Block bootstrapping for a panel mean break test

Abstract

We consider block bootstrappings for panel mean change test of the squared CUSUM test of Horváth and Hušková (J Time Ser Anal 33:631–648, 2012): the circular block bootstrapping and stationary bootstrapping. First order asymptotic null validity of the test is proved under serial and/or cross-sectional correlation. Consistency of the test under an alternative hypothesis is also proved. A Monte-Carlo experiment reveals that the existing tests of Horváth and Hušková (2012) and others have severe size distortions for serially and/or cross-sectionally correlated panels, and the block bootstrappings remedy this size distortion problem. A real data analysis illustrates the proposed method.

Proofs

Proof of Theorem 1

Under $H_0$, we have, $Z_{i,T}(z) = \frac{1}{{\hat{\sigma }}_i\sqrt{T}} \sum _{t=1}^{[Tz]} (X_{i,t} - {\bar{X}}_i) = \frac{1}{{\hat{\sigma }}_i \sqrt{T}} (\sum _{t=1}^{[Tz]}e_{i,t} - \frac{[Tz]}{T}\sum _{t=1}^Te_{i,t}),$ and the result holds by (C1)–(C2) and the continuous mapping theorem. $\square$

Lemma A.1

Under (C1), (C2) and (B1), (B2), both for $Z^{*}_T=Z_T^{*CB}$ and $Z_T^*=Z_T^{*SB}$, under $H_0$, as $T\rightarrow \infty$, $Z_T^{*}(\cdot ) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} {\hat{\Sigma }}^{-1/2} B^e(\cdot ),$ where $Z_T^{*}(z) = [Z_{1,T}^{*}(z), \ldots , Z_{N,T}^{*}(z)]'$, $Z_{i,T}^{*}(z) = \frac{1}{\sqrt{T}} \sum _{t=1}^{[Tz]} \frac{(X_{i,t}^* - {\bar{X}}_i)}{{\hat{\sigma }}_i^*}$.

Proof

We first consider $Z_T^*=Z_T^{*CB}$. Let $X_t^*$ be the circular block bootstrap sample $X_t^{*CB},~t=1,\ldots ,T$. Let $L=L_{CB}$. Let ${\hat{u}}_t = {\hat{\Sigma }}^{-1/2}(X_t - {\bar{X}})$ and let ${\hat{u}}_t^* ={\hat{\Sigma }}^{-1/2}(X_t^*-{\bar{X}})$. Define the sum in the j-th block ${\hat{U}}_{I_j,L} = \sum _{s=0}^{L-1} {\hat{u}}_{I_j+s}$. Let

$$\begin{aligned} S_T^{*}(z) =\frac{1}{\sqrt{T}} \sum _{t=1}^{[Tz]} {\hat{u}}_{t}^*,~~ {\hat{W}}_T^*(z) = \frac{1}{\sqrt{T}} \sum _{j=1}^{m_{[Tz]}-1} {\hat{U}}_{I_j,L},~~ R_T^*(z) = \frac{1}{\sqrt{T}} \sum _{t=L(m_{[Tz]}-1)}^{[Tz]-L(m_{[Tz]}-1)} {\hat{u}}_t. \end{aligned}$$

Then $S_T^*(z) = {\hat{W}}_T^*(z) + R_T^*(z),~ 0<z<1$.

Note that ${\hat{W}}_T^*(z)$ is partial sum process of a triangular array

$$\begin{aligned} {\hat{U}}_{I_1,L},\ldots , {\hat{U}}_{I_{m_{[Tz]}},L}, \text { which are conditionally i.i.d. with}~E^*({\hat{U}}_{I_j,L})=0 \end{aligned}$$

(4)

and, by (14) of Politis and Romano (1994) and $L^{-1} + \frac{L}{T} = o(1)$ of (B1),

$$\begin{aligned} L^{-1} Var^*({\hat{U}}_{I_j,L}) = {L}^{-1}{T}^{-1} \sum _{t=1}^T({\hat{U}}_{t,L}{\hat{U}}_{t,L}') = {\hat{\Omega }}^u {\mathop {\longrightarrow }\limits ^\mathrm{p}} \Omega ^u, \end{aligned}$$

