Estimating multiple breaks in mean sequentially with fractionally integrated errors

Abstract

This paper studies the estimation issue for multiple breaks in mean when the model errors are fractionally integrated processes of order d with $-0.5<d<0.5$. For this problem with $0<d<0.5$, (Lavielle and Moulines in J Time Ser Anal 21(1):33–59, 2000) estimated the break fractions simultaneously by least squares method. The computational complexity of this method is of order $O(T^2)$ even employing the dynamic programming algorithm (Bai and Perron in J Appl Econom 18(1):1–22, 2003), where T is the sample size. It is well-known that the computational complexity of the sequential method is of order O(T) (Bai in Econom Theory 13(3):315–352, 1997), which is obviously more efficient than the simultaneous method in terms of the computational cost. Therefore, in this paper, we revisit the issue of estimating multiple breaks in mean with fractionally integrated errors, and examine the sequential method for the above issue. It is found that: (1) when the break magnitudes are fixed, the convergence rates of the estimators for the break fractions are all 1 / T, which is invariant to the fractionally differencing parameter d over the entire range $d\in (-0.5, 0.5)$; (2) when the break magnitudes shrink to zero as $T\rightarrow \infty$, both the convergence rates and the asymptotic distributions of the estimators for the break fractions rely on the fractionally differencing parameter d for all $0\le d<0.5$. The convergence rates of the estimators for the break fractions in the case of $-0.5<d<0$ in this paper are not optimal due to the absence of optimal Hájek-Rényi inequalities. Our theoretical results extend and improve the results established in Bai (1997) and Kuan and Hsu (J Time Ser Anal 19(6):693–708, 1998), respectively. Monte Carlo simulations are conducted to examine the finite-sample performance for the estimators. Our theoretical findings are supported by the Monte Carlo simulations.

Appendix

1.1 Appendix A

In this subsection, we provide the proofs of the results in Sect. 2. In the next lemma, we present some results on bounding the maximum of weighted partial sums of I(d) processes with $-0.5<d<0.5$.

Lemma 6.1

Suppose $\{{x}_{t}\}$ is an I(d) process, that is, (2.1) is satisfied, where $-0.5<d<0.5$ and $\{u_t\}$ is a sequence of i.i.d. random variables with mean zero, finite variance and $E(|u_t|^{2+\delta })<\infty$ for some $\delta >0$. Then, for lager positive integer m,

$$\begin{aligned}&\max _{k\ge m}\frac{1}{k}|S_k|=O_p \left( \frac{1}{\sqrt{m}}\right) ,~~-0.5<d\le 0, \end{aligned}$$

(6.1)

$$\begin{aligned}&\max _{k\ge m}\frac{1}{k}|S_k|=O_p \left( \frac{1}{m^{0.5-d}}\right) ,~~0<d<0.5, \end{aligned}$$

(6.2)

$$\begin{aligned}&\max _{1\le k\le m}\frac{1}{\sqrt{k}}|S_k|=O_p (\sqrt{\log m}),~~-0.5<d\le 0, \end{aligned}$$

(6.3)

$$\begin{aligned}&\max _{1\le k\le m}\frac{1}{\sqrt{k}}|S_k|=O_p(m^d),~~0<d<0.5. \end{aligned}$$

(6.4)

Proof

These results are consequences of some Hájek-Rényi inequalities, where (6.1) and (6.3) with $-0.5<d<0$ are consequences of Theorem 2.1 in Hu et al. (2011), (6.1) and (6.3) with $d=0$ are consequences of classical Hájek-Rényi inequalities for i.i.d. random variables, see Lin and Bai (2010), and (6.2) and (6.4) are consequences of Theorem 1 and Lemma 2.2 in Lavielle and Moulines (2000). $\square$

To prove the first two lemmas in Sect. 2, we need the following equations which are taken from Bai (1997)¹,

$$\begin{aligned} U_T(k/T)= & {} \frac{k_1^0-k}{T}a_T^2(k)+\frac{k_2^0-k_1^0}{T}b_T^2(k)+\frac{T-k_2^0}{T}c_T^2(k)+\frac{1}{T}\sum _{t=1}^Tx_t^2\nonumber \\&+R_{1T}(k), \quad k\in [1,k_1^0], \end{aligned}$$

(6.5)

$$\begin{aligned} U_T(k/T)= & {} \frac{k_1^0}{T}d_T^2(k)+\frac{k-k_1^0}{T}e_T^2(k)+\frac{k_2^0-k}{T}f_T^2(k)\nonumber \\&\quad +\frac{T-k_2^0}{T}g_T^2(k)+\frac{1}{T}\sum _{t=1}^Tx_t^2 \nonumber \\&+R_{2T}(k), \quad k\in [k_1^0+1,k_2^0], \end{aligned}$$

(6.6)

$$\begin{aligned} U_T(k/T)= & {} \frac{k_1^0}{T}h_T^2(k)+\frac{k_2^0-k_1^0}{T}p_T^2(k)+\frac{k-k_2^0}{T}q_T^2(k)+\frac{1}{T}\sum _{t=1}^Tx_t^2\nonumber \\&+R_{3T}(k), \quad k \in [k_2^0+1,T], \end{aligned}$$

(6.7)

where

$$\begin{aligned}&{\left\{ \begin{array}{ll}a_T(k)=\frac{1}{T-k}[(T-k_1^0)(\mu _1-\mu _2)+(T-k_2^0)(\mu _2-\mu _3)],~~k\in [1,k_1^0]\\ b_T(k)=\frac{1}{T-k}[(k_1^0-k)(\mu _2-\mu _1)+(T-k_2^0)(\mu _2-\mu _3)],~~k\in [1,k_1^0] \\ c_T(k)=\frac{1}{T-k}[(k_1^0-k)(\mu _2-\mu _1)+(k_2^0-k)(\mu _3-\mu _2)],~~k\in [1,k_1^0] \\ d_T(k)=\frac{1}{k}[(k-k_1^0)(\mu _1-\mu _2)],~~ k\in [k_1^0+1,k_2^0]\\ e_T(k)=\frac{1}{k}[k_1^0(\mu _2-\mu _1)],~~ k\in [k_1^0+1,k_2^0]\\ f_T(k)=\frac{1}{T-k}[(T-k_2^0)(\mu _2-\mu _3)],~~ k\in [k_1^0+1,k_2^0]\\ g_T(k)=\frac{1}{T-k}[(k_2^0-k)(\mu _3-\mu _2)],~~ k\in [k_1^0+1,k_2^0] \\ h_T(k)=\frac{1}{k}[(k-k_1^0)(\mu _1-\mu _2)+(k-k_2^0)(\mu _2-\mu _3)],~~k \in [k_2^0+1,T] \\ p_T(k)=\frac{1}{k}[k_1^0(\mu _2-\mu _1)+(k-k_2^0)(\mu _2-\mu _3)],~~k \in [k_2^0+1,T] \\ q_T(k)=\frac{1}{k}[k_1^0(\mu _2-\mu _1)+k_2^0(\mu _3-\mu _2)],~~k \in [k_2^0+1,T] \end{array}\right. }, \end{aligned}$$

(6.8)

$$\begin{aligned} R_{1T}(k)= & {} \frac{1}{T}\left[ 2a_T(k)\sum _{t=k+1}^{k_1^0}x_t+2b_T(k)\sum _{t=k_1^0+1}^{k_2^0}x_t+2c_T(k)\sum _{t=k_2^0+1}^Tx_t\right] \nonumber \\&\quad -\frac{k}{T}(A_T(k))^2-\frac{T-k}{T}(A_T^*(k))^2,~~ k\in [1,k_1^0], \end{aligned}$$

(6.9)

$$\begin{aligned} R_{2T}(k)= & {} \frac{1}{T}\left[ 2d_T(k)\sum _{t=1}^{k_1^0}x_t+2e_T(k)\sum _{t=k_1^0+1}^{k}x_t \right. \nonumber \\&\quad \left. +2f_T(k)\sum _{t=k+1}^{k_2^0}x_t+2g_T(k)\sum _{t=k_2^0+1}^Tx_t\right] \nonumber \\&\quad -\frac{k}{T}(A_T(k))^2-\frac{T-k}{T}(A_T^*(k))^2,~~k\in [k_1^0+1,k_2^0], \end{aligned}$$

(6.10)

$$\begin{aligned} R_{3T}(k)= & {} \frac{1}{T}\left[ 2h_T(k)\sum _{t=1}^{k_1^0}x_t+2p_T(k)\sum _{t=k_1^0+1}^{k_2^0}x_t+2q_T(k)\sum _{t=k_2^0+1}^kx_t\right] \nonumber \\&\quad -\frac{k}{T}(A_T(k))^2-\frac{T-k}{T}(A_T^*(k))^2,~~ k \in [k_2^0+1,T] \end{aligned}$$

(6.11)

with

$$\begin{aligned} A_T(k)=\frac{1}{k}\sum _{t=1}^kx_t~~\mathrm{and}~~ A_T^*(k)=\frac{1}{T-k}\sum _{t=k+1}^Tx_t. \end{aligned}$$

In the next lemma, the orders of $R_{iT}(k),~i=1,2,3$ are characterized.

