d | n=400 | n=600 |
0.1 | 0.025 | 0.037 |
0.2 | 0.036 | 0.045 |
0.3 | 0.048 | 0.051 |
0.4 | 0.056 | 0.055 |
In this paper, the sieve bootstrap test for multiple change points in the mean of long memory sequence is studied. Firstly, the ANOVA test statistics for change points detection is obtained. Secondly, sieve bootstrap statistics is constructed and the consistency under the Mallows measure is proved. Finally, the effectiveness of the method was illustrated by simulation and example analysis. Simulation results show that our method can not only control the empirical size well but also have reasonable good power.
Citation: Wenzhi Zhao, Dou Liu, Huiming Wang. Sieve bootstrap test for multiple change points in the mean of long memory sequence[J]. AIMS Mathematics, 2022, 7(6): 10245-10255. doi: 10.3934/math.2022570
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In this paper, the sieve bootstrap test for multiple change points in the mean of long memory sequence is studied. Firstly, the ANOVA test statistics for change points detection is obtained. Secondly, sieve bootstrap statistics is constructed and the consistency under the Mallows measure is proved. Finally, the effectiveness of the method was illustrated by simulation and example analysis. Simulation results show that our method can not only control the empirical size well but also have reasonable good power.
The presence of change point can easily mislead the conventional time series analysis and result in erroneous conclusions. One of the problems in change point analysis is to detect whether there are change points in the statistical sequence.
The statistical literature on change point problem start with Page (1954) [1], who publish an article on quality inspection, which has attracted the attention of experts in various fields. Horváth and Kokoszka et al. (1997, 1998) [2,3] study the CUSUM estimator of mean change point and obtain the limiting distribution for the estimator. Kuan et al. (1998) [4] propose least square method to estimate mean change point in fractionally integrated process. Shao (2011) [5] use ratio statistics to solve the testing problem of mean change point. In the actual statistical test, the data sequence may have more than one change points. The detection method designed for at-most-one change point problem performs poor under multiple change point detection problem. So the study of the multiple change points detection is of great significance. Among them, Bai et al. (1997, 1998, 2003) [6,7,8,9] consider the estimation and detection problem of the multiple change points in linear process. Bardet et al. (2010) [10] and Kejriwal et al. (2013) [11] use different methods to solve the same problem. Lijing et al. (2020) [12] and Macneill et al. (2020) [13,14] consider the detection problem of multiple change points in linear process. Noriah et al. (2014) [15] propose ANOVA statistics to test the multiple change points in i.i.d. sequence. The limiting distribution of test statistics is derived under i.i.d. assumption. While, it is not easy to obtain limiting distribution for long range dependence sequence. Fortunately, bootstrap method give a chance to solve this problem more conveniently.
Long memory processes are prevalent in many areas, for example, in geophysical sciences, microeconomic, asset pricing, stock returns and exchange rates. Hidalago and Robinson et al. (1996) [16] propose the Wald method to test mean change point in long memory sequence. Lazarova (2005) [17] study the change point detection problem in linear regression models with long memory error. Wang (2008) [18] give the estimation method for change point in non-parametric regression models with long memory error. The above literatures focus on the single change point of long memory sequence. However, in practice, due to the interference of various factors, the statistical property of long memory sequence may change not only once, but many times. Thus, it is very necessary to study the multiple point problem of long memory sequence. In this paper, we study the multiple change point detection problem for long memory sequence.
Sieve bootstrap method was introduced by Bühlmann (1997) [19] firstly. Alonso et al. (2002, 2003, 2004) [20,21,22] and Mukhopadhyay et al. (2010) [23] use sieve bootstrap to study the forecasting problems for time series. Poskitt (2008) [24] prove the properties of the sieve bootstrap method and point out that sieve bootstrap method is very useful to analyze long memory sequence. So, in this paper, we consider sieve bootstrap test for multiple change points in the mean of long memory sequence.
