Testing structural changes in panel data with small fixed panel size and bootstrap

Abstract

Panel data of our interest consist of a moderate or relatively large number of panels, while the panels contain a small number of observations. This paper establishes testing procedures to detect a possible common change in means of the panels. To this end, we consider a ratio type test statistic and derive its asymptotic distribution under the no change null hypothesis. Moreover, we prove the consistency of the test under the alternative. The main advantage of such an approach is that the variance of the observations neither has to be known nor estimated. On the other hand, the correlation structure is required to be calculated. To overcome this issue, a bootstrap technique is proposed in the way of a completely data driven approach without any tuning parameters. The validity of the bootstrap algorithm is shown. As a by-product of the developed tests, we introduce a common break point estimate and prove its consistency. The results are illustrated through a simulation study. An application of the procedure to actuarial data is presented.

Appendices

Appendix 1: Supporting theorems

Suppose that $\{\varvec{\xi }_n\}_{n=1}^{\infty }$ is a sequence of random variables/vectors existing on a probability space $(\varOmega ,\mathcal {F},\mathsf {P})$. A bootstrap version of $\varvec{\xi }\equiv [\varvec{\xi }_1,\ldots ,\varvec{\xi }_n]^{\top }$ is its (randomly) resampled sequence with replacement—denoted by $\varvec{\xi }^*\equiv [\varvec{\xi }_1^*,\ldots ,\varvec{\xi }_n^*]^{\top }$—with the same length, where for each $i\in \{1,\ldots ,n\}$ it holds that $\mathsf {P}_{\varvec{\xi }}^*[\varvec{\xi }_i^*=\varvec{\xi }_j]\equiv \mathsf {P}[\varvec{\xi }_i^*=\varvec{\xi }_j|\varvec{\xi }] =1/n,\,j=1,\ldots ,n$. In the sequel, $\mathsf {P}_{\varvec{\xi }}^*$ denotes the conditional probability given ${\varvec{\xi }}$. So, $\varvec{\xi }_i^*$ has a discrete uniform distribution on $\{\varvec{\xi }_1,\ldots ,\varvec{\xi }_n\}$ for every $i=1,\ldots ,n$. The conditional expectation and variance given ${\varvec{\xi }}$ are denoted by $\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}$ and $\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}$.

If a statistic has an approximate normal distribution, one may be interested in the asymptotic comparison of the bootstrap distribution with the original one. A tool for assessing such an approximate closeness can be a bootstrap central limit theorem for triangular arrays.

Theorem 6

(Bootstrap CLT for triangular arrays) Let $\{\xi _{n,k_n}\}_{n=1}^{\infty }$ be a triangular array of zero mean random variables on the same probability space such that the elements of the vector $[\xi _{n,1},\ldots ,\xi _{n,k_n}]^{\top }$ are iid for every $n\in \mathbb {N}$ satisfying

$$\begin{aligned} \sup _{n\in \mathbb {N}}\mathsf {E}_{\mathsf {P}}\xi _{n,1}^4<\infty \end{aligned}$$

(5)

and $k_n\rightarrow \infty$ as $n\rightarrow \infty$. Suppose that $\varvec{\xi }^*\equiv [\xi _{n,1}^*,\ldots ,\xi _{n,k_n}^*]^{\top }$ is the bootstrapped version of $\varvec{\xi }\equiv [\xi _{n,1},\ldots ,\xi _{n,k_n}]^{\top }$ and denote

$$\begin{aligned} \bar{\xi }_n:=k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i},\quad \bar{\xi }_n^*:=k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}^*, \quad \text{ and }\quad \varsigma _n^2:=\mathsf {Var}\,_{\mathsf {P}}\xi _{n,1}. \end{aligned}$$

If

$$\begin{aligned} \liminf _{n\rightarrow \infty }\varsigma _n^2=\varsigma ^2>0, \end{aligned}$$

(6)

then

$$\begin{aligned} \sup _{x\in \mathbb {R}}\left| \mathsf {P}_{\varvec{\xi }}^*\left[ \frac{\sqrt{k_n}}{\sqrt{\varsigma _n^2}} \left( \bar{\xi }_n^*-\bar{\xi }_n\right) \le x\right] -\mathsf {P}\left[ \frac{\sqrt{k_n}}{\sqrt{\varsigma _n^2}}\bar{\xi }_n\le x\right] \right| \xrightarrow [n\rightarrow \infty ]{\mathsf {P}}0. \end{aligned}$$

Theorem 7

(Bootstrap multivariate CLT for triangular arrays) Let $\{\varvec{\xi }_{n,k_n}\}_{n=1}^{\infty }$ be a triangular array of zero mean $q$-dimensional random vectors on the same probability space such that the elements of the vector sequence $\{\varvec{\xi }_{n,1},\ldots ,\varvec{\xi }_{n,k_n}\}$ are iid for every $n\in \mathbb {N}$ satisfying

$$\begin{aligned} \sup _{n\in \mathbb {N}}\mathsf {E}_{\mathsf {P}}|\xi _{n,1}^{(j)}|^4<\infty ,\quad j\in \{1,\ldots ,q\}, \end{aligned}$$

