a goodness-of-fit test for arch models

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Journal of Econometrics 141 (2007) 835–875

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A goodness-of-fit test for ARCHð1Þ models

Javier Hidalgoa,�, Paolo Zaffaronib

aEconomics Department, London School of Economics, Houghton Street, London WC2A 2AE, UKbTanaka Business School, Imperial College London, London SW7 2AZ, UK

Available online 5 January 2007

Abstract

A goodness-of-fit test in the class of conditional heteroscedastic time series models is examined.

Due to the nonstandard limiting distribution of the test, we propose to bootstrap the test, showing its

asymptotic validity. Moreover, we illustrate the finite sample performance of the test by a small

Monte Carlo study.

r 2006 Elsevier B.V. All rights reserved.

JEL classification: C22; C23

Keywords: GARCH models; Model specification; Bootstrap tests

1. Introduction

We shall consider an observable process xt satisfying

xt ¼ stzt; t 2 Z, (1.1)

where fztgt2Z is an independent identically distributed (iid) sequence of random variableswith zero mean and unit variance, and s2t ¼ Eðx2

t jJt�1Þ, where Jt is the sigma-algebragenerated by fx2

s ; sptg. Existing literature deals with parametric modeling of theconditional heteroskedasticity s2t . One very general model is

s2t ¼ s2t ðy0Þ ¼ m0 þX1j¼1

bjðz0Þx2t�j,

see front matter r 2006 Elsevier B.V. All rights reserved.

.jeconom.2006.11.005

nding author. Tel.: +4420 7955 7503; fax: +4420 7831 1840.

dresses: [email protected] (J. Hidalgo), [email protected] (P. Zaffaroni).

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dx.doi.org/10.1016/j.jeconom.2006.11.005

mailto:[email protected]

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ARTICLE IN PRESSJ. Hidalgo, P. Zaffaroni / Journal of Econometrics 141 (2007) 835–875836

m040; bjðz0ÞX0 ðjX1Þ;X1j¼1

bjðz0Þo1, ð1:2Þ

where y00 ¼ ðm0; z00Þ0 is an rþ 1 dimensional parameter vector.

The ARCHð1Þ model (1.1)–(1.2) was introduced by Robinson (1991) as a class ofalternatives for testing serial independence of xt, and it is a generalization of Bollerslev’s(1986) GARCHðp; qÞ model defined as

s2t ¼ em0 þXp

i¼1

ai0x2t�i þ

Xq

j¼1

bj0s2t�j,

em040; ai0X0 ð1pippÞ; bj0X0 ð1pjpqÞ. ð1:3Þ

The latter model is a further generalization of Engle’s (1982) original ARCHðpÞ model.However, contrary to the previous models, the model given in (1.1)–(1.2) allows for longmemory behavior in fx2

t gt2Z, being many fractional models nested into (1.1)–(1.2). Amongothers, we can cite Ding and Granger’s (1996) long memory GARCH model and thefractionally integrated GARCH (FIGARCH) of Baillie et al. (1996); see also Robinsonand Zaffaroni (2006), henceforth abbreviated as RZ, who discuss these and otherparameterizations of interest covered by (1.1)–(1.2).However, the previous class of models are only one possibility among many alternatives,

some of them nonnested, considered in the theoretical and empirical literature of volatilitymodelling. For instance, one can consider ARCH-type asymmetric models where, unlike in(1.2), s2t is an asymmetric function of fx2

s ; sotg. Some examples are Engle’s (1990)asymmetric GARCH, the exponential GARCH model of Nelson (1991), the linear ARCHmodel of Robinson (1991) or the power GARCH model of Ding and Granger (1996). Seealso the work of Linton and Mammen (2005) and Glosten et al. (1993). Another class ofmodels are the stochastic volatility (SV) models introduced by Taylor (1986) and explored byHarvey et al. (1994). The SV model is a popular approach for modelling dynamicconditional heteroskedasticity, and in particular after the work of Hull and White (1987), asa way to approximate continuous time diffusion underlying option pricing models featuringchanging volatility. See also the surveys by Ghysels et al. (1995) and Shephard (2004).The previous discussion suggests that when testing the adequacy of (1.2), a sensible way

to proceed is to leave the alternative model unspecified. That is, to provide a goodness-of-fit test. The latter type of test, also known as omnibus test, traces back to Kolmogorov’s(1933) pioneer work, when testing for a specific probability distribution function, orGrenander and Rosenblatt (1957), see their Chapter 6, for testing the hypothesis of whitenoise dependence. More recently, these type of tests have gained growing interest, see forexample Stute (1997) or Delgado et al. (2005).In the framework of model (1.3), there are several rival procedures to test the adequacy

of s2t . Among them, the Box–Pierce–Ljung’s Portmanteau test, see Ljung and Box (1978),which resembles Neyman’s (1937) smooth test. In our context, they are defined as

BT ¼XnT

j¼1

br2j ,where brj is an estimator of the jth correlation coefficient of fz2t gt2Z and nTX1 is aparameter to be chosen by the practitioner. The Portmanteau test has been relatively

ARTICLE IN PRESSJ. Hidalgo, P. Zaffaroni / Journal of Econometrics 141 (2007) 835–875 837

explored, see Li and Mak (1994) and Berkes et al. (2003a) for GARCHðp; qÞ models withfinite p and q, although their validity for the general models considered in this paper is anopen question. This test is regarded as a compromise between omnibus and directionaltests. On the other hand, as discussed in Delgado et al. (2005), our omnibus test, describedin (2.7) below, as well as BT can be obtained as particular functionals of (2.5) given inSection 2. However, contrary to the goodness-of-fit test that we propose, the power of BT

depends very much on the choice of nT . For instance, they have only trivial power in thedirection of local alternatives converging to the null at the rate T�1=2, though they are ableto detect local alternatives converging to the null at the rate T�1=2n

1=4T . Therefore, for size

accuracy we will require to choose a fairly large nT , although a smaller nT may be desirablefor power improvements; that is, the choice of nT induces a trade-off between size andpower. Finally, there still exists the issue of how to choose nT in empirical applications.

On the other hand, contrary to the test based on BT , which has a standard distribution,the distribution of our test is not standard, but model-based, whose critical values, ifpossible, are difficult to compute. So, the paper provides valid bootstrap-based tests forgoodness of fit for the ARCHð1Þ model (1.1), being our result valid for both (shortmemory) GARCHðp; qÞ as well as for (long memory) FIGARCHðp; d; qÞ models, amongmany others. Furthermore, we will not require finite second moments for the observableprocess fxtgt2Z. Finally, it should be mentioned that the results of the paper follow if (1.1)is modified to

yt ¼ gðZt; xÞ þ xt,

where x is some finite dimensional parameter and fZtgt2Z is a stationary strong mixingsequence. However, to simplify the already lengthy arguments of our results, we havedecided to use (1.1)–(1.2) instead.

The remainder of the paper is organized as follows. Section 2 describes the test and itsproperties. Because of its nonstandard limiting distribution, Section 3 presents a bootstrapprocedure, showing its asymptotic validity. A small Monte Carlo experiment to examinethe performance of our test in small samples is given in Section 4, whereas Section 5 givesthe proofs of our main results of Sections 2 and 3, which employ a series of Lemmasconfined into Section 6.

2. The test

We will describe and examine a test for the adequacy of model (1.1)–(1.2). To that end,let

MðyÞ ¼ fs2t ðyÞ ¼ mþX1j¼1

bjðzÞx2t�j; y ¼ ðm; z0Þ0 2 Yg,

where Y � Rrþ1 is a compact parameter space. Hence, our null hypothesis H0 becomes tocheck whether s2t in (1.1) belongs to MðyÞ for some value y0 of the parameter space Y.That is, the null hypothesis becomes

H0: s2t 2MðyÞ.

The alternative hypothesis is the negation of H0.Before we describe the test we need some preliminaries. Given a stretch of data fxtg

Tt¼1�‘,

where ‘ ¼ ‘ðTÞ increases to infinity with T, we shall estimate y ¼ ðm; z0Þ0 as follows. Denote


by IðAÞ the indicator function and consider

QT ðyÞ ¼1

T

XT

t¼1

qtðyÞ; QT ðyÞ ¼1

T

XT

t¼1

qtðyÞ, (2.1)

where

qtðyÞ ¼x2

t

s2t ðyÞþ ln s2t ðyÞ; qtðyÞ ¼

x2t

s2t ðyÞþ ln s2t ðyÞ,

with s2t ðyÞ ¼ mþPtþ‘�1

j¼1 bjðzÞx2t�jIðtX2� ‘Þ. We then define the pseudo-maximum

likelihood (PMLE) and (observed) PMLE asey ¼ arg miny2Y

QT ðyÞ; by ¼ arg miny2Y

QT ðyÞ, (2.2)

respectively. As usual a subscript 0 to a parameter vector, say y0, indicates the true valuewhereas y is any admissible value of the parameter in the compact set Y.It is worth giving a brief discussion on the reason why we start the observations at the

time t ¼ 1� ‘ instead of at t ¼ 1 as is usually written. The main reason is because ofnotational simplicity. In fact, what we are doing is to discard the first ‘ observations.Nevertheless, ‘ can be chosen to increase with T as slow as we wish, so that we do notbelieve that this adjustment is needed in practice. For example, we can choose‘2 ¼ log log T , so that with T ¼ 107, we have that ‘o2, indicating that indeed inempirical applications we can take ‘ ¼ 0. The motivation is because otherwise we cannotcontrol the order of magnitude of (6.15) to be opðT

�1=2Þ as we need in the proof of Lemma6.6. The latter is true regardless of the number of finite moments assumed for the observeddata xt. However, ‘ can be taken equal to zero for the results given in (2.3) below. Finally,it is worth mentioning that this issue does not appear with finite ARCHðpÞ models.Under suitable regularity conditions, see below, RZ have shown that ey and by obey the

central limit theorem property, e.g.

T1=2ðey� y0Þ!dNð0;V Þ; T1=2ðby� y0Þ!

dNð0;V Þ, (2.3)

where

V ¼ 12ðEz4t � 1ÞH�1ðy0Þ; HðyÞ ¼ E

qqy

ln s2t ðyÞqqy0

ln s2t ðyÞ� �

. (2.4)

Note that if fztgt2Z were Gaussian random variables, then V ¼ H�1ðy0Þ.

We now envisage the following procedure to test H0. Consider by in (2.2) and the

estimated conditional variance s2t ðbyÞ. Compute the standardized residuals

zt ¼ ezt �1

T þ ‘

XT

t¼1�‘

ezt ðt ¼ 1� ‘; . . . ;TÞ,

where ezt ¼ s�1t ðbyÞxt, t ¼ 1� ‘; . . . ;T . Then, under H0, we should expect that the sequence

fz2t gTt¼1�‘ behaves as if they were a constant mean iid sequence of random variables with

finite variance. In particular, we make use of the fact that if, say, fytgt2Z is an iid sequenceof random variables, then its spectral distribution function is a straight line through theorigin with slope Eðy2

t Þ=2p. Hence, following ideas in Grenander and Rosenblatt’s (1957,Chapter 6), we can test for H0 by comparing the estimate of the spectral distribution


function with its theoretical value. More specifically, denote

IyðlÞ ¼1

T

XT

t¼1

yteitl

��2

as the periodogram for a generic sequence fytgTt¼1. Then, after observing that the more

natural estimator of the spectral distribution function is given byR l0

IyðmÞdm and the

estimator of the variance is given by eT�1PeTj¼1 IyðljÞ, where for integer j, we write lj ¼

ð2pjÞ=T and where henceforth eT ¼ ½T=2� with ½a� denoting the integer part of a. The testthen proceeds by deciding whether

TðsÞ ¼1eT X

s

j¼1

Iz2 ðljÞeT�1PeTk¼1 Iz2ðlkÞ

� 1

0@ 1A; s ¼ 1; . . . ; eT , (2.5)

is not significantly different than zero for all s ¼ 1; . . . ; eT . Observe that because under the

alternative hypothesis, ez2t ’ s2t =s2t ðy�Þz2t , where y� denotes the pseudo-value, we have that

Cov ðez2t ;ez2s Þa0, which will imply that fez2t gTt¼1�‘ is a correlated sequence of random

variables. So, under the alternative hypothesis, we can conclude that the spectral density

function of z2t will not be constant in ½0;p�, implying thatR l0 Iz2ðmÞdm!

P R l0 f z2 ðmÞdma

Eðs2t =s2t ðy�ÞÞl=2p.On the other hand, because TðsÞ is the Riemann approximation toR l

0Iz2ðmÞdm=

R p0

Iz2 ðmÞdm, we deduce that TðsÞ will have a mean different than zero and

hence when normalized by T1=2, T1=2jTðsÞj ! 1, implying the consistency of the test.

Observe that because (2.5) is evaluated at the Fourier frequencies lj, 1pjp eT , then the

periodogram, and so TðsÞ, entails sample mean correction.

Remark 2.1. From the definition of TðsÞ, we note that its value is scale invariant. That is,TðsÞ is invariant if instead of using zt we would have used the standardized residuals

bzt ¼ ðT þ ‘Þ�1XT

t¼1�‘

z2t

!�1=2zt ðt ¼ 1� ‘; . . . ;TÞ (2.6)

in its computation.

To decide if TðsÞ is significantly different than zero, a common procedure is to employthe Kolmogorov–Smirnov or Cramer–von Mises functionals

KS ¼ sup

s¼1;...;eT eT1=2jTðsÞj; CvM ¼

1eT XeT

s¼1

j eT1=2TðsÞj2, (2.7)

respectively. Of course, it goes without saying that any other functional jð�Þ of TðsÞ willsuffice, as Corollary 2.2 indicates. However, the functionals given in (2.7) are the onesemployed in the Monte-Carlo experiment, but more importantly there are the moststandard functional employed in this type of tests with empirical data.


Before we state the asymptotic properties of eT1=2TðsÞ and, as a consequence of the

continuous mapping theorem, the limit distributions of both KS and CvM, we introducethe following regularity conditions.Condition C1. fztgt2Z is an iid sequence with Ez0 ¼ 0, Ez20 ¼ 1, Ejz0j

io1, for some i44,and probability density function, f ðzÞ, satisfying

f ðzÞ ¼ OðLðz�1ÞzcÞ as z! 0þ ,

for c4� 1 and a function L that is slowly varying at the origin.Condition C2. There exist mL and mU such that 0omLomUo1 and a compact set U 2 Rr,

such that Y ¼ ½mL;mU � � U.Condition C3. y0 ¼ ðm0; z

00Þ0 is an interior point of the compact set Y.

