k−1j=1 j

PT−1s=k

1s

2-8 Ei³PT

t=2 xit

´³PTt=2

1t−1

Pt−1s=1 xis

´= σ2

2 (T − 1)2 T− σ2PT−2

k=1

PT−k−1j=1 j

PT−1s=k

1s − σ2 (T − 1)

+ σ2PT−1

j=1 jPT−1

s=j1s

Through simulation, each formulae is varified.

8.1 Appendix A: Proof of Proposition 2

8.1.1 The case of unknown constant

It is straightforward to see there is no asymptotic bias. To obtain the asymptotic variance of

ρpr, we consider the following limit.

plimN→∞1

N

NXi=1

TXt=1

(yit−1 − yit−1)2

= plimN→∞1

N

NXi=1

"TXt=1

x2it−1 −1

T

ÃTXt=1

xit−1

!ÃTXt=1

1

t

tXs=1

xis

!#

= σ2µ1

2T 2 +

1

2T

¶− σ2

12(T + 1)2 − 1

T

T−1Xk=1

T−kXj=1

jTXs=k

1

s

29

Note that the Hamonic number HT =PT

s=11s can be approximated to be lnT + γ where γ is

Euler constant. Similarly

TXs=k

1

s=

Z T

k

1

sds+ γ = lnT − ln k + γ

This approximation also provides

T−1Xk=1

T−kXj=1

jTXs=k

1

s'

Z T−1

1

µZ T−k

1j (lnT − ln k) dj

¶dk

= (T − k) k lnT − (T − k) k ln k + (T − k) k − (k lnT − k ln k + k) ,

and

T−1Xk=1

T−kXj=1

jTXs=k

1

s' 11

36T 3 − 11

12T 2 − 1

2(lnT )T +

1

6T +

5

6lnT +

8

9

Finally, the asymptotic limit is given by

plimN→∞1

N

NXi=1

TXt=1

(yit−1 − yit−1)2 = σ2½11

36T 2 − 17

12T − 1

2lnT +O (1)

¾Meanwhile the probability limit of the square of the regression residuals is given by

plimN→∞1

N

NXi=1

1

T

TXt=1

u2it = plimN→∞1

N

NXi=1

1

T

TXt=1

¡wcit − ρwc

it−1¢2

= plimN→∞1

N

NXi=1

1

T

TXt=1

ε2it−1 = σ2

Hence the asymptotic variance of ρpr for the case of unknown constant is given by

plimN→∞

PNi=1

1T

PTt=1 u

2itPN

i=1

PTt=1 (yit−1 − yit−1)2

=

·11

36T 2 − 17

12T − 1

2lnT +O (1)

¸−18.1.2 For the case of linear trend:

The bias of ρpr is given by

plimN→∞¡ρpr − 1

¢= plimN→∞

CNT

DNT

where

DNT =1

N

NXi=1

TXt=1

(yit−1 − yi·−1 − 2 [yit−1 − yi·−1])2

30

and

CNT =1

N

NXi=1

TXt=1

(yit−1 − yi·−1 − 2 [yit−1 − yi·−1]) (uit − ui·)

We consider the denominator term first. From Lemma 2-1 through 2-12, we have

plimN→∞1

N

NXi=1

TXt=2

(xit−1 − xi·−1)2 =σ2

6(T − 2)T,

TXt=1

(yit−1 − yi·−1 − 2 [yit−1 − yi·−1])2

=TXt=1

(yit−1 − yi·−1)2 + 4TXt=1

[yit−1 − yi·−1]2 − 4TXt=1

(yit−1 − yi·−1) [yit−1 − yi·−1]

plimN→∞1

N

NXi=1

TXt=2

(xit−1 − xi·−1)2 =σ2

6(T − 2)T,

plimN→∞1

N

NXi=1

TXt=2

[yit−1 − yi·−1]2

= plimN→∞1

N

NXi=1

TXt=2

µ1

t− 1¶2Ãt−1X

s=1

yis

!2− 1

T − 1

"TXt=2

µ1

t− 1¶Ãt−1X

s=1

yis

!#2=

TXt=2

µ1

t− 1¶2µ1

6t+

1

3t3 − 1

2t2¶

− 1

T − 1

T−1Xj=1

j

T−1Xs=j

1

s

2 + 2 T−1Xk=2

T−kXj=1

j

T−1Xs=j

1

s

T−1Xs=j+(k−1)

1

s

plimN→∞1

N

NXi=1

TXt=1

(yit−1 − yi·−1) [yit−1 − yi·−1]

= plimN→∞1

N

NXi=1

(TXt=1

Ãyit−1

1

t− 1t−1Xs=1

yis

!− 1

T − 1TXt=1

Ã1

t− 1t−1Xs=1

yis

!TXt=1

yit−1

)