Lemma 6.2

If $x_t\thicksim I(d)$ with $-0.5<d<0.5$, then

$$\begin{aligned} R_{iT}(k)=O_p(T^{-0.5+d})~~{uniformly~ in}~ k,~ i=1,2,3. \end{aligned}$$

Proof

It is easy to see that the terms $a_T(k), \ldots , q_T(k)$ in (6.8) are all uniformly bounded both in T and in k. Consider the term $R_{1T}(k)$ first. From the property P3, it is true that $\sum _{t=k_1^0+1}^{k_2^0}x_t=O_p(T^{0.5+d})$ and $\sum _{t=k_2^0+1}^{T}x_t=O_p(T^{0.5+d})$, which yield

$$\begin{aligned} \frac{1}{T}\sum _{t=k_1^0+1}^{k_2^0}x_t=O_p(T^{-0.5+d}),~~\frac{1}{T}\sum _{t=k_2^0+1}^{T}x_t=O_p(T^{-0.5+d}). \end{aligned}$$

(6.12)

In addition, applying the functional central limit theorem to $\sum _{t=k+1}^{k_1^0}x_t$ (applied with the data order reversed by treating $k_1^0$ as 1), one has

$$\begin{aligned}\frac{1}{T}\sup _{1\le k\le k_1^0}\sum _{t=k+1}^{k_1^0}x_t=O_p(T^{-0.5+d}).\end{aligned}$$

For the rest terms in $R_{1T}(k)$, applying Lemma 6.1 leads to

$$\begin{aligned} \sup _{1\le k\le k_1^0}\left| \frac{k}{T}(A_T(k))^2\right|= & {} \frac{1}{T}\left( \sup _{1\le k\le k_1^0}\frac{1}{\sqrt{k}}\left| \sum _{t=1}^k x_t\right| \right) ^2\nonumber \\= & {} \left\{ \begin{array}{ll} O_p(\frac{\log T}{T}),~~&{}-0.5<d\le 0, \\ O_p(T^{-1+2d}),~~&{}0<d<0.5, \end{array} \right. \end{aligned}$$

(6.13)

and applying the functional central limit theorem to $\sum _{t=k+1}^T x_t$ (applied with the data order reversed by treating T as 1) yields

$$\begin{aligned} \sup _{1\le k\le k_1^0}\left| \frac{T-k}{T}(A_T^{*}(k))^2\right|= & {} \frac{1}{T}\left( \sup _{1\le k\le k_1^0}\frac{1}{\sqrt{T-k}}\left| \sum _{t=k+1}^T x_t\right| \right) ^2\nonumber \\\le & {} \frac{1}{T(T-k_1^0)}\left( \sup _{1\le k\le T-1}\left| \sum _{t=k+1}^T x_t\right| \right) ^2\nonumber \\= & {} O_p(T^{-1+2d}). \end{aligned}$$

(6.14)

Combining (6.12)–(6.14) leads to

$$\begin{aligned} \sup _{1\le k\le k_1^0}|R_{1T}(k)|=O_p(T^{-0.5+d}). \end{aligned}$$

Similarly, it can be proved that

$$\begin{aligned} \sup _{k_1^0+1\le k\le k_2^0}|R_{2T}(k)|=O_p(T^{-0.5+d})~~\mathrm{and}~~\sup _{k_2^0+1\le k\le T}|R_{3T}(k)|=O_p(T^{-0.5+d}). \end{aligned}$$

The proof is complete. $\square$

Proof of Lemma 2.1

Since the non-stochastic terms $a_T(k), \ldots , q_T(k)$ are all uniformly bounded, $\frac{1}{T}\sum _{t=1}^Tx_t^2$ converges to its expectation in probability by the property P5, $R_{iT}(k)$ converges uniformly in probability to zero by Lemma 6.2, the uniform limit of $U_T(\tau )$ is acquired easily. $\square$

Proof of Lemma 2.2

$R_{iT}(k), i=1,2,3$ are the only stochastic terms of $U_T(k/T)-\frac{1}{T}\sum _{t=1}^{T}x_t^2$ and they are all of order $O_p(T^{-0.5+d})$ uniformly in k by Lemma 6.2. Moreover, it follows from the property P3 that $ER_{iT}(k),~i=1,2,3$ are of order smaller than $O(T^{-0.5+d})$ uniformly in k by some algebra. These arguments lead to Lemma 2.2. $\square$

The next four lemmas are useful for proving Lemma 2.3.

Lemma 6.3

If $x_t\thicksim I(d)$ with $-0.5<d<0.5$, then for any positive integer i and for any positive integer $j>i$, one has

$$\begin{aligned} \left| E\left[ \frac{1}{j-i}\left( \sum _{t=1}^ix_t\right) \left( \sum _{s=i+1}^jx_s\right) \right] \right| \le \left\{ \begin{array}{ll} O(1),~~&{} d<0,\\ 0,~~ &{} d=0, \\ O(i^{2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

Proof

When $d=0$, it is trivial that $|E[\frac{1}{j-i}(\sum _{t=1}^ix_t)(\sum _{s=i+1}^jx_s)]|=0$. When $d\ne 0$, since $\gamma _x(h)=E(x_tx_{t+h}) =O(h^{2d-1})$ according to the property P2, it is true that

$$\begin{aligned} \left| E\left[ \frac{1}{j-i}\left( \sum _{t=1}^ix_t\right) \left( \sum _{s=i+1}^jx_s\right) \right] \right|= & {} \left| \frac{1}{j-i}\sum _{t=1}^i\sum _{s=i+1}^j\gamma _x(s-t)\right| \\\le & {} C\left| \frac{1}{j-i}\sum _{t=1}^i\sum _{s=i+1}^j(i+1-t)^{2d-1}\right| \\= & {} C\left| \frac{j-i}{j-i}\sum _{t=1}^i(i+1-t)^{2d-1}\right| \\= & {} \left\{ \begin{array}{ll} O(1),~~&{} d<0,\\ O(i^{2d}),~~ &{} d>0, \end{array} \right. \end{aligned}$$

as desired. $\square$

In addition, applying the property P3, the following result is also true,

$$\begin{aligned} E\left[ \frac{1}{j-i}\left( \sum _{t=i+1}^jx_t\right) ^2\right] =O((j-i)^{2d}). \end{aligned}$$

(6.15)

Lemma 6.4

In Model (2.2), under Assumptions A1 and A2, we have for $i=1,2,3$,

$$\begin{aligned} T|ER_{iT}(k)-ER_{iT}(k_1^0)|=\left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} d\le 0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

Proof

We only prove the result for $i=1$ since the proofs for $i=1, 2, 3$ are similar. Write

$$\begin{aligned} T|ER_{1T}(k)-ER_{1T}(k_1^0)|&=\left| E\left[ k(A_T(k))^2-k_1^0(A_T(k_1^0))^2\right] \right. \nonumber \\&\quad \left. -E\left[ (T-k)(A_T^*(k))^2-(T-k_1^0)(A_T^*(k_1^0))^2\right] \right| \nonumber \\&\le \left| E\left[ k(A_T(k))^2-k_1^0(A_T(k_1^0))^2\right] \right| \nonumber \\&\quad +\left| E\left[ (T-k)(A_T^*(k))^2-(T-k_1^0)(A_T^*(k_1^0))^2\right] \right| . \end{aligned}$$

(6.16)

Note that, when $k<k_1^0$, we have

$$\begin{aligned}&k(A_T(k))^2-k_1^0(A_T(k_1^0))^2 \nonumber \\&\quad =(k_1^0-k)\left\{ \frac{1}{k_1^0k}\left( \sum _{t=1}^kx_t\right) ^2- \frac{2}{(k_1^0-k)k_1^0}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right. \nonumber \\&\qquad \left. -\frac{1}{k_1^0(k_1^0-k)}\left( \sum _{t=k+1}^{k_1^0}x_t\right) ^2\right\} . \end{aligned}$$

(6.17)

Thus, applying Lemma 6.3 and (6.15) we have

$$\begin{aligned}&\left| E\left[ k(A_T(k))^2-k_1^0(A_T(k_1^0))^2\right] \right| \nonumber \\&\quad =\left\{ \begin{array}{ll} (k_1^0-k)\left| O(\frac{k^{2d}}{T})-O(\frac{1}{T})-O(\frac{(k_1^0-k)^{2d}}{T})\right| ,~~ &{} d<0\\ (k_1^0-k)\left| O(\frac{k^{2d}}{T})-O(\frac{(k_1^0-k)^{2d}}{T})\right| ,~~ &{} d=0\\ (k_1^0-k)\left| O(\frac{k^{2d}}{T})-O(\frac{k^{2d}}{T})-O(\frac{(k_1^0-k)^{2d}}{T})\right| ,~~ &{} d>0 \end{array} \right. \nonumber \\&\quad \le \left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} d<0,\\ |k_1^0-k|\cdot O(T^{-1}),~~ &{} d=0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

