Detection of multiple change-points in high-dimensional panel data with cross-sectional and temporal dependence

Abstract

We consider the detection of multiple change-points in a high-dimensional time series exhibiting both cross-sectional and temporal dependence. Several test statistics based on the celebrated CUSUM statistic are used and discussed. In particular, we propose a novel block wild bootstrap method to address the presence of cross-sectional and temporal dependence. Furthermore, binary segmentation and the moving sum algorithm are considered to detect and locate multiple change-points. We also give some theoretical justifications for the moving sum method. An extensive numerical study provides insights on the performance of the proposed methods. Finally, our proposed procedures are used to analyze financial stock data in Korea.

Appendix A: Technical results and their proofs

In this appendix, we generalize a result in Leadbetter et al. (2012). Before we formally state our result, the following paragraph attempts to clarify how our result complements those in Leadbetter et al. (2012). Leadbetter et al. (2012) consider stationary Gaussian processes $\{\xi (t)\}_{0 \le t \le T}$ with covariance function $r(v) = \mathbb {E}(\xi (t)\xi (t+v))$ which is assumed to satisfy

$$\begin{aligned} r(v) = 1 - \lambda |v|^{\alpha } + o(|v|^{\alpha }) \hspace{0.2cm} \text { as } \hspace{0.2cm} v \rightarrow 0, \end{aligned}$$

(A.1)

for some $\alpha \in (0,2]$. Chapter 12.3 in Leadbetter et al. (2012) establishes convergence results for the supremum over an increasing interval $M(T) = \sup _{0 \le t \le T} \xi (t)$. One aims to find functions a(T), b(T) such that

$$\begin{aligned} \textrm{P}(a(T)(M(T)-b(T)) \le x) \rightarrow \exp \left( - e^{-x} \right) , \hspace{0.2cm} \text { as } \hspace{0.2cm} T \rightarrow \infty . \end{aligned}$$

However, results for multivariate Gaussian processes are only established for the case $\alpha =2$ in (A.1); see Chapter 11.2 in Leadbetter et al. (2012). For our results, we need to consider multivariate Gaussian processes in the case $\alpha = 1$.

Let $\{{\varvec{\xi }}(t) = (\xi _{1}(t),\dots ,\xi _{p}(t))' \}_{0 \le t \le T}$ be a jointly Gaussian process with zero means, variances one and covariance function $r_{ij}(v) = \textrm{Cov}(\xi _{i}(t),\xi _{j}(t+v))$. In order to establish results for the joint probability of the suprema $M_{k}(T) = \sup _{0 \le t \le T} \xi _{k}(t)$, we suppose

$$\begin{aligned} r_{ii}(v) =1 - \lambda _{k} |v| + o(|v|) \hspace{0.2cm} \text { as } \hspace{0.2cm} v \rightarrow 0, \end{aligned}$$

(A.2)

$$\begin{aligned} r_{ij}(v) \log (v) \rightarrow 0, \hspace{0.2cm} \text { as } \hspace{0.2cm} v \rightarrow \infty , \end{aligned}$$

(A.3)

$$\begin{aligned} \max _{i \ne j} \sup _{v} |r_{ij}(v)| < 1. \end{aligned}$$

(A.4)

The final result can then be stated as follows.

Theorem A.1

Let $u_{k} = u_{k}(T) \rightarrow \infty$ as $T \rightarrow \infty$, so that

$$\begin{aligned} T\mu _{k}:= T \lambda _{k} \frac{1}{\sqrt{2\pi }} \exp \left( - \frac{u_{k}^2}{2}\right) \rightarrow \tau _{k} >0, \hspace{0.2cm} 1 \le k \le p \end{aligned}$$

(A.5)

and suppose that the jointly stationary Gaussian process $\{{\varvec{\xi }}(t)\}_{0 \le t \le T}$ satisfies (A.2)–(A.4). Then,

$$\begin{aligned} \textrm{P}(M_{k}(T) \le u_{k}, 1 \le k \le p) \rightarrow \exp \left( - \sum _{k=1}^{p}\tau _{k} \right) , \hspace{0.2cm} \text { as } \hspace{0.2cm} T \rightarrow \infty . \end{aligned}$$

Most of the required results and techniques to prove Theorem A.1 are already provided in Leadbetter et al. (2012). For completeness, we present its proof here. The proof is based on a series of lemmas which provide an approximation of the probability of suprema over an increasing interval. Following the notation in Leadbetter et al. (2012), we introduce for a fixed $h>0$, $n = [T/h]$.