(7)

where $\varvec{\xi }_{n,1}\equiv [\xi _{n,1}^{(1)},\ldots ,\xi _{n,1}^{(q)}]^{\top }\in \mathbb {R}^q,\,n\in \mathbb {N}$ and $k_n\rightarrow \infty$ as $n\rightarrow \infty$. Assume that $\varvec{\varXi }^*\equiv [\varvec{\xi }_{n,1}^*,\ldots ,\varvec{\xi }_{n,k_n}^*]^{\top }$ is the bootstrapped version of $\varvec{\varXi }\equiv [\varvec{\xi }_{n,1},\ldots ,\varvec{\xi }_{n,k_n}]^{\top }$. Denote

$$\begin{aligned} \bar{\varvec{\xi }}_n:=k_n^{-1}\sum _{i=1}^{k_n}\varvec{\xi }_{n,i},\quad \bar{\varvec{\xi }}_n^*:=k_n^{-1}\sum _{i=1}^{k_n}\varvec{\xi }_{n,i}^*, \quad \text{ and }\quad \varvec{\varGamma }_n:=\mathsf {Var}\,_{\mathsf {P}}\varvec{\xi }_{n,1}. \end{aligned}$$

If

$$\begin{aligned} \liminf _{n\rightarrow \infty }\varvec{\varGamma }_n=\varvec{\varGamma }>\mathbf {0}, \end{aligned}$$

(8)

then

$$\begin{aligned} \mathsf {P}_{\varvec{\varXi }}^*\left[ \sqrt{k_n}\varvec{\varGamma }_n^{-1/2}\left( \bar{\varvec{\xi }}_n^*-\bar{\varvec{\xi }}_n\right) \le \mathbf {x}\right] -\mathsf {P}\left[ \sqrt{k_n}\varvec{\varGamma }_n^{-1/2}\bar{\varvec{\xi }}_n\le \mathbf {x}\right] \xrightarrow [n\rightarrow \infty ]{\mathsf {P}}0,\quad \forall \mathbf {x}\in \mathbb {R}^q. \end{aligned}$$

Appendix 2: Proofs

Proof (of Theorem 1)

Let us define

$$\begin{aligned} U_N(t):=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{s=1}^t(Y_{i,s}-\mu _i). \end{aligned}$$

Using the multivariate Lindeberg-Lévy CLT for a sequence of $T$-dimensional iid random vectors $\{[\sum _{s=1}^1\varepsilon _{i,s},\ldots , \sum _{s=1}^T\varepsilon _{i,s}]^{\top }\}_{i\in \mathbb {N}}$, we have under $H_0$

$$\begin{aligned}{}[U_N(1),\ldots ,U_N(T)]^{\top }\xrightarrow [N\rightarrow \infty ]{\fancyscript{D}}[X_1,\ldots ,X_T]^{\top }, \end{aligned}$$

since $\mathsf {Var}\,[\sum _{s=1}^1\varepsilon _{1,s},\ldots ,\sum _{s=1}^T\varepsilon _{1,s}]^{\top }=\varvec{\varLambda }$. Indeed, the $t$-th diagonal element of the covariance matrix $\varvec{\varLambda }$ is

$$\begin{aligned} \mathsf {Var}\,\sum _{s=1}^t\varepsilon _{1,s}=r(t) \end{aligned}$$

and the upper off-diagonal element on position $(t,v)$ is

$$\begin{aligned} \mathsf {Cov}\,\left( \sum _{s=1}^t\varepsilon _{1,s},\sum _{u=1}^v\varepsilon _{1,u}\right)&=\mathsf {Var}\,\sum _{s=1}^t\varepsilon _{1,s}+\mathsf {Cov}\,\left( \sum _{s=1}^t\varepsilon _{1,s},\sum _{u=t+1}^v\varepsilon _{1,u}\right) \\&=r(t)+R(t,v),\quad t<v. \end{aligned}$$

Moreover, let us define the reverse analogue to $U_N(t)$, i.e.,

$$\begin{aligned} V_N(t):=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{s=t+1}^T(Y_{i,s}-\mu _i)=U_N(T)-U_N(t). \end{aligned}$$

Hence,

$$\begin{aligned} U_{N}(s)-\frac{s}{t}U_{N}(t)&=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\left\{ \sum _{r=1}^s\left[ \left( Y_{i,r}-\mu _i\right) -\frac{1}{t}\sum _{v=1}^t \left( Y_{i,v}-\mu _i\right) \right] \right\} \\&=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{r=1}^s\left( Y_{i,r}-\bar{Y}_{i,t}\right) \end{aligned}$$

and, consequently,

$$\begin{aligned} V_{N}(s)\!-\!\frac{T\!-\!s}{T\!-\!t}V_{N}(t)&=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N \left\{ \sum _{r=s+1}^T\left[ \left( Y_{i,r}\!-\!\mu _i\right) \!-\!\frac{1}{T\!-\!t}\sum _{v=t+1}^T \left( Y_{i,v}\!-\!\mu _i\right) \right] \right\} \\&=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{r=s+1}^T\left( Y_{i,r}-\widetilde{Y}_{i,t}\right) . \end{aligned}$$

