Javier Hidalgo

Goodness of fit for lattice processes

Keywords: JEL Classi…cation: C21, C23 …t tests, Spatial linear processes, Spectral domain, Bootstrap tests

The paper discusses tests for the correct speci…cation of a model when data is observed in a d-dimensional lattice, extending previous work when the data is collected in the real line. As it happens with the latter type of data, the asymptotic distribution of the tests are functionals of a Gaussian sheet process, say B ( ), 2 [0; ] d . Because it is not easy to …nd a time transformation h ( ) such that B (h ( )) becomes the standard Brownian sheet, a consequence is that the critical values are di¢ cult, if at all possible, to obtain. So, to overcome the problem of its implementation, we propose to employ a bootstrap approach, showing its validity in our context.

INTRODUCTION

The paper is concerned with testing the goodness of …t of a parametric family of models for data collected in a lattice. More speci…cally, we are concerned with the correct speci…cation (or model selection) of the dynamic structure with time series and/or spatial stationary processes fx (t)g t2Z de…ned on a d-dimensional lattice. The key idea of the test is to compare how close is the parametric and nonparametric …ts of the data to provide support for the null hypothesis. In the paper, we shall explicitly consider data for which d 3. The motivation lies in the fact that the most often type of data available in economics is when d = 2, say with agricultural or environmental data, or when d = 3. An important example of the latter is the spatial-temporal data sets, that is data collected in a lattice during a number of periods. However, we ought to mention that extensions to higher index lattice processes can be adapted under suitable modi…cations.

All throughout the paper we will assume that the (spatial) process fx (t)g t2Z d can be represented by the multilateral model (1.1) x (t) = X j2Z d (j) " (t j) ; X

j2Z d 2 (j) < 1 (0) = 1,
for some sequence f" (t)g t2Z d satisfying E (" (t)) = 0 and E (" (0) " (t)) = 2 " I (t = 0), where I ( ) denotes the indicator function. Notice that because our model is multilateral, the sequence f" (t)g t2Z d loses its interpretation as the "prediction"error or that it can be regarded as innovations. Under (1:1), the spectral density function of fx (t)g t2Z d , f ( ), can be factorized as

f ( ) = 2 "
(2 ) d j ( )j 2 , 2 d , where = ( ; ] and with ( ) = P j2Z d (j) e ij . The function ( ) summarizes the covariogram structure of fx (t)g t2Z d , which is the main feature to obtain accurate prediction/extrapolation and/or interpolation (kriging) in the case of spatial data. Notice that the ultimate aim when modelling data is nothing but to predict the future. Henceforth the notation "j " means the inner product of multi-indices j and of dimensional d. Also, any element a that belongs to Z d (or d ), the d-fold Cartesian product of the set Z (or ), will be referred to as a multi-index of dimension d. Also, we shall write, say, a = (a [START_REF] An | The maximum of the periodogram[END_REF] ; :::; a [d]) with the square brackets used to denote the components of a.

The aim of the paper is to test whether the data support the null hypothesis that ( ) belongs to a speci…c parametric family (1.2) H = f ( ) : 2 g , where R p is a proper compact parameter set. That is, we are interested on the null hypothesis (1.3) H 0 : 8 2 [ ; ] d and for some 0 2 , j ( )j 2 = j 0 ( )j 2 .

The alternative hypothesis H 1 is the negation of H 0 . Alternatively we could have formulated the null hypothesis in terms of the covariogram f (s)g s2Z d , where (s) = Cov (x (t) ; x (t + s)). That is, the null hypothesis is that the covariogram follows a particular parametric family, say f (s)g s2Z d = f # (s)g s2Z d , where from now on we denote # = 0 ; 2 " 0 . This is the case after observing that for any stationary spatial lattice process fx (t)g t2Z d , the spectral density f ( ) and the covariogram (s) are related through the expression (s)

# (s) = R d f ( ) e is d R d f # ( ) e is d ; s 2 Z d .
One parameterization of (1:1), or (2:12) below, is the ARM A …eld model

P (L) (x (t) ) = Q (L) " (t) ,
where denoting henceforth for muti-indeces z and j, z j = Y d `=1

z [`] j [`] with the convention that 0 0 = 1,

P (z) = X j2Z d (j) z j ; (0) = 1; Q (z) = X j2Z d (j) z j ; (0) = 1,
are …nite series in Z d . That is, only a …nite number of the (j) 0 s and (j) 0 s coe¢ cients are non-zero. For instance the ARM A …eld model given by k2 X j= k1

(j) (x (t j) ) = `2 X j= `1 (j) " (t j) (0) = (0) = 1
whose spectral density function is Notice that the ARM A …eld model becomes a causal representation if the polynomials Q (L) and P (L) are both unilateral. It is worth mentioning that [START_REF] Whittle | On stationary processes in the plane[END_REF] showed that, almost any given stationary bilateral scheme on a plane lattice, there corresponds a unilateral autoregression having the same spectral scheme although not necessarily of …nite order as is the case when d = 1. See also Guyon (1982a).

f ( ) = 2 " (2 )
Another parametric model of interest is the extension of the classical Bloom…eld (1973) exponential model, see also [START_REF] Whittle | On stationary processes in the plane[END_REF] Section 6, to processes in a lattice. These models are characterized as having a spectral density function

f # ( ) = 2 " exp ( X ` 0
(`; ) cos (` )

)

,

where " " denotes the lexicographical (dictionary) ordering which is de…ned as

j k , (9`> 0) (8i < `) (j [i] = k [i] ^j [`] < k [`]) ,
that is, if one of the terms j [`] < k [`] and all the proceeding ones are equal. For instance, when d = 2, we would then have that, say, ` 0 corresponds to the half plane of Z 2 , Z 2 = (` [START_REF] An | The maximum of the periodogram[END_REF] ; ` [START_REF] Anderson | Goodness-of-…t for autoregressive processes[END_REF]) 2 Z d : (` [START_REF] An | The maximum of the periodogram[END_REF] = 0 ^` [START_REF] Anderson | Goodness-of-…t for autoregressive processes[END_REF] < 0) _ ` [START_REF] An | The maximum of the periodogram[END_REF] < 0 . Observe that if we allowed `in the last displayed equality to belong to Z d the model would not be then identi…ed as cos (` ) = cos ( ` ).

When d = 1, the problem of testing a speci…c dependence structure of the data is very exhaustive and prominent. Di¤erent tests have been formulated using either the spectral density or the autocorrelation functions. Regarding the former, we can cite among others, the pioneer work by [START_REF] Grenander | Statistical Analysis of Stationary Time Series[END_REF] to test for the null hypothesis of white noise dependence. A classical test using the autocorrelation function is the Box and Pierce (1970) statistic. For a latter reference, see Delgado, Hidalgo and Velasco (2005) and references therein. In the paper, we have chosen to employ frequency domain techniques or to base the test in terms of the spectral density function, contrary to a "time domain" approach based on the covariance/variogram structure of the data.

Our tests fall into the category of goodness of …t tests as we do not specify any particular alternative model or family. The tests are based on a direct comparison between two estimates of the spectral density function in a way similar to the well known Hausman-Durbin-Wu's test. That is, they rely on the comparison of two estimates: one which is only consistent under the null, whereas the second (less e¢ cient) estimator is consistent under the maintained hypothesis. Although the literature when d > 1 is not very vast and exhaustive, some work has already been done, see for instance [START_REF] Diblasi | On the use of the variogram in checking the independence in spatial data[END_REF] or [START_REF] Crujeiras | Gooodnessof-…t tests for the spatial spectral density[END_REF]. However, our work di¤ers from theirs in that contrary to [START_REF] Diblasi | On the use of the variogram in checking the independence in spatial data[END_REF] we do test for general speci…cations and that contrary to [START_REF] Crujeiras | Gooodnessof-…t tests for the spatial spectral density[END_REF] our test does not involve any bandwidth or smoothing parameter. In fact, the latter approach uses the distance between a smooth estimator of the spectral density function and its parametric estimator under H 0 . This approach provides asymptotically distribution free tests under suitable conditions on the smoothing parameter, see for instance [START_REF] Hong | Consistent testing for serial-correlation of unknown form[END_REF] or [START_REF] Paparoditis | Spectral density based goodness-of-…t tests for time series models[END_REF] among others. However, the latter approach seems to be a mere artifact when testing for a particular parametric family and the …nal outcome of all these tests may depend on the arbitrary choice of the bandwidth parameter(s) for which no relevant theory is available for testing purposes. That is, there are not rules available on how to choose the bandwidth parameter with empirical data. In fact, we might face the strange situation that with the same data set two di¤erent practitioners might conclude di¤erently. The latter is clearly not very desirable from both theoretical or applied stand point of view. So, in this context, one of our main motivation is to extend goodness-of-…t tests examined and described when d = 1 to d 1, where we do not require the choice of any bandwidth parameter. For that purpose, we rely on the periodogram which although it is not a consistent estimator for f ( ), its integral is a consistent estimator of the spectral distribution function as the integral is the most natural smoothing algorithm.

The remainder of the paper is organized as follows. In the next section, we present the test and examine its asymptotic properties when the true value of the parameter 0 is known, whereas Section 3 extends these results to more realistic situations where we need to estimate the parameters of the model. Because, the asymptotic distribution of the test in the latter scenario is not pivotal and model dependent, Section 4 describes the bootstrap test showing its validity. Section 5 gives the proof of a series of lemmas employed in the proof of our main results in Section 6.