= −14T 2 +

3

4T − 1

2+

1

T − 1

T−2Xk=1

T−k−1Xj=1

jT−1Xs=k

1

s

Hence

plimN→∞1

N

NXi=1

TXt=1

(yit−1 − yi·−1 − 2 [yit−1 − yi·−1])2

= AT + 4BT − 4CT =1

9T 2 +O (T )

31

where

AT =1

6(T − 2)T = 1

6T 2 +O (T )

BT =TXt=2

µ1

t− 1¶2µ1

6t+

1

3t3 − 1

2t2¶− 1

T − 1T−1Xj=1

j

T−1Xs=j

1

s

2

− 2

T − 1T−1Xk=2

T−kXj=1

j

T−1Xs=j

1

s

T−1Xs=j+(k−1)

1

s

(18)

=1

6T 2 − 1

8T 2 +O (T )

CT = −14T 2 +

3

4T − 1

2+

1

T − 1

T−2Xk=1

T−k−1Xj=1

jT−1Xs=k

1

s

=

1

18T 2 +O (T )

Since

TXt=2

µ1

t− 1¶2µ1

6t+

1

3t3 − 1

2t2¶

=1

6T 2 +O (T )

1

T − 1T−1Xj=1

j

T−1Xs=j

1

s

2 =1

4T +O (1)

2

T − 1T−1Xk=2

T−kXj=1

j

T−1Xs=j

1

s

T−1Xs=j+(k−1)

1

s

=1

8T 2 +O (T )

1

T − 1

T−2Xk=1

T−k−1Xj=1

jT−1Xs=k

1

s

=11

36T 2 +O(T )

The nominator term is also similarly obtained by

plimN→∞CNT = plimN→∞1

N

NXi=1

TXt=2

(yit−1 − yi·−1) (uit − ui·)

−2plimN→∞1

N

NXi=1

TXt=2

[yit−1 − yi·−1] (uit − ui·)

From the direct calculation,

plimN→∞1

N

NXi=1

TXt=2

(xit−1 − xi·−1) (uit − ui·) = −12T + 1

32

while

plimN→∞1

N

NXi=1

TXt=2

[yit−1 − yi·−1] (uit − ui·)

= − 1

T − 1plimN→∞1

N

NXi=1

ÃÃTXt=2

1

t− 1t−1Xs=1

xis

!ÃTXt=1

uit

!!

= − 1

T − 1T−1Xk=2

T−1Xj=k

T−1Xs=j

1

s

since

plimN→∞1

N

NXi=1

ÃÃTXt=2

1

t− 1t−1Xs=1

xis

!ÃTXt=2

uit

!!

= Eu2

"x1

T−1Xs=1

1

s+ x2

T−1Xs=2

1

s+ · · ·+ xT−1

T−1Xs=T−1

1

s

#...

+EuT

"x1

T−1Xs=1

1

s+ x2

T−1Xs=2

1

s+ · · ·+ xT−1

T−1Xs=T−1

1

s

#

=T−1Xj=2

T−1Xs=j

1

s+ ...+

1

T − 1

=T−1Xk=2

T−1Xj=k

T−1Xs=j

1

s

After numerical calculation, we have

plimN→∞CNT = 0

Hence we have

plimN→∞¡ρpr − 1

¢= plimN→∞

CNT

DNT= 0

To obtain the asymptotic variance, recall that uit − ui· = εit − εi· and ρpr = ρ+Op

¡T−2

¢.

Observe this.

plimN→∞1

N

NXi=1

1

T

TXt=1

(uit − ui·)2 = σ2 −O¡T−1

¢Hence the asymptotic variance of ρpr for the case of unknown constant is given by

plimN→∞

PNi=1

1T

PTt=1 (uit − ui·)2PN

i=1

PTt=1 (yit−1 − yi·−1 − 2 [yit−1 − yi·−1])2

=9

T 2+O

¡T−3

¢

33

8.1.3 Asymptotic Normality

Following Harris and Tzavalis (1999), we assume xi0 = 0. For fixed T, as N →∞, the

asymptotic normality holds due to the Linderberg-Lecy CLT.

8.2 Appendix B: Proofs of Propostion 3 and 4

8.2.1 Proof of Proposition 3

Let σ2 = 1 for notational convinience. Note that when ρ < 1, from lemma 1-3

plimN→∞1

N

NXi=1

tXt=1

Ã1

s

sXk=1

xikxi1

!2=

µ2a +σ2

(1−ρ)2 ln t+O (1)

µ2a +σ2

(1−ρ)2(19)

= ln (t) +O (1) (20)

since µ2a is finite and is independent of t.

