Estimating SPARMA Models with Dependent Error Terms

Yacouba Boubacar Maïnassara; Abdoulkarim Ilmi Amir

doi:10.1515/jtse-2021-0022

Published by De Gruyter January 31, 2022

Estimating SPARMA Models with Dependent Error Terms

Yacouba Boubacar Maïnassara and Abdoulkarim Ilmi Amir

From the journal Journal of Time Series Econometrics

https://doi.org/10.1515/jtse-2021-0022

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Abstract

We are interested in a class of seasonal autoregressive moving average (SARMA) models with periodically varying parameters, so-called seasonal periodic autoregressive moving average (SPARMA) models under the assumption that the errors are uncorrelated but non-independent (i.e. weak SPARMA models). Relaxing the classical independence assumption on the errors considerably extends the range of application of the SPARMA models, and allows one to cover linear representations of general nonlinear processes. We establish the asymptotic properties of the quasi-generalized least squares (QLS) estimator of these models. Particular attention is given to the estimation of the asymptotic variance matrix of the QLS estimator, which may be very different from that obtained in the standard framework. A set of Monte Carlo experiments are presented.

Keywords: quasi-generalized least squares; seasonality; weak PARMA models; weak SARMA; weak SPARMA models

AMS 2000 Subject Classifications: Primary 62M10; 62F03; 62F05; secondary 91B84; 62P05

JEL Classification: C02; C13; C22

Corresponding author: Yacouba Boubacar Maïnassara, Laboratoire de mathématiques de Besançon, Université Bourgogne Franche-Comté, UMR CNRS 6623, 16 route de Gray, 25030, Besançon, France, E-mail: yacouba.boubacar_mainassara@univ-fcomte.fr

Article note: This article is linked with another article which can be found here: https://doi.org/10.1515/jtse-2022-0002.

Acknowledgments

We sincerely thank the anonymous referee and Editor for helpful remarks.

Appendix A: Proofs

The proofs of Theorem 1 and 2 are quite technical. These are adaptations of the arguments used in Francq, Roy, and Saidi (2011). In order to shorten our paper and to point out the novelties of our proofs, only the new arguments will be developed. Let K and ρ be generic constants, whose values will be modified along the proofs, such that K > 0 and ρ ∈ 0,1 .

A.1 Reminder on Technical Issues on WLS Method for SPARMA Models

We recall that, given a realization X ₁, …, X _NT, the periodic noise ϵ _nT+ν(α) is approximated by e _nT+ν(α) which is defined in (8).

The starting point in the asymptotic analysis, is the property that ϵ _nT+ν(α) − e _nT+ν(α) converges uniformly to 0 (almost-surely) as n goes to infinity. Similar properties also holds for the derivatives with respect to α of ϵ _nT+ν(α) − e _nT+ν(α). We sum up the fact that we shall need in the sequel. We refer to the appendix of Francq, Roy, and Saidi (2011) for a more detailed treatment.

Under Assumption (A1), for any α ∈ Δ_δ and any ( l , m ) ∈ { 1 , … , k 0 T } 2 , there exists absolutely summable and deterministic sequences ( C i ( θ ) ) i ≥ 0 , ( C i , l ( θ ) ) i ≥ 0 and ( C i , l , m ( θ ) ) i ≥ 0 such that, almost surely,

(22) ϵ n ( α ) = ∑ i = 0 ∞ C i ( α ) X n − i , ∂ ϵ n ( α ) ∂ α m = ∑ i = 0 ∞ C m , i ( α ) X n − i and ∂ 2 ϵ n ( α ) ∂ α m ∂ α l = ∑ i = 0 ∞ C m , l , i ( α ) X n − i

(23) e n ( α ) = ∑ i = 0 N − 2 C i ( α ) X n − i , ∂ e n ( α ) ∂ α m = ∑ i = 0 N − 2 C m , i ( α ) X n − i and ∂ 2 e n ( α ) ∂ α m ∂ α l = ∑ i = 0 N − 2 C m , l , i ( α ) X n − i .

A useful property of the matrices sequences C, is that they are asymptotically exponentially small. Indeed, there exists ρ ∈ 0,1 such that, for all i ≥ 0,

(24) sup α ∈ Δ δ C i ( α ) + C m , i ( α ) + C m , l , i ( α ) ≤ K ρ i ,

where K is a positive constant. Finally, from the above estimates, we are able to deduce that for any ( l , m ) ∈ { 1 , … , k 0 T } 2 , it holds

(25) sup α ∈ Δ δ ϵ n T + ν ( α ) − e n T + ν ( α ) ≤ K ρ n ,

(26) ρ n sup α ∈ Δ δ e n T + ν ( α ) = ρ n sup α ∈ Δ δ ϵ n T + ν ( α ) ⟶ n → ∞ a.s. 0 ,

(27) sup σ 2 > σ ̲ sup α ∈ Δ δ O N σ 2 ( α ) − Q N σ 2 ( α ) = O ( 1 / N ) a.s. N → ∞ ,

for any constant σ ̲ > 0 . Analogous estimates to (25), (26) and (27) are satisfied for first and second order derivatives of ϵ _nT+ν(α) and e _nT+ν(α).

