Determination of the Number of Breaks in Heterogeneous Panel Data Models

Lu Wang; Shuke Hu

doi:10.1515/snde-2024-0018

Article

Determination of the Number of Breaks in Heterogeneous Panel Data Models

Lu Wang and Shuke Hu

Published/Copyright: August 8, 2024

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From the journal Studies in Nonlinear Dynamics & Econometrics

Abstract

This paper considers a heterogeneous panel data model with an unknown number of breaks. We propose a so-called two-stage procedure to determine the number of breaks and detect the location of break points. The consistency of the estimated number of breaks and the estimated break points are established under fairly general conditions. Monte Carlo simulations and two empirical applications are provided to demonstrate the finite sample performance of the proposed method.

Keywords: heterogeneous panel data models; structural changes; the number of breaks; the location of break points

JEL Classification: C23

Corresponding author: Lu Wang, School of Mathematics and Statistics, Huanghuai University, Zhumadian 463000, China, E-mail: Lwang008@163.com

Funding source: Henan Provincial Science and Technology Research Project

Award Identifier / Grant number: 232102310200

Award Identifier / Grant number: 242102110145

Research funding: This research is partially supported in part by the Science and Technology Research Project of Henan Province (Grant No. 232102310200, 242102110145).

Appendix: Technical Details

Let S 1 = {S _j: S _j can not contain any break }, S 2 = {S _j:S _j contains a break point}. The break point may locate at the interior or boundary of some S _j. We denote S 2 a = S j : S j contains a break point k _l such that lim ω j → ∞ ω j , 1 ω j ∈ ( 0,1 ) , S 2 b = S j : S j contains a break point k _l such that lim ω j → ∞ ω j , 1 ω j = 0 , S 2 c = S j : S j contains a break point k _l such that lim ω j → ∞ ω j , 1 ω j = 1 , where ω _j,1 = k _l − v _j and ω _j,2 = v _j+1 − k _l. Define S = { 0,1 , … , J } , S 2 = S 2 a ∪ S 2 b ∪ S 2 c , η N ω j = min ( N , ω j ) . Let c j = lim T → ∞ ω j , 1 ω j when j ∈ S 2 . Let Θ j * = b l , β j * = β l and a j = η N ω j − 1 when j ∈ S 1 and S j ⊂ k l , k l + 1 . Let Θ j * = b l − 1 , β j * = β l − 1 and a j = ω j , 2 ω j + η N ω j − 1 when j ∈ S 2 a and k _l ∈ S _j. Let Θ j * = b l , β j * = β l and a j = ω j , 1 ω j + η N ω j − 1 when j ∈ S 2 b and k _l ∈ S _j. Let Θ j * = c j b l − 1 + ( 1 − c j ) b l , β j * = c j β l − 1 + ( 1 − c j ) β l and a j = ω j , 1 ω j − c j + η N ω j − 1 when j ∈ S 2 c and k _l ∈ S _j.

Lemma 1.

Under Assumptions 1 – 6, 1 N ∑ i = 1 N β ̇ i , j − β j * = O p ( a j ) .

Proof of Lemma 1. For case (i) j ∈ S 1 , we focus on the subinterval S j ⊂ k l , k l + 1 . Noting that Y i , S j = X i , S j β i ( l ) + ε i , S j and β ̇ i , j = X i , S j ′ X i , S j − 1 X i , S j ′ Y i , S j . Then we have

(10) 1 N ∑ i = 1 N β ̇ i , j = β j * + 1 N ∑ i = 1 N ξ i , l + 1 N ∑ i = 1 N X i , S j ′ X i , S j − 1 X i , S j ′ ε i , S j .

Since E ( ε i t ) = 0 , ɛ _it are cross-sectionally independent and independent with x _it, then we have

(11) E 1 N ∑ i = 1 N X i , S j ′ X i , S j − 1 X i , S j ′ ε i , S j = 0 ,

(12) Var 1 N ω j ∑ i = 1 N X i , S j ′ X i , S j ω j − 1 1 ω j X i , S j ′ ε i , S j = 1 N ω j ∑ i = 1 N X i , S j ′ X i , S j ω j − 1 1 ω j X i , S j ′ Var ( ε i , S j ) X i , S j X i , S j ′ X i , S j ω j − 1 = O p 1 ω j .