(5)

where ${\hat{\Omega }}^u = {\hat{\gamma }}^u(0) +\sum _{h=-L}^L ( 1- \frac{|h|}{L}) {\hat{\gamma }}^u(h)$, ${\hat{\gamma }}^u(h) = T^{-1}\sum _{t=h+1}^T ( {\hat{u}}_t - {\bar{u}}) ({\hat{u}}_{t-h} - {\bar{u}})',$${\hat{\gamma }}^u(-h) = {\hat{\gamma }}^u(h)'$, $h \ge 0$, ${\bar{u}} = \frac{1}{T} \sum _{t=1}^T {\hat{u}}_t$. Let ${\hat{\Sigma }}^*=diag( {\hat{\sigma }}_1^{*2},\ldots , {\hat{\sigma }}_N^{*2})$. We will show

$$\begin{aligned} \sup _z ||R_T^{*}(z)|| {\mathop {\longrightarrow }\limits ^\mathrm{p^*}} 0,~~ {\hat{W}}^*_T(z) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} {\hat{\Sigma }}^{-1/2} B^e(\cdot ),~~{\hat{\Sigma }}^*-{\hat{\Sigma }} {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0. \end{aligned}$$

Then, we have $\sup _z||Z_T^*(z)-S_T^*(z)||=o_p(1),$ and hence have the result.

We first show $\sup _z ||R_T^{*}(z)|| {\mathop {\longrightarrow }\limits ^\mathrm{p^*}} 0$. For given $\varepsilon >0$, applying the Markov inequality, we have the first result because, by (B1),

$$\begin{aligned} \begin{aligned} P^*[\sup _z||R_T^{*}(z)||>\varepsilon ]&= P^*\left[ \sup _z \frac{1}{\sqrt{T}} ||\sum _{s=[Tz] -L(m_{[Tz]}-1)}^{L-1} {\hat{u}}_{ I_{m_{[Tz]}}+s} ||>\varepsilon \right] \\&\le ~P^*\left[ \frac{L}{\sqrt{T}} \max _t ||{\hat{u}}_{t}||>\varepsilon \right] \le \frac{L}{\sqrt{T}\varepsilon } E^*[\max _t ||{\hat{u}}_{t}||] \\&= \frac{L}{\sqrt{T}\varepsilon }\max _t||{\hat{\Sigma }}^{-1/2}(e_t-{\bar{e}}_T)|| =o_p(1),~~{\bar{e}}_T=\frac{1}{T}\sum _{t=1}^Te_t. \end{aligned} \end{aligned}$$

We second show ${\hat{W}}_T^*(z) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} {\hat{\Sigma }}^{-1/2} B^e(\cdot )$. This invariance principle for ${\hat{W}}_T^*(z)$ is obtained by applying Theorem 13.5 on p. 142 of Billingsley (1999) for which it suffices to show (6) and (7) below: for any integer $k>0$, $0 \le z_1< z_2< \cdots <z_k \le 1$,

$$\begin{aligned} \begin{aligned} {[{\hat{W}}_T^*(z_1), \ldots , {\hat{W}}_T^*(z_k)]}^{'} {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} N_k(0, \Omega ^z),~\text { in probability}, \end{aligned} \end{aligned}$$

(6)

where $\Omega ^z = diag(z_1, \ldots , z_k)\otimes {\tilde{\Omega }}^u,$${\tilde{\Omega }}^u={\hat{\Sigma }}^{-1/2}\Omega ^e{\hat{\Sigma }}^{-1/2}$ and $\otimes$ is the Kronecker product; and for $0 \le z<r<s\le 1$, for some $0<C< \infty$ not depending on T, z, r, s, conditionally on the data $X_{1}, \ldots , X_{T}$,

$$\begin{aligned} \begin{aligned} E^*[||{\hat{W}}_T^*(r)-{\hat{W}}_T^*(z)||^2||{\hat{W}}_T^*(s)-{\hat{W}}_T^*(r)||^2] \le C(s-z)^2. \end{aligned} \end{aligned}$$