Using the Cramér–Wold device, we end up with

$$\begin{aligned}&\displaystyle \max _{t=2,\ldots ,T-2}\frac{\max _{s=1,\ldots ,t}\left| U_{N}(s)-\frac{s}{t}U_{N}(t)\right| }{\max _{s=t,\ldots ,T-1}\left| V_{N}(s)-\frac{T-s}{T-t}V_{N}(t)\right| }\\&\displaystyle \xrightarrow [N\rightarrow \infty ]{\fancyscript{D}}\max _{t=2,\ldots ,T-2}\frac{\max _{s=1,\ldots ,t}\left| X_s-\frac{s}{t}X_t\right| }{\max _{s=t,\ldots ,T-1}\left| (X_T-X_s)-\frac{T-s}{T-t}(X_T-X_t)\right| }. \end{aligned}$$

$\square$

Proof (of Theorem 2)

Let $t=\tau +1$. Then, under alternative $H_1$

$$\begin{aligned}&\frac{1}{\sigma \sqrt{N}}\max _{s=1,\ldots ,\tau +1}\left| \sum _{i=1}^N\left[ \sum _{r=1}^s\left( Y_{i,r}-\bar{Y}_{i,\tau +1}\right) \right] \right| \nonumber \\&\quad \ge \frac{1}{\sigma \sqrt{N}}\left| \sum _{i=1}^N\sum _{r=1}^{\tau }\left( Y_{i,r}-\bar{Y}_{i,\tau +1}\right) \right| \\&\quad =\frac{1}{\sigma \sqrt{N}}\left| \sum _{i=1}^N\sum _{r=1}^{\tau }\left( \mu _i+\sigma \varepsilon _{i,r}-\frac{1}{\tau +1}\sum _{v=1}^{\tau +1}(\mu _i+\sigma \varepsilon _{i,v})-\frac{1}{\tau +1}\delta _i\right) \right| \\&\quad =\frac{1}{\sqrt{N}}\left| \sum _{i=1}^N\sum _{r=1}^{\tau }\left( \varepsilon _{i,r}-\bar{\varepsilon }_{i,\tau +1}\right) -\frac{\tau }{\sigma (\tau +1)}\sum _{i=1}^N\delta _i\right| \\&\quad =\mathcal {O}_{\mathsf {P}}(1)+\frac{\tau }{\sigma (\tau +1)\sqrt{N}}\left| \sum _{i=1}^N\delta _i\right| \xrightarrow []{\mathsf {P}}\infty ,\quad N\rightarrow \infty , \end{aligned}$$

where $\bar{\varepsilon }_{i,\tau +1}=\frac{1}{\tau }\sum _{v=1}^{\tau +1}\varepsilon _{i,v}$.

Since there is no change after $\tau +1$ and $\tau \le T-3$, then by Theorem 1 we have

$$\begin{aligned} \frac{1}{\sigma \sqrt{N}}\max _{s=\tau +1,\ldots ,T\!-\!1} \left| \sum _{i=1}^N\sum _{r=s+1}^T\left( Y_{i,r}\!-\! \widetilde{Y}_{i,\tau +1}\right) \right| \xrightarrow [N\rightarrow \infty ]{\fancyscript{D}} \max _{s=\tau +1,\ldots ,T\!-\!1} \left| Z_s\!-\!\frac{T\!-\!s}{T\!-\!\tau }Z_{\tau +1}\right| . \end{aligned}$$

$\square$

Proof (of Theorem 3)

Let us define $S_N^{(i)}(t):=\frac{1}{t}\sum _{s=1}^t(Y_{i,s}-\bar{Y}_{i,t})^2$ and, consequently, $S_N(t):=\frac{1}{N}\sum _{i=1}^N S_N^{(i)}(t)$. Then,

$$\begin{aligned} S_N^{(i)}(t)=\left\{ \begin{array}{ll} \frac{\sigma ^2}{t}\sum _{s=1}^t(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})^2, &{}\quad t\le \tau ,\\ \frac{1}{t}\left[ \sum _{s=1}^{\tau }(\sigma \varepsilon _{i,s}-\sigma \bar{\varepsilon }_{i,t}-\frac{t-\tau }{t}\delta _i)^2\right. \\ \quad \left. +\sum _{s=\tau +1}^{t}(\sigma \varepsilon _{i,s}-\sigma \bar{\varepsilon }_{i,t} +\frac{\tau }{t}\delta _i)^2\right] , &{}\quad t>\tau ; \end{array} \right. \end{aligned}$$

where $\bar{\varepsilon }_{i,t}=\frac{1}{t}\sum _{s=1}^t\varepsilon _{i,s}$. By the definition of the cumulative autocorrelation function, we have for $2\le t\le \tau$

$$\begin{aligned} \mathsf {E}S_N^{(i)}(t)&=\frac{\sigma ^2}{t}\sum _{s=1}^t \mathsf {E}(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})^2=\frac{\sigma ^2}{t}\sum _{s=1}^t\left[ 1-\frac{2}{t}\sum _{r=1}^t\mathsf {E}\varepsilon _{i,s} \varepsilon _{i,r}+\frac{1}{t^2}r(t)\right] \\&=\sigma ^2\left( 1-\frac{r(t)}{t^2}\right) . \end{aligned}$$