This implies that the sequences N ∂ Q N σ 2 ( α 0 ) / ∂ α and N ∂ O N σ 2 ( α 0 ) / ∂ α have the same asymptotic distribution. More precisely we have

(28) N ∂ ∂ α Q N σ 2 ( α 0 ) − ∂ ∂ α O N σ 2 ( α 0 ) = o P ( 1 ) .

We also have

N ∂ O N σ ̂ 2 ( α ) ∂ α − ∂ O N σ 0 2 ( α ) ∂ α α = α 0 = ∑ ν = 1 T 1 σ ̂ ν 2 − 1 σ ν 2 1 N ∑ n = 0 N − 1 ϵ n T + ν ( α 0 ) ∂ ϵ n T + ν ∂ α α = α 0 = o P ( 1 ) ,

by using the fact that

(29) max ν = 1 , … , T 1 σ ̂ ν 2 − 1 σ ν 2 < ϵ almost surely N large enough ,

for any ɛ > 0 and N large enough. And that N − 1 ∑ n = 0 N − 1 ϵ n T + ν ( α 0 ) ∂ ϵ n T + ν / ∂ α α = α 0 N is a tight sequence.

A.2 Proof of Theorem 3

The proof of Theorem 3 is based on series lemmas. We use the multiplicative matrix norm defined by: ∥ A ∥ = sup ∥ x ∥ ≤ 1 ∥ A x ∥ = ϱ 1 2 ( A ′ A ) , where A is a d ₁ × d ₂ matrix, ∥x∥ is the Euclidean norm of the vector x ∈ R d 2 , and ϱ(.) denotes the spectral radius. This norm satisfies

(30) ∥ A ∥ 2 = ∑ i = 1 k 1 ∑ j = 1 k 2 a i , j 2

with a _i,j the entries of A ∈ R k 1 × k 2 . The choice of the norm is crucial for the following results to hold(with e.g. the Euclidean, this result is not valid). We denote

Σ ϒ , ϒ ̲ r = E ϒ n ϒ ̲ r , n ′ Σ ϒ = E ϒ n ϒ n ′ Σ ϒ ̲ r = E ϒ ̲ r , n ϒ ̲ r , n ′

where ϒ ̲ r , n = ( ϒ n − 1 , … , ϒ n − r ) . For any N ≥ 0 we have

I ̂ N SP = Φ ̂ r − 1 ( 1 ) Σ ̂ u ̂ r Φ ̂ r − 1 ( 1 ) = ( Φ ̂ r − 1 ( 1 ) − Φ − 1 ( 1 ) ) Σ ̂ u ̂ r Φ ̂ r − 1 ( 1 ) + Φ − 1 ( 1 ) ( Σ ̂ u ̂ r − Σ u ) Φ ̂ r − 1 ( 1 ) ′ + Φ − 1 ( 1 ) Σ u ( Φ ̂ r − 1 ( 1 ) − Φ − 1 ( 1 ) ′ ) + ϕ − 1 ( 1 ) Σ u Φ − 1 ( 1 ) ′ .

We the obtain

(31) ∥ I ̂ N S P − I ( α 0 ) ∥ ≤ ∥ Φ ̂ r − 1 ( 1 ) − ϕ − 1 ( 1 ) ∥ ∥ Σ ̂ u ̂ r ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ + ∥ Φ − 1 ( 1 ) ∥ × ∥ Σ ̂ u ̂ r − Σ u ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ + ∥ Φ − ∥ ∥ Σ u ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ − Φ − 1 ( 1 ) ′ ∥ ≤ ∥ Φ ̂ − 1 ( 1 ) r − ϕ − 1 ( 1 ) ∥ ∥ Σ ̂ u ̂ r ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ + ∥ Φ − 1 ( 1 ) ∥ ∥ Σ u ∥ + ∥ Σ ̂ u ̂ r − Σ u ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ ∥ Φ − 1 ∥ ≤ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ ∥ Φ − Φ ̂ r ( 1 ) ∥ ∥ Φ − 1 ∥ ∥ Σ ̂ u ̂ r ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ + ∥ Φ − 1 ∥ ∥ Σ u ∥ ∥ Σ ̂ u ̂ r − Σ u ∥ ∥ Φ ̂ r − 1 ( 1 ) ′ ∥ ∥ Φ − 1 ∥ .

It suffices to show that Φ ̂ r ( 1 ) ′ → Φ and Σ ̂ u ̂ r → Σ u in probability, to have the proof of Theorem 3. Let the r × 1 vector 1_r = (1, …, 1)′ and the rT(p + q + P + Q) × T(p + q + P + Q) matrix E r = I T ( p + q + P + Q ) ⊗ 1 r , where ⊗ denotes the matrix kronecker product and I _m the m ⊗ m identity matrix. Write Φ ̲ * = ( ϕ 1 , … , ϕ r ) where the ϕ _i’s are defined by (15). We have

(32) ∥ Φ ̂ r ( 1 ) − Φ ( 1 ) ∥ = ∑ k = 1 r Φ ̂ r , k − ∑ k = 1 r Φ r , k + ∑ k = 1 r Φ r , k − ∑ k = 1 ∞ Φ r , k ≤ ∑ k = 1 r ( Φ ̂ r ( 1 ) − Φ r , k ) + ∑ k = 1 r ( Φ r , k − Φ k ) + ∑ k = r + 1 ∞ Φ k ≤ ( Φ ̂ r ̲ − Φ ̲ r ) E r + Φ ̲ r * − ϕ ̲ r E r + ∑ k = r + 1 ∞ Φ k ≤ r Φ ̂ ̲ r − Φ ̲ r + Φ ̲ r * − Φ ̲ r + ∑ k = r + 1 ∞ Φ k .