Proof

By Lemma 2 below

$$\begin{aligned} \textrm{P}(M_{k}(nh) \le u_{k}, k =1,\dots , p) - \textrm{P}^{n}(M_{k}(h) \le u_{k}, k=1,\dots ,p) \rightarrow 0. \end{aligned}$$

Then, since $nh \le T \le (n+1)h$, it follows

$$\begin{aligned} \textrm{P}(M_{k}(T) \le u_{k}, k=1,\dots ,p)&= \textrm{P}^{n}(M_{k}(h) \le u_{k}, k=1,\dots ,p) + o(1) \nonumber \\&= \left( 1 - \sum _{k=1}^{p} \textrm{P}(M_{k}(h) > u_{k}) + o\left( \sum _{k=1}^{p} \mu _{k} \right) \right) ^{n} \end{aligned}$$

(A.6)

$$\begin{aligned}&= \left( 1 - \sum _{k=1}^{p} (\mu _{k} h + o(\mu _{k})) + o\left( \sum _{k=1}^{p} \mu _{k} \right) \right) ^{n} \end{aligned}$$

(A.7)

$$\begin{aligned}&\rightarrow \exp \left( - \sum _{k=1}^{p}\tau _{k} \right) , \end{aligned}$$

(A.8)

where (A.6) follows by (A.23) in Lemma 5 and (A.7) by Theorem 12.2.9 in Leadbetter et al. (2012). Furthermore, the last relation (A.8) is valid since by (A.5) and $n = [T/h]$ we get $\mu _{k} \sim \frac{\tau _{k}}{T}$ and $n \sim \frac{T}{h}$. $\square$

The following Lemma 2 is crucial in proving Theorem A.1. Its proof will be given by a series of lemmas below the statement.

Lemma 2

Suppose (A.2)–(A.4). Then,

$$\begin{aligned} \textrm{P}(M_{k}(nh) \le u_{k}, k =1,\dots , p) - \textrm{P}^{n}(M_{k}(h) \le u_{k}, k=1,\dots ,p) \rightarrow 0. \end{aligned}$$

Proof

The result follows by Lemmas 3 and 4 below. $\square$

Recall that for a fixed $h>0$, $n = [T/h]$, and divide [0, nh] into n intervals of length h. We further split those intervals into subintervals $I_{r}$ and $I^{*}_{r}$ of length $h-\varepsilon$ and $\varepsilon$, respectively. The following lemma corresponds to Lemma 12.3.2 in Leadbetter et al. (2012).

Lemma 3

Suppose $u \rightarrow \infty$, $q \rightarrow 0$, $u^2q \rightarrow a>0$, (A.2) and (A.5) hold. Then,

$$\begin{aligned} 0\le & {} \textrm{P}\left( M_{k} \left( \bigcup _{r=1}^{n} I_{r} \right) \le u_{k}, k =1,\dots , p \right) - \textrm{P}(M_{k}(nh) \le u_{k}, k =1,\dots , p ) \nonumber \\\le & {} \frac{\varepsilon }{h} \sum _{k=1}^{p} \tau _{k}, \end{aligned}$$

(A.9)

$$\begin{aligned} 0\le & {} \textrm{P}\left( \xi _{k}(jq) \le u_{k}, jq \in \bigcup _{r=1}^{n} I_{r}, k =1,\dots , p \right) \nonumber \\{} & {} \quad - \textrm{P}\left( M_{k} \left( \bigcup _{r=1}^{n} I_{r} \right) \le u_{k}, k =1,\dots , p \right) \le \rho (a) \sum _{k=1}^{p} \tau _{k} \end{aligned}$$

(A.10)

with $\tau _{k}$ as defined in (A.5) and $\rho (a) \rightarrow 0$ as $a \rightarrow 0$.