In the other case when $t>\tau$, one can calculate

$$\begin{aligned} \mathsf {E}S_N^{(i)}(t)&=\sigma ^2\left( 1-\frac{r(t)}{t^2}\right) +\frac{\tau }{t}\left( \frac{t-\tau }{t}\right) ^2\delta _i^2+ \frac{t-\tau }{t}\left( \frac{\tau }{t}\right) ^2\delta _i^2\\&=\sigma ^2\left( 1-\frac{r(t)}{t^2}\right) +\frac{\tau (t-\tau )}{t^2}\delta _i^2. \end{aligned}$$

Realize that $S_N^{(i)}(t)-\mathsf {E}S_N^{(i)}(t)$ are independent with zero mean for fixed $t$ and $i=1,\ldots ,N$. Due to Assumption C2, for $2\le t\le \tau$ it holds

$$\begin{aligned} \mathsf {Var}\,S_N(t)=\frac{1}{N^2}\sum _{i=1}^N\frac{\sigma ^4}{t^2}\mathsf {Var}\,\left[ \sum _{s=1}^t(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})^2\right] = \frac{1}{N}C_1(t,\sigma ), \end{aligned}$$

where $C_1(t,\sigma )>0$ is some constant not depending on $N$. If $t>\tau$, then

$$\begin{aligned} \mathsf {Var}\,S_N(t)&=\frac{1}{N^2}\sum _{i=1}^N\frac{1}{t^2}\mathsf {Var}\,\Bigg [\sigma ^2\sum _{s=1}^{\tau }(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})^2\\&\quad -2\frac{t-\tau }{t}\sigma \delta _i\sum _{s=1}^{\tau }(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})+\left( \frac{t-\tau }{t}\right) ^2\delta _i^2\\&\quad +\sigma ^2\sum _{s=\tau +1}^t(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})^2+2\frac{\tau }{t}\sigma \delta _i\sum _{s=\tau +1}^t(\varepsilon _{i,s}-\bar{\varepsilon }_{i,t})+\left( \frac{\tau }{t}\right) ^2\delta _i^2\Bigg ]\\&\le \frac{1}{N}C_2(t,\tau ,\sigma )+\frac{1}{N^2}C_3(t,\tau ,\sigma )\sum _{i=1}^N\delta _i^2+\frac{1}{N^2}C_4(t,\tau ,\sigma )\left| \sum _{i=1}^N\delta _i\right| , \end{aligned}$$

where $C_j(t,\tau ,\sigma )>0$ does not depend on $N$ for $j=2,3,4$.

The Chebyshev inequality provides $S_N(t)-\mathsf {E}S_N(t)=\mathcal {O}_{\mathsf {P}}\left( \sqrt{\mathsf {Var}\,S_N(t)}\right)$ as $N\rightarrow \infty$. According to Assumption C1 and the Cauchy-Schwarz inequality, we have

$$\begin{aligned} \frac{1}{N^2}\left| \sum _{i=1}^N\delta _i\right| \le \frac{1}{N}\sqrt{\frac{1}{N}\sum _{i=1}^N\delta _i^2}\rightarrow 0,\quad N\rightarrow \infty . \end{aligned}$$

Since the index set $\{1,\ldots ,T\}$ is finite and $\tau$ is finite as well, then

$$\begin{aligned} \max _{1\le t\le T}\mathsf {Var}\,S_N(t)\le \frac{1}{N}K_1(\sigma )+K_2(\sigma )\frac{1}{N^2}\sum _{i=1}^N\delta _i^2+K_3(\sigma )\frac{1}{N^2}\left| \sum _{i=1}^N\delta _i\right| \le \frac{1}{N}K_4(\sigma ), \end{aligned}$$

where $K_j(\sigma )>0$ are constants not depending on $N$ for $j=1,2,3,4$. Thus, we also have uniform stochastic boundedness, i.e.,

$$\begin{aligned} \max _{1\le t\le T}|S_N(t)-\mathsf {E}S_N(t)|=\mathcal {O}_{\mathsf {P}}\left( \frac{1}{\sqrt{N}}\right) ,\quad N\rightarrow \infty . \end{aligned}$$

Adding and subtracting, one has

$$\begin{aligned} S_N(\tau )-S_N(t)&=S_N(\tau )-\mathsf {E}S_N(\tau )-[S_N(t)-\mathsf {E}S_N(t)]+\mathsf {E}S_N(\tau )-\mathsf {E}S_N(t)\\&\ge -2\max _{1\le r\le T}|S_N(r)-\mathsf {E}S_N(r)|+\mathsf {E}S_N(\tau )-\mathsf {E}S_N(t)\\&= -2\max _{1\le r\le T}|S_N(r)-\mathsf {E}S_N(r)|+\sigma ^2\left( \frac{r(t)}{t^2}-\frac{r(\tau )}{\tau ^2}\right) \\&\quad +\mathcal {I}\{t>\tau \}\frac{\tau (t-\tau )}{t^2}\frac{1}{N}\sum _{i=1}^N\delta _i^2. \end{aligned}$$