Thus, the third term on the right hand of Equation (10) is O p ( 1 N ω j ) .

Note that ξ _i,l ∼ IID(0, Ω_ξ,l), then we have 1 N ∑ i = 1 N ξ i , l = O p ( 1 N ) .

Thus,

(13) 1 N ∑ i = 1 N β ̇ i , j − β j * = O p 1 N .

For case (ii) j ∈ S _2a, without loss of generality, we assume k _l ∈ S _j. The model in S _j can be written as

Y i , S j = X i , S j , 1 β i ( l − 1 ) X i , S j , 2 β i ( l ) + ε i , S j ,

where S _j,1 = (v _j, k _l], S _j,2 = (k _l, v _j+1], X i , S j , 1 = ( x i v j + 1 , … , x i k l ) ′ , X i , S j , 2 = ( x i k l + 1 , … , x i v j + 1 ) ′ .

(14) 1 N ∑ i = 1 N β ̇ i , j = 1 N ∑ i = 1 N β i ( l − 1 ) + 1 N ∑ i = 1 N X i , S j ′ X i , S j − 1 X i , S j , 2 ′ X i , S j , 2 δ i , l + 1 N ∑ i = 1 N X i , S j ′ X i , S j − 1 X i , S j ′ ε i , S j = 1 N ∑ i = 1 N β i ( l − 1 ) + ω j , 2 ω j 1 N ∑ i = 1 N X i , S j ′ X i , S j ω j − 1 X i , S j , 2 ′ X i , S j , 2 ω j , 2 δ i , l + O p 1 N ω j = β j * + 1 N ∑ i = 1 N ξ i , l − 1 + O p ω j , 2 ω j + O p 1 N ω j = β j * + O p ω j , 2 ω j + O p 1 N .

For case (iii) j ∈ S _2b, the similar proof technical of case (ii) can be used to prove that

(15) 1 N ∑ i = 1 N β ̇ i , j = β j * + O p ω j , 1 ω j + O p 1 N .

For case (iv) j ∈ S 2 c , we have

(16) 1 N ∑ i = 1 N β ̇ i , j = 1 N ∑ i = 1 N β i ( l − 1 ) + 1 N ∑ i = 1 N X i , S j ′ X i , S j − 1 X i , S j , 2 ′ X i , S j , 2 δ i , l + 1 N ∑ i = 1 N X i , S j ′ X i , S j − 1 X i , S j ′ ε i , S j ≕ A 1 + A 2 + A 3 .

For A ₂,

A 2 = ω j , 2 ω j 1 N ∑ i = 1 N Σ i + 1 ω j X i , S j ′ X i , S j − Σ i − 1 Σ i + 1 ω j , 2 X i , S j , 2 ′ X i , S j , 2 − Σ i δ i , l .

By the Equation 5.8.1 in Horn and Johnson (1985) and Assumption 5, we have

(17) A 2 = 1 N ∑ i = 1 N ( 1 − c j ) δ i , l + O p ω j , 1 ω j − c j + O p 1 ω j .

Using the proof technical of the second and third term in Equation (10), we have ‖ A 1 ‖ = O p ( 1 N ) and ‖ A 3 ‖ = O p ( 1 N ω j ) . Thus, 1 N ∑ i = 1 N β ̇ i , j = c j β l − 1 + ( 1 − c j ) β l + O p ( ω j , 1 ω j − c j ) + η N , ω j − 1 . □

Lemma 2.

Under Assumptions 1 – 6, 1 N Θ j * − Θ ̂ j = O p ( a j ) , j = 0, 1, …, J.

Proof of Lemma 2. Define Θ j = Θ j * + a j η j , j = 0, 1, …, J. Next we will prove that Θ ̂ j falls in a sphere Θ j * + a j η j : 1 N ‖ η j ‖ ≤ ς j M , where ς _j is constant.