TESTS WHEN THE PARAMETERS ARE KNOWN

This section discusses and examines how we can test the null hypothesis H 0 given in (1:3). That is, the hypothesis

H 0 : f ( ) = 2 "
(2 ) d j 0 ( )j

H 0 : G 0 ( ) G 0 ( ) = d Y `=1 [`] for all 2 [0; ] d , where G ( ) = 2 R j (!)j 2 f (!) d! with the notation (2.2) Z = Z [1] ( [1]^0) Z [2]
[2]

:::

Z [d] [d]
.

Under H 0 , G 0 ( ) is the spectral distribution function of the lattice process f" (t)g t2Z and G 0 ( ) = 2 " . Notice that by symmetry of f ( ), it is irrelevant which coordinate we choose to belong only to [0; ] as the choice does not a¤ect the value of G ( ) and so the value of the test given below.

Given a record fx (t)g n t=1 and denoting henceforth N =

Y d `=1 n [`], a natural estimator of G 0 ( ) is e G ;N ( ) = 2 1 N [ñ = ] X j= [ñ = ] I x ( j ) j ( j )j 2 ,
where, for a generic sequence fv (t)g n t=1 , I v ( ) denotes the periodogram

I v ( ) = 1 N n X t=1 v (t) e it 2 ;
2 e d and similarly to the de…nition of R , we employ henceforth the notation

(2.3) [ñ = ] X j=[ñ = ] = [ñ[1] [1]= ] X j[1]=[ñ[1] [1]= ] + [ñ[2] [2]= ] X j[2]=[ñ[2] [2]= ]
:::

[ñ[d] [d]= ] X j[d]=[ñ[d] [d]= ]
,

where [q] + = max fjqj ; 1g. Also we have abbreviated [n [`] =2] by ñ [`] for `= 1; :::; d.

As usual we have excluded the frequency j = 0 from the sum

P [ñ = ] j=[ñ = ]
, so that we can take Ex (t) = 0 or assume that x (t) has been already centered around its sample mean. It is often the case that in real applications, in order to make use of the fast Fourier transform, the periodogram is evaluated at the Fourier frequencies, that is

k = k[1] ; :::; k[d]
0 , where with k [1] = 0; 1; :::; ñ [START_REF] An | The maximum of the periodogram[END_REF] and k [`] = 0; 1; :::

; ñ [`]
for `= 2; :::; d,

k[`] = 2 k [`] n [`] ; `= 1; :::; d.
Unfortunately, as noted by Guyon (1982b), due to nonnegligible end e¤ects, the bias of I x ( j ) does not converge to zero fast enough when d > 1, so that it would have unwanted consequences. One of these is that the Whittle estimator of #, see Guyon (1982b), does not have the standard asymptotic properties as when d = 1. Because of that, in the paper, we shall employ the taper periodogram de…ned as (2.4)

I T v ( j ) = w T v ( j ) 2 ,
for a generic sequence fv (t)g n t=1 and

w T v ( j ) = 1 ( P n t=1 h 2 (t)) 1=2 n X t=1 h (t) v (t) e it j
is the taper discrete Fourier transform. Tapering is primarily a technique employed to reduce the bias of the "standard"periodogram I v ( ). Notice that when h (t) = 1, we have that the taper discrete Fourier transform w T v ( j ) becomes the standard discrete Fourier transform (DFT). It is worth mentioning that to alleviate the bias problem, alternative procedures to tapering have been proposed. One of these proposals was due to Guyon (1982b), who replaced the periodogram by

I v ( k ) = 1 (2 ) d X h2D b v (h) e ih k , where b v (h) = 1 N jhj P t(h) v (t) v (t + h) and D ={h : n [`] < h [`]
< n [`] ; `= 1; :::; d}. However, Dahlhaus and Künsch (1987) have criticized the use of I v ( k ) on the grounds that the Whittle estimator, see (3:1) below, loses its minimum distance interpretability and that the objective function possesses several local maxima. The latter implies that to obtain the maximum of the Whittle function becomes more strenuous. Another possibility is the one described by [START_REF] Robinson | Modi…ed Whittle estimation of multilateral models on a lattice[END_REF]. The latter proposal will be helpful when d 4. However as we only consider explicitly the most common scenario d 3, it su¢ ces for our results to hold true to employ the taper periodogram I T v ( j ). The bene…ts of tapering can be seen following the properties of the cosine-bell (or Hanning) taper, which is de…ned as

(2.5) h (t) = 1 2 d d Y `=1 h `(t [`]) ; h `(t [`]) = 1 cos 2 t [`] n [`] .
Indeed, denoting the taper Dirichlet kernel by

D T `( [`]) = P n[`] t[`]=1 h `(t [`]) e it[`] [`]
, we have that (2.6) sup

n[`]; [`]>0 D T `( [`]) = O min n n [`] ; n [`] 2 j [`]j 3 o .
The immediate consequence of property (2:6) is that the bias of the taper periodogram is of smaller order of magnitude than the one of the standard periodogram. Observe that

(2.7) D T ` j[`] = 1 6 1=2 D ` j[`] 1 + 2D ` j[`] D ` j[`]+1 , where D `( [`]) = P n[`] t[`]=1 e it[`] [`]
is the Dirichlet kernel. It is worth observing that the standard DFT and the cosine-bell taper DFT are related by the equality

(2.8) w T x ( j ) = 1 6 d=2 d Y `=1 w x j[`] 1 + 2w x j[`] w x j[`]+1 .
In the paper we shall explicitly consider the cosine-bell, although the same results follow employing other taper functions such as Parzen or Kolmogorov's tapers.

The formulation of H 0 given in (2:1) suggests to use the Bartlett's T p process as a basis for testing H 0 . The T p process is de…ned as (2.9)

;N ( ) = 2 1=2 N 1=2 " G ;N ( ) G ;N ( ) d Y `=1 [`] # , 2 [0; ] d , where (2.10) G ;N ( ) = 2 1 N [ñ = ] X j= [ñ = ] I T x ( j ) j ( j )j 2 .
It is worth mentioning that similarly we might have employed the U p process as Grenander and Rosenblatt (1957) did. The latter is de…ned as

U ;N ( ) = 2 1 N [ñ = ] X j= [ñ = ] n I T x ( j ) 2 " j ( j )j 2 o .
One motivation to employ ;N ( ) instead of U ;N ( ) is that the latter statistic is not invariant to the variance of f" (t)g t2Z d as is the former statistic ;N ( ) in (2:9). Notice that because we have excluded the frequency j = 0 from the de…nition of

P [ñ = ] j= [ñ =
] and ;N ( ) is scale invariant, it is easy to show that a linear transformation of the data does not change the value of ;N and therefore we can assume, without loss of generality, that Ex (t) = 0 and Var (" (t)) = 1.

One rational of the statistic ;N ( ) follows from the observation (see Lemma 4 in Section 5) that under H 0 , we have that

max ñ j ñ E I T x ( j ) j 0 ( j )j 2 I T " ( j ) = o (1) ,
where "a b" means that a [`] b [`] for all `= 1; :::; d. Also, observe that 0 < j [START_REF] An | The maximum of the periodogram[END_REF] ñ [START_REF] An | The maximum of the periodogram[END_REF] whereas ñ [`] < j [`] ñ [`] for `= 2; :::; d.

Thus, from the previous observation, we can expect that 0;N will be asymptotically equivalent to Bartlett's U p process for f" (t)g t2Z d , i.e.

(2.11)

0 N ( ) = 2 1=2 N 1=2 " G 0 N ( ) G 0 N ( ) d Y `=1 [`] # , with G 0 N ( ) = 2 1 N P [ñ = ] j= [ñ = ] I T " ( j ) ; 2 [0; ] d .
Observe that the U p process 0 N and the T p process 0;N are identical when fx (t)g t2Z d is a "white noise" lattice process.

Let us introduce the following regularity conditions. Condition C1: (a) f" (t)g t2Z d in (1:1) is a zero mean independent identically distributed sequence of random variables with variance 2 " = 1 and …nite 4th moments with " denoting the fourth cumulant of f" (t)g t2Z d .

(b) The multilateral Moving Average representation of fx (t)g t2Z d in (1:1) can be written (or it has a representation) as a multilateral Autocorrelation model (2.12)

X j2Z d (j) x (t j) = " (t) (0) = 1,
where (j) is the coe¢ cient of z j in the Fourier expansion of L 1 (z), where

L (z) = L (z [1] ; :::; z [d]) = X j2Z d (j) z j . Condition C2: N = Y d `=1 n [`],
where n [`] ñ for `= 1; :::; d, and "a b" means that C 1 a=b C for some …nite positive constant C. Condition C3: fh (t)g n t=1 is the cosine-bell taper function in (2:5). We now comment on Conditions C1 to C3. Part (a) of Condition C1 seems to be a minimal condition for Proposition 1 below to hold true. Observe that due to the quadratic nature of 0 N , for the latter to have …nite second moments, we require …nite fourth moments of f" (t)g t2Z d . Also we have assumed that the true value of 2 " is 1. The latter follows from our comments made after the de…nition of G ;N ( ) in (2:10). However, we shall emphasize that we are not saying or suggesting that the true value of 2 " is known, only that it is equal to 1. Su¢ cient regularity conditions required for the validity of the expansion in (2:12) is (z) be no zero for any z [`], `= 1; :::d, which simultaneously satisfy jz [1]j = 1; :::; jz [d]j = 1 at least when the Moving Average representation is of …nite order. The latter implies that f ( ) is a positive function.