The above inequality holds for each $t\in \{2,\ldots ,T\}$ and, particularly, it holds for $\widehat{\tau }_N$. Note that $\widehat{\tau }_N=\arg \max _tS_N(t)$. Hence, $S_N(\tau )-S_N(\widehat{\tau }_N)\le 0$. Therefore,

$$\begin{aligned}&2\sqrt{N}\max _{1\le r\le T}|S_N(r)-\mathsf {E}S_N(r)|\nonumber \\&\quad \ge \sqrt{N}\left[ \sigma ^2\left( \frac{r(\widehat{\tau }_N)}{\widehat{\tau }_N^2}-\frac{r(\tau )}{\tau ^2}\right) +\mathcal {I}\{\widehat{\tau }_N>\tau \}\frac{\tau (\widehat{\tau }_N-\tau )}{\widehat{\tau }_N^2}\frac{1}{N}\sum _{i=1}^N\delta _i^2\right] . \end{aligned}$$

(9)

If $\widehat{\tau }_N>\tau$, then the left hand side of (9) is $\mathcal {O}_{\mathsf {P}}(1)$ as $N\rightarrow \infty$, but the right hand side is unbounded because of Assumption C1. So, if $\widehat{\tau }_N\le \tau$, then

$$\begin{aligned} 0\xleftarrow [N\rightarrow \infty ]{\mathsf {P}} 2\max _{1\le r\le T}|S_N(r)-\mathsf {E}S_N(r)|\ge \sigma ^2\left( \frac{r(\widehat{\tau }_N)}{\widehat{\tau }_N^2}-\frac{r(\tau )}{\tau ^2}\right) , \end{aligned}$$

which yields due to the monotonicity of $r(t)/t^2$ that $\mathsf {P}[\widehat{\tau }_N=\tau ]\rightarrow 1$ as $N\rightarrow \infty$. $\square$

Proof (of Theorem 4)

Let us define $\widehat{\epsilon }_{i,t}:=\sigma ^{-1}\sum _{s=1}^t\widehat{e}_{i,s}$, $\widehat{\epsilon }_{i,t}^*:=\sigma ^{-1}\sum _{s=1}^t\widehat{e}_{i,s}^*$,

$$\begin{aligned} \widehat{U}_N(t):=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{s=1}^t\widehat{e}_{i,s}=\frac{1}{\sqrt{N}}\sum _{i=1}^N\widehat{\epsilon }_{i,t}, \end{aligned}$$

and

$$\begin{aligned} \widehat{U}_N^*(t)&:=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{s=1}^t\widehat{Y}_{i,s}^*=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{s=1}^t\left( \widehat{e}_{i,s}^*-\frac{1}{N}\sum _{i=1}^N\widehat{e}_{i,s}\right) \\&=\frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{s=1}^t\left( \widehat{e}_{i,s}^*-\widehat{e}_{i,s}\right) =\frac{1}{\sqrt{N}}\sum _{i=1}^N\left( \widehat{\epsilon }_{i,t}^*-\widehat{\epsilon }_{i,t}\right) . \end{aligned}$$

Realize that $\widehat{\epsilon }_{i,t}$ depends on $\widehat{\tau }_N$ and, hence, it depends on $N$. Thus, $\widehat{\epsilon }_{i,t}\equiv \widehat{\epsilon }_{i,t}(N)$. Since Assumption C2 holds, then according to the bootstrap multivariate CLT for triangular arrays (Theorem 7) of $T$-dimensional vectors $\varvec{\xi }_{N,i}=[\widehat{\epsilon }_{i,1}(N),\ldots ,\widehat{\epsilon }_{i,T}(N)]^{\top }$ with $k_N=N$, we have

$$\begin{aligned}&\mathsf {P}\left[ \varvec{\varGamma }_N^{-1/2}[\widehat{U}_N^*(1),\ldots ,\widehat{U}_N^*(T)]^{\top }\le \mathbf {x}\big |\mathbb {Y}\right] -\mathsf {P}\left[ \varvec{\varGamma }_N^{-1/2}[\widehat{U}_N(1),\ldots ,\widehat{U}_N(T)]^{\top }\le \mathbf {x}\right] \\&\xrightarrow [N\rightarrow \infty ]{\mathsf {P}}0,\quad \forall \mathbf {x}\in \mathbb {R}^T, \end{aligned}$$

where $\varvec{\varGamma }_N=\mathsf {Var}\,[\widehat{\epsilon }_{i,1},\ldots , \widehat{\epsilon }_{i,T}]^{\top }$.