We assume that n observations X1,X2,⋯,Xn are given by:
Xt=μ(t)+ et,t=1,2,⋯,n, |
μ(t)={μ1,1≤t≤n1,μ2,n1+1≤t≤n2,⋯μk+1,nk+1≤t≤n, | (2.1) |
where et=φ(B)εt=∞∑j=0φjεt−j,φj∼c0jd−1(0<c0<∞,j→∞,0<d<0.5), and ∞∑j=0φ2j<∞,εt is an i.i.d. process with mean zero and finite variance σ2. The symbol "∼" indicates that the ratio of left- and right-hand sides tends to 1. The sequence {et,t=1,2,⋯,n} is a linear stationary sequence with long memory, so is {Xt,t=1,2,⋯,n}. Parameters μ1,μ2,⋯,μk+1 are the finite constant and n1,n2,⋯,nk are the unknown change point. We consider here the problem of testing the null hypothesis of no change point:
H0:μ1=μ2=⋯=μk+1 | (2.2) |
against the multiple change points alternative:
H0:μ1=μ2=⋯=μk+1. | (2.3) |
Let τ=(0<τ1<τ2<⋯<τk<1) be any partition of [0, 1], [nτi]≥[nτi−1]+2,i=1,2,⋯,k+1 with τ0=0,τk+1=1. Define di,n=[nτi]−[nτi−1],i=1,2,⋯,k+1. Let S0=0,Sr=r∑j=1Xj, r=1,2,⋯,n and ˉX=1nSn. For i=1,2,⋯,k+1, the mean of X[nτi−1]+1,⋯,X[nτi] is ¯Xi=1di,n[nτi]∑t=[nτi−1]+1Xt.
The one-way ANOVA-type test statistic proposed by Noriah et al. (2014) [15] is:
Zn(k):=∫⋯∫τVn(τ)dτ, | (2.4) |
where
Vn(τ)=c−20n−2dk+1∏i=1ai.nSSTr(τ), |
and
SSTr(τ)=k+1∑i=1di,n(¯Xi−ˉX)2. |
Often, the number of change points k is unknown and we assume that it has an upper bound K. Test statistic is defined as Zn(K)=max1≤k≤KZn(k) and the limiting distribution is approximated by sieve bootstrap method.
{Xt} is invertible and has an AR(∞) representation. The specific steps of the sieve bootstrap method are as follows:
Step 1. Having observed the samples x1,x2,⋯,xn, we start estimating ˆd. Specific estimation method can be referred to Hurst (1951) [25];
Step 2. Make ˆd-order difference on {xt} to obtain the sequence {yt}. To fit the AR(p) autoregressive process of {yt}, that is
^yt=ˆϕ0+ˆϕ1y1+ˆϕ2y2+⋯+ˆϕpyt−p. | (2.5) |
Given a maximum AR order p=p(n) of the autoregressive approximation, then choose the optimal ˆp using the BIC criterion, where p(n)=o(n), and p(n)→∞ as the sample size n→∞. We obtain the residuals ˆε1,ˆε2,⋯,ˆεt. The residual is re-sampled to obtain new residual sequence, and the autoregressive process y∗t=^yt+ˆε∗t is constructed for the new residual sequence.
Step 3. The new long memory sequence {x∗t} is constructed by x∗t=(1−B)−ˆdy∗t. Calculates the corresponding sieve bootstrap test statistics value z∗n(K):
z∗n(K):=max1≤k≤K∫⋯∫τc−20n−2dk+1∏i=1di.nk+1∑i=1di,n(¯xi∗−¯x∗)2dτ. | (2.6) |
Step 4. Repeat Steps 2–3 B times and obtain B values z∗i(K),i=1,2,⋯,B. The sieve bootstrap approximation of the p value is p∗=1B#{i:z∗i(K)≥zn(K)}, where # denotes the number of elements in the set and zn(K):=max1≤k≤K∫⋯∫τc−20n−2dk+1∏i=1di.nk+1∑i=1di,n(¯xi−ˉx)2dτ. We reject H0 if p∗<α.
In order to prove that the asymptotic results are consistent, the following assumption and lemmas are required [24].