Condition C4. For all jX1 and g41þ d for some d41,

infz2U

bjðzÞX0; supz2U

bjðzÞpKj�g; bjðz0ÞpKbkðz0Þ for some1pkpj,

where K throughout denotes a generic, positive constant.Condition C5. For all jX1, bjðzÞ has continuous kth derivative on U such that, with zi

denoting the ith element of z,

qkbjðzÞqzi1 � � � qzik

��pKb

1�Zj ðzÞ

for all Z40 and all ik ¼ 1; . . . ; r; kp3.Condition C6. There exists a strictly stationary and ergodic solution xt to (1.1) and

Ex2rt o1 for some ð1þ dÞ=goro1.Condition C7. For each z 2 U there exist integers ji ¼ jiðzÞ, i ¼ 1; . . . ; r, such that

1pjiðzÞo � � �ojrðzÞo1 and rankfCðj1;...;jrÞðzÞg ¼ r, where

Cðj1;...;jrÞðzÞ ¼ fbð1Þj1

ðzÞ; . . . ; bð1ÞjrðzÞg; b

ð1Þj ðzÞ ¼

qbjðzÞqz

.

Conditions C1–C7 are the same as those in RZ. A formal proof of the fact that suchconditions are satisfied by various parameterizations, such as (1.3), can be found there,whose comments equally apply here. In particular, note that Condition C1 allows for someasymmetry in the distribution of the zt, which does not make model (1.1)–(1.2) necessarilysymmetric. Condition C4 indicates the generality of the ARCHð1Þ, since models withboth exponentially and hyperbolically decaying coefficients bjðzÞ are allowed for. In thelatter case, though, the constraint g42 is stronger than that in RZ for the (observed)PMLE by not to have an asymptotic bias of order OpðT

1=2Þ making (2.3) invalid. However,the constraint g42 will guarantee that the other of magnitude of (6.15) is opðT

�1=2Þ. Itseems possible to allow for the slightly weaker condition g43

2, but this would imply plenty

of notational and mathematical complications when obtaining the order of magnitude of(6.15) to be opðT

�1=2Þ. Because to assume g42 it is not significantly a stronger conditioncompared with g43

2, we have preferred to leave the condition as it stands. Concerning C6,

conditions for unique stationary solutions of model (1.3) were given in Nelson (1990) forthe GARCHð1; 1Þ model and latter generalized to the GARCHðp; qÞ model by Bougeroland Picard (1992); see also Berkes et al. (2003b). Giraitis et al. (2000) show thatP1

k¼1 bjðzÞo1 is sufficient for Condition C6, and RZ provide an alternative weaker


condition. Condition C7 is a crucial identification condition, necessary to establishconsistency and it guarantees that Hðy0Þ in (2.4) is nonsingular.

We now introduce some notation. Let vt ¼ z2t � 1 for all t ¼ 1� ‘; . . . ;T and

gðlÞ ¼ 2X1r¼1

gðrÞ cosðrlÞ, (2.8)

where gðrÞ ¼ Eðv1ðq=qyÞ log s2rþ1ðy0ÞÞ. Also, define

GðWÞ ¼Z W

0

gðpuÞdu� WZ 1

0

gðpuÞdu.

Theorem 2.1. Define W ¼ ½s= eT �. Assuming C1–C7, as T !1, in D½0; 1�,

eT1=2Tð½ eTW�Þ ) GðWÞ; W 2 ½0; 1�,

where GðWÞ is a Gaussian process with covariance structure given by

KðW1;W2Þ ¼ minðW1;W2Þ þ5

2Varðz20ÞG0ðW1ÞH�1ðy0ÞGðW2Þ. (2.9)

The results of Theorem 2.1 are somehow expected and similar to those obtainedelsewhere, for instance in Hidalgo and Kreiss (2006). In particular, the first term on theright of (2.9) corresponds to the covariance structure if y0 were known a priori, whereas thesecond one is due to the fact that y0 is replaced by by given in (2.2). However, there are somedifferences between the aforementioned paper and the present one. The first one is that inHidalgo and Kreiss (2006), it was employed the ratio between the periodogram of the dataand the estimated spectral density function, whereas here we have employed theperiodogram of the innovations. A second difference is technical. Indeed, the mainproblem is that as we can only employ the truncation of s2t ðyÞ, that is s

2t ðyÞ, to control for

such a difference creates a considerable technical difficulty as our assumptions rule out thedata xt to be Near Epoch dependent. More specifically, the technical difficulty comes fromthe fact that we need to show that the contribution due to the first term on the right of

x2t

s2t ðy0Þ� 1 ¼

s2ts2t ðy0Þ

� 1

� �z2t þ ðz

2t � 1Þ

is negligible. Note that the same technical difficulty will be present if we employ thePortmanteau test BT or any other smooth test. In fact, our proofs indicate that also for theasymptotic theory to hold, that is that BT , after suitable normalization, converges to a w2

distribution, we need to discard the first ‘ observations, as was done to prove the validityof our omnibus test based on a functional of Tð½ eTW�Þ.

Corollary 2.2. Assuming C1–C7, for any continuous functional j in ½0; 1�, we have that, as

T !1,

jð eT1=2Tð½ eTW�ÞÞ!

djðGðWÞÞ; W 2 ½0; 1�.

Proof. The proof is standard by Theorem 2.1 and the continuous mapping theorem, so it isomitted. &

In an asymptotic test, we would reject H0 if jð eT1=2Tð½ eTW�ÞÞ4ca, where ca is the

ð1� aÞth quantile of jðGðWÞÞ. Thence, the question is how to obtain ca in practice. Because


of the complicated nature of the covariance structure of GðWÞ in (2.9), to find a (time)transformation that would lead to a (standard) Brownian process appears to be adifficult task, if at all possible. In principle, the asymptotic distribution, say jðGðWÞÞfor some functional jð�Þ, could be simulated and so their quantiles. However, thecovariance structure KðW1;W2Þ is model-dependent and not free of nuisance parameters.The latter implies that we need to compute new critical values everytime a new modeland/or data are under consideration. Thus, under these circumstances, it appearsthat bootstrap algorithms are appropriate. We shall then describe and examine abootstrap approach to implement the test in the next section. Also, as a by product, thebootstrap can be used to provide an estimate of the asymptotic covariance matrix V ofT1=2ðby� y0Þ.

3. The bootstrap test

The basic idea of the bootstrap is, given a stretch of data AT ¼ fatgTt¼1 say, to treat the

data as if it were the true population, and to carry out Monte-Carlo experiments in whichpseudo-data is drawn from AT . Since Efron’s (1979) work on the bootstrap, an immenseeffort has been devoted to its development. We can cite two main motivations/reasons.First, bootstrap methods are capable of approximating the finite sample distribution ofstatistics better than those based on their asymptotic counterparts. And secondly, andperhaps the most important, it allows computing valid asymptotic quantiles of the limitingdistribution in situations where (a) the limiting distribution is unknown or (b) even ifknown, the practitioner is unable to compute its quantiles. In the present paper we face thelatter situation.In this section we shall propose a bootstrap algorithm to test for H0. Recall that the first

requirement of the bootstrap to be valid is that when the null hypothesis is true, we needthat the bootstrap analogue of (2.7), e.g. (3.4) below, converges in bootstrap sense to thesame limiting distribution as with the original data. (See (3.5) below for what it is meant bythe latter concept.) A second requirement for (3.4) to be valid, with good power properties,is that when the null hypothesis is false, the bootstrap statistic must also converge inbootstrap distribution although, possibly, to a different one. It is worth mentioning andobserving that strictly speaking for the test to have power, we only need that under thealternative hypothesis the bootstrap statistic diverges slower than when the null hypothesisis true. The latter is the case when subsampling methods, see Politis and Romano (1994),are employed. However, as Corollary 3.4 below indicates, the (asymptotic) distribution of(3.4) is the same under both the null and alternative hypotheses, which is guaranteed if thebootstrap sample/model is obtained under H0. This is expected because as we can see fromStep 2 described below, the bootstrap data fx�t g

Tt¼1�‘ satisfy or follow the model

hypothesized under the null hypothesis. That is, irrespective of whether or not fxtgt2Zfollows (1.1)–(1.2), the bootstrap sample does by construction, so that the bootstrapstatistic will have the same asymptotic distribution under the maintained hypothesis. It isin this sense how we shall consider the results of Corollary 3.4. So, from the previouscomments, we can expect that our procedure will have better power properties whencompared to subsampling methods.We now describe the bootstrap in the following five steps.Step 1: Compute bzt, t ¼ 1� ‘; . . . ;T , as in (2.6). Then, draw a random sample of size

T þ ‘ from ZT ¼ fbztgTt¼1�‘, denote as ðz�1�‘; . . . ; z

�0; z�1; . . . ; z

�T Þ.


The standardization of bzt guarantees that the first two (bootstrap) moments of fz�t gTt¼1�‘

are exactly equal to those of fztgt2Z.Step 2: Obtain the bootstrap sample ðx�1�‘; . . . ;x

�0; x�1; . . . ;x

�T Þ from

x�t ¼ s�t ðbyÞz�t ; x��‘ ¼ 0,

s�2t ðbyÞ ¼ bmþ Xtþ‘�1

j¼1

bjðbzÞx�2t�jIðtX2� ‘Þ.

Step 3: Compute the (bootstrap) PMLE of by asby� ¼ arg miny2Y

Q�

T ðyÞ, (3.1)

where

Q�

T ðyÞ ¼1

T

XT

t¼1

q�t ðyÞ; q�t ðyÞ ¼x�2t

s�2t ðyÞþ ln s�2t ðyÞ. (3.2)

Step 4: For t ¼ 1� ‘; . . . ;T , compute the centered bootstrap residuals

z�t ¼ ez�t � 1

T þ ‘

XT

t¼1�‘

ez�t ,where ez�t ¼ x�t =s

�t ðby�Þ.

And finally,Step 5: Compute the bootstrap analogue of (2.5). That is,

T�ðsÞ ¼1eT X

s

j¼1

Iz�2 ðljÞeT�1PeTk¼1 Iz�2 ðlkÞ

� 1

0@ 1A; s ¼ 1; . . . ; eT . (3.3)

Remark 3.1. Because T�ðsÞ is, as was TðsÞ, invariant to the scale of z�t , in what follows weshall assume that E�z�2t ¼ 1 without loss of generality. That is, ðT þ ‘Þ�1

PTt¼1�‘ z�2t ¼ 1.

From here, the bootstrap analogues of (2.7) are given by

KS� ¼ sup

s¼1;...;eT j eT1=2

T�ðsÞj; CvM� ¼1eT XeT

s¼1

j eT1=2T�ðsÞj2. (3.4)

Before we establish our main result, see Theorem 3.3 below, we give two propositions. In the

first of them, we show the consistency of by�, that is by� � by ¼ op� ð1Þ, where op� ð1Þ means that

Prfjby� � byj4djZT g!P0 for all d40. In the second one, we shall show that T1=2ðby� � byÞ!d�

Nð0;V Þ (in probability), where V is given in (2.4) and ‘‘!d�

’’ means convergence in distribution

in the bootstrap sense, that is for each continuity point z of GðzÞ ¼ limT!1 PrfT1=2ðby�y0Þpzg we have that

PrfT1=2ðby� � byÞpzjZT g!p

GðzÞ. (3.5)

Proposition 3.1. Assuming C1–C7, under the maintain hypothesis, by� � by ¼ op� ð1Þ.


Proposition 3.2. Assuming C1–C7, under the maintain hypothesis, as T !1,

T1=2ðby� � byÞ!d� Nð0;V Þ in probability.

Once, we have shown that the bootstrap PMLE by� possesses the same asymptoticproperties of by, we shall state the main result of this section.

Theorem 3.3. Assuming C1–C7, under the maintain hypothesis, as T !1,

eT1=2T�ð½ eTW�Þ ) GðWÞ; ðW 2 ½0; 1�Þ in probability,

where GðWÞ is a Gaussian process with covariance structure given in (2.9).

Corollary 3.4. Assuming C1–C7, for any continuous functional j in ½0; 1�, we have that, as

T !1,

jð eT1=2T�ð½ eTW�ÞÞ!

d�

jðGðWÞÞ in probability.

Proof. The proof is standard by Theorem 3.3 and the continuous mapping theorem, so it isomitted. &

The results of Theorem 3.3 and Corollary 3.4 allow us to implement the (bootstrap) testas follows. Let cT ;1�a and c1�a be such that

Prfjð eT1=2Tð½ eTW�ÞÞ4cT ;1�ag ¼ a; Prfjð eT1=2

Tð½ eTW�ÞÞ4c1�ag !T!1

a,

respectively. Then, Corollary 2.2 indicates that cT ;1�a! c1�a, whereas Corollary 3.4indicates that c�1�a satisfies c�1�a!

Pc1�a, where c�1�a is such that

Prfjð eT1=2T�ð½ eTW�ÞÞ4c�1�ajZT g ¼ a.

However, since the finite sample distribution of jð eT1=2T�ð½ eTW�ÞÞ is not available, we rely

on Monte-Carlo algorithms to approximate, as accurate as desired, the value c�1�a. To that

end, for b ¼ 1; . . . ;B, consider the bootstrap samples x�ðbÞ ¼ ðx�ðbÞ1�‘; . . . ; x

�ðbÞT Þ

0 and compute

jð eT1=2T�ðbÞð½ eTW�ÞÞ for each b. Then, c�1�a is approximated by the value c�B1�a that satisfies

1

B

XB

b¼1

Iðjð eT1=2T�ðbÞð½ eTW�ÞÞ4c�B1�aÞ ¼ a,

where Ið�Þ denotes the indicator function. We finish this section indicating how the

bootstrap can be used to give an estimate of V. Indeed, bV� ¼ B�1PB

b¼1 Tðby�ðbÞ � y�Þ2,

where y�¼ B�1

PBb¼1by�ðbÞ, is such that bV� � V ¼ op� ð1Þ.