Now, it is sufficient to realize that $[\widehat{U}_N(1),\ldots ,\widehat{U}_N(T)]^{\top }$ has an approximate multivariate normal distribution with zero mean and covariance matrix $\varvec{\varGamma }=\lim _{N\rightarrow \infty }\varvec{\varGamma }_N$. Using the law of total variance,

$$\begin{aligned} \mathsf {Var}\,\widehat{\epsilon }_{i,t}&=\mathsf {E}[\mathsf {Var}\,\{\widehat{\epsilon }_{i,t}|\widehat{\tau }_N\}]+\mathsf {Var}\,[\mathsf {E}\{\widehat{\epsilon }_{i,t}|\widehat{\tau }_N\}]\\&=\sum _{\pi =1}^T\mathsf {P}[\widehat{\tau }_N=\pi ]\mathsf {Var}\,[\widehat{\epsilon }_{i,t}|\widehat{\tau }_N=\pi ]+\sum _{\pi =1}^T\mathsf {P}[\widehat{\tau }_N=\pi ]\{\mathsf {E}[\widehat{\epsilon }_{i,t}|\widehat{\tau }_N=\pi ]\}^2\\&\quad -\left\{ \sum _{\pi =1}^T\mathsf {P}[\widehat{\tau }_N=\pi ]\mathsf {E}[\widehat{\epsilon }_{i,t}|\widehat{\tau }_N=\pi ]\right\} ^2. \end{aligned}$$

Since $\lim _{N\rightarrow \infty }\mathsf {P}[\widehat{\tau }_N=\tau ]=1$ and $\mathsf {E}[\widehat{e}_{i,t}|\widehat{\tau }_N=\tau ]=0$, then

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathsf {Var}\,\widehat{\epsilon }_{i,t}= \lim _{N\rightarrow \infty }\mathsf {Var}\,[\widehat{\epsilon }_{i,t}|\widehat{\tau }_N=\tau ]. \end{aligned}$$

Similarly with the covariance, i.e., after applying the law of total covariance, we have

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathsf {Cov}\,\left( \widehat{\epsilon }_{i,t},\widehat{\epsilon }_{i,v}\right) =\lim _{N\rightarrow \infty }\mathsf {Cov}\,\left( \widehat{\epsilon }_{i,t}, \widehat{\epsilon }_{i,v}|\widehat{\tau }_N=\tau \right) \!. \end{aligned}$$

Note that

$$\begin{aligned} \left( \widehat{e}_{i,t}|\widehat{\tau }_N=\tau \right) =\left\{ \begin{array}{l@{\quad }l} \sigma (\varepsilon _{i,t}-\bar{\varepsilon }_{i,\tau }),&{} t\le \tau ;\\ \sigma (\varepsilon _{i,t}-\widetilde{\varepsilon }_{i,\tau }),&{} t>\tau ; \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} \bar{\varepsilon }_{i,t}=\frac{1}{t}\sum _{s=1}^t \varepsilon _{i,s}\quad \text{ and }\quad \widetilde{\varepsilon }_{i,t}=\frac{1}{T-t}\sum _{s=t+1}^T \varepsilon _{i,s}. \end{aligned}$$

Taking into account the definitions of $r(t)$, $R(t,v)$, and $S(t,v,d)$ together with some simple algebra, we obtain that $\mathsf {Var}\,[\widehat{\epsilon }_{i,s}|\widehat{\tau }_N=\tau ]=\gamma _{t,t}(\tau )$ and $\mathsf {Cov}\,\left( \widehat{\epsilon }_{i,t},\widehat{\epsilon }_{i,v}|\widehat{\tau }_N=\tau \right) =\gamma _{t,v}(\tau )$ for $t<v$, where the elements $\gamma _{t,t}(\tau )$ and $\gamma _{t,v}(\tau )$ are as in the statement of Theorem 4.

Then the sum in the nominator of $\mathcal {R}_N^*(T)$ can be alternatively rewritten as

$$\begin{aligned} \frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\sum _{r=1}^s\left( \widehat{Y}_{i,r}^*-\bar{\widehat{Y}}_{i,t}^*\right)&= \frac{1}{\sigma \sqrt{N}}\sum _{i=1}^N\left\{ \left[ \sum _{r=1}^s \widehat{Y}_{i,r}^*\right] -\frac{s}{t}\sum _{v=1}^t \widehat{Y}_{i,v}^*\right\} \\&= \widehat{U}_{N}^*(s)-\frac{s}{t}\widehat{U}_{N}^*(t). \end{aligned}$$

Concerning the denominator of $\mathcal {R}_N^*(T)$, one needs to perform a similar calculation as in the proof of Theorem 1 with $V_N(t)$, i.e., to define $\widehat{V}_N(t)$ and $\widehat{V}_N^*(t)$ analogously to $\widehat{U}_N(t)$ and $\widehat{U}_N^*(t)$ as $V_N(t)$ is to $U_N(t)$. Applying the Cramér–Wold theorem completes the proof. $\square$

Proof (of Theorem 5)

Recall the notation from the proof of Theorem 4. Under $H_0$, B2, and C2 it holds

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathsf {P}[\widehat{\tau }_N=T]=1. \end{aligned}$$

Then in view of (4),

$$\begin{aligned} \lim _{N\rightarrow \infty }\mathsf {P}\left[ \widehat{U}_N(s) -\frac{s}{t}\widehat{U}_N(t)=U_N(s)-\frac{s}{t}U_N(t)\right] =1,\quad 1\le s\le t\le T. \end{aligned}$$

