In-Fill Asymptotic Distribution of the Change Point Estimator when Estimating Breaks One at a Time

Toshikazu Tayanagi; Eiji Kurozumi

doi:10.1515/jtse-2022-0013

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In-Fill Asymptotic Distribution of the Change Point Estimator when Estimating Breaks One at a Time

Toshikazu Tayanagi und Eiji Kurozumi

Veröffentlicht/Copyright: 21. März 2023

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Abstract

In this study, we investigate the least squares (LS) estimator of a structural change point by the in-fill asymptotic theory, which has been recently used by Jiang, Wang, and Yu (2018. “New Distribution Theory for the Estimation of Structural Break Point in Mean.” Journal of Econometrics 205 (1): 156–76; 2020. “In-Fill Asymptotic Theory for Structural Break Point in Autoregressions.” Econometric Reviews 40 (4): 359–86), when the model with two structural changes is estimated as the model with only a one-time structural change. We, hence, show that the finite sample distribution of the estimator of the first break has four peaks, which is different from the classical long-span asymptotic distribution, which contains only one peak. Conversely, the in-fill asymptotic distribution of the estimator has four peaks and can approximate the finite sample distribution very well. We also demonstrate that the estimator is consistent in the in-fill asymptotic framework with a relatively large magnitude of the break. In the latter case, the finite sample distribution of the estimator has only one peak and is well approximated by both the in-fill and long-span asymptotic theory.

Keywords: structural break; in-fill asymptotics; time series

JEL Classification: C13; C22

Corresponding author: Toshikazu Tayanagi, Department of Economics, Hitotsubashi University, 2-1 Naka, Kunitachi, Tokyo 186-8601, Japan, E-mail: ed215004@g.hit-u.ac.jp

Funding source: Japan Society for the Promotion of Science

Award Identifier / Grant number: 19K01585 and 22K01422

Acknowledgement

We thank the anonymous referee for constructive comments and the participants at economics and statistics seminar held at Hitotsubashi university for helpful discussions and comments. All errors are our responsibility.

Research funding: Eiji Kurozumi’s research was supported by JSPS KAKENHI Grant Number 19K01585 and 22K01422.

Appendix

Proof of Theorem 1

Following Bai (1994), (2.2) is expressed as

(A.1) k ̂ = arg min k { S T ( k ) } = arg max k T V T ( k ) 2 ,

where

(A.2) V T ( k ) = k ( T − k ) T 2 Y ̄ k − Y ̄ k * , Y ̄ k = 1 k ∑ t = 1 k y t , Y ̄ k * = 1 T − k ∑ t = k + 1 T y t .

Under Assumptions 1 and 2, we have, by the functional central limit theorem (FCLT),

T k ∑ t = 1 k X t = T k 1 T ∑ t = 1 k X t ⇒ 1 τ a ( 1 ) σ B ( τ ) , T T − k ∑ t = k + 1 T X t = T T − k 1 T ∑ t = k + 1 T X t ⇒ 1 1 − τ a ( 1 ) σ ( B ( 1 ) − B ( τ ) ) ,

for k = [τT] with a given τ ∈ (0, 1), where B(⋅) is a standard Brownian motion on [0, 1]. We show that the limiting distribution of (A.2) depends on the regime wherein k is located.

For k = [ τ T ] ≤ k 1 0 , we have k ( T − k ) / T 2 → τ ( 1 − τ ) , while, by the FCLT,

T Y ̄ k − Y ̄ k * = T 1 k ∑ t = 1 k y t − 1 T − k ∑ t = k + 1 T y t = T 1 k ∑ t = 1 k y t − 1 T − k ∑ t = k + 1 k 1 0 y t + ∑ t = k 1 0 + 1 k 2 0 y t + ∑ t = k 2 0 + 1 T y t = T k ∑ t = 1 k X t + T μ + δ 1 ε h − T T − k ∑ t = k + 1 T X t + ( k 1 0 − k ) μ + δ 1 ε h + ( k 2 0 − k 1 0 ) × μ + δ 2 ε h + ( T − k 2 0 ) μ + δ 3 ε h = T k ∑ t = 1 k X t − T T − k ∑ t = k + 1 T X t + δ 1 ε h T − T T − k ( k 1 0 − k ) δ 1 ε h − T T − k ( k 2 0 − k 1 0 ) δ 2 ε h − T T − k ( T − k 2 0 ) δ 3 ε h ⇒ 1 τ a ( 1 ) σ B ( τ ) − 1 1 − τ a ( 1 ) σ ( B ( 1 ) − B ( τ ) ) + δ 1 ε − τ 1 0 − τ 1 − τ δ 1 ε − τ 2 0 − τ 1 0 1 − τ δ 2 ε − 1 − τ 2 0 1 − τ δ 3 ε .

Similarly, for k 1 0 < k ≤ k 2 0 , we have

T Y ̄ k − Y ̄ k * = T 1 k ∑ t = 1 k 1 0 y t + ∑ t = k 1 0 + 1 k y t − 1 T − k ∑ t = k + 1 k 2 0 y t + ∑ t = k 2 0 + 1 T y t = T k ∑ t = 1 k X t + k μ h + k 1 0 δ 1 h ε + ( k − k 1 0 ) δ 2 h ε − T T − k ∑ t = k + 1 T X t + ( T − k ) μ h + ( k 2 0 − k ) × δ 2 h ε + ( T − k 2 0 ) δ 3 h ε = T k ∑ t = 1 k X t − T T − k ∑ t = k + 1 T X t + T k k 1 0 δ 1 h ε + T k ( k − k 1 0 ) × δ 2 h ε − T T − k ( k 2 0 − k ) δ 2 h ε − T T − k ( T − k 2 0 ) δ 3 h ε ⇒ 1 τ a ( 1 ) σ B ( τ ) − 1 1 − τ a ( 1 ) σ ( B ( 1 ) − B ( τ ) ) + τ 1 0 τ δ 1 ε + τ − τ 1 0 τ δ 2 ε − τ 2 0 − τ 1 − τ δ 2 ε − 1 − τ 2 0 1 − τ δ 3 ε ,

and for k 2 0 < k

T Y ̄ k − Y ̄ k * = T k ∑ t = 1 k 1 0 y t + ∑ t = k 1 0 + 1 k 2 0 y t + ∑ t = k 2 0 + 1 k y t − T T − k ∑ t = k + 1 T y t = T k ∑ t = 1 k X t + k μ h + k 1 0 δ 1 ε h + ( k 2 0 − k 1 0 ) × δ 2 h ε + ( k − k 2 0 ) δ 3 h ε − T T − k ∑ t = k + 1 T X t + ( T − k ) μ h + ( T − k ) δ 3 h ε ⇒ 1 τ a ( 1 ) σ B ( τ ) − 1 1 − τ a ( 1 ) σ ( B ( 1 ) − B ( τ ) ) + τ 1 0 τ δ 1 ε + τ 2 0 − τ 1 0 τ δ 2 ε + τ − τ 2 0 τ δ 3 ε − δ 3 ε .

Therefore, we have

τ ̂ → d arg max 0 < τ < 1 τ ( 1 − τ ) J ( τ ) 2 ,

where J(τ) is defined in the main statement of Theorem 1. ■

We need several lemmas to prove Propositions 1, 2, and Theorem 2.