(6.18)

When $k\ge k_1^0$, we have

$$\begin{aligned}&k(A_T(k))^2-k_1^0(A_T(k_1^0))^2\\&\quad =(k-k_1^0)\left\{ -\frac{1}{k_1^0k}\left( \sum _{t=1}^{k_1^0}x_t\right) ^2+ \frac{2}{(k-k_1^0)k}\left( \sum _{t=1}^{k_1^0} x_t\right) \left( \sum _{t=k_1^0+1}^{k}x_t\right) \right. \nonumber \\&\quad \qquad \left. +\frac{1}{k(k-k_1^0)}\left( \sum _{t=k_1^0+1}^{k}x_t\right) ^2\right\} . \end{aligned}$$

Thus, applying Lemma 6.3 and (6.15) again we have

$$\begin{aligned}&\left| E\left[ k(A_T(k))^2-k_1^0(A_T(k_1^0))^2\right] \right| \nonumber \\&\quad =\left\{ \begin{array}{ll} (k-k_1^0)\left| -O(\frac{T^{2d}}{T})+O(\frac{1}{T})+O(\frac{(k-k_1^0)^{2d}}{T})\right| ,~~ &{} d<0\\ (k-k_1^0)\left| -O(\frac{T^{2d}}{T})+O(\frac{(k-k_1^0)^{2d}}{T})\right| ,~~ &{} d=0\\ (k-k_1^0)\left| -O(\frac{T^{2d}}{T})+O(\frac{T^{2d}}{T})+O(\frac{(k-k_1^0)^{2d}}{T})\right| ,~~ &{} d>0 \end{array} \right. \nonumber \\&\quad \le \left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} d<0,\\ |k_1^0-k|\cdot O(T^{-1}),~~ &{} d=0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

(6.19)

Combining (6.18) and (6.19) together yields

$$\begin{aligned} \left| E\left[ k(A_T(k))^2-k_1^0(A_T(k_1^0))^2\right] \right| =\left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} d\le 0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

(6.20)

Similarly, one has

$$\begin{aligned} \left| E\left[ (T-k)(A_T^*(k))^2-(T-k_1^0)(A_T^*(k_1^0))^2\right] \right| =\left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} d\le 0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

(6.21)

Combining (6.16), (6.20) and (6.21), one immediately has

$$\begin{aligned} T|ER_{1T}(k)-ER_{1T}(k_1^0)|=\left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} d\le 0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} d>0. \end{array} \right. \end{aligned}$$

The proof is complete. $\square$

Lemma 6.5

In Model (2.2), under Assumptions A1 and A2, there exists a positive constant $M<\infty$ such that

$$\begin{aligned} ES_T(k)-ES_T(k_1^0)\ge & {} T[ER_{1T}(k)-ER_{1T}(k_1^0)]\\\ge & {} \left\{ \begin{array}{ll} -M|k_1^0-k|/T,~~ &{} d\le 0,\\ -M|k_1^0-k|/T^{1-2d},~~ &{} d>0, \end{array} \right. \\ ES_T(k)-ES_T(k_2^0)\ge & {} T[ER_{3T}(k)-ER_{3T}(k_2^0)]\\\ge & {} \left\{ \begin{array}{ll} -M|k_2^0-k|/T,~~ &{} d\le 0,\\ -M|k_2^0-k|/T^{1-2d},~~ &{} d>0. \end{array} \right. \end{aligned}$$

Proof

The proof is similar to that of Lemma 13 in Bai (1997), where Lemma 6.4 above is used. The details are therefore omitted. $\square$

Lemma 6.6

In Model (2.2), under Assumptions A1–A3, there exists a constant $C>0$ such that for all large T,

$$\begin{aligned} ES_T(k)-ES_T(k_1^0)\ge C|k-k_1^0|,\quad for ~k\le k_1^0. \end{aligned}$$

Proof

The proof is similar to that of Lemma 14 in Bai (1997), where Lemma 6.5 and the following observation are used: $M|k_1^0-k|/T^{1-2d}=o(|k_1^0-k|)$ for any $-0.5<d<0.5$. $\square$

Proof of Lemma 2.3

For $k\le k_1^0$, Lemma 2.3 is implied straightforwardly by Lemma 6.6. For $k>k_1^0$, we follow the proof of Lemma 3 in Bai (1997), but some details are different. When $k\in [k_1^0+1, k_2^0]$, it has been proved in Bai (1997) that

$$\begin{aligned} ES_T(k)-ES_T(k_1^0)\ge & {} (k-k_1^0)\frac{k_2^0}{k}C^{*}-(k-k_1^0)O(T^{-1})\\&+T\left[ ER_{2T}(k)-ER_{2T}(k_1^0)\right] , \end{aligned}$$

where $C^{*}$ is a positive constant and its definition can be found in Bai (1997). Note that $k_2^0/k\ge 1$ for $k\in [k_1^0+1, k_2^0]$, and

$$\begin{aligned} T\left[ ER_{2T}(k)-ER_{2T}(k_1^0)\right] =o(|k-k_1^0|)~~\mathrm{for~any}~-0.5<d<0.5 \end{aligned}$$

by Lemma 6.4. Thus, for all large T,

$$\begin{aligned} ES_T(k)-ES_T(k_1^0)\ge (k-k_1^0)C^{*}/2. \end{aligned}$$

(6.22)

When $k\in [k_2^0+1, T]$, from Lemma 6.5 and (6.22) with $k=k_2^0$, it is true that for all large T,

$$\begin{aligned}&ES_T(k)-ES_T(k_1^0)\nonumber \\&\quad =ES_T(k)-ES_T(k_2^0)+ES_T(k_2^0)-ES_T(k_1^0)\nonumber \\&\quad \ge ES_T(k_2^0)-ES_T(k_1^0)-\left\{ \begin{array}{ll} M|T-k_2^0|/T,~~ &{} -0.5<d\le 0\\ M|T-k_2^0|/T^{1-2d},~~ &{} 0<d<0.5. \end{array} \right. \end{aligned}$$

Note that $M|T-k_2^0|/T=O(1)$ and $M|T-k_2^0|/T^{1-2d}=O(T^{2d})$ and $ES_T(k)-ES_T(k_1^0)\ge (k_2^0-k_1^0)C^{*}/2$ which tends to infinity at rate T. Hence, we have

$$\begin{aligned} ES_T(k)-ES_T(k_1^0)\ge & {} \left( ES_T(k_2^0)-ES_T(k_1^0)\right) (1-o(1))\nonumber \\\ge & {} \left( (k-k_1^0)\frac{k_2^0-k_1^0}{T-k_1^0}\frac{ES_T(k_2^0)-ES_T(k_1^0)}{k_2^0-k_1^0}\right) (1-o(1))\nonumber \\\ge & {} (k-k_1^0)\frac{\tau _2^0-\tau _1^0}{1-\tau _1^0}\frac{C^{*}}{4}. \end{aligned}$$

The proof is complete. $\square$

Proof of Lemma 2.4

For all $k\in [1,T]$ and large T, we have

$$\begin{aligned} \begin{aligned} S_T(k)-S_T(k_1^0)&= \left[ S_T(k)-\sum _{t=1}^{T}x_t^2\right] -\left[ ES_T(k)-\sum _{t=1}^{T}Ex_t^2\right] \\&\quad -\left\{ \left[ S_T(k_1^0)-\sum _{t=1}^{T}x_t^2\right] -\left[ ES_T(k_1^0)-\sum _{t=1}^{T}Ex_t^2\right] \right\} +ES_T(k)-ES_T(k_1^0) \\&\ge -2\sup _{1\le j\le T}\left| \left[ S_T(j)-\sum _{t=1}^Tx_t^2\right] -\left[ ES_T(j)-\sum _{t=1}^{T}Ex_t^2\right] \right| +ES_T(k)-ES_T(k_1^0) \\&\ge -2\sup _{1\le j\le T}\left| \left[ S_T(j)-\sum _{t=1}^{T}x_t^2\right] -\left[ ES_T(j)-\sum _{t=1}^{T}Ex_t^2\right] \right| +C|k-k_1^0|, \end{aligned} \end{aligned}$$

where the last inequality is implied by Lemma 2.3. The preceding inequality still holds if $k={\hat{k}}_1$. Since $S_T({\hat{k}}_1)-S_T(k_1^0)\le 0$, we then have

$$\begin{aligned} |{\hat{k}}_1-k_1^0|\le \frac{2}{C}\sup _{1\le j\le T}\left| \left\{ S_T(j)-\sum _{t=1}^{T}x_t^2\right\} -\left\{ ES_T(j)-\sum _{t=1}^{T}Ex_t^2\right\} \right| , \end{aligned}$$

that is

$$\begin{aligned} |{\hat{\tau }}_1-\tau _1^0|\le O_p\left( \frac{1}{T}\right) +\frac{2}{C}\sup _{1\le j\le T}\left| \left\{ U_T(j/T)-\frac{1}{T}\sum _{t=1}^{T}x_t^2\right\} -\left\{ EU_T(j/T)-\frac{1}{T}\sum _{t=1}^{T}Ex_t^2\right\} \right| . \end{aligned}$$