$\square$

Proof (of Theorem 6)

The Lyapunov condition (Billingsley 1986, [p. 371]) for a triangular array of random variables $\{\xi _{n,k_n}\}_{n=1}^{\infty }$ is satisfied due to (5) and (6), i.e., for $\omega =2$:

$$\begin{aligned} \frac{1}{\sqrt{k_n\varsigma _n^2}^{2+\omega }}\sum _{i=1}^{k_n}\mathsf {E}| \xi _{n,i}|^{2+\omega }\le \frac{k_n^{-\omega /2}}{\varsigma _n^{2+\omega }} \sup _{\iota \in \mathbb {N}}\mathsf {E}|\xi _{\iota ,1}|^{2+\omega }\rightarrow 0,\quad n\rightarrow \infty . \end{aligned}$$

Therefor, the CLT for $\{\xi _{n,k_n}\}_{n=1}^{\infty }$ holds and

$$\begin{aligned} \sup _{x\in \mathbb {R}}\left| \mathsf {P}\left[ \frac{\sqrt{k_n}}{\sqrt{\varsigma _n^2}}\bar{\xi }_n\le x\right] -\int _{-\infty }^x\frac{1}{\sqrt{2\pi }}\exp \left\{ -\frac{t^2}{2}\right\} \text{ d }t\right| \xrightarrow [n\rightarrow \infty ]{}0. \end{aligned}$$

Now, to prove the theorem, it suffices to show the following three statements:

• (i) $\sup _{x\in \mathbb {R}}\left| \mathsf {P}_{\varvec{\xi }}^*\left[ \frac{\sqrt{k_n}}{\sqrt{\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*}}\left( \bar{\xi }_n^*-\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\bar{\xi }_n^*\right) \le x\right] \!-\!\int _{-\infty }^x\frac{1}{\sqrt{2\pi }}\exp \left\{ -\frac{t^2}{2}\right\} \text{ d }t\right| \xrightarrow [n\rightarrow \infty ]{\mathsf {P}}0$;

• (ii) $\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*-\varsigma _n^2\xrightarrow [n\rightarrow \infty ]{\mathsf {P}}0$;

• (iii) $\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\bar{\xi }_n^*=\bar{\xi }_n,\, [\mathsf {P}]-a.s.$

Proving (iii) is trivial, because $\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\bar{\xi }_n^*=\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*=k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}=\bar{\xi }_n,\, [\mathsf {P}]$-$a.s.$

Let us calculate the conditional variance of the bootstrapped variable $\xi _{n,1}^*$: $\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*=\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^{*2}-(\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*)^2=k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}^2-\left( k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}\right) ^2,\, [\mathsf {P}]$-$a.s.$ The weak law of large numbers together with (5) provides

$$\begin{aligned} \bar{\xi }_n-n^{-1}\sum _{i=i}^{k_n}\mathsf {E}_{\mathsf {P}}\xi _{n,i}=\bar{\xi }_n\xrightarrow [n\rightarrow \infty ]{\mathsf {P}}0 \end{aligned}$$

and

$$\begin{aligned} 0\xleftarrow [n\rightarrow \infty ]{\mathsf {P}}k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}^2-\left( k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}\right) ^2-\mathsf {E}_{\mathsf {P}}\xi _{n,1}^2=\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*-\varsigma _n^2. \end{aligned}$$

The last result of the WLLN is true, because (5) implies

$$\begin{aligned} k_n^{-2}\sum _{i=1}^{k_n}\mathsf {Var}\,_{\mathsf {P}}\xi _{n,i}^2\le k_n^{-2}\sum _{i=1}^{k_n}\mathsf {E}_{\mathsf {P}}\xi _{n,i}^4\le k_n^{-1}\sup _{\iota \in \mathbb {N}} \mathsf {E}_{\mathsf {P}}\xi _{\iota ,1}^4\xrightarrow [n\rightarrow \infty ]{} 0. \end{aligned}$$

Thus (ii) is proved.

The Berry–Esseen–Katz theorem (see Katz 1963) with $g(x)=|x|^{\epsilon },\,\epsilon >0$ for the bootstrapped sequence of $iid$ (with respect to $\mathsf {P}^*$) random variables $\{\xi _{n,i}^*\}_{i=1}^{k_n}$ results in

$$\begin{aligned}&\sup _{x\in \mathbb {R}}\left| \mathsf {P}_{\varvec{\xi }}^*\left[ \frac{\sqrt{k_n}}{\sqrt{\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*}}\left( \bar{\xi }_n^*-\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\bar{\xi }_n^*\right) \le x\right] -\int _{-\infty }^x\frac{1}{\sqrt{2\pi }}\exp \left\{ -\frac{t^2}{2}\right\} \text{ d }t\right| \nonumber \\&\quad \le Ck_n^{-\epsilon /2}\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\left| \frac{\xi _{n,1}^*-\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*}{\mathsf {Var}\,_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*}\right| ^{2+\epsilon }\quad [\mathsf {P}]-a.s., \end{aligned}$$

(10)

for all $n\in \mathbb {N}$ where $C>0$ is an absolute constant.

