Estimating non-simultaneous changes in the mean of vectors

Abstract

A sequence of independent vectors with correlated components is considered. It is supposed that there is one change point in the mean of each component and changes need not occur simultaneously. The asymptotic distribution of the change point estimators is studied. If the true change points are well separated, the explicit asymptotic distribution of the change point estimators is presented. In the case the true change points coincide, it is shown that the limit distribution of properly standardized change points estimates exists. It depends not only on the underlying time series dependence structure, but also on the ratio of the sizes of the changes. The asymptotic distribution function is not known, but due to the invariance principle it can be obtained by simulations.

Appendix

After transforming $\big (X_1(i),X_2(i)\big )$ into $\big (X_1(i), Z(i)\big )$ our problem is a special case of change point estimation in linear regression where the regression parameters are connected by some linear constrains. We have to prove that under the assumptions of Theorem 1 the asymptotic distribution of the estimates of $k_1^*$ and $k_2^*$ is the same for known and unknown parameters $a_n$ and $b_n$. Steps of the proof are similar as in Hušková (1995).

We prove Theorem 1 for Model II. The proof for Model I and similarly for Model I and II, with $L=a_n/b_n$ known, goes along the same lines.

To see how far $\widehat{k}_1$ is from $k_1^*$, resp. $\widehat{k}_2$ from $k_2^*$, we will study $\chi ^2(k_1^*,k_2^*)-\chi ^2(k_1,k_2)$. Similarly as in Hušková (1995) we can express

$$\begin{aligned} \chi ^2(k_1,k_2)=A_1(k_1,k_2)+2\,A_2(k_1,k_2)+A_3(k_1,k_2), \end{aligned}$$

where for Model II and $k_1 \le k_2$ we define:

$$\begin{aligned} A_1(k_1,k_2)&=\frac{k_2}{k_1(k_2-\rho ^2k_1)}\left( S_{ev}(k_1,k_2)\right) ^2+\frac{1}{k_2} \left( S_v(k_2)\right) ^2,\\ A_2(k_1,k_2)&= \frac{k_2}{k_1(k_2-\rho ^2k_1)}S_{ev}(k_1,k_2)m_{xz}(k_1,k_2)+\frac{1}{k_2} S_v(k_2) m_z(k_2),\\ A_3(k_1,k_2)&= \frac{k_2}{k_1(k_2-\rho ^2k_1)}\left( m_{xz}(k_1,k_2)\right) ^2+\frac{1}{k_2} \left( m_z(k_2)\right) ^2,\\ S_{ev}(k_1,k_2)&=\sqrt{1-\rho ^2}S_{e}(k_1)-\rho \left( S_v(k_1)-\frac{k_1}{k_2}S_v(k_2)\right) ,\\ m_{xz}(k_1,k_2)&=\sqrt{1-\rho ^2}m_{x1}(k_1)-\rho \left( m_z(k_1)-\frac{k_1}{k_2}m_z(k_2)\right) , \end{aligned}$$

and for $k_2 < k_1$ we define:

$$\begin{aligned} A_1(k_1,k_2)&=\frac{1}{(k_1-\rho ^2k_2)}\left( S_{ev}(k_1,k_2)\right) ^2+\frac{1}{k_2} \left( S_v(k_2)\right) ^2,\\ A_2(k_1,k_2)&= \frac{1}{(k_1-\rho ^2 k_2)}S_{ev}(k_1,k_2)m_{xz}(k_1,k_2)+\frac{1}{k_2} S_v(k_2) m_z(k_2),\\ A_3(k_1,k_2)&= \frac{1}{(k_1-\rho ^2 k_2)}\left( m_{xz}(k_1,k_2)\right) ^2+\frac{1}{k_2} \left( m_z(k_2)\right) ^2,\\ S_{ev}(k_1,k_2)&= \sqrt{1-\rho ^2}S_{e}(k_1)-\rho \left( S_v(k_1)-S_v(k_2)\right) ,\\ m_{xz}(k_1,k_2)&= \sqrt{1-\rho ^2}m_{x1}(k_1)-\rho \left( m_z(k_1)-m_z(k_2)\right) . \end{aligned}$$

The proof of Theorem 1 is split into several lemmas.

Lemma A.1

As $n \rightarrow \infty$, it holds

$$\begin{aligned} \max _{\begin{array}{c} {[}\beta n{]} \le k_1 \le n\\ {[}\beta n{]} \le k_2 \le n \end{array}} |A_1(k_1,k_2)| = O_P(1), \quad \max _{\begin{array}{c} [\beta n] \le k_1 \le n\\ {[}\beta n{]} \le k_2 \le n \end{array}} |A_2(k_1,k_2)| = O_P(\Delta _n \sqrt{n}). \end{aligned}$$

(A-1)

Proof

The equalities

$$\begin{aligned} \max _{[\beta n] \le k \le n} \frac{1}{\sqrt{k}} \sum _{i=1}^k e_i(1)=O_P(1)\quad \text {and}\quad \max _{[\beta n] \le k \le n} \frac{1}{\sqrt{k}} \sum _{i=1}^k v_i=O_P(1) \end{aligned}$$

yield the assertion of Lemma A.1.

Now, we need to estimate the difference $A_3(k_1,k_2)-A_3(k_1^*,k_2^*)$. The set of all possible values $\mathcal {A}=\{(k_1,k_2), [\beta n] \le k_1 \le n, [\beta n] \le k_2 \le n\}$ can be expresses as: $\mathcal {A}= \mathcal {A}_1 \cup \ldots \cup \mathcal {A}_6 \cup \mathcal {B}_1 \cup \cdots \cup \mathcal {B}_6$, where