In view of Assumptions of Theorem 3 we have

∑ k = r + 1 ∞ Φ k ≤ ∑ k = r + 1 ∞ Φ k ⟶ N → ∞ 0 .

To obtain the convergence in probability of Φ ̂ r ( 1 ) towards Φ(1), one must show the convergence in probability towards 0 of r Φ ̂ ̲ r − Φ ̲ r and r Φ ̲ r * − Φ ̲ r .

By (17)) we obtain

(33) ϒ n ( α 0 ) = Φ ̲ r ϒ ̲ r , n ( α 0 ) + u r , n ,

and Σ u r = Var ( u r , n ) = E u r , n ( ϒ n ( α 0 ) − Φ ̲ r ϒ ̲ r , n ( α 0 ) ) ′ . The vector u _r,n is orthogonal to ϒ ̲ r , n ( α 0 ) . Therefore

Var ( u r , n ) = E ( ϒ n ( α 0 ) − Φ ̲ r ϒ ̲ r , n ( α 0 ) ) ϒ n ′ ( α 0 ) = Σ ϒ − Φ ̲ r Σ ϒ , ϒ r ′ .

Consequently the least squares estimator of Σ u r can be rewritten in the form:

(34) Σ ̂ u ̂ r = Σ ̂ ϒ ̂ − Φ ̂ ̲ r Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ ,

where

(35) Σ ̂ ϒ ̂ = 1 N ∑ n = 0 N − 1 ϒ ̂ n ϒ ̂ n ′ .

Similar arguments combined with (15) yied

Σ u = E u n u n ′ = E u n ϒ n ( α 0 ) ′ = E ϒ n ( α 0 ) ϒ n ( α 0 ) ′ − ∑ k = 1 r Φ k E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ − ∑ k = r + 1 ∞ Φ k E ϒ n − k ( α 0 ) ϒ t ( α 0 ) ′ = Σ ϒ − Φ ̲ r * Σ ϒ , ϒ ̲ r ′ − ∑ k = r + 1 ∞ Φ k E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ .

By (34) we obtain

(36) Σ ̂ u ̂ r − Σ u = Σ ̂ w ̂ − Φ ̂ ̲ r Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ − Σ ϒ + Φ ̲ r * Σ ϒ , ϒ ̲ r ′ + ∑ k = r + 1 ∞ Φ k E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ = Σ ̂ ϒ ̂ − Σ ϒ − Φ ̂ r ̲ − Φ ̲ r * Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ − Φ ̲ r * Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ − Σ ϒ , ϒ ̲ r ′ + ∑ k = r + 1 ∞ Φ k E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ ≤ Σ ̂ ϒ ̂ − Σ ϒ + Φ ̂ r ̲ − Φ ̲ r * Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ − Σ ϒ , ϒ ̲ r ′ + Φ ̂ r ̲ − Φ ̲ r * Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ + Φ ̲ r * Σ ̂ ϒ ̂ , ϒ ̂ ̲ r ′ − Σ ϒ , ϒ ̲ r ′ + ∑ k = r + 1 ∞ Φ k E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ .

In view of Assumptions of Theorem 3 and (30) we obtain

∑ k = r + 1 ∞ Φ k E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ ≤ Φ k ∑ k = r + 1 ∞ E ϒ n − k ( α 0 ) ϒ n ( α 0 ) ′ ≤ K ∑ k = r + 1 ∞ 1 k 2 ⟶ N → ∞ 0

and we also have

Φ ̲ r * 2 ≤ ∑ k ≥ Tr Φ k Φ k ′ < ∞ .

Lemma 1

Under the assumptions of the Theorem 3 we have

sup r ≥ 1 Σ ϒ , ϒ ̲ r , Σ ϒ , ϒ ̲ r , Σ , ϒ ̲ r < ∞ .

Proof

See Lemma 1 in the supplementary material of Boubacar Mainassara, Carbon, and Francq (2012). □

Lemma 2

Under the assumptions of the Theorem 3 there exists a finite positive constant K such that, for 1 ≤ r ₁, r ₂ ≤ r and 1 ≤ m ₁, m ₂ ≤ T(p + q + P + Q) we have

sup n ∈ Z ∑ h = − ∞ ∞ | cov ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r − 2 , m 2 ( α 0 ) , ϒ n − r 1 − h , m 1 ( α 0 ) w n − r 2 − h , m 2 ( α 0 ) | < K .

