2.1 Cauchy Limit Theory for General Linear Processes

Consider a linear process $\{u_k\}_{ k\in \mathbb {Z}} $ defined by $u_k=\sum _{j=0}^\infty \psi _j \epsilon _{k-j}$ , where ${\sum _{j=0}^\infty \psi _j^2<\infty }$ . Denote by $\gamma (j), j\ge 0, $ the co-variances of $\{u_k\}_{ k\in \mathbb {Z}} $ , i.e.,

$$ \begin{align*} \gamma (j) = \mathbb{E}\big(u_0 u_j\big)=\sigma^2\,\sum_{k=0}^\infty \psi_k\psi_{j+k}, \quad j=0,1,\ldots \end{align*} $$

Write $\Gamma _n=\frac 12 {\gamma (0)}+ \sum _{j=1}^n \rho _n^{-j}\gamma (j)$ and $\sigma _n^2= \tau ^{-1} \, k_n\, \Gamma _n=(\rho _n-1)^{-1}\, \Gamma _n,$ where ${\tau>0}$ , $ 0<k_n \to \infty $ satisfying $\lim _{n\to \infty }n/k_n=\infty $ and $\rho _n=1+\tau /k_n$ as given in model (1.1). Our first result on the Cauchy limit theory is as follows.

Theorem 2.1 provides a quite general framework for a Cauchy limit theory when $u_k$ is a general linear process with martingale difference innovations. Condition $\sigma _n^2\to \infty $ is necessary to establish (1.2). Indeed, as seen in Lemma 5.2, $ var \big (\sum _{j=1}^{n} \rho _n^{-j}u_j\big )=\big [1+o(1)\big ]\,\sigma _n^2$ under minor additional conditions, indicating that (1.2) will fail if $\sigma _n^2$ is finite. Since $k_n/n\to 0$ , condition (b) is trivially true for many empirical examples such as the co-variances $\gamma (j)=\mathbb {E} (u_0u_j), j\ge 0, $ satisfying $\sum _{j=0}^\infty |\gamma (j)|<\infty $ or $\gamma (j)\sim C\,j^{-\beta }$ for any $\beta>0$ . Condition (c) implies that $n\rho _n^{-n}=o(\sigma _n^2)$ due to $\gamma (0)>0$ , providing a trade-off between $\rho _n$ (or equivalently $k_n$ ) and $\gamma (j)$ . The Cauchy limit theory in (1.2) depends on the asymptotic independence between $\sum _{j=1}^{n} \rho _n^{-j}u_j$ and $\sum _{j=1}^{n} \rho _n^{j-n}u_j$ , which in turn requires the fact:

(2.1) $$ \begin{align} {cov} \big(\sum_{j=1}^{n} \rho_n^{-j}u_j, \sum_{j=1}^{n} \rho_n^{j-n}u_j\big) \sim \sum_{j=0}^{n}(n-j) \rho_n^{j-n}\,\gamma(j) = o(\sigma_n^2). \end{align} $$

See the proof of Lemma 5.2 for details. In terms of (2.1), condition (c) is necessary in case that $\gamma (j)\ge 0 $ for all $ j\ge 0$ . To illustrate the applications of Theorem 2.1, we introduce the following conditions on the co-variances $\gamma (j)=\mathbb {E} (u_0u_j), j\ge 0. $

1. C3.

   1. (i) $\gamma (j) \sim j^{-\beta }\,l(j)$ for some $0<\beta <1$ , where $l(x)$ is a slowly varying function at infinity and eventually continuous as $x\to \infty $ ;

   2. (ii) $\sum _{j=0}^\infty |\gamma (j)|<\infty $ and $\sum _{j\in \mathbb {Z}} \, \gamma (j) \not =0$ ;

   3. (iii) $\sum _{j\in \mathbb {Z}} \, \gamma (j)=0$ , $\sum _{j=0}^\infty |\gamma (j)|<\infty $ and $\gamma (j) \sim C_\beta \, j^{-1-\beta }$ for some ${0<\beta <1}$ .

The proof of Corollary 2.1 depends on the estimation of $\sigma _n^2$ or equivalently $\Gamma _n$ under C3. The required results are summarized in the following proposition, which is of interest in its own right.

Proposition 2.1. If C3(i) holds, then

(2.2) $$ \begin{align} \frac {l^{-1}(k_n)}{\, k_n^{1-\beta}}\,\,\Gamma_n \to \int_{0}^\infty e^{-\tau x} x^{-\beta}dx. \end{align} $$

If C3(ii) holds, then

(2.3) $$ \begin{align} \Gamma_n\to \frac 12 \sum_{j\in \mathbb{Z}} \, \gamma(j) =\frac 12\gamma(0)+ \sum_{j=1}^\infty \gamma(j). \end{align} $$

If C3(iii) holds, then

(2.4) $$ \begin{align} k_n^{\beta}\,\,\Gamma_n \to \,C_\beta\, \int_{0}^\infty (e^{-\tau x}-1) \, x^{-1-\beta} dx. \end{align} $$

Moreover, if $\gamma (j)$ satisfies one of C3(i)–(iii), then

(2.5) $$ \begin{align} \frac 12 \gamma(0)+\frac 1n\, \sum_{j=1}^n (n-j)\gamma(j) = o(n \Gamma_n/k_n). \end{align} $$

Remark 2.1. It is well-known that $u_k$ is a long-memory, a short memory, and an anti-persistent process if $\gamma _k=\mathbb {E} (u_0u_k), k\ge 0,$ satisfies C3(i), C3(ii), and C3(iii), respectively. The Cauchy limit theory for long and short memory processes has been considered with $k_n=n^{\alpha }$ for some $\alpha \in (0,1)$ in PM, Magdalinos (Reference Magdalinos2012) and Arvanitis and Magdalinos (Reference Arvanitis and Magdalinos2018). Corollary 2.1 generalizes these existing results by allowing for any $ 0<k_n \to \infty $ satisfying $\lim _{n\to \infty }n/k_n=\infty $ . Note that, for the long memory process $u_k=\sum _{i=0}^\infty \psi _i \epsilon _{k-i}$ with $\psi _i=i^{-\kappa }L(i),$ where $\kappa \in (1/2, 1)$ ,

$$ \begin{align*} \gamma(j) =\sigma^2\, \sum_{i=1}^\infty \psi_i \psi_{i+j}\ \sim\ \sigma^2\, j^{1-2\kappa} L^2(j) \int_0^\infty (y+1)^{-\kappa}y^{-\kappa} dy. \end{align*} $$

An application of (2.2) yields Lemma 1(ii) of Magdalinos (Reference Magdalinos2012),Footnote ¹ i.e.,