4. Monte-Carlo experiment

In order to investigate how well the bootstrap test given in (3.4) performs in finitesamples, a small Monte Carlo experiment was carried out. All throughout our MonteCarlo experiment we have employed 1,000 replications with samples sizes n ¼ 256, 512 and1024. To calculate the bootstrap statistics, for all the models and sample sizes considered,999 bootstrap samples were employed, that is we have chosen B ¼ 999.


To examine the empirical size of the test, we have considered the GARCHðp; qÞ model

xt ¼ stzt,

s2t ¼ 1þXp

i¼1

ai0x2t�i þ

Xq

j¼1

bj0s2t�j, (4.1)

with several combinations of p and/or q, where except for the GARCHð0; 1Þ, wherebj0 ¼ 0:5, the remaining models are such that the sum of the coefficients is equal to 0:9,which is approximately what it is observed in applications with real data. The results forsizes 5% and 1% are given in Table 1, being the design of each experiment described in thetable. Each entry of Table 1 represents the proportion of rejections of H0, when H0 is true,across the 1,000 replications, whereas Tables 2 and 3 contains the proportions of rejectionof H0, when H1 is true.

Table 1

1,000 Montecarlo iterations and 999 Bootstrap replications, Model: (4.1)

KS CvM

T

256 512 1024 256 512 1024

Size 5%

(a) 2.6 1.4 1.1 2.6 1.6 1.1

(b) 4.9 6.0 5.0 4.8 5.7 5.1

(c) 6.3 6.1 4.1 6.4 6.2 4.2

(d) 6.1 5.2 4.2 6.1 5.1 4.2

(e) 4.8 5.6 4.9 4.4 5.1 4.4

(f) 5.8 6.1 4.2 5.9 5.8 4.6

(g) 5.9 6.1 4.2 6.1 6.2 4.2

KS CvM

T

256 512 1024 256 512 1024

Size 1%

(a) 1.6 0.8 0.6 1.4 0.8 0.6

(b) 2.6 2.6 2.4 2.4 3.3 2.5

(c) 3.8 3.2 1.7 3.7 3.3 1.6

(d) 3.3 2.6 1.2 3.1 2.6 1.2

(e) 2.4 3.1 2.4 2.3 2.8 2.3

(f) 2.7 2.7 1.4 2.6 2.8 1.4

(g) 3.2 3.3 1.6 3.3 6.2 1.6

The dgp listed in column 1 are:

(a) p ¼ 0; q ¼ 1, a1 ¼ 0:5.(b) p ¼ 1; q ¼ 1, a1 ¼ 0:3, b1 ¼ 0:6.(c) p ¼ 1; q ¼ 1, a1 ¼ 0:6, b1 ¼ 0:3.(d) p ¼ 1; q ¼ 1, a1 ¼ 0:8, b1 ¼ 0:1.(e) p ¼ 2; q ¼ 1, a1 ¼ 0:3, b1 ¼ 0:5, b2 ¼ 0:1.(f) p ¼ 1; q ¼ 2, a1 ¼ 0:1, a2 ¼ 0:5, b1 ¼ 0:3.(g) p ¼ 1; q ¼ 2, a1 ¼ 0:5, a2 ¼ 0:1, b1 ¼ 0:3.

ARTICLE IN PRESS

Table 2

1,000 Montecarlo iterations; 999 Bootstrap replications

KS CvM

T

256 512 1024 256 512 1024

GARCHð1; 1ÞTrue model (4.1)

(a) 98.8 99.5 99.7 99.0 99.5 99.7

(b) 96.9 98.2 99.1 97.1 98.2 99.2

The estimated model is ARCH(1).

The true dgp model is GARCHð1; 1Þ.(a) a10 ¼ 0:3, b10 ¼ 0:6.(b) a10 ¼ 0:6, b10 ¼ 0:3.

Table 3

1,000 Montecarlo iterations; 999 Bootstrap replications

KS CvM

T

256 512 1024 256 512 1024

EGARCH

True Model (4.2)

(a) 98.9 99.5 99.4 98.9 99.5 99.4

(b) 98.9 99.5 99.4 98.9 99.5 99.4

SV

True Model (4.3)

(a) 99.4 99.6 99.8 99.4 99.6 99.8

(b) 99.4 99.6 99.8 99.4 99.6 99.8

The estimated dgp listed in column 1 is:

(a) r ¼ 1; s ¼ 0.

(b) r ¼ 1; s ¼ 1.

J. Hidalgo, P. Zaffaroni / Journal of Econometrics 141 (2007) 835–875846

We first observe that the finite sample performance of the KS and CvM tests is verysimilar. Except for the first experiment, that is model (a), the results at the 5% level arevery good even for sample sizes as small as T ¼ 512. So, this indicates that with real datasample sizes, the table suggests that their performance is very accurate and reliable.Although the results at the 1% are not as good as at the 5%, the results for models (c), (d),(f) and (g) are very encouraging. However, it is clear that the outcome improves across allthe models as the sample size increases, suggesting that, with the typical sizes we encounterwith real data, the tests should work according to the theory. Also, we should mention thatonly 1,000 replications have been performed, so the results in terms of accuracy at the 1%are not expected to be as good as those we obtain at the 5%.


Table 2 presents the power of the test when the true model is

s2t ¼ 1þ a10x2t�i þ b10s

2t�j,

with a10 ¼ 0:3, b10 ¼ 0:6 and a10 ¼ 0:6, b10 ¼ 0:3, but we estimate an ARCH(1) model.The two specifications considered are labeled as (a) and (b), respectively, in Table 2.Meanwhile, Table 3 presents the power of the test when the true model is given by xt ¼ stzt

and s2t given by

log s2t ¼ 1þX1i¼0

0:5ið�0:3zt�i�1 þ 0:5ðjzt�i�1j �ffiffiffiffiffiffiffiffi2=p

pÞÞ (4.2)

log s2t ¼ 1þX1i¼0

0:5iut�i, (4.3)

where, as in the specification (4.1), fztgt2Z is a sequence of independent standard normalrandom variables. Model (4.2) is the EGARCHmodel of Nelson (1991), whereas specification(4.3) is the SV model of Taylor (1986). The model estimated is the GARCH model

s2t ¼ 1þXr

i¼1

ai0x2t�i þ

Xs

j¼1

bj0s2t�j

for two combinations of r and s.The results in Tables 2 and 3 indicate that the power of our tests are very good and that

they are capable to discriminate between GARCH and EGARCH or SV models. All thecomputations have been carried out in Cþþ and the codes are available upon requestfrom the second author.

5. Proofs

In several places of the proofs, we make use of the following expansion obtained byrecursive substitution

s2t ðyÞ ¼ mþXtþ‘�2k¼1

X1j1;...;jk¼1

bj1ðzÞ � � � bjkðzÞz2t�j1

� � � z2t�Pk

s¼1js

IXk

s¼1

jsotþ ‘ � 1

!þ btþ‘�1

1 ðzÞz2t�1 � � � z22�‘x

21�‘. ð5:1Þ

Observe that

s2t ðyÞ � s2t ðyÞ ¼X1j¼‘

bjþtðzÞx2�j. (5.2)

Let us introduce some notation used in this and next sections. Denote

atðyÞ ¼qqy

mþXtþ‘�1j¼1

qqy

bjðzÞ� �

z2t�js2t�jðyÞ,

ytðyÞ ¼ ðq=qyÞs2t ðyÞ; eytðyÞ ¼ ðq

2=qyqy0Þ log s2t ðyÞ,

ytðyÞ ¼ ðq=qyÞs2t ðyÞ; eytðyÞ ¼ ðq

2=qyqy0Þ log s2t ðyÞ,


whereas the bootstrap counterparts will be denoted with an ‘‘�’’ upperscript. So, for

instance a�t ðyÞ ¼qqy mþ

Ptþ‘�1j¼1 ð

qqy bjðzÞÞz�2t�js

�2t�jðyÞ. Finally, we shall write

s�2t ðq; yÞ ¼ mþXtþ‘�2k¼1

X1j1;...; jk¼1

bj1ðzÞ � � � bjkðzÞz�2t�j1

� � � z�2t�Pk

s¼1js

IXk

s¼1

jsoq

!þ btþ‘�1

1 ðzÞz�2t�1 � � � z�22�‘x

�21�‘IðtoqÞ,

a�t ðq; yÞ ¼qqy

mþXminðq;tþ‘�1Þ

j¼1

qqy

bjðzÞ� �

z�2t�js�2t�jðq; yÞ (5.3)

dropping reference to by and /or bz when evaluated at by. Notice that a�t ðtþ ‘ � 1; yÞ¼: a�t ðyÞis equal to y�t ðyÞ.

Finally henceforth, we shall abbreviate fjðz0Þ and fjðbzÞ by fj and

bfj, respectively, for a

generic sequence of coefficients fjðzÞ, and gðy0Þ and gðbyÞ by g and bg, respectively, for a genericfunction g. Likewise, for the bootstrap expressions, we shall abbreviate f�j ðbzÞ by f�j , for a

generic sequence of (stochastic) coefficients f�j ðzÞ and g�ðbyÞ by g� for a generic function g�.

5.1. Proof of Theorem 2.1

By definition of Tð½ eTW�Þ given in (2.5), we have that for all W 2 ½0; 1�,

eT1=2Tð½ eTW�Þ ¼

1dVarðz2t ÞeT1=2

X½eTW�

j¼1

Iz2;j �1eT XeT

k¼1

Iz2;k

0@ 1A,

where dVarðz2t Þ ¼eT�1PeTk¼1 Iz2;k with hðljÞ ¼ hj for any generic function hð�Þ.

First, Lemma 6.1 part ðbÞ implies that dVarðz2t Þ!P

Varðz2t Þ, which together with Lemma

6.7 imply that it suffices to show the weak convergence of

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

Iz2;j �1eTXeT

k¼1

Iz2;k

0@ 1A� 1eTX½eTW�

j¼1

g0j �WeT XeT

j¼1

g0j

0@ 1AeT1=2ðby� y0Þ

8<:9=;

to GðWÞ. Next, because gðrÞ ¼ Oðr�d Þ by Lemma 6.4, we have that gðlÞ given in (2.8) is

continuous and then supW2½0;1�k eT�1P½eTW�j¼1 gj �

R W0 gðpuÞduk ¼ oð1Þ by Brillinger (1981,

p. 15). So, the limit distribution of eT1=2Tð½ eTW�Þ is governed by that of

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

Iz2;j �1eTXeT

k¼1

Iz2;k

0@ 1A� G0ðWÞ eT1=2ðby� y0Þ

8<:9=;.

On the other hand, using the linearization

eT1=2ðby� y0Þ ¼ �H�1 eT1=2

hT þ opð1Þ,


see RZ, where hT ðyÞ ¼ ðq=qyÞQT ðyÞ and QT ðyÞ is given in (2.1), it suffices to examine theweak convergence in the Skorohod’s metric space of

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

Iz2;j �1eT XeT

k¼1

Iz2;k

0@ 1Aþ G0ðWÞH�1 eT1=2hT

8<:9=;. (5.4)

The convergence of the finite dimensional distributions follows by standard arguments,

after we observe that by RZ, eT1=2hT converges to a Normal random variable, and that C1

implies that

1eT1=2

X½eTW�

j¼1

ðIz2;j � Varðz2t ÞÞ!d

Nð0;WVar2ðz2t ÞÞ,

by Brillinger’s (1981) Theorem 7.8.3 for all W 2 ½0; 1�. Also notice that the limiting

covariance of the second term inside the braces of (5.4) is G0ðWÞVarðz2t ÞH�1GðWÞ=2. Next,

because by C1, vt ¼ z2t � 1 are iid random variables, and EðIz2;jhT Þ ¼ Varðz2t ÞeT�1PT

t¼1

Eðv0yt=s2t Þ cosðtljÞ, we have then that by standard arguments,

E1eT1=2

X½eTW�

j¼1

Iz2;j �1eT XeT

k¼1

Iz2;k

0@ 1AeT1=2hT ! Varðz2t ÞG

0ðWÞ

and thus the limiting covariance structure of the process (5.4) is KðW1;W2Þ given in (2.9).So, to complete the proof of the theorem, we are left to examine the tightness condition

of the process defined in (5.4). First of all, by Billingsley’s (1968) Theorem 16.7 and

Cauchy–Schwarz’s inequality, G0ðWÞH�1 eT1=2hT is tight since kGðW2Þ � GðW1Þk2pK jW2 � W1j2

and Eðk eT1=2hTk

2Þo1. On the other hand, the tightness condition of eT�1=2P½eTW�j¼1 ðIz2;j �eT�1PeTk¼1 Iz2;kÞ follows proceeding as with Brillinger’s (1981) Theorem 7.8.1. In fact, the latter

theorem indicates that

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

ðIz2;j � Varðz2t ÞÞ )eBðWÞ; W 2 ½0; 1�

in the Skorohod space, where eBðWÞ denotes the standard Brownian motion, so that

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

Iz2;j �1eT XeT

k¼1

Iz2;k

0@ 1A) BðWÞ; W 2 ½0; 1�,

where BðWÞ is the standard Brownian bridge. This completes the proof. &

5.2. Proof of Proposition 3.1

We shall first show that

E� supy2YjQ�

T ðyÞ �QT ðyÞj ¼ opð1Þ, (5.5)


where QT ðyÞ and Q�

T ðyÞ are given in (2.1) and (3.2), respectively. By Lemmas 6.14 and 6.15,

we have that E�jQ�

T ðyÞ �QT ðyÞj ¼ opð1Þ. So, to complete the proof, it suffices to show the

equicontinuity condition for Q�

T ðyÞ �QT ðyÞ because Y is a compact set. But this is the case

because Lemma 6.16 implies that

E�1

jy1 � y2jj ln s�2t ðy1Þ � ln s�2t ðy2Þj þ

x�2t

s�2t ðy1Þ�

x�2t

s�2t ðy2Þ

�� ¼ Kt,

where Kt is a sequence of Opð1Þ random variables. Hence, (5.5) holds true.