Proof

We can take the supremum over the integers n > 0, and write the proof in the case m ₁ = m ₂ = m. We have

∑ h = − ∞ ∞ | cov ( ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) , ϒ n − r 1 − h , m 1 ( α 0 ) ϒ n − r 2 − h , m 2 ( α 0 ) ) | ≤ K 8 ∑ h 1 , … , h 8 = 0 ∞ cov ( Y 1 , h 1 , h 2 Y 1 + s , h 3 , h 4 , Y 1 + h , h 5 , h 6 Y 1 + s + h , h 7 , h 8 )

where

Y t , h 1 , h 2 = ϵ t − h 1 ϵ t − h 2 − m − E ϵ t − h 1 ϵ t − h 2 − m

A slight extension of (Francq and Zakoïan 2019, Corollary A.3) concludes. □

Lemma 3

Under the assumptions of the Theorem 3, r Σ ̂ ϒ ̲ r − Σ ϒ ̲ r , r Σ ̂ ϒ , ϒ ̲ r − Σ ϒ , ϒ ̲ r and r Σ ̂ ϒ − Σ ϒ tend to 0 in probability as N → ∞ when r = o ( N 1 3 ) .

Proof

For 1 ≤ m ₁, m ₂ ≤ T(p + q + P + Q) and 1 ≤ r ₁, r ₂ ≤ r, the ( { ( r 1 − 1 ) ( T ( p + q + P + Q ) + m 1 } , ( r 2 − 1 ) ( T ( p + q + P + Q ) + m 2 ) − the element of Σ ̂ ϒ ̲ r is given by:

1 N ∑ n = 0 N − 1 ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) .

For all β > 0 we use (30) and we obtain

P r Σ ̂ ϒ ̲ r − Σ ϒ ̲ r ≥ β ≤ r β 2 E Σ ̂ ϒ ̲ r − Σ ϒ ̲ r 2 ≤ r β 2 E 1 N ∑ n = 0 N − 1 ϒ ̲ r , n ϒ ̲ r , n ′ − E ϒ ̲ r , n ϒ ̲ r , n 2 ≤ r β 2 ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) E × 1 N ∑ n = 0 N − 1 ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) − E ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) 2 .

The stationarity of the process ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) n and Lemma 2 imply

P r Σ ̂ ϒ ̲ r − Σ ϒ ̲ r ≥ β ≤ r β 2 ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) Var × 1 N ∑ n = 0 N − 1 ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) ≤ r N β 2 ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) ∑ n = 0 N − 1 ∑ n ′ = 0 N − 1 × cov ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) , ϒ n ′ − r 1 , m 1 ( α 0 ) × ϒ n ′ − r 2 , m 2 ( α 0 ) ≤ r N β 2 ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) ∑ h = 2 − N N − 2 × ( N − | h | ) cov ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) , ϒ n ′ − r 1 , m 1 ( α 0 ) ϒ n ′ − r 2 , m 2 ( α 0 ) ≤ r N β 2 ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) sup n ∈ Z × ∑ h = − ∞ ∞ | c o v ϒ n − r 1 , m 1 ( α 0 ) ϒ n − r 2 , m 2 ( α 0 ) , ϒ n − r 1 − h , m 1 ( α 0 ) ϒ n − r 2 − h , m 2 ( α 0 ) | ≤ K ( T ( p + q + P + Q ) ) 2 r 3 N β 2 .

Finaly we have

E r Σ ̂ ϒ − Σ ϒ 2 ≤ E r Σ ̂ ϒ , ϒ ̲ r − Σ ϒ , ϒ ̲ r 2 ≤ E r Σ ̂ ϒ ̲ r − Σ ϒ ̲ r 2 ≤ K ( T ( p + q + P + Q ) ) 2 r 3 N ⟶ N → ∞ 0 ,

when r = o ( N 1 3 ) . The result follows. □

We show in the following lemma that the previous lemma remains valid when we replace ϒ_n(α ₀) by ϒ ̂ n .

Lemma 4

Under the assumptions of the Theorem 3, r Σ ̂ ϒ ̂ ̲ r − Σ ϒ ̲ r , r Σ ̂ ϒ ̂ , ϒ ̲ r − Σ ϒ , ϒ ̲ r and r Σ ̂ ϒ ̂ − Σ ϒ tend to 0 in probability as N → ∞ when r = o ( N 1 3 ) .

Proof

We denote Σ ̂ ϒ ̲ r , N the matrix obtained by replacing e n ( α ̂ ) by ϵ n ( α ̂ ) in Σ ̂ ϒ ̂ ̲ r . We have

r Σ ̂ ϒ ̂ ̲ r − Σ ϒ ̲ r ≤ r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N + r Σ ̂ ϒ ̲ r , N − Σ ̂ ϒ ̲ r + r Σ ̂ ϒ ̲ r − Σ ϒ ̲ r

By Lemma 3 the term r Σ ̂ ϒ ̂ ̲ r − Σ ϒ ̲ r converges in probability. The lemma will be proved as soon as we show that

(37) r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N = o P ( 1 )

(38) and r Σ ̂ ϒ ̲ r , N − Σ ̂ ϒ ̲ r = o P ( 1 )

when r = o ( N 1 3 ) . This is done in two separate steps.

Step 1: proof of (37) .