(2.6) $$ \begin{align} \frac {\sigma_n^2}{k_n^{3-2\kappa}L^2(k_n)} &= \frac {\tau^{-1}\Gamma_n}{k_n^{2-2\kappa}L^2(k_n)} \nonumber\\ &\to \sigma^2\, \tau^{2\kappa-3}\,\int_0^{\infty} e^{- x} x^{1-2\kappa}dx \int_0^\infty (y+1)^{-\kappa}y^{-\kappa} dy. \end{align} $$

Remark 2.2. Using the two-step approach developed in Phillips et al. (Reference Phillips, Magdalinos and Giraitis2010), Liu et al. (Reference Liu, Xiao and Yu2021) considered the Cauchy limit theory for anti-persistent innovations under C3(iii), but requiring an additional proof as explained in Section 1. There is an additional restriction for the $k_n\to \infty $ under C3(iii), i.e., ${nk_n^{\beta -1}e^{-\tau n/k_n}\to 0}$ or equivalently $n\rho _n^{-n} =o(\sigma _n^2)$ due to (2.4). Note that, under C3(iii), $n\rho _n^{-n} =o(\sigma _n^2)$ if and only if $\sum _{j=0}^{n}(n-j) \rho _n^{j-n}\,\gamma (j) = o(\sigma _n^2)$ .Footnote ² In terms of (2.1), this additional restriction $nk_n^{\beta -1}e^{-\tau n/k_n}\to 0$ or

$$ \begin{align*} \lim\inf_{n\to\infty}n\log ^{-1}n /k_n\ge (1+\tau_0)/\tau \qquad \text{ for any } \tau_0>1-\beta \end{align*} $$

is necessary to establish (1.2), indicating that the Cauchy limit theory for anti-persistent innovations under C3(iii) is violated when $\rho _n$ is very close to the local to unity range: $k_n =n/m_n$ with $m_n=\frac {1+\tau _0}\tau \log n$ for any $-1<\tau _0< 1-\beta $ . This fact seems new to the literature.

Remark 2.3. A typical example for the co-variances $\gamma (j)=\mathbb {E} (u_0u_j), j\ge 0, $ satisfying C3 is the $ARIMA(p,d,q)$ process $\{u_k\}_{k\ge 1}$ defined by

(2.7) $$ \begin{align} (1-B)^{d}u_k=\eta_k,\qquad \phi(B)\eta_k=\theta(B)\epsilon_k, \end{align} $$

where $d\in (-\frac 12, \frac 12)$ ; B is a back-shift operator and $\epsilon _k$ are i.i.d. random variables with zero mean and finite variance; $\phi (B)$ and $\theta (B)$ are polynomial functions of B with order p and q, respectively, and both of them only have roots outside the unit circle, i.e., the $ARMA(p,q)$ process $\eta _k$ is taken to be stationary and invertible. The fractional difference operator $(1-B)^{\gamma }$ is defined by its Maclaurin series (by its binomial expansion, if $\gamma $ is an integer):

$$ \begin{align*} (1-B)^{\gamma}=\sum_{j=0}^{\infty}\frac {\Gamma(-\gamma+j)}{\Gamma(-\gamma)\Gamma(j+1)}B^j\quad \text{where} \ \ \Gamma(z)=\Big \{\begin{array}{ll} \int_0^{\infty}s^{z-1}e^{-z}ds &\text{if } z>0\\ \infty &\text{if } z=0. \end{array} \Big. \end{align*} $$

If $z<0$ , $\Gamma (z)$ is defined by the recursion formula $ z\Gamma (z)=\Gamma (z+1). $ It is well-known that if $d=0$ , $\gamma (k)=\mathbb {E} (u_0u_k)$ satisfies $\sum _{k=0}^\infty |\gamma (k)|<\infty $ , i.e., C3(ii) is satisfied. Moreover, for some constant $C_d$ depending only on d, we have

$$ \begin{align*} \gamma(k) \sim C_{d}\,\Big \{\begin{array}{@{}ll} k^{2d-1} &\text{if } 0<d<1/2\\ k^{d-1} &\text{if } -1/2<d<0, \end{array} \Big. \end{align*} $$

and $\sum _{j\in \mathbb {Z}} \, \gamma (j)=0$ if $d<0$ (see, e.g., Hosking (Reference Hosking1981) or Kokoszka and Taqqu (Reference Kokoszka and Taqqu1995)). Based on these facts, the following result is a direct corollary of Corollary 2.1.

Corollary 2.2. Suppose the innovation sequence $\{u_k\}_{k\ge 1}$ in model (1.1) is given by (2.7).

1. (a) If $0\le d<1/2$ , we have (1.2) for any $ 0<k_n \to \infty $ satisfying $\lim _{n\to \infty }n/k_n=\infty $ .

2. (b) If $-1/2< d<0$ , we have (1.2) for any $ 0<k_n \to \infty $ satisfying

   $$ \begin{align*}\lim\inf_{n\to\infty}n\log ^{-1}n /k_n> (2+d)/\tau.\end{align*} $$

3. (c) Moreover, the Cauchy limit theory in (1.2) is invariant with $d\in (-\frac 12, \frac 12)$ for any $0<k_n\to \infty $ satisfying $\lim \inf _{n\to \infty }n\log ^{-1}n /k_n>3\tau ^{-1}/2$ .

2.2 Cauchy Limit Theory for Other Dependent Processes

While Theorem 2.1 is sufficiently general in allowing $u_k$ in model (1.1) to be a linear process defined as $u_k=\sum _{j=0}^\infty \psi _j \epsilon _{k-j}$ , where $\sum _{j=0}^\infty \psi _j^2<\infty $ , it does not encompass many important practical models, such as those with innovations exhibiting nonlinear structures. To fill the gap, this section considers the model (1.1) with $u_k$ defined by

(2.8) $$ \begin{align} u_k=\epsilon_k+z_{k-1}-z_k, \ \end{align} $$

where $\big \{\epsilon _k, {\cal F}_k\big \}_{k\ge 1}$ is a sequence of martingale differences satisfying one of C1 and C2, and $z_k$ is an arbitrary random variable satisfying $\sup _{k\ge 0}\mathbb {E} z_k^2<\infty $ . The following theorem shows that the Cauchy limit theory still holds for such dependent processes. This result seems new to the literature.