Proof

We consider (A.9) and (A.10) separately.

Proof of (A.9): By Boole’s inequality

$$\begin{aligned} 0&\le \textrm{P}\left( M_{k} \left( \bigcup _{r=1}^{n} I_{r} \right) \le u_{k}, k =1,\dots , p \right) - \textrm{P}(M_{k}(nh) \le u_{k}, k =1,\dots , p ) \nonumber \\ {}&\le n \textrm{P}(M_{k}(I_{1}^{*})> u_{k}, k =1,\dots , p ) \nonumber \\ {}&\sim n \sum _{k=1}^{p} \textrm{P}(M_{k}(I_{1}^{*}) > u_{k} ) \nonumber \\ {}&= n \sum _{k=1}^{p} \mu _{k} \varepsilon \rightarrow \sum _{k=1}^{p} \tau _{k} \frac{\varepsilon }{h}, \end{aligned}$$

(A.11)

where (A.11) follows under assumption (A.2) by Theorem 12.2.9 in Leadbetter et al. (2012) and since $n \mu _{k} \sim T \frac{\mu _{k}}{h} \rightarrow \frac{\tau _{k}}{T}$.

Proof of (A.10): Following the proof of Lemma 12.3.2 (ii) in Leadbetter et al. (2012), we get

$$\begin{aligned} 0&\le \textrm{P}\left( \xi _{k}(jq) \le u_{k}, jq \in \bigcup _{r=1}^{n} I_{r}, k =1,\dots , p \right) - \textrm{P}\left( M \left( \bigcup _{r=1}^{n} I_{r} \right) \le u_{k}, k =1,\dots , p \right) \nonumber \\ {}&\le n \max _{r=1,\dots ,n} \left( \textrm{P}\left( \xi _{k}(jq) \le u_{k}, jq \in I_{r}, k =1,\dots , p \right) - \textrm{P}\left( M_{k} \left( I_{r} \right) \le u_{k}, k =1,\dots , p \right) \right) \nonumber \\ {}&\le n \max _{r=1,\dots ,n} \big ( \textrm{P}(\xi _{k}(0)> u_{k}, k =1,\dots , p ) + \textrm{P}\left( \xi _{k}(jq) \le u_{k}, jq \in [0,h], k =1,\dots , p \right) \nonumber \\ {}&\hspace{1cm}- \textrm{P}\left( M_{k}(h) \le u_{k}, k =1,\dots , p \right) \big ) \nonumber \\ {}&= n \max _{r=1,\dots ,n} \big ( \textrm{P}(\xi _{k}(0)> u_{k}, k =1,\dots , p ) - \textrm{P}\left( \xi _{k}(jq)> u_{k}, jq \in [0,h], k =1,\dots , p \right) \nonumber \\ {}&\hspace{1cm}+ \textrm{P}\left( M_{k}(h) > u_{k}, k =1,\dots , p \right) \big ), \end{aligned}$$

(A.12)

$$\begin{aligned}&\le n \left( (1-H_{1}(a)) h \sum _{k=1}^{p} \mu _{k} + o\left( \sum _{k=1}^{p} \mu _{k} \right) \right) , \end{aligned}$$

(A.13)

where $H_{1}(a)$ is a constant depending on the limit $u^2q \rightarrow a>0$ and satisfying $\rho (a):= 1-H_{1}(a) \rightarrow 0$ as $a \rightarrow 0$; see the proof of Lemma 12.3.2 in Leadbetter et al. (2012). Each probability in (A.12) can be dealt with separately to get (A.13). With explanations given below, we get for the first probability in (A.12)

$$\begin{aligned} \begin{gathered} \textrm{P}(\xi _{k}(0)> u_{k}, k =1,\dots , p ) \le \sum _{k=1}^{p} \textrm{P}(\xi _{k}(0) > u_{k}) \le \frac{1}{\sqrt{2\pi }} \sum _{k=1}^{p} \frac{1}{u_{k}} \exp \left( - \frac{u_{k}^{2}}{2} \right) = o\left( \sum _{k=1}^{p} \mu _{k} \right) , \end{gathered}\nonumber \\ \end{aligned}$$