For all β > 0 we have

P r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N ≥ β ≤ r β E Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N ≤ r β E 1 N ∑ n = 0 N − 1 ϒ ̂ ̲ r , n ϒ ̂ ̲ r , n ′ − 1 N ∑ n = 0 N − 1 ϒ ̲ r , n ( N ) ϒ ̲ r , n ′ ( N ) ≤ K r β ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) E × 1 N ∑ n = 0 N − 1 ϒ ̂ ̲ n − r 1 , m 1 ϒ ̂ ̲ n − r 2 , m 2 − 1 N ∑ n = 0 N − 1 ϒ ̲ n − r 1 , m 1 ( N ) ϒ ̲ n − r 2 , m 2 ( N ) ,

where

ϒ ̲ n , m ( N ) = ∑ ν = 1 T σ ν − 2 ϵ n T + ν ( α ̂ ) ∂ ∂ α m ϵ n T + ν ( α ̂ ) and ϒ ̲ r , n ( N ) = ϒ n − 1 ′ ( N ) , … , ϒ n − r ′ ( N ) ′ .

It is follow that

(39) P r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N ≥ β ≤ K r N β ∑ ν = 1 T 1 σ ̂ ν 2 − 1 σ ν 2 ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) × ∑ m 2 = 1 T ( p + q + P + Q ) E ∑ n = 0 N − 1 A ̂ n , r 1 A ̂ n , r 2 − A n , r 1 A n , r 2

where

A n , r i = ϵ n T + ν − r i ( α ̂ ) ∂ ∂ α m i ϵ n T + ν − r i ( α ̂ ) , for i = { 1,2 } .

Now notice that

e n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 e n T + ν − r 1 ( α ̂ ) e n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) − ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) ϵ n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) = e n T + ν − r 1 ( α ̂ ) − ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 e n T + ν − r 1 ( α ̂ ) e n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 × e n T + ν − r 2 ( α ̂ ) + ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 e n T + ν − r 1 ( α ̂ ) − ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) e n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) + ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) e n T + ν − r 2 ( α ̂ ) − ϵ n T + ν − r 2 ( α ̂ ) × ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) + ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) ϵ n T + ν − r 2 × ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) − ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) .

We replace the above identity in (39) and we obtain by Hölder’s inequality that

(40) P r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N ≥ β ≤ K r N β ∑ ν = 1 T 1 σ ̂ ν − 1 σ ν ∑ r 1 = 1 r ∑ r 2 = 1 r ∑ m 1 = 1 T ( p + q + P + Q ) × ∑ m 2 = 1 T ( p + q + P + Q ) E D N , 1 + D N , 2 + D N , 3 + D N , 4

where

D N , 1 = ∑ n = 0 N − 1 e n T + ν − r 1 ( α ̂ ) − ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 e n T + ν − r 1 ( α ̂ ) e n T + ν − r 2 ( α ̂ ) × ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) D N , 2 = ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 e n T + ν − r 1 ( α ̂ ) − ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) × e n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) D N , 3 = ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) e n T + ν − r 2 ( α ̂ ) − ϵ n T + ν − r 2 ( α ̂ ) × ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) D N , 4 = ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) ϵ n T + ν − r 2 × ∂ ∂ α m 2 e n T + ν − r 2 ( α ̂ ) − ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) .

We have

| D N , 1 | ≤ K ∑ n = 0 N − 1 | e n T + ν − r 1 ( α ̂ ) − ϵ n T + ν − r 1 ( α ̂ ) | ≤ K r + ∑ n = r 1 + 1 N − 1 ρ n .

Then we obtain

(41) D N , 1 ≤ K r .

The same calculations hold for the terms D _N,2, D _N,3, and D _N,4. Thus

(42) | D N , 1 | + | D N , 2 | + | D N , 3 | + | D N , 4 | ≤ K r

and reporting this estimation in (40) and using (29), we have

P r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N ≥ β ≤ ϵ K r N β r T 2 ( p + q + P + Q ) 2 ≤ K r 5 2 N .

Then the sequence r Σ ̂ ϒ ̂ ̲ r − Σ ̂ ϒ ̲ r , N converge in probability to 0 as N → ∞ when r = r ( N ) = o ( N 5 2 ) .

Step 2: proof of (38) .

First we follow the same approach than in the previous step. We have

Σ ̂ ϒ ̲ r , N − Σ ̂ ϒ ̲ r 2 = 1 N ∑ n = 0 N − 1 ϒ ̲ r , n ( n ) ϒ ̲ r , n ′ ( n ) − ϒ ̲ r , n ϒ ̲ r , n ′ 2 ≤ K r β ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) × 1 N ∑ n = 0 N − 1 ϒ n − r 1 , m 1 ( n ) ϒ n − r 2 , m 2 ( n ) − ϒ n − r 1 , r 1 ϒ n − r 2 , m 2 2 ≤ K r β ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) ∑ m 2 = 1 T ( p + q + P + Q ) ∑ ν = 1 T 1 σ ̂ ν 2 − 1 σ ν 2 2 × 1 N ∑ n = 0 N − 1 A n , r 1 , m 1 ( α ̂ ) A n , r 1 A n , r 2 , ( α ̂ ) − A n , r 1 ( α 0 ) A n , r 2 ( α 0 ) 2 .