Lemma 1

Under Assumption 1, there exists an M < ∞ such that, for all i and all j > i

E ∑ t = 1 i X t ∑ s = i + 1 j X s ≤ M ,
E 1 j − i ∑ t = i + 1 j X t 2 ≤ M .

Proof

(a) and (b) are given by Lemma 11 and (A.12) in Bai (1997), respectively.■

Lemma 2

Under Assumptions 1 and 2, there exists an M < ∞ such that, for i = 1, 2, and 3,

T | E [ R i T ( k ) ] − E [ R i T ( k 1 0 ) ] | ≤ | k 1 0 − k | T M ,

where R _iT(k) are defined by (A.4), (A.12), and (A.14).

Proof

These relations are given by (A.14) in Bai (1997).■

Lemma 3

Under Assumptions 1, 2, 4, and 5 in which ɛ → 0 and h → 0, the following relations hold:

sup 1 ≤ k ≤ T U T ( k / T ) − E [ U T ( k / T ) ] − T − 1 ∑ t = 1 T ( X t 2 − E [ X t 2 ] ) = O p ( h / ε ) .
There exists C > 0 for all large T such that

E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] ≥ C ( h / ε ) 2 | k − k 1 0 | .

Proof

A T k = 1 k ∑ t = 1 k X t and A T k * = 1 T − k ∑ t = k + 1 T X t .

To prove (a), we consider the three cases; (i) k ≤ k 1 0 , (ii) k 1 0 < k ≤ k 2 0 , and (iii) k 2 0 < k . For k ≤ k 1 0 , it can be shown that

U T ( k / T ) − E [ U T ( k / T ) ] − T − 1 ∑ t = 1 T ( X t 2 − E [ X t 2 ] ) = R 1 T ( k ) − E [ R 1 T ( k ) ] ,

where, as given by (A5) in Bai (1997),

(A.3) U T ( k / T ) = 1 T S T ( k ) = ( k 1 0 − k ) T a T k 2 + ( k 2 0 − k 1 0 ) T b T k 2 + ( T − k 2 0 ) T c T k 2 + 1 T ∑ t = 1 T X t 2 + R 1 T ( k ) ,

(A.4) R 1 T ( k ) = 1 T 2 a T k ∑ t = k + 1 k 1 0 X t + 2 b T k ∑ t = k 1 0 + 1 k 2 0 X t + 2 c T k ∑ t = k 2 0 + 1 T X t − 2 T [ ( k 1 0 − k ) a T k + ( k 2 0 − k 1 0 ) b T k + ( T − k 2 0 ) c T k ] A T k * − k T ( A T k ) 2 − T − k T ( A T k * ) 2 ,

(A.5) a T k = 1 T − k { ( T − k 1 0 ) ( μ 1 − μ 2 ) + ( T − k 2 0 ) ( μ 2 − μ 3 ) } ,

(A.6) b T k = 1 T − k { ( k 1 0 − k ) ( μ 2 − μ 1 ) + ( T − k 2 0 ) ( μ 2 − μ 3 ) } ,

(A.7) c T k = 1 T − k { ( k 1 0 − k ) ( μ 2 − μ 1 ) + ( k 2 0 − k ) ( μ 3 − μ 2 ) } .

Note that a _Tk, b _Tk, and c _Tk are O ( h / ε ) from the definitions of μ _i for i = 1, 2, and 3, respectively.

To investigate R _1T(k), we observe, by the FCLT, that

1 T ∑ t = k + 1 k 1 0 X t = O p ( T − 1 / 2 ) , 1 T ∑ t = k 1 0 + 1 k 2 0 X t = O p ( T − 1 / 2 ) , and 1 T ∑ t = k 2 0 + 1 T X t = O p ( T − 1 / 2 ) .

Thus, the first term on the right-hand side of R _1T(k) in (A.4) is O p ( T − 1 / 2 h / ε ) = O p ( h / ε ) .

To evaluate the second term, because A T k * is O _p(T ^−1/2) = O _p(h ^1/2) uniformly in k ≤ k 1 0 by the FCLT, we can observe that the second term on the right-hand side of (A.4) is O _p(h/ϵ) uniformly in k ≤ k 1 0 .

For the third term, we use Hájek–Rényi inequality given in Proposition 1 in Bai (1994); for a given α > 0, there exists some constant C ₁ > 0 such that

(A.8) P sup k = 1 , … , T 1 k ∑ t = 1 k X t > α ≤ C 1 α 2 log ⁡ T .

(A.8) implies that sup k 1 k ∑ t = 1 k X t = O p log ⁡ T and thus

k T ( A T k ) 2 = 1 T 1 k ∑ t = 1 k X t 2 = 1 T O p log ⁡ T 2 = o p ( h / ε )

uniformly in k ∈ [1, T] under Assumption 4.

Because A T k * = O p ( T − 1 / 2 ) , the fourth term on the right-hand side of (A.4) becomes

T − k T A T k * 2 = O p ( T − 1 ) = o p ( h / ε )

uniformly in k ∈ [ 1 , k 1 0 ] .

Combining these results, we have

(A.9) R 1 T ( k ) = O p ( h / ε ) uniformly in k ≤ k 1 0 .

Next, we investigate E[R _1T(k)]. Note that the expectations of the first and second terms on the right-hand side of (A.4) are zero, while those of the last two terms are O(T ⁻¹) by Lemma 1(b). Then, we observe that

(A.10) E [ R 1 T ( k ) ] = O ( T − 1 ) = O ( h ) = o ( h / ε ) uniformly in k ≤ k 1 0 .

From (A.9) and (A.10), we observe that

R 1 T ( k ) − E [ R 1 T ( k ) ] = O p ( h / ε ) uniformly in k ≤ k 1 0 .

In the case where k 1 0 < k ≤ k 2 0 , U _T(k/T) can be expressed as (see (A.9) in Bai (1997)),

(A.11) U T ( k / T ) = k 1 0 ( k − k 1 0 ) k T ( μ 2 − μ 1 ) 2 + ( k 2 0 − k ) ( T − k 2 0 ) T ( T − k ) ( μ 3 − μ 2 ) 2 + 1 T ∑ t = 1 T X t 2 + R 2 T ( k ) ,

where

(A.12) R 2 T ( k ) = 1 T 2 d T k ∑ t = 1 k 1 0 X t + 2 e T k ∑ t = k 1 0 + 1 k X t + 2 f T k ∑ t = k + 1 k 2 0 X t + 2 g T k ∑ t = k 2 0 + 1 T X t − 2 T k 1 0 d T k + ( k − k 1 0 ) e T k A T k − 2 T ( k 2 0 − k ) f T k + ( T − k 2 0 ) g T k A T k * − k T ( A T k ) 2 − T − k T ( A T k * ) 2 , d T k = [ ( k − k 1 0 ) / k ] ( μ 1 − μ 2 ) , e T k = ( k 1 0 / k ) ( μ 2 − μ 1 ) , f T k = [ ( T − k 2 0 ) / ( T − k ) ] ( μ 2 − μ 3 ) , g T k = [ ( k 2 0 − k ) / ( T − k ) ] ( μ 3 − μ 2 ) ,

and we consider

U T ( k / T ) − E [ U T ( k / T ) ] − T − 1 ∑ t = 1 T ( X t − E [ X t 2 ] ) = R 2 T ( k ) − E [ R 2 T ( k ) ] .

In exactly the same manner as in the case of k ≤ k 1 0 , we can show that (a) holds.