L Θ 0 , … , Θ J − L Θ 0 * , … , Θ J * = 1 2 N ∑ j = 0 J 1 ω j Y S j − X S j Θ j 2 + γ ∑ j = 1 J ω j Θ j − Θ j − 1 − 1 2 N ∑ j = 0 J 1 ω j Y S j − X S j Θ j * 2 − γ ∑ j = 1 J ω j Θ j * − Θ j − 1 * = 1 2 N ∑ j = 0 J 1 ω j Y S j − X S j Θ j * + X S j Θ j * − X S j Θ j 2 − 1 2 N ∑ j = 0 J 1 ω j Y S j − X S j Θ j * 2 + γ ∑ j = 1 J ω j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * = 1 N ∑ j = 0 J 1 ω j X S j Θ j * − Θ j 2 + 1 N ∑ j = 0 J 1 ω j Y S j − X S j Θ j * ′ X S j Θ j * − Θ j + γ ∑ j = 1 J ω j Θ j − Θ j − 1 − Θ j * − Θ j − 1 *

= 1 2 N ∑ j ∈ S 1 1 ω j X S j Θ j * − Θ j 2 + 2 Y S j − X S j Θ j * ′ X S j Θ j * − Θ j + 1 2 N ∑ j ∈ S 2 a 1 ω j X S j Θ j * − Θ j 2 + 2 Y S j − X S j Θ j * ′ X S j Θ j * − Θ j + 1 2 N ∑ j ∈ S 2 b 1 ω j X S j Θ j * − Θ j 2 + 2 Y S j − X S j Θ j * ′ X S j Θ j * − Θ j + 1 2 N ∑ j ∈ S 2 c 1 ω j X S j Θ j * − Θ j 2 + 2 Y S j − X S j Θ j * ′ X S j Θ j * − Θ j + γ ∑ j = 0 J ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * ≕ B 1 + B 2 + B 3 + B 4 + B 5 .

For B ₁,

B 1 = 1 N ∑ j ∈ S 1 1 ω j X S j Θ j * − Θ j 2 + 1 N ∑ j ∈ S 1 1 ω j Y S j − X S j Θ j * ′ X S j Θ j * − Θ j ≔ B 1,1 + B 1,2 .

For B _1,1,

B 1,1 = 1 2 ∑ j ∈ S 1 1 N Θ j * − Θ j ′ X S j ′ X S j ω j 1 N Θ j * − Θ j ≥ 1 2 min j ∈ S 1 μ min X S j ′ X S j ω j ∑ j ∈ S 1 1 N Θ j * − Θ j 2 .

For B _1,2,

B 1,2 = 1 N ∑ j ∈ S 1 1 ω j Y S j − X S j Θ j * ′ X S j Θ j * − Θ j = 1 2 N ∑ j ∈ S 1 1 ω j ε S j ′ X S j Θ j * − Θ j ≤ 1 2 N ∑ j ∈ S 1 1 ω j ε S j ′ X S j Θ j * − Θ j ≤ 1 2 ω j ∑ j ∈ S 1 1 N ∑ i = 1 N 1 ω j ∑ t = v j + 1 v j + 1 ε i t x i t 2 1 / 2 1 N Θ j * − Θ j = ∑ j ∈ S 1 O p 1 ω j 1 N Θ j * − Θ j .

Thus, B 1 = ∑ j ∈ S 1 1 2 μ * a j 2 1 N η j 2 + O p ( 1 ω j ) a j 1 N ‖ η j ‖ .

For B ₂,

B 2 = 1 N ∑ j ∈ S 2 a 1 ω j X S j Θ j * − Θ j 2 + 1 2 N ∑ j ∈ S 2 a ω j , 2 ω j D 1 * Θ j * − Θ j + 1 2 N ∑ j ∈ S 2 a 1 ω j ε S j ′ X S j Θ j * − Θ j ≔ B 2,1 + B 2,2 + B 2,3 ,

where

D 1 * = X 1 , S j , 2 ′ X 1 , S j , 2 ω j , 2 δ 1 , l , … , X N , S j , 2 ′ X N , S j , 2 ω j , 2 δ N , l ′ .

For B _2,1,

B 2,1 = 1 N ∑ j ∈ S 2 a 1 ω j X S j Θ j * − Θ j 2 ≥ 1 2 min j ∈ S 2 a μ min X S j ′ X S j ω j ∑ j ∈ S 2 a 1 N Θ j * − Θ j 2 .