Since

ϵ n T + ν − r 1 ( α ̂ ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) ϵ n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) − ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) ϵ n T + ν − r 2 ( α 0 ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α 0 ) = ϵ n T + ν − r 1 ( α ̂ ) − ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) ϵ n T + ν − r 2 ( α ̂ ) × ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) + ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) − ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) ϵ n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) + ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) ϵ n T + ν − r 2 ( α ̂ ) − ϵ n T + ν − r 2 ( α 0 ) × ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) + ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) ϵ n T + ν − r 2 × ( α 0 ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) − ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α 0 )

one has

(43) Σ ̂ ϒ ̲ r , N − Σ ̂ ϒ ̲ r ≤ ϵ K r β ∑ r 1 = 0 r ∑ r 2 = 0 r ∑ m 1 = 1 T ( p + q + P + Q ) E N , 1 + E N , 2 + E N , 3 + E N , 4

where

E N , 1 = 1 N ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α ̂ ) − ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) × ϵ n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) E N , 2 = 1 N ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) − ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) × ϵ n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) E N , 3 = 1 N ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) × ϵ n T + ν − r 2 ( α ̂ ) − ϵ n T + ν − r 2 ( α 0 ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) E N , 4 = 1 N ∑ n = 0 N − 1 ϵ n T + ν − r 1 ( α 0 ) ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) ϵ n T + ν − r 2 ( α 0 ) × ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) − ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α 0 ) .

Taylor expansions around α ₀ yield that there exists α ̲ and α ̄ between α ̂ and α ₀ such that

ϵ n T + ν − r 1 ( α ̂ ) − ϵ n T + ν − r 1 ( α 0 ) ≤ r n T + ν α ̂ − α 0

and

∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) − ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α 0 ) ≤ s n T + ν α ̂ − α 0

with r n T + ν = ‖ ∂ ϵ n T + ν ( α ̲ ) / ∂ α ‖ and s n T + ν = ‖ ∂ 2 ϵ n T + ν ( α ̄ ) / ∂ α ′ ∂ α m ‖ . Using the fact that

E r n T + ν − r 1 ∂ ∂ α m 1 ϵ n T + ν − r 1 ( α ̂ ) ϵ n T + ν − r 2 ( α ̂ ) ∂ ∂ α m 2 ϵ n T + ν − r 2 ( α ̂ ) < ∞

and that ( N ( α ̂ − α 0 ) ) N is a tight sequence (which implies that α ̂ − α 0 = O P ( 1 / N ) ), we deduce that

E N , 1 = O P 1 N .

The same arguments are valid for E _N,2, E _N,3 and E _N,4. Consequently E N , 1 + E N , 2 + E N , 3 + E N , 4 = O P 1 / N and (43) yields

Σ ̂ ϒ ̲ r , N − Σ ̂ ϒ ̲ r 2 = O P r 2 N

when r = o ( N 1 3 ) we finaly obtain r Σ ̂ ϒ ̲ r , N − Σ ̂ ϒ ̲ r = O P ( 1 ) . □

Lemma 5

Under the assumptions of the Theorem 3, we have

r Φ ̲ r * − Φ ̲ r = O P ( 1 ) as r → ∞ .

Proof

Racall that (15) and (33) we have

ϒ n ( α 0 ) = Φ ̲ r ϒ ̲ r , n + u r , n = Φ ̲ r * ϒ ̲ r , n + ∑ k = r + 1 ∞ Φ r ϒ n − k ( α 0 ) + u n ≔ Φ ̲ r * ϒ ̲ r , n + u r , n * .

By the orthogonality conditions in (15) and (33), one has

Σ u r * , ϒ ̲ r ≔ E u r , n * ϒ ̲ r , n ′ = E ϒ n ( α 0 ) − Φ ̲ r * ϒ ̲ r , n ϒ ̲ r , n ′ = E Φ ̲ r ϒ ̲ r , n + u r , n − Φ ̲ r * ϒ ̲ r , n ϒ ̲ r , n ′ = Φ ̲ r − Φ ̲ r * Σ ϒ ̲ r

and consequently

(44) Φ ̲ r * − Φ ̲ r = Σ u r * , ϒ ̲ r Σ ϒ ̲ r − 1

Using Lemmas 1, 2, and (44) implies that

P r Φ ̲ r * − Φ ̲ r ≥ β ≤ r β Σ u r * , ϒ ̲ r Σ ϒ ̲ r − 1 ≤ K r β E ∑ ≥ r + 1 Φ r ϒ n − k ( α 0 ) + u n ϒ ̲ r , n ′ ≤ K r β ∑ k ≥ r + 1 Φ k E ϒ n − k ( α 0 ) ϒ ̲ r , n ′ ≤ K r β ∑ l ≥ 1 Φ l + r E ϒ n − k ( α 0 ) ϒ n − k ′ ( α 0 ) , … , ϒ n − r ′ ≤ K r β ∑ l ≥ 1 Φ l + r ∑ j = 1 T ( p + q + P + Q ) ∑ k = 1 T ( p + q + P + Q ) × ∑ r 1 r E ϒ n − r − l , j ( α 0 ) ϒ n − r − l , k ( α 0 ) 2 1 2 ≤ K r β ∑ l ≥ 1 Φ l + r ∑ j = 1 T ( p + q + P + Q ) ∑ k = 1 T ( p + q + P + Q ) × ∑ r 1 r E ϒ n − r 1 − l , j 2 ( α 0 ) ϒ n − r 1 − l , k 2 ( α 0 ) 1 2 ≤ K T ( p + q + P + Q ) r β ∑ l ≥ 1 Φ l + r .