For k 2 0 < k , as given by (A.11) in Bai (1997),

(A.13) U T ( k / T ) = k 1 0 T h T k 2 + k 2 0 − k 1 0 T p T k 2 + k − k 2 0 T q T k 2 + 1 T ∑ t = 1 T X t 2 + R 3 T ( k ) ,

where

(A.14) R 3 T ( k ) = 1 T 2 h T k ∑ t = 1 k 1 0 X t + 2 p T k ∑ t = k 1 0 + 1 k 2 0 X t + 2 q T k ∑ t = k 2 0 + 1 k X t − 2 T [ k 1 0 h T k + ( k 2 0 − k 1 0 ) p T k + ( k − k 2 0 ) q T k ] A T k − k T ( A T k ) 2 − ( T − k ) T ( A T k * ) 2 , h T k = 1 k [ ( k − k 1 0 ) ( μ 1 − μ 2 ) + ( k − k 2 0 ) ( μ 2 − μ 3 ) ] , p T k = 1 k [ k 1 0 ( μ 2 − μ 1 ) + ( k − k 2 0 ) ( μ 2 − μ 3 ) ] , q T k = 1 k [ k 1 0 ( μ 2 − μ 1 ) + k 2 0 ( μ 3 − μ 2 ) ] ,

and we consider

U T ( k / T ) − E [ U T ( k / T ) ] − T − 1 ∑ t = 1 T ( X t − E [ X t 2 ] ) = R 3 T ( k ) − E [ R 3 T ( k ) ] .

The order of the preceding equality is obtained in the same manner.

For k ≤ k 1 0 , the left-hand side of Lemma 3(b) becomes, as given by (A.17) in Bai (1997),

(A.15) E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] = k 1 0 − k ( 1 − k / T ) ( 1 − k 1 0 / T ) ( 1 − k 1 0 / T ) ( μ 1 − μ 2 ) + ( 1 − k 2 0 / T ) ( μ 2 − μ 3 ) 2 + T E [ R 1 T ( k ) ] − E [ R 1 T ( k 1 0 ) ] .

We first note that

( 1 − k 1 0 / T ) ( μ 1 − μ 2 ) + ( 1 − k 2 0 / T ) ( μ 2 − μ 3 ) ≠ 0 ,

which can be obtained in the same manner as (A.18) in Bai (1997). Therefore, there exists a C > 0 such that

k 1 0 − k ( 1 − k / T ) ( 1 − k 1 0 / T ) ( 1 − k 1 0 / T ) ( μ 1 − μ 2 ) + ( 1 − k 2 0 / T ) ( μ 2 − μ 3 ) 2 = k 1 0 − k ( 1 − τ ) ( 1 − τ 1 0 ) h ε 2 ( 1 − τ 1 0 ) ( δ 1 − δ 2 ) + ( 1 − τ 2 0 ) ( δ 2 − δ 3 ) 2 ≥ | k 1 0 − k | h ε 2 C .

On the other hand, for the second term on the right-hand side of (A.15), we observe by Lemma 2 that

T | E R 1 T ( k ) − E [ R 1 T ( k 1 0 ) ] | ≤ | k 1 0 − k | T M = | k 1 0 − k | M h = | k 1 0 − k | × o h ε 2 .

Thus, we have

E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] ≥ C h ε 2 | k − k 1 0 | − | k 1 0 − k | M h ≥ C h ε 2 | k − k 1 0 | / 2 .

For k 1 0 < k ≤ k 2 0 , it can be shown that, as given by inequality below (A.20) in Bai (1997),

E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] ≥ ( k − k 1 0 ) k 2 0 k k 1 0 k 2 0 ( μ 2 − μ 1 ) 2 − ( T − k 2 0 ) ( T − k 1 0 ) ( μ 3 − μ 2 ) 2 + T E [ R 2 T ( k ) ] − E [ R 2 T ( k 1 0 ) ] .

Because

k 1 0 k 2 0 ( μ 2 − μ 1 ) 2 − ( T − k 2 0 ) ( T − k 1 0 ) ( μ 3 − μ 2 ) 2 = τ 1 0 τ 2 0 ( μ 2 − μ 1 ) 2 − ( 1 − τ 2 0 ) ( 1 − τ 1 0 ) ( μ 3 − μ 2 ) 2 > 0

as implied by (4.1), and using Lemma 2, we have

E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] ≥ ( k − k 1 0 ) h ε 2 τ 2 0 τ τ 1 0 τ 2 0 ( δ 2 − δ 1 ) 2 − ( 1 − τ 2 0 ) ( 1 − τ 1 0 ) ( δ 3 − δ 2 ) 2 + T E [ R 2 T ( k ) ] − E [ R 2 T ( k 1 0 ) ] ≥ ( k − k 1 0 ) h ε 2 C − M | k 1 0 − k | h ≥ ( k − k 1 0 ) h ε 2 C / 2

for some M > 0 and C > 0.

For k > k 2 0 , S _T(k) becomes, from (A.13),

S T ( k ) = k 1 0 h T k 2 + ( k 2 0 − k 1 0 ) p T k 2 + ( k − k 2 0 ) q T k 2 + ∑ t = 1 T X t 2 + T R 3 T ( k ) .

Similarly to the case k < k 1 0 , it can be shown that

E S T ( k ) − E S T ( k 1 0 ) ≥ ( k − k 1 0 ) h ε 2 C

for some constant C. ■

Lemma 4

Under Assumptions 1, 2, 4, and 5 in which ɛ → 0 and h → 0, for every ϵ, there exists an M < ∞ such that

P min k ∈ D T , M c S T ( k ) − S T ( k 1 0 ) ≤ 0 < ϵ ,

where D T , M c = k : T η ≤ k ≤ T τ 2 0 ( 1 − η ) , | k − k 1 0 | > M h ε − 2 .

Proof

The proof proceeds similarly to that of Lemma 4 of Bai (1997). For k ≤ k 1 0 ,

(A.16) S T ( k ) − S T ( k 1 0 ) = S T ( k ) − E [ S T ( k ) ] − ( S T ( k 1 0 ) − E [ S T ( k 1 0 ) ] ) + E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] ≥ S T ( k ) − E [ S T ( k ) ] − ( S T ( k 1 0 ) − E [ S T ( k 1 0 ) ] ) + C | k − k 1 0 | h ε 2

for some C > 0, where the inequality holds by Lemma 3(b). Then, S T ( k ) − S T ( k 1 0 ) ≤ 0 implies that

C h ε 2 ≤ 1 | k − k 1 0 | S T ( k ) − E [ S T ( k ) ] − S T ( k 1 0 ) − E [ S T ( k 1 0 ) ] .

By Lemma 2, we have, for some M < ∞,

S T ( k ) − E [ S T ( k ) ] − ( S T ( k 1 0 ) − E [ S T ( k 1 0 ) ] ) = T ( R 1 T ( k ) − E [ R 1 T ( k ) ] ) − T ( R 1 T ( k 1 0 ) − E [ R 1 T ( k 1 0 ) ] ) ≤ T ( R 1 T ( k ) − R 1 T ( k 1 0 ) ) + M | k − k 1 0 | T .

Therefore, it is sufficient to show that for every η > 0 and ϵ > 0, there exists an M > 0 such that

P η h ε 2 < sup k ∈ D T M c 1 | k − k 1 0 | T ( R 1 T ( k ) − E [ R 1 T ( k ) ] ) + M T < ϵ ,

but, because M / T = O ( h ) = o ( h / ε ) 2 , we shall show that

P sup k ∈ D T M c 1 | k − k 1 0 | T ( R 1 T ( k ) − E [ R 1 T ( k ) ] ) > η h ε 2 < ϵ .