On the other hand, by definition by� ¼ arg miny2YQ�

T ðyÞ and by ¼ arg miny2Y QT ðyÞ, so

that we conclude that by� � by ¼ op� ð1Þ by standard arguments. &

5.3. Proof of Proposition 3.2

By Taylor’s expansion of Q�

T ðyÞ and that h�T ðby�Þ ¼ 0, we have that

T1=2ðby� � byÞ ¼ �H��1T ðey�ÞT1=2h�T ,

where ey� is an intermediate point between by� and by and where h�T ðyÞ ¼ T�1PT

t¼1h�t ðyÞ and

H�T ðyÞ ¼ T�1PT

t¼1 H�t ðyÞ, with

h�t ðyÞ ¼ �1

2

y�t ðyÞs�2t ðyÞ

� z�2t

s�2t

s�2t ðyÞy�t ðyÞs�2t ðyÞ

� �,

H�t ðyÞ ¼ �1

2ey�t ðyÞ þ 2

s�2t

s�2t ðyÞy�t ðyÞy

�0t ðyÞ

s�4t ðyÞ� z�2t

s�2t

s�2t ðyÞ

qqy0 y

�t ðyÞ

s�2t ðyÞ

( ). ð5:6Þ

On the other hand, proceeding as with the proof of Proposition 3.1, we obtain that

H�T ðyÞ �HT ðyÞ ¼ op� ð1Þ

uniformly in a neighborhood of by, say NðbyÞ. The last displayed equality and C7 imply that

H�T ðyÞ is a positive definite matrix for all y 2NðbyÞ and hence H��1T ðey�Þ � bH�1T ¼ op� ð1Þ by

a routine application of Slutzky’s theorem.Next, we already know that HT ðyÞ converges uniformly to HðyÞ in a neighborhood of y0,

say Nðy0Þ. So, the proof is completed if we show that

T1=2h�T ¼1

T1=2

XT

t¼1

h�t !d�

Nð0;Varðz2t ÞHÞ ðin probabilityÞ. (5.7)

It is obvious that E�ðT1=2h�T Þ ¼ 0. Now, because fv�t ¼ z�2t � 1gTt¼1 (see Remark 3.1) is azero mean iid sequence of random variables, the second (bootstrap) moment of T1=2h�T is

TE�ðh�T h�0T Þ ¼ fE�ðz�4t Þ � 1gE�

1

T

XT

t¼1

y�t y�0ts�4t

, (5.8)

whose first factor on the right of (5.8) converges in probability to Eðz4t Þ � 1 by Lemma 6.8.Next, since Eðs�4t yty

0tÞ �H ¼ oð1Þ by an obvious application of Lemma 6.2, it suffices to


show that

E�1

T

XT

t¼1

y�t y�0ts�4t

� E1

T

XT

t¼1

yty0t

s4t¼ opð1Þ. (5.9)

To that end, by Lemma 6.9 and proceeding as with the proof of Lemma 6.2, it is easy toobserve that the left side of (5.9) is

E�1

T

XT

t¼2qþ1

a�t ðqÞa�0t ðqÞ

s�4t ðqÞ� E

1

T

XT

t¼2qþ1

atðqÞa0tðqÞ

s4t ðqÞþOp

1

qd

� �,

since E�P2q

t¼1a�t ðqÞa

�0t ðqÞ

s�4t ðqÞ

�� ¼ Op� ðqÞ, where s�2t ðq; yÞ and a�t ðq; yÞ are given in (5.3) and

s2t ðq; yÞ ¼ mþXtþ‘�2k¼1

X1j1;...;jk¼1

bj1ðzÞ:::bjkðzÞz2t�j1

� � � z2t�j1��jkI

Xk

s¼1

jsoq

!þ btþ‘�1

1 ðzÞz2t�1 � � � z22�‘x

21�‘IðtoqÞ,

atðq; yÞ ¼qqy


j¼1

qqy

bjðzÞ� �

z2t�js2t�jðq; yÞ.

But, after observing that a�t ðqÞ, say, depends only on z�t�1; . . . ; z�t�2q, we have that (5.9)

holds true by Lemma 6.13 with

hðz�t�1; . . . ; z�t�2q;

byÞ ¼ a�t ðqÞa�0t ðqÞ

s�4t ðqÞ

there. So, we conclude that (5.8) converges in probability to fEðz4t Þ � 1gH, choosing q largeenough.

Hence it remains to show the Lindeberg’s condition, for which a sufficient condition is

1

T i=4

XT

t¼1

E�kh�t ki=2 ¼ opð1Þ. (5.10)

However, this is the case because proceeding as before, we have that

E�v�i=2t E�

y�t y�0ts�4t

�� i=4!P Evi=2t E

yty0t

s4t

�� i=4

o1,

which implies that the left side of (5.10) is OpðT1�i=4Þ ¼ opð1Þ since i44. &

5.4. Proof of Theorem 3.3

By definition of T�ð½ eTW�Þ given in (3.3), we have that for all W 2 ½0; 1�,

eT1=2T�ð½ eTW�Þ ¼

1dVarðz�2t ÞeT1=2

X½eTW�

j¼1

Iz�2;j �1eT XeT

k¼1

Iz�2;k

0@ 1A,

where dVarðz�2t Þ ¼eT�1PeTk¼1 Iz�2;k ¼ T�1

PTt¼1 z�4t � ðT

�1PT

t¼1 z�2t Þ2 is (in bootstrap sense) a

consistent estimator of Varðz2t Þ by Lemma 6.8.


Proceeding as with the proof of Theorem 2.1 but using Lemma 6.20 instead of Lemma6.7 there, we conclude that it suffices to show the weak convergence of

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

Iz�2;j �1eT XeT

k¼1

Iz�2;k

0@ 1A� G0ðWÞ eT1=2ðby� � byÞ

8<:9=;.

Now, by Proposition 3.2, we have the linearization of eT1=2ðby� � byÞ given by

eT1=2ðby� � byÞ ¼ � bH�1T

eT1=2h�T þ op� ð1Þ,

where HT ðyÞ ¼ E�fðq2=qyqy0ÞQ�

T ðyÞg with Q�

T ðyÞ given in (3.2), it suffices to examine theweak convergence, in the Skorohod’s metric space, of

1

Varðz2t Þ

1eT1=2

X½eTW�

j¼1

Iz�2;j �1eT XeT

k¼1

Iz�2;k

0@ 1Aþ G0ðWÞ bH�1TeT1=2

h�T

8<:9=;. (5.11)

The convergence of the finite dimensional distributions follows by standard arguments

after we observe that Proposition 3.2 implies that eT1=2bh�T converges in bootstrap sense to a

Normal random variable. On the other hand, by Hidalgo and Kreiss (2006), we obtain that

1eT1=2

X½eTW�

j¼1

ðIz�2;j � Var�ðz�2t ÞÞ!d�

Nð0;WVar2ðz2t ÞÞ ðin probabilityÞ.

Likewise proceeding as in the proof of (6.42), we have that

E�1eT1=2

X½eTW�

j¼1

Iz�2;jeT1=2

h�T

0@ 1A� Varðz2t Þ1eT X½eTW�

j¼1

gj!P0,

because the left side is proportional to

1eT X½eTW�

j¼1

XT

t¼1

E� v�0y�ts�2t

� �cosðtljÞ �

1

2gj

!ð1þ op� ð1ÞÞ

and following as with the proof of (6.17), eT�1P½eTW�j¼1 ð

PTt¼1Eðv0

yt

s2tÞ cosðtljÞ �

12

gjÞ ¼ oð1Þ.

From here now it follows that the covariance structure of (5.11) converges in probability toKðW1;W2Þ.So, to complete the proof of the theorem we need to examine the tightness condition of

the process defined in (5.11). The tightness condition of G0ðWÞ bH�1TeT1=2

h�T holds true

because TE�ðh�T h�0T Þ ¼bHT40, so that

E�jðG0ðW2Þ � G0ðW1ÞÞ bH�1TeT1=2

h�T j2pKTkGðW2Þ � GðW1Þk2

pKT jW2 � W1j2,

where KT is a triangle array of Opð1Þ random variables, whereas the tightness condition ofeT�1=2P½eTW�j¼1 ðIz�2;j �

eT�1PeTk¼1Iz�2;kÞ follows because, after standard algebra, the latter


expression is equal to

2eT1=2

X½eTW�

j¼1

Iz�2;j �1eT XeT

k¼1

Iz�2;k

0@ 1A,

where Iz�2 ðljÞ ¼ T�1PT

1¼tosv�t v�s cosððt� sÞljÞ and

E�1eT1=2

X½eTW2�

j¼½eTW1�þ1

Iz�2;j

��i=2

p1eT i=4

X½eTW2�

j¼½eTW2�þ1

KT

0B@1CA

i=4

pKT jW2 � W1ji=4.

The latter inequality holds because v�tPT

s¼tþ1v�s cosððt� sÞljÞ is a martingale difference sequence

with finite i=2 moments and that E�ðT�1PT

t¼1jz�t jiÞ ¼ Opð1Þ by Lemma 6.8. Note that, we only

need to consider the situation eT�1=2pðW2 � W1Þ. This completes the proof of the theorem. &

6. Lemmas

Lemma 6.1. Assuming C1–C7, we have that for all p42 and some d40,

ðaÞ supt¼1;...;T

bs2t � s2ts2t

�� ¼ OpðT

�ðp�2Þ=2pÞ; ðbÞ1

T

XT

t¼1

bz4t ¼ Ez4t 1þOp1

Td

� �� . (6.1)

Proof. We begin with part (a). First, Taylor’s expansion implies that

bs2t � s2t ¼ ðby� y0Þ0yt þ ð

by� y0Þ0 qqy0

ytðeyÞðby� y0Þ, (6.2)

where ey is an intermediate point between y0 and by. So, the left side of part (a) of (6.1) isbounded by

kby� y0k supt¼1;...;T

yt

s2t

�� þ kby� y0k2 supt¼1;...;T

qqy0 ytð

eyÞs2t

��.

Using the inequality

supk

jckj

� �p

pX

k

jckjp; ðp40Þ, (6.3)

and proceeding as in the proof of Lemma 6 of RZ, for all p40, we have that

Efsupt¼1;...;T ðkyt

s2tk þ supy2Yk

qqy0

ytðyÞ

s2tkÞgp ¼ OðTÞ. But, by� y0 ¼ OpðT

�1=2Þ, so that the

conclusion follows by standard arguments. Observe that the latter aforementioned lemmaimplies that

sup

y2fkby�y0koK=T1=2g

qqy0 ytðyÞ

s2t

��

p

p sup

z2fkbz�z0koK=T1=2g

Ptþ‘�1j¼1 b

ð2Þpj ðzÞb

r�pj x

2rt�j

s2t

�� (6.4)

and Eðsupz2Nðz0ÞP1

j¼1bð2Þpj ðzÞb

r�pj x

2rt�jÞoK for T large enough by C4–C6 with Z small

enough.


Next part (b), whose left side is

1

T

XT

t¼1

z4t 1þs2t � bs2tbs2t

!2

þ 2s2t � bs2tbs2t

8<:9=;.

Because Ez4t oK , part (a) with pX4 implies that the last displayed expression is

1

T

XT

t¼1

z4t 1þs2t � s2t

s2t

� �2

þ 2s2t � s2t

s2tþOpðT

�1=4Þ

( ), (6.5)

where the OpðT�1=4Þ is uniform in t. However, using (5.2) we have that

Eðs2t � s2t Þr¼ E

X1j¼‘

bjþtx2�j

��r

¼ KX1j¼‘

brjþtx

2r�j ¼ Oððtþ ‘Þ�d

Þ, (6.6)

by C4–C6. On the other hand, for p ¼ 1; 2,

KEXT

t¼1

z4ts2t � s2t

s2t

� �p��

��r=p

pKXT

t¼1

Ez4r=pt E

s2t � s2ts2t

� �r

¼ Oð1Þ. (6.7)

So, by Markov’s inequality and that T�1PT

t¼1 z4t � Ez4t ¼ OðmaxðT�1þ4=i;T�1=2ÞÞ, we have

that (6.5) is T�1PT

t¼1 z4t ð1þOpðT�1=4ÞÞ ¼ Ez4t ð1þOpðT

�dÞÞ, where d ¼ minf1� 4=i;1=4g. &

To simplify the notation, we shall denote ðq=qzÞbj and ðq=qz0Þbð1Þj by bð1Þj and b

ð2Þj ,

respectively.

Lemma 6.2. Assuming C1–C7, for any p40 and q ¼ 1; 2,

E

P1j¼tþ‘ b

ðqÞj x2

t�j

s2t

��p

þ Es2t � s2t

s2t

�� p þ Eyt � at

s2t

�� p ¼ O1

ðtþ ‘Þd

� �. (6.8)

Proof. Proceeding as in RZ’s Lemma 6 and that s2t 40, for any p40, the first term on theleft of (6.8) is bounded by

KEX1

j¼tþ‘

bðqÞpj b

r�pj x

2rt�j

�� ¼ Oððtþ ‘Þ�d

Þ

by C4–C6. On the other hand, using (5.2) and that jðs2t � s2t Þ=s2t joK , the contribution due

to the second term on the left of (6.8) is bounded by EjP1

j¼‘bjþtx2�jj

r ¼ Oððtþ ‘Þ�dÞ

because by C6, brj oKj�1�d .

Finally, by definition, the third term on the left of (6.8) is bounded by

KE

Ptþ‘�1j¼1 b

ð1Þj z2t�jðs

2t�j � s2t�jÞ

s2t

��p

.

By Holder’s inequality, the last displayed expression is bounded by

KEXtþ‘�1j¼1

bð1Þpj b

r�pj z

2rt�jðs


r

! Ptþ‘�1j¼1 bjz

2t�jðs


s2t

��

( ).


The second factor inside the braces is bounded by 2, whereas the expectation of the first

factor is Oððtþ ‘Þ�dÞ because jb

ð1Þpj b

r�pj jpKj�1�d and (6.6). Recall that by C1, z2t is

independent of fs2s � s2s gspt. &

Henceforth, we define s2t;�s and at;�s as s2t and at, respectively, but without the termsinvolving fz2r gspr.

Lemma 6.3. Assuming C1–C7, for any p40,

Es2t � s2t;�1

s2t

��p

þ Eat � at;�1

s2t

�� p ¼ O‘Ið‘ptÞ

t1þdþ

Iðto‘Þ

td

� �.