By the assumptions of Theorem 3, ∑ l ≥ 1 Φ l + r = o ( 1 ) as r → ∞. The proof of the lemma follows. □

Lemma 6

Under the assumptions of the Theorem 3 we have

r Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 = o P ( 1 )

as N → ∞ when r = o ( N 1 3 ) and r → ∞.

Proof

We have

Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 ≤ Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 + Σ ϒ ̲ r − 1 Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r Σ ϒ ̲ r − 1 .

and by induction we obtain

Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 ≤ Σ ϒ ̲ r − 1 ∑ k = 1 ∞ Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r k Σ ϒ ̲ r − 1 k .

We have

P r Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 > β ≤ P Σ ϒ ̲ r − 1 ∑ k = 1 ∞ Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r k Σ ϒ ̲ r − 1 k > β ≤ P Σ ϒ ̲ r − 1 ∑ k = 1 ∞ Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r k Σ ϒ ̲ r − 1 k > β a n d Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r Σ ϒ ̲ r − 1 < 1 + P Σ ϒ ̲ r − 1 ∑ k = 1 ∞ Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r k Σ ϒ ̲ r − 1 k > β a n d Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r Σ ϒ ̲ r − 1 ≥ 1 ≤ P r Σ ϒ ̲ r − 1 2 Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r 1 − Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r Σ ϒ ̲ r − 1 + P r Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r Σ ϒ ̲ r − 1 ≥ 1 ≤ P r Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r > β Σ ϒ ̲ r − 1 2 + β r − 1 2 Σ ϒ ̲ r − 1 + P r Σ ϒ ̲ r − Σ ̂ ϒ ̂ ̲ r ≥ Σ ϒ ̲ r − 1

Lemmas 1 and 4 imply the result. □

Lemma 7

Under the assumptions of the Theorem 3, we have

r Φ ̂ ̲ r − Φ ̲ r = O P ( 1 ) as r → ∞ and r = o N 1 3 .

Proof

Lemmas 1 and 6 yield

(45) Σ ̂ ϒ ̂ ̲ r − 1 ≤ Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 + Σ ϒ ̲ r − 1 = O P ( 1 ) .

By (33), we have

0 = E u r , n ϒ ̲ r , n ′ = E ϒ n ( α 0 ) − Φ ̲ r ϒ ̲ r , n ′ ϒ ̲ r , n ′ = Σ ϒ , ϒ ̲ r − Φ ̲ r Σ ϒ ̲ r

and so we have Φ ̲ r = Σ ϒ , w ̲ r Σ ϒ ̲ r − 1 . Lemmas 1, 4, 6 and (45) imply

r Φ ̂ ̲ r − Φ ̲ r = r Σ ̂ ϒ ̂ , ϒ ̂ ̲ r Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ , ϒ ̲ r Σ ϒ ̲ r − 1 = r Σ ̂ ϒ ̂ , ϒ ̂ ̲ r − Σ ϒ , ϒ ̲ r Σ ̂ ϒ ̂ ̲ r − 1 + Σ ϒ , ϒ ̲ r Σ ̂ ϒ ̂ ̲ r − 1 − Σ ϒ ̲ r − 1 = o P ( 1 ) .

The lemma is proved □

Proof of Theorem 3

Proof

Since by Lemma 4 we have Σ ̂ ϒ ̂ − Σ ϒ = o P ( r − 1 2 ) = o P ( 1 ) and Σ ̂ ϒ ̂ , ϒ ̂ ̲ r − Σ ϒ , ϒ ̲ r = o P ( r − 1 2 ) = o P ( 1 ) , and by Lemma 5: Φ ̂ ̲ r − Φ ̲ r = o P ( r − 1 2 ) = o P ( 1 ) . The Theorem 3 is then proved. □

Appendix B: Verification of Assumption (A1) on Example

Example 1

We now examine the causality and stationarity conditions on the following SPARMA(1, 1)(1, 1) model:

X n T + ν = a 1 ( ν ) X n T + ν − 1 + a 1 ( ν ) X ( n − 1 ) T + ν + a 1 ( ν ) a 1 ( ν ) X ( n − 1 ) T + ν − 1 − b 1 ( ν ) ϵ n T + ν − 1 − b 1 ( ν ) ϵ ( n − 1 ) T + ν + b 1 ( ν ) b 1 ( ν ) ϵ ( n − 1 ) T + ν − 1 + ϵ n T + ν .

This SPARMA(1, 1)(1, 1) model admits the following PARMA(1 + T, 1 + T) representation:

X n T + ν − ∑ i = 1 1 + T a i * ( ν ) X n T + ν − i = − ∑ i = 1 1 + T b i * ( ν ) ϵ n T + ν − i + ϵ n T + ν .