$$\begin{aligned} \begin{array}{llll} &{}\mathcal {A}_{1}=\{(k_1,k_2) \in \mathcal {A}, k_1 \le k_2 \le k_1^*< k_2^*\}, &{}&{}\, \mathcal {A}_{4}=\{(k_1,k_2) \in \mathcal {A}, k_1^* \le k_1 \le k_2 \le k_2^*\}, \\ &{}\mathcal {A}_{2}=\{(k_1,k_2) \in \mathcal {A}, k_1 \le k_1^* \le k_2 \le k_2^*\}, &{}&{}\, \mathcal {A}_{5}=\{(k_1,k_2) \in \mathcal {A}, k_1^* \le k_1 \le k_2^*< k_2\}, \\ &{}\mathcal {A}_{3}=\{(k_1,k_2) \in \mathcal {A}, k_1 \le k_1^*< k_2^* \le k_2\}, &{}&{}\, \mathcal {A}_{6}=\{(k_1,k_2) \in \mathcal {A}, k_1^*< k_2^* \le k_1 \le k_2\}, \\ &{}\mathcal {B}_{1}=\{(k_1,k_2) \in \mathcal {A}, k_2 \le k_1 \le k_1^*< k_2^*\}, &{}&{}\, \mathcal {B}_{4}=\{(k_1,k_2) \in \mathcal {A}, k_1^* \le k_2 \le k_1 \le k_2^*\}, \\ &{}\mathcal {B}_{2}=\{(k_1,k_2) \in \mathcal {A}, k_2 \le k_1^* \le k_1 \le k_2^*\}, &{}&{}\, \mathcal {B}_{5}=\{(k_1,k_2) \in \mathcal {A}, k_1^* \le k_2 \le k_2^*< k_1\}, \\ &{}\mathcal {B}_{3}=\{(k_1,k_2) \in \mathcal {A}, k_2 \le k_1^*< k_2^* \le k_1\}, &{}&{}\, \mathcal {B}_{6}=\{(k_1,k_2) \in \mathcal {A}, k_1^* < k_2^* \le k_2 \le k_1\}. \end{array} \end{aligned}$$

Lemma A.2

There exists a constant $C_1>0$ such that for $(k_1,k_2) \in \mathcal {B}_1 \cup \cdots \cup \mathcal {B}_6 \cup \mathcal {A}_1 \cup \mathcal {A}_6$ it holds

$$\begin{aligned} A_3(k_1^*,k_2^*)-A_3(k_1,k_2) \ge C_1\, \Delta _n^2\, n, \end{aligned}$$

(A-2)

and there exists a constant $C_2>0$ such that for $(k_1,k_2) \in \mathcal {A}_2 \cup \cdots \cup \mathcal {A}_5$ it holds

$$\begin{aligned} A_3({k}_1^*,{k}_2^*)-A_3 (k_1,k_2) \ge C_2 \,\Delta _n^2\,\left( |k_1-k_1^*| +|k_2-k_2^*|\right) . \end{aligned}$$

(A-3)

Proof

First , it holds

$$\begin{aligned} A_3(k_1^*,k_2^*)=\frac{a^2k_1^* +b^2k_2^*-2\rho a b k_1^*}{1-\rho ^2 }. \end{aligned}$$

The assertion (A-2) is a consequence of the inequalities

$$\begin{aligned}&A_3(k_1,k_2) \le A_3(k_1^*,k_1^*) \qquad \text {for} \,(k_1,k_2) \in \mathcal {A}_1 \cup \mathcal {B}_1 \cup \mathcal {B}_2,\\&A_3(k_1,k_2) \le A_3(k_2^*,k_1^*) \qquad \text {for} \,(k_1,k_2) \in \mathcal {B}_3,\\&A_3(k_1,k_2) \le \max \big (A_3(k_1^*,k_1^*),A_3(k_2^*,k_2^*)\big ) \qquad \text {for} \,(k_1,k_2) \in \mathcal {B}_4,\\&A_3(k_1,k_2) \le A_3(k_2^*,k_2^*) \qquad \text {for} \, (k_1,k_2) \in \mathcal {A}_6 \cup \mathcal {B}_5 \cup \mathcal {B}_6,\\ \end{aligned}$$

and the equalities

$$\begin{aligned}&A_3(k_1^*,k_2^*)-A_3(k_1^*,k_1^*)= \frac{b^2(k_2^*-k_1^*)}{1-\rho ^2},\\&A_3(k_1^*,k_2^*)-A_3(k_2^*,k_2^*)= \frac{a^2k_1^*(k_2^*-k_1^*)\rho ^2}{k_2^*(1-\rho ^2)},\\&A_3(k_1^*,k_2^*)-A_3(k_2^*,k_1^*)=\frac{k_1^*(k_2^*-k_1^*)(a^2+b^2+2\rho ab)+b^2(k_2^*-k_1^*)^2}{k_2^*-\rho ^2 k_1^*}. \end{aligned}$$

The assertion (A-3) has to be proved for all $\mathcal {A}_2, \ldots , \mathcal {A}_4$ separately. We obtain it using

$$\begin{aligned}&A_3(k_1,k_2)\\&\quad =\frac{1}{1-\rho ^2}\left( \frac{(k_1k_2-2k_1k_1^*\rho ^2+k_1^{*2}\rho ^2)a^2}{(k_2-\rho ^2k_1)}+k_2b^2- 2\,k_1^*\rho a b\right) \quad \text {for} \, (k_1,k_2) \in \mathcal {A}_2,\\&\quad =\frac{1}{1-\rho ^2}\Bigg (\frac{k_1k_2-2\rho ^2k_1k_1^*+\rho ^2k_1^{*2}}{k_2-\rho ^2k_1}\,a^2+ \frac{k_2^{*2}-2\rho ^2k_1k_2^*+k_1k_2\rho ^2}{k_2-\rho ^2 k_1}\,b^2\\&\qquad -2\,\frac{k_1k_2-\rho ^2k_1k_1^*-k_1k_2^*+k_1^*k_2^*}{k_2-\rho ^2 k_1}\rho a b\Bigg ) \quad \text {for} \, (k_1,k_2) \in \mathcal {A}_3,\\&\quad =\frac{1}{1-\rho ^2}\left( \frac{k_1^{*2} a^2}{k_1}+k_2b^2-2k_1^* \rho ab\right) \quad \text {for} \, (k_1,k_2) \in \mathcal {A}_4,\\&\quad =\frac{1}{1-\rho ^2}\left( \frac{k_1^{*2}a^2}{k_1}+\frac{\rho ^2k_1(k_2-k_2^*)^2+k_2^{*2}(k_2-\rho ^2k_1)}{k_2(k_2-\rho ^2k_1)}b^2 -2k_1^*\rho ab \right) \quad \text {for} \,(k_1,k_2) \in \mathcal {A}_5. \end{aligned}$$