The proof is then finished by applying Lemma 2.2. $\square$

Proof of Lemma 2.5

One can complete the proof by following the lines in the proof of Lemma 4 in Bai (1997), where Lemmas 2.3 and 6.4 and the property P3 are used. $\square$

Proof of Theorem 2.1

The proof is easy by using Lemmas 2.4 and 2.5. The reader can refer to the proof of Proposition 2 in Bai (1997) for more details. $\square$

Proof of Lemma 2.6

First, from (6.5), it is easy to see that

$$\begin{aligned}&\left| \left[ U_T(k/T)-\frac{1}{T}\sum _{t=1}^Tx_t^2\right] -\left[ EU_T(k/T)-\frac{1}{T}\sum _{t=1}^T Ex_t^2\right] \right| \\&\quad =\left\{ \begin{array}{ll} |R_{1T}(k)-ER_{1T}(k)|,~~ &{} k\in [1,k_1^0]\\ |R_{2T}(k)-ER_{2T}(k)|,~~ &{} k\in [k_1^0+1,k_2^0]\\ |R_{3T}(k)-ER_{3T}(k)|,~~ &{} k\in [k_2^0+1, T] \end{array} \right. \le \left\{ \begin{array}{ll} |R_{1T}(k)|+|ER_{1T}(k)|,~~ &{} k\in [1,k_1^0],\\ |R_{2T}(k)|+|ER_{2T}(k)|,~~ &{} k\in [k_1^0+1,k_2^0],\\ |R_{3T}(k)|+|ER_{3T}(k)|,~~ &{} k\in [k_2^0+1, T]. \end{array} \right. \end{aligned}$$

Note that the terms $a_T(k),\ldots ,q_T(k)$ in (6.8) are all of order $O(v_T)$, then by following the lines in the proof of Lemma 6.2 and noticing Assumption A4 it is easy to have

$$\begin{aligned} \sup _{k}|R_{iT}(k)|=O_p(v_T T^{-0.5+d}),~~i=1,2,3. \end{aligned}$$

(6.23)

In addition, by the property P3,

$$\begin{aligned} |ER_{iT}(k)|\le O\left( \frac{k^{2d}}{T}\right) +O\left( \frac{(T-k)^{2d}}{T}\right) ,~~i=1,2,3, \end{aligned}$$

which further implies

$$\begin{aligned} \sup _{k}|ER_{iT}(k)|\le \left\{ \begin{array}{ll} O(T^{-1}),~~ &{} -0.5<d\le 0,\\ O(T^{-1+2d}),~~ &{} 0<d<0.5, \end{array} \right. ~~~~i=1,2,3. \end{aligned}$$

This together with Assumption A4 imply

$$\begin{aligned} \sup _{k}|ER_{iT}(k)|=o(v_T T^{-0.5+d}),~~i=1,2,3. \end{aligned}$$

(6.24)

Combining (6.23) and (6.24) leads to the desired result. $\square$

Proof of Lemma 2.7

We only prove the result for the case of $k\le k_1^0$ since the case of $k>k_1^0$ can be handled similarly. It follows from the proof of Lemma 13 in Bai (1997) that

$$\begin{aligned} ES_T(k)-ES_T(k_1^0)= & {} \frac{k_1^0-k}{(1-k/T)(1-k_1^0/T)}[(1-k_1^0/T)(\mu _{1T}-\mu _{2T})\nonumber \\&\quad +(1-k_2^0/T)(\mu _{2T}-\mu _{3T})]^2+T[ER_{1T}(k)-ER_{1T}(k_1^0)].\nonumber \\ \end{aligned}$$

(6.25)

Since $\mu _{iT}={\tilde{\mu }}_i v_T$, it is easy to see that the first term on the right-hand side of (6.25) is at least as large as $C(k_1^0-k)v_T^2$, where C is a positive constant depending on $\tau _i^0~(i=1,2)$ and ${\tilde{\mu }}_j~(j=1,2,3)$. For the second term on the right-hand side of (6.25), applying the same arguments as those used in the proof of Lemma 6.4, it can be proved that

$$\begin{aligned} T|ER_{1T}(k)-ER_{1T}(k_1^0)|=\left\{ \begin{array}{ll} |k_1^0-k|\cdot O(T^{-1}),~~ &{} -0.5<d\le 0,\\ |k_1^0-k|\cdot O(T^{-1+2d}),~~ &{} 0<d<0.5. \end{array} \right. \end{aligned}$$

In view of Assumption A4 again, one has

$$\begin{aligned} T|ER_{1T}(k)-ER_{1T}(k_1^0)|=o(|k_1^0-k|v_T^2). \end{aligned}$$

Then, the proof is complete. $\square$

Proof of Lemma 2.8

The proof of Lemma 2.8 is similar to that of Proposition 1 in Bai (1997), where Lemmas 2.6 and 2.7 are used. $\square$

Proof of Lemma 2.9

The proof is similar to that of Lemma 9 in Bai (1997). The key of this proof is to verify for any given $\eta >0$ and $\varepsilon >0$, there exists a large positive constant $M<\infty$ such that for all large T,

$$\begin{aligned} P\left( \sup _{k\in D_{T,M}^*}\frac{T|R_{1T}(k)-R_{1T}(k_1^0)|}{v_T^2|k-k_1^0|}>\eta \right) <\varepsilon . \end{aligned}$$

Note that when $k\in D_{T,M}^*$, we have either $T\eta<k<k_1^0-Mv_T^{-2}$ or $k_1^0+Mv_T^{-2}<k<T\tau _2^0(1-\eta )$ when $-0.5<d<0$, and either $T\eta<k<k_1^0-Mv_T^{-2/(1-2d)}$ or $k_1^0+Mv_T^{-2/(1-2d)}<k<T\tau _2^0(1-\eta )$ when $0\le d<0.5$. To save space, we only consider the cases of $k<k_1^0-Mv_T^{-2}$ when $-0.5<d<0$ and $k<k_1^0-Mv_T^{-2/(1-2d)}$ when $0\le d<0.5$.

Write

$$\begin{aligned}&T\left[ R_{1T}(k)-R_{1T}(k_1^0)\right] \nonumber \\&\quad =2\Big (a_T(k)\sum _{t=k+1}^{k_1^0}x_t\Big ) +2\left( (b_T(k)-b_T(k_1^0))\sum _{t=k_1^0+1}^{k_2^0}x_t\right) +2\left( (c_T(k)-c_T(k_1^0))\sum _{t=k_2^0+1}^{T}x_t\right) \nonumber \\&\quad \quad +\left[ k_1^0\left( A_T(k_1^0)\right) ^2-k\left( A_T(k)\right) ^2\right] +\left[ (T-k_1^0)\left( A_T^*(k_1^0)\right) ^2-(T-k)\left( A_T^*(k)\right) ^2\right] . \end{aligned}$$

(6.26)

Thus, is suffices to prove that, when M and T are large enough, every term on the right-hand side of (6.26) divided by $v_T^2(k_1^0-k)$ is arbitrarily small in probability, uniformly in $T\eta \le k<k_1^0-Mv_T^{-2}$ when $-0.5<d<0$ and in $T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}$ when $0\le d<0.5$. Note that, when $d=0$, $T\eta \le k<k_1^0-Mv_T^{-2}$ is equivalent to $T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}$.