(A.14)

where the second inequality in (A.14) is a consequence of $1-\Phi (u) \sim \varphi (u)/u$, where $\Phi$ and $\varphi$ denote the Gaussian distribution and density functions respectively. The last relation in (A.14) follows by (12.2.18) in Leadbetter et al. (2012).

For the second probability in (A.12), one can write

$$\begin{aligned} \textrm{P}\left( \xi _{k}(jq)> u_{k}, jq \in [0,h], k =1,\dots , p \right)&\le \textrm{P}\left( \max _{jq \in [0,h]} \xi _{k}(jq)> u_{k}, k =1,\dots , p \right) \nonumber \\ {}&\le \sum _{k=1}^{p} \textrm{P}\left( \max _{jq \in [0,h]} \xi _{k}(jq) > u_{k} \right) \nonumber \\ {}&= H_{1}(a) h \sum _{k=1}^{p} \mu _{k} + o\left( \sum _{k=1}^{p} \mu _{k} \right) , \end{aligned}$$

(A.15)

where (A.15) follows by Lemma 12.2.4 (i) in Leadbetter et al. (2012) with $\alpha =1$.

Similarly, the third probability in (A.12) satisfies,

$$\begin{aligned} \begin{gathered} \textrm{P}\left( M_{k}(h)> u_{k}, k =1,\dots , p \right) \le \sum _{k=1}^{p} \textrm{P}\left( M_{k}(h) > u_{k} \right) = h \sum _{k=1}^{p} \mu _{k} + o\left( \sum _{k=1}^{p} \mu _{k} \right) , \end{gathered} \end{aligned}$$

by (12.2.18) in Leadbetter et al. (2012). $\square$

The following lemma corresponds to Lemma 12.3.3 in Leadbetter et al. (2012).

Lemma 4

Let $r(v) \rightarrow 0$ as $v \rightarrow \infty$, $u^2q \rightarrow a>0$, (A.2), (A.3) and (A.5) be satisfied. Then, as $T \rightarrow \infty$,

$$\begin{aligned}{} & {} \textrm{P}\left( \xi _{k}(jq) \le u_{k}, jq \in \bigcup _{r=1}^{n} I_{r}, k =1,\dots , p \right) -\prod _{r=1}^{n} \textrm{P}(\xi _{k}(jq) \nonumber \\{} & {} \le u_{k}, jq \in I_{r}, k =1,\dots , p ) \rightarrow 0, \end{aligned}$$

(A.16)

$$\begin{aligned} \begin{aligned}&\limsup _{u \rightarrow \infty } \left| \prod _{r=1}^{n} \textrm{P}(\xi _{k}(jq) \le u_{k}, jq \in I_{r}, k =1,\dots , p ) - \textrm{P}^{n}(M(h) \le u_{k}, k =1,\dots , p ) \right| \\ {}&\le \left( \rho (a)+\frac{\varepsilon }{h}\right) \sum _{k=1}^{p} \tau _{k}. \end{aligned} \end{aligned}$$

(A.17)

Proof

We prove (A.16) and (A.17) separately.

Proof of (A.16): Follows by the proof of relation (11.2.4) in Leadbetter et al. (2012) which is part of the proof of Lemma 11.2.1.