$\square$

Lemma A.3

Let $\{\varepsilon _n\}$ be a sequence such that $\varepsilon _n \rightarrow 0$ and $\Delta _n \sqrt{n} \varepsilon _n \rightarrow \infty$. Then,

$$\begin{aligned}&\max _{\begin{array}{c} |k_1-k_1^*| \ge n \varepsilon _n \,\cup \, |k_2-k_2^*| \,\ge \,n \varepsilon _n \end{array}} \frac{A_1(k_1,k_2)}{\left( A_3(k_1,k_2)-A_3(k_1^*,k_2^*)\right) } \rightarrow 0, \end{aligned}$$

(A-4)

$$\begin{aligned}&\max _{\begin{array}{c} |k_1-k_1^*|\, \ge \,n \varepsilon _n \, \cup \, |k_2-k_2^*|\, \ge \, n \varepsilon _n \end{array}} \frac{A_2(k_1,k_2)}{\big (A_3(k_1,k_2)-A_3(k_1^*,k_2^*)\big )} \rightarrow 0. \end{aligned}$$

(A-5)

Consequently,

$$\begin{aligned} P\left( \left| \widehat{k}_1-k_1^*\right| \le \varepsilon _n\,n\right) \rightarrow 1 \quad \mathrm {and} \quad P\left( \left| \widehat{k}_2-k_2^*\right| \le \varepsilon _n\,n\right) \rightarrow 1. \end{aligned}$$

(A-6)

Proof

The equalities (A-4) and (A-5) follow from (A-1), (A-2) and (A-3). $\square$

In what follows we will use the equalities:

$$\begin{aligned}&A_1(k_1,k_2)-A_1\left( k_1^*,k_2^*\right) \nonumber \\&\quad =\frac{k_2n}{k_1(k_2-\rho ^2k_1)}\left( \frac{S_{ev}(k_1,k_2)}{\sqrt{n}}+\frac{S_{ev}(k_1^*,k_2^*)}{\sqrt{n}}\right) \left( \frac{S_{ev}(k_1,k_2)}{\sqrt{n}}-\frac{S_{ev}(k_1^*,k_2^*)}{\sqrt{n}}\right) \nonumber \\&\qquad +\left( \frac{S_{ev}(k_1^*,k_2^*)}{\sqrt{n}}\right) ^2\left( \frac{k_2n}{k_1(k_2-\rho ^2 k_1)}-\frac{k_2^* n}{k_1^*(k_2^*-\rho ^2 k_1^*)}\right) \nonumber \\&\qquad +\frac{n}{k_2}\left( \frac{S_v(k_2)}{\sqrt{n}}+\frac{S_v(k_2^*)}{\sqrt{n}}\right) \left( \frac{S_v(k_2)}{\sqrt{n}}-\frac{S_v(k_2^*)}{\sqrt{n}}\right) \nonumber \\&\qquad +\left( \frac{S_v(k_2^*)}{\sqrt{n}}\right) ^2\left( \frac{n}{k_2}-\frac{n}{k_2^*}\right) , \end{aligned}$$

(A-7)

$$\begin{aligned}&A_2(k_1,k_2)-A_2\left( k_1^*,k_2^*\right) \nonumber \\&\quad =\frac{k_2n}{k_1(k_2-\rho ^2k_1)}\left( \frac{m_{xz}(k_1,k_2)}{\sqrt{n}}- \frac{m_{xz}(k_1^*,k_2^*)}{\sqrt{n}}\right) \frac{S_{ev}(k_1,k_2)}{\sqrt{n}}\nonumber \\&\qquad +\frac{k_2n}{k_1(k_2-\rho ^2k_1)} \frac{m_{xz}(k_1^*,k_2^*)}{\sqrt{n}}\left( \frac{S_{ev}(k_1,k_2)}{\sqrt{n}}-\frac{S_{ev}(k_1^*,k_2^*)}{\sqrt{n}}\right) \nonumber \\&\qquad +\left( \frac{k_2n}{k_1(k_2-\rho ^2k_1)} - \frac{k_2^*n}{k_1^*(k_2^*-\rho ^2k_1^*)}\right) \frac{m_{xz}(k_1^*,k_2^*)}{\sqrt{n}}\frac{S_{ev}(k_1^*,k_2^*)}{\sqrt{n}} \nonumber \\&\qquad +\frac{n}{k_2}\left( \frac{m_{z}(k_2)}{\sqrt{n}}- \frac{m_{z}(k_2^*)}{\sqrt{n}}\right) \frac{S_v(k_2)}{\sqrt{n} }+ \frac{n}{k_2}\frac{m_{z}(k_2^*)}{\sqrt{n}}\left( \frac{S_v(k_2)}{\sqrt{n}}-\frac{S_v(k_2^*)}{\sqrt{n}}\right) \nonumber \\&\qquad +\left( \frac{n}{k_2} - \frac{n}{k_2^*}\right) \frac{m_{z}(k_2^*)}{\sqrt{n}}\frac{S_v(k_2^*)}{\sqrt{n}}. \end{aligned}$$

(A-8)

Lemma A.4

Let $\{\varepsilon _n\}$ be a sequence such that $\varepsilon _n \rightarrow 0$ and $\Delta _n \sqrt{n} \varepsilon _n \rightarrow \infty$.

$$\begin{aligned}&\max _{\begin{array}{c} |k_1-k_1^*| \le \varepsilon _n n\\ |k_2-k_2^*| \le \varepsilon _n n \end{array}} \left| \chi ^2(k_1,k_2)-\chi ^2(k_1^*,k_2^*)\nonumber \right. \\&\left. \quad -\left( A_3(k_1,k_2)-A_3(k_1^*,k_2^*)\right) -\left( A_2(k_1,k_2)-A_2(k_1^*,k_2^*\right) \big )\right| = o_p(1) \end{aligned}$$