Since the largest eigenvalue of X i , S j , 2 ′ X i , S j , 2 ω j , 2 and 1 N ∑ i = 1 N δ i , l 2 are upper bounded, then

1 N D 1 * 2 = 1 N ∑ i = 1 N X i , S j , 2 ′ X i , S j , 2 ω j , 2 δ i , l 2 = O p ( 1 ) .

Therefore,

B 2,2 = ∑ j ∈ S 2 a O p ω j , 2 ω j 1 N Θ j * − Θ j .

Using the proof technical of B _1,2, we have

B 2,3 = ∑ j ∈ S 2 a O p 1 ω j 1 N Θ j * − Θ j .

Thus, B 2 = ∑ j ∈ S 2 a 1 2 μ * a j 2 1 N η j 2 + O p ( 1 ω j + ω j , 2 ω j ) a j 1 N η j .

For B ₃,

B 3 = 1 N ∑ j ∈ S 2 b 1 ω j X S j Θ j * − Θ j 2 + 1 2 N ∑ j ∈ S 2 b ω j , 1 ω j D 2 * Θ j * − Θ j + 1 2 N ∑ j ∈ S 2 b 1 ω j ε S j ′ X S j Θ j * − Θ j ≕ B 3,1 + B 3,2 + B 3,3 ,

where D 2 * = X 1 , S j , 1 ′ X 1 , S j , 1 ω j , 1 δ 1 , l , … , X N , S j , 1 ′ X N , S j , 1 ω j , 1 δ N , l .

For B _3,1,

B 3,1 = 1 N ∑ j ∈ S 2 b 1 ω j X S j Θ j * − Θ j 2 ≥ 1 2 min j ∈ S 2 b μ min X S j ′ X S j ω j ∑ j ∈ S 2 b 1 N Θ j * − Θ j 2 .

Since the largest eigenvalue of X i , S j , 1 ′ X i , S j , 1 ω j , 1 and 1 N ∑ i = 1 N ‖ δ i , l ‖ 2 are upper bounded, then

‖ 1 N D 2 * ‖ 2 = 1 N ∑ i = 1 N X i , S j , 1 ′ X i , S j , 1 ω j , 1 2 δ i , l 2 = O p ( 1 ) .

Then, we have B 3,2 = ∑ j ∈ S 2 b O p ( ω j , 1 ω j ) 1 N Θ j * − Θ j .

The proof technical of B _1,2 can be used to prove that

B 3,3 = O p 1 ω j 1 N ∑ j ∈ S 1 Θ j * − Θ j 2 1 / 2 .

Thus, we have

B 3 = ∑ j ∈ S 2 b 1 2 μ * a j 2 1 N η j 2 + O p 1 ω j + ω j , 1 ω j a j 1 N η j ,

where μ * = min 0 ≤ j ≤ J ( μ min X S j ′ X S j ω j ) .

For B ₄,

B 4 = 1 N ∑ j ∈ S 2 c 1 ω j X S j Θ j * − Θ j 2 + 1 2 N ∑ j ∈ S 2 c 1 ω j X S j , 1 b l − 1 X S j , 2 b l − X S j Θ j * ′ X S j Θ j * − Θ j + 1 2 N ∑ j ∈ S 2 c 1 ω j ε S j ′ X S j Θ j * − Θ j ≕ B 4,1 + B 4,2 + B 4,3 .

For B _4,1,

B 4,1 = 1 N ∑ j ∈ S 2 c 1 ω j X S j Θ j * − Θ j 2 ≥ 1 2 min j ∈ S 2 c μ min X S j ′ X S j ω j ∑ j ∈ S 2 c 1 N Θ j * − Θ j 2 .

For B _4,2,

B 4,2 = 1 2 N ∑ j ∈ S 2 c 1 ω j D 3 * Θ j * − Θ j

where D 3 * = ( X 1 , S j , 1 ′ X 1 , S j , 1 β 1 ( l − 1 ) + X 1 , S j , 2 ′ X 1 , S j , 2 β 1 ( l ) , … , X N , S j , 1 ′ X N , S j , 1 β N ( l − 1 ) + X N , S j , 2 ′ X N , S j , 2 β N ( l ) ) ′ . Note that

D 3 * ω j N 2 = 1 ω j 2 N ∑ i = 1 N X i , S j , 1 ′ X i , S j , 1 β i ( l − 1 ) − c j X i , S j ′ X i , S j β i ( l − 1 ) 2 + 1 ω j 2 N ∑ i = 1 N X i , S j , 2 ′ X i , S j , 2 β i ( l ) − ( 1 − c j ) X i , S j ′ X i , S j β i ( l ) 2 .