A wide class of dependent processes $u_k$ can be expressed in a form as in (2.8). To illustrate, let $ \eta _i, i\in \mathbb {Z},$ be i.i.d. random variables with $\mathbb {E} \eta _0=0$ and $\mathbb {E} \eta _0^2=1$ . Consider a stationary causal process $u_k$ defined by

$$ \begin{align*} u_k=F(\eta_k, \eta_{k-1}, \ldots), \quad k=0, 1,2,\ldots, \end{align*} $$

where F is a measurable function such that $\mathbb {E}u_0=0$ and $0<\mathbb {E}u_0^2<\infty $ . Write ${\cal F}_k=\sigma (\eta _i, i\le k)$ , $|| Z ||_p=(\mathbb {E}|Z|^p)^{1/p}<\infty $ and denote ${\cal P}_k Z= \mathbb {E} (Z|{\cal F}_k)-\mathbb {E} (Z|{\cal F}_{k-1})$ for any $\mathbb {E}|Z|<\infty $ . Set

$$ \begin{align*} \epsilon_{k}=\sum\limits_{i=0}^\infty {\cal P}_k u_{i+k}, \quad z_{k}=\sum\limits_{i=1}^\infty \mathbb{E} (u_{i+k}|\mathcal{F}_k). \end{align*} $$

Utilizing Proposition 2.2, we can conclude from Theorem 2.2 that the Cauchy limit theory is valid in the model (1.1), where the innovations $u_k$ exhibit the nonlinear structures illustrated in the following examples. For a comprehensive discussion on the causal processes and related examples, we refer to Wu (Reference Wu2005, Reference Wu2007) and Wu and Min (Reference Wu and Min2005).

Example 2.1. (TAR model)

$$ \begin{align*} u_k=\phi_1 \max(u_{k-1},0)+\phi_2 \max(-u_{k-1},0)+\eta_k \end{align*} $$

with $\max (|\phi _1|,|\phi _2|)<1$ and $\mathbb {E} (|\eta _0|^q)<\infty $ for some $q>0$ .

Example 2.2. (Bilinear model)

$$ \begin{align*} u_k=(\alpha_1+\beta_1\eta_{k-1})u_{k-1}+\eta_k, \end{align*} $$

where $\alpha _1$ and $\beta _1$ are real parameters so that $\mathbb {E}|\alpha _1+\beta _1 \eta _0|^q<1$ for some $q>0$ .

Example 2.3. (GARCH model)

$$ \begin{align*} u_k=\sqrt{h_k}\eta_k ,\quad h_k=\alpha_0+\sum_{i=1}^m \alpha_i u_{k-i}^2+\sum_{j=1}^l \beta_j h_{t-j}, \end{align*} $$

where $\alpha _0>0,~\alpha _j\geq 0$ for $1\leq j\leq m$ , $\beta _i\geq 0$ for $1\leq i\leq l$ , and $\sum \limits _{i=1}^m \alpha _i +\sum \limits _{j=1}^l \beta _j<1$ .

5.1 Preliminary Lemmas

As in Section 2.1, suppose that $\big \{\epsilon _k, {\cal F}_k\big \}_{k\in \mathbb {Z}}$ is a sequence of martingale differences satisfying one of C1 or C2, and $u_k=\sum _{j=0}^\infty \psi _j \epsilon _{k-j}$ , where $\sum _{j=0}^\infty \psi _j^2<\infty $ . For each $1\le l\le m,$ let

$$ \begin{align*} Z_{n, l} = \sum_{k=1}^n b_{nk, l} u_k \quad \text{and}\quad \sigma_{n, l}^2 =var(Z_{n, l}), \end{align*} $$

where $b_{nk, l}$ is a sequence of constants. First, Lemma 5.1 is a corollary of Theorem 2.3 in Abadir et al. (Reference Abadir, Distaso, Giraitis and Koul2014).

Lemma 5.1. Suppose that, for each $1\le l\le m,$

(5.1) $$ \begin{align} |b_{n1, l}|+\sum_{k=2}^n |b_{nk, l}-b_{nk-1, l}| =o(\sigma_{n, l}), \end{align} $$

and there exists a positive-definite matrix $\Omega $ so that

(5.2) $$ \begin{align} \Big(cov (\sigma_{n,l}^{-1}\, Z_{n, l}, \ \sigma_{n,l_1}^{-1}\,Z_{n, l_1}) \Big)_{l, l_1=1, \ldots, m} \to \Omega. \end{align} $$

Then, as $n\to \infty $ , we have

(5.3) $$ \begin{align} \Big(\sigma_{n,1}^{-1}\,Z_{n, 1}, \ldots, \sigma_{n,m}^{-1}\, Z_{n, m} \Big) \to_D N_m (0, \Omega). \end{align} $$

We next introduce our second lemma. Its proof is given in Section 5.2. Recall that $\rho _n=1+\frac {\tau }{k_n}$ , where $\tau>0$ and $ 0<k_n\to \infty $ satisfying $\lim _{n\to \infty }n/k_n=\infty $ and we still use the notations $\gamma (k), \Gamma _n$ and $\sigma _n^2$ as in Section 2.1. Write

$$ \begin{align*} S_{n, 1} = \sum_{k=1}^{M_{1n}} \rho_n^{-k}\, u_k \quad \text{and}\quad S_{n, 2}=\sum_{k=M_{2n}}^{n} \rho_n^{k-n}\, u_k , \end{align*} $$

where $M_{1n}$ and $M_{2n}$ are two sequences of positive integers satisfying that $1\le M_{in}\le n$ for $i=1$ and $2$ .

Lemma 5.2. Under the conditions of Theorem 2.1, for any $M_{1n}/k_n\to \infty $ and $M_{2n}/n\to 0$ , we have

(5.4) $$ \begin{align} var(S_{n, 1}); \quad var(S_{n, 2}) = \big[1+o(1)\big]\, \sigma_n^2 \end{align} $$

and

(5.5) $$ \begin{align} \Big(\sigma_{n}^{-1}\,S_{n, 1}, \ \sigma_{n}^{-1}\, S_{n, 2} \Big) \to_D (S_1, S_2), \end{align} $$

where $S_1, S_2\sim N(0,1)$ and $S_1$ is independent of $S_2$ .

Remark 5.1. (5.4) implies that for any $m_n$ satisfying $m_n/k_n\to \infty $ and $m_n/n\to 0$ ,

(5.6) $$ \begin{align} var (\widehat S_{n, 1}) =o(\sigma_n^2) \quad \text{and} \quad var (\widehat S_{n, 2}) =o(\sigma_n^2), \end{align} $$

where $ \widehat S_{n, 1} =\sum _{k=m_n}^{n} \rho _n^{-k}\, u_k $ and $ \widehat S_{n, 2}=\sum _{k=1}^{m_n} \rho _n^{k-n}\, u_k. $ Furthermore, the same arguments given in the proof of (5.4) yield that, uniformly for $1\le j\le n$ ,

(5.7) $$ \begin{align} \mathbb E a_j^2 + \mathbb Eb_j^2 \le A_0\, \sigma_n^2, \end{align} $$

where $a_j=\sum _{k=1}^{j} \rho _n^{-k}\, u_k$ , $b_j=\sum _{k=j}^{n} \rho _n^{-k}\, u_k$ , and $A_0$ is an absolute constant.