(7)

The first one (6) is shown for $k=2$ because it straightforwardly extends to k-variate cases:

$$\begin{aligned} \begin{aligned} {[{\hat{W}}_T^*(z_1), {\hat{W}}_T^*(z_2)-{\hat{W}}_T^*(z_1)] }^{'} {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} N\left( 0,\left( {\begin{matrix} z_1 &{}\quad 0 \\ 0&{}\quad z_2-z_1\end{matrix}}\right) \otimes {\tilde{\Omega }}^u\right) , \end{aligned} \end{aligned}$$

(8)

Recall that, given T, ${\hat{U}}_{I_j,L}, j=1,2,\ldots$ are iid having $E^*[{\hat{U}}_{I_j,L}]=0$ and $Var^*[{\hat{U}}_{I_j,L}]=Var^*[{\hat{U}}_{I_1,L}] = L{\hat{\Omega }}^u$. Therefore, for ${\hat{W}}_T^*(z_1) = \frac{1}{\sqrt{T}} \sum _{j=1}^{m_{[Tz_1]}} {\hat{U}}_{I_{j}, L}$, we have $E^*[{\hat{W}}_T^*(z_1)]=0$ and $Var^*[{\hat{W}}_T^*(z_1)] = \frac{m_{[Tz_1]}}{T} L{\hat{\Omega }}^u = {\tilde{\Omega }}^uz_1 + o_p(1)$ and similarly, $E^*[{\hat{W}}_T^*(z_2) - {\hat{W}}_T^*(z_1)] = 0$ and $Var^*[{\hat{W}}_T^*(z_2) - {\hat{W}}_T^*(z_1)] = {\tilde{\Omega }}^u (z_2-z_1) + o_p(1)$, $Cov^*[{\hat{W}}_T^*(z_1), {\hat{W}}_T^*(z_2) - {\hat{W}}_T^*(z_1)] = o_p(1).$ We first establish asymptotic normality for ${\hat{W}}_T^*(z_1)$. Note that $\frac{\sqrt{T}}{\sqrt{Lm}}{\hat{W}}_T^*(z_1) = \frac{1}{\sqrt{Lm}} \sum _{j=1}^m {\hat{U}}_{I_j,L} = \sum _{j=1}^m U_j^*\sqrt{\frac{L}{m}},$$m = m_{[Tz_1]},$ is a sum of conditionally iid triangular array elements $U_j^*\sqrt{L/m}$, where $U_j^* = {\hat{U}}_{I_j,L}/L,~ j=1,\ldots ,m$. By the same argument in the proof of Theorem 3.2 of Lahiri (2003) for asymptotic normality of circular block average, the Linderberg’s condition, as $T \rightarrow \infty$,

$$\begin{aligned} \Delta _T(a) = \frac{L}{m} \sum _{j=1}^m E^*[||U_j^*||^2] I(\sqrt{L}||U_j^*||>2a) {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0 \end{aligned}$$

(9)

is verified for $a = (T/L)^{1/4}$. Applying the same subsequence argument in the proof of Lahiri (2003, p. 57) with (9) and $Var^*[{\hat{W}}_T^*(z_1)] = z_1 {\tilde{\Omega }}^u + o_p(1)$, we get ${\hat{W}}_T^*(z_1) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} N(0, z_1{\tilde{\Omega }}^u)$. Now, we apply the above argument for asymptotic normality of ${\hat{W}}_T^*(z_1)$ to show $\lambda _1 {\hat{W}}_T^*(z_1) + \lambda _2({\hat{W}}_T^*(z_2)-{\hat{W}}_T^*(z_1)) = \lambda _1 \sum _{j=1}^{m_{[Tz_1]}} {\hat{U}}_{I_j,L} + \lambda _2 \sum _{j=m_{[Tz_1]}+1}^{m_{[Tz_2]}} {\hat{U}}_{I_j,L} {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} N(0, \lambda _1^2 {\tilde{\Omega }}^u z_1 + \lambda _2^2{\tilde{\Omega }}^u(z_2-z_1))$. Then, by the Cramer–Wold device, we get (8) and hence (6). We have (7) because, conditionally on $X_1,\ldots ,X_T$,

