Nothing Special   »   [go: up one dir, main page]
Skip to main content
Springer Nature Link
Log in

Adjusted blockwise empirical likelihood for long memory time series models

  • Original Paper
  • Published:
Statistical Methods & Applications Aims and scope Submit manuscript

Abstract

In this paper, we introduce an adjusted blockwise empirical likelihood (ABEL) method for long memory time series models. By dividing time series into blocks and by adding an appropriate adjustment term, we construct the ABEL ratio and the confidence interval for the mean of the process. Under mild conditions, we show that Wilks’ theorem still holds for the ABEL ratio by choosing a specific block correction factor. The Monte Carlo simulation studies are reported to assess the finite sample performance of the proposed ABEL method.

This is a preview of subscription content, log in via an institution to check access.

Access this article

Log in via an institution

Subscribe and save

Springer+ Basic
$34.99 /Month
  • Get 10 units per month
  • Download Article/Chapter or eBook
  • 1 Unit = 1 Article or 1 Chapter
  • Cancel anytime
Subscribe now

Buy Now

Price excludes VAT (USA)
Tax calculation will be finalised during checkout.

Instant access to the full article PDF.

Institutional subscriptions

Similar content being viewed by others

References

Download references

Acknowledgements

We sincerely wish to thank the two referees for their queries and many insightful remarks and suggestions which have led to improving the presentation of the results.

Author information

Authors and Affiliations

  1. Department of Mathematics, Nanjing University, Nanjing, 210093, People’s Republic of China

    Feifan Jiang & Lihong Wang

Authors
  1. Feifan Jiang
    View author publications

    You can also search for this author in PubMed Google Scholar

  2. Lihong Wang
    View author publications

    You can also search for this author in PubMed Google Scholar

Corresponding author

Correspondence to Lihong Wang.

Additional information

This work was supported by National Natural Science Foundation of China (NSFC) Grants 11671194, 11171147 and 11501287.

Appendix

Appendix

Proof of Theorem 1

Let \(Z_{li}=T_{li}-\mu _0\), \(1\le i \le N\), \(Z_{l,N+1}=-{A_N}{N}^{-1}\sum ^{N}_{i=1}Z_{li}\), and define \(M_{n{\mu }_0} \equiv \max _{1 \le i \le N} |Z_{li}|\). First we examine the magnitude of \(M_{n{\mu }_0}\). Note that

$$\begin{aligned} \max _{1 \le i \le N}\frac{|Z_{li}|}{\sqrt{Var(Z_{l1})}}=\left( \max _{1\le i\le N}\frac{|Z_{li}|^q}{(Var(Z_{l1}))^{q/2}}\right) ^{1/q}\le \left( \sum ^{N}_{i=1}\frac{{|Z_{li}|}^q}{(Var(Z_{l1}))^{q/2}}\right) ^{1/q}. \end{aligned}$$

Moreover, by Assumption (A3) and Lemma 4 of Davydov (1970),

$$\begin{aligned} E|Z_{l1}|^q \le C(Var(Z_{l1}))^{q/2}. \end{aligned}$$

This, together with \(l^2/n=o(1)\), implies that

$$\begin{aligned}&\left( \frac{l}{n}\right) ^{1/2-d}\max _{1 \le i \le N}\frac{|Z_{li}|}{\sqrt{Var(Z_{l1})}} \le \left( \frac{l}{n}\right) ^{1/2-d}\left( \sum ^{N}_{i=1}\frac{{|Z_{li}|}^q}{(Var(Z_{l1}))^{q/2}}\right) ^{1/q}\nonumber \\&\ \ =\left( \frac{l}{n}\right) ^{1/2-d}n^{1/q}\left( \frac{1}{n}\sum ^{N}_{i=1}\frac{{|Z_{li}|}^q}{(Var(Z_{l1}))^{q/2}}\right) ^{1/q} =\left( \frac{l}{n}\right) ^{1/2-d}n^{1/q}O_p(1)\nonumber \\&\ \ =o(n^{-1/4+d/2+1/q})O_p(1)=o_p(1). \end{aligned}$$
(3)