For k ≤ k 1 0 , note that by (A.4),

(A.17) T | R 1 T ( k ) − R 1 T ( k 1 0 ) | = 2 a T k ∑ t = k + 1 k 1 0 X t + 2 ( b T k − b T k 1 0 ) ∑ t = k 1 0 + 1 k 2 0 X t + 2 ( c T k − c T k 1 0 ) ∑ t = k 2 0 + 1 T X t − 2 ( k 1 0 − k ) a T k A T k * − 2 ( k 2 0 − k 1 0 ) ( b T k A T k * − b T k 1 0 A T k 1 0 * ) − 2 ( T − k 2 0 ) ( c T k A T k * − c T k 1 0 A T k 1 0 * ) + k 1 0 A T k 1 0 2 − k A T k 2 + ( T − k 1 0 ) A T k 1 0 * 2 − ( T − k ) A T k * 2 .

Subsequently, we shall show that each term on the right-hand side of (A.17) divided by k 1 0 − k is of smaller order than ( h / ε ) 2 uniformly.

As | a T k | ≤ h ε C 1 for some C ₁ < ∞ uniformly, the first term on the right-hand side of (A.17) is evaluated as, by Hájek–Rényi inequality,

(A.18) P sup k ≤ k 1 0 − M h ε − 2 a T k k 1 0 − k ∑ t = k + 1 k 1 0 X t > η h ε 2 ≤ P sup k ≤ k 1 0 − M h ε − 2 1 k 1 0 − k ∑ t = k + 1 k 1 0 X t > η h ε C 1 − 1 ≤ C 1 2 C η 2 M

for some C < ∞. By taking a large value of M, the right-hand side of (A.18) becomes small.

For the second term on the right-hand side of (A.17), because T − k 2 0 < T − k 1 0 , we have

| b T k − b T k 1 0 | = 1 T − k ( k 1 0 − k ) ( μ 2 − μ 1 ) + ( T − k 2 0 ) ( μ 2 − μ 3 ) − 1 T − k 1 0 ( T − k 2 0 ) ( μ 2 − μ 3 ) = k 1 0 − k T − k ( μ 2 − μ 1 ) + k − k 1 0 ( T − k ) ( T − k 1 0 ) ( T − k 2 0 ) ( μ 2 − μ 3 ) ≤ k 1 0 − k T − k ( δ 2 − δ 1 ) h ε + k 1 0 − k T − k ( δ 2 − δ 3 ) h ε ≤ k 1 0 − k T − k C 2 h ε ,

for some C ₂ > 0. Then, we have

b T k − b T k 1 0 k 1 0 − k ∑ t = k 1 0 + 1 k 2 0 X t ≤ C 2 1 T − k h ε ∑ t = k 1 0 + 1 k 2 0 X t ≤ C × O p h ε = o p h ε 2 .

For the third term on the right-hand side of (A.17), we note that

| c T k − c T k 1 0 | = 1 T − k ( k 1 0 − k ) ( μ 2 − μ 1 ) + ( k 2 0 − k ) ( μ 3 − μ 2 ) − 1 T − k 1 0 ( k 2 0 − k 1 0 ) ( μ 3 − μ 2 ) = k 1 0 − k T − k ( μ 2 − μ 1 ) + k 2 0 − k T − k − k 2 0 − k 1 0 T − k 1 0 ( μ 3 − μ 2 ) ≤ k 1 0 − k T − k ( δ 2 − δ 1 ) h ε + k 1 0 − k T − k ( δ 3 − δ 2 ) h ε ≤ k 1 0 − k T − k C 3 h ε ,

for some C ₃ > 0, where the first inequality holds because

k 2 0 − k T − k − k 2 0 − k 1 0 T − k 1 0 = ( T − k 2 0 ) ( k 1 0 − k ) ( T − k ) ( T − k 1 0 ) ≤ k 1 0 − k T − k ,

as T − k 2 0 T − k 1 0 < 1 . Then, it can be seen that

c T k − c T k 1 0 k 1 0 − k ∑ t = k 2 0 + 1 T X t ≤ C 3 T − k h ε ∑ t = k 2 0 + 1 T X t ≤ C × O p h ε = o p h ε 2 .

The fourth term on the right-hand side of (A.17) divided by ( k 1 0 − k ) h ε 2 is

( k 1 0 − k ) − 1 h ε − 2 ( k 1 0 − k ) a T k A T k * = o p ( 1 ) ,

because | a T k | = O p h ε and A T k * = O p ( T − 1 / 2 ) = O p ( h 1 / 2 ) .

For the fifth term on the right-hand side of (A.17), we observe that

(A.19) ( k 2 0 − k 1 0 ) ( b T k A T k * − b T k 1 0 A T k 1 0 * ) = ( k 2 0 − k 1 0 ) [ b T k − b T k 1 0 ] A T k * − ( k 2 0 − k 1 0 ) b T k 1 0 [ A T k 1 0 * − A T k * ] .

The first term on the right-hand side of (A.19) divided by ( k 1 0 − k ) ( h / ε ) 2 becomes

h ε − 2 k 2 0 − k 1 0 k 1 0 − k [ b T k − b T k 1 0 ] A T k * = h ε − 2 O p h ε T − 1 / 2 = O p ( ε ) ,

while the second term is given by

h ε − 2 k 2 0 − k 1 0 k 1 0 − k b T k 1 0 [ A T k 1 0 * − A T k * ] = h ε − 2 k 2 0 − k 1 0 k 1 0 − k 1 T − k 1 0 ( T − k 2 0 ) ( μ 2 − μ 3 ) [ A T k 1 0 * − A T k * ] = h ε − 1 k 2 0 − k 1 0 k 1 0 − k T − k 2 0 T − k 1 0 ( δ 2 − δ 3 ) [ A T k 1 0 * − A T k * ] .

Note that

(A.20) k 2 0 − k 1 0 k 1 0 − k [ A T k 1 0 * − A T k * ] = k 2 0 − k 1 0 ( T − k ) ( T − k 1 0 ) ∑ t = k 1 0 + 1 T X t − k 2 0 − k 1 0 ( T − k ) ( k 1 0 − k ) ∑ t = k + 1 k 1 0 X t = O p h − o p h ε .

Here, the first term on the last expression holds by FCLT, and the order of the second term is obtained from (A.18). Hence, the fifth term on the right-hand side of (A.17) divided by ( k 1 0 − k ) ( h / ε ) 2 is o _p(1).

The sixth term on the right-hand side of (A.17) divided by ( k 1 0 − k ) ( h / ε ) 2 is treated similar to the fifth term.

For the seventh term on the right-hand side of (A.17), we note that k A T k 2 = O p ( 1 ) uniformly on D T , M c because k is proportional to T on D T , M c . Thus, we have

( k 1 0 − k ) − 1 h ε − 2 k 1 0 A T k 1 0 2 − k A T k 2 ≤ 1 M h ε 2 h ε − 2 O p ( 1 ) = 1 M O p ( 1 ) ,

which can be small for all large values of T by choosing a large value of M.

