A parametric stationarity test with smooth breaks

Ching-Chuan Tsong; Cheng-Feng Lee; Li Ju Tsai

doi:10.1515/snde-2015-0091

Article

A parametric stationarity test with smooth breaks

Ching-Chuan Tsong , Cheng-Feng Lee and Li Ju Tsai

Published/Copyright: September 26, 2018

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From the journal Studies in Nonlinear Dynamics & Econometrics Volume 23 Issue 2

Abstract

We propose a test to investigate the stationarity null against the unit-root alternative where a Fourier component is employed to approximate nonlinear deterministic trend of unknown form. A parametric adjustment is also adopted to accommodate possible stationary error. The asymptotic distribution of the test under the null is derived, and the asymptotic critical values are tabulated. We also show that it is a consistent test. Even with small sample sizes often encountered in empirical applications, our parametric stationarity test employing Fourier term has good size and power properties when trend breaks are gradual. The validity of the Fisher hypothesis for 15 OECD countries is investigated to illustrate the usefulness of our test.

Keywords: Fourier component; nonlinear trend; stationarity test

JEL Classification: C12; C22; E43

Acknowledgment

The authors are deeply indebted to the referees for their helpful suggestions and insightful comments on an earlier version of this paper. We alone, however, are responsible for the views and any remaining shortcomings herein.

Funding
Tsong acknowledges financial support from Taiwan Ministry of Science and Technology (Grant No. MOST 104-2410-H-260-001-).

Appendix

Mathematical Proofs

In this appendix, we only prove the case of m = 1, the proofs for m = 0 are similar and omitted. Note also that to conserve space, we only prove the case of p = 1. The counterpart of p > 1 can be obtained in the same way but with more tedious algebra. The following lemmas are required for the proof of Theorem 1.

Lemma 1

Let β^=(δ^0, δ^1, α^k, β^k)′ represent the LS estimator of β = (δ₀, δ₁, α_k, β_k)′ in Eqs. (1) and (2). Under the null hypothesis, we have

(18)DT(β^−β)⇒σv(∫01g(r)g(r)′dr)−1(∫01g(r)dW(r)),

where DT=diag(T, T3/2, T, T),σv2=(1−ϕ1)−2σε2 is the long run variance of v_t.

Proof:

The LS estimators for β can be written as follows:

DT(β^−β)=(DT−1(∑t=1Tztzt′)DT−1)−1DT−1∑t=1Tztvt,

where zt=(1, t, sin⁡(2kπtT), cos⁡(2kπtT))′.

Note that with some algebra, we have

DT−1∑tztzt′DT−1⇒(∫01g(r)g(r)′dr)−1,DT−1∑t=1Tztvt⇒σv∫01g(r)dW(r).

The lemma is done with continuous mapping theorem.

Lemma 2

Let ϕ^1 be the ML estimator for the AR(1) coefficient ϕ₁ when Δv^t is fitted with an ARMA(1,1) model. Under the null hypothesis, we have ϕ^1−ϕ1=op(1).

Proof:

Let ϕ~1 denote the ML estimator for ϕ₁ when Δv_t is fitted with an ARMA(1,1) model. According to Potscher (1991), ϕ~1−ϕ1=op(1) under the null of θ=1 and under the alternative θ < 1, with θ being the MA coefficient. Let L_T(Δv₁, …, Δv_T) and LT(Δv^1,...,Δv^T) denote the log-likelihood functions when Δv_t and Δv^t are fitted with an ARMA(1,1) model, respectively. With first Taylor expansion around Δv_t, we have

(19)T−1/2(LT(Δv^1, ..., Δv^T)−LT(Δv1, ..., ΔvT))=T−1/2∑t=1TLTt∗(Δv^t−Δvt)+op(1),

with LTt∗ being the derivative of the log-likelihood function evaluated at Δv^t. Note that v^t=vt−DT−1zt′DT(β^−β) as v^t being the LS residual in Eq. (6). Therefore, Δv^t−Δvt=−DT−1Δzt′DT(β^−β). Plug it into Eq. (19), and it is easy to show that T−1/2(LT(Δv^1, ..., Δv^T)−LT(Δv1, ..., ΔvT))=op(1), and then ϕ^1−ϕ~1=op(1) can be obtained. This is because DT(β^−β)=Op(1) as shown in Lemma 1 and both LTt∗ and Δz_t are I(0) process. With the results of ϕ~1−ϕ1=op(1) and ϕ^1−ϕ~1=op(1),Lemma 2 is done since ϕ^1−ϕ1=(ϕ^1−ϕ~1)+(ϕ~1−ϕ1).

Lemma 3

Let S^[Tr]=∑t=1[Tr]v^t, with v^t being the LS residual in Eq. (6), and [Tr] denoting the integer part of Tr, 0 ≤ r ≤ 1. Under the null, we have

(20)1TS^[Tr]⇒σvW¯(r),

where W¯(r) is defined in Theorem 1.

Proof:

Note that 1TS^[Tr]=1T∑t=1[Tr]v^t=1T∑t=1[Tr]vt−1T∑t=1[Tr](DT−1zt)′(DT(β^−β)). With Lemma 1 and T−1/2∑t=1[Tr]vt⇒σvW(r) we have:

1TS^[Tr]⇒σv[W(r)−(∫0rg(s)′ds)(∫01g(r)g(r)′dr)−1(∫01g(r)dW(r))]=σv[W(r)−∫0r(∫01h(r, s)dW(r))ds]=σvW¯(r).