(A-9)

Proof

For any sequence of zero mean unit variance sequence of i.i.d. random variables $\{e(i)\}$ it holds it holds

$$\begin{aligned} \max _{ |l-l^*| \le \varepsilon _n n} \frac{1}{\sqrt{\varepsilon _n n} } \left| \sum _{i=\min (l,l^*)+1}^{\max (l,l^*)} e(i)\right| = O_p(1). \end{aligned}$$

Using this bound for $\{e_1(i)\}$ and $\{v(i)\}$ with $k_1^*$ and $k_2^*$ we get that

$$\begin{aligned} \max _{\begin{array}{c} |k_1-k_1^*| \le \varepsilon _n n\\ |k_2-k_2^*| \le \varepsilon _n n \end{array}} \left| A_1(k_1,k_2) - A_1(k_1^*,k_2^*)\right| = O_P(\sqrt{\varepsilon _n}) \end{aligned}$$

(A-10)

which proves (A-9). $\square$

Let $\{q_n\}$ be a sequence such that $q_n \rightarrow \infty$. We introduce a set $\mathcal {C}=\{(k_1,k_2);( |k_1-k_1^*| \le \varepsilon _n n, |k_2-k_2^*| \le \varepsilon _n n\, \mathrm {such\, that}\, (|k_1-k_1^*| \ge q_n/\Delta _n^2\, \mathrm {or}\, |k_2-k_2^*| \ge q_n/\Delta _n^2)\}$. The set $\mathcal {C}= \mathcal {C}_1 \cup \mathcal {C}_2 \cup \mathcal {C}_3$ where $\mathcal {C}_1=\{(k_1,k_2) \in \mathcal {A}, q_n/\Delta _n^2 \le |k_1-k_1^*| \le \varepsilon _n n, |k_2-k_2^*| < q_n/\Delta _n^2\}$, $\mathcal {C}_2=\{(k_1,k_2) \in \mathcal {A}, q_n/\Delta _n^2 \le |k_2-k_2^*| \le \varepsilon _n n, |k_1-k_1^*| < q_n/\Delta _n^2\}$, $\mathcal {C}_3=\{(k_1,k_2) \in \mathcal {A}, q_n/\Delta _n^2 \le |k_1-k_1^*| \le \varepsilon _n n, q_n/\Delta _n^2 \le |k_2-k_2^*| \le \varepsilon _n n\}$.

Lemma A.5

It holds

$$\begin{aligned}&\max _{(k_1,k_2) \in \mathcal {C}} \frac{\left| A_1(k_1,k_2)-A_1(k_1^*,k_2^*)\right| }{\Delta _n^2\left( \left| k_1-k_1^*\right| +\left| k_2-k_2^*\right| \right) }\,=\,o_P(1), \end{aligned}$$

(A-11)

$$\begin{aligned}&\max _{(k_1,k_2) \in \mathcal {C}} \frac{|A_2(k_1,k_2)-A_2(k_1^*,k_2^*)|}{\Delta _n^2\left( \left| k_1-k_1^*\right| +\left| k_2-k_2^*\right| \right) }\,=\,o_P(1). \end{aligned}$$

(A-12)

Proof

We consider $(k_1,k_2) \in \mathcal {C}_1$, then

$$\begin{aligned}&\frac{1}{\Delta _n} \max _{q_n/\Delta _n^2 \le |k_1-k_1^*| \le \varepsilon _n n} \frac{1}{|k_1-k_1^*|} \sum _{i=\min (k_1,k_1^*)+1}^{\max (k_1,k_1^*)} e_1(i) = o_p(1),\\&\frac{1}{\Delta _n} \max _{|k_2-k_2^*| \le q_n/\Delta _n^2} \frac{\Delta _n^2}{q_n} \sum _{i=\min (k_2,k_2^*)+1}^{\max (k_2,k_2^*)} v(i) = o_p(1). \end{aligned}$$

The similar equalities hold for $\mathcal {C}_2$ and $\mathcal {C}_3$. Therefore, it holds

$$\begin{aligned} \max _{(k_1,k_2) \in \mathcal {C}} \frac{\left| S_{ev}(k_1,k_2)-S_{ev}(k_1^*,k_2^*)\right| }{\Delta _n\left( |k_1-k_1^*|+|k_2-k_2^*|\right) } \,=\, o_p(1). \end{aligned}$$

(A-13)

The (A-13) yields (A-11). The (A-13) together with

$$\begin{aligned} \max _{(k_1,k_2) \in \mathcal {C}} \frac{\big |m_{xz}(k_1,k_2)-m_{xz}(k_1^*,k_2^*)\big |}{\Delta _n\big (|k_1-k_1^*|+|k_2-k_2^*|\big )} =O(1). \end{aligned}$$

(A-14)

yields (A-12). $\square$

Lemma A.6

It holds

$$\begin{aligned} \Delta _n^2\left| \widehat{k}_1 - k_1^*\right| = O_P(1) \quad \mathrm {and} \quad \Delta _n^2\left| \widehat{k}_2 - k_2^*\right| = O_P(1). \end{aligned}$$

Proof

The assertions are consequence of the previous lemmas. $\square$

For any $C>0$ we introduce a set $\mathcal {M}_C=\{(k_1,k_2); |k_1-k_1^*|\le C/\Delta _n^2 \cap |k_2-k_2^*|\le C/\Delta _n^2\}$.