Proof. We shall consider only the contribution due to the first term on the left, being thatfor the second term identically handled after proceeding as in RZ’s Lemma 6. First, using

(5.1), it is clear that s2t � s2t;�1 only involves terms for which t� 2pPk

s¼1 jsotþ ‘ � 1 in

the expression (5.1). So, by Holder’s inequality and because ðs2t � s2t;�1Þ=s2toK , that term

is bounded by

KX½log t1þd �

k¼1

þXtþ‘�2

k¼1þ½log t1þd �

8<:9=;ðEz

2rt Þ

kXtþ‘�2j1¼1

� � �Xtþ‘�2�Pk�2

s¼1js

jk�1¼1

Xtþ‘�2�Pk�1

s¼1js

jk¼t�2�Pk�1

s¼1js

brj1� � � b

rjk

þ Kðb1Ez2r1 Þ

tþ‘�1Ex2r1�‘.

The contribution due toP½log t1þd �

k¼1 is Oð‘Ið‘ptÞt1þd þ

Iðto‘Þtd Þ because b

rj pKj�1�d by C6. On the

other hand, the contribution due toPtþ‘�2

k¼1þ½log t1þd � is Oðt�1�dÞ because it is bounded byPtþ‘�2k¼1þ½log t1þd �ðEz

2rt�j

Pj b

rj Þ

k and Ez2rt�j

Pj b

rj o1. Finally, the latter implies that the

contribution due to ðb1Ez2r1 Þ

tþ‘�1 is clearly Oðt�1�d Þ. &

Lemma 6.4. Assuming C1–C7, for all tXt0 for some t041,

ðaÞ E v1ytþ1

s2tþ1

!�� ¼ O

1

td

� �; ðbÞ E v1

ytþ1

s2tþ1

!�� ¼ O

1

td

� �,

whereas for any tot0, E v1ytþ1

s2tþ1

�� þ E v1ytþ1

s2tþ1

�� oK .

Proof. We begin with part (a), which left side is bounded by

E v1atþ1

s2tþ1

!��þ E v1

ytþ1 � atþ1

s2tþ1

( )��. (6.9)

By standard algebra, the second term of (6.9) is bounded by

K Ev1

Ptþ‘�1j¼1 b

ð1Þj z2tþ1�jðs

2tþ1�j � s2tþ1�jÞ

s2tþ1;�1

��þ Ev1

ytþ1 � atþ1

s2tþ1;�1

!s2tþ1;�1 � s2tþ1

s2tþ1

!��

¼ KE v1bð1Þt z21

s21 � s21s2tþ1;�1

��þO

1

td

� �,


because by C1, v1 is independent of z2tþ1�jðs2tþ1�j � s2tþ1�jÞ for jat using (5.2) and the

definition of s2tþ1;�1, and then because Lemmas 6.2 and 6.3 imply that Ejv1ðytþ1 �

atþ1Þ=s2tþ1;�1j ¼ Oðt�dÞ using (5.2). Note that Lemma 6.3 implies that with probability

approaching one, s2t;�1Xs2t ð1� Kt�d ÞXs2t =2. Then observing that bð1Þt z21ðs21 � s21Þ=s

2tþ1

oK ,we obtain that the right side of the last displayed equality is

KE v1bð1Þt z21

s21 � s21s2tþ1

��r

þO1

td

� �¼ O

1

td

� �

by C5–C6 and Lemma 6.2 after observing that Ejs21 � s21jr ¼ Oð‘�dÞ, which by (5.2) does

not involve any term with z1.Next, the expression inside the absolute value of the first term of (6.9) is

E v1atþ1;�1

s2tþ1;�1

!þ E v1

atþ1 � atþ1;�1

s2tþ1

!�

atþ1;�1

s2tþ1;�1

s2tþ1 � s2tþ1;�1s2tþ1

" # !.

The first term of the last displayed expression is zero by C1, whereas the second term isOðt�dÞ by Lemma 6.3. This concludes the proof of part (a).Part (b) follows proceeding as in part (a) and observing that using (5.2) yt � yt, say, is

independent of v1. Finally, proceeding as above, it is clear that for any tot0,jEðv1ytþ1=s

2tþ1Þj þ jEðv1ytþ1=s

2tþ1ÞjoK . &

Denote jsjþ ¼ maxf0; jsjg and

wj ¼1

T

XT

t;r¼1;tar

vtvr

yr

s2reiðt�rÞlj ; ewj ¼

1

T

XT

t;r¼1;tar

vtvr

ar

s2reiðt�rÞlj .

Lemma 6.5. Assuming C1–C7, for 0oj; kp eT ,

ðaÞ Ejwj � ewjj ¼ OðT�1=2 logTÞ; ðbÞ jEðewjew0�kÞj ¼ Oðjj � kj�1þ Þ.

Proof. To simplify the notation we shall denote wj and ewj a typical element of wj and ewj,respectively. By definition, Ejwj � ewjj is bounded by

E1

T

XT

t;r¼1

vtvr

yr � ar

s2r

� �eiðt�rÞlj

��þ E

1

T

XT

t¼1

v2tyt � at

s2t

� �� ¼ O

logT

T1=2

� �

because C1 implies that E supjjPT

t¼1vteitlj j ¼ OðT1=2 logTÞ by An et al. (1983), and then by

Lemma 6.2. Recall that by C1, vrðyr�ar

s2rÞ is a martingale difference with finite second

moments. This completes the proof of part (a).


Next, we prove part (b). By C1, the left side is bounded by

2 E1

T2

XT

t1pt2

vt1vt2

XT

t2or

ar

s2r

� �2

eiðt1�rÞlj�iðt2�rÞlk

��

p2 E1

T2

XT

1ot1ot2

vt1vt2

XT

t2or

ar

s2r

� �2

�ar;�t1

s2r;�t1

!28<:

9=;eiðt1�rÞlj�iðt2�rÞlk

��

þ 21

T2

XT

t¼1

XT

tor

Ev2tar

s2r

� �2

eiðt�rÞlj�k

��.

By Lemma 6.3, the first term on the right of the last displayed inequality is bounded by

K‘

T2

XT

1ot1ot2or;‘or

jt1 � rj�1�d þOð‘2=T2Þ ¼ oð1=T1=2Þ

because ‘ is arbitrarily small. Next, the second term on the right of the last displayedinequality is bounded by

1

T2

XT

t¼1

XT

tor

Ev2tar

s2r

� �2

�at

s2t

� �2( )

eiðt�rÞlj�k

��þ Ev2t

1

T2

XT

t¼1

Eat

s2t

� �XT

tor

eiðt�rÞlj�k

��.

Proceeding as in the proof of Lemma 6.3, the first term is OðT�1Þ, whereas the second termis proportional to

1

1� eilj�k

1

T2

XT

t¼1

Eat

s2t

� �2

ð1� eitlj�k Þ

�� ¼ Oðjj � kj�1þ Þ

because Eðat=s2t Þ2¼ Oð1Þ. &

Lemma 6.6. Assuming C1–C7, we have that for all j ¼ 1; . . . ; eT ,

Iz2;j � Iz2;j þ 2ðby� y0Þ0Qj ¼ opðT

�1=2Þ,

where

Qj ¼ T�1XT

t;r¼1

vtvr

yr

s2rcosððt� rÞljÞ þ T�1

XT

t;r¼1

vt

yr

s2rcosððt� rÞljÞ.

Proof. First, by definition of z2t and z2t , Iz2;j � Iz2;j is

1

T

XT

t¼1

z2ts2t � bs2tbs2t þ

s2t � s2tbs2t !

eitlj

��2

þ2

T

XT

t;r¼1

z2ts2t � bs2tbs2t þ

s2t � s2tbs2t !

z2r cosððt� rÞljÞ.

(6.10)

Now, by standard algebra,

XT

t¼1

z2tbs2t � s2tbs2t eitlj ¼

XT

t¼1

z2tbs2t � s2t

s2teitlj �

XT

t¼1

z2tðs2t � bs2t Þ2

s2tbs2t eitlj , (6.11)


which right side, using (6.2), is

ðby� y0Þ0XT

t¼1

z2tyt

s2teitlj þ ðby� y0Þ

0XT

t¼1

z2t

qqy0 ytð

eyÞs2t

eitlj ðby� y0Þ

� ðby� y0Þ0XT

t¼1

z2tyty0t

s2tbs2t eitlj ðby� y0Þ �XT

t¼1

z2tððby� y0Þ

0 qqy0 ytð

eyÞðby� y0ÞÞ2

s2t bs2t eitlj

� 2ðby� y0Þ0XT

t¼1

z2t

qqy0 ytð

eyÞðby� y0Þy0t

s2t bs2t eitlj ðby� y0Þ.

Next, using Lemma 6.1 part (a) with pX4 there, say, (2.3) and that Ekeyt þ yt=s2t k

poK

using Lemma 6.2 in an obvious way and that Ekeyt þ yt=s2t k

poK , we have that the lastdisplayed expression, and thus the left side of (6.11), is for all j

ðby� y0Þ0XT

t¼1

z2tyt

s2teitlj þ ðby� y0Þ

0XT

t¼1

z2teyte

itlj ðby� y0Þ þOpðT�1=4Þ. (6.12)

Notice that proceeding as with (6.4), by C5 and (2.3) for pX1,

qqy0 ytð

eyÞ � qqy0 yt

s2t

��

p

pkby� y0kp

Pt�1j¼1 supzjb

ð3Þpj ðzÞjb

r�pj x

2rt�j

s2t

�� ¼ OpðT

�p=2Þ.

Also, recall that by definition eyt ¼ ðq2 log s2t =qyqy

0Þ ¼ s�2t

qqy0 yt � s�4t yty

0t.

Next, we show that

1

T1=2

XT

t¼1

z2tyt

s2teitlj ¼ opðT

1=4Þ. (6.13)

To that end, write the left side of (6.13) as

1

T1=2

XT

t¼1

vt

yt

s2teitlj þ

1

T1=2

XT

t¼1

yt � at

s2teitlj þ

1

T1=2

XT

t¼1

at

s2teitlj . (6.14)

The first term on the right of (6.14) is Opð1Þ because fvtgt2Z is a zero mean iid sequence withfinite second moments and Ejyt=s

2t j2oK , while the second term is opð1Þ by Lemma 6.2

choosing p ¼ 1 there, and then because d41. Finally, because for ja0;T ,PT

t¼1 eitlj ¼ 0,

the third term on the right of (6.14) is

1

T1=2

XT

t¼1

at

s2t� E

at

s2t

� �� eitlj þ

1

T1=2

XT

t¼1

Eat

s2t�

yt

s2t

� �eitlj .

The second term of the last displayed expression is oð1Þ by Lemma 6.2 and then becaused41. On the other hand, the second moment of the first term is

1

T

XT

t¼1

Eat

s2t� E

at

s2t

� �� 2

þ2

T

XT

1¼tor

Eat

s2t� E

at

s2t

� �� ar

s2r� E

ar

s2r

� �� ,


which first term is obviously Oð1Þ, whereas the second term is

2

T

XT

1¼tor

Eat

s2t� E

at

s2t

� �� ar

s2r�

ar;�t

s2r;�t

( ) !.

But, proceeding and arguing as in the proof of Lemma 6.4, it is easy to see that Ej ar

s2r�

ar;�t

s2r;�t

jp ¼ Oðjt� rj�dÞ for any p40. So, because d41, choosing p large enough, we have that

the last expression is oðT1=2Þ. Hence we have shown that (6.13) holds by Markov’sinequality.

Proceeding similarly, T�1PT

t¼1 z2teyte

itlj ¼ opð1Þ and so by (2.3), ð6:11Þ ¼ opðT1=4Þ. So, to

show that the first term of (6.10) is opðT�1=2Þ, it suffices to show that

XT

t¼1

z2ts2t � s2tbs2t eitlj ¼ opð1Þ. (6.15)

By Lemma 6.1 part (a), it suffices to show that

XT

t¼1

z2ts2t � s2t

s2teitlj

�� ¼ opð1Þ. (6.16)

However, by (5.2) and C4 and because js2t � s2t j=s2toK , for any pX1

E supt¼1;...;T

s2t � s2ts2t

�� ppKE supt¼1;...;T

X1j¼‘

ðj þ tÞ�gx2�j

��r

pKX1j¼‘

j�gr ¼ Oð‘�dÞ

by C6. So, with probability approaching one, 12os2t =s

2t . Hence, (6.16) holds ifPT

t¼1 Ejz2ts2t�s

2t

s2tj ¼ opð1Þ, where the left side is bounded by

KXT

t¼1

s2t � s2ts2t

�� rpKXT

t¼1

ðtþ ‘Þ�d¼ oð1Þ

by (6.8) and that s2t4K�1, and then because ‘!1 and d41. So, we conclude that the

first term of (6.10) is opðT�1=2Þ.

To complete the proof, it remains to show that the second term of (6.10) is �2ðby� y0Þ0

QjþopðT�1=2Þ. Because the previous arguments indicate that j 1

T

PTt;r¼1 z2t

s2t�s2tbs2t z2r cosððt� rÞ

ljÞj ¼ opðT�1=2Þ, we only need to examine

2

T

XT

t;r¼1

z2ts2t � bs2tbs2t z2r cosððt� rÞljÞ.


By (6.12), thatPT

r¼1 z2reirlj ¼ OpðT

1=2Þ and that s2t � bs2t ¼ OpðT�1=2Þ by Lemma 6.1, we

have that proceeding as with (6.10), the last displayed expression is

2ðy0 � byÞ0

T

XT

t;r¼1

z2tyt

s2tz2r cosððt� rÞljÞ

� 2ðby� y0Þ

0

T

XT

t;r¼1

z2teytz

2r cosððt� rÞljÞð

by� y0Þ þ op1

T1=2

� �.

The first term of the last displayed expression is easily seen to be �2ðby� y0Þ0Qj , whereas

the second term is opðT�1=2Þ by (2.3) and because by C1,

PTr¼1 z2r e

irlj ¼ OpðT1=2Þ and

T�1PT

t¼1 z2teyte

itlj ¼ opð1Þ proceeding similarly as in the proof of (6.13) . This completes the

proof of the lemma. &

Lemma 6.7. Assuming C1–C7,

sup

s¼1;...;eT1eT1=2

Xs

j¼1

Iz2;j � Iz2;j þ 2Varðz2t ÞEyt

s2t

� �þ gj

� �0ðby� y0Þ

� �� ¼ opð1Þ,

where gj ¼ gðljÞ is given in (2.8) .