Consider the term $a_T(k)\sum _{t=k+1}^{k_1^0}x_t/(v_T^2(k_1^0-k))$. First, note that $|a_T(k)|\le L v_T$ with a positive constant L uniformly in $k\in [1,T]$. When $-0.5<d\le 0$, applying (6.1) in Lemma 6.1 (applied with the data order reversed by treating $k_1^0$ as 1), it is true that

$$\begin{aligned}&\sup _{T\eta \le k<k_1^0-Mv_T^{-2}}a_T(k)\sum _{t=k+1}^{k_1^0}x_t/(v_T^2(k_1^0-k))\nonumber \\&\quad =O_p\left( \frac{1}{v_T\sqrt{Mv_T^{-2}}}\right) =O_p(M^{-0.5})=o_p(1)~~\mathrm{as}~M\rightarrow \infty ,~-0.5<d\le 0, \end{aligned}$$

and

$$\begin{aligned}&\sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}a_T(k)\sum _{t=k+1}^{k_1^0}x_t/(v_T^2(k_1^0-k))\nonumber \\&\quad =O_p\left( \frac{1}{v_T(Mv_T^{-2/(1-2d)})^{0.5-d}}\right) \nonumber \\&\quad =O_p(M^{-0.5+d})=o_p(1)~~\mathrm{as}~M\rightarrow \infty ,~0<d<0.5. \end{aligned}$$

Consider the term $(b_T(k)-b_T(k_1^0))\sum _{t=k_1^0+1}^{k_2^0}x_t/(v_T^2(k_1^0-k))$. Similar to the results in (A.26) in Bai (1997), it is easy to establish

$$\begin{aligned} |b_T(k)-b_T(k_1^0)|\le \left| \frac{k_1^0-k}{T-k}\right| v_T C~~\mathrm{and}~~|c_T(k)-c_T(k_1^0)|\le \left| \frac{k_1^0-k}{T-k}\right| v_T C \end{aligned}$$

for some $0<C<\infty$. Hence, using the property P3, we have

$$\begin{aligned}&\sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\left| \frac{(b_T(k)-b_T(k_1^0))\sum _{t=k_1^0+1}^{k_2^0}x_t}{v_T^2(k_1^0-k)}\right| \nonumber \\&\quad \le \sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\left| \frac{(b_T(k)-b_T(k_1^0))}{v_T^2(k_1^0-k)}\right| \cdot \left| \sum _{t=k_1^0+1}^{k_2^0}x_t\right| \nonumber \\&\quad =O_p\left( \frac{1}{Tv_T}\cdot T^{0.5+d}\right) =o_p(1),~~-0.5<d\le 0 \end{aligned}$$

by Assumption A4, and similarly,

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{(b_T(k)-b_T(k_1^0))\sum _{t=k_1^0+1}^{k_2^0}x_t}{v_T^2(k_1^0-k)}\right| =o_p(1),~~0<d<0.5. \end{aligned}$$

Applying the above arguments, it can analogously be proved that

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\left| \frac{(c_T(k)-c_T(k_1^0))\sum _{t=k_2^0+1}^{T}x_t}{v_T^2(k_1^0-k)}\right| =o_p(1),~~-0.5<d\le 0 \end{aligned}$$

and

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{(c_T(k)-c_T(k_1^0))\sum _{t=k_2^0+1}^{T}x_t}{v_T^2(k_1^0-k)}\right| =o_p(1),~~0<d<0.5. \end{aligned}$$

Consider the term $\left[ k_1^0\left( A_T(k_1^0)\right) ^2-k\left( A_T(k)\right) ^2\right] /(v_T^2(k_1^0-k))$. From (6.17), we have

$$\begin{aligned}&\frac{k_1^0\left( A_T(k_1^0)\right) ^2-k\left( A_T(k)\right) ^2}{v_T^2(k_1^0-k)}\nonumber \\&\quad =-\frac{1}{k_1^0kv_T^2}\left( \sum _{t=1}^kx_t\right) ^2+ \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \nonumber \\&\quad +\frac{1}{k_1^0(k_1^0-k)v_T^2}\left( \sum _{t=k+1}^{k_1^0}x_t\right) ^2. \end{aligned}$$

By Lemma 6.1 and Assumption A4, it is true that: (1)

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\frac{1}{k_1^0kv_T^2}\left( \sum _{t=1}^kx_t\right) ^2\le & {} \frac{1}{k_1^0T\eta v_T^2}\sup _{1\le k<k_1^0}\left( \sum _{t=1}^kx_t\right) ^2\nonumber \\= & {} O_p\left( \frac{T^{1+2d}}{T^2v_T^2}\right) =o_p(1),~~-0.5<d\le 0\nonumber \\ \end{aligned}$$

(6.27)

by the functional central limith theorem and Assumption A4, and similarly,

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}\frac{1}{k_1^0kv_T^2}\left( \sum _{t=1}^kx_t\right) ^2=O_p\left( \frac{T^{1+2d}}{T^2v_T^2}\right) =o_p(1),~~0<d<0.5;\nonumber \\ \end{aligned}$$

(6.28)

(2)

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\frac{1}{k_1^0(k_1^0-k)v_T^2}\left( \sum _{t=k+1}^{k_1^0}x_t\right) ^2\le & {} \frac{1}{k_1^0 v_T^2}\sup _{1\le k<k_1^0}\left( \frac{1}{\sqrt{k_1^0-k}}\sum _{t=k+1}^{k_1^0}x_t\right) ^2\nonumber \\= & {} O_p\left( \frac{\log T}{Tv_T^2}\right) =o_p(1),~~-0.5<d\le 0\nonumber \\ \end{aligned}$$

(6.29)

by Lemma 6.1 and Assumption A4, and similarly,

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}\frac{1}{k_1^0(k_1^0-k)v_T^2}\left( \sum _{t=k+1}^{k_1^0}x_t\right) ^2= & {} O_p\left( \frac{T^{2d}}{Tv_T^2}\right) \nonumber \\= & {} o_p(1),~~0<d<0.5;\quad \qquad \end{aligned}$$

(6.30)

(3) note that

$$\begin{aligned}&\sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\left| \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\quad \le \sup _{T\eta \le k\le k_1^0/2}\left| \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\qquad +\sup _{k_1^0/2<k<k_1^0-Mv_T^{-2}}\left| \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\quad \le \sup _{T\eta \le k\le k_1^0/2}\left| \frac{2}{k_1^0v_T^2}\left( \frac{1}{\sqrt{k}}\sum _{t=1}^kx_t\right) \left( \frac{1}{\sqrt{k_1^0-k}}\sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\qquad +\sup _{k_1^0/2<k<k_1^0-Mv_T^{-2}}\left| \frac{2}{\sqrt{k_1^0}v_T^2}\left( \frac{1}{\sqrt{k_1^0}}\sum _{t=1}^kx_t\right) \left( \frac{1}{k_1^0-k}\sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\quad \le \frac{1}{k_1^0 v_T^2}\sup _{T\eta \le k\le k_1^0/2}\left( \frac{1}{\sqrt{k}}\sum _{t=1}^{k}x_t\right) ^2+\frac{1}{k_1^0 v_T^2}\sup _{T\eta \le k<k_1^0/2}\left( \frac{1}{\sqrt{k_1^0-k}}\sum _{t=k+1}^{k_1^0}x_t\right) ^2\nonumber \\&\qquad +\sup _{k_1^0/2<k<k_1^0-Mv_T^{-2}}\left| \frac{2}{\sqrt{k_1^0}v_T^2}\left( \frac{1}{\sqrt{k_1^0}}\sum _{t=1}^kx_t\right) \left( \frac{1}{k_1^0-k}\sum _{t=k+1}^{k_1^0}x_t\right) \right| ,~~-0.5<d\le 0, \end{aligned}$$

where Cauchy-Schwarz inequality is used in the last inequality. Moreover, in view of Lemma 6.1 and the functional central limit theorem, it is true that

$$\begin{aligned}&\sup _{k_1^0/2<k<k_1^0-Mv_T^{-2}}\left| \frac{2}{\sqrt{k_1^0}v_T^2}\left( \frac{1}{\sqrt{k_1^0}}\sum _{t=1}^kx_t\right) \left( \frac{1}{k_1^0-k}\sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\quad =O_p\left( \frac{T^{d}}{\sqrt{T}v_T^2}\frac{1}{\sqrt{Mv_T^{-2}}}\right) \nonumber \\&\quad =O_p\left( \frac{1}{T^{0.5-d}v_T\sqrt{M}}\right) =o_p(1),~~-0.5<d\le 0 \end{aligned}$$

by Assumption A4. This together with (6.27) and (6.29) imply that

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2}}\left| \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right| =o_p(1),~~-0.5<d\le 0. \end{aligned}$$

Similarly, it is true that

$$\begin{aligned}&\sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\quad \le \frac{1}{k_1^0 v_T^2}\sup _{T\eta \le k\le k_1^0/2}\left( \frac{1}{\sqrt{k}}\sum _{t=1}^{k}x_t\right) ^2+\frac{1}{k_1^0 v_T^2}\sup _{T\eta \le k<k_1^0/2}\left( \frac{1}{\sqrt{k_1^0-k}}\sum _{t=k+1}^{k_1^0}x_t\right) ^2\nonumber \\&\quad \quad +\sup _{k_1^0/2<k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{2}{\sqrt{k_1^0}v_T^2}\left( \frac{1}{\sqrt{k_1^0}}\sum _{t=1}^kx_t\right) \left( \frac{1}{k_1^0-k}\sum _{t=k+1}^{k_1^0}x_t\right) \right| ,~~0<d<0.5 \end{aligned}$$

and

$$\begin{aligned}&\sup _{k_1^0/2<k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{2}{\sqrt{k_1^0}v_T^2}\left( \frac{1}{\sqrt{k_1^0}}\sum _{t=1}^kx_t\right) \left( \frac{1}{k_1^0-k}\sum _{t=k+1}^{k_1^0}x_t\right) \right| \nonumber \\&\quad =O_p\left( \frac{T^{d}}{\sqrt{T}v_T^2}\frac{1}{(Mv_T^{-2/(1-2d)})^{0.5-d}}\right) =O_p\left( \frac{1}{T^{0.5-d}v_TM^{0.5-d}}\right) =o_p(1),~~0<d<0.5. \end{aligned}$$