Assumption 1. Let ξt denote the σ-algebra of events determined by εs, s≤t. Also, assume that {εt}t∈Z are i.i.d and that
E[εt|ξt−1]=0,E[ε2t|ξt−1]=σ2,t∈Z. |
Furthermore, assume that E[ε4t]<∞,t∈Z.
Lemma 1. Assume that n observations X1,X2,⋯,Xn satisfy Eq (2.1). Then, for t∈Z,0<d<0.5,p=p(n)→∞, we have
1nn∑t=1(ˆε∗t−εt)2=Oa.s{(pλmin(Γp))(lognn)1−2d}, | (3.1) |
where λmin(Γp)=O{p−q}, q≥0.
Proof. See Lemma 2 in reference [24], the proof is omitted.
Lemma 2. Assume that n observations X1,X2,⋯,Xn satisfy Eq (2.1). We obtain the represents e∗t=∞∑j=0ˆφjˆε∗t−j, et=∞∑j=0φjεt−j, then, for t∈Z,0<d<0.5,p(n)→∞, we have
∞∑j=0|ˆφj−φj|=Oa.s{(p5λmin(Γp))12(lognn)12−d}, | (3.2) |
where λmin(Γp)=O{p−q}, q≥0.
Proof. See Lemma 3 in reference [24], the proof is omitted.
Theorem 1. Let η(FX,FY) denotes Mallow's measure of the distance between two probability distribution FX and FY, defined as inf{E||X−Y||2}12 where the infimum is taken over all square integrable random variables X and Y in Rm with marginal distributions FX and FY. Assume that n observations X1,X2,⋯,Xn satisfy Eq (2.1). So if the null hypothesis H0 is true, when n→∞, for 0<d<0.5 and p=p(n)→∞. Then with probability one
η(FZ∗n(k),FZn(k))=O{(p5λmin(Γ2p))12(lognn)12−d}, | (3.3) |
where Zn(k) denotes for the test statistic, Z∗n(k) denotes for the sieve bootstrap test statistic, FZn(k) denotes for the true distribution, FZ∗n(k) denotes for the sieve bootstrap approximate distribution, λmin(Γp)=O{p−q},q≥0.
Proof. From the definition of e∗t=∞∑j=0ˆφjˆε∗t−j and et=∞∑j=0φjεt−j, we have
¯Xi=1di,n[nτi]∑t=[nτi−1]+1Xt=1di,n[nτi]∑t=[nτi−1]+1(μt+et) |
and
¯X∗i=1di,n[nτi]∑t=[nτi−1]+1X∗t=1di,n[nτi]∑t=[nτi−1]+1(μt+e∗t). | (3.4) |
It follows that
|¯Xi−¯X∗i|2≤1di,n[nτi]∑t=[nτi−1]+1|Xt−X∗t|2=1di,n[nτi]∑t=[nτi−1]+1|et−e∗t|2. | (3.5) |
Where Z∗n(k) denotes for the sieve bootstrap test statistic, that is
Zn∗(k):=∫⋯∫τc−20n−2dk+1∏i=1di.nk+1∑i=1di,n(¯Xi∗−¯X∗)2dτ. |
So applying the mean value theorem of calculus, we have
Zn(k)−Z∗n(k)=n∑t=1∂Zi(k)∂Xt(Xt−X∗t)=n∑t=1∂Zi(k)∂Xt(et−e∗t), |
and
||Zn(k)−Z∗n(k)||≤n∑t=1||∂Zi(k)∂Xt|||Xt−X∗t|=n∑t=1||∂Zi(k)∂Xt|||et−e∗t|. | (3.6) |
By assumption, n∂Zi(k)∂Xt are continuous on the domain R and hence uniformly bounded, so there is M>0. We can therefore conclude that n||∂Zi(k)∂Xt||≤M and Zn(k) will satisfy the Lipschitz condition, that is
||Zn(k)−Z∗n(k)||2≤1nn∑t=1M2|et−e∗t|2. | (3.7) |
Similar to the proof of Theorem 1 in literature [22], from the definition of Mallows metric and applying the Cauchy-Schwartz inequality, we have