Finally we deduce the following VARMA(2, 2) model:

A 0 X n = A 1 X n − 1 + A 1 X n − 2 − B 1 ϵ n − 1 − B 2 ϵ n − 2 + B 0 ϵ n

where the matrices A ₀, A ₁, A ₂, B ₀, B ₁, and B ₂ are given for simplify for T = 4 by

A 0 = 1 0 0 0 − a 1 * ( 2 ) 1 0 0 − a 2 * ( 3 ) − a 1 * ( 3 ) 1 0 − a 3 * ( 4 ) − a 2 * ( 4 ) − a 1 * ( 4 ) 1 = 1 0 0 0 − a 1 ( 2 ) 1 0 0 0 − a 1 ( 3 ) 1 0 0 0 − a 1 ( 4 ) 1

A 1 = a 4 * ( 1 ) a 3 * ( 1 ) a 2 * ( 1 ) a 1 * ( 1 ) a 5 * ( 2 ) a 4 * ( 2 ) a 3 * ( 2 ) a 2 * ( 2 ) a 6 * ( 3 ) a 5 * ( 3 ) a 4 * ( 3 ) a 3 * ( 3 ) a 7 * ( 4 ) a 6 * ( 4 ) a 5 * ( 4 ) a 4 * ( 4 ) = a 1 ( 1 ) 0 0 a 1 ( 1 ) − a 1 ( 2 ) a 1 ( 2 ) a 1 ( 2 ) 0 0 0 − a 1 ( 3 ) a 1 ( 3 ) a 1 ( 3 ) 0 0 0 − a 1 ( 4 ) a 1 ( 4 ) a 1 ( 4 )

A 2 = a 8 * ( 1 ) a 7 * ( 1 ) a 6 * ( 1 ) a 5 * ( 1 ) a 9 * ( 2 ) a 8 * ( 2 ) a 7 * ( 2 ) a 6 * ( 2 ) a 10 * ( 3 ) a 9 * ( 3 ) a 8 * ( 3 ) a 7 * ( 3 ) a 11 * ( 4 ) a 10 * ( 4 ) a 9 * ( 4 ) a 8 * ( 4 ) = 0 0 0 − a 1 ( 1 ) a 1 ( 1 ) 0 0 0 0 0 0 0 0 0 0 0 0

A 0 − A 1 z − A 2 z 2 = 1 − a 1 ( 1 ) z 0 0 a 1 ( 1 ) a 1 ( 1 ) z 2 − a 1 ( 1 ) z a 1 ( 2 ) a 1 ( 2 ) z − a 1 ( 2 ) z 1 − a 1 ( 2 ) z 0 0 0 a 1 ( 3 ) a 1 ( 3 ) z − a 1 ( 3 ) z 1 − a 1 ( 3 ) z 0 0 0 a 1 ( 4 ) a 1 ( 4 ) z − a 1 ( 4 ) z 1 − a 1 ( 4 ) z

we have det A 0 − A 1 z − A 2 z 2 = ∏ ν = 1 T 1 − a 1 ( ν ) z 1 − ∏ ν = 1 T a 1 ( ν ) z ≠ 0 and

B 0 = 1 0 0 0 − b 1 * ( 2 ) 1 0 0 − b 2 * ( 3 ) − b 1 * ( 3 ) 1 0 − b 3 * ( 4 ) − b 2 * ( 4 ) − b 1 * ( 4 ) 1 = 1 0 0 0 − b 1 ( 2 ) 1 0 0 0 − b 1 ( 3 ) 1 0 0 0 − b 1 ( 4 ) 1

B 1 = b 4 * ( 1 ) b 3 * ( 1 ) b 2 * ( 1 ) b 1 * ( 1 ) b 5 * ( 2 ) b 4 * ( 2 ) b 3 * ( 2 ) b 2 * ( 2 ) b 6 * ( 3 ) b 5 * ( 3 ) b 4 * ( 3 ) b 3 * ( 3 ) b 7 * ( 4 ) b 6 * ( 4 ) b 5 * ( 4 ) b 4 * ( 4 ) = b 1 ( 1 ) 0 0 b 1 ( 1 ) − b 1 ( 2 ) b 1 ( 2 ) b 1 ( 2 ) 0 0 0 − b 1 ( 3 ) b 1 ( 3 ) b 1 ( 3 ) 0 0 0 − b 1 ( 4 ) b 1 ( 4 ) b 1 ( 4 )

B 2 = b 8 * ( 1 ) b 7 * ( 1 ) b 6 * ( 1 ) b 5 * ( 1 ) b 9 * ( 2 ) b 8 * ( 2 ) b 7 * ( 2 ) b 6 * ( 2 ) b 10 * ( 3 ) b 9 * ( 3 ) b 8 * ( 3 ) b 7 * ( 3 ) b 11 * ( 4 ) b 10 * ( 4 ) b 9 * ( 4 ) b 8 * ( 4 ) = 0 0 0 b 1 ( 1 ) b 1 ( 1 ) 0 0 0 0 0 0 0 0 0 0 0 0 .

We have det B 0 − B 1 z − B 2 z 2 = ∏ ν = 1 T 1 − b 1 ( ν ) z 1 − ∏ ν = 1 T b 1 ( ν ) z ≠ 0 .

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Received: 2021-03-28

Revised: 2021-09-07

Accepted: 2021-12-27

Published Online: 2022-01-31

Estimating SPARMA Models with Dependent Error Terms

Abstract

Acknowledgments

Appendix A: Proofs

A.1 Reminder on Technical Issues on WLS Method for SPARMA Models

A.2 Proof of Theorem 3

Lemma 1

Proof

Lemma 2

Proof

Lemma 3

Proof

Lemma 4

Proof

Lemma 5

Proof

Lemma 6

Proof

Lemma 7

Proof

Proof

Appendix B: Verification of Assumption (A1) on Example

Example 1

References

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