Lemma A.7

For any C it holds

$$\begin{aligned}&\max _{(k_1,k_2) \in \mathcal {M}_C}\Bigg |\chi ^2(k_1,k_2)-\chi ^2(k_1^*,k_2^*)-\Bigg (-\frac{a_n^2|k_1-k_1^*|}{1-\rho ^2}-\frac{b_n^2|k_2-k_2^*|}{1-\rho ^2}\nonumber \\&\quad +\frac{2 a_n}{\sqrt{1-\rho ^2}}{\text {sign}}(k_1-k_1^*)\sum _{i=\min (k_1,k_1^*)+1}^{\max (k_1,k_1^*)} \Big (\sqrt{1-\rho ^2}e_1(i) -\rho v(i)\Big ) \nonumber \\&\quad +\frac{2 b_n}{\sqrt{1-\rho ^2}} {\text {sign}}(k_2-k_2^*)\sum _{i=\min (k_2,k_2^*)+1}^{\max (k_2,k_2^*)} v(i)\Bigg )\Bigg |=o_P(1). \end{aligned}$$

(A-15)

Proof

For $C>0$ it holds

$$\begin{aligned}&\max _{(k_1,k_2) \in \mathcal {M}_C}\left| A_3(k_1,k_2)-A_3(k_1^*,k_2^*)- \left( -\frac{a_n^2|k_1-k_1^*|}{1-\rho ^2}-\frac{b_n^2|k_2-k_2^*|}{1-\rho ^2}\right) \right| \\&\quad = O\left( \Delta _n^2 \frac{(k_1-k_1^*)^2+(k_2-k_2^*)^2}{n}\right) = o(1).\\&\max _{(k_1,k_2) \in \mathcal {M}_C}\left| A_2(k_1,k_2)-A_2(k_1^*,k_2^*) \right. \\&\left. \qquad -\frac{a_n}{\sqrt{1-\rho ^2}}{\text {sign}}(k_1-k_1^*)\sum _{i=\min (k_1,k_1^*)+1}^{\max (k_1,k_1^*)} \left( \sqrt{1-\rho ^2}e_1(i) -\rho v(i)\right) \right. \nonumber \\&\left. \qquad -\frac{b_n}{\sqrt{1-\rho ^2}} {\text {sign}}(k_2-k_2^*)\sum _{i=\min (k_2,k_2^*)+1}^{\max (k_2,k_2^*)} v(i)\right| =o_p(1). \end{aligned}$$

$\square$

Proof of Theorem 1.

As it is explained in Csörgő and Horváth (1997), the assertion of Theorem 1 follows from the convergence of partial sums and the fact that the variables $\{\sqrt{1-\rho ^2}e_1(i) -\rho v(i), i=\lfloor k_1^*-C/\Delta _n^2\rfloor ,\ldots ,\lceil k_1^*+C/\Delta _n^2 \rceil \}$ and $\{v(i), i=\lfloor k_2^*-C/\Delta _n^2\rfloor ,\ldots ,\lceil k_2^*+C/\Delta _n^2 \rceil \}$ are independent for n large enough.

Proof of Theorem 2.

Theorem 2 is a consequence of the following two lemmas. For any $C>0$ denote $\mathcal {M}_C=\{(k_1,k_2); |k_1-k^*| \le C/\Delta _n^2 \cap |k_2-k^*| \le C/\Delta _n^2\}$.

Lemma A.8

Denote $D_{A3}(k_1,k_2,k^*)=A_3(k_1,k_2)-A_3(k^*,k^*)$. For any $C>0$ it holds

$$\begin{aligned}&\max _{\begin{array}{c} (k_1,k_2) \in \mathcal {M}_C\\ k_1 \le k_2 \le k^* \end{array}} \left| D_{A3}(k_1,k_2,k^*) - \left( -\frac{a_n^2(k^*-k_1)}{1-\rho ^2}-\frac{(b_n^2-2\rho a_nb_n)(k^*-k_2)}{1-\rho ^2}\right) \right| =o_p(1),\\&\max _{\begin{array}{c} (k_1,k_2) \in \mathcal {M}_C\\ k_2 \le k_1 \le k^* \end{array}} \left| D_{A3}(k_1,k_2,k^*) - \left( -\frac{(a_n^2-2\rho a_n b_n)(k^*-k_1)}{1-\rho ^2}-\frac{b_n^2(k^*-k_2)}{1-\rho ^2}\right) \right| =o_p(1),\\&\max _{\begin{array}{c} (k_1,k_2) \in \mathcal {M}_C\\ k_2 \le k^* \le k_1 \end{array}} \left| D_{A3}(k_1,k_2,k^*) - \left( -\frac{a_n^2(k_1-k^*)}{1-\rho ^2}-\frac{b_n^2(k^*-k_2)}{1-\rho ^2}\right) \right| =o_p(1),\\&\max _{\begin{array}{c} (k_1,k_2) \in \mathcal {M}_C\\ k^* \le k_2 \le k_1 \end{array}} \left| D_{A3}(k_1,k_2,k^*) - \left( -\frac{a_n^2(k_1-k^*)}{1-\rho ^2}-\frac{(b_n^2-2\rho a_nb_n)(k_2-k^*)}{1-\rho ^2}\right) \right| =o_p(1),\\&\max _{\begin{array}{c} (k_1,k_2) \in \mathcal {M}_C\\ k^* \le k_1 \le k_2 \end{array}} \left| D_{A3}(k_1,k_2,k^*) - \left( -\frac{(a_n^2-2\rho a_n b_n)(k_1-k^*)}{1-\rho ^2}-\frac{b_n^2(k_2-k^*)}{1-\rho ^2}\right) \right| =o_p(1),\\&\max _{\begin{array}{c} (k_1,k_2) \in \mathcal {M}_C\\ k_1 \le k^* \le k_2 \end{array}} \left| D_{A3}(k_1,k_2,k^*) - \left( -\frac{a_n^2(k^*-k_1)}{1-\rho ^2}-\frac{b_n^2(k_2-k^*)}{1-\rho ^2}\right) \right| =o_p(1). \end{aligned}$$

Lemma A.9

For any $C>0$ it holds

$$\begin{aligned}&\max _{(k_1,k_2) \in \mathcal {M}_C}\Big |A_2(k_1,k_2)-A_2(k^*,k^*) \\&\quad -\frac{a_n}{\sqrt{1-\rho ^2}}{\text {sign}}(k_1-k^*)\sum _{i=\min (k_1,k^*)+1}^{\max (k_1,k^*)} \Big (\sqrt{1-\rho ^2}e_1(i) -\rho v(i)\Big ) \nonumber \\&\quad -\frac{b_n}{\sqrt{1-\rho ^2}} {\text {sign}}(k_2-k^*)\sum _{i=\min (k_2,k^*)+1}^{\max (k_2,k^*)} v(i)\Big |=o_p(1). \nonumber \end{aligned}$$

Proof of Theorem 3.