For the first term of above equation,

1 ω j 2 N ∑ i = 1 N X i , S j , 1 ′ X i , S j , 1 β i ( l − 1 ) − c j X i , S j ′ X i , S j β i ( l − 1 ) 2 ≤ 1 N ∑ i = 1 N ω j , 1 ω j X S i , j , 1 ′ X S i , j , 1 ω j , 1 − Σ i + ω j , 1 ω j − c j Σ i + Σ i − X i , S j ′ X i , S j ω j c j 2 ⋅ ‖ β i ( l − 1 ) ‖ 2 ≤ 3 N ∑ i = 1 N ω j , 1 ω j X i , S j , 1 ′ X i , S j , 1 ω j , 1 − Σ i 2 ⋅ β i ( l − 1 ) 2 + ( ω j , 1 ω j − c j ) 2 3 N ∑ i = 1 N ‖ Σ i ‖ 2 ⋅ β i ( l − 1 ) 2 + 3 N ∑ i = 1 N Σ i − X i , S j ′ X i , S j ω j c j 2 ⋅ β i ( l − 1 ) 2 = O p 1 ω j + O p ω j , 1 ω j − c j 2 .

Using the proof technical of the first term, the second term is O p ( 1 ω j ) + O p ( ω j , 1 ω j − c j ) 2 .

Thus, B 4,2 = ∑ j ∈ S 2 c O p 1 ω j + ω j , 1 ω j − c j 1 N Θ j * − Θ j .

The proof technical of B _1,2 is used to prove that

B 4,3 = ∑ j ∈ S 1 O p 1 ω j 1 N ∥ Θ j * − Θ j ∥ .

Therefore, B 4 = ∑ j ∈ S 2 c 1 2 μ * a j 2 1 N η j 2 + O p ( 1 ω j + ω j , 1 ω j − c j ) a j 1 N η j .

B 5 = γ ∑ j ∈ S 1 , j − 1 ∈ S 1 ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * + γ ∑ j ∈ S 1 , j − 1 ∈ S 2 b ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * + γ ∑ j ∈ S 2 a , j − 1 ∈ S 1 ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * + γ ∑ j ∈ S 2 b ∪ S 2 c , j − 1 ∈ S 1 ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * + γ ∑ j ∈ S 1 , j − 1 ∈ S 2 a ∪ S 2 c ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * ≕ B 5,1 + B 5,2 + B 5,3 + B 5,4 + B 5,5 .

Since Θ j * = Θ j − 1 * in equations B _5,1, B _5,2 and B _5,3, then B _5,1 ≥ 0, B _5,2 ≥ 0 and B _5,3 ≥ 0.

For B _5,4,

B 5,4 = γ ∑ j ∈ S 2 b ∪ S 2 c , j − 1 ∈ S 1 ρ j Θ j − Θ j − 1 − Θ j * − Θ j − 1 * ≤ γ ∑ j ∈ S 2 b ∪ S 2 c , j − 1 ∈ S 1 ρ j Θ j − Θ j * + Θ j − 1 − Θ j − 1 * ≤ 2 γ max j ∈ S 2 b ∪ S 2 c , j − 1 ∈ S 1 ρ j ∑ j = 0 J Θ j * − Θ j .

Note that ρ j = ∑ i = 1 N ( β ̇ i , j − β ̇ i , j − 1 ) N − κ , j ∈ S 2 b ∪ S 2 c and j − 1 ∈ S 1 . By Lemma 1, we have max j ∈ S 2 b ∪ S 2 c , j − 1 ∈ S 1 ρ j = O p ( 1 ) . Thus, B 5,4 = O p ( γ ) ∑ j = 0 J Θ j * − Θ j .