Remark 5.2. Results (5.4)–(5.7) still hold if $u_k$ is replaced by $w_k=u_k+z_{k-1}-z_k$ , where $z_k$ is an arbitrary random variable satisfying $\sup _{k\ge 1}\mathbb {E} z_k^2<\infty $ . To illustrate, we consider (5.4). In fact, for any $M\ge 1$ , we have

$$ \begin{align*} A_M:=\sum_{k=1}^{M} \rho_n^{-k} (z_{k-1}-z_k) = \rho_n^{-M} z_M +(\rho_n^{-1}-1)\sum_{k=1}^{M-1} \rho_n^{-k} z_k. \end{align*} $$

Note that, by Hólder’s inequality,

$$ \begin{align*} \mathbb EA_M^2&\le \mathbb{E} z_1^2 + (\rho_n^{-1}-1)^2 \sum_{k=1}^{M-1}\rho_n^{-k} \sum_{k=1}^{M-1} \rho_n^{-k} \mathbb E z_k^2 \nonumber\\ &\le \mathbb{E} z_1^2 + C\, (\rho_n^{-1}-1)^2 \big(\sum_{k=1}^{M-1}\rho_n^{-k}\big)^2 \le C\, \end{align*} $$

uniformly for $1\le M\le n$ . It follows from $\sum _{k=1}^{M_{1n}} \rho _n^{-k}\, w_k = S_{n,1}+A_{M_{1n}}$ that

$$ \begin{align*} var\big(\sum_{k=1}^{M_{1n}} \rho_n^{-k}\, w_k\big) &= var (S_{n,1}) +var (A_{M_{1n}}) +2 cov (S_{n1}, A_{M_{1n}}) \nonumber\\ &= var (S_{n, 1}) +o(\sigma_n^2) =\big[1+o(1)\big]\, \sigma_n^2. \end{align*} $$

Similarly, we have

$$ \begin{align*} var\big(\sum_{k=M_{2n}}^n \rho_n^{k-n}\, w_k \big) = var(S_{n,2})+o(\sigma_n^2)=\big[1+o(1)\big]\, \sigma_n^2. \end{align*} $$

Hence, (5.4) holds if $u_k$ is replaced by $w_k=u_k+z_{k-1}-z_k$ .

5.2 Proof of Lemma 5.2

First note that, for any $M_{1n}/k_n\to \infty $ and $M_{2n}/n\to 0$ ,

(5.8) $$ \begin{align} \frac 1{k_n}\sum_{k=1}^{M_{1n}} \rho_n^{-2k} =\frac {1- \rho_n^{-2M_{1n}} }{k_n ( \rho_n^2-1)} \to (2\tau)^{-1} \end{align} $$

and

(5.9) $$ \begin{align} \frac 1{k_n}\sum_{k=M_{2n}}^n \rho_n^{2(k-n)} =\frac 1{k_n}\sum_{k=0}^{n-M_{2n}} \rho_n^{-2k} \to (2\tau)^{-1}. \end{align} $$

Without loss of generality, assume that $M_{1n}=n$ and $M_{2n}=1$ in the proof for the convenience of notation. It is readily seen from (5.8) that

(5.10) $$ \begin{align} var(S_{n, 1}) &= \sum_{k=1}^{n}\rho_n^{-2k}\, \gamma(0)+2\sum_{k=1}^{n} \sum_{j=k+1}^{n} \rho_n^{-k}\rho_n^{-j} \gamma(j-k) \nonumber\\ &= \sum_{k=1}^{n}\rho_n^{-2k}\, \gamma(0)+2\sum_{k=1}^{n} \rho_n^{-2k}\sum_{j=1}^{{n}-k} \rho_n^{-j} \gamma(j) \nonumber\\ &=2\, \Gamma_n\,\sum_{k=1}^{n}\rho_n^{-2k}\,-R_n = \big[1+o(1)\big]\, \sigma_n^2 -R_n, \end{align} $$

where $R_n = 2\, \sum _{k=1}^{n} \rho _n^{-2k} \,\sum _{j={n}-k+1}^{n} \rho _n^{-j} \gamma (j).$ We may write

$$ \begin{align*} R_n &=2\,\sum_{j=1}^{n} \rho_n^{-j} \gamma(j) \sum_{k=n-j+1}^n \rho_n^{-2k} =2 \rho_n^{-2n}\, \sum_{j=1}^{n} \rho_n^{-j} \gamma(j) \sum_{k=0}^{j-1} \rho_n^{2k} \nonumber\\ &=\frac {2\rho_n^{-2n}}{\rho_n^2-1}\,\big[\sum_{j=1}^{n} \rho_n^{j} \gamma(j) - \sum_{j=1}^{n} \rho_n^{-j} \gamma(j) \big]. \end{align*} $$

It follows from conditions (a) and (b) of Theorem 2.1 and $k_n/n\to 0$ that

$$ \begin{align*} |R_n| &\le C\, k_n\, \rho_n^{-n}\sum_{j=1}^{n} |\gamma(j)| \le C_1\,k_n\, \rho_n^{-n}\sum_{j=1}^{n-k_n} |\gamma(j)| \\ &\le C\, \sum_{j=1}^{n-k_n} (n-j)\rho_n^{j-n}|\gamma(j)| =o(\sigma_n^2). \end{align*} $$

Taking this estimate into (5.10), we obtain $ var(S_{n, 1})= \big [1+o(1)\big ]\, \sigma _n^2$ . Similarly, we have $ var(S_{n, 2}) = \big [1+o(1)\big ]\, \sigma _n^2$ . This proves (5.4).

We next prove (5.5). Let $b_{nk, 1}=\rho _n^{-k}$ and $b_{nk, 2}=\rho _n^{k-n}$ . It is routine to see that, for $l=1$ and $2$ ,

$$ \begin{align*} |b_{n1, l}|+\sum_{k=2}^n |b_{nk, l}-b_{nk-1, l}| \le \rho_n^{-1}+ (1-\rho_n^{-1})\, \sum_{k=1}^n \rho_n^{-k} \le C<\infty. \end{align*} $$

By using Lemma 5.1, result (5.5) will follow if we prove

(5.11) $$ \begin{align} cov (\sigma_{n}^{-1}\,S_{n, 1}, \sigma_{n}^{-1}\, S_{n, 2} ) = \sigma_n^{-2} E \big(S_{n, 1}S_{n, 2} \big)\to 0. \end{align} $$

In fact, by noting $\gamma (j)=\gamma (-j)$ and recalling $\sum _{k=0}^{n} (n-k) \rho _n^{k-n}\,|\gamma (k)|=o(\sigma _n^2)$ , simple calculation shows that

$$ \begin{align*} E \big(\ S_{n, 1} S_{n, 2} \big) &=\sum_{k=1}^{n} \sum_{j=1}^{n} \rho_n^{k-j-n}\,\gamma(k-j)= \sum_{j=1}^{n} \sum_{k=1-j}^{n-j} \rho_n^{k-n}\,\gamma(k)\nonumber\\ &= \sum_{j=1}^n \sum_{k=1}^{n-j} \rho_n^{k-n}\,\gamma(k)+ \sum_{j=1}^n \sum_{k=0}^{j-1} \rho_n^{-k-n}\,\gamma(k)\nonumber\\ &= \sum_{k=0}^{n-1}(n-k) \rho_n^{k-n}\,\gamma(k)+ \sum_{k=0}^{n-1} (n-k)\rho_n^{-k-n}\,\gamma(k) \nonumber\\ &= o(\sigma_n^2). \end{align*} $$

This proves (5.11) and also completes the proof of Lemma 5.2.