Lemma 4

Let v^t denote the LS residual in Eq. (6), and σv2=(1−ϕ1)−2σε2. Under the null we have σv−2T−2∑t=1TS^t2⇒∫01W¯(r)2dr.

Proof:

σv−2T−2∑t=1TS^t2=σv−2T−1∑t=1T(S^tT)2⇒∫01W¯(r)2dr by Lemma 3.

Lemma 5

Let σ^ε2=T−1∑t=1Tε^t2, where ε^t=v^t−ϕ^1v^t−1, with v^t denoting the LS residual in Eq. (6), and ϕ^1 being the MLE for the AR(1) parameter ϕ₁ when v^t is fitted with an ARIMA(1,1,1) model. Under the null, we have σ^ε2=σε2+op(1).

Proof:

First note that ε^t can be written as ε^t=(v^t−ϕ1v^t−1)−(ϕ^1−ϕ1)v^t−1, and

(21)T−1∑t=1Tε^t2=T−1∑t=1T(v^t−ϕ1v^t−1)2+(ϕ^1−ϕ1)2T−1∑t=1Tv^t−12−2(ϕ^1−ϕ1)T−1∑t=1Tv^t−1(v^t−ϕ1v^t−1).

Moreover, we have v^t−ϕ1v^t−1=εt−(1−ϕ1)DT−1zt′DT(β^−β)−ϕ1DT−1Δzt′DT(β^−β) because v^t=vt−DT−1zt′DT(β^−β). We first consider the first term T−1∑t=1T(v^t−ϕ1v^t−1)2 of the right-hand side in Eq. (21), and it can be written as:

(22)T−1∑t=1T(v^t−ϕ1v^t−1)2=T−1∑t=1Tεt2+(1−ϕ1)2T−1∑t=1T(DT−1zt′DT(β^−β))2+ϕ12T−1∑t=1T(DT−1Δzt′DT(β^−β))2+cross - product terms.

Obviously, T−1∑t=1Tεt2=σε2+op(1). With the result of DT(β^−β)=Op(1) in Lemma 1 and Δz_t is an I(0) process, we have

T−1∑t=1T(DT−1zt′DT(β^−β))2=op(1),T−1∑t=1T(DT−1Δzt′DT(β^−β))2=op(1) and so are the cross product terms. Hence, T−1∑t=1T(v^t−ϕ1v^t−1)2=σε2+op(1). With the similar technique and ϕ^1−ϕ1=op(1) in Lemma 2, the rest of the right-hand side terms in Eq. (21) can be easily shown to converge to zero in probability. Therefore, σ^ε2=σε2+op(1).

Proof of Theorem 1:

Let St=∑i=1tε^i and S^t=∑i=1tv^i. With ε^i=v^i−ϕ^1v^i−1, we have St=S^t−ϕ^1S^t−1. Therefore, T−2∑t=1TSt2 can be expressed as:

(23)T−2∑t=1TSt2=T−2∑t=1T(S^t−ϕ^1S^t−1)2=T−2∑t=1TS^t2−2ϕ^1T−2∑t=1TS^tS^t−1+T−2ϕ^12∑t=1TS^t−12=(1−ϕ^1)2T−2∑t=1TS^t2+op(1).

Now, consider the FLMC^m test.

(24)FLMCm=σ^ε−2T−2∑t=1TSt2=σ^ε−2(1−ϕ^1)2T−2∑t=1TS^t2+op(1)=σ^ε−2(1−ϕ^1)2σv2(σv−2T−2∑t=1TS^t2)+op(1)⇒∫01W¯(r)2dr.

Eq. (24) holds with Lemma 2, Lemma 4, Lemma 5, and σv2=(1−ϕ1)−2σε2.

Proof of Theorem 2:

It is worth noting that with similar methods in the proofs of Lemma 1, Lemma 2, Lemma 3, Lemma 5, and under the alternative we have the following results because the I(1) process γ_t = γ_t₋₁ + η_t dominates.

T−1DT(β^−β)=Op(1).
ϕ^1−ϕ1=Op(1). (Since ϕ^1−ϕ~1=Op(1) with Eq. (19) being O_p(1) in this case.)
T−3/2S[Tr]=T−3/2∑t=1[Tr]ε^t=Op(1).
T−1σ^ε2=T−2∑t=1Tε^t2=Op(1).

Therefore,

(25)FLMCm=∑t=1TSt2T2σ^ε2=T(T−4∑t=1TSt2T−1σ^ε2)=Op(T).

Proof of Theorem 3:

First note that under the assumption of Theorem 3, we have

(a) ϕ^1−ϕ1=op(1), (b) T⁻¹S_[Tr] = O_p(1), and (c) σ^ε2=T−1∑t=1Tε^t2=Op(1). This is because

(26)T−1/2D~T(β~−β∗)=(D~T−1(∑t=1Tz~tz~′t)D~T−1)−1(T−1/2D~T−1∑t=1Tz~tξt),

where D~T=diag(T, T3/2),ξt=vt+αksin⁡(2kπtT)+βkcos⁡(2kπtT), and β~ is the LS estimator for β*=(δ₀, δ₁)′. Note that with the similar arguments in the proof of Lemma 1 we have

(27)T−1/2D~T(β~−β∗)=Op(1).

Hence, with (27) and based on the same arguments in the proofs of Lemma 2, and Lemma 5, the results of (a), (b) and (c) hold. Therefore, we have

LMCm=∑t=1TSt2T2σ^ε2=T(T−3∑t=1TSt2σ^ε2)=Op(T).

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Keywords for this article

Fourier component; nonlinear trend; stationarity test