η(FZ∗n(k),FZn(k))2≤E[E∗[||Z∗n(k)−Zn(k)||2]≤E[E∗[1nn∑t=1M2|et−e∗t|2]≤1nn∑t=1E[E∗[M2]]1nn∑t=1E[E∗[(et−e∗t)2]]. | (3.8) |
We obtain the representation e∗t=∞∑j=0ˆφjˆε∗t−j,et=∞∑j=0φjεt−j, from which it follows that
et−e∗t=∞∑j=0(φj−ˆφj)εt−j+∞∑j=0ˆφj(εt−j−ˆε∗t−j)=u(t)+v(t). |
Where 1nn∑t=1E[E∗[M2]]≤∞, thus we are faced with the task of evaluating E[E∗[(u(t)+v(t))2]. Consider E[E∗[v(t)2] firstly. By construction, εt−ˆε∗t are i.i.d with respect to all the observations X1,X2,⋯,Xt. Hence
E∗[v(t)2]=1nn∑t=1(ˆε∗t−εt)2∞∑j=0|ˆφj|2. | (3.9) |
Using Lemma 1 we get that,
1nn∑t=1(ˆε∗t−εt)2=Oa.s{(pλmin(Γp))(lognn)1−2d}. | (3.10) |
Since ∞∑j=0|φj|≤∞, using Lemma 2 we can conclude that
∞∑j=0|ˆφj|≤∞∑j=0|φj|+∞∑j=0|ˆφj−φj|=O(1)+O{(p5λmin(Γ2p))12(lognn)12−d}. |
Thus,
E[E∗[v(t)2]]=O{(pλmin(Γp))(lognn)1−2d}. | (3.11) |
Now consider E[E∗[u(t)2]]. Since u(t)=∞∑j=0(φj−ˆφj)εt−j is a constant relative to all the observations X1,X2,⋯,Xt, we have
E[E∗[u(t)2]]=E[u(t)2E∗(1)]. |
According to Poskitt (2008) [24], under H0, Xt has the following spectral density
f(ω)=σ2|φ(eiω)|22ϕ. |
Thus,
E[u(t)2]=σ22π∫π−π|φ(eiω)−ˆφ(eiω)|2|ϕ(eiω)φ(eiω)|2dω. | (3.12) |
For any δ>0,ω∈(−π,π], from Lemma 4 in literature [24], we have
|ϕ(eiω)−φ(eiω)|≤1+|ϕ(eiω)φ(eiω)−1|≤1+δ. |
Hence
E[u(t)2]≤σ22π∫π−π|φ(eiω)−ˆφ(eiω)|2(1+δ)2dω=[δ(1+δ)]22π∞∑j=0|ˆφj−φj|2. |
And by Lemma 2 this equals
∞∑j=0|ˆφj−φj|=Oa.s{(p5λmin(Γp))12(lognn)12−d}. |
We can therefore conclude that
E[u(t)2]=O{(p5λmin(Γp))(lognn)1−2d}. | (3.13) |
To sum up, we have
η(FZ∗n(k),FZn(k))=O{(p5λmin(Γ2p))12(lognn)12−d}. |
Which completes the proof of the theorem.
In this section, we evaluate the performance of the test statistics trough a simulation. Experiments are conducted based on sample sizes n=400 and n=600 with 1000 replications. Consider the following data generation process:
Xt=μ(t)+ et,t=1,2,⋯,n, |
μ(t)={μ1,1≤t≤n1,μ2,n1+1≤t≤n2,⋯μk+1,nk+1≤t≤n, |
where et is a FARIMA(0, d, 0) process. Simulation studies are based on B = 5000 and α=0.05. Take d=0.1,0.2,0.3,0.4. Under H0, μ1=μ2=⋯=μk+1=0, the empirical sizes of Z∗n(K) are summarized in Table 1. Under H1, the number of change points K=2 and 3. When K=2, the change point combinations (n1,n2) are divided into three situations: (18n,38n), (38n,58n) and (58n,78n) and (μ1,μ2,μ3)=(0,1,2). When K=3, the change point combinations (n1,n2,n3) are divided into two situations: (18n,38n,58n) and (38n,58n,78n). The mean parameters (μ1,μ2,μ3,μ4) are taken as (0,1,2,3) and (0,1,−1,2). The empirical powers of Z∗n(K) are shown from Table 2 to Table 5.