Looking at the proof of Proposition 1 below, and then that of Theorem 1, it appears that we do not need to assume …nite fourth moments of the sequence f" (t)g t2Z . The reason is similar to the work of [START_REF] Anderson | On the asymptotic distribution of the autocorrelations of a sample from a linear stochastic process[END_REF]. However, as in the more realistic situation when we need to estimate the unknown parameters of the model, we require …nite fourth moments to obtain the asymptotic properties of the estimates, we have just preferred to leave the condition as it stands.

Condition C2 can be generalized to the case where the rate of convergence to zero of n 1 [`] di¤ers for di¤erent `= 1; :::; d. However, for notational simplicity we prefer to leave it as it stands. On the other hand, in C3 the taper function employed for the asymptotics to follow can be more general, as those given by Kolmogorov's or Parzen's tapers. In fact, in situations where d > 3, it might be needed for the results of the paper to follow. However, as the most important cases in empirical applications are covered in the paper, we shall leave the cosine-bell taper explicitly as the taper function to be employed.

The empirical processes 0 N ( ) and 0;N ( ) given in (2:11) and (2:9) respectively are random elements in D [0; ] d . The functional space D [0; ] d is endowed with the Skorohod's metric (see [START_REF] Billingsley | Convergence of probability measures[END_REF] or [START_REF] Bickel | Convergence criteria for multiparameter stochastic processes and its applications[END_REF]) and convergence in distribution in the corresponding topology is denoted by ")". Proposition 1. Under C1 C3, we have that

(2.13) 0 N ( ) ) e B ( ) = B d Y `=1 [`] B (1) 2 [0; ] d , where n B (u) : u 2 [0; 1] d o
is the standard Brownian sheet.

Remark 1. Recall that the covariance structure of the standard Brownian sheet is

Cov (B (u) ; B (v)) = d Y `=1 (u [`] ^v [`]) , for u; v 2 [0; 1] d .
Proposition 1 extends Grenander and Rosenblatt's (1957) results when d = 1, although under stronger conditions than the ones we have assumed in this paper. In particular, we do not need to assume eighth bounded moments.

To establish the asymptotic equivalence between 0;N and 0 N , we introduce the following smoothness condition. Our main result of this section is the following theorem.

Condition C.4: j ( )j 2 = P j2Z d (j) e ij
Theorem 1. Consider (1:1) and assume C1 C4. Then, under H 0 ,

0;N ( ) ) e B ( ) 2 [0; ] d .
Proof. The proof is an immediate consequence of Proposition 1 and Lemma 4 after we observe that Lemma 4, with ( ) = 1 there, implies that

N 1=2 sup 2[0; ] d G 0;N ( ) G 0 N ( ) = o p (1) so that N 1=2 sup 2[0; ] d G 0 ;N ( ) G 0 ;N ( ) G 0 N ( ) G 0 N ( ) = o p (1)
by standard algebra. Remark 2. An immediate conclusion from Theorem 1 and Proposition 1 is that

(2.14) G 0;N ( ) 2 " = O p N 1=2 .
We now comment on the result of Theorem 1. The theorem indicates that 0;N is asymptotically pivotal. One consequence is that critical regions of tests based on a continuous functional : D [0; ] d 7 ! R + can be easily obtained. Di¤erent functionals lead to tests with di¤erent power properties. Among them are omnibus, directional and/or Portmanteau-type tests. For example, classical functionals which lead to omnibus tests are the Kolmogorov-Smirnov ( (g) = sup 2[0; ] d jg ( )j) and the Cramér-von Mises (

(g) = d R g ( ) 2 d ).
In fact we have the following corollary.

Corollary 1. Under H 0 and C1 C4, we have that for any continuous functional

( ), ( 0;N ) d ! e B .
Proof. The proof follows from Theorem 1 and the continuous mapping theorem.

Unfortunately, the results of Theorem 1 and Corollary 1 are only valid when the "true" value of 0 is known, which in practical situations is unrealistic. The question is then how are our previous results a¤ected when we estimate 0 ? This is the topic of the next section.

TESTS WHEN THE PARAMETERS ARE UNKNOWN

This section extends the results of Section 2 to the more realistic situation where we need to estimate the parameters 0 to implement the test. That is, we replace 0 in ;N ( ) by an estimator, for example b given in (3:1) below. In this scenario, drawing the terminology from [START_REF] Durbin | Weak convergence of the sample distribution function when parameters are estimated[END_REF], we say that our null hypothesis H 0 becomes a composite hypothesis.

A popular estimator of

# 0 0 = 0 0 ; 2 " is the Whittle (1954) estimator de…ned as b # c = arg min #2 R + Q c (#) , where Q c (#) = R n log f # ( ) + I T x ( ) (2 ) d f # ( ) o d or in its discrete version (3.1) b # = arg min #2 R + Q N (#) , where (3.2) Q N (#) = 1 N ñ X j= ñ ( log f # ( j ) + I T x ( j ) (2 ) d f # ( j )
)

with f # ( j ) = 2 " j ( j )j = (2 )
d and R p is a compact set. Recall our notation given in (2:3) and that the true value of the variance of " (t) is unknown.

In this scenario, the T p process 0;N ( ) becomes

(3.3) b ;N ( ) = 2 1=2 N 1=2 " G b ;N ( ) G b ;N ( ) d Y `=1 [`] # , 2 [0; ] d ,
where G ;N ( ) is given in (2:10).

It is worth noticing that, contrary to the standard causal models, as Whittle (1954) …rst noticed, the estimator of # 0 obtained by

= arg min 2 2 N ñ X j= ñ I T x ( j ) j ( j )j 2 , 2 " = 2 N ñ X j= ñ I T x ( j ) ( j )
2 is inconsistent. The main reason for the lack of consistency of is that when the model is not causal R ' ( ) d 6 = 0, where from now on we write

(3.4) ' ( ) = @ @ log j ( )j 2 and # ( ) = @ @# log f # ( ) = ' 0 ( ) ; 2 " 0 .
Let's introduce the following regularity conditions.

q #;N = 1 N ñ X j= ñ # ( j ) ( I T x ( j ) 2 " j ( j )j 2 1
)

; Q #;N = 1 N ñ X j= ñ # ( j ) 0 # ( j ) ,
and also, recalling our notation in (2:2),

# = (2 ) d Z # ( ) d and # = (2 ) d Z # ( ) 0 # ( ) d .
Notice that we write explicitly 2 " as it is a parameter in itself. Condition C8: #0 is a continuously positive de…nite matrix.

Theorem 2. Under C1-C3 and C5 C8, we have that

N 1=2 b # # 0 d ! N 0; 2 1 #0 V #0 1 #0 ,
where

V #0 = 2 #0 + " (35=18) d #0 0 #0 .
Proof. First, by de…nition, we know that

b # # 0 = Q 1 e #;N q #0;N ,
where e

# is an intermediate point between # 0 and b #, q #;N is given in (3:5) and Q #;N is given by

Q #;N + 1 N ñ X j= ñ 2 # ( j ) 0 # ( j ) @ 2 f # ( j ) @#@# 0 ( I T x ( j ) 2 " j ( j )j 2 1 ) = Q #;N + o p (1)
by

q #0;N = 1 N ñ X j= ñ #0 ( j ) I T " ( j ) 1 + o p (1)
.

From here the proof proceeds as in [START_REF] Robinson | Modi…ed Whittle estimation of multilateral models on a lattice[END_REF].

Looking at the proof of Theorem 2, and denoting in what follows

e ' ( ) = ' ( ) 2 (2 ) d Z ' ( ) d , e # ( ) = e ' 0 ( ) ; 0 0 e ' ;N ( j ) = ' ( j ) 2 N ñ X j= ñ ' ( j )
# # 0 = Q 1 #0;N 8 < : Z e 0 ( ) 0;N (d ) + Z #0 ( ) d 1 N ñ X j= ñ I T x ( j ) (2 ) d f #0 ( j ) 1 ! 9 = ; + o p N 1=2 . (3.6)
Now using (3:6) and de…ning

1 ( ) = e B ( ) 1
(2 ) Proof. The proof follows from Theorem 3 and the continuous mapping theorem.

d Z e ' 0 0 d ! e 1 ( 0 ) Z e ' 0;N e B d ,
( ) = 0 N ( ) 0 @ 1 N [ñ = ] X j= [ñ = ] e ' 0 0;N ( j ) 1 A e 1 0;N 1 N ñ X j= ñ e ' 0 0;N ( j ) I T " ( j ) +o p (
The main conclusion that we draw from Theorem 3 is that the T p -process b ;N is no longer asymptotically pivotal, so that the immediate consequence is that tests based on , for example the Kolmogorov or Cramér-von Mises's statistics, are not useful for practical purposes as its asymptotic critical values are di¢ cult, if at all possible, to obtain. To compute the critical values of the asymptotic distribution of b N , several approaches have been described and examined. A …rst approach makes use of a bandwidth parameter that must behave in some required sense. This procedure makes the asymptotic distribution of the statistics b N pivotal, so that its critical values are readily available. Among them, the popular Portmanteau test. [START_REF] Box | Distribution of residual autocorrelations in autoregressive-integrated moving average time series models[END_REF] showed that the partial sum of the residuals squared autocorrelations of a stationary ARMA process is approximately chi-squared distributed assuming that the number of autocorrelations considered diverges to in…nity with the sample size at an appropriate rate. Alternatively we could employ a frequency domain approach as in [START_REF] Hong | Consistent testing for serial-correlation of unknown form[END_REF] or [START_REF] Paparoditis | Spectral density based goodness-of-…t tests for time series models[END_REF], who compared a nonparametric estimator of f ( ) with the parametric one. The …rst shortcoming of the latter method is that the power of the test is smaller than the one proposed in the paper, that is if we denote by b N the bandwidth parameter, their test has a local power of order N b d N 1=2 whereas ours is N 1=2 . A second potential drawback is that the choice of b N seems an artifact when testing for a particular parametric family and the …nal outcome of all these tests may depend on the arbitrary choice of the bandwidth parameter for which no relevant theory is available. That is, there are not rules available on how to choose b N for the purpose of testing.