Theorem 3 is a multivariate version of Theorem 1. We need again to express $\chi ^2(k_1,\ldots ,k_p)$ as a sum of three terms.

We decompose the matrix $\varvec{\Sigma }^{-1}=||d_{jl}||=\varvec{R}\varvec{R}^T$ with $\varvec{R}=||r_{jl}||$ being an upper triangular matrix and introduce vectors $\varvec{Z}=\big (Z_1(i),\ldots ,Z_p(i)\big )^T=\varvec{R}^T\varvec{X}(i)$ and $\varvec{v}(i)=(v_1(i),\ldots ,v_p(i)\big )^T=\varvec{R}^T\varvec{e}(i)$ with uncorrelated components. The estimates $\widehat{k}_1,\ldots ,\widehat{k}_p$ maximize

$$\begin{aligned} \chi ^2(k_1,\ldots ,k_p)=\widetilde{\varvec{Z}}^T \varvec{B}^{-1} \widetilde{\varvec{Z}}, \end{aligned}$$

where $\widetilde{\varvec{Z}}=\left( r_{11}\sum _{i=1}^{k_1}Z_1(i)+\cdots +r_{1p}\sum _{i=1}^{k_1}Z_p(i),\ldots ,r_{pp}\sum _{i=1}^{k_p} Z_p(i)\right) ^T$ and the matrix $\varvec{B}=\varvec{B}(\varvec{k})=||k_{jl}d_{jl}||$ with $k_{jl}=\min (k_j,k_l)$. Denoting $\widetilde{\varvec{m}}=E \widetilde{\varvec{Z}}= \left( \widetilde{m}_1(k_1),\ldots ,\widetilde{m}_p(k_p)\right) ^T$ and $\widetilde{\varvec{v}}=\left( \widetilde{v}_1(k_1),\ldots ,\widetilde{v}_p(k_p)\right) ^T$, where $\widetilde{v}_1(k_1)=r_{11}\sum _{i=1}^{k_1} v_1(i) +\cdots + r_{1p}\sum _{i=1}^{k_1} v_p(i)$, ...,$\widetilde{v}_p(k_p)=r_{pp}\sum _{i=1}^{k_p} v_p(i)$, we may express

$$\begin{aligned} \chi ^2(k_1,\ldots ,k_p)=A_1(k_1,\ldots ,k_p)+2\,A_2(k_1,\ldots ,k_p)+A_3(k_1,\ldots ,k_p), \end{aligned}$$

with $A_1(k_1,\ldots ,k_p)=\widetilde{\varvec{v}}^T\varvec{B}^{-1}\widetilde{\varvec{v}}$, $A_2(k_1,\ldots ,k_p)=\widetilde{\varvec{m}}^T\varvec{B}^{-1}\widetilde{\varvec{v}}$, $A_3(k_1,\ldots ,k_p)=\widetilde{\varvec{m}}^T\varvec{B}^{-1}\widetilde{\varvec{m}}$.

We notice that for $|B_{qq}(k_1,\ldots ,k_q)|=\det (\,||k_{jl}\,d_{jl}||_{j,l=1}^q\,)$, $q=1,\ldots ,p$, it holds that

$$\begin{aligned} \min _{\begin{array}{c} {[\beta n]} \le k_1 \le n\\ \vdots \\ {[\beta n]} \le k_q \le n\\ \end{array}} |B_{qq}(k_1,\ldots ,k_q)| \ge const\, n^q. \end{aligned}$$

We may express $\varvec{B}=\varvec{U}^T\varvec{U}$, where $\varvec{U}=||u_{jl}||$ is an upper triangle matrix with $u_{jj}^2=|B_{jj}|/|B_{j-1,j-1}|.$ An inverse $\varvec{B}^{-1}=\varvec{U}^{-1}\big (\varvec{U}^T\big )^{-1}$ where $\big (\varvec{U}^T\big )^{-1}=||t_{jl}(k_1,\ldots ,k_p)||$ is a lower triangular matrix. From the algorithm of the Cholesky decomposition we obtain that all $t_{jl}(k_1,\ldots ,k_p)$, $1 \le j \le p$, $1 \le l \le j$, may be expressed in the form

$$\begin{aligned} t_{jl}(k_1,\ldots ,k_l)=\frac{P_{jl}^h(k_1,\ldots ,k_p)}{\sqrt{P_{jl}^d(k_1,\ldots ,k_p)}}, \end{aligned}$$

where $P_{jl}^h(k_1,\ldots ,k_p)$ and $P_{jl}^d(k_1,\ldots ,k_p)$ are both polynomials of $k_1,\ldots ,k_p$ such that ${\text {deg}}(P_{jl}^d) \ge 2\,{\text {deg}}(P_{jl}^h)+1$, where ${\text {deg}}(P_{jl}^h)$ is a degree of the polynomial $P_{jl}^h(k_1,\ldots ,k_p)$ and ${\text {deg}}(P_{jl}^d)$ is a degree of the polynomial $P_{jl}^d(k_1,\ldots ,k_p)$. Moreover, $P_{jl}^d$ is a product of $|B_{11}|^{q_1},\ldots , |B_{pp}|^{q_p}$ for some $q_1,\ldots , q_p \in \{0\} \cup \varvec{N}$.