d | n=400 | n=600 |
0.1 | 0.025 | 0.037 |
0.2 | 0.036 | 0.045 |
0.3 | 0.048 | 0.051 |
0.4 | 0.056 | 0.055 |
d | (18n,38n) | (38n,58n) | (58n,78n) |
0.1 | 0.699 | 0.721 | 0.704 |
0.2 | 0.686 | 0.709 | 0.737 |
0.3 | 0.715 | 0.753 | 0.828 |
0.4 | 0.853 | 0.714 | 0.895 |
d | (18n,38n) | (38n,58n) | (58n,78n) |
0.1 | 0.684 | 0.643 | 0.776 |
0.2 | 0.688 | 0.715 | 0.675 |
0.3 | 0.722 | 0.825 | 0.870 |
0.4 | 0.885 | 0.909 | 0.916 |
d | (μ1,μ2,μ3,μ4) | (18n,38n,58n) | (38n,58n,78n) |
0.1 | (0,1,2,3) | 0.614 | 0.693 |
(0,1,-1,2) | 0.559 | 0.685 | |
0.2 | (0,1,2,3) | 0.640 | 0.708 |
(0,1,-1,2) | 0.668 | 0.690 | |
0.3 | (0,1,2,3) | 0.713 | 0.734 |
(0,1,-1,2) | 0.682 | 0.699 | |
0.4 | (0,1,2,3) | 0.775 | 0.877 |
(0,1,-1,2) | 0.717 | 0.850 |
d | (μ1,μ2,μ3,μ4) | (18n,38n,58n) | (38n,58n,78n) |
0.1 | (0,1,2,3) | 0.702 | 0.685 |
(0,1,-1,2) | 0.694 | 0.696 | |
0.2 | (0,1,2,3) | 0.698 | 0.690 |
(0,1,-1,2) | 0.709 | 0.689 | |
0.3 | (0,1,2,3) | 0.732 | 0.877 |
(0,1,-1,2) | 0.786 | 0.792 | |
0.4 | (0,1,2,3) | 0.814 | 0.901 |
(0,1,-1,2) | 0.827 | 0.874 |
Table 1 displays the empirical size of Z∗n(K) under H0. With the increase of sample size n, the empirical size is close to the significant level α=0.05.
It can be seen from Tables 2–5 that the empirical power of Z∗n(K) under H1 increases significantly when the sample size n increases. The larger d is, the bigger the value of the empirical power.
In this subsection, the method in this paper is used to detect mean change in monthly average temperature of Northern Hemisphere (1854–1989). The data comes from Beran (1994) [26] and ˆd=0.37.
Wang (2008) [18] suggests that there is one change point in the data. Thus, we take K=1 and repeat Steps 1–4 to generate 5000 values of sieve bootstrap statistics. Figure 1 shows these values. The horizontal solid line is significant level α=0.05. p∗=15000{#i:z∗i(K)≥zn(K)}=0.0020<0.05. Thus, we reject H0.
Further, the least square method in Kuan (1998) [4] is used to estimate location of change point. The change point estimator is in the 860 observation, corresponding to July 1925. Figure 2 shows these results. The vertical line in the figure indicates the position of the change point. The two horizontal lines in the figure indicate the mean value before and after the change point. The mean value before the change point is -0.34, and the mean value after the change point is 0.02, which means that the monthly average temperature in the Northern Hemisphere has increased.
This paper considers the sieve bootstrap test for multiple change points in the mean of long memory sequence. The consistency of sieve bootstrap approximation is proved. Numerical simulation and application results support the conclusion.
This work was supported by the National Natural Foundation of China (Grant nos. 11771353, 12171391) and the Natural Science Basic Research Program of Shaanxi (No. 2022JM-024).
The authors declare that they have no conflicts of interest.