A second alternative is in the spirit of Durbin, Knott and Taylor (1976) for the classical empirical process, and it was the route followed by [START_REF] Anderson | Goodness-of-…t for autoregressive processes[END_REF], who proposed to approximate the critical value of the Cramér-von Mises test for a stationary AR model. The method considers a truncated version of the spectral representation of b ;N with estimated orthogonal components. The number of estimated orthogonal components must suitably increase with the sample size. However, its implementation is quite cumbersome even for the rather simpler case when d = 1. See for instance [START_REF] Anderson | Goodness-of-…t for autoregressive processes[END_REF] for details.

So, in view of the preceding arguments, we consider a third approach based on bootstrap algorithms. This is the route employed, among others, by Chen and Romano (2000) or Hainz and Dahlhaus (2000) for short-range models using the U p process and by [START_REF] Hidalgo | Bootstrap speci…cation tests for linear covariance stationary processes[END_REF], who allow also long-range dependence using the T p process. Of course all those articles were for d = 1. Also, we will see that bootstraps employed when d = 1 are not valid in our context.

BOOTSTRAP TEST FOR THE TEST

Since Efron (1979), bootstrap algorithms have become a common tool in applied work and thus considerable e¤ort has been devoted to its development. The primary motivation for this e¤ort is that they have proved to be a very useful statistical tool. We can cite two main examples/reasons. First, bootstrap methods are capable of approximating the …nite sample distribution of statistics better than those based on their asymptotic counterparts. And secondly, and perhaps the most important, they allow computing valid asymptotic quantiles of the limiting distribution in situations when the practitioner is unable to compute its quantiles.

In the present paper we face the latter situation. Following our comments at the end of the previous section, the aim of this section is to propose a bootstrap procedure for b ;N given in

H a : f # ( ) 1 + 1 N 1=2 g ( ) for some # 2 R +
where g ( ) is some symmetric, non-constant continuous function in

[0; ] d such that 1 N 1=2 g ( ) > 1 for all N
1, b N must also converge, in bootstrap distribution to ( 1 ), whereas under the alternative hypothesis, we only require that b N is bounded in probability to have good power properties.

Remark 3. We should point out that H a could have been written as

H a : f # ( ) + 1 N 1=2 e g ( ) for some # 2 R +
where e g ( ) is a positive integrable function. However, since we are concerned with the relative error of I T x ( j ) compared to f # ( j ) j ( j )j 2 , we found notationally more convenient to write H a as given in (4:1).

When d = 1, Hidalgo and Kreiss (2006) examined a bootstrap algorithm based on an approach in [START_REF] Hidalgo | An alternative bootstrap to moving blocks for time series regression models[END_REF] showing its validity and consistency. This bootstrap consists on the following 3 STEPS.

e I T x ( j ) = f b # ( j ) I T x ( j ) ,
where

f b # ( j ) = G b ;N ( ) (2 )
#2 R + e Q N (#) ,
where, with

f # ( j ) = 2 " j ( j )j = (2 ) d , e Q N (#) = 1 N ñ X j= ñ ( log f # ( j ) + e I T x ( j ) (2 ) d f # ( j )
)

.

STEP 3: Compute the bootstrap T p process

;N ( ) = 2 1=2 N 1=2 2 4 e G ;N ( ) G ;N ( ) d Y `=1 [`] 3 5 , 2 [0; ] d , where e G ;N ( ) = 2 1 N P [ñ = ] j= [ñ = ] e I T x ( j ) = j ( j )j 2 .
Other procedures are possible as that based on that of [START_REF] Franke | On bootstrapping kernel spectral estimate[END_REF], where the bootstrap periodogram e

I T x ( j ) = b ( j ) 2 I T x ( j ) is replaced by I T x ( j ) = f b # ( j ) j , where
ñ; :::; ñ are independent exponential random variables. However, unlike in the case of d = 1, the previous bootstrap algorithm will not be valid. The reason is because the bootstrap does not correctly "estimate" the fourth cumulant " . More speci…cally the asymptotic distribution of the bootstrap estimator # in (4:2) will not have the same asymptotic variance as that of b # in (3:1). So to overcome this problem, following [START_REF] Hidalgo | Uniform con…dence bands for kernel regression estimates with dependent data[END_REF], see also [START_REF] Hidalgo | Bootstrap in time series regression models[END_REF], we propose in the paper an alternative algorithm, as described in the next 4 STEPS. is asymptotically independent of the mean and variance of f" (t)g t2Z d , we do not need to standardize b " (t) to obtain the bootstrap sample. (b) The motivation to compute the residuals as in STEP 1 comes from the observation that, for t = 1; :::; n,

" (t) ' (2 ) d=2 1 N 1=2 ñ X j= ñ e it j 1 0 ( j ) w x ( j ) .
STEP 2: For t = 1; :::; n, compute the bootstrap observations

(4.3) x (t) = (2 ) d=2 1 N 1=2 ñ X
j= ñ e it j b ( j ) w " ( j ) , where w " ( j ) is the standard DFT of f" (t)g n t=1 , and the taper periodogram I T x ( j ) as de…ned in (2:4). STEP 3: The bootstrap analogue of b # is given by

(4.4) b # = arg min #2 R + Q N (#) , where (4.5) Q N (#) = 1 N ñ X j= ñ ( log f # ( j ) + I T x ( j ) (2 ) d f # ( j )
)

.

STEP 4: Compute the bootstrap T p process

(4.6) b ;N ( ) = 2 1=2 N 1=2 " G b ;N ( ) G b ;N ( ) d Y `=1 [`] # , 2 [0; ] d , with G ;N ( ) = 2N 1 P [ñ = ] j= [ñ = ] j ( j )j 2 I T
x ( j ). Theorem 4. Under C1 C3 and C5 C8, we have that

N 1=2 b # b # d ! N 0; 2 1 #0 V #0 1 #0 ,
# b # = Q 1 b #;N 8 < : Z e b ( ) b ;N (d ) + Z b # ( ) d 1 N ñ X j= ñ I T x ( j ) (2 ) d f b # ( j ) 1 ! 9 = ; + o p N 1=2 . (4.7) Denote G 0 N ( ) = 2 1 N P [ñ = ] j= [ñ = ] I T " ( j ) and let (4.8) 0 N ( ) = 2 1=2 N 1=2 " G 0 N ( ) G 0 N ( ) d Y `=1 [`] # , 2 [0; ] d .
Theorem 5. Under H 0 and assuming C :1 C :3 and C5 C8 , uniformly in

2 [0 ; ], (a) b ;N ( ) = 0 N ( ) 0 @ 1 N [ñ = ] X j= [ñ = ] e ' 0 b ;N ( j ) 1 A e 1 b ;N 1 N ñ X j= ñ e ' 0 b ;N ( j ) I T " ( j ) +o p (1) . (b) b ;N d ) 1 .
A conclusion from Theorem 5 is the following corollary.

) ! c a (1
) and c (1 ) ) , respectively, where c (1 ) is de…ned as Pr

p ! c a (1
n jb N j > c (1 ) o = .
Typically, the …nite sample distribution of b N is not available, although the critical values c (1 ) can be approximated, as accurately as desired, by standard Monte-Carlo simulation. To that end, consider the bootstrap samples

LEMMAS

First, we introduce some notation. We denote the conjugate of a complex number a by a. Also, for a generic function ( ), we abbreviate ( j ) by j = j [START_REF] An | The maximum of the periodogram[END_REF] ; :::; j[d] 0 and C will denote a generic positive and …nite constant.