Proof. First, we shall show that

EðQjÞ ¼ Varðz2t ÞEyr

s2r

� �þ gj þ oð1Þ. (6.17)

By C1, we have that the first moment of the first term on the right of Qj is Varðz2t Þ

T�1PT

t¼1 Eðyt=s2t Þ ¼ Varðz2t ÞEðyt=s

2t Þ þ oðT�1=2Þ by Lemma 6.2 and that d41. Next, C1

implies that the first moment of the second term of Qj is

1

T

XT

t¼1

XT

r¼tþ1

E vtyr

s2r

� �cosððt� rÞljÞ �

1

T

XT

t¼1

XT

r¼tþ1

E vtyr

s2r�

yr

s2r

� �� cosððt� rÞljÞ,

and hence (6.17) is shown because by Brillinger (1981, p. 15) and Lemma 6.4,PT

t¼1

Efv1ytþ1

s2tþ1

g cosðtljÞ �12gj ¼ oð1Þ, whereas the second term of the last displayed expression is

o(1) as we now show. For some aominf1; d=2g, this term is

1

T

X½Ta�

t¼1

XT

r¼tþ1

þXT

t¼½Ta�þ1

Xtþ½Ta�

r¼tþ1

þXT

t¼½Ta�þ1

XT

r¼tþ½Ta�þ1

( )E vt

yr

s2r�

yr

s2r

� �� cosððt� rÞljÞ.

By Lemma 6.2 and the Cauchy–Schwarz’s inequality, the contribution of the first twoterms of the latter displayed expression are bounded by

K

T

X½Ta�

t¼1

XT

r¼tþ1

þXT

t¼½Ta�þ1

Xtþ½Ta�

r¼tþ1

( )1

ðrþ ‘Þd=2¼ oð1Þ,

whereas the contribution due toPT

t¼½Ta�þ1

PTr¼tþ½Ta�þ1 is bounded by

K

T

XT

t¼½Ta�þ1

XT

r¼tþ½Ta�þ1

1

j r� tjð1þdÞ=2¼ oð1Þ


by Lemma 6.4 and because d41. (Notice that ‘o½Ta�.) Then, Lemma 6.6 and (2.3) implythat to complete the proof it suffices to show that

sup

s¼1;...;eT1eTX

s

j¼1

1

T

XT

t¼1

XT

r¼1

vtvryr

s2reiðt�rÞlj � Varðz2t ÞE

yr

s2r

� � !��, (6.18)

sup

s¼1;...;eT1eTX

s

j¼1

1

T

XT

t¼1

XT

r¼1

vt

yr

s2reiðt�rÞlj �

1

2gj

!�� (6.19)

are opð1Þ. First, by the triangle inequality, (6.18) is bounded by

K1

T

XT

t¼1

v2tyt

s2t� Varðz2t ÞE

yt

s2t

� �� þ K sup

s¼1;...;eT1eT X

s

j¼1

wj

��.

The first term of the last displayed expression is clearly opð1Þ by ergodicity of v2t yt=s2t , by

Lemma 6.2 and Markov’s inequality.To complete the proof that (6.18) is opð1Þ, we need to show that the second term of the

last displayed expression is also opð1Þ. First, that term is bounded by

K1eT XeT

j¼1

kwj � ewjk þ K sup

s¼1;...;eT1eT X

s

j¼1

ewj

��. (6.20)

The expectation of the first term of (6.20) is o(1) by Lemma 6.5 part (a). Next, let k ¼

0; . . . ; ½ eT$� � 1 with 0o$o1, and denote W ¼ 1�$. Then, by standard inequalities, the

second moment of the second term of (6.20) is bounded by

KE1eT2

sup

s¼1;...;eTXs

j¼1

�XkðsÞ½eTW

�

j¼1

8><>:9>=>;ewj

��2

þ KE1eT2

sup

s¼1;...;eTXkðsÞ½eTW

�

j¼1

ewj

��2

, (6.21)

where kðsÞ denotes the value of k ¼ 0; . . . ; ½ eT$� � 1 such that kðsÞ½ eTW

� is the largest integer

smaller than or equal to s, and using the conventionPd

c � 0 if doc.

From the definition of kðsÞ, we obtain that the second term of (6.21) is bounded by

EKeT2

sup

k¼0;...;½eT$

��1

Xk½eTW�

j¼1

ewj

��2

pKeT2

X½eT$

�

k¼1

EXk½eTW�

j¼1

ewj

��2

by (6.3) with p ¼ 2 there. But, by Lemma 6.5 part (b) and then because 0o$o1, the right-hand side of the last displayed inequality is bounded by

KeT2

X½eT$

�

k¼1

Xk½eTW�

j;l¼1

jj � lj�1þ ¼ OðT$�1 logTÞ ¼ oð1Þ.


To complete the proof that ð6:20Þ ¼ opð1Þ, we need to show that the first term in (6.21) iso(1). To that end, we note that this term is bounded by

EKeT2

sup

k¼0;...;½eT$

��1

sup

s¼1þk½eTW�;...;ðkþ1Þ½eTW

�

Xs

j¼1þk½eTW�

ewj

��2

.

By (6.3) and then Lemma 6.5 part (b), the last displayed expression is bounded by

KeT2

X½eT$

��1

k¼0

Xðkþ1Þ½eTW�

s¼1þk½eTW�

EXs

j¼1þk½eTW�

ewj

��2

pK

T2

X½eT$

��1

k¼0

Xðkþ1Þ½eTW�

s¼1þk½eTW�

ðs� k½ eTW�Þ ¼ oð1Þ,

because 0o$. So, the second term of (6.21) is o(1), and hence by Markov’s inequality, thesecond term of (6.20) is opð1Þ, which concludes the proof of (6.18) .Finally (6.19) follows similarly. First, it is bounded by

1eT XT

j¼1

1

T

XT

t¼1

vt

yt

s2t

��þ sup

s¼1;...;eT1eT X

s

j¼1

1

T

XT

1¼tar

vt

yr

s2r� E vt

yr

s2r

� �� cos ððt� rÞljÞ

!��

þ sup

s¼1;...;eT1eT X

s

j¼1

1

T

XT

1¼tor

E vt

yr

s2r

� �cosððt� rÞljÞ

!�

1

2gj

��.

The first term is clearly OpðT�1=2Þ by C1 and that Ekyt=s

2t k

2oK . The proof that the secondterm is opð1Þ follows step by step that of Lemma 6.5 and so it is omitted. The third term isalso o(1) because Lemma 6.4 and standard arguments imply that

PTt¼1 Eðv1

ytþ1

s2tþ1

Þeitlj�12gj ¼ oð1Þ. &

The following Lemmas are used in the proof of the validity of the bootstrap. Inparticular, Lemmas 6.9–6.16 will be employed to show the consistency of the bootstrap

estimator by�, e.g. Proposition 3.1, whereas the remaining ones are used for the weak

convergence of the (bootstrap) process T�ð½ eTW�Þ. In what follows, E�ðxtÞ denotes thebootstrap expectation of the random variable xt. That is, E�ðxtÞ ¼ EðxtjZT Þ andPr�fxtpxg ¼ Prfxtpx jZT g.

Lemma 6.8. Assuming C1–C7, for any 0oWoi, 1oBp2, and some d40,

ðaÞ E�bz�Wt ¼ EzWt ð1þOpðT�dÞÞ; ðbÞ E�x�B1�‘ ¼ EzB1�‘ðm

B0 þOpðT

�dÞÞ,

where the OpðT�dÞ is independent of tpT .

Proof. By definition, E�ðbz�Wt Þ ¼ T�1PT

t¼1bzWt . Now proceed as with the proof of Lemma 6.1part (b), cf. (6.7), to conclude part (a). Part (b), follows because by definition ofx�1�‘,E

�x�B1�‘ ¼ bmBT�1PTt¼1 bzBt . The conclusion now follows by part (a) and (2.3). &

As in (5.1), we can write s�2t ðyÞ as

mþXtþ‘�2k¼1

X1j1;...;jk¼1

bj1 ðzÞ � � � bjkðzÞz�2t�j1

� � � z�2t�Pk

s¼1js

IXk

s¼1

jsotþ ‘ � 1

!þ btþ‘�1

1 ðzÞz�2t�1 � � � z�22�‘x

�21�‘. ð6:22Þ


For easy reference, we recall our definitions of s�2t ðq; yÞ and a�t ðq; yÞ. That is,

s�2t ðq; yÞ ¼ mþXtþ‘�2k¼1

X1j1;...;jk¼1

bj1 ðzÞ � � � bjkðzÞz�2t�j1

� � � z�2t�Pk

s¼1js

IXk

s¼1

jsoq

!þ btþ‘�1

1 ðzÞz�2t�1 � � � z�22�‘x

�21�‘IðtoqÞ,

a�t ðq; yÞ ¼qqy


j¼1

qqy

bjðzÞ� �

z�2t�js�2t�jðq; yÞ,

dropping reference to by and/or bz when evaluated at by. Notice that a�t ðtþ ‘ � 1; yÞ¼: a�t ðyÞ isequal to y�t ðyÞ.


ðaÞ E� js�2t � s�2t ðqÞjr ¼ Opðq

�dÞ; ðbÞ E�a�t � a�t ðqÞ

s�2t

�� p ¼ Opðq�d log qÞ.

Proof. We begin with part (a). First, by definition s�2t � s�2t ðqÞ isXtþ‘�2k¼1

X1j1;...;jk¼1;qo

Pk

s¼1jsotþ‘�1

bbj1 � � �bbjk

z�2t�j1� � � z�2t�j1��jk

þ bbtþ‘�1

1 z�2t�1 � � � z�22�‘x

�21�‘IðtXqÞ.

Hence, E�js�2t � s�2t ðqÞjr is bounded by

KXtþ‘�2

k¼1þ½log q1þd �

E�jz�2rt j

XT

j¼1

bbr

j

!k��

��þ ðbbr

1E�z�2r1 Þ

tþ‘�2E�x�2r1�‘IðtXqÞ

þ KX½log q1þd �

k¼1

E�ðz�2rt Þ

kXT

j1;...;jk¼1;qoPk

s¼1jsotþ‘�2

bb2r

j1� � � bb2r

jk

0B@1CA

��. ð6:23Þ

Now because for some d40,

E�z�2rt ¼ Ez

2r1 ð1þOpðT

�dÞÞ,

by Lemma 6.8 and that Ejz2r1 jP1

j¼1 supz2Ubrj ðzÞo1, we have that with probability

approaching one

E�jz�2r1 j

X1j¼1

supz2U

brj ðzÞo1. (6.24)

So, (6.23) and hence E�js�2t � s�2t ðqÞjr ¼ Opðq

�dÞ because supz2U brj ðzÞpKj�1�d .

Next, we show part (b). Proceeding as in Lemma 6.2, e.g. Holder’s inequality as in RZ’sLemma 6, a typical element of the left side is bounded by

KE�Xtþ‘�1

j¼qþ1

bbð1Þpjbbr�p

j z�2rt�j s

�2rt�j

�� ¼ Opð1Þ

Xtþ‘�1j¼qþ1

bbð1Þpjbbr�p

j

��,


by part (a) with q ¼ 1 there. The conclusion now follows because (2.3) and C5 imply that,for k ¼ 0; 1 and with b

ð0Þj ¼ bj,

bbðkÞj ¼ bðkÞj þ b

ðkþ1Þj ðezÞðbz� z0Þ ¼ b

ðkÞj ð1þOpðT

�dÞÞ (6.25)

when jpT , and then because supzbð1Þpj ðzÞb

r�pj ðzÞ ¼ Oðj�ð1þdÞð1�ZÞÞ by C4–C6 with Z small

enough. &

Henceforth, we define s�2t;�s and a�t;�s as s�2t and a�t , respectively, but without the terms

involving fbz�2r gspr.


E�s�2t � s�2t;�1

s�2t

��p

þ E�a�t � a�t;�1

s�2t

�� p ¼ Op‘Ið‘ptÞ

t1þdþ

Iðto‘Þ

td

� �.

Proof. We shall consider only the first term on the left, being the proof of the secondterm identically handled. Arguing as in the proof of Lemma 6.3 and becausejðs�2t � s�2t;�1Þ=s

�2t joK, we have that the first term on the left is bounded by

KX½log t1þd �

k¼1

þXtþ‘�2

k¼1þ½log t1þd �

8<:9=;E�ðz

�2rt Þ

kXtþ‘�2j1¼1

� � �Xtþ‘�2�Pk�2

s¼1js

jk�1¼1

Xtþ‘�2�Pk�1

s¼1js

jk¼t�2�Pk�1

s¼1js

bbr

j1. . . bbr

jk

þ ðbbr

1E�z�2r1 Þ

tþ‘�1E�x�2r1�‘.

Because (6.25), the contribution due toP½log t1þd �

k¼1 is Opð‘Ið‘ptÞ

t1þd þIðto‘Þ

td Þ by Lemma 6.8 and

proceeding as in Lemma 6.9. On the other hand, the contribution due toPtþ‘�2

k¼1þ½log t1þd � is

bounded byPtþ‘�2

k¼1þ½log t1þd �ðE�z�2rt�j

Pjbbr

j Þk¼ Opðt

�1�dÞ by (6.24) . Finally, clearly the third

term is also Opðt�1�d Þ using (6.24). &

Henceforth fKtgtX1 will denote a sequence of Opð1Þ random variables.

Lemma 6.11. Assuming C1–C7, for any 0ovoi=2,

E� supy2Y

s�2t

s�2t ðyÞ

� �v

¼ Kt. (6.26)

Proof. Proceeding as in with the proof of Berkes et al.’s (2003b) Lemma 6.6, we have thatfor any MX1,

s�2t

s�2t ðyÞpKM

YMk¼1

ð1þ z�2t�kÞ

!1PM

k¼1z�2t�k

.