In proceed,

B 1 + B 2 + B 3 + B 4 + B 5,4 + B 5,5 = 1 2 ∑ j ∈ S 1 μ * a j 2 1 N η j 2 + O p 1 ω j + γ N a j 1 N ‖ η j ‖ + 1 2 ∑ j ∈ S 2 a μ * a j 2 1 N η j 2 + O p 1 ω j + ω j , 2 ω j + γ N a j 1 N ‖ η j ‖ + 1 2 ∑ j ∈ S 2 b μ * a j 2 1 N η j 2 + O p 1 ω j + ω j , 1 ω j + γ N a j 1 N ‖ η j ‖ + 1 2 ∑ j ∈ S 2 c μ * a j 2 1 N η j 2 + O p 1 ω j + ω j , 1 ω j − c j + γ N a j 1 N ‖ η j ‖ = 1 2 ∑ j = 0 J μ * N η j 2 + O p 1 + γ N a j − 1 1 N ‖ η j ‖ a j 2 .

Since B _5,1, B _5,2, B _5,3 ≥ 0 and L ( Θ ̂ 0 , … , Θ ̂ J ) − L Θ 0 * , … , Θ J * ≤ 0 , then Θ ̂ 0 , … , Θ ̂ J satisfy the following inequality with probability approaching to one

(18) B 1 + B 2 + B 3 + B 4 + B 5,4 + B 5,5 ≤ 0 .

Note that 1 a j ≤ ω j and γ N ω j = O ( 1 ) , then γ N a j − 1 = O p ( 1 ) . Thus, there exist a constant ς _j such that 1 N ‖ η j ‖ ≤ ς j by the properties of quadratic functions; Otherwise, it is in contradiction with Equation 18. □

Lemma 3.

Under Assumptions 1 – 6,

for j, j − 1 ∈ S ₁, lim N , T → ∞ P ( Θ ̂ j = Θ ̂ j − 1 ) = 1 ;
for j ∈ S _2c, lim N , T → ∞ P ( Θ ̂ j + 1 ≠ Θ ̂ j , Θ ̂ j ≠ Θ ̂ j − 1 ) = 1 ;
for j ∈ S _2a, lim N , T → ∞ P ( Θ ̂ j + 1 ≠ Θ ̂ j ) = 1 ;
for j ∈ S _2b, lim N , T → ∞ P ( Θ ̂ j ≠ Θ ̂ j − 1 ) = 1 .

Proof of Lemma 3. (i) The model with at least one change point. If Δ ̂ j ≠ 0 , let ϑ j = Δ ̂ j ‖ Δ ̂ j ‖ . Otherwise, ϑ _j is a constant that satisfy ‖ϑ _j‖ ≤ 1. Next we will consider three cases: (a) 2 ≤ j ≤ J − 1, (b) j = J and (c) j = 1.

For case (a), we consider three subcases a ( 1 ) j + 1 ∈ S 2 b ∪ S 2 c , a ( 2 ) j + 1 ∈ S 2 a and a ( 3 ) j + 1 ∈ S 1 .

For a(1), we can apply the subdifferential calculus to obtain the Karush-Kuhn-Tucker (KKT) condition of minimizing the objective function (4) with respect to Θ_j

(19) ∂ L ( Θ 0 , … , Θ J ) ∂ Θ j = 1 2 N ω j X S j ′ ( Y S j − X S j Θ ̂ j ) + γ ρ j Δ ̂ j ‖ Δ ̂ j ‖ + γ ρ j + 1 Δ ̂ j + 1 ‖ Δ ̂ j + 1 ‖ = 0 .

That is,

(20) 0 = a j N X S j * ′ Y S j * − X S j * Θ ̂ j + a j N ω j γ ρ j Δ ̂ j ‖ Δ ̂ j ‖ + a j N ω j γ ρ j + 1 Δ ̂ j + 1 ‖ Δ ̂ j + 1 ‖ = a j N X S j ′ Y S j − X S j Θ j * + X S j Θ j * − X S j Θ ̂ j + a j N ω j γ ρ j Δ ̂ j ‖ Δ ̂ j ‖ + a j N ω j γ ρ j + 1 Δ ̂ j + 1 ‖ Δ ̂ j + 1 ‖ = a j X S j ′ N Y S j − X S j Θ j * + a j X S j ′ X S j N Θ j * − Θ ̂ j + a j N ω j γ ρ j Δ ̂ j ‖ Δ ̂ j ‖ + a j N ω j γ ρ j + 1 Δ ̂ j + 1 ‖ Δ ̂ j + 1 ‖ = a j N X S j ′ ε S j + a j N X S j ′ X S j Θ j * − Θ ̂ j + a j N ω j γ ρ j Δ ̂ j ‖ Δ ̂ j ‖ + a j N ω j γ ρ j + 1 Δ ̂ j + 1 ‖ Δ ̂ j + 1 ‖ = D j , 1 + D j , 2 + D j , 3 + D j , 4 .