For the next two lemmas, we shall assume that f (t)g t2Z d and f (t)g t2Z d are two stationary spatial processes with a representation as that in (1:1) and whose respective errors satisfy C1. Also f ( ) = ( 2

f ;j = O ñ 2 ; (b) E w T ;j w T ;j = O d Y `=1 j [`] 3 ! .
Proof. We begin with part (a). By de…nition, the left side of the equality in (a) is

Z dñ (f ( ) f ( j )) d Y `=1 K T [`] j[`] d ,
suppressing any reference to `in K T `and/or D T `for notational simplicity. Now, because f ( ) is twice continuously di¤erentiable and R K T ( ) d = 0, we have that the last displayed expression is bounded in modulus by

C Z d d X `=1 d X p=1 [`] j[`] [p] j[p] d Y `=1 K T [`] j[`] d C Z d d X `=1 [`] j[`] 2 d Y `=1 K T [`] j[`] d
by the Cauchy-Schwarz inequality. Now, using (2:6), that the Fejer's kernel integrates 1 and that

P n[`] t[`]=1 h `(t [`]) 2
Cñ, we obtain that the right side of the last displayed inequality is, by C2 and standard algebra, bounded by

C N Z d d X `=1 [`] j[`] 2 d Y `=1 min n ñ2 ; ñ 4 [`] j[`] 6 o d = O ñ 2 .
Next we show part (b). Again by de…nition and that jf ( )j < C, we obtain that E w T ;j w T ;j is bounded by

C Z d jf ( )j d Y `=1 n 1 D T [`] j[`] D T [`] + j[`] d C d Y `=1 j [`]
3 by standard arguments after using (2:6).

E w T ;j w T ;k = 6 d Z d f ( ) d Y `=1 n [`] 1 D T [`] j[`] D T k[`]
[`] d .

Because jf ( )j < C, the modulus of the right side of (5:1) is bounded by

C d Y `=1 ñ 1 ( Z (j[`]+k[`])=2 + Z (j[`]+k[`])=2 ) D T [`] j[`] D T k[`] [`] d [`]
using C2. Now using (2:6) and because R 0 D T ( ) d < C, the contribution due to a factor of the type

R (j[`]+k[`])=2 when k[`] < j[`] is bounded by C jj [`] k [`]j 3 + Z (j[`]+k[`])=2 D T k[`] [`] d [`] = O jj [`] k [`]j 3 + , whereas if j[`] < k[`] by C jj [`] k [`]j 3 + R (j[`]+k[`])=2 D T [`] j[`] d [`] = O jj [`] k [`]j 3 +
. Recall that k j so we have for some `= 1; ::

; d 1, j[`] < k[`]
. Finally, proceeding similarly the contribution due to a factor of the type R

(j[`]+k[`])=2 is O jj [`] k [`]j 3 +
. Now conclude by Hölder's inequality and that

E w T ;j w T ;k = O Y d `=1 jj [`] k [`]j 3 +
. On the other hand, because when

jj [`] k [`]j 6
= 2 for some `= 1; :::; d,

Z d d Y `=1 D T [`] j[`] D T k[`] [`] d = 0, (5.2)
we have that in this case, except multiplicative constants, the left side of (5:1) is

N 1 Z d (f ( ) f ( j )) d Y `=1 D [`] j[`] D k[`] [`] d ,
which, by the mean value theorem, is bounded in absolute value by

N 1 Z d d X `=1 [`] j[`] d Y `=1 D [`] j[`] D k[`] [`] d = O log ñ ñ , because j D ( )j < C, R 0 jD ( )j d = O (log ñ) and the Cauchy-Schwarz inequality imply that R D j[`] D k[`] d = O (ñ)
. Now, when for all `= 1; :::; d, jj [`] k [`]j = 2, because the left side of (5:2) is 1, we have that proceeding as above, the left side of (5:1) is f ;j + O ñ 1 log ñ . This concludes the proof.

In what follows, we shall abbreviate w T

x ( ) = ( ) and w T " ( ) by u ( ) and v ( ) respectively for all 2 d . Lemma 3. Let ( ) be a continuously di¤ erentiable function in d . Under C1-C4, we have that for all r s 2 e Z d

(5.3)

E s X j=r j v j (u j v j ) 2 = O log ñ ñ d Y `=1 js [`] r [`]j + ! .
Proof. Denote % j = u j v j . By standard arguments, the left side of (5:3)

is s X j=r 2 j E v j v j % j % j + s X j6 =k=r j k E v j v k % j % k = s X j=r 2 j fa j1 + a j2 g + s X j6 =k=r j k fb jk;1 + b jk;2 g , where a j1 = E (v j v j ) E % j % j + E v j % j 2 + E v j % j 2 a j2 = cum (v j ; v j ; u j ; u j ) + cum (v j ; v j ; v j ; v j ) cum (v j ; v j ; u j ; v j ) cum (v j ; v j ; u j ; v j ) b jk;1 = E (v j v k ) E % j % k + E v j % j E (v k % k ) + E (v j % k ) E v k % j b jk;2 = cum (v j ; v k ; u j ; u k ) + cum (v j ; v k ; v j ; v k ) cum (v j ; v k ; u j ; v k ) cum (v j ; v k ; u j ; v k ) .
After observing that E (v j u j ) = 1 + O ñ 2 and E v j % j = E (v j u j ) E (v j v j ), we have that Lemma 1 implies that a j1 = O ñ 2 , whereas Lemmas 1 and 2 imply that b jk;1 = O c 2 jk + ñ 1 log nI (jj [`] k [`]j = 2; `= 1; :::; d) , with c jk as de…ned there. From here it is immediate to conclude that the contribution due to a j1 and b jk;1 into the left of (5:3) is its right side.

Finally we examine a j2 and b jk;2 . Using formulae in Brillinger [(1981), (2:6:3), page 26, and (2:10:3), page 39], we deduce after standard algebra that

b jk;2 = " N 2 Z d Z d ( ) j 1 ( ) k 1 D T ( j ) D T ( + k ) D T ( j k ) d d .
By the Cauchy-Schwarz inequality, we have that jb jk;2 j 2 is bounded by Proof. We shall consider the proof in the positive quadrant

CN 1 times Z d ( ) j 1 2 K T ( j ) d Z d ( ) k 1 2 K T ( + k ) K T ( j k ) d d .
P [ñ = ]
j=1 , being the proof for the remaining 2 d 1 1 quadrants similarly handled. By the Cauchy-Schwarz and the triangle inequalities, it su¢ ces to show that

(5.4) E sup s s X j=1 j ( I T x;j j j j 2 I T ";j ) E sup s s X j=1 j % j 2 + 2E sup s s X j=1 j v j % j is o N 1=2
, where we abbreviate "sup s=1;:::;ñ " by "sup s " and % j = u j v j . The …rst term on the right of (5:4) is bounded by

C ñ X j=1 n E ju j j 2 1 (E (u j v j ) 1) (E (u j v j ) 1) + E jv j j 2 1 o = o N 1=2 ,
because j C, d < 4 and by Lemma 1, for instance

E u j v j u j 1 2 " j j j 2 E w T x;j w T ";j w T x;j 2 " j j j j 2 = O 1 ñ2 .
Next, we examine the second term of (5:4). To that end, let q = 0; : : : ; [ñ & ] 1 for some 0 < & < 1=d. (Recall that ñ = ñ [START_REF] An | The maximum of the periodogram[END_REF] ; ::: ñ [d] for any > 0.) Standard inequalities imply that the square of the second term on the right of (5:4) is bounded by

(5.5) E max s 8 > < > : s X j=1 q(s)[ñ 1 & ] X j=1 9 > = > ; j v j % j 2 + E max s q(s)[ñ 1 & ] X j=1 j v j % j 2 ,
where herewith q(s) denotes the value of q = 0; : : : ; [ñ & ] 1 such that q(s) ñ1 & is the largest vector s 1 such that s 1 s, and using the convention

P d j=c 0 if d < c. From now on, we abbreviate (ñ [1] = [ñ & [1]] ; :::; ñ [d] = [ñ & [d]]) by ñ1 & .
From the de…nition of q (s) and sup p jc p j 2 = sup p jc p j 2 P p jc p j 2 , the second term of (5:5) is bounded by

P [ñ & ] 1 q=1 E Pq [ñ 1 & ] j=1 j v j % j 2 = O N 1+& ñ 1 log 2 ñ = o (N ) by Lemma 3 and because & < 1=d.
To complete the proof we need to show that the …rst term in (5:5) is o (N ). To that end, we note that it is bounded by

E max q=1;:::;[ñ & ] 1 max s=1+q[ñ 1 & ];:::;(q+1)[ñ 1 & ] s X j=1+q[ñ 1 & ] j v j % j 2 which is O (N & ) E max s=1;:::;[ñ 1 & ] P s j=1 j v j % j 2
. So, we have that the square of the second term on the right of (5:4) is

o (N ) + O (N & ) E max s=1;:::;[ñ 1 & ] s X j=1 j v j % j 2 .
Observe that the second factor of the second term of the last displayed expression is similar to the second term on the right of (5:4) but with s = 1; :::; ñ1 & instead of s = 1; :::; ñ. So, repeating the same steps, the last displayed expression, and so the square of the second term on the right of (5:4), is

o (N ) + N & P 1 p=0 (1 &) p E max s=1;:::;([ñ 1 & ]) s X j=1 j v j % j 2 = o (N ) + O N & P 1 p=0 (1 &) p ñ(1 &) X s=1 E s X j=1 j v j % j 2 = o (N )
after choosing large enough because & < 1=d. This completes the proof.