$$\begin{aligned} \begin{aligned}&E^* [||{\hat{W}}_T^*(r) - {\hat{W}}_T^*(z)||^2||{\hat{W}}_T^*(s)-{\hat{W}}_T^*(r)||^2] \\&\quad =E^* [||{\hat{W}}_T^*(r) - {\hat{W}}_T^*(z)||^2]E^*[||{\hat{W}}_T^*(s)-{\hat{W}}_T^*(r)||^2] \\&\quad = \frac{1}{T^2} E^*\left[ \sum _{j=m_{[Tz]}+1}^{m_{[Tr]}} {\hat{U}}_{I_j,L}' {\hat{U}}_{I_j,L}\right] E^*\left[ \sum _{j=m_{[Tr]}+1}^{m_{[Ts]}} {\hat{U}}_{I_j,L}' {\hat{U}}_{I_j,L}\right] \\&\quad = (r-z)(s-r) \left\{ \frac{E^*\left[ {\hat{U}}_{I_1,L}'{\hat{U}}_{I_1,L}\right] }{L}\right\} ^2 \le C (s-z)^2, \end{aligned} \end{aligned}$$

by $L^{-1}E^*[{\hat{U}}_{I_1,L}'{\hat{U}}_{I_1,L}] = tr(L^{-1} E^*[{\hat{U}}_{I_1,L}{\hat{U}}_{I_1,L}'])=tr({\tilde{\Omega }}^u)$ of (5).

We next consider $Z_T^*=Z_T^{*SB}$. The proof is similar to the above proof under (C1), (C2) and (B1) because modifications of (4)–(5) for stationary bootstrapping also hold as

$$\begin{aligned} {\hat{U}}_{I_j,L_j},~j=1,2,\ldots , \text { which are conditionally i.i.d. with}~E^*({\hat{U}}_{I_j,L_j})=0, \nonumber \\\end{aligned}$$

(10)

$$\begin{aligned} p_{SB} Var^*({\hat{U}}_{I,L}) = {\tilde{\Omega }}^u +o_p(1) {\mathop {\longrightarrow }\limits ^\mathrm{p}} \Omega ^u, \text { by Remark 3 of Politis and Romano (1994)}.\nonumber \\ \end{aligned}$$

(11)

We finally show ${\hat{\Sigma }}^*-{\hat{\Sigma }} \rightarrow 0$. It suffices to show ${\hat{\sigma }}_i^{2*} {\mathop {\longrightarrow }\limits ^\mathrm{p}} {\hat{\sigma }}_i^2.$ Let ${\hat{\gamma }}^*_i(h)$ be the bootstrap versions of ${\hat{\gamma }}_i(h)$. We have the final result because, thanks to (B1),

$$\begin{aligned} \begin{aligned} |{\hat{\sigma }}_i^2 - {\hat{\sigma }}_i^{2*}|&= |K(0)({\hat{\gamma }}_i(0)-{\hat{\gamma }}_i^*(0)) \\&\quad +\,2K\left( \frac{1}{l}\right) ({\hat{\gamma }}_i(1)-{\hat{\gamma }}_i^*(1)) + \cdots + 2K\left( \frac{l}{l}\right) ({\hat{\gamma }}_i(l) - {\hat{\gamma }}_i^*(l))| \\&\le 2l \max _{0 \le h \le l}|{\hat{\gamma }}_i(h) -{\hat{\gamma }}_i^*(h)| = O_p\left( \frac{l\sqrt{lnT}}{\sqrt{T}}\right) = o_p(1) \end{aligned} \end{aligned}$$