Thus, by Holder’s inequality and that fz�t gTt¼1�‘ is a sequence of iid random variables, we

have that for any 0ovoi=2, the left side of (6.26) is bounded by

KM E�YMk¼1

ð1þ z�2t�kÞ

!i=20@ 1A2v=i

E�1PM

k¼1z�2t�k

!vi=ði�2vÞ0@ 1Aði�2vÞ=i

¼ KMðE�ð1þ z�2t Þi=2Þ2Mv=i E�

1PMk¼1z

�2t�k


¼ KM 1

T

XT

t¼1

ð1þ bz2t Þi=2 !2Mv=i

1

TM

XT

t1;...;tM¼1

1PMk¼1bz2tk


.

So, using that bz2t ¼ z2tbs�2t ðs

2t �

bs2t Þ þ z2t , Lemma 6.1 implies that we have that

1

T

XT

t¼1

jbztji ¼

1

T

XT

t¼1

j ztji þ opð1Þ; E

XMk¼1

z2tk

!�vi=ði�2vÞ

oK

by C1 and that ergodicity of z2t implies that T�1PT

t¼1 j ztji!P E j ztj

i. So, (6.26) holdstrue. &

Lemma 6.12. Assuming C1–C7, for any pX1,

E� supz2U

P1pkptþ‘�1b

ð1Þk ðzÞx

�2t�k

1þP

1pkptþ‘�1bkðzÞx�2t�k

��p

¼ Kt. (6.27)

Proof. Arguments in Lemma 6 of RZ indicate that the left side of (6.27) is bounded by

KE�Xtþ‘�1k¼1

supz2Uj bð1Þk ðzÞj

pbkðzÞx�2rt�k

!¼ K

Xtþ‘�1k¼1

k�grð1�ZÞE�ðs�2rt�kÞ1

T

XT

t¼1

bz2rt

!.

Then, choose Kt as the right side of the last displayed equality to conclude by summabilityof k�grð1�ZÞ and since Lemma 6.9 implies that

T�1XT

t¼1

bz2rt sup1pkotþ‘�2

E�s�2rt�k ¼ T�1XT

t¼1

bz2rt E�ðs�2rt ð2ÞÞð1þOp� ð1ÞÞ

and that clearly E�ðs�2rt ð2ÞÞ ¼ Opð1Þ and that by Lemma 6.1, T�1PTt¼1

bz2rt is also Opð1Þ. &

Lemma 6.13. Let hðat; . . . ; at�p; yÞ be a continuous differentiable function in all its arguments

with finite second moments. Then, under C1–C7,

E�1

T

XT

t¼1

ðhðz�2t ; . . . ; z�2t�p;

byÞ � hðz2t ; . . . ; z2t�p; y0ÞÞ

�� ¼ Op

p1=2

T1=2

� �.


Proof. It suffices to show that

ðaÞ E�1

T

XT

t¼1

ðhðz�2t ; . . . ; z�2t�p;

byÞ � E�hðz�2t ; . . . ; z�2t�p;

byÞÞ��2

¼ Opp

T

,

ðbÞ E1

T

XT

t¼1

ðhðz2t ; . . . ; z2t�p; y0Þ � E�hðz�2t ; . . . ; z

�2t�p;

byÞÞ�� ¼ O

p1=2

T1=2

� �.

We begin with part (a). Because fz�2t gTt¼1�‘ is an iid sequence, it implies that

hðz�2t ; . . . ; z�2t�p;

byÞ is a p-dependent process, so that the left side is bounded by

Kp

TE�h2ðz�2t ; . . . ; z

�2t�p;

byÞ ¼ Kp

Tpþ2

XT

t1;...;tpþ1¼1

h2ðbz2t1 ; . . . ;bz2tpþ1

;byÞ.Now using that bz2t ¼ z2t

bs�2t ðs2t �

bs2t Þ þ z2t , an obvious extension of Lemma 6.1 and thecontinuous differentiability of h2, it is easily observed that

1

Tpþ1

XT

t1;...;tpþ1¼1

h2ðbz2t1 ; . . . ;bz2tpþ1

;byÞ ¼ 1

Tpþ1

XT

t1;...;tpþ1¼1

h2ðz2t1 ; . . . ; z

2tpþ1; y0Þ þ ðby� y0ÞZT ,

where ZT is a vector of random variables such that E jZT joK . So, the proof of part ðaÞ iscompleted if the first term on the right of the last displayed expression is Opð1Þ, whichfollows by standard V-statistics theory, see Serfling (1980).Next we show part (b). Proceeding as with part (a), the left side is bounded by

1

Tpþ1

XT

t1;...;tpþ1¼1

hðz2t1 ; . . . ; z2tpþ1; y0Þ �

1

T

XT

t¼1

hðz2t ; . . . ; z2t�p; y0Þ

��þ jby� y0 j jZT j,

which is OpðT�1=2Þ, because by V- and U-statistics theory, we have that

E1

Tpþ1

XT

t1;...;tpþ1¼1

hðz2t1 ; . . . ; z2tpþ1; y0Þ �

T

pþ 1

!�1 XT

t1a��atpþ1¼1

hðz2t1 ; . . . ; z2tpþ1; y0Þ

��2

¼ o1

T

� �

ET

pþ 1

!�1 XT

t1a��atpþ1¼1

ðhðz2t1 ; . . . ; z2tpþ1; y0Þ � Ehðz2t1 ; . . . ; z

2tpþ1; y0ÞÞ

��2

¼ O1

T

� �.

On the other hand,

E1

T

XT

t¼1

hðz2t ; . . . ; z2t�p; y0Þ � Ehðz2t ; . . . ; z

2t�p; y0Þ

��2

¼ O1

T

� �by standard arguments. Notice that because fz2t gt2Z is a sequence of iid random variables,Ehðz2t1 ; . . . ; z

2tpþ1; y0Þ ¼ Ehðz2t ; . . . ; z

2t�p; y0Þ for t1at2a � � �atpþ1. &


Lemma 6.14. Assuming C1–C7, for all y 2 Y,

E�1

T

XT

t¼1

ðln s�2t ðyÞ � ln s2t ðyÞÞ

�� ¼ opð1Þ. (6.28)

Proof. First, denoting by �s�2t ðyÞ and �s2t ðyÞ the first two terms of (6.22) and (5.1),respectively, we have that ln s�2t ðyÞ � ln s2t ðyÞ is

ðln s�2t ðyÞ � ln �s�2t ðyÞÞ þ ðln �s�2t ðyÞ � ln �s2t ðyÞÞ þ ðln �s2t ðyÞ � ln s2t ðyÞÞ. (6.29)

Next, because j ðs�2t ðyÞ � �s�2t ðyÞÞ=s�2t ðyÞ joK , for any p40,

E�s�2t ðyÞ � �s�2t ðyÞ

s�2t ðyÞ

�� ppK supz2U

br1ðzÞE

�ðz�2rt Þ

��tþ‘�1

E�jx�2r1�‘ j.

Hence by the mean value theorem, we have that the contribution of the first term of(6.29) into the left of (6.28) is, for some b 2 ð0; 1Þ,

E�1

T

XT

t¼1

s�2t ðyÞ � �s�2t ðyÞbs�2t ðyÞ þ ð1� bÞ �s�2t ðyÞ

��p 1

T

XT

t¼1

supz2U

br1ðzÞE

�ðz�2rt Þ

��tþ‘�1

E� jx�2r1�‘ j,

which is OpðT�1Þ because by Lemma 6.8 part ðbÞ, E� jx

�2r1�‘ j ¼ Opð1Þ and (6.24). Proceeding

similarly we can show that the contribution of the third term of (6.29) into the left of (6.28)is also OpðT

�1Þ by Markov’s inequality.So, it remains to examine the contribution of the second term of (6.29) into the left of

(6.28), that is for all ao1,

E�1

T

X½Ta�

t¼1

þXT

t¼½Ta�þ1

( )ðln �s�2t ðyÞ � ln �s2t ðyÞÞ

��.

It is obvious that the contribution of the first sum is opð1Þ. So we only need to examine thecontribution of the second sum. To that end, write �s�2t ðyÞ as

mþXm

l¼1

XMk1;...;kl¼1

bk1ðzÞ � � � bkl

ðzÞz�2t�k1� � � z�2t�k1��kl

IðmMotÞ þ Z�t .

First, by C4 and using Lemma 6 of RZ, we have that E� jZ�t = �s�2t ðyÞ j is, for some d40,

bounded by

KXtþ‘�2

l¼mþ1

E�z�2rt

X1k¼1

supz2U

brkðzÞ

!l

þ KM�1Xm

l¼1

E�z�2rt

X1k¼1

supz2U

brkðzÞ

!l

¼ opð1Þ

choosing M and m large enough, by (6.24) . Recall that t4½Ta�. Likewise with the first twoterms of (5.1). So, to complete the proof it suffices to show that

E�1

T

XT

t¼½Ta�þ1

ln

mþPm

l¼1

PMk1;...;kl¼1


ðzÞz�2t�k1� � � z�2

t�Pl

s¼1ks

IðmMotÞ

mþPm

l¼1

PMk1;...;kl¼1


ðzÞz2t�k1� � � z2

t�Pl

s¼1ks

IðmMotÞ

0B@1CA

��


is opð1Þ, which follows by Lemma 6.13 choosing

hðxt�1; . . . ; xt�mM Þ ¼ ln mþXm

l¼1

XMk1;...;kl¼1


ðzÞxt�k1� � � x

t�Pl

s¼1ks

IðmMotÞ

!

there for a generic sequence fxtgt2Z. &


E�1

T

XT

t¼1

x�2t

s�2t ðyÞ�

x2t

s2t ðyÞ

� �� ¼ opð1Þ. (6.30)

Proof. By definition, the left side of (6.30) is bounded by

E�1

T

XT

t¼1

s�2t

s�2t ðyÞz�2t �

s2ts2t ðyÞ

z2t

� ��þ 1

T

XT

t¼1

s2t � s2ts2t ðyÞ

z2t

� ��.

Because by Lemma 6.11 and the proof of Theorem 2 of RZ, s�2t =s�2t ðyÞ and s2t =s

2t ðyÞ have

finite second moments and z�2t and z2t are both iid sequences with unit mean and finitevariance, it suffices to show that

E�1

T

XT

t¼1

s�2t

s�2t ðyÞ�

s2ts2t ðyÞ

� �� ¼ opð1Þ, (6.31)

because using the arguments given after (6.16) s2t =s2to2 with probability approaching 1

and then by Berkes et al.’s (2003b) Lemma 6.6, we have that

XT

t¼1

Es2t � s2ts2t ðyÞ

�� pXT

t¼1

Es2t � s2t

s2t

�� 2 !1=2

Es2t

s2t ðyÞ

� �2 !1=2

¼ oðT1=2Þ

by Lemma 6.2. But because (6.25) and Lemma 6.12 imply that

E� supy2Y

qs�2t ðyÞ=qys�2t ðyÞ

�� pKE� supz2U

Ptþ‘�2k¼1 b

ð1Þk ðzÞx

�2t�k

1þPtþ‘�2

k¼1 bkðzÞx�2t�k

�� ¼ Kt,

we have that (6.31) holds true if

E�1

T

XT

t¼1

s�2t ðy0Þs�2t ðyÞ

�s2t

s2t ðyÞ

� �� ¼ opð1Þ,

which is the case proceeding as with the proof of Lemma 6.14. &

The next lemma shows the (bootstrap) equicontinuity condition, which together with theprevious two lemmas allow us to obtain the uniform convergence of the bootstrapobjective function in (3.2) to that in (2.1) .


E� supy1;y22Y

1

j y1 � y2 jx�2t

s�2t ðy1Þ�

x�2t

s�2t ðy2Þ

�� !

¼ Kt, (6.32)


E� supy1;y22Y

1

j y1 � y2 jj log s�2t ðy1Þ � log s�2t ðy2Þ j

!¼ Kt. (6.33)

Proof. We begin showing (6.32) . By the mean value theorem,

x�2t

s�2t ðy1Þ�

x�2t

s�2t ðy2Þ

�� pK j y1 � y2 jx�2t

s�2t ðyÞ

�� y�t ðyÞ

s�2t ðyÞ

��,

where y is an intermediate point between y1 and y2. Next for 0ovoi, by Lemmas 6.8 and6.11 and the Cauchy–Schwarz inequality,

E� supy2Y

x�2t

s�2t ðyÞ

�� v=2p E� supy2Y

s�2t

s�2t ðyÞ

�� v� �1=2

E� supy2Y

z�2t

s�2t ðyÞ

�� v� �1=2

¼ Kt.

On the other hand, Lemma 6.12 implies that E�supy2Y j ðs�2t ðyÞÞ

�1y�t ðyÞ j ¼ Kt. So, the leftside of (6.32) is Kt. To complete the proof we have to show the inequality in (6.33) whichfollows by similar arguments and so it is omitted. &

For all t ¼ 1; . . . ;T , we shall abbreviate z�2t � 1 by v�t . Also, denote

ew�j ¼ T�1XT

t;r¼1;tar

v�t v�ra�rs�2r

eiðt�rÞlj .

Lemma 6.17. Assuming C1–C7, for 0oj; kp eT ,

E�ðew�j ew�0�kÞ ¼ Opðjj � kj�1þ Þ.

Proof. Because v�t is a zero mean iid sequence, the left side of the last displayed equality is

E�1

T2

XT

t1pt2

v�t1v�t2

XT

t2or

a�rs�2r

� �2

eiðt1�rÞlj�iðt2�rÞlk

��

¼ E�1

T2

XT

1ot1ot2

v�t1v�t2

XT

t2or

a�rs�2r

� �2

�a�r;�t1

s�2r;�t1

!28<:

9=;eiðt1�rÞlj�iðt2�rÞlk

��

þ1

T2

XT

t¼1

XT

tor

E�v�2t

a�rs�2r

� �2

eiðt�rÞlj�k

��.

By Lemma 6.10, the first term on the right of the last displayed inequality is bounded by

K‘

T2

XT

1ot1ot2or;‘or

j t1 � rj�1�dOpð1Þ þOpð‘2=T2Þ ¼ opð1=T1=2Þ

because ‘ ¼ oðlog2 log TÞ is arbitrarily small. Now, the second term on the right of the lastdisplayed inequality is bounded by

1

T2

XT

t¼1

XT

tor

E�v�2t

a�rs�2r

� �2

�a�ts�2t

� �2( )

eiðt�rÞlj�k

��þ E�v�2t

1

T2

XT

t¼1

E�a�ts�2t

� �2XT

tor

eiðt�rÞlj�k

��.