This together with (6.28) and (6.30) imply that

$$\begin{aligned} \sup _{T\eta \le k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{2}{(k_1^0-k)k_1^0v_T^2}\left( \sum _{t=1}^kx_t\right) \left( \sum _{t=k+1}^{k_1^0}x_t\right) \right| =o_p(1),~~0<d<0.5. \end{aligned}$$

Combining the above results, it is true that

$$\begin{aligned} \sup _{1\le k<k_1^0-Mv_T^{-2}}\left| \frac{k_1^0\left( A_T(k_1^0)\right) ^2-k\left( A_T(k)\right) ^2}{v_T^2(k_1^0-k)}\right| =o_p(1),~~-0.5<d\le 0 \end{aligned}$$

and

$$\begin{aligned} \sup _{1\le k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{k_1^0\left( A_T(k_1^0)\right) ^2-k\left( A_T(k)\right) ^2}{v_T^2(k_1^0-k)}\right| =o_p(1),~~0<d<0.5. \end{aligned}$$

Similarly, it can be proved

$$\begin{aligned} \sup _{1\le k<k_1^0-Mv_T^{-2}}\left| \frac{(T-k_1^0)\left( A_T^*(k_1^0)\right) ^2-(T-k)\left( A_T^*(k)\right) ^2}{v_T^2(k_1^0-k)}\right| =o_p(1),~~-0.5<d\le 0 \end{aligned}$$

and

$$\begin{aligned} \sup _{1\le k<k_1^0-Mv_T^{-2/(1-2d)}}\left| \frac{(T-k_1^0)\left( A_T^*(k_1^0)\right) ^2-(T-k)\left( A_T^*(k)\right) ^2}{v_T^2(k_1^0-k)}\right| =o_p(1),~~0<d<0.5. \end{aligned}$$

The proof is complete. $\square$

Proof of Theorem 2.2

The proof is similar to that of Theorem 2.1 by using Lemma 2.9 instead of Lemma 2.5. Hence the details are omitted. $\square$

Proof of Theorem 2.3

We study the process

$$\begin{aligned} \Lambda _T(s)=v_T^{4d/(1-2d)}\left\{ S_T(k_1^0+\lfloor sv_T^{-2/(1-2d)}\rfloor )-S_T(k_1^0)\right\} ,~~0\le d<0.5, \end{aligned}$$

where $s\in [-M,M]$ for an arbitrary large constant $0<M<\infty$. Let $l=\lfloor sv_T^{-2/(1-2d)}\rfloor$,

$$\begin{aligned} \Delta _T(l)=v_T^{4d/(1-2d)}\left\{ S_T(k_1^0+l)-S_T(k_1^0)\right\} , \end{aligned}$$

and

$$\begin{aligned} {\hat{l}}=\underset{l}{\mathop {\arg \min }}\Delta _T(l). \end{aligned}$$

Clearly, $P({\hat{l}}={\hat{k}}-k_1^0)\ge 1-\varepsilon$ for any small $\varepsilon >0$ and all large $T=T(\varepsilon )$. Hence, we only need to study $\Delta _T(l)$ to acquire the asymptotic distribution of ${\hat{\tau }}_1$.

Firstly, we consider the case of $0\le l\le Mv_T^{-2/(1-2d)}$. We introduce some new notations. Denote

$$\begin{aligned} {\hat{\mu }}_1^*= & {} \frac{1}{k_1^0+l}\sum _{t=1}^{k_1^0+l}y_t, \quad {\hat{\mu }}_2^*=\frac{1}{T-k_1^0-l}\sum _{t=k_1^0+l+1}^{T}y_t,\\ {\hat{\mu }}_1= & {} \frac{1}{k_1^0}\sum _{t=1}^{k_1^0}y_t, \quad {\hat{\mu }}_2=\frac{1}{T-k_1^0}\sum _{t=k_1^0+1}^{T}y_t. \end{aligned}$$

It is not difficult to show, by the property P3 and Assumption A4, that

$$\begin{aligned} \left\{ \begin{array}{ll} {\hat{\mu }}_1^*-\mu _{1T}=O_p(T^{-0.5+d}), \\ {\hat{\mu }}_2-\mu _{2T}-\frac{1-\tau _2^0}{1-\tau _1^0}(\mu _{3T}-\mu _{2T})=O_p(T^{-0.5+d}). \end{array} \right. \end{aligned}$$

(6.31)

Then, we can write

$$\begin{aligned}&v_T^{4d/(1-2d)}S_T(k_1^0+l)\nonumber \\&\quad = v_T^{4d/(1-2d)}\left\{ \sum _{t=1}^{k_1^0}(y_t-{\hat{\mu }}_1^*)^2 +\sum _{t=k_1^0+1}^{k_1^0+l}(y_t-{\hat{\mu }}_1^*)^2 +\sum _{t=k_1^0+l+1}^{T}(y_t-{\hat{\mu }}_2^*)^2\right\} \nonumber \\ \end{aligned}$$

(6.32)

and

$$\begin{aligned}&v_T^{4d/(1-2d)}S_T(k_1^0)\nonumber \\&\quad =v_T^{4d/(1-2d)}\left\{ \sum _{t=1}^{k_1^0}(y_t-{\hat{\mu }}_1)^2 +\sum _{t=k_1^0+1}^{k_1^0+l}(y_t-{\hat{\mu }}_2)^2 +\sum _{t=k_1^0+l+1}^{T}(y_t-{\hat{\mu }}_2)^2\right\} .\nonumber \\ \end{aligned}$$

(6.33)

The differences between the two first and two third terms of the right-hand sides of (6.32) and (6.33) are

$$\begin{aligned} v_T^{4d/(1-2d)}\left\{ \sum _{t=1}^{k_1^0}(y_t-{\hat{\mu }}_1^*)^2-\sum _{t=1}^{k_1^0}(y_t-{\hat{\mu }}_1)^2\right\} =v_T^{4d/(1-2d)}k_1^0({\hat{\mu }}_1^*-{\hat{\mu }}_1)^2\qquad \end{aligned}$$

(6.34)

and

$$\begin{aligned}&v_T^{4d/(1-2d)}\left\{ \sum _{t=k_1^0+l+1}^{T}(y_t-{\hat{\mu }}_2^*)^2-\sum _{t=k_1^0+l+1}^{T}(y_t-{\hat{\mu }}_2)^2\right\} \nonumber \\&\quad =-v_T^{4d/(1-2d)}(T-k_1^0-l)({\hat{\mu }}_2^*-{\hat{\mu }}_2)^2,\qquad \end{aligned}$$

(6.35)

respectively. Notice that $l=o(T)$ by Assumption A4, we then have

$$\begin{aligned}&{\hat{\mu }}_1^*-{\hat{\mu }}_1\nonumber \\&\quad =\frac{-l}{k_1^0(k_1^0+l)}\sum _{t=1}^{k_1^0}x_t+\frac{1}{k_1^0+l}\sum _{t=k_1^0+1}^{k_1^0+l}x_t +\frac{lv_T}{k_1^0+l}({\tilde{\mu }}_2-{\tilde{\mu }}_1)\nonumber \\&\quad =-O_p\left( \frac{l}{T^{1.5-d}}\right) +O_p\left( \frac{l^{0.5+d}}{T}\right) +O_p\left( \frac{lv_T}{T}\right) \nonumber \\&\quad =O_p\left( \frac{1}{Tv_T^{(1+2d)/(1-2d)}}\right) . \end{aligned}$$

Similarly, we have

$$\begin{aligned} {\hat{\mu }}_2^*-{\hat{\mu }}_2=O_p\left( \frac{1}{Tv_T^{(1+2d)/(1-2d)}}\right) .\qquad \end{aligned}$$

Hence, it follows from (6.34) and (6.35) that

$$\begin{aligned} v_T^{4d/(1-2d)}\left\{ \sum _{t=1}^{k_1^0}(y_t-{\hat{\mu }}_1^*)^2-\sum _{t=1}^{k_1^0}(y_t-{\hat{\mu }}_1)^2\right\} =O_p\left( \frac{1}{Tv_T^{2/(1-2d)}}\right) =o_p(1)\nonumber \\ \end{aligned}$$

(6.36)

and

$$\begin{aligned} v_T^{2/(1-2d)}\left\{ \sum _{t=k_1^0+l+1}^{T}(y_t-{\hat{\mu }}_2^*)^2-\sum _{t=k_1^0+l+1}^{T}(y_t-{\hat{\mu }}_2)^2\right\} =O_p\left( \frac{1}{Tv_T^{2/(1-2d)}}\right) =o_p(1)\nonumber \\ \end{aligned}$$

(6.37)

by Assumption A4.