It follows that there exists a constant C such that for $k_1,k_1^*,\ldots ,k_p,k^*_p \in [\beta n,n]$ and all $j=1,\ldots ,p$, $l=1,\ldots ,j$ it holds

$$\begin{aligned}&|t_{jl}(k_1,\ldots ,k_p)| \le \frac{C}{\sqrt{n}},\\&|t_{jl}(k_1,\ldots ,k_p)-t_{jl}(k^*_1,\ldots ,k^*_p)| \le \frac{C}{n^{3/2}} \Big (|k_1-k_1^*|+\cdots +|k_p-k^*_p|\Big ). \end{aligned}$$

For proving a consistency of the estimates $\widehat{k}_1,\ldots ,\widehat{k}_p$ we prove Lemma A.2 for a multidimensional case. The components of the vectors $\varvec{Z}(i),i=1,\ldots ,n$, may be expressed as

$$\begin{aligned} \begin{array}{llll} &{}\phantom {:} Z_1(i) = \beta _{11} +\, v_1(i), &{}&{} \quad i=1,\ldots , k^*_1,\, \\ &{}\phantom {:} Z_1(i) = \, v_1(i), &{}&{} \quad i=k^*_1+1,\ldots , n, \nonumber \\ &{}\phantom {:} \vdots &{}&{} \\ &{}\phantom {:} Z_p(i) = \beta _{p1} + \, v_p(i), &{}&{} \quad i=1,\ldots , k^*_1,\, \nonumber \nonumber \\ &{}\phantom {:} \vdots &{}&{} \\ &{}\phantom {:} Z_p(i) = \beta _{pp}+ v_p(i),&{}&{} \quad i=k^*_{p-1}+1,\ldots , k^*_p, \nonumber \\ &{}\phantom {:} Z_p(i) = \, v_p(i), &{}&{} \quad i=k^*_p+1,\ldots , n, \end{array} \end{aligned}$$

where the parameters $\{\beta _{jl}, j=1,\ldots ,p,\,l=1,\ldots ,j\}$ are connected by $p(p-1)/2$ linear constrains. Denote $RSS_w(k_1,\ldots ,k_p)$ an unrestricted minimum with respect to all $\beta _{11},\ldots ,\beta _{pp}$ of the sum of squares

$$\begin{aligned} \sum _{i=1}^{k_1} \big (Z_1(i)-\beta _{11}\big )^2+\cdots +\sum _{i=1}^{k_1} \big (Z_p(i)-\beta _{11}\big )^2+\cdots + \sum _{i=k_{p-1}+1}^{k_p} \big (Z_p(i)-\beta _{pp}\big )^2 \end{aligned}$$

(A-16)

and denote $RSS(k_1,\ldots ,k_p)$ a restricted minimum of (A-16) under the condition that those $p(p-1)/2$ linear constrains are fulfilled. We denote

$$\begin{aligned} \chi ^2_w(k_1,\ldots ,k_p)=\sum _{j=1}^p \sum _{i=1}^n Z_j^2(i) - RSS_w. \end{aligned}$$

For all $k_1,\ldots ,k_p$

$$\begin{aligned} \chi ^2(k_1,\ldots ,k_p)=\chi ^2_w(k_1,\ldots ,k_p)-d^2(k_1,\ldots ,k_p), \end{aligned}$$

(A-17)

where $d^2(k_1,\ldots ,k_p)$ attains non-negative values. For all $k_1,\ldots , k_p$ we may express the difference

$$\begin{aligned} \chi ^2(k_1^*,\ldots ,k_p^*)-\chi ^2(k_1,\ldots ,k_p)&=\chi _w^2(k_1^*,\ldots ,k_p^*)-\chi _w^2(k_1,\ldots ,k_p)-\\&\quad -\big (d^2(k_1^*,\ldots ,k_p^*)-d^2(k_1,\ldots ,k_p)\big ). \end{aligned}$$

Denote $A_{3w}(k_1,\ldots ,k_p)$ the value of $\chi ^2_w(k_1,\ldots ,k_p)$ computed for the sequences $\{E Z_j(i), 1 \le i \le n\}$, $j=1,\ldots ,p$, instead for the sequences $\{Z_j(i),1 \le i \le n\}$, $j=1,\ldots ,p$, i.e. put all $v_j(i)=0$, $1 \le j \le p$, $1 \le i \le n$. Similarly indeed, $A_3(k_1,\ldots ,k_p)$ is the value of $\chi ^2(k_1,\ldots ,k_p)$ for all $v_j(i)=0$, $1 \le j \le p$, $1 \le i \le n$. In agreement with (A-17) we may write

$$\begin{aligned} A_3(k_1,\ldots ,k_p)=A_{3w}(k_1,\ldots ,k_p)-d_3^2(k_1,\ldots ,k_p), \end{aligned}$$

where $d_3^2(k_1,\ldots ,k_p)$ attains non-negative values. For the model (14) the considered constrains are fulfilled and therefore $d_3^2(k_1^*,\ldots ,k_p^*)=0$ which yields

$$\begin{aligned} A_3(k_1^*,\ldots ,k_p^*)-A_3(k_1,\ldots ,k_p) \ge A_{3w}(k_1^*,\ldots ,k_p^*)-A_{3w}(k_1,\ldots ,k_p). \end{aligned}$$

It may be easily proved that

$$\begin{aligned} A_{3w}(k_1^*,\ldots ,k_p^*)-A_{3w}(k_1,\ldots ,k_p) \ge const\,\Delta _{wn}^2 \big (|k_1-k_1^*|+\cdots +|k_p-k_p^*|\big ) \end{aligned}$$

with $\Delta _{wn}^2=\beta _{11}^2+(\beta _{22}-\beta _{21})^2+\cdots +(\beta _{pp}-\beta _{p,p-1})^2+\beta _{pp}^2.$ As we may find a constant C such that $a_1^2+\cdots +a_p^2 \le C\,\Delta _{wn}^2$, we get the assertion of Lemma A.2:

$$\begin{aligned} A_3(k_1^*,\ldots ,k_p^*)-A_3(k_1,\ldots ,k_p) \ge C\,(a_1^2+\cdots +a_p^2) \left( |k_1-k_1^*| + \cdots + |k_p-k_p^*|\right) . \end{aligned}$$

The assertions of Lemma A.1, A.3, A.4, A.5, A.6 hold true as

$$\begin{aligned} A_1(k_1,\ldots ,k_p)&=\left( t_{11}\widetilde{v}_1(k_1)\right) ^2+\cdots +\left( t_{p1}\widetilde{v}_1(k_1)+\cdots +t_{pp}\widetilde{v}_p(k_p)\right) ^2,\\ A_2(k_1,\ldots ,k_p)&=\left( t_{11}\widetilde{v}_1(k_1)\right) \left( t_{11}\widetilde{m}_1(k_1)\right) +\cdots +\\&\quad +\left( t_{p1}\widetilde{v}_1(k_1)+\cdots +t_{pp}\widetilde{v}_p(k_p)\right) \left( t_{p1}\widetilde{m}_1(k_1)+\cdots +t_{pp}\widetilde{m}_p(k_p)\right) . \end{aligned}$$

Finally, we prove Lemma A.7. The assertion of Theorem 3 is then an easy consequence.