Note that ‖ D j , 1 ‖ = 1 N ∑ i = 1 N ‖ a j ∑ t = v j + 1 v j + 1 x i t ε i t ‖ 2 1 / 2 and a j = η N ω j − 1 = 1 ω j , then we have ‖D _j,1‖ = O _p(1). For 1 ≤ i ≤ N, the largest eigenvalue of X i , S j ′ X i , S j ω j are uniformly upper bounded, then ‖D _j,2‖ = O _p(1) by Lemma 2. Since j + 1 ∈ S 2 b ∪ S 2 c , j ∈ S 1 , then ρ _j+1 is upper bound by Lemma 1. Noting that ‖D _j,4‖ = O _p(1) under Assumption 4. If Δ ̂ j ≠ 0 , then ‖D _j,3‖ diverges; This conclusion is in contradiction with (20). Thus, Δ ̂ j lies in a position where ‖Θ_j − Θ_j−1‖ is not differentiable with respect to Θ_j. That is, Δ ̂ j = 0 .

For a(2), if Δ ̂ j ≠ 0 , D _j,3 and D _j,4 have the same divergence rate. Considering the Karush-Kuhn-Tucker (KKT) condition of minimizing the objective function (4) with respect to Θ_j+1,

(21) 0 = a j + 1 N X S j + 1 ′ e S j + 1 + a j + 1 N X S j + 1 ′ X S j + 1 Θ j * − Θ ̂ j + a j + 1 N ω j + 1 γ ρ j + 1 ϑ j + 1 + a j + 1 N ω j + 1 γ ρ j + 2 ϑ j + 2 ≕ D j + 1,1 + D j + 1,2 + D j + 1,3 + D j + 1,4 .

Using the proof technical of a(1), we have D _j+1,1 = O _p(1) and D _j+1,2 = O _p(1). Note that D j + 1,3 = a j + 1 a j D j , 4 , a _j+1 ≥ a _j and D _j+1,3 diverges. But D _j+1,4 = O _p(1) is in contradiction with Equation (21). Thus, we reach a conclusion that Δ ̂ j lies in a position where ‖Θ_j − Θ_j−1‖ is not differentiable with respect to Θ_j. That is, Δ ̂ j = 0 .

For a(3), if Δ ̂ j ≠ 0 , D _j,3 and D _j,4 have the same divergence speed by the Karush-Kuhn-Tucker(KKT) condition of minimizing the objective function (4) with respect to Θ_j. Note that D j + 1,3 = a j + 1 a j D j , 4 and a _j+1 = a _j, then D _j+1,3 and D _j,4 diverge at the same speed. If j + 2 ∈ S 1 , then D j + 2,3 = a j + 2 a j D j , 4 , that is, D _j+2,3 and D _j,4 diverge at the same speed. And so on, until j + l ∈ S 1 and j + l + 1 ∈ S 2 , then we have D j + l , 3 = a j + l a j D j , 4 . As the discussion of subcases a(1) and a(2), a contradiction would arise unless there is no point after v _j+1.

In proceed, we consider the Karush-Kuhn-Tucker (KKT) condition of minimizing the objective function (4) with respect to Θ_j−1

(22) 0 = a j − 1 N X S j − 1 ′ e S j − 1 + a j − 1 N X S j − 1 ′ X S j − 1 Θ j − 1 * − Θ ̂ j − 1 + a j − 1 N ω j − 1 γ ρ j − 1 ϑ j − 1 + a j − 1 N ω j − 1 γ ρ j ϑ j ≕ D j − 1,1 + D j − 1,2 + D j − 1,3 + D j − 1,4

If Δ ̂ j ≠ 0 , then D _j−1,3 and D _j−1,4 have the same divergence speed. If j − 2 ∈ S 1 , then D j − 1,3 = a j − 1 a j − 2 D j − 2,4 . Thus, D _j−2,4 and D _j−1,3 have the same divergence speed. And so on, until j − l ∈ S 1 and j − l − 1 ∈ S 2 , we have D j − 1,3 = a j − 1 a j − l D j − l , 4 . As the discussion of cases a(1) and a(2), a contradiction would arise unless there is no point after v _j. Note that there are at least one change point, then Δ ̂ j lies in a position where ‖Θ_j − Θ_j−1‖ is not differentiable with Θ_j. Thus, Δ ̂ j = 0 .