5.3 Proof of Theorem 2.1

Without loss of generality, assume $y_0=0$ . First note that, by squaring (1.1) and summing over $k\in \{1, \ldots ,n\}$ ,

(5.12) $$ \begin{align} (\rho_n^2-1 )\, \sum_{k=1}^n y_{k-1}^2 = y_n^2 - 2 \sum_{k=1}^n y_{k-1} u_k- \sum_{k=1}^n u_{k}^2. \end{align} $$

Furthermore, by taking $M=m_n$ so that $m_n/k_n\to \infty $ and $m_n/n\to 0$ , we may write

(5.13) $$ \begin{align} \sum_{k=1}^n y_{k-1} u_k &= \sum_{k=1}^M y_{k-1} u_k + \sum_{k=M}^n \rho_n^{k-1} u_k \sum_{i=M+1}^{k-1} \rho_n^{-i} u_i + \sum_{k=M}^n \rho_n^{k-1} u_k \sum_{i=0}^M \rho_n^{-i} u_i \nonumber\\ &:=\Delta_{n1, M}+\Delta_{n2, M}+\Delta_{n3, M}. \end{align} $$

Since $y_n=\rho _n^n\, \sum _{i=1}^n\rho _n^{-i} u_i$ and

(5.14) $$ \begin{align} \frac 1 {\sigma_n}\sum_{i=1}^n\rho_n^{-i} u_i = \frac 1{\sigma_n}\sum_{i=0}^M \rho_n^{-i} u_i +o_P(1), \quad \frac 1 {\sigma_n}\sum_{k=1}^n \rho_n^{k-n} u_k=\frac 1 {\sigma_n} \sum_{k=M}^n \rho_n^{k-n} u_k+o_P(1), \end{align} $$

by using (5.6), it follows from Lemma 5.2 and the continuous mapping theorem that

(5.15) $$ \begin{align} \big( \rho_n^{-n} \sigma_n^{-2}\, \Delta_{n3, M}, \ \rho_n^{-2n}\, \sigma_n^{-2} y_n^2\big) \to_D (XY,\ Y^2), \end{align} $$

where $X, Y\sim N(0,1)$ and X and Y are independent. In terms of (5.12)–(5.15), to show (1.2), it suffices to show that

(5.16) $$ \begin{align} \frac {\rho_n^{-n}}{\sigma_n^2} \,\big(\Delta_{1n, M}+\Delta_{2n, M}\big) = o_P(1). \end{align} $$

Indeed, by noting $\sum _{k=1}^nu_k^2=O_P(n)$ and recalling $n\rho _n^{-n}=o(\sigma _n^2)$ , simple calculation from (5.15) and (5.16) shows that

(5.17) $$ \begin{align} \rho_n^{-n}\sigma_n^{-2} \sum_{k=1}^n y_{k-1} u_k = \rho_n^{-n} \sigma_n^{-2}\, \Delta_{n3, M} +o_P(1)=O_P(1) \end{align} $$

and $ \rho _n^{-2n}\, \sigma _n^{-2} (\rho _n^2-1 )\, \sum _{k=1}^n y_{k-1}^2 =\rho _n^{-2n}\, \sigma _n^{-2} y_n^2 +o_P(1). $ Consequently, we have

$$ \begin{align*} \frac {\rho_n^{n}}{\rho_n^2-1} (\widehat \rho_n-\rho_n) &= \big[\rho_n^{-2n}\, \sigma_n^{-2} (\rho_n^2-1 )\,\sum_{k=1}^n y_{k-1}^2\big]^{-1}\,\rho_n^{-n}\sigma_n^{-2} \sum_{k=1}^n y_{k-1} u_k \nonumber\\ &= \frac { \rho_n^{-n} \sigma_n^{-2}\, \Delta_{n3, M} +o_P(1)}{\rho_n^{-2n}\, \sigma_n^{-2} y_n^2 +o_P(1)} \to_D X/Y=_D{\cal C}, \end{align*} $$

as required.

We next prove (5.16). Let $a_{j}=\sum _{i=1}^j \rho _n^{-i}\, u_i$ and $a_0=0$ . Note that

$$ \begin{align*} 2\Delta_{1n, M} &= \sum_{k=1}^M \rho_n^{2k} \big(a_k^2-a_{k-1}^2\big) -\sum_{k=1}^M u_k^2 \nonumber\\ &= \rho_n^{2M}a_M^2 +\sum_{k=1}^{M-1} \big(\rho_n^{2k}-\rho_n^{2k+2}\big)a_k^2-\sum_{k=1}^M u_k^2. \end{align*} $$

It follows from (5.7), $m_n/n\to 0$ and $n\rho _n^{-n}/\sigma _n^2\to 0$ that

$$ \begin{align*} \sigma_n^{-2}\,\rho_n^{-n}\, E |\Delta_{1n, M}| \le A_0\, \rho_n^2 \rho_n^{2m_n-n} +C\,m_n\rho_n^{-n} \mathbb{E}u_1^2/\sigma_n^2\to 0, \end{align*} $$

i.e., $\Delta _{1n, M}=o_P\big (\sigma _n^2\rho _n^n\big )$ .