When ρ = 1, from the direct calculation (see lemma 2-3)

plimN→∞1

N

NXi=1

tXt=1

Ã1

s

sXk=1

xikxi1

!2=

µ2a + σ2¡16 t2 + 1

3 t+16 ln t

¢+O (1)

µ2a + σ2(21)

=1

6

¡t2 + 2t+ ln t

¢+O (1) (22)

8.2.2 Proof of Proposition 4

Expanding qt for every t gives tXs=1

yis − 2s

sXq=1

yiq

2

=

ÃtX

s=1

yis

!2+ 4

tXs=1

1

s

sXq=1

yiq

2 − 4Ã tXs=1

yis

! tXs=1

1

s

sXq=1

yiq

(23)

Take probability limit as N →∞. The limit of the first term is given in (19) and (21). The

limit of the second term is rather complicated. To find the maximum order of the second

term, consider

T−1Pt=1

1

t

tPs=1

x2is =

·xi1 +

1

2(xi1 + xi2) + · · ·+ 1

T − 1 (xi1 + · · ·+ xiT−1)¸2

=

"xi1

T−1Xs=1

1

s+ xi2

T−1Xs=2

1

s+ · · ·+ xiT−1

T−1Xs=T−1

1

s

#2

34

Stationary xit

First we consider the case of stationary xit. The probability limit of the second term in (23)

is given by

plimN→∞1

N

NXi=1

µT−1Pt=1

1

t

tPs=1

xs

¶2

= σ2x

ÃT−1Xs=1

1

s

!2+

ÃT−1Xs=2

1

s

!2+ · · ·+

ÃT−1X

s=T−1

1

s

!2+2ρσ2x

"ÃT−1Xs=1

1

s

!ÃT−1Xs=2

1

s

!+ · · ·+

ÃT−1X

s=T−2

1

s

!ÃT−1X

s=T−1

1

s

!#(24)

+2ρσ2x

"ÃT−1Xs=1

1

s

!ÃT−1Xs=3

1

s

!+ · · ·+

ÃT−1X

s=T−3

1

s

!ÃT−1X

s=T−1

1

s

!#...

+2ρT−2σ2x

ÃT−1Xs=1

1

s

!ÃT−1X

s=T−1

1

s

!

The second row in (24) is given byÃT−1Xs=1

1

s

!ÃT−1Xs=2

1

s

!+ · · ·+

ÃT−1X

s=T−2

1

s

!ÃT−1X

s=T−1

1

s

!

=

ÃT−1Xs=1

1

s

!ÃT−1Xs=1

1

s− 1!+ · · ·+

ÃT−1X

s=T−2

1

s

!ÃT−1X

s=T−2

1

s− 1

T − 2

!

=

ÃT−1Xs=1

1

s

!2+ · · ·+

ÃT−1X

s=T−2

1

s

!2−ÃT−1Xs=1

1

s

!−ÃT−1Xs=2

1

s

!µ1

2

¶− · · ·−

ÃT−1X

s=T−2

1

s

!µ1

T − 2¶

=T−1Xj=1

T−1Xs=j

1

s

2 −Ã T−1Xs=T−1

1

s

!2−

T−2Xj=1

1

j

T−1Xs=j

1

s

Note that

T−2Xj=1

1

j

T−1Xs=j

1

s

'Z1

jln(T − 1)−

Z1

jln j = ln (T − 2) ln(T − 1)− 1

2ln2 (T − 2)

35

The third row in (24) is given byÃT−1Xs=1

1

s

!ÃT−1Xs=3

1

s

!+ · · ·+

ÃT−1X

s=T−3

1

s

!ÃT−1X

s=T−1

1

s

!

=

ÃT−1Xs=1

1

s

!ÃT−1Xs=1

1

s− 1− 1

2

!+ · · ·+

ÃT−1X

s=T−3

1

s

!ÃT−1X

s=T−3

1

s− 1

T − 2 −1

T − 3

!

=

ÃT−1Xs=1

1

s

!2+ · · ·+

ÃT−1X

s=T−3

1

s

!2−

T−2Xj=1

1

j

T−1Xs=j

1

s

− T−3Xj=1

1

j

T−1Xs=j

1

s

=

T−1Xj=1

T−1Xs=j

1

s

2 −Ã T−1Xs=T−2

1

s

!2−

T−2Xj=1

1

j

T−1Xs=j

1

s

− T−3Xj=1

1

j

T−1Xs=j

1

s

T−3Xj=1

1

j

T−1Xs=j

1

s

= ln (T − 3) ln (T − 1)− 12ln2 (T − 3)

and so on. Collect the second terms in every row.