Lemma 5. Let ( ; #) be as in Lemma 3 for all # 2 R + , and continuously di¤ erentiable in # for all . Assuming C1 C4,

(5.6) 1 N sup #2 R + ñ X j= ñ j (#) I T x;j j 0;j j 2 1 ! = o p (1) .
Proof. By the triangle inequality, the left side of (5:6) is bounded by (5.7)

C N sup #2 R + ñ X j= ñ j (#) I T x;j j 0;j j 2 I T ";j ! + C N sup #2 R + ñ X j= ñ j (#) I T ";j 1 .
Now, because by assumption j ( ; #)j < C, the …rst term of (5:7) is bounded by

C N ñ X j= ñ I T x;j j 0;j j 2 I T ";j C N ñ X j= ñ % j 2 + C N ñ X j= ñ v j % j = o p (1)
by Markov's inequality because by the Cauchy-Schwarz inequality E v j % j 2 E jv j j 2 E % j 2 and then proceeding as in Lemma 3. Next, we show that the second term of (5:7) is o p [START_REF] An | The maximum of the periodogram[END_REF]. But this follows by standard arguments (see also Lemma 15) and because j (#) is continuously di¤erentiable in #. Proof. The proof follows very easily using Lemma 5. Indeed, (3:2) is

(5.8) 1 N ñ X j= ñ f #0;j f #;j I T x;j (2 ) d f #0;j 1 ! + 1 N 8 < : ñ X j= ñ f #0;j f #;j log f #0;j f #;j + log f #0;j 9 =
; . Now the second term of (5:8) converges using Brillinger (1981, p:15) to

Z f #0 ( ) f # ( ) log f #0 ( ) f # ( ) d + Z log f #0 ( ) d (2 )
d 2 + Z log f #0 ( ) d
with equality when f #0 ( ) = f # ( ) which is the case only if # = # 0 by C7. On the other hand, the …rst term of (5:8) converges to zero uniformly in # by Lemma 5 because f 1 #;j f #0;j satis…es the same conditions as ( ; #) there by C6. From here the conclusion of the lemma is standard, so we omit its details. Lemma 7. Assume C1 C3 and C5 C8. Under H 0 , uniform in 2 [0; ] d , (5.9)

1 N 1=2 [ñ = ] X j= [ñ = ] j 0 B @ I T x;j b ;j 2 I T ";j 1 C A = 0 @ 1 N [ñ = ] X j= [ñ = ] j ' 0 0;j 1 A N 1=2 b 0 +o p (1)
, where ( ) is as in Lemma 3.

Proof. The di¤erence between the left side of (5:9) and the …rst term on its right side is

1 N 1=2 [ñ = ] X j= [ñ = ] j I T x;j j 0;j j 2 2 6 4 j 0;j j 2 b ;j 2 1 + ' 0 0;j b 0 3 7 5
(5.10)

+ 1 N 1=2 [ñ = ] X j= [ñ = ] j I T x;j j 0;j j 2 I T ";j ! 1 N 1=2 [ñ = ] X j= [ñ = ] j ' 0 0;j I T x;j j 0;j j 2 b 0 .
First, because each component of the vector ( ) ' 0 ( ) satis…es the same conditions of ( ) in Lemma 4, Markov's inequality implies that the second term of (5:10) is o p (1), whereas the third term is Finally, by mean value theorem, the norm of the …rst term of (5:10) is bounded by (5.11)

N 1 P [ñ = ] j= [ñ = ] j ' 0 0;j N 1=2 b 0 + o p (
CN 1=2 b 0 2 1 N [ñ = ] X j= [ñ = ] I T x;j j 0;j j 2 = O p N 1=2 ,
by Theorem 2 and proceeding as with the third term of (5:10). This concludes the proof.

We now introduce the following notation. "

T (t) = h (t) " (t) and for v 1 < v 2 2 [0; ] d ,
(5.12)

E 1;N (v 1 ; v 2 ) = 0 @ 1 N [ñv2= ] X j=[ñv1= ] j 1 A N 1=2 P n t=1 h 2 (t) n X t=1 " T (t) 2 1 ! (5.13) E 2;N (v 1 ; v 2 ) = 1 N [ñv2= ] X j=[ñv1= ] j N 1=2 P n t=1 h 2 (t) n X t16 =t2=1
" T (t 1 ) " T (t 2 ) e i(t1 t2) j .

Notice that

E 1;N (v 1 ; v 2 ) + E 2;N (v 1 ; v 2 ) = N 1=2 P [ñv2= ] j=[ñv1= ] j I T ";j 1 . Lemma 8. Let v 1 < v < v 2 2 [0; ] d and
( ) as in Lemma 3. Then, assuming C1 C3, for some > 0,

(5.14) E jE j;N (v 1 ; v)j jE j;N (v; v 2 )j C Y d `=1 (v 2 [`] v 1 [`]) 2 , j = 1; 2.
Proof. The proof follows proceeding as that of Lemma 6 of [START_REF] Delgado | Distribution free goodness-of-…t tests for linear processes[END_REF] and observing that by continuity of ( ),

N 1 P [ñv2= ] p=[ñv1= ] q p C Y d `=1 (v 2 [`] v 1 [`]) for any q 1.
Next we will show that the processes E 1;N (0; ) and E 2;N (0; ) are tight. From [START_REF] Bickel | Convergence criteria for multiparameter stochastic processes and its applications[END_REF] it su¢ ces to show the following lemma. Lemma 9. Assuming C1, we have that

(5.15) (a) E Y d `=1 E (`) 1;N 0; 1[`] E (`) 1;N 0; 2[`] 2 C Y d `=1 2[`] 1[`] 2 (5.16) (b) E Y d `=1 E (`) 2;N 0; 1[`] E (`) 2;N 0; 2[`] 4 C Y d `=1 2[`] 1[`] 2
for all 1[`] < 2[`] 2 [0; ], `= 1; :::; d, and where, say,

E (`) 1;N 1[`] ; 2[`] = 0 B @ 1 n [`] [ñ 2[`] = ] X j[`]=[ñ 1[`] = ] j 1 C A N 1=2 P n t=1 h 2 (t) n X t=1 " T (t) 2 1 ! E (`) 2;N 1[`] ; 2[`] = 1 n [`] [ñ 2[`] = ] X j[`]=[ñ 1[`] = ] j N 1=2 P n t=1 h 2 (t) n X t16 =t2=1
" T (t 1 ) " T (t 2 ) e i(t1 t2) j .

Proof. The proof follows after observing that E Proof. We shall handle the case when both T x ;j and T x ;j are u j , being the other cases identically handled. We begin with part (a). By de…nition and using (2:8), it is easy to show that (5.17)

w T x ;j = 1 6 d=2 ñ X k= ñ b ;k w " ;k Y d `=1 I `(j; k) , where I `(j; k) = 2I (j [`] = k [`]) I (j [`] 1 = k [`]) I (j [`] + 1 = k [`]).
E s X j=r j v j u j v j 2 = O p 1 ñ d Y `=1 js [`] r [`]j + ! .
Proof. Denote % j = u j v j . By standard arguments, the left side of (5:17) is

s X j=r 2 j E v j v j % j % j + s X j6 =k=r j k E v j v k % j % k = s X j=r 2 j a j1 + a j2 + s X j6 =k=r j k b jk;1 + b jk;2 ,
where, by Lemmas 11 and 12 part (b),

a j1 = E v j v j E % j % j + E v j % j 2 ;
a j2 = cum v j ; v j ; u j ; u j + cum v j ; v j ; v j ; v j cum v j ; v j ; u j ; v j cum v j ; v j ; u j ; v j

b jk;1 = E v j v k E % j % k + E v j % j E (v k % k ) b jk;2 = cum v j ; v k ; u j ; u k + cum v j ; v k ; v j ; v k cum v j ; v k ; u j ; v k cum v j ; v k ; u j ; v k .
After observing that E v j u j = b 2 " + O p ñ 2 and E v j % j = E v j u j E v j v j , we have that Lemma 11 implies that a j1 = O p ñ 2 , whereas Lemmas 11 and 12 imply that b jk;1 = O p ñ 2 + ñ 1 I (jj [`] k [`]j = 2; `= 1; :::; d) . From here it is immediate to conclude that the contribution due to a j1 and b jk;1 into the left of (5:17) is its right side.