From Proposition 3.3.1 and Theorem 4.3.1 of Giraitis et al. (2012),

$$\begin{aligned}&l^{1/2-d}Z_{l1} \rightarrow _D c_0 Z, \qquad l^{1-2d} Var(Z_{l1})=c_0^2+o(1), \end{aligned}$$

where \(c_0^2=c^2B(d, 1-2d)/(d(1+2d))\). Let \(a_t^2=t^{2d-1}c_0^2\), \(1\le t\le n\). Thus, by (3),

$$\begin{aligned} \frac{a_nM_{n\mu _0}}{a^2_l}\sim \Big (\frac{l}{n}\Big )^{1/2-d} \frac{M_{n\mu _0}}{\sqrt{Var(Z_{l1})}}=o_p(1). \end{aligned}$$

That is,

$$\begin{aligned} M_{n\mu _0}=o_p\Big (\frac{a^2_l}{a_n}\Big ). \end{aligned}$$
(4)

Next, we consider \(|\lambda _{\mu _0}|\), which is the solution of the equation

$$\begin{aligned} \frac{1}{N+1}\sum ^{N+1}_{i=1}\frac{Z_{li}}{1+\lambda Z_{li}}=0. \end{aligned}$$
(5)

Let \(\bar{Z}_{n\mu _0}=(N+1)^{-1}\sum ^{N+1}_{i=1}Z_{li}\) and \(\hat{S}^2_{l\mu _0}=N^{-1}\sum ^N_{i=1} Z_{li}^2\). Since

$$\begin{aligned} \bar{Z}_{n\mu _0}=\frac{1}{N+1}\Big (1-\frac{A_N}{N}\Big ) \sum ^{N}_{i=1}Z_{li}. \end{aligned}$$

Proposition 2(a) of Nordman et al. (2007) and \(A_N=O(N^{\frac{1}{2}-d})\) imply that

$$\begin{aligned} a_n^{-1}\bar{Z}_{n\mu _0}\rightarrow _D Z. \end{aligned}$$
(6)

Moreover, it follows from Proposition 2(b) of Nordman et al. (2007) that

$$\begin{aligned} \hat{S}^2_{l\mu _0}/a_l^2\rightarrow _P 1. \end{aligned}$$
(7)

From(5),

$$\begin{aligned} 0= & {} \Big |\frac{1}{N+1}\sum ^{N+1}_{i=1}\frac{Z_{li}}{1+\lambda Z_{li}}\Big |\\= & {} \Big |\frac{1}{N+1}\sum ^{N+1}_{i=1}\Big (Z_{li}-\frac{\lambda Z^2_{li}}{1+\lambda Z_{li}}\Big )\Big |\\\ge & {} \frac{|\lambda |}{N+1}\sum ^{N+1}_{i=1}\frac{Z^2_{li}}{|1+\lambda Z_{li}|}-\frac{1}{N+1}\Big |\sum ^{N+1}_{i=1}Z_{li}\Big |\\\ge & {} \frac{|\lambda |}{N+1}\sum ^{N}_{i=1}\frac{Z^2_{li}}{|1+\lambda Z_{li}|}-\frac{1}{N+1}\Big |\sum ^{N+1}_{i=1}Z_{li}\Big |\\\ge & {} \frac{|\lambda |}{1+|\lambda | M_{n\mu _0}}\cdot \frac{1}{N+1}\sum ^{N}_{i=1}{Z^2_{li}}-\frac{1}{N+1}\Big |\sum ^{N+1}_{i=1}Z_{li}\Big |. \end{aligned}$$

Combining this with (6) and (7), we arrive at

$$\begin{aligned} \frac{|\lambda _{\mu _0}|}{1+|\lambda _{\mu _0}| M_{n\mu _0}}\le \frac{|\bar{Z}_{n\mu _0}|}{{\frac{1}{N+1}}\sum ^N_{i=1} Z^2_{li}}= O_p\Big (\frac{a_n}{a^2_l}\Big ), \end{aligned}$$

and therefore by (4),

$$\begin{aligned} |\lambda _{\mu _0}|= O_p\Big (\frac{a_n}{a^2_l}\Big ). \end{aligned}$$
(8)