by Theorem 3 of An and Chen (1982). $\square$

Proof of Theorem 2

Let $X_t^*$ be the circular block bootstrap sample $X_t^{*CB},~t=1,\ldots ,T$. Let $S_T^{**}(z)=\frac{1}{\sqrt{T}}\sum _{t=1}^{[Tz]}{\hat{u}}_t^{**}$ where ${\hat{u}}_t^{**}={\hat{\Sigma }}^{-1/2}(X_t^*-{\bar{X}}^*),~{\bar{X}}^*=\frac{1}{T}\sum _{t=1}^TX_t^*$. If we show $S_T^{**}(\cdot ) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} {\hat{\Sigma }}^{-1/2}B^{e0}(\cdot ),~B^{e0}(z)=B^e(z) - zB^e(1)$, in probability, then the result follows from the continuous mapping theorem. By Lemma A.1,

$$\begin{aligned} S_T^{**}(z) -S_T^{*}(z)= & {} \frac{1}{\sqrt{T}} \sum _{t=1}^{[Tz]} {\hat{\Sigma }}^{-1/2}({\bar{X}}- {\bar{X}}^*) \\= & {} \frac{[Tz]}{\sqrt{T}}{\hat{\Sigma }}^{-1/2}({\bar{X}}- {\bar{X}}^*) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} -{\hat{\Sigma }}^{-1/2}zB^e(1), \text { in probability}. \end{aligned}$$

Since $S_T^*(z) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} {\hat{\Sigma }}^{-1/2} B^e(z)$, by Lemma A.1, we have $S_T^{**} (z) {\mathop {\longrightarrow }\limits ^\mathrm{d^*}} {\hat{\Sigma }}^{-1/2} (B^e(z) - zB^e(1)) = {\hat{\Sigma }}^{-1/2} B^{e0}(z)$, in probability. We thus have the result because of $({\hat{\Sigma }}^*-{\hat{\Sigma }} {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0)$. $\square$

Proof of Theorem 3

We get the result easily by replacing circular block bootstrapping sample $X^{*}$ with $X^{*SB}$ in Proof of Theorem 2. $\square$

Proof of Theorem 4

It is enough to show, conditionally on $X_1,\ldots ,X_T$, as $T \rightarrow \infty$, (i) $\sup _{0 \le z \le 1}|{V}_{N,T}^*(z)| =O_p(1),$ (ii) $\sup _{0 \le z \le 1} |{V}_{N,T}(z)| {\mathop {\longrightarrow }\limits ^\mathrm{p}} \infty$. We first show (i). Note that, under the alternative hypothesis, conditionally on the sample, $X_{i,t}^*$ do not have mean change because of random block selection. We therefore have (i) by the same arguments as that in the proof of Theorem 2.

We next show (ii). Choose an i such that $\delta _i\ne 0$. It suffices to show that $\sup _{0 \le z \le 1} {Z}_{i,T}^2(z) {\mathop {\longrightarrow }\limits ^\mathrm{p}} \infty .$ We have ${Z}_{i,T}(z) = \frac{1}{{\hat{\sigma }}_i \sqrt{T}} (\sum _{t=1}^{[Tz]}e_{i,t} - \frac{[Tz]}{T} \sum _{t=1}^T e_{i,t}) + \frac{1}{{\hat{\sigma }}_i\sqrt{T}}\lambda _{i,T}(z),$ where $\lambda _{i,T}(z) = -\delta _i[Tz](\frac{T-t_0}{T}),$ if $[Tz] \le t_0\le T$; $\lambda _{i,T}(z) = -\delta _it_0(\frac{T-[Tz]}{T}),$ otherwise. By Theorem 1, we have

$$\begin{aligned} \begin{aligned} {Z}_{i,T}^2(z)&= \left\{ \frac{1}{{\hat{\sigma }}_i}B_i^{e0}(z)\right\} ^2 + \frac{2}{{\hat{\sigma }}_i^{2}\sqrt{T}}B_i^{e0}(z) \lambda _{i,T}(z) + \frac{1}{{\hat{\sigma }}_i^{2}T}\lambda _{i,T}^2(z)+o_p(1)\\&= \frac{A_1+A_2+A_3}{{\hat{\sigma }}_i^{2}}+o_p(1). \end{aligned} \end{aligned}$$