Proof of (A.17): Simple application of triangular inequality yields

$$\begin{aligned}&\left| \prod _{r=1}^{n} \textrm{P}(\xi _{k}(jq) \le u_{k}, jq \in I_{r}, k =1,\dots , p ) - \textrm{P}^{n}(M(h) \le u_{k}, k =1,\dots , p ) \right| \nonumber \\ {}&\le \left| \prod _{r=1}^{n} \textrm{P}(\xi _{k}(jq) \le u_{k}, jq \in I_{r}, k =1,\dots , p ) - \prod _{r=1}^{n} \textrm{P}(M_{k}(I_{r}) \le u_{k}, k =1,\dots , p ) \right| \nonumber \\ {}&\hspace{1cm}+ \left| \prod _{r=1}^{n} \textrm{P}(M_{k}(I_{r}) \le u_{k}, k =1,\dots , p ) - \textrm{P}^{n}(M(h) \le u_{k}, k =1,\dots , p ) \right| . \nonumber \\ \end{aligned}$$

(A.18)

The first probability in (A.18) can be bounded as

$$\begin{aligned} 0&\le \prod _{r=1}^{n} \textrm{P}(\xi _{k}(jq) \le u_{k}, jq \in I_{r}, k =1,\dots , p ) - \prod _{r=1}^{n} \textrm{P}(M_{k}(I_{r}) \le u_{k}, k =1,\dots , p ) \nonumber \\ {}&\le n \max _{r =1,\dots ,n}( \textrm{P}(\xi _{k}(jq) \le u_{k}, jq \in I_{r}, k =1,\dots , p ) - \textrm{P}(M_{k}(I_{r}) \le u_{k}, k =1,\dots , p ) ) \nonumber \\ {}&\le n\left( (1-H_{1}(a)) h \sum _{k=1}^{p} \mu _{k} + o\left( \sum _{k=1}^{p} \mu _{k} \right) \right) , \nonumber \\ \end{aligned}$$

(A.19)

where we used for (A.19) the same arguments as in the proof of (A.10). The second probability in (A.18) satisfies

$$\begin{aligned} 0&\le \prod _{r=1}^{n} \textrm{P}(M_{k}(I_{r}) \le u_{k}, k =1,\dots , p ) - \textrm{P}^{n}(M(h) \le u_{k}, k =1,\dots , p ) \nonumber \\ {}&= \textrm{P}^{n}(M_{k}(I_{1}) \le u_{k}, k =1,\dots , p ) - \textrm{P}^{n}(M(h) \le u_{k}, k =1,\dots , p ) \end{aligned}$$

(A.20)

$$\begin{aligned}&\le n \left( \textrm{P}(M_{k}(I_{1}) \le u_{k}, k =1,\dots , p ) - \textrm{P}(M(h) \le u_{k}, k =1,\dots , p ) \right) \nonumber \\ {}&\le n \textrm{P}(M_{k}(I_{1})> u_{k}, k =1,\dots , p ) \nonumber \\ {}&\le n \sum _{k=1}^{p} \textrm{P}(M_{k}(I_{1}) > u_{k}) \nonumber \\ {}&\sim n \sum _{k=1}^{p} \mu _{k} \varepsilon \rightarrow \sum _{k=1}^{p} \tau _{k} \frac{\varepsilon }{h}, \end{aligned}$$

(A.21)

where (A.20) is due to stationarity and the relation (A.21) follows under assumption (A.2) by Theorem 12.2.9 in Leadbetter et al. (2012). $\square$

Lemma 5

Suppose (A.4). Then,

$$\begin{aligned} \textrm{P}(M_{k}(h)> u_{k}, M_{l}(h) > u_{l}) = o(\mu _{k} + \mu _{l}) \hspace{0.2cm} \text { for } \hspace{0.2cm} k \ne l, \nonumber \\ \end{aligned}$$

(A.22)

$$\begin{aligned} \textrm{P}(M_{k}(h) \le u_{k}, k=1,\dots , p) = 1 - \sum _{k=1}^{p} \textrm{P}(M_{k}(h) > u_{k}) + o\left( \sum _{k=1}^{p} \mu _{k} \right) . \nonumber \\ \end{aligned}$$

(A.23)

Proof

The statement coincides with Lemma 11.2.2 in Leadbetter et al. (2012) and can be proven without assuming (A.2). Therefore, we omit the details. Note that (A.22) is only necessary to prove (A.23). $\square$