The Jensen inequality and Minkowski inequality provide an upper bound for the nominator from the right-hand side of (10):

$$\begin{aligned} \mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}|\xi _{n,1}^*-\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}\xi _{n,1}^*|^{2+\epsilon }&=k_n^{-1}\sum _{i=1}^{k_n}\left| \xi _{n,i}-k_n^{-1}\sum _{j=1}^{k_n}\xi _{n,j}\right| ^{2+\epsilon }\\&\le k_n^{-1}\left\{ \left( \sum _{i=1}^{k_n}|\xi _{n,i}|^{2+\epsilon }\right) ^{1/(2+\epsilon )}+k_n^{-(1+\epsilon )/(2+\epsilon )}\left| \sum _{j=1}^{k_n}\xi _{n,j}\right| \right\} ^{2+\epsilon }\\&\le 2^{1+\epsilon }k_n^{-1}\sum _{i=1}^{k_n}|\xi _{n,i}|^{2+\epsilon }+2^{1+\epsilon }\left| k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}\right| ^{2+\epsilon }\quad [\mathsf {P}]-a.s. \end{aligned}$$

The right-hand side of the previously derived upper bound is uniformly bounded in probability $\mathsf {P}$, because of Markov’s inequality and (5). Indeed, for fixed $\eta >0$

$$\begin{aligned} \mathsf {P}\left[ k_n^{-1}\sum _{i=1}^{k_n}|\xi _{n,i}|^{2+\epsilon } \ge \eta \right]&\le \eta ^{-1}k_n^{-1}\sum _{i=1}^{k_n}\mathsf {E}_{\mathsf {P}}|\xi _{n,i}|^{2+\epsilon }\\&\le \eta ^{-1}\sup _{\iota \in \mathbb {N}}\mathsf {E}_{\mathsf {P}}|\xi _{\iota ,1}|^{2+\epsilon }<\infty ,\quad \forall n\in \mathbb {N} \end{aligned}$$

and

$$\begin{aligned} \mathsf {P}\left[ \left| k_n^{-1}\sum _{i=1}^{k_n}\xi _{n,i}\right| \ge \eta \right]&\le \eta ^{-1}k_n^{-1}\mathsf {E}_{\mathsf {P}}\left| \sum _{i=1}^{k_n}\xi _{n,i}\right| \\&\le \eta ^{-1}\sup _{\iota \in \mathbb {N}}\mathsf {E}_{\mathsf {P}}|\xi _{\iota ,1}|<\infty ,\quad \forall n\in \mathbb {N}. \end{aligned}$$

Since $\mathsf {E}_{\mathsf {P}_{\varvec{\xi }}^*}|\xi _{n,1}^*-\mathsf {E}_{\mathsf {P}^*}\xi _{n,1}^*|^{2+\epsilon }$ is bounded in probability $\mathsf {P}$ uniformly over $n$ and the denominator of the right-hand side of (10) is uniformly bounded away from zero due to (6), then the left-hand side of (10) converges in probability $\mathsf {P}$ to zero as $n$ tends to infinity. So, (i) is proved as well. $\square$

Proof (of Theorem 7)

According to the Cramér–Wold theorem, it is sufficient to ensure that all assumptions of one-dimensional bootstrap CLT (6) for triangular arrays are valid for any linear combination of the elements of the random vector $\varvec{\xi }_{n,1},\,n\in \mathbb {N}$.

For arbitrary fixed $\mathbf {t}\in \mathbb {R}^q$ using the Jensen inequality, we get

$$\begin{aligned} \sup _{n\in \mathbb {N}}\mathsf {E}_{\mathsf {P}}|\mathbf {t}^{\top }\varvec{\xi }_{n,1}|^4\le q^3\sup _{n\in \mathbb {N}}\sum _{j=1}^q t_j^4\mathsf {E}_{\mathsf {P}}|\xi _{n,1}^{(j)}|^4\le q^4\max _{j=1,\ldots ,q}t_j^4\sup _{n\in \mathbb {N}}\mathsf {E}_{\mathsf {P}}|\xi _{n,1}^{(j)}|^4<\infty . \end{aligned}$$

Hence, assumption (7) implies assumption (5) for the random variables $\{\mathbf {t}^{\top }\varvec{\xi }_{n,k_n}\}_{n\in \mathbb {N}}$.

Similarly, assumption (8) implies assumption (6) for such an arbitrary linear combination, i.e., positive definiteness of the matrix $\varvec{\varGamma }$ yields

$$\begin{aligned} \liminf _{n\rightarrow \infty }\mathsf {Var}\,_{\mathsf {P}}\mathbf {t}^{\top }\varvec{\xi }_{n,1}=\liminf _{n\rightarrow \infty }\mathbf {t}^{\top }\left( \mathsf {Var}\,_{\mathsf {P}}\varvec{\xi }_{n,1}\right) \mathbf {t}\ge \mathbf {t}^{\top }\left( \liminf _{n\rightarrow \infty }\varvec{\varGamma }_n\right) \mathbf {t}=\mathbf {t}^{\top }\varvec{\varGamma }\mathbf {t}>0. \end{aligned}$$

$\square$