Similarly, the eighth term on the right-hand side of (A.17) can be evaluated as

( k 1 0 − k ) − 1 h ε − 2 ( T − k 1 0 ) A T k 1 0 * 2 − ( T − k ) A T k * 2 ≤ 1 M O p ( 1 )

on D T , M c .

As all the terms on the right-hand side of (A.17) divided by ( k 1 0 − k ) ( h / ε ) 2 converge in probability to 0, we have

P sup k ∈ D T M c 1 | k − k 1 0 | T ( R 1 T ( k ) − E [ R 1 T ( k ) ] ) > η h ε 2 < ϵ .

The case where k 1 0 < k is proved in the same manner and thus omitted. ■

Proof of Proposition 1

The proof proceeds similarly to that of Corollary 1 in Bai (1997). For some C > 0, we have

S T ( k ) − S T ( k 1 0 ) = S T ( k ) − E [ S T ( k ) ] − [ S T ( k 1 0 ) − E [ S T ( k 1 0 ) ] ] + E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] + ∑ t = 1 T X t 2 − E [ X t 2 ] − ∑ t = 1 T X t 2 − E [ X t 2 ] ≥ − 2 sup 1 ≤ j ≤ T S T ( j ) − E [ S T ( j ) ] − ∑ t = 1 T X t 2 − E [ X t 2 ] + E [ S T ( k ) ] − E [ S T ( k 1 0 ) ] ≥ − 2 sup 1 ≤ j ≤ T S T ( j ) − E [ S T ( j ) ] − ∑ t = 1 T X t 2 − E [ X t 2 ] + C h ε 2 | k − k 1 0 | ,

where the last inequality holds by Lemma 3(b). As S T ( k ̂ ) − S T ( k 1 0 ) ≤ 0 , the above inequality implies

| k ̂ − k 1 0 | ≤ C − 1 h ε − 2 2 sup 1 ≤ j ≤ T S T ( j ) − E [ S T ( j ) ] − ∑ t = 1 T X t 2 − E [ X t 2 ] .

Dividing both sides by T, we have, by Lemma 3(a),

| τ ̂ − τ 1 0 | ≤ 2 C − 1 h ε − 2 O p h ε = O p ε . ■

Proof of Proposition 2

As τ ̂ is consistent for τ ₀ by Proposition 1, we can observe that, for any given value of ϵ > 0, P ( k ̂ ∉ D T ) ≤ ϵ for all large T. Thus, we have, using Lemma 4,

P T | τ ̂ − τ 1 0 | > M h ε − 2 ≤ P ( k ̂ ∉ D T ) + P k ̂ ∈ D T , | k ̂ − k 1 0 | > M h ε − 2 ≤ ϵ + P min k ∈ D T M c { S T ( k ) − S T ( k 1 0 ) } ≤ 0 ≤ 2 ϵ . ■

Proof of Theorem 2

The proof proceeds similarly to Proposition 8 in Bai (1997). Given the convergence order of τ ̂ obtained in Proposition 2, we focus on the O ( ( h / ε ) − 2 ) neighborhood of k 1 0 . More precisely, let M be an arbitrary large positive value and k be given by, for s ∈ [−M, M],

k = k 1 0 + ℓ , ℓ = s h ε − 2 .

Then, we can observe that k ̂ = k 1 0 + ℓ ̂ and

(A.21) ℓ ̂ = k ̂ − k 1 0 = arg min ℓ { S T ( k 1 0 + ℓ ) − S T ( k 1 0 ) } .

Thus, we investigate the asymptotic behavior of S T ( k 1 0 + ℓ ) − S T ( k 1 0 ) .

First, we consider the case where k > k 1 0 . This implies ℓ > 0 and s > 0.

For ℓ > 0, define

μ ̂ 1 = 1 k 1 0 ∑ t = 1 k 1 0 y t , μ ̂ 2 = 1 T − k 1 0 ∑ t = k 1 0 + 1 T y t , μ ̂ 1 * = 1 k 1 0 + ℓ ∑ t = 1 k 1 0 + ℓ y t , μ ̂ 2 * = 1 T − k 1 0 − ℓ ∑ t = k 1 0 + ℓ + 1 T y t .

It is not difficult to show that

μ ̂ 1 * − μ 1 = O p ( T − 1 / 2 ) = O p ( h 1 / 2 ) , μ ̂ 1 − μ 1 = O p ( T − 1 / 2 ) = O p ( h 1 / 2 ) , μ ̂ 2 − μ 2 − 1 − τ 2 0 1 − τ 1 0 ( μ 3 − μ 2 ) = O p ( T − 1 / 2 ) = O p ( h 1 / 2 ) , μ ̂ i * − μ ̂ i = O p ( ε h 1 / 2 ) for i = 1,2 .

We decompose the sums of the squared residuals into

(A.22) S T ( k 1 0 + ℓ ) = ∑ t = 1 k 1 0 ( y t − μ ̂ 1 * ) 2 + ∑ t = k 1 0 + 1 k 1 0 + ℓ ( y t − μ ̂ 1 * ) 2 + ∑ t = k 1 0 + ℓ + 1 T ( y t − μ ̂ 2 * ) 2 ,

(A.23) S T ( k 1 0 ) = ∑ t = 1 k 1 0 ( y t − μ ̂ 1 ) 2 + ∑ t = k 1 0 + 1 k 1 0 + ℓ ( y t − μ ̂ 2 ) 2 + ∑ t = k 1 0 + ℓ + 1 T ( y t − μ ̂ 2 ) 2 .

The differences between the two first and third terms on the right-hand side of (A.22) and (A.23) are given by

∑ t = 1 k 1 0 ( y t − μ ̂ 1 * ) 2 − ∑ t = 1 k 1 0 ( y t − μ ̂ 1 ) 2 = k 1 0 ( μ ̂ 1 * − μ ̂ 1 ) 2 = O p ( T ) ( O p ( ε h ) ) 2 = O p ( ε 2 ) ,

and

∑ t = k 1 0 + ℓ + 1 T ( y t − μ ̂ 2 * ) 2 − ∑ t = k 1 0 + ℓ + 1 T ( y t − μ ̂ 2 ) 2 = − ( T − k 1 0 − ℓ ) ( μ ̂ 2 − μ ̂ 2 * ) 2 = O p ( T ) ( O p ( ε h ) ) 2 = O p ( ε 2 ) .

Meanwhile, the difference between the two second terms on the right-hand side of (A.22) and (A.23) becomes

∑ t = k 1 0 + 1 k 1 0 + ℓ ( y t − μ ̂ 1 * ) 2 − ∑ t = k 1 0 + 1 k 1 0 + ℓ ( y t − μ ̂ 2 ) 2 = 2 ( μ ̂ 2 − μ ̂ 1 * ) ∑ t = k 1 0 + 1 k 1 0 + ℓ X t + ℓ ( μ 2 − μ ̂ 1 * ) 2 − ( μ 2 − μ ̂ 2 ) 2 .

