Adjusted blockwise empirical likelihood for long memory time series models

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.

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. $\square$