ρ

ÃT−1X

s=T−1

1

s

!2+ ρ2

ÃT−1X

s=T−2

1

s

!2+ · · ·+ ρT−2

ÃT−1Xs=2

1

s

!2

=

µ1

T − 1¶2 ¡

ρ+ ρ2 + · · ·+ ρT−2¢+

µ1

T − 2¶2 ¡

ρ2 + · · ·+ ρT−2¢+

· · ·+ 12ρT−2 + cross terms of O

¡T−2

¢= O

¡T−2

¢by setting ρT = o (1) .

Next, the third terms in (24) are given by

ρ

½ln (T − 2) ln(T − 1)− 1

2ln2 (T − 2)

¾+

ρ2½ln (T − 3) ln (T − 1)− 1

2ln2 (T − 3)

¾+ρ2

½ln (T − 2) ln(T − 1)− 1

2ln2 (T − 2)

¾+

ρ3½ln (T − 4) ln (T − 1)− 1

2ln2 (T − 4)

¾+ · · ·

· · ·+ρT−2

½ln (1) ln (T − 1)− 1

2ln2 (1)

¾+

...+ ρT−2½ln (T − 2) ln(T − 1)− 1

2ln2 (T − 2)

¾< O

¡ln2 T

¢36

Hence we have

E

µT−1Pt=1

1

t

tPs=1

xis

¶2= 2Tσ2x + 4Tσ

2x

T−1Xk=2

ρk−1 +O¡ln2 T

¢=

2

(1− ρ)2T +O

¡ln2 T

¢(25)

Next consider the limit of the third term in (23). Note thatµTPt=2

xit−1¶µ

T−1Pt=1

1

t

tPs=1

xis

¶= (xi1 + · · ·+ xiT−1)

"xi1

T−1Xs=1

1

s+ · · ·+ xiT−1

T−1Xs=T−1

1

s

#

Hence its probability limit is given by

plimN→∞1

N

NXi=1

µT−1Pt=1

xit

¶µT−1Pt=1

1

t

tPs=1

xis

¶

= σ2x

T − 1 +T−2Xk=1

T−k−1Xj=1

ρjT−1Xs=k+1

1

s

+ T−2Xk=1

T−k−1Xj=1

ρT−k−jT−k−jXs=1

1

s

Observe this

T−2Xk=1

T−k−1Xj=1

ρjT−1Xs=k+1

1

s

=ρ− ρT−2

1− ρs1 +

ρ− ρT−3

1− ρs2 + ...+ ρsT−1

=ρ

1− ρ

T−1Xj=1

T−1Xs=j

1

s

− 1

1− ρ

T−1Xj=1

ρT−jT−1Xs=j

1

s

=ρ

1− ρ(T − 1)−O (lnT )

since

TXj=1

ρT−jTXs=j

1

s< ρ

TXj=1

ρT−jTXs=1

1

s=

ρ

1− ρlnT

While

T−2Xk=1

T−k−1Xj=1

ρT−k−jT−k−jXs=1

1

s

= ρ1− ρT−1

1− ρsT−1 + ρ

1− ρT−2

1− ρsT−2 + ...+ ρ

1− ρ

1− ρs2

= ρ1

1− ρ

T−1Xj=2

T−1Xs=j

1

s

− ρ21

1− ρ

T−1Xj=2

ρj−1T−1Xs=j

1

s

=ρ

1− ρ(T − 2)−O (lnT )

37

Hence we have

E

µT−1Pt=1

xt

¶µT−1Pt=1

1

t

tPs=1

xs

¶=

σ2

(1− ρ)2T +O (lnT ) (26)

Combining the results in (25), (26) with (19) providesÃTXt=1

xit

!2+ 4

ÃTXt=1

Ã1

t

tXs=1

xis

!!2− 4

ÃTXt=1

xit

!ÃTXt=1

Ã1

t

tXs=1

xis

!!

= σ2µ

T

1− ρ2+ 8

T

(1− ρ)2 (1 + ρ)− 4 T

(1− ρ)2

¶+O (lnT )

= 5σ2

1− ρ2T +O (lnT )

Replace T by t, and divide it by t2. This results in (11) for stationary xit.

Nonstationary xit

From the proof of Propostion 2, we already know the maximum order of all three terms.

From (18), the direct calculation showsÃTXt=1

xit

!2+ 4

ÃTXt=1

Ã1

t

tXs=1

xis

!!2− 4

ÃTXt=1

xit

!ÃTXt=1

Ã1

t

tXs=1

xis

!!

=1

3T 3 +

4

8T 3 − 4

µ7

36

¶T 3 +O

¡T 2¢

=1

18T 3 +O (lnT )

38