Note that

μ ̂ 2 − μ ̂ 1 * = { ( μ 2 − μ 1 ) + ( μ 1 − μ ̂ 1 * ) } − ( μ 2 − μ ̂ 2 ) = { ( μ 2 − μ 1 ) + ( μ 1 − μ ̂ 1 * ) } + 1 − τ 2 0 1 − τ 1 0 ( μ 3 − μ 2 ) + O p ( h 1 / 2 ) = ( μ 2 − μ 1 ) + λ 1 ( μ 2 − μ 1 ) + O p ( h 1 / 2 ) = ( 1 + λ 1 ) ( δ 2 − δ 1 ) h ε + O p ( h 1 / 2 ) ,

and

( μ 2 − μ ̂ 1 * ) 2 − ( μ 2 − μ ̂ 2 ) 2 = { ( μ 2 − μ 1 ) − ( μ ̂ 1 * − μ 1 ) } 2 − ( μ 2 − μ ̂ 2 ) 2 = { ( μ 2 − μ 1 ) − O p ( h 1 / 2 ) } 2 − { λ 1 ( μ 2 − μ 1 ) + O p ( h 1 / 2 ) } 2 = ( μ 2 − μ 1 ) 2 + O p ( h ) − 2 ( μ 2 − μ 1 ) O p ( h 1 / 2 ) − ( λ 1 ( μ 2 − μ 1 ) ) 2 − O p ( h / ε ) = ( μ 2 − μ 1 ) 2 ( 1 − λ 1 2 ) + O p ( h / ε ) = ( δ 2 − δ 1 ) h ε 2 ( 1 − λ 1 2 ) + O p ( h / ε ) .

Thus,

(A.24) 2 ( μ ̂ 2 − μ ̂ 1 * ) ∑ t = k 1 0 + 1 k 1 0 + ℓ X t + ℓ ( μ 2 − μ ̂ 1 * ) 2 − ( μ 2 − μ ̂ 2 ) 2 = 2 ( 1 + λ 1 ) ( δ 2 − δ 1 ) h ε + O p ( h 1 / 2 ) ∑ t = k 1 0 + 1 k 1 0 + ℓ X t + ℓ ( δ 2 − δ 1 ) h ε 2 ( 1 − λ 1 2 ) + O p ( h / ε ) ⇒ 2 ( 1 + λ 1 ) ( δ 2 − δ 1 ) σ a ( 1 ) B 1 ( s ) + s ( δ 2 − δ 1 ) 2 ( 1 − λ 1 2 ) ,

where B ₁(⋅) is a standard Brownian motion on 0 , ∞ . Then, we obtain

(A.25) s ̂ = arg min s > 0 2 ( 1 + λ 1 ) ( δ 2 − δ 1 ) σ a ( 1 ) B 1 ( s ) + s ( δ 2 − δ 1 ) 2 ( 1 − λ 1 2 ) = arg min s > 0 ( 1 + λ 1 ) { 2 σ a ( 1 ) B 1 ( s ( δ 2 − δ 1 ) 2 ) + s ( δ 2 − δ 1 ) 2 ( 1 − λ ) } = ( δ 2 − δ 1 ) − 2 σ 2 a ( 1 ) 2 arg min u > 0 2 σ a ( 1 ) B 1 ( σ 2 a ( 1 ) 2 u ) + σ 2 a ( 1 ) 2 u ( 1 − λ ) = ( δ 2 − δ 1 ) − 2 σ 2 a ( 1 ) 2 arg min u > 0 σ 2 a ( 1 ) 2 { 2 B 1 ( u ) + u ( 1 − λ ) } = d ( δ 2 − δ 1 ) − 2 σ 2 a ( 1 ) 2 arg min u > 0 Γ ( u , λ 1 ) ,

where the last equality in distribution is obtained by letting u = s ( δ 2 − δ 1 ) 2 σ − 2 a ( 1 ) − 2 .

Then, we consider the case where k = k 1 0 − ℓ and ℓ = s ( h / ε ) 2 ≥ 0 . As in the case of k > k 1 0 , we observe that

(A.26) S T ( k 1 0 − ℓ ) = ∑ t = 1 k 1 0 − ℓ ( y t − μ ̂ 1 * ) 2 + ∑ t = k 1 0 − ℓ + 1 k 1 0 ( y t − μ ̂ 2 * ) 2 + ∑ t = k 1 0 + 1 T ( y t − μ ̂ 2 * ) 2 ,

(A.27) S T ( k 1 0 ) = ∑ t = 1 k 1 0 − ℓ ( y t − μ ̂ 1 ) 2 + ∑ t = k 1 0 − ℓ + 1 k 1 0 ( y t − μ ̂ 1 ) 2 + ∑ t = k 1 0 + 1 T ( y t − μ ̂ 2 ) 2 .

The difference between the two first and third terms on the right-hand side of (A.26) and (A.27) is shown to be o _p(1) in the same manner as in the case of k > k 1 0 , whereas the difference between the two second terms converges in distribution to

(A.28) ∑ t = k 1 0 − ℓ + 1 k 1 0 ( y t − μ ̂ 2 * ) 2 − ∑ t = k 1 0 − ℓ + 1 k 1 0 ( y t − μ ̂ 1 ) 2 = 2 ( μ ̂ 1 − μ ̂ 2 * ) ∑ t = k 1 0 − ℓ + 1 k 1 0 X t + ℓ ( ( μ ̂ 2 * − μ 1 ) 2 − ( μ ̂ 1 − μ 1 ) 2 ) ⇒ 2 ( δ 2 − δ 1 ) ( 1 + λ 1 ) σ a ( 1 ) B 2 ( s ) + | s | ( δ 2 − δ 1 ) 2 ( 1 + λ 1 ) 2 ,

where B ₂(⋅) is a standard Brownian motion on 0 , ∞ independent of B ₁(⋅).Thus, we have

(A.29) s ̂ = arg min s > 0 2 ( δ 2 − δ 1 ) ( 1 + λ 1 ) σ a ( 1 ) B 2 ( s ) + | s | ( δ 2 − δ 1 ) 2 ( 1 + λ 1 2 ) = arg min s < 0 2 ( δ 2 − δ 1 ) ( 1 + λ 1 ) σ a ( 1 ) B 2 ( − s ) + | s | ( δ 2 − δ 1 ) 2 ( 1 + λ 1 2 ) = d ( δ 2 − δ 1 ) − 2 σ 2 a ( 1 ) 2 arg min u < 0 Γ ( u , λ 1 ) .

By definition, s ̂ = ℓ ̂ h ε 2 = T h ε 2 ( τ − τ 1 0 ) and thus, we obtain the theorem.■

Proof of Proposition 3

We define the partial means and the objective functions with the subsample of { y t ; t ∈ k ̂ , T } and { y t ; t ∈ k 1 0 , T } as follows:

T ′ = T − k ̂ , Y ̄ k ′ = 1 k − k ̂ ∑ t = k ̂ + 1 k y t , T 0 ′ = T − k 1 0 , Y ̄ 0 , k ′ = 1 k − k 1 0 ∑ t = k 1 0 + 1 k y t , V T ′ ′ ( k ) = ( k − k ̂ ) ( T − k ) T ′ 2 ( Y ̄ k ′ − Y ̄ k * ) , V T 0 ′ ′ ( k ) = ( k − k 1 0 ) ( T − k ) T 0 ′ 2 ( Y ̄ 0 , k ′ − Y ̄ k * ) .

It is easily shown that the minimization of S T ′ ( k ) is equivalent to the maximization of V T ′ ′ ( k ) .