Again from (5),

$$\begin{aligned} 0= & {} -\frac{1}{N+1}\sum ^{N+1}_{i=1}\frac{Z_{li}}{1+\lambda _{\mu _0} Z_{li}}\\= & {} \frac{1}{N+1}\sum ^{N+1}_{i=1}\Big (\frac{\lambda _{\mu _0}Z^2_{li}}{1+\lambda _{\mu _0} Z_{li}}-Z_{li}\Big )\\= & {} \frac{1}{N+1}\sum ^N_{i=1}\Big (\lambda _{\mu _0}Z^2_{li}-\frac{\lambda ^2_{\mu _0}Z^3_{li}}{1+\lambda _{\mu _0}Z_{li}}-Z_{li}\Big )\\&\quad +\frac{1}{N+1}\Big (\frac{\lambda _{\mu _0}Z^2_{l,N+1}}{1+\lambda _{\mu _0} Z_{l, N+1}}-Z_{l,N+1}\Big ), \end{aligned}$$

we obtain

$$\begin{aligned} \lambda _{\mu _0}=\frac{\bar{Z}_{n\mu _0} +I_1+I_2}{\frac{1}{N+1}\sum ^N_{i=1}Z^2_{li}}, \end{aligned}$$
(9)

where

$$\begin{aligned} I_1=\frac{1}{N+1}\sum ^N_{i=1}\frac{\lambda ^2_{\mu _0}Z^3_{li}}{1+\lambda _{\mu _0}Z_{li}},\ \ I_2=-\frac{1}{N+1}\cdot \frac{\lambda _{\mu _0}Z^2_{l,N+1}}{1+\lambda _{\mu _0}Z_{l, N+1}}. \end{aligned}$$

Let \(\xi _i=\lambda _{\mu _0}Z_{li}\), \(1 \le i \le N+1\). (4) and (8) yield that

$$\begin{aligned} \max _{1 \le i \le N}|\xi _i|=|\lambda _{\mu _0}M_{n\mu _0}|=o_p(1). \end{aligned}$$

In addition,

$$\begin{aligned} \small {|\xi _{N+1}|\!=\!|\lambda _{\mu _0}| A_N \cdot \frac{1}{N}\left| \sum ^N_{i=1}Z_{li}\right| \!=\!O_p(a_n^2/a_l^2)O(N^{\frac{1}{2}-d})=o_p((l^2/n)^{1/2-d})=o_p(1).} \end{aligned}$$

This means that

$$\begin{aligned} \max _{1 \le i \le N+1}{|\xi _i|}=o_p(1). \end{aligned}$$
(10)

Thus we find hat

$$\begin{aligned} |I_1|\le \max _{1 \le i \le N}\Big |\frac{1}{1+\xi _i}\Big |\cdot M_{n\mu _0} \lambda ^2_{\mu _0} \frac{1}{N+1} \sum ^{N}_{i=1}Z_{li}^2=o_p(a_n). \end{aligned}$$

In a similar way, by Schwartz inequality,

$$\begin{aligned} |I_2|\le \frac{1}{N+1}\cdot \frac{1}{|1+\xi _{N+1}|}\cdot |\lambda _{\mu _0}|\cdot \frac{A^2_N}{N}\sum ^{N}_{i=1}Z^2_{li}=o_p(a_n). \end{aligned}$$

Let \(I=I_1+I_2\), we arrive at

$$\begin{aligned} |I|=o_p(a_n). \end{aligned}$$
(11)