For any $C>0$ we consider $k_1,\ldots ,k_p$ satisfying $|k_1-k_1^*| \le C/\Delta _n^2, \ldots , |k_p-k_p^*| \le C/\Delta _n^2$. It is supposed that $k_1^*=[n\tau _1^*],\ldots ,k_p^*=[n\tau _p^*]$ and $\tau _1^*< \cdots < \tau _p^*$, and therefore $k_1< k_2<\cdots < k_p$ for large enough n. As any $k_j$ may be situated on both sides of $k_j^*$ there exist $2^p$ mutual positions of $k_1,\ldots ,k_p$ and $k_1^*,\ldots , k_p^*$, e.g. $k_1 \le k_1^*< k_2 \le k_2^*< \cdots < k_p \le k_p^*$ or $k_1^*< k_1< k_2 \le k_2^* < \cdots k_p \le k_p^*$ etc. In what follows we prove Lemma A.7 for $k_1 \le k_1^*< k_2 \le k_2^*< \cdots < k_p \le k_p^*$ and $|k_1-k_1^*| \le C/\Delta _n^2 \cap \cdots \cap |k_p-k_p^*| \le C/\Delta _n^2.$ For all other positions one may proceed similarly.

Denote $\varvec{a}=(a_1,\ldots ,a_p)^T$. Recall that $\varvec{\Sigma }^{-1}=\varvec{R}\varvec{R}^T$ where $\varvec{R}$ is an upper triangular matrix. Let $\varvec{r}_j=(0,\ldots ,0,r_{jj},\ldots , r_{jp})$ be the j-th row of $\varvec{R}$, $\varvec{R}_1=\varvec{R}$ and $\varvec{R}_j$, $j=2,\ldots ,p$, be a matrix arising from the matrix $\varvec{R}$ by replacing the first $j-1$ rows by zero rows. We may express

$$\begin{aligned} \varvec{B}= & {} \big (k_1\varvec{R}_1\varvec{R}_1^T+(k_2-k_1)\varvec{R}_2\varvec{R}_2^T+\cdots + (k_p-k_{p-1})\varvec{R}_p\varvec{R}_p^T\big ),\\ \varvec{B}^*= & {} \big (k_1^*\varvec{R}_1\varvec{R}_1^T+(k_2^*-k_1^*)\varvec{R}_2\varvec{R}_2^T+\cdots + (k_p^*-k_{p-1}^*)\varvec{R}_p\varvec{R}_p^T\big ). \end{aligned}$$

Then $A_3(k_1,\ldots ,k_p)=\varvec{a}^T\varvec{F}^T\varvec{B}^{-1}\varvec{F}\varvec{a}$ with $\varvec{F}=\varvec{B}+(k_2^*-k_1)\varvec{R}_2\varvec{R}_1^T+(k_1-k_1^*)\varvec{R}_2\varvec{R}_2^T+\cdots +(k_{p-1}-k_{p-1}^*)\varvec{R}_p\varvec{R}_p^T$, so that

$$\begin{aligned}&A_3(k_1,\ldots ,k_p)-A_3(k_1^*,\ldots ,k_p^*)=(k_1^*-k_1)\varvec{a}^T\big (-\varvec{R}_1\varvec{R}_1^T- \varvec{R}_2\varvec{R}_2^T+\varvec{R}_1\varvec{R}_2^T \\&\qquad + \varvec{R}_2\varvec{R}_1^T\big )\varvec{a}+ (k_2^*-k_2)\varvec{a}^T\big (-\varvec{R}_2\varvec{R}_2^T- \varvec{R}_3\varvec{R}_3^T+\varvec{R}_2\varvec{R}_3^T + \varvec{R}_3\varvec{R}_2^T\big )\varvec{a}+\cdots \\&\qquad +(k_p^*-k_p)\varvec{a}^T\big (-\varvec{R}_p\varvec{R}_p^T\big )\varvec{a}+o(1)\\&\quad =-(k_1^*-k_1)a_1^2d_{11}-\cdots -(k_p^*-k_p)a_p^2d_{pp}+o(1),\\&A_2(k_1,\ldots ,k_p)-A_2(k_1^*,\ldots ,k_p^*)= - \varvec{a}^T\big (\varvec{R}_1-\varvec{R}_2\big )\sum _{i=k_1+1}^{k_1^*} \varvec{v}(i)\\&\qquad -\varvec{a}^T\big (\varvec{R}_2-\varvec{R}_3\big )\sum _{i=k_2+1}^{k_2^*}\varvec{v}(i)-\cdots - \varvec{a}^T\big (\varvec{R}_{p-1}-\varvec{R}_p\big )\sum _{i=k_p+1}^{k_p^*}\varvec{v}(i)+o_P(1)\\&\quad =-a_1\varvec{r}_1 \sum _{i=k_1+1}^{k_1^*} \varvec{v}(i)-a_2\varvec{r}_2 \sum _{i=k_2+1}^{k_2^*} \varvec{v}(i)-\cdots - a_p\varvec{r}_p \sum _{i=k_p+1}^{k_p^*} \varvec{v}(i) + o_P(1)\\&\quad =-a_1\sqrt{d_{11}}\sum _{i=k_1+1}^{k_1^*} \frac{\varvec{r}_1\varvec{v}(i)}{\sqrt{d_{11}}}-\cdots - a_p\sqrt{d_{pp}}\sum _{i=k_p+1}^{k_p^*} \frac{\varvec{r}_p\varvec{v}(i)}{\sqrt{d_{pp}}} + o_P(1). \end{aligned}$$