First, we show that k in the neighborhood of k ̂ , k 1 0 and T does not maximize V T ′ ′ ( k ) . Suppose an increasing sequence M _T such that M _T → ∞ with ε 2 / h M T → 0 and hM _T → 0. If k ̂ < k ≤ max k 1 0 , k ̂ + M T , we have

( k − k ̂ ) ( T − k ) ( T − k ̂ ) 2 Y ̄ k ′ = O p 1 T 1 k − k ̂ ∑ t = k ̂ + 1 k μ + δ i ε h + X t = O p M T h ε + O p h , ( k − k ̂ ) ( T − k ) ( T − k ̂ ) 2 Y ̄ k * = O p M T T T 2 1 T − k ∑ t = k + 1 T μ + δ i ε h + X t = O p M T h ε ,

because | k − k ̂ | ≤ M T and | k 1 0 − k ̂ | ≤ M T . Then, we observe that

V T ′ ′ ( k ) = O p M T h ε + O p h ,

whereas it is not difficult to see that

V T ′ ′ ( k 2 0 ) = O p h ε .

These imply that V T ′ ′ ( k 2 0 ) > V T ′ ′ ( k ) over k ̂ < k ≤ max k 1 0 , k ̂ + M T for large values of T. Similarly, it can be shown that V T ′ ′ ( k 2 0 ) > V T ′ ′ ( k ) over T − M _T ≤ k ≤ T − 1.

Next, we consider the case where max k 1 0 , k ̂ + M T < k < T − M T . By (11) in Bai (1994), we have

V T ′ ′ ( k ) − V T ′ ′ ( k 2 0 ) ≤ V T ′ ′ ( k ) − V T 0 ′ ′ ( k ) + V T ′ ′ ( k 2 0 ) − V T 0 ′ ′ ( k 2 0 ) + V T 0 ′ ′ ( k ) − V T 0 ′ ′ ( k 2 0 ) ,

and then,

(A.30) | τ ̂ 2 − τ 2 0 | ≤ 1 C ε h 2 sup V T 0 ′ ′ ( k ) − E V T 0 ′ ′ ( k ) + sup V T ′ ′ ( k ) − V T 0 ′ ′ ( k ) + V T ′ ′ ( k 2 0 ) − V T 0 ′ ′ ( k 2 0 ) ,

because E V T 0 ′ ′ ( k ) − E V T 0 ′ ′ ( k 2 0 ) ≥ C k 2 0 T − k T h ε holds. By Bai (1994), we can show that the first term on the right-hand side of (A.30) is

(A.31) sup V T 0 ′ ′ ( k ) − E [ V T 0 ′ ′ ( k ) ] = O p log ⁡ T T .

The second term on the right-hand side of (A.30) is

(A.32) V T ′ ′ ( k ) − V T 0 ′ ′ ( k ) = ( k − k 1 0 ) ( T − k ) T − k 1 0 Y ̄ k ′ − Y ̄ 0 , k ′ + o p h ε

uniformly.

If k 1 0 < k ̂ , we have

Y ̄ k ′ − Y ̄ 0 , k ′ = 1 k − k ̂ − 1 k − k 1 0 ∑ t = k ̂ + 1 k y t + 1 k − k 1 0 ∑ t = k 1 0 + 1 k ̂ y t = k ̂ − k 1 0 k − k 1 0 1 k − k ̂ ∑ t = k ̂ + 1 k y t + k ̂ − k 1 0 k − k 1 0 O p h ε + 1 k − k 1 0 O p M T = o p h ε

uniformly.

We have the same order in the case of k ̂ < k 1 0 and thus, (A.32) is o p h ε uniformly.

By (A.30)–(A.32), τ ̂ 2 is shown to be consistent. ■

Proof of Proposition 4

The proof is fundamentally the same as Proposition 3 in Bai (1994) and thus, we omit the proof. ■

Proof of Theorem 3

The proof proceeds in the same manner as that of Theorem 2. Note that k ̂ 2 is given by

k ̂ 2 = arg min k ̂ ≤ k ≤ T − 1 S T ′ ( k ) = arg min k ̂ ≤ k ≤ T − 1 S T ′ ( k ) − S T ′ ( k 2 0 ) ,

and thus, we investigate the asymptotic behavior of S T ′ ( k ) − S T ′ ( k 2 0 ) . Since τ ̂ 2 − τ 2 0 = O p ( ε 2 ) by Proposition 4, let us consider k = k 2 0 + ℓ where ℓ = s h ε − 2 with s ∈ [−M, M] for some large value of M. If k ≥ k 2 0 , we define

μ ̃ 2 = 1 k 2 0 + ℓ − k ̂ ∑ t = k ̂ + 1 k 2 0 + ℓ y t , μ ̃ 2 * = 1 k 2 0 − k ̂ ∑ t = k ̂ + 1 k 2 0 y t , μ ̃ 3 = 1 T − k 2 0 − ℓ ∑ t = k 2 0 + ℓ + 1 T y t , μ ̃ 3 * = 1 T − k 2 0 ∑ t = k 2 0 + 1 T y t ,

and

(A.33) S T ′ ( k 2 0 + ℓ ) = ∑ t = k ̂ + 1 k 2 0 ( y t − μ ̃ 2 ) 2 + ∑ t = k 2 0 + 1 k 2 0 + ℓ ( y t − μ ̃ 2 ) 2 + ∑ t = k 2 0 + ℓ + 1 T ( y t − μ ̃ 3 ) 2 ,

(A.34) S T ′ ( k 2 0 ) = ∑ t = k ̂ + 1 k 2 0 ( y t − μ ̃ 2 * ) 2 + ∑ t = k 2 0 + 1 k 2 0 + ℓ ( y t − μ ̃ 3 * ) 2 + ∑ t = k 2 0 + ℓ + 1 T ( y t − μ ̃ 3 * ) 2 .

The difference between the two third terms on the right-hand side of (A.33) and (A.34) is

∑ t = k 2 0 + ℓ + 1 T ( y t − μ ̃ 3 ) 2 − ∑ t = k 2 0 + ℓ + 1 T ( y t − μ ̃ 3 * ) 2 = − ( T − k 2 0 − ℓ ) μ ̃ 3 2 + 2 ( T − k 2 0 − ℓ ) μ ̃ 3 μ ̃ 3 * 2 − ( T − k 2 0 − ℓ ) μ ̃ 3 * 2 = − ( T − k 2 0 − ℓ ) μ ̃ 3 − μ ̃ 3 * 2 = O p ε 2 ,

because the difference between the partial means is

μ ̃ 3 − μ ̃ 3 * = 1 T − k 2 0 − ℓ ∑ t = k 2 0 + ℓ + 1 T y t − 1 T − k 2 0 ∑ t = k 2 0 + 1 T y t = 1 T − k 2 0 − ℓ ∑ t = k 2 0 + ℓ + 1 T X t − 1 T − k 2 0 ∑ t = k 2 0 + 1 T X t = 1 T − k 2 0 − ℓ − 1 T − k 2 0 ∑ t = k 2 0 + ℓ + 1 T X t − 1 T − k 2 0 ∑ t = k 2 0 + 1 k 2 0 + ℓ X t = ℓ ( T − k 2 0 − ℓ ) ( T − k 2 0 ) ∑ t = k 2 0 + ℓ + 1 T X t − 1 T − k 2 0 ∑ t = k 2 0 + 1 k 2 0 + ℓ X t = 1 T 2 h ε − 2 O p T + 1 T O p h ε − 1 = O p ( ε h ) .