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d | n=400 | n=600 |
0.1 | 0.025 | 0.037 |
0.2 | 0.036 | 0.045 |
0.3 | 0.048 | 0.051 |
0.4 | 0.056 | 0.055 |
d | (18n,38n) | (38n,58n) | (58n,78n) |
0.1 | 0.699 | 0.721 | 0.704 |
0.2 | 0.686 | 0.709 | 0.737 |
0.3 | 0.715 | 0.753 | 0.828 |
0.4 | 0.853 | 0.714 | 0.895 |
d | (18n,38n) | (38n,58n) | (58n,78n) |
0.1 | 0.684 | 0.643 | 0.776 |
0.2 | 0.688 | 0.715 | 0.675 |
0.3 | 0.722 | 0.825 | 0.870 |
0.4 | 0.885 | 0.909 | 0.916 |
d | (μ1,μ2,μ3,μ4) | (18n,38n,58n) | (38n,58n,78n) |
0.1 | (0,1,2,3) | 0.614 | 0.693 |
(0,1,-1,2) | 0.559 | 0.685 | |
0.2 | (0,1,2,3) | 0.640 | 0.708 |
(0,1,-1,2) | 0.668 | 0.690 | |
0.3 | (0,1,2,3) | 0.713 | 0.734 |
(0,1,-1,2) | 0.682 | 0.699 | |
0.4 | (0,1,2,3) | 0.775 | 0.877 |
(0,1,-1,2) | 0.717 | 0.850 |
d | (μ1,μ2,μ3,μ4) | (18n,38n,58n) | (38n,58n,78n) |
0.1 | (0,1,2,3) | 0.702 | 0.685 |
(0,1,-1,2) | 0.694 | 0.696 | |
0.2 | (0,1,2,3) | 0.698 | 0.690 |
(0,1,-1,2) | 0.709 | 0.689 | |
0.3 | (0,1,2,3) | 0.732 | 0.877 |
(0,1,-1,2) | 0.786 | 0.792 | |
0.4 | (0,1,2,3) | 0.814 | 0.901 |
(0,1,-1,2) | 0.827 | 0.874 |
d | n=400 | n=600 |
0.1 | 0.025 | 0.037 |
0.2 | 0.036 | 0.045 |
0.3 | 0.048 | 0.051 |
0.4 | 0.056 | 0.055 |
d | (18n,38n) | (38n,58n) | (58n,78n) |
0.1 | 0.699 | 0.721 | 0.704 |
0.2 | 0.686 | 0.709 | 0.737 |
0.3 | 0.715 | 0.753 | 0.828 |
0.4 | 0.853 | 0.714 | 0.895 |
d | (18n,38n) | (38n,58n) | (58n,78n) |
0.1 | 0.684 | 0.643 | 0.776 |
0.2 | 0.688 | 0.715 | 0.675 |
0.3 | 0.722 | 0.825 | 0.870 |
0.4 | 0.885 | 0.909 | 0.916 |
d | (μ1,μ2,μ3,μ4) | (18n,38n,58n) | (38n,58n,78n) |
0.1 | (0,1,2,3) | 0.614 | 0.693 |
(0,1,-1,2) | 0.559 | 0.685 | |
0.2 | (0,1,2,3) | 0.640 | 0.708 |
(0,1,-1,2) | 0.668 | 0.690 | |
0.3 | (0,1,2,3) | 0.713 | 0.734 |
(0,1,-1,2) | 0.682 | 0.699 | |
0.4 | (0,1,2,3) | 0.775 | 0.877 |
(0,1,-1,2) | 0.717 | 0.850 |
d | (μ1,μ2,μ3,μ4) | (18n,38n,58n) | (38n,58n,78n) |
0.1 | (0,1,2,3) | 0.702 | 0.685 |
(0,1,-1,2) | 0.694 | 0.696 | |
0.2 | (0,1,2,3) | 0.698 | 0.690 |
(0,1,-1,2) | 0.709 | 0.689 | |
0.3 | (0,1,2,3) | 0.732 | 0.877 |
(0,1,-1,2) | 0.786 | 0.792 | |
0.4 | (0,1,2,3) | 0.814 | 0.901 |
(0,1,-1,2) | 0.827 | 0.874 |