Proceeding as in the proof of Lemma 6.10, the first term is OpðT�1Þ, whereas the second

term is

E�v�2t

1

1� eilj�k

1

T2

XT

t¼1

E�a�ts�2t

� �2

ð1� eitlj�k Þ

�� ¼ Opðj j � k j�1þ Þ,

since E�ða�t =s�2t Þ

2¼ Opð1Þ. This completes the proof. &

Lemma 6.18. Assuming C1–C7, we have that for all p42 and some d40,

supt¼1;...;T

bs�2t � s�2t

s�2t

�� ¼ Op� ðT

�ðp�2Þ=2pÞ. (6.34)

Proof. Proceeding as in Lemma 6.1, the left side of (6.34) is bounded by

kby� � byk supt¼1;...;T

y�ts�2t

�� þ kby� � by k2 supt¼1;...;T

qqy0 y

�t ðeyÞ

s�2t

��,

where ey is an intermediate point between by� and by. Using (6.3), for all p40, we have that

E�fðsupt¼1;...;Tky�ts�2t

k þ supy2Ykqqy0

y�t ðyÞ

s�2t

kÞgp ¼ OpðTÞ. But, by� � by ¼ Op� ðT�1=2Þ by Proposition

3.2. So, the conclusion follows by standard arguments. Observe that by Lemma 6.9, say,

y�ts�2t

�� p

pOpðq�d log qÞ þ

a�t ðqÞ

s�2t ðqÞ

�� p

¼ Opð1Þ (6.35)

because E�ka�t ðqÞ

s�2t ðqÞkppKE�k

a�t ðqÞ

s�2t ðqÞkr ¼ Opð1Þ proceeding as in Lemma 6.9. &

Define Q�j as the bootstrap analogue of Qj given in the statement of Lemma 6.6.

Lemma 6.19. Assuming C1–C6, we have that for all j ¼ 1; . . . ; eT ,

Iz�2;j � Iz�2;j þ 2ðby� � byÞ0Q�j ¼ op� ðT�1=2Þ.

Proof. First, by definition of z�2t and z�2t , Iz�2;j � Iz�2;j is

1

T

XT

t¼1

z�2t

s�2t �bs�2tbs�2t

eitlj

��2

þ2

T

XT

t;r¼1

z�2r z�2t

s�2t �bs�2tbs�2t

cos ððt� rÞljÞ. (6.36)

On the other hand,XT

t¼1

z�2t

bs�2t � s�2tbs�2t

eitlj ¼XT

t¼1

z�2t

bs�2t � s�2t

s�2t

eitlj �XT

t¼1

z�2t

ðs�2t �bs�2t Þ

2

s�2tbs�2t

eitlj . (6.37)

Using an expansion similar to that in (6.2), we have that the right side of (6.37) is

ðby� � byÞ0XT

t¼1

z�2t

y�ts�2t

eitlj þ ðby� � byÞ0XT

t¼1

z�2t

qqy0 y

�t ðeyÞ

s�2t

eitlj ðby� � byÞ� ðby� � byÞ0XT

t¼1

z�2t

y�t y�0t

s�2tbs�2t

eitlj ðby� � byÞ �XT

t¼1

z�2t

ððby� � byÞ0 qqy0 y

�t ðeyÞðby� � byÞÞ2

s�2tbs�2t

eitlj

� 2ðby� � byÞ0XT

t¼1

z�2t

qqy0 y

�t ðeyÞðby� � byÞy�0ts�2tbs�2t

eitlj ðby� � byÞ.


So, using the previous Lemma 6.18 with pX4 there, that ðby� � byÞ ¼ Op� ðT�1=2Þ by

Proposition 3.2 and that E� jey�t þ y�t =s�2t j ¼ Opð1Þ proceeding as with (6.35), we have that

the last displayed expression, and thus the left side of (6.37), is

ðby� � byÞ0XT

t¼1

z�2t

y�ts�2t

eitlj þ ðby� � byÞ0XT

t¼1

z�2tey�t eitlj ðby� � byÞ þOp� ðT

�1=4Þ,

because by Lemma 6.8 and that fv�t gTt¼1 is iid sequence, we have that E� j

PTt¼1 v�t e

itlj j ¼

OpðT1=2Þ and that, say, proceeding as in the proof of Lemmas 6.2 and 6.8, we have that

E� jPT

t¼1 s��4t a�t

qqy0 a

�t e

itlj j ¼ opðTÞ. Notice that proceeding as with (6.4), by C5 and (2.3)for pX1,

qqy0 y

�t ðeyÞ � q

qy0 y�t

s�2t

��

p

pkby� � bykp

Ptþ‘�1j¼1 supz b

ð2Þpj ðzÞb

r�pj ðzÞx

�2rt�j

s�2t

��

p

¼ Op� ðT�p=2Þ,

by summability of supzbð2Þpj ðzÞb

r�pj ðzÞ and arguing similarly as with Lemma 6.18. Recall

that by definition ey�t ¼ s��2tqqy0 a

�t ðeyÞ þ s��4t a�t a0�t .

We next show that

1

T1=2

XT

t¼1

z�2t

y�ts�2t

eitlj ¼ op� ðT1=4Þ. (6.38)

Recalling that y�t ¼ a�t , we have that the left side of (6.38) is

1

T1=2

XT

t¼1

v�ty�ts�2t

eitlj þ1

T1=2

XT

t¼1

a�ts�2t

�a�t ðqÞ

s�2t ðqÞ

� �eitlj þ

1

T1=2

XT

t¼1

a�t ðqÞ

s�2t ðqÞeitlj . (6.39)

The first term of (6.39) is Op� ð1Þ because v�t is iid with finite second moments and by (6.35),E� jey�t þ y�t =s

�2t j

2 is Opð1Þ, while the second term is OpðT1=6Þ by Lemma 6.9 and choosing

q ¼ Oðt1=3Þ there. Finally, the third term is

1

T1=2

X½Ta�

t¼1

a�t ðqÞ

s�2t ðqÞeitlj þ

1

T1=2

XT

t¼½Ta�þ1

a�t ðqÞ

s�2t ðqÞ�

a�t ðTaÞ

s�2t ðTaÞ

� �eitlj þ

1

T1=2

XT

t¼½Ta�þ1

a�t ðTaÞ

s�2t ðTaÞeitlj .

The first term is clearly Opð1Þ, whereas the second term is op� ð1Þ proceeding as in Lemma6.10. Finally the third term is opðT

1=4Þ, because for ja0;T ,PT

t¼1 eitlj ¼ 0 and that

a�t ðTaÞ=s�2t ðT

aÞ is a Ta-dependent process with finite second moments and choosing ao12.

Proceeding similarly, we have that T�1PT

t¼1 z�2tey�t ¼ op� ð1Þ and so (6.38) is op� ðT

1=4Þ andtherefore ð6:37Þ ¼ op� ðT

1=4Þ.To complete the proof, it remains to show that the second term of (6.36) is�2ðby� � byÞ0Q�j þ op� ðT

�1=2Þ. Proceeding as above, the second term of (6.36) is

2ðby� by�Þ0

T

XT

t;r¼1

z�2t

y�ts�2t

z�2r cos ððt� rÞljÞ

� 2ðby� � byÞ0

T

XT

t;r¼1

z�2tey�t z�2r cos ððt� rÞljÞð

by� � byÞ þ op�1

T1=2

� �.

The first term of the last displayed expression is easily seen to be �2ðby� � byÞ0Q�jbecause for ja0;T ,

PTt¼1 e

itlj ¼ 0, whereas the second term is op� ðT�1=2Þ because


supj jPT

t¼1 z�2t eitlj j ¼ OpðT1=2 log TÞ by An et al. (1983) and then Lemma 6.8, and then

because proceeding as with the proof of (6.38) we have thatPT

t¼1y�ts�2t

z�2t eitlj ¼ OpðT3=4Þ.

This completes the proof of the lemma. &


sup

s¼1;...;eT1eT1=2

Xs

j¼1

Iz�2;j � Iz�2;j þ 2Varðz2t ÞEyt

s2t

� �þ gj

� �0ðby� � byÞ� ��

�� ¼ op� ð1Þ.

Proof. First, we notice that

1

T

XT

t¼1

E�v�2t E�y�ts�2t

� �� Varðz2t ÞE

yt

s2t

� �� ¼ opð1Þ (6.40)

because by Lemma 6.8, E�v�2t � Varðz2t Þ ¼ opð1Þ and then because Lemma 6.9 implies that

E�y�ts�2t

� �¼ E�

a�t ðqÞ

s�2t ðqÞ

� �þOpðq

�1Þ ¼ EatðqÞ

s2t ðqÞ

� �þOpðq

�1Þ

by Lemma 6.13 choosing hðz�2t ; . . . ; z�2t�qÞ ¼ a�t ðqÞ=s

�2t ðqÞ. (Recall that y�t ðyÞ ¼ a�t ðyÞ.) Then,

choosing q large enough, by Lemma 6.18, that by� � by ¼ Op� ðT�1=2Þ and proceeding as with

the proof of (6.17), it suffices to show that

sup

s¼1;...;eT1eT X

s

j¼1

1

T

XT

t¼1

XT

r¼1

v�t v�ry�rs�2r

a�t cosððt� rÞljÞ � Varðz20ÞEyt

s2t

� � !��, (6.41)

sup

s¼1;...;eT1eT X

s

j¼1

1

T

XT

t¼1

XT

r¼1

v�ty�rs�2r

cos ððt� rÞljÞ �1

2gj

!�� (6.42)

are op� ð1Þ. First, by the triangle inequality, (6.41) is bounded by

K1

T

XT

t¼1

v�2t

y�ts�2t

� Varðz2t ÞEyt

s2t

� �� þ K sup

s¼1;...;eT1eT X

s

j¼1

ew�j��

��. (6.43)

The first term of the last displayed expression is op� ð1Þ by (6.40). So, to complete the proof

of the contribution due to the second term into the left of (6.41) . To that end, let

k ¼ 0; . . . ; ½ eT$� � 1 with 0o$o1. Then, denoting W ¼ 1�$, by the triangle inequality,

the second moment of that term is bounded by

KE�1eT2

sup

s¼1;...;eTXs

j¼1

�XkðsÞ ½eTW

�

j¼1

8><>:9>=>;ew�j

��2

þ KE�1eT2

sup

s¼1;...;eTXkðsÞ½eTW

�

j¼1

ew�j��

��2

, (6.44)

where kðsÞ denotes the value of k ¼ 0; . . . ; ½ eT$� � 1 such that kðsÞ½ eTW

� is the largest integersmaller than or equal to s.


From the definition of kðsÞ, the second term of (6.44) is bounded by

E�1eT2

maxk¼0;...;½eT$

��1

Xk½eTW�

j¼1

ew�j��

��2

p1eT2

X½eT$

��1

k¼0

E�Xk½eTW�

j¼1

ew�j��

��2

by (6.3) with p ¼ 2 there. But, by Lemma 6.17 and then because $o1, the right hand sideof the last displayed inequality is bounded by

log TeT2

X½eT$

��1

l¼0

Xl½eTW�

j;k¼1

j j � k j�1þ KT ¼ OpðT$�1 log TÞ ¼ opð1Þ.

To complete the proof we need to show that the first term in (6.44) is opð1Þ. To that end,we note that this term is bounded by

E�1eT2

maxk¼0;...;½eT$

��1

max

s¼1þk½eTW�;...;ðkþ1Þ½eTW

�

Xs

j¼1þk½eTW�

ew�j��

��2

.

By (6.3), the last displayed expression is bounded by

1eT2

X½eT$

��1

k¼0

Xðkþ1Þ ½eTW�

s¼1þk½eTW�

E�Xs

j¼1þkeT=½eT$

�

ew�j��

��2

pKT

T2

X½eT$

��1

k¼0

Xðkþ1Þ½eTW�

s¼1þk½eTW�

ðs� k½ eTW�Þ

¼ opð1Þ,

because 0o$. So, the second term of (6.44) is opð1Þ, and hence by Markov’s inequality, thesecond term of (6.43) is opð1Þ, which concludes the proof of (6.41).

Finally (6.42) follows similarly. First, the left side is bounded by

1eT XT

j¼1

1

T

XT

1¼t

v�ty�ts�2t

��þ sup

s¼1;...;eT1eTX

s

j¼1

1

T

XT

1¼tar

v�ty�rs�2r

� E� v�ty�rs�2r

� �� eiðt�rÞlj

!��

þ sup

s¼1;...;eT1eT X

s

j¼1

1

T

XT

t;r¼1;tar

E� v�ty�rs�2r

� �cos ððt� rÞljÞ

!�

1

2gj

��.

The first term is clearly Op� ðT�1=2Þ because fv�t g

Tt¼1 is a zero mean iid random variables

and E�ky�t =s�2t k

2oK . The proof that the second term is opð1Þ follows step by stepthat of Lemma 6.17 and so it is omitted. The third term is also opð1Þ as we now show.BecauseX1

k¼½TK�

E v1yk

s2k

� �eiklj

��þ XT

k¼½TK�

E� v�1y�ks�2k

� �eiklj

�� ¼ OðT�KÞ þOpðT

�KÞ

by Lemmas 6.3 and 6.10, respectively, it suffices to show thatX½TK�

k¼1

E� v�1y�ks�2k

� �� E v1

yk

s2k

� �� eiklj

�� ¼ opð1Þ.


But this is the case because, for all k, Lemma 6.13 indicates that E jE�ðv�1y�ks�2k

Þ � Eðv1yk

s2kÞ j ¼

OðkT�1=2Þ and then choosing Ko1=4. Recall that by definition,y�ks�2k

depends only on

ðz�21�‘; . . . ; z�2k Þ and ‘ ¼ oðlog2 log TÞ is arbitrarily small. &

Acknowledgments

The first author’s research was supported by ESRC Grant R000239936. Also, we thankthe comments of two referees on a previous version of the paper. Of course, all remainingerrors are our sole responsibility.

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a goodness-of-fit test for arch models

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