The difference between the two first terms on the right-hand side of (A.33) and (A.34) can be shown to be O p ε 2 similarly. On the contrary, the difference between the two second terms on the right-hand side of (A.33) and (A.34) becomes

(A.35) ∑ t = k 2 0 + 1 k 2 0 + ℓ ( y t − μ ̃ 2 ) 2 − ∑ t = k 2 0 + 1 k 2 0 + ℓ ( y t − μ ̃ 3 * ) 2 = ℓ μ ̃ 2 2 − ℓ μ ̃ 3 * 2 − 2 μ ̃ 2 ∑ t = k 2 0 + 1 k 2 0 + ℓ y t + 2 μ ̃ 3 * ∑ t = k 2 0 + 1 k 2 0 + ℓ y t = 2 ( μ ̃ 3 * − μ ̃ 2 ) ∑ t = k 2 0 + 1 k 2 0 + ℓ X t + ℓ ( μ 3 − μ ̃ 2 ) 2 − ( μ 3 − μ ̃ 3 * ) 2 .

The difference between the partial means of the first term on the right-hand side of (A.35) is

(A.36) μ ̃ 3 * − μ ̃ 2 = ( μ 3 − μ 2 ) + ( μ 2 − μ ̃ 2 ) − ( μ 3 − μ ̃ 3 * ) .

The third term on the right-hand side of (A.36) is, by FCLT,

(A.37) μ 3 − μ ̃ 3 * = − 1 T − k 2 0 ∑ t = k 2 0 + 1 T X t = O p ( T − 1 / 2 ) .

If k ̂ ≤ k 1 0 , the second term on the right-hand side of (A.36) is

(A.38) μ 2 − μ ̃ 2 = μ 2 − 1 k 2 0 + ℓ − k ̂ ∑ t = k ̂ + 1 k 2 0 + ℓ μ t − 1 k 2 0 + ℓ − k ̂ ∑ t = k ̂ + 1 k 2 0 + ℓ X t = − ℓ k 2 0 + ℓ − k ̂ μ 3 + ℓ + k 1 0 − k ̂ k 2 0 + ℓ − k ̂ μ 2 − k 1 0 − k ̂ k 2 0 + ℓ − k ̂ μ 1 − 1 k 2 0 + ℓ − k ̂ ∑ t = k ̂ + 1 k 2 0 + ℓ X t = k 1 0 − k ̂ k 2 0 + ℓ − k ̂ ( μ 2 − μ 1 ) − ℓ k 2 0 + ℓ − k ̂ ( μ 3 − μ 2 ) − 1 k 2 0 + ℓ − k ̂ ∑ t = k ̂ + 1 k 2 0 + ℓ X t = O p 1 T ε 2 h h ε + O p ( T − 1 / 2 ) = O p ( ε h ) + O p ( T − 1 / 2 ) .

Similarly, it is shown that μ 2 − μ ̃ 2 = O p ( ε h ) if k 1 0 < k ̂ . By (A.36)–(A.38) becomes

(A.39) μ ̃ 3 * − μ ̃ 2 = ( μ 3 − μ 2 ) + ( μ 2 − μ ̃ 2 ) − ( μ 3 − μ ̃ 3 * ) = ( μ 3 − μ 2 ) + O p ( ε h ) + O p ( T − 1 / 2 ) = ( δ 3 − δ 2 ) h ε ( 1 + o p ( 1 ) ) .

The second term on the right-hand side of the (A.35) is, using (A.37) and (A.38),

(A.40) ( μ 3 − μ ̃ 2 ) 2 − ( μ 3 − μ ̃ 3 * ) 2 = ( μ 3 − μ 2 ) − ( μ 2 − μ ̃ 2 ) 2 − ( μ 3 − μ ̃ 3 * ) 2 = ( μ 3 − μ 2 ) − O p ( ε h ) − O p ( T − 1 / 2 ) 2 + O p ( T − 1 ) = ( δ 3 − δ 2 ) h ε ( 1 + o p ( 1 ) ) 2 + O p ( T − 1 ) .

Therefore, by (A.39) and (A.40), we have

(A.41) 2 ( μ ̃ 3 * − μ ̃ 2 ) ∑ t = k 2 0 + 1 k 2 0 + ℓ X t + ℓ ( μ 3 − μ ̃ 2 ) 2 − ( μ 3 − μ ̃ 3 * ) 2 = 2 ( δ 3 − δ 2 ) h ε ( 1 + o p ( 1 ) ) ∑ t = k 2 0 + 1 k 2 0 + ℓ X t + ℓ ( δ 3 − δ 2 ) h ε ( 1 + o p ( 1 ) ) 2 + O p ( T − 1 ) ⇒ 2 ( δ 3 − δ 2 ) a ( 1 ) σ B 1 ( s ) + s ( δ 3 − δ 2 ) 2 ,

where B ₁(⋅) is a standard Brownian motion on 0 , ∞ .

If k < k 2 0 , by defining k = k 2 0 − ℓ for ℓ > 0 and changing ℓ in μ ̃ 2 , μ ̃ 2 * , μ ̃ 3 , and μ ̃ 3 * to −ℓ, we can express the SSR as follows:

(A.42) S T ′ ( k 2 0 − ℓ ) = ∑ t = k ̂ + 1 k 2 0 − ℓ ( y t − μ ̃ 2 ) 2 + ∑ t = k 2 0 − ℓ + 1 k 2 0 ( y t − μ ̃ 3 ) 2 + ∑ t = k 2 0 + 1 T ( y t − μ ̃ 3 ) 2 ,

(A.43) S T ′ ( k 2 0 ) = ∑ t = k ̂ + 1 k 2 0 − ℓ ( y t − μ ̃ 2 * ) 2 + ∑ t = k 2 0 − ℓ + 1 k 2 0 ( y t − μ ̃ 2 * ) 2 + ∑ t = k 2 0 + 1 T ( y t − μ ̃ 3 * ) 2 .

In the same way as the case of k ≥ k 2 0 , the difference between (A.42) and (A.43) weakly converges to

(A.44) S T ′ ( k 2 0 − ℓ ) − S T ′ ( k 2 0 ) ⇒ − 2 ( δ 3 − δ 2 ) a ( 1 ) σ B 2 ( | s | ) + s ( δ 3 − δ 2 ) 2 ,

where B ₂(⋅) is a standard Brownian motion on 0 , ∞ independent of B ₁(⋅). Therefore, we have, by (A.41) and (A.44),

S T ′ ( k ) − S T ′ ( k 2 0 ) ⇒ 2 ( δ 3 − δ 2 ) a ( 1 ) σ B 1 ( s ) + s ( δ 3 − δ 2 ) 2 if s ≥ 0 2 ( δ 3 − δ 2 ) a ( 1 ) σ B 2 ( | s | ) + | s | ( δ 3 − δ 2 ) 2 if s < 0 .

By the continuous mapping theorem, we finally obtain

( δ 3 − δ 2 ) 2 h ε 2 ( k ̂ 2 − k 2 0 ) ⇒ arg min u ∈ ( − ∞ , ∞ ) W ( u ) + | u | 2 ,

where

W ( u ) = a ( 1 ) σ B 1 ( u ) if u ≥ 0 a ( 1 ) σ B 2 ( | u | ) if u < 0 ,

and the theorem is proved. ■

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Received: 2022-05-11

Revised: 2023-01-26

Accepted: 2023-02-22

Published Online: 2023-03-21

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Artikel in diesem Heft

https://doi.org/10.1515/jtse-2022-0013

Schlagwörter für diesen Artikel

structural break; in-fill asymptotics; time series