Now, consider the difference between the two second terms on the right-hand sides of (6.32) and (6.33). For $t\in [k_1^0+1,k_1^0+l]$, $y_t=\mu _{2T}+x_t$. Then,

$$\begin{aligned}&\sum _{t=k_1^0+1}^{k_1^0+l}(y_t-{\hat{\mu }}_1^*)^2-\sum _{t=k_1^0+1}^{k_1^0+l}(y_t-{\hat{\mu }}_2)^2\nonumber \\&\quad =2[\mu _{2T}-{\hat{\mu }}_1^*-(\mu _{2T}-{\hat{\mu }}_2)]\sum _{t=k_1^0+1}^{k_1^0+l}x_t+l[(\mu _{2T}-{\hat{\mu }}_1^*)^2-(\mu _{2T}-{\hat{\mu }}_2)^2]. \end{aligned}$$

From (6.31), it is not difficult to have

$$\begin{aligned} \mu _{2T}-{\hat{\mu }}_1^*-(\mu _{2T}-{\hat{\mu }}_2)= & {} \mu _{2T}-\mu _{1T}+\mu _{1T}-{\hat{\mu }}_1^*-(\mu _{2T}-{\hat{\mu }}_2)\nonumber \\= & {} v_T({\tilde{\mu }}_2-{\tilde{\mu }}_1)(1+\lambda _1)+O_p(T^{-0.5+d})\nonumber \\= & {} v_T({\tilde{\mu }}_2-{\tilde{\mu }}_1)(1+\lambda _1)(1+o_p(1)) \end{aligned}$$

and

$$\begin{aligned} (\mu _{2T}-{\hat{\mu }}_1^*)^2-(\mu _{2T}-{\hat{\mu }}_2)^2= & {} v_T^2({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1-\lambda _1^2)+O_p(v_T T^{-0.5+d})+O_p(T^{-1+2d})\nonumber \\= & {} v_T^2({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1-\lambda _1^2)(1+o_p(1)), \end{aligned}$$

where

$$\begin{aligned} \lambda _1=\frac{1-\tau _2^0}{1-\tau _1^0}\left( \frac{{\tilde{\mu }}_3-{\tilde{\mu }}_2}{{\tilde{\mu }}_2-{\tilde{\mu }}_1}\right) . \end{aligned}$$

Thus, based on the above arguments, we have

$$\begin{aligned}&v_T^{4d/(1-2d)}\left\{ \sum _{t=k_1^0+1}^{k_1^0+l}(y_t-{\hat{\mu }}_1^*)^2-\sum _{t=k_1^0+1}^{k_1^0+l}(y_t-{\hat{\mu }}_2)^2\right\} \nonumber \\&\quad =2({\tilde{\mu }}_2-{\tilde{\mu }}_1)(1+\lambda _1)v_T^{(1+2d)/(1-2d)} \sum _{t=k_1^0+1}^{k_1^0+l}x_t \cdot (1+o_p(1))\nonumber \\&\quad \quad +v_T^{2/(1-2d)} l ({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1-\lambda _1^2)\cdot (1+o_p(1))\nonumber \\&\quad \Rightarrow 2\kappa ({\tilde{\mu }}_2-{\tilde{\mu }}_1)(1+\lambda _1)B_d^{(1)}(s)+s({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1-\lambda _1^2) \end{aligned}$$

by the fact

$$\begin{aligned} \sum _{t=k_1^0+1}^{k_1^0+l}x_t=(1-B)^{-d}\sum _{t=k_1^0+1}^{k_1^0+l}u_t{\mathop {=}\limits ^{d}}(1-B)^{-d}\sum _{t=1}^{l}u_t=\sum _{t=1}^{l}x_t \end{aligned}$$

and the functional central limit theorem (see Assumption A1), where $B_d^{(1)}(\cdot )$ is a two-sided fractional Brownian motion, which together with (6.36) and (6.37) yield

$$\begin{aligned} \Lambda _T(s)\Rightarrow 2\kappa ({\tilde{\mu }}_2-{\tilde{\mu }}_1)(1+\lambda _1)B_d^{(1)}(s)+s({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1-\lambda _1^2), ~~ s>0. \end{aligned}$$

Similarly, it can be proved that

$$\begin{aligned} \Lambda _T(s)\Rightarrow 2\kappa ({\tilde{\mu }}_2-{\tilde{\mu }}_1)(1+\lambda _1)B_d^{(1)}(s)-s({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1+\lambda _1)^2, ~~ s<0. \end{aligned}$$

Define

$$\begin{aligned} \Gamma _1(s,\lambda )={\left\{ \begin{array}{ll}2\kappa ({\tilde{\mu }}_2-{\tilde{\mu }}_1)B_d^{(1)}(s)-s({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1+\lambda ), \qquad &{}\text {if s<0},\\ 0, \qquad &{}\text {if s=0},\\ 2\kappa ({\tilde{\mu }}_2-{\tilde{\mu }}_1)B_d^{(1)}(s)+s({\tilde{\mu }}_2-{\tilde{\mu }}_1)^2(1-\lambda ), \qquad &{}\text {if s>0}.\end{array}\right. } \end{aligned}$$

Then, we have

$$\begin{aligned} \Lambda _T(s)\Rightarrow (1+\lambda _1)\Gamma _1(s,\lambda _1). \end{aligned}$$

This implies that

$$\begin{aligned} \begin{aligned} Tv_T^{2/(1-2d)}({\hat{\tau }}_1-\tau _1^0)&{\mathop {\rightarrow }\limits ^{d}} \underset{s}{\mathop {\arg \min }}(1+\lambda _1)\Gamma _1(s,\lambda _1) \\&{\mathop {=}\limits ^{d}} \underset{s}{\mathop {\arg \min }}\Gamma _1(s,\lambda _1) \end{aligned} \end{aligned}$$

by applying the continuous mapping theorem for argmax/argmin functionals (Kim and Pollard 1990) and the fact that $1+\lambda _1>0$ (see page 347 in Bai 1997).

The proof is complete. $\square$

Proof of Theorem 2.4

The proof is similar to that of Theorem 2.3, thus the details are omitted for saving space. $\square$

1.2 Appendix B

In this subsection, we provide the proofs of the results in Sect. 3.

Proof of Theorem 3.1

The proof of Theorem 3.1 is routine and very similar to that of Theorem 2.1, but the details are very long and tedious. Hence, this proof is omitted for saving space. $\square$

Proof of Theorem 3.2

First, under the case of shrinking breaks (that is, Assumption A4 is satisfied), it can be proved that for any $\varepsilon >0$, there exists a positive constant $M<\infty$ such that for all large T,

$$\begin{aligned} P(T|{\hat{\tau }}_i-\tau _i^0|>Mv_T^{-2/(1-2d)})<\varepsilon ,~~0\le d<0.5. \end{aligned}$$

Then, we study the process

$$\begin{aligned} \Lambda _T^{\prime }(s)=v_T^{4d/(1-2d)}\left\{ S_T(k_i^0+\lfloor sv_T^{-2/(1-2d)}\rfloor )-S_T(k_i^0)\right\} ,~~0\le d<0.5, \end{aligned}$$

where $s\in [-M,M]$. Let $l=\lfloor sv_T^{-2/(1-2d)}\rfloor$,

$$\begin{aligned} \Delta _T^{\prime }(l)=v_T^{4d/(1-2d)}\left\{ S_T(k_i^0+l)-S_T(k_i^0)\right\} ,~~{\hat{l}}=\underset{l}{\mathop {\arg \min }}\Delta _T^{\prime }(l). \end{aligned}$$

As before, we only need to study $\Delta _T^{\prime }(l)$ to acquire the asymptotic distribution of ${\hat{\tau }}_i$.