Finally we examine a j2 and b jk;2 . By de…nition of, for example, w T x ( ) and that cum (" (t 1 ) ; :::; " (t 4 )) = b " I (t 1 = :::

= t 4 ), it is obvious that b jk;2 is O p n 1 I (jj [`] k [`]j
I T x ;j b ;j 2 I T " ;j 9 > = > ; E sup s s X j=1 j % j 2 + 2E sup s s X j=1 j v j % j is o p N 1=2
, where we abbreviate "sup s=1;:::;ñ " by "sup s " and % j = u j v j . The …rst term on the right of (5:18) is bounded by Next, we examine the second term of (5:18). With same notation as in Lemma 4, the square of the second term on the right of (5:18) is bounded by

C ñ X j=1 n E u j 2 b 2 " E u j v j b 2 " E u j v j b 2 " + E v j 2 b 2 " o = o p N 1=2 ,
(5.19) E max s 8 > < > : s X j=1 q(s)[ñ 1 & ] X j=1 9 > = > ; j v j % j 2 + E max s q(s)[ñ 1 & ] X j=1 j v j % j 2 .
From the de…nition of q (s) and sup p jc p j 2 = sup p jc p j 2 P p jc p j 2 , the second term of (5:19) is bounded by

P [ñ & ] 1 q=1 E Pq [ñ 1 & ] j=1 j v j % j 2 = O p N 1+& ñ 1 log 2 ñ = o p (N )
;(q+1)[ñ 1 & ] s X j=1+q[ñ 1 & ] j v j % j 2 which is O p (N & ) E max s=1;:::;[ñ 1 & ] P s j=1 j v j % j 2
. So, we have that the square of the second term on the right of (5:18) is

o p (N ) + O p (N & ) E max s=1;:::;[ñ 1 & ] s X j=1 j v j % j 2 .
Now proceeding as in Lemma 4, the square of the second term on the right of (5:18) is

o p (N ) + O p N & P 1 p=0 (1 &) p ñ(1 &) X s=1 E s X j=1 j v j % j 2 = o p (N )
after choosing large enough because & < 1=d. This completes the proof. Proof. By the triangle inequality, the left side of (5:20) is bounded by (5.21)

C N sup #2 R + ñ X j= ñ j (#) 0 B @ I T x ;j b ;j 2 I T " ;j 1 C A + C N sup #2 R + ñ X j= ñ j (#) I T " ;j b 2 "
. Now, because by assumption j ( ; #)j < C, the …rst term of (5:21) is bounded by

C N ñ X j= ñ I T x ;j b ;j 2 I T " ;j C N ñ X j= ñ u j v j 2 + C N ñ X j= ñ v j u j v j = o p (1)
by Markov's inequality because by the Cauchy-Schwarz inequality E v j u j v j 2 E v j 2 E u j v j 2 and then proceeding as in Lemma 13. Next, the second term of (5:21) is o p (1). First, the …nite dimensional distributions of S (#) = N 1 P ñ j= ñ j (#) I T " ;j b 2 " converge to zero in probability. Indeed, the second bootstrap moment is

1 N 2 ñ X j;k= ñ j (#) k (#) E n I T " ;j b 2 " I T " ;k b 2 " o = O p 1 N
by standard algebra after observing that E (w " ;j w " ;k ) = b 2 " I (j = k), E (w " ;j w " ;k ) = 0 and cum (w " ;j ; w " ;j ; w " ;k ; w " ;k ) = O N 1 b " I (j = k). To …nish, we need to show the tightness of the process S (#). But this is immediate because proceeding as with the last displayed equality

E jS (# 2 ) S (# 1 )j 2 = j# 2 # 1 j 2 O p (1)
because continuous di¤erentiability of ( ; #) for all 2 d implies that j (# 2 ) j (# 1 ) C j# 2 # 1 j. Proof. The proof follows very easily using Lemma 15. Indeed, (4:5) is

(5.22) 1 N ñ X j= ñ f b #;j f #;j I T x ;j (2 ) d f b #;j 1 ! + 1 N 8 < : ñ X j= ñ f b #;j f #;j log f b #;j f #;j + log f b #;j 9 =
; . Now, the di¤erence between the second term of (5:22) and

Z f b # ( ) f # ( ) log f b # ( ) f # ( ) d + Z log f b # ( ) d
converges to zero in probability using Brillinger (1981, p:15) and that uniformly in ,

f b # ( ) f #0 ( ) = o p (1)
. Moreover, the last displayed expression is greater than or equal to (2 )

d 2 + R log f b # ( ) d with equality when f b # ( ) = f # ( ) which is the case only if # = b
# by C7. On the other hand, the …rst term of (5:22) converges to zero uniformly in # by Lemma 15 because f 1 #;j f b #;j satis…es the same conditions as ( ; #) there by C6. From here the conclusion of the lemma is standard proceeding as in Theorem 1 of [START_REF] Hannan | The asymptotic theory of linear time series models[END_REF], so we omit its details.

1 N 1=2 [ñ = ] X j= [ñ = ] j 0 B @ I T x ;j b ;j 2 I T " ;j 1 C A = 0 @ 1 N [ñ = ] X j= [ñ = ] j ' 0 b ;j 1 A N 1=2 b b +o p (1)
, where ( ) is as in Lemma 3.

Proof. The di¤erence between the left side of (5:23) and the …rst term on its right side is

1 N 1=2 [ñ = ] X j= [ñ = ] j I T x ;j b ;j 2 2 6 4 b ;j 2 b ;j 2 1 + ' 0 b ;j b b 3 7 5
(5.24)

+ 1 N 1=2 [ñ = ] X j= [ñ = ] j 0 B @ I T x ;j b ;j 2 I T " ;j 1 C A 1 N 1=2 [ñ = ] X j= [ñ = ] j ' 0 b ;j I T x ;j b ;j 2 b b .
First, because each component of the vector ( ) ' b ( ) satis…es the same conditions of ( ) in Lemma 14, Markov's inequality implies that the second term of (5:24) is o p (1), whereas the third term is

N 1 P [ñ = ] j=1 j ' 0 b ;j N 1=2 b b + o p (1)
by Lemma 14 and because proceeding as in the proof of Theorem 4

1 N 1=2 ñ X j= ñ j ' b ;j I T " ;j b 2 " = O p (1)
Finally, by mean value theorem, the norm of the …rst term of (5:24) is bounded by

CN 1=2 b b 2 1 N [ñ = ] X j= [ñ = ] I T x ;j b ;j 2 = O p N 1=2 ,
by Theorem 4 and proceeding as with the second term of (5:24). This concludes the proof.

We now introduce the following notation. For

v 1 ; v 2 2 [0; ] d , denote E 1;N (v 1 ; v 2 )
and E 2;N (v 1 ; v 2 ) as in (5:12) and (5:13) but with " T (t) there replaced by " T (t) = h (t) " (t) and also let

H N = H N (v 1 ; v 2 ) a sequence bounded in probability. Lemma 18. Let v 1 < v < v 2 2 [0; ] d .
Then, assuming C1 C3, and for some > 0, with ( ) as in Lemma 3,

E E j;N (v 1 ; v) E j;N (v; v 2 ) H N (v 1 ; v 2 ) Y d `=1 (v 2 [`] v 1 [`]) 2 , j = 1; 2.
Proof. The proof proceeds as in Lemma 8 but instead of using Delgado et al.' Next we will show that the processes E 1;N (0; ) and E 2;N (0; ) are tight. To that end, it su¢ ces to show the following lemma. " T (t 1 ) " T (t 2 ) e i(t1 t2) j 1 C A .

Proof. The proof follows after observing that E We shall be a bit more general. In particular, for a vector function ( ) as in Lemma 3, we will show that To that end, it su¢ ces to show that (a) for all 2 [0; ] d , (6.1) = I (j = k 2) + cum (w " ;j ; w " ;j ; w " ;k ; w " ;k ) . Now, because cum (" (t 1 ) ; " (t 2 ) ; " (t 3 ) ; " (t 4 )) = b " I (t 1 = ::: = t 4 ), we have that the left side in (a) is

S N ( ) = 1 N 1=2 [ñ = ]
1 N 1=2 [ñ = ]
1 N ñ X j= ñ b #;j 0 b #;j 2 + b " 35 18 .
From here we conclude (a) by Lemma 10 and that # ( ) is continuous by C6.

We now show (b). By standard inequalities, the left side is bounded by

2

  is a positive and continuously di¤erentiable function on [ ; ] d .

  Lemma 5 and that b # # 0 = o p (1) by Lemma 6. On the other hand, by Brillinger (1981, p.15) and standard arguments, since e # # 0 = o p (1), we have that Q e #;N #0 = o p (1). Next, by Lemma 4 with ( ) = #0 ( j ) there,

  , e #;N ( ) = e ' 0 ;N ( ) ; 0 0 with ' ( ) given in (3:4), standard algebra establishes that the Whittle estimator b # in (3:1) satis…es the asymptotic linearization b

where e = 1 (

 1 obtain the following result. Theorem 3. Under H 0 and assuming C1 C3 and C5 C8 , uniformly in 2 [0 ; ], (a) b ;N

Corollary 2 .

 2 1) , and e ;N = 1 N P ñ j= ñ e ' ;N ( j ) e ' 0 ;N ( j ). (b) b ;N ) 1 . Let b N := b ;N , where ( ) be a continuous functional : D [0; ] d ! R. Under H 0 and the same conditions of Theorem 3, we have that b N d ! ( 1 ) .

( 3 : 3 )

 33 and thus for b N = b ;N . The resampling method must be such that the bootstrap statistic, say b N , is such that b N ! d ( 1 ) in probability under H 0 , where "! d " denotes Pr h b N zj x i p ! G (z) , at each continuity point z of G (z) = Pr ( ( 1 ) z). Moreover, under local alternatives (4.1)

STEP 1 :

 1 Let e x (t) = (x (t) x) =b x , where x = N 1 P n t=1 x (t) and b 2 x = N 1 P N t=1 (x (t) x)2 , and a random sample of size N with replacement from the empirical distribution of e x (t), denoted by x = fx (t)g n t=1 . STEP 2: For j = 1; :::; ñ, compute the bootstrap periodogram

  d b ( j ) 2 and I T x ( j ) as de…ned in (2:4) and then the bootstrap analogue of b # by (4.2) # = arg min

STEP 1 :ñ e it j 1 b

 11 We …rst obtain the residuals b " (t) = (2 ) ( j ) w x ( j ) , for t = 1; :::; n. From here as usual, we obtain a random sample of size N with replacement from the empirical distribution function of fb " (t)g n t=1 . Let's denote the bootstrap sample by f" (t)g n t=1 . Remark 4. (a) Notice that because b N = b ;N

!

  where d denotes convergence in bootstrap distribution. As with b # in Section 3, b # in (4:4) satis…es the asymptotic linearization b

Corollary 3 .N;( 1 ) and c a ( 1 )

 311 Under the maintained hypothesis and assuming C1 C3 and C5 C8 , we have that for any continuous functional , The proof follows from Theorem 5 and the continuous mapping theorem. Thus, Theorem 5 and Corollary 3 indicate that the bootstrap statistic b N is consistent. That is, let c f

for `= 1 ; 1 )Corollary 4 .