Next we consider case (b). Noting that only one term contains Θ_J in Equation (4), the Karush-Kuhn-Tucker (KKT) condition of minimizing the objective function (4) with respect to Θ_J is given by

(23) 0 = a J N X S J ′ e S J + a J N X S J ′ X S J Θ J * − Θ ̂ J + a J N ω J γ ρ J ϑ J .

Using the same proof technical of D _j,1, D _j,2 in Equation (20), the first two terms in the right hand of Equation (23) are O _p(1). If Δ ̂ J ≠ 0 , the third term in the right hand of Equation (23) diverges. This conclusion is in contradiction with (23). Thus, Δ ̂ J lies in a position where ‖Θ_J − Θ_J−1‖ is not differentiable with Θ_J, that is, Δ ̂ J = 0 .

For case (c), considering the Karush-Kuhn-Tucker (KKT) condition of minimizing the objective function (4) with respect to Θ₀; Using the proof technical of case (b), we have Δ ̂ 1 = 0 .

For the case of no change point. If there exist j such that Θ ̂ j − Θ ̂ j − 1 ≠ 0 , it is in contradiction with case (b) or (c).

(ii) For j ∈ S _2c,

(24) 1 N ‖ Θ ̂ j − Θ ̂ j − 1 ‖ = 1 N ‖ Θ ̂ j − Θ j * + Θ j * − Θ j − 1 * + Θ j − 1 * − Θ ̂ j − 1 ‖ = 1 N ‖ Θ j * − Θ j − 1 * ‖ + O p ( a j ) + O p ( a j − 1 ) .

Noting that 1 N ‖ Θ j * − Θ j − 1 * ‖ 2 = ( 1 − c j ) 2 1 N ∑ i = 1 N δ i , l ′ δ i , l , then Θ ̂ j ≠ Θ ̂ j − 1 . By analogy, we have Θ ̂ j + 1 ≠ Θ ̂ j .

For (iii) and (iv), the proof technical of (ii) can be adopted. □

Proof of Theorem 1. For Δ ̃ j ≠ 0 , Δ ̂ j + 1 ≠ 0 , if S _j can not contain any break, then at most one of S _j−1 and S _j contain a change point. By Lemma 3(i), at most one of Δ ̂ j and Δ ̂ j + 1 is equal to zero. This is in contradiction with Δ ̂ j ≠ 0 , Δ ̂ j + 1 ≠ 0 . Thus, S _j contains a change point.

For Δ ̂ j ≠ 0 , Δ ̂ j + 1 = 0 , Δ ̂ j − 1 = 0 , since Δ ̂ j ≠ 0 , then one of S _j and S _j−1 contains a change point by Lemma 3(i). By Lemma 3(ii), this change point cannot be located in the interior or right end of S _j. Otherwise, a contradiction would arise. Therefore, this change point is located in S _j−1 or the left end of S _j.

For Δ ̂ j = 0 , Δ ̂ j + 1 ≠ 0 , Δ ̂ j + 2 = 0 , the above proof technical can be adopted.

For Δ ̂ j = 0 , Δ ̂ j + 1 = 0 , if j ∈ S 2 , this is in contradiction with Lemma 3(ii)–(iv). Thus, S _j can not contain any break. □

Proof of Theorem 2. The proof technical of Theorem 1 in Baltagi, Feng, and Kao (2016) can be used to prove this Theorem. □

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Supplementary Material

This article contains supplementary material (https://doi.org/10.1515/snde-2024-0018).

Received: 2024-03-12

Accepted: 2024-07-03

Published Online: 2024-08-08

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Supplementary Material Details

https://doi.org/10.1515/snde-2024-0018

Keywords for this article

heterogeneous panel data models; structural changes; the number of breaks; the location of break points