Firstly, we consider the case of $0\le l\le Mv_T^{-2/(1-2d)}$. Denote

$$\begin{aligned} {\hat{\mu }}_{i1}^*= & {} \frac{1}{k_i^0+l}\sum _{t=1}^{k_i^0+l}y_t, \quad {\hat{\mu }}_{i2}^*=\frac{1}{T-k_i^0-l}\sum _{t=k_i^0+l+1}^{T}y_t,\\ {\hat{\mu }}_{i1}= & {} \frac{1}{k_i^0}\sum _{t=1}^{k_i^0}y_t, \quad {\hat{\mu }}_{i2}=\frac{1}{T-k_i^0}\sum _{t=k_i^0+1}^{T}y_t. \end{aligned}$$

It is straightforward to obtain, by the property P3 and Assumption A4, that,

$$\begin{aligned} \mu _{iT}-{\hat{\mu }}_{i1}^*= & {} \mu _{iT}-\frac{1}{k_i^0+l}\sum _{t=1}^{k_i^0+l}x_t\nonumber \\&-\frac{1}{k_i^0+l}\big [k_1^0\mu _{1T}+(k_2^0-k_1^0)\mu _{2T}+\cdots +(k_i^0-k_{i-1}^0)\mu _{iT}+l\mu _{i+1,T}\big ]\nonumber \\= & {} \frac{v_T}{\tau _i^0}\sum _{j=1}^{i-1}\tau _j^0({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_{j})+O_p(T^{-0.5+d})+O(lv_T/T)\nonumber \\= & {} \frac{v_T}{\tau _i^0}\sum _{j=1}^{i-1}\tau _j^0({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_{j})+O_p(T^{-0.5+d}), \end{aligned}$$

(6.38)

and

$$\begin{aligned}&\mu _{i+1,T}-{\hat{\mu }}_{i2}\nonumber \\&\quad =\mu _{i+1,T}-\frac{1}{T-k_i^0}\sum _{t=k_i^0+1}^Tx_t\nonumber \\&\quad -\frac{1}{T-k_i^0}\big [(k_{i+1}^0-k_i^0)\mu _{i+1,T}+\cdots +(T-k_m^0)\mu _{m+1,T}\big ]\nonumber \\&\quad =-\frac{v_T}{1-\tau _i^0}\sum _{j=i+1}^m(1-\tau _j^0)({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_j)+O_p(T^{-0.5+d}). \end{aligned}$$

(6.39)

It is easy to see that the expression of $\Delta _T^{\prime }(l)$ can be written as follows,

$$\begin{aligned} \Delta _T^{\prime }(l)= & {} v_T^{4d/(1-2d)}\left\{ \sum _{t=1}^{k_i^0}\big [(y_t-{\hat{\mu }}_{i1}^*)^2-(y_t-{\hat{\mu }}_{i1})^2\big ]\right. \nonumber \\&+\sum _{t=k_i^0+1}^{k_i^0+l}\big [(y_t-{\hat{\mu }}_{i1}^*)^2-(y_t-{\hat{\mu }}_{i2})^2\big ]\nonumber \\&+\left. \sum _{t=k_i^0+l+1}^{T}\big [(y_t-{\hat{\mu }}_{i2}^*)^2-(y_t-{\hat{\mu }}_{i2})^2\big ]\right\} . \end{aligned}$$

Similar to the proof of Theorem 2.3, it can be proved that the second term on the right hand side of equation is the leading term. For this leading term, when $t\in [k_i^0+1,k_i^0+l]$, $y_t=\mu _{i+1,T}+x_t$, and

$$\begin{aligned}&\sum _{t=k_i^0+1}^{k_i^0+l}\big [(y_t-{\hat{\mu }}_{i1}^*)^2-(y_t-{\hat{\mu }}_{i2})^2\big ]\nonumber \\&\quad =2[\mu _{i+1,T}-{\hat{\mu }}_{i1}^*-(\mu _{i+1,T}-{\hat{\mu }}_{i2})]\sum _{t=k_i^0+1}^{k_i^0+l}x_t\nonumber \\&\qquad +\,l[(\mu _{i+1,T}-{\hat{\mu }}_{i1}^*)^2-(\mu _{i+1,T}-{\hat{\mu }}_{i2})^2]. \end{aligned}$$

From (6.38) and (6.39), we have

$$\begin{aligned}&[\mu _{i+1,T}-{\hat{\mu }}_{i1}^*-(\mu _{i+1,T}-{\hat{\mu }}_{i2})]\nonumber \\&\quad =\mu _{i+1,T}-\mu _{iT}+\mu _{iT}-{\hat{\mu }}_{i1}^*-(\mu _{i+1,T}-{\hat{\mu }}_{i2})\nonumber \\&\quad =v_T\left[ ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)+\frac{1}{\tau _i^0}\sum _{j=1}^{i-1}\tau _j^0({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_{j})\right. \nonumber \\&\quad \quad \left. +\frac{1}{1-\tau _i^0}\sum _{j=i+1}^m(1-\tau _j^0)({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_j)\right] +O_p(T^{-0.5+d}), \end{aligned}$$

and

$$\begin{aligned}&(\mu _{i+1,T}-{\hat{\mu }}_{i1}^*)^2-(\mu _{i+1,T}-{\hat{\mu }}_{i2})^2\nonumber \\&\quad =[\mu _{i+1,T}-\mu _{iT}+\mu _{iT}-{\hat{\mu }}_{i1}^*]^2-(\mu _{i+1,T}-{\hat{\mu }}_{i2})^2\nonumber \\&\quad =v_T^2\left\{ \left[ ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)+\frac{1}{\tau _i^0}\sum _{j=1}^{i-1}\tau _j^0({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_{j})\right] ^2\right. \nonumber \\&\quad \quad \left. -\left[ \frac{1}{1-\tau _i^0}\sum _{j=i+1}^m(1-\tau _j^0)({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_j)\right] ^2\right\} \nonumber \\&\quad +O_p(v_TT^{-0.5+d})+O_p(T^{-1+2d}). \end{aligned}$$

Denote

$$\begin{aligned} \rho _i= & {} \frac{1}{\tau _i^0({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)}\sum _{j=1}^{i-1}\tau _j^0({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_{j}),\\ \omega _i= & {} \frac{1}{(1-\tau _i^0)({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)}\sum _{j=i+1}^m(1-\tau _j^0)({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_j). \end{aligned}$$

Then, from the functional central limith theorem and the above arguments, we have

$$\begin{aligned}&v_T^{4d/(1-2d)}\left\{ \sum _{t=k_i^0+1}^{k_i^0+l}\big [(y_t-{\hat{\mu }}_{i1}^*)^2-(y_t-{\hat{\mu }}_{i2})^2\big ]\right\} \nonumber \\&\quad =2({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)(1+\rho _i+\omega _i)v_T^{(1+2d)/(1-2d)} \sum _{t=k_i^0+1}^{k_i^0+l}x_t \cdot (1+o_p(1))\nonumber \\&\quad \quad +\, v_T^{2/(1-2d)} l ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)^2[(1+\rho _i)^2-\omega _i^2)]\cdot (1+o_p(1))\nonumber \\&\quad \Rightarrow 2\kappa ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)(1+\rho _i+\omega _i)B_d^{(i)}(s)+s({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)^2[(1+\rho _i)^2-\omega _i^2)], \end{aligned}$$

where $B_d^{(i)}(\cdot )$ is a two-sided fractional Brownian motion. Hence,

$$\begin{aligned} \Lambda _T^{\prime }(s)\Rightarrow & {} 2\kappa ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)(1+\rho _i+\omega _i)B_d^{(i)}(s)\\&+s({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)^2[(1+\rho _i)^2-\omega _i^2)],~~s>0. \end{aligned}$$

It can be proved similarly that,

$$\begin{aligned} \Lambda _T^{\prime }(s)\Rightarrow & {} 2\kappa ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)(1+\rho _i+\omega _i)B_d^{(i)}(s)\\&-s({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)^2[(1+\omega _i)^2-\rho _i^2)],~~s<0. \end{aligned}$$

Define

$$\begin{aligned} \Gamma _i(s,\lambda )={\left\{ \begin{array}{ll}2\kappa ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)B_d^{(i)}(s)-s({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)^2(1+\lambda ), \qquad &{}\text {if s<0},\\ 0, \qquad &{}\text {if }s=0,\\ 2\kappa ({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)B_d^{(i)}(s)+s({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)^2(1-\lambda ), \qquad &{}\text {if }s>0.\end{array}\right. } \end{aligned}$$

Then, we have

$$\begin{aligned} \Lambda _T^{\prime }(s)\Rightarrow (1+\omega _i+\rho _i)\Gamma _i(s,\lambda _i) \end{aligned}$$

with

$$\begin{aligned} \lambda _i=\omega _i-\rho _i=\frac{1}{({\tilde{\mu }}_{i+1}-{\tilde{\mu }}_i)} \left[ \frac{1}{(1-\tau _i^0)}\sum _{j=i+1}^m(1-\tau _j^0)({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_j)-\frac{1}{\tau _i^0}\sum _{j=1}^{i-1}\tau _j^0({\tilde{\mu }}_{j+1}-{\tilde{\mu }}_{j})\right] . \end{aligned}$$

This implies that

$$\begin{aligned} \begin{aligned} Tv_T^{2/(1-2d)}({\hat{\tau }}_i-\tau _i^0) {\mathop {\rightarrow }\limits ^{d}} \underset{s}{\mathop {\arg \min }}(1+\omega _i+\rho _i)\Gamma _i(s,\lambda _i) {\mathop {=}\limits ^{d}} \underset{s}{\mathop {\arg \min }}\Gamma _i(s,\lambda _i) \end{aligned} \end{aligned}$$

by applying the continuous mapping theorem for argmax/argmin functionals (Kim and Pollard 1990) and the fact that $1+\omega _i+\rho _i>0$ which can be derived by Assumption A8.

The proof is complete. $\square$