The proof for $\Delta _{2n, M}=o_P\big (\sigma _n^2\rho _n^n\big )$ is more laborious. To this end, let $ b_k=\sum _{i=k+1}^n \rho _n^{-i} u_i$ and $\widetilde b_k=\sum _{i=M+1}^k \rho _n^{-i} u_i$ . We have

$$ \begin{align*} 2 \rho_n\, \Delta_{2n, M} &= \sum_{k=M+1}^n \rho_n^{2k} \big(\widetilde b_k^2-\widetilde b_{k-1}^2\big)-\sum_{k=M+1}^n u_k^2 \nonumber\\ &= \rho_n^{2n} \,\widetilde b_n^2 +\sum_{k=M+1}^{n-1} \big(\rho_n^{2k}- \rho_n^{2k+2}\big)\widetilde b_k^2-\sum_{k=M+1}^n u_k^2 \nonumber\\ &= \rho_n^{2{M+1}} \,\widetilde b_n^2 +\sum_{k=M+1}^{n-1} \big(\rho_n^{2k}- \rho_n^{2k+2}\big) (\widetilde b_k^2-\widetilde b_n^2)-\sum_{k=M+1}^n u_k^2. \end{align*} $$

Note that $\rho _n^{2{M+1}} \,\widetilde b_n^2-\sum _{k=M+1}^n u_k^2=o_P\big (\sigma _n^2\rho _n^n\big )$ as in the proof of $\Delta _{1n, M}=o_P\big (\sigma _n^2\rho _n^n\big )$ . It suffices to show that

(5.18) $$ \begin{align} {\cal L}_n :=\sum_{k=M+1}^{n-1} \big(\rho_n^{2k}- \rho_n^{2k+2}\big) (\widetilde b_k^2-\widetilde b_n^2) =o_P\big(\sigma_n^2\rho_n^n\big). \end{align} $$

Let ${\cal L}_{1n}=(1-\rho _n^2)\sum _{k=M+1}^{n-1} \rho _n^{2k} \, b_k^2$ . We may write

(5.19) $$ \begin{align} {\cal L}_n &=(1-\rho_n^2)\sum_{k=M+1}^{n-1} \rho_n^{2k} \big[(\widetilde b_n- b_k)^2-\widetilde b_n^2\big]\nonumber\\ &= -2\widetilde b_n\, (1-\rho_n^2)\,\sum_{k=M+1}^{n-1} \rho_n^{2k}\, b_k \, +{\cal L}_{1n}\nonumber\\ &= -2\widetilde b_n\,(1-\rho_n^2)\sum_{i=M+2}^n \rho_n^{-i} u_i \sum_{k=M+1}^{i-1}\rho_n^{2k}+{\cal L}_{1n} \nonumber\\ &=-2\widetilde b_n\,\sum_{i=M+2}^n \rho_n^{-i} u_i \big[\rho_n^{2(M+1)}-\rho_n^{2i}\big]+{\cal L}_{1n} \nonumber\\ &=-2\,\rho_n^{2(M+1)}\, \widetilde b_n \, b_{M+1} +2 \rho_n^n \, \widetilde b_n\, \widehat b_M +{\cal L}_{1n} , \end{align} $$

where $\widehat b_M=\sum _{i=M+2}^n \rho _n^{i-n} u_i$ . Using (5.4) and (5.6), we have

$$ \begin{align*} \sigma_n^{-2}\, E |\widetilde b_n \, b_{M+1}| &\le\sigma_n^{-2}\, (E\widetilde b_n^2)^{1/2} (Eb_{M+1}^2)^{1/2} \to 0, \nonumber\\ \sigma_n^{-2}\,E |\widetilde b_n\, \widehat b_M | &\le\sigma_n^{-2}\, (E\widetilde b_n^2)^{1/2} (E \widehat b_{M}^2)^{1/2} \to 0. \end{align*} $$

On the other hand, it follows from (5.7) that, for each $1\le k\le n$ ,

$$ \begin{align*} \rho_n^{2k} Eb_k^2 = E \big(\sum_{i=1}^{n-k}\rho_n^{-i}u_{i+k}\big)^2=E \big(\sum_{i=1}^{n-k}\rho_n^{-i}u_{i}\big)^2 \le C\sigma_n^2, \end{align*} $$

indicating that

$$ \begin{align*} \sigma_n^{-2}\rho_n^{-n}E |{\cal L}_{1n}| \le C\, n(1-\rho_n^2)\rho_n^{-n}\le C_1 (n/k_n)e^{-\tau n/k_n}\to 0. \end{align*} $$

Taking these estimates into (5.19), we obtain

$$ \begin{align*} \sigma_n^{-2}\rho_n^{-n}\, \mathbb{E}|{\cal L}_n| \le C\,\sigma_n^{-2} \big(E |\widetilde b_n \, b_{M+1}|+ E |\widetilde b_n\, \widehat b_M |\big) +\sigma_n^{-2}\rho_n^{-n}E |{\cal L}_{1n}| \to 0, \end{align*} $$

implying (5.18). This proves $\Delta _{2n, M}=o_P\big (\sigma _n^2\rho _n^n\big )$ and hence completes the proof of (5.16). The proof of Theorem 2.1 is now complete.

5.4 Proof of Theorem 2.2

As noticed in Remark 4.2, (5.4)–(5.7) still hold if $u_k=\sum _{j=0}^\infty \psi _j\epsilon _{k-j}$ is replaced by $u_k=\epsilon _k+z_{k-1}-z_k$ , the proof following the same lines as in the proof of Theorem 2.1, and hence the details are omitted.

5.5 Proof of Proposition 2.1

Recall that $\Gamma _n = \frac 12 {\gamma (0)}+ \sum _{j=1}^n \rho _n^{-j}\gamma (j)$ , where $\gamma (j)=\gamma (-j)$ and $\rho _n=1+\tau /k_n$ with $\tau>0$ and $ 0<k_n \to \infty $ satisfying $\lim _{n\to \infty }n/k_n=\infty $ . We start with (2.2) and, without loss of generality, assume that $\gamma (j)= j^{-\beta }\,l(j)$ . First note that $l(a k_n)/l(k_n)\to 1$ for any $0<a<\infty $ and, as $M\to \infty $ ,

$$ \begin{align*} \sum_{j=Mk_n}^n \rho_n^{-j}|\gamma(j)| &\le C\, \sum_{j=Mk_n}^n e^{-\tau j/k_n} |l(j)|j^{-\beta} \le C\, l(k_n) k_n^{-\beta} \sum_{j=Mk_n}^n e^{-\tau j/k_n} \nonumber\\ &\le Ce^{-\tau M} l(k_n) k_n^{1-\beta}=o\big[ l(k_n) k_n^{1-\beta}\big], \end{align*} $$

and

$$ \begin{align*} \sum_{j=1}^{k_n/M} \rho_n^{-j}|\gamma(j)| &\le C\, \sum_{j=1}^{k_n/M} |l(j)|j^{-\beta} \le C\, l(k_n/M) (k_n/M)^{1-\beta} \nonumber\\ &\le CM^{\beta-1} |l(k_n)| k_n^{1-\beta}=o\big[ l(k_n) k_n^{1-\beta}\big]. \end{align*} $$