 114 :::; B, and compute b ;N ( ) as in (4:5) for each `. Then, c (1 ) is approximated by the value c B (that satis…es B 1 P B Next we study the behaviour of the bootstrap tests under the alternative hypothesis. Assuming C.1-C.8, under H 1 , b N d ! (e 1 ) in probability, where e 1 is a centered Gaussian process with covariance structure as 1 but with 0 replaced by 1 =plim b . Proof. The proof proceeds exactly as that of Theorem 5 and then Corollary 3 but instead of writing b 0 = o p (1) we write b 1 = o p (1) and 1 instead of 0 .

Lemma 2 .=

 2 Let k j 2 e Z d and c jk = min Y f ;j I (jj [`] k [`]j = 2; `= 1; :::; d) + O (c jk ) (b) E w T ;j w T ;k = O (c jk ) . Proof. We shall handle part (a) only, being part (b) identical. By de…nition, (5.1)

Proceeding as in Lemma 2 Lemma 4 .

 24 and by C4, we then obtain that b jk;2 = O ñ 2 N 1=2 . Likewise a j2 = O ñ 2 N 1=2 . From here, the conclusion of the lemma easily follows by observing that Y d `=1 js [`] r [`]j + N . Let ( ) be a function as in Lemma 3. Then, under C1 C4 ,

Lemma 6 .

 6 Assume C1 C3 and C5 C8. Then, b # # 0 = o p (1).

1 )

 1 by Lemma 4 and because proceeding as in Robinson and Vidal-Sanz (2006) 1 N 1=2 ñ X j= ñ j ' 0;j I T ";j 1 = O p (1) .

b " 2

 2 ] ; 2[`] for `= 1; 2 and then by Lemma 8. Lemma 10. Under C1 C3 and C5 C8, (t) and b 4;" = 1 N n X t=1 b " 4 (t) are consistent estimators of 2 " and 4;" , respectively. Proof. See Theorem 4 of Robinson and Vidal-Sanz (2006).In what follows, we shall abbreviate w T x ( ) = b ( ) and w T " ( ) by u ( ) and v ( ) respectively for all 2 d . Lemma 11. Consider j 2 e Z d . Then, for T x ;j and T x ;j equal to u j or v j ,

2 " 1 !b 2 " 1 + 2 "I 2 = b ;j 2 = 1 +

 21212221 So, because E (w " ;j w " ;k ) = b 2 " I (j = k), the left side of the equality in (a) is b From here the conclusion is standard because j ( )j 2 is twice di¤erentiable uniformly in 2 for all 2 d and that b 0 = o p (1). Next we show part (b). That follows immediately because, say, E (w " ;k w " ;k ) = 0. Lemma 12. Let k j 2 e Z d . Then,under C1 C4, O p ñ 1 I (jj [`] k [`]j = 2; `= 1; :::; d) Again, we shall handle the case when T x ;j and T x ;j are u j . We shall examine part (a) only, being part (b) identical. Proceeding as with the proof of part (a) of the previous lemma, the left side of the equality in (a) is b `(j; p) I `(k; p) .From here, we see that the last expression is zero except when jj [`] k [`]j = 2, for all `= 1; :::; d, in which case is b ;j 1 O p ñ 1 as j ( )j 2 is twice di¤erentiable uniformly in 2 for all 2 d and b 0 = o p (1). Lemma 13. Let ( ) be as in Lemma 3. Under C1 C4, we have that for all r s 2 e Z d

  = 2; `= 1; :::; d) . Notice that b " = b 4;" 3b 4 " = O p (1) by Lemma 10. Lemma 14. Let ( ) be a function as in Lemma 3. Then, under C1 C4 , o p N 1=2 .Proof. We shall consider the proof in the positive quadrantP [ñ = ]j=1 , being the proof for the remaining 2 d 1 1 quadrants similarly handled. By the Cauchy-Schwarz and triangle inequalities, it su¢ ces to show that (

because j C, d < 4 u j v j u j b 2 " b 2 "

 422 and by Lemma 11, for instance E

Lemma 15 .

 15 Let ( ; #) be is as in Lemma 5. Assuming C1 C4,

Lemma 16 .

 16 Assume C1 C3 and C5 C8. Then, b # b # = o p (1).

Lemma 17 .

 17 Assume C1 C8. Under H 0 , uniform in 2 [0; ] d (5.23)

  2005) Lemma 5 we use Lemma 7.3 of Hidalgo and Kreiss (2006).

Lemma 19 . 2 for all 1 [

 1921 Assuming C1 we have that (5.25) `] < 2[`] 2 [0; ], `= 1; :::; d, and where E

PROOFS 6 . 1 .

 61 ] ; 2[`] for `= 1; 2 and then by Lemma 18.6. Proof of Proposition 1.

v 2 P n t=1 h 4 t = P n t=1 h 2 t 2 =

 222 [0; ] d , Cov (B ( ) ; B (v)) = 2 + " 35 18 d R ( ) 0 ( ) d ,as n35 18 .

E 6 . 2 . 2 " j ;j j 2 1 )=b 2 " 1 o

 622121 We begin with (a). Its proof follows directly by that in[START_REF] Robinson | Modi…ed Whittle estimation of multilateral models on a lattice[END_REF] and observing that because ( ) is continuously di¤erentiable, then byBrillinger (1981, p:15),N 1 P [ñv= ] j= [ñv= ] j 0 j (2 ) d R v v ( ) 0 ( ) d = O p ñ 1, and thus it is omitted.Part (b) follows after observing that EI T ";j = 1 by C1 and E I T ";j I T "" T (t) " T (s)" T (r) " T (s) e i(t s) j +i(r u) k = 2I (j k = 0; n) + I (j + k = 0; n) ]=1 h `(t [`]) e ip[`]( j[`] k[`]) = n [`] I (j [`] k [`] = 0; n [`]) and that byBrillinger (1981, p.15) we have thatN 1 P ñ t=1 h p (t) ! 2 d R [0;1] d h p (u)du for all p 0. Finally, part (c) follows by Lemma 9. Proof of Theorem 3. Part (a). The proof is identical to that of Theorem 5, but instead of using Lemmas 14 and 17 we employ respectively Lemmas 4 and 7. Next the proof of part (b) is identical to that of Theorem 5, but instead of Proposition 1 we employ Theorem 4 and instead of Lemmas 18 and 19 we employ Lemmas 8 and 9. 6.3. Proof of Theorem 4. First, by standard algebra b # b and Q #;N is as de…ned in the proof of Theorem 2 but with I T x;j replaced by I T x ;j there. Now, because f #;j is twice continuously di¤erentiable by C6, and b # b # = o p (1) by Lemma 16, we easily conclude that Q e # ;N Q e # ;N = o p (1), and that Q e # ;N Q #0;N = o p (1). On the other hand, by Lemma 14 with j = b #;j there, q b #;N = N 1=2 ñ + o p (1) .So, to complete the proof it su¢ ces to showWe begin with (a). By Lemmas 11 and 12, we have that E

2 " 2 byj =b 2 " 2 = 6 . 4 . 2 ( 6 . 2 )X

 222264262 An et al. (1983) because f" (t)g n t=1 is a random sample. Now conclude part (b) since E I T " ;O p (1) by Lemma 10. This concludes the proof. Proof of Theorem 5. Part (a). By Lemma 17 with ( ) = 1 there and the de…nitions of G ;N ( ) and G 0 N ( ), we have that by Theorem 4 uniform in 2 [0; ] d , N 1=2 G b ;N ( ) G 0 N + o p (1) . Now because G b ;N ( ) G 0 N ( ) = o p N 1=2 by Lemma 14, by (6:2), we obtain that uniformly in , b ;N ( ) is 0 N ( ) + N 1=2 G b ;N ( ) G 0 k= ñ e ' b ;N (k) I T " ;k + o p (1) .

  Before we introduce and describe the test, we notice that we can alternatively state the null hypothesis H 0 as(2.1)

2

8 2 e d for some value 0 , when the "true" value of 0 is known, and where herewith e d denotes [0; ]

[ ; ] d 1 , that is 2 e d if [1] 2 [0; ] and [`]

2

[ ; ] for `= 2; :::; d.

  by Lemma 13 and because & < 1=d.To complete the proof we need to show that the …rst term in (5:19) is o p (N ).

	To that end, we note that it is bounded by
	E	max q=1;:::;[ñ & ] 1	max s=1+q[ñ 1 & ];:::