By (10), we can assume that \(|\xi _i|<1,1 \le i \le N+1\). Then Taylor expansion yields that

$$\begin{aligned} \log (1+\xi _i)=\xi _i-\frac{\xi ^2_i}{2}+v_i, \ \ 1\le i\le N, \end{aligned}$$
(12)

and

$$\begin{aligned} \log (1+\xi _{N+1})=\xi _{N+1}+v_{N+1}, \end{aligned}$$
(13)

where \(|v_i|\le \frac{1}{3}|\lambda _{\mu _0}|^3M_{n\mu _0}Z^2_{li}\) for \(1\le i\le N\) and

$$\begin{aligned} |v_{N+1}|\le \frac{1}{2}\lambda _{\mu _0}^2Z^2_{l, N+1}\le \frac{1}{2}\lambda ^2_{\mu _0} A^2_N N^{-1}\sum ^N_{i=1}Z^2_{li}. \end{aligned}$$

Since \(B_n=N^{-1}a_l^2a_n^{-2}\), we obtain

$$\begin{aligned} 2B_n\sum ^{N+1}_{i=1}v_i= & {} 2B_n\sum ^{N}_{i=1}v_i+2B_n v_{N+1}\nonumber \\\le & {} a^2_la^{-2}_n|\lambda _{\mu _0}|^3M_{n\mu _0}N^{-1}\sum ^{N}_{i=1}Z^2_{li} +a^2_la^{-2}_n\lambda _{\mu _0}^2A^2_N N^{-2}\sum ^{N}_{i=1}Z^2_{li}\nonumber \\= & {} o_p(1). \end{aligned}$$
(14)

Now using (9), (12) and (13), and by (6), (7), (11) and (14), we find that

$$\begin{aligned} -2B_n\log (R^A_n(\mu _0))= & {} 2B_n\sum ^{N+1}_{i=1}\log (1+\lambda _{\mu _0} Z_{li})\\= & {} 2B_n\left( \sum ^{N+1}_{i=1}\lambda _{\mu _0}Z_{li}-\sum ^N_{i=1}\frac{\lambda ^2_{\mu _0}Z^2_{li}}{2}+\sum ^{N+1}_{i=1}v_i\right) \\= & {} 2\left( 1+\frac{1}{N}\right) \frac{a^2_l}{a^2_n}\lambda _{\mu _0}\bar{Z}_{n\mu _0}- \frac{a^2_l}{a_n^2}\lambda ^2_{\mu _0}\hat{S}^2_{l\mu _0}+2B_n\sum ^{N+1}_{i=1}v_i\\= & {} \left( 1+\frac{1}{N}\right) ^2\left\{ \frac{2a^2_l}{a^2_n}\frac{(\bar{Z}_{n\mu _0}+I)\bar{Z}_{n\mu _0}}{\hat{S}^2_{l\mu _0}} -\frac{a^2_l}{a_n^2}\frac{(\bar{Z}_{n\mu _0}+I)^2}{\hat{S}^2_{l\mu _0}}\right\} \\&+2B_n\sum ^{N+1}_{i=1}v_i\\= & {} \left( 1+\frac{1}{N}\right) ^2\left\{ \frac{a^2_l}{a^2_n}\cdot \frac{\bar{Z}^2_{n\mu _0}}{\hat{S}^2_{l\mu _0}} -\frac{a^2_l}{a^2_n}\cdot \frac{I^2}{\hat{S}^2_{l\mu _0}}\right\} +2B_n\sum ^{N+1}_{i=1}v_i\\= & {} \frac{a^2_l}{a^2_n}\cdot \frac{\bar{Z}^2_{n\mu _0}}{\hat{S}^2_{l\mu _0}}\left( 1+O_p(N^{-1})\right) +o_p(1)\\\rightarrow & {} _D \, \chi ^2 (1). \end{aligned}$$

This completes the proof of Theorem 1. \(\square \)

Rights and permissions

Reprints and permissions

About this article

Check for updates. Verify currency and authenticity via CrossMark

Cite this article

Jiang, F., Wang, L. Adjusted blockwise empirical likelihood for long memory time series models. Stat Methods Appl 27, 319–332 (2018). https://doi.org/10.1007/s10260-017-0403-1

Download citation

  • Accepted:

  • Published:

  • Issue Date:

  • DOI: https://doi.org/10.1007/s10260-017-0403-1

Keywords

Mathematics Subject Classification

Search

Navigation