It suffices to show that, for each $M\ge 1$ ,

(5.20) $$ \begin{align} \frac {l^{-1}(k_n)}{\, k_n^{1-\beta}}\,\,\sum_{j=k_n/M}^{Mk_n} \rho_n^{-j}\gamma(j) \to \int_{1/M}^M e^{-\tau x} x^{-\beta}dx , \quad \text{as } n\to\infty. \end{align} $$

This is simple. Indeed, by noting $[y]$ is right continuous where $[y]$ is the integer part of y, we have $\rho _n^{-[k_n x]}\to e^{-\tau x}$ and

$$ \begin{align*}\frac { l([k_n x]) [k_nx]^{-\beta} }{l(k_n) k_n^{-\beta} x^{-\beta}} \to 1,\end{align*} $$

uniformly for $1/M\le x\le M$ . As a consequence, simple calculation shows that

$$ \begin{align*} \frac {l^{-1}(k_n)}{\, k_n^{1-\beta}} \sum_{j=k_n/M}^{Mk_n} \rho_n^{-j}\gamma(j) &= \int_{1/M}^M \rho_n^{-[k_n x]}\, \frac { l([k_n x]) [k_nx]^{-\beta} }{l(k_n) k_n^{-\beta} x^{-\beta}}\, x^{-\beta} dx + o(1) \nonumber\\ &\to \int_{1/M}^M e^{-\tau x} x^{-\beta}dx, \end{align*} $$

as required.

The proof of (2.3) is simple and hence the details are omitted.

We next prove (2.4). Since $\sum _{j\in \mathbb {Z}} \, \gamma (j)=0$ and $n/k_n\to \infty $ , we have

$$ \begin{align*} \gamma(0)+2 \sum_{j=1}^n \gamma(j) = -2C_\beta\, \sum_{j=n+1}^\infty j^{-1-\beta}= o(k_n^\beta). \end{align*} $$

It suffices to show that

(5.21) $$ \begin{align} \widetilde \Gamma_n &:= k_n ^\beta\, \big[ \Gamma_n- \frac 12 \gamma(0)- \sum_{j=1}^n \gamma(j) \big] =k_n^{\beta}\, \sum_{j=1}^n \gamma(j)\big( \rho_n^{-j}-1\big) \nonumber\\ &\to C_\beta\, \int_{0}^\infty (e^{-\tau x}-1) \, x^{-1-\beta} dx. \end{align} $$

For $M\ge 1$ , we may write

(5.22) $$ \begin{align} \widetilde \Gamma_n &= C_\beta\, k_n^\beta\, \Big(\sum_{j=1}^{k_n/M}+\sum_{j=k_n/M+1}^{Mk_n}+\sum_{j=Mk_n+1}^n \Big)\, \big( \rho_n^{-j}-1\big) \, j^{-1-\beta}\nonumber\\ &:= \widetilde \Gamma_{1n} + \widetilde \Gamma_{2n} +\widetilde \Gamma_{3n}. \end{align} $$

As in the proof of (2.2), for each $M\ge 1$ , we have

$$ \begin{align*} \widetilde \Gamma_{2n} \to C_\beta\int_{1/M}^M (e^{-\tau x}-1) \, x^{-1-\beta} dx, \quad \text{as } n\to\infty. \end{align*} $$

On the other hand, it is readily seen that

$$ \begin{align*} |\widetilde \Gamma_{3n}| \le C\,k_n^\beta\, \sum_{j=Mk_n+1}^n \, j^{-1-\beta} \le C/M^{\beta}. \end{align*} $$

As for $\Gamma _{1n}$ , by noting that $ 1-\rho _n^{-j} \le 2j\tau /k_n $ whenever M is large enough and $1\le j\le k_n/M$ , we have

$$ \begin{align*} \widetilde \Gamma_{1n} \le C\, k_n^{-1+\beta}\, \sum_{j=1}^{k_n/M} j^{-\beta} \le CM^{\beta-1}. \end{align*} $$

Taking these estimates into (5.22), we establish (5.21) by $n\to \infty $ first and then $M\to \infty $ .

We finally prove (2.5). We only verify (2.5) under $\sum _{j\in \mathbb {Z}} \, \gamma (j)=0$ and $\gamma (j) \sim C_\beta \, j^{-1-\beta }$ for some $0<\beta <1$ . Others are obvious and the details are omitted. In fact, under given conditions, it follows from $\gamma (j)=\gamma (-j)$ and (2.4) that

$$ \begin{align*} \frac 12 \gamma(0)+\frac 1n \sum_{j=1}^n (n-j)\gamma(j) &\le C\,\sum_{j={n+1}}^\infty j^{-1-\beta}+\frac Cn\sum_{j=1}^n j^{-\beta} \nonumber\\ & \le C\, n^{-\beta} \le C\, (n \Gamma_n/k_n)\, (k_n/n)^{1-\beta} =o (n \Gamma_n/k_n), \end{align*} $$

as required. The proof of Proposition 2.1 is now complete.

5.6 Proof of Corollary 2.1

We only verify the condition (c) of Theorem 2.1 under C3(i) and C3(iii). In terms of Proposition 2.1, others conditions in Theorem 2.1 are obvious and hence the details are omitted. We start with the verification under C3(i) and, without loss of generality, assume that $\gamma (j)= j^{-\beta }\,l(j)$ for $j\ge 1$ . In fact, for each $1\le M<n/k_n$ , standard arguments as in the proof of Proposition 2.1 yield

(5.23) $$ \begin{align} R_{1M}&:=\sum_{j=Mk_n}^{n} (n-j) \rho_n^{j-n}\,j^{-\beta}|l(j)| \le C\, M^{-\beta} k_n^{-\beta}l(M\,k_n) \sum_{j=1}^{n}j \rho_n^{-j} \nonumber\\ &\le C\, M^{-\tilde \beta} k_n^{-\beta}l(k_n)\int_1^n x e^{-\tau x/k_n} dx \quad (\text{where } 0<\tilde \beta<\beta)\nonumber\\ &\le CM^{-\tilde \beta} k_n^{2-\beta}l(k_n). \end{align} $$

Similarly, by noting $\rho _n^{M k_n}\le C\,e^{M\tau }$ , we have

(5.24) $$ \begin{align} R_{2M} &:=\sum_{j=1}^{Mk_n} (n-j) \rho_n^{j-n}\,j^{-\beta}|l(j)| \le C\, n\rho_n^{-n}\, e^{M\tau}\, \sum_{j=1}^{Mk_n} j^{-\beta}|l(j)| \nonumber\\ &\le C\,M\,e^{M\tau}\, (n/k_n)\, e^{-\tau n/k_n}\, k_n^{2-\beta} l(k_n). \end{align} $$

Recall that $\sigma _n^2 \asymp k_n^{2-\beta } l(k_n)$ under C3(i) and $(n/k_n)\, e^{-\tau n/k_n}\to 0$ due to $n/k_n\to \infty $ . It follows from (5.23) and (5.24) by taking $M=M_n\to \infty $ but slow enough so that $M\,e^{M\tau }\, (n/k_n)\, e^{-\tau n/k_n}\to 0$ that

$$ \begin{align*} \sum_{j=0}^{n} (n-j) \rho_n^{j-n}\,|\gamma(j)| \le n\rho_n^{-n}+R_{1M}+R_{2M} =o(\sigma_n^2), \end{align*} $$

as required.