We have $\sup _z A_1=O_p(1)$. The result follows if we show $\frac{1}{\sqrt{T}}\sup _z|A_2|=O_p(1)$, $\liminf _{T\rightarrow \infty }$$\sup _z\frac{A_3}{T}>0$, and $\lim {\hat{\sigma }}_i^{2}/T {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0$. We denote ${\underline{\tau }} := \liminf _{T\rightarrow \infty } \frac{t_0}{T} \text { and } {\overline{\tau }}:= \limsup _{T\rightarrow \infty } \frac{t_0}{T}.$ In case of $[Tz] \le t_0 \le T$, we have

$$\begin{aligned} \frac{1}{T^2} \lambda _{i,T}^2(z) = \frac{1}{T^2} \delta _i^2\left( \frac{[Tz](T-t_0)}{T}\right) ^2 \ge \frac{[Tz]^2}{T^2} (1- {\overline{\tau }})^2 \delta _i^2=D_1(z) \end{aligned}$$

and in case of $0 \le t_0 \le [Tz]$,

$$\begin{aligned} \frac{1}{T^2} \lambda _{i,T}^2(z) = \frac{1}{T^2} \delta _i^2 \left( \frac{t_0(T-[Tz])}{T}\right) ^2 \ge \frac{(T-[Tz])^2}{T^2} {\underline{\tau }}^2 \delta _i^2=D_2(z). \end{aligned}$$

Therefore,

$$\begin{aligned} \begin{aligned} \liminf _{T\rightarrow \infty } \sup _z\frac{A_3}{T}&= \liminf _{T \rightarrow \infty }\sup _{0 \le z \le 1}\frac{1}{T^2}\lambda _{i,T}^2(z) \ge \lim _{T \rightarrow \infty } \max \left( D_1\left( \frac{t_0}{T}\right) , D_2\left( \frac{t_0}{T}\right) \right) \\&= \delta _i^2{\underline{\tau }}^2(1-{\overline{\tau }})^2>0. \\ \end{aligned} \end{aligned}$$

From the above analysis for $A_3$, we have $\frac{1}{\sqrt{T}}\sup _z|A_2|=O_p(1).$ We complete the proof by showing ${\hat{\sigma }}_i^{2}/T {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0$. Let ${\hat{\sigma }}_i^2(L,X)$ be the estimator of the long run variance $\sigma _i^2$ constructed with bandwidth L from data X. Then ${\hat{\sigma }}_i^2 = {\hat{\sigma }}_i^2 (l,X_i)$. Let $e_i = (e_{i1}, \ldots , e_{iT})'$. Then, ${\hat{\sigma }}_i^2 (l,e_i) {\mathop {\longrightarrow }\limits ^\mathrm{p}} \sigma _i^2$ because $l^{-1} + l/T=o(1)$. We need to show $T^{-1}{\hat{\sigma }}_i^2 = T^{-1} {\hat{\sigma }}_i^2(l,X_i) {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0$. Let ${\hat{\gamma }}_i^e(h) = \frac{1}{T} \sum _{t=h+1}^T (e_{it} - {\bar{e}}_i)(e_{i,t-1}-{\bar{e}}_i)$, ${\bar{e}}_i = \frac{1}{T}\sum _{t=1}^T e_{it}$. We have $|{\hat{\gamma }}_i(h) - {\hat{\gamma }}_i^e(h)| \le M$, uniformly in $t_0$, h and T, because $X_{it} = e_{it} + \delta _i I_{\{t \ge t_0\}}.$ We have $|{\hat{\sigma }}_i^2(l, X_i) - {\hat{\sigma }}_i^2(l,e_i)| \le \sum _{h=-l}^l |{\hat{\gamma }}_i(h) - {\hat{\gamma }}_i^e(h)| \le M(2l+1),$ uniformly in i, $t_0$, T. Therefore, $\frac{1}{T}{\hat{\sigma }}_i^2(l,X_i) = \frac{1}{T} {\hat{\sigma }}_i^2(l,e_i) + O_p(l/T) {\mathop {\longrightarrow }\limits ^\mathrm{p}} 0$. $\square$