We next consider the verification under C3(iii). In this case, for each $1\le M<n/k_n$ , we have

(5.25) $$ \begin{align} R_{3M}&:=\sum_{j=Mk_n}^{n} (n-j) \rho_n^{j-n}\,j^{-1-\beta} \le C\, M^{-1-\beta} k_n^{-1-\beta}\, \sum_{j=1}^{n}j \rho_n^{-j} \nonumber\\ &\le C\,M^{-1-\beta} k_n^{-1-\beta}\, \, \int_1^n x e^{-\tau x/k_n} dx \le CM^{-1} k_n^{1-\beta}, \end{align} $$

and

(5.26) $$ \begin{align} R_{4M} &:=\sum_{j=1}^{Mk_n} (n-j) \rho_n^{j-n}\,j^{-1-\beta} \le C\, n\rho_n^{Mk_n-n} \nonumber\\ &\le C\,e^{M\tau}\,n\rho_n^{-n}. \end{align} $$

It follows from (5.25) and (5.26) by taking $M=M_n\to \infty $ but slow enough so that ${e^{M\tau }}\,n\rho _n^{-n}=o(\sigma _n^2)$ , i.e, ${e^{M\tau }}\, nk_n^{\beta -1}e^{-\tau n/k_n}\to 0$ that

$$ \begin{align*} \sum_{j=0}^{n} (n-j) \rho_n^{j-n}\,|\gamma(j)| \le n\rho_n^{-n}+R_{3M}+R_{4M} =o(\sigma_n^2), \end{align*} $$

under C3(iii) and the additional condition $n\rho _n^{-n}=o(\sigma _n^2)$ or $nk_n^{\beta -1}e^{-\tau n/k_n}\to 0$ . The proof of Corollary 2.1 is now complete.

5.7 Proof of Theorem 3.1

Without loss of generality, assume $y_0=0$ . We may write

$$ \begin{align*} \begin{bmatrix} \widehat \alpha_n\\ \widehat \rho_n \end{bmatrix}-\begin{bmatrix} \alpha_n\\ \rho_n \end{bmatrix} = \begin{bmatrix} n & \sum_{t=1}^ny_{t-1}\\ \sum_{t=1}^ny_{t-1} & \sum_{t=1}^ny_{t-1}^2 \end{bmatrix}^{-1}\, \begin{bmatrix} \sum_{t=1}^nu_{t}\\ \sum_{t=1}^ny_{t-1}u_t \end{bmatrix}, \end{align*} $$

indicating that

(5.27) $$ \begin{align} \widehat \rho_n -\rho_n =\frac {n \sum_{k=1}^n y_{k-1} u_{k} -\sum_{k=1}^n y_{k-1}\,\sum_{k=1}^n u_{k}}{ n\sum_{k=1}^n y_{k-1}^2 -\big(\sum_{k=1}^n y_{k-1}\big)^2}. \end{align} $$

To prove (3.2), first note that $var\big (\sum _{k=1}^n u_{k}\big )=2n \Gamma _n^*$ . This, together with the conditions that $\sigma _n^2=\tau ^{-1}k_n\Gamma _n$ and $k_n\Gamma _n^*=o(n \Gamma _n)$ , implies that

(5.28) $$ \begin{align} \frac {k_n}{n\sigma_n}\sum_{k=1}^n u_{k} =o_P(1). \end{align} $$

On the other hand, by summing (3.1) over $k\in \{1, \ldots ,k\}$ , we have

(5.29) $$ \begin{align} \frac {\tau}{k_n}\,\sum_{k=1}^n y_{k-1} =(\rho_n-1)\,\sum_{k=1}^n y_{k-1} = y_n -n\alpha_n- \sum_{k=1}^n u_k \end{align} $$

and, for $k\ge 1$ ,

(5.30) $$ \begin{align} y_k = \alpha_n \sum_{j=0}^{k-1}\rho_n + \sum_{j=1}^{k}\rho_n^{k-j}u_j =\tilde \alpha_{nk}+ \tilde y_{k} , \end{align} $$

where $ \tilde y_k =\rho _n \tilde y_{k-1}+ u_k$ and $\tilde \alpha _{nk}=\tau ^{-1}\,k_n\,\alpha _n (\rho _n^k-1)=l_n\sigma _n(\rho _n^k-1). $ Since $\tilde y_n=O_P(\sigma _n\rho _n^n)$ by (5.4), it follows from (5.28)–(5.30) and $(n/k_n)\rho _n^{-n}\to 0$ that

(5.31) $$ \begin{align} \frac 1{k_n}\, \frac 1{\rho_n^n\sigma_n}\,\sum_{k=1}^n y_{k-1} &= \frac {\tilde y_n}{\rho_n^n\sigma_n}+l_n -\frac 1{k_n}\,\frac 1{\rho_n^n\sigma_n}\,\Big(n\alpha_n+ \sum_{k=1}^n u_k\Big) \nonumber\\ &=O_P(1+|l_n|). \end{align} $$

Since (5.28) and (5.31) imply that

(5.32) $$ \begin{align} \frac 1n \, \frac 1{\rho_n^n\sigma_n^2}\, \sum_{k=1}^n y_{k-1}\,\sum_{k=1}^n u_{k}=o_P (1+|l_n|) \end{align} $$

and

(5.33) $$ \begin{align} \frac { \rho_n^2-1 }{\rho_n^{2n}\, \sigma_n^{2}}\, \frac 1 n\,\big(\sum_{k=1}^n y_{k-1}\big)^2 &= \frac {k_n^2 (\rho_n^2-1 )}n\, \Big[\frac 1{k_n}\, \frac 1{\rho_n^n\sigma_n}\,\sum_{k=1}^n y_{k-1}\Big]^2\nonumber\\ &=o_P \big[(1+|l_n|)^2\big] \end{align} $$

due to $\rho _n^2-1\le Ck_n^{-1}$ and $k_n/n\to 0$ , result (3.2) will follow if we prove

(5.34) $$ \begin{align} \rho_n^{-n}\, \sigma_n^{-2}\, \sum_{k=1}^n y_{k-1} u_{k} &=\Big[ \sigma_n^{-1} \sum_{i=0}^n \rho_n^{-i} u_{i} +l_n\Big]\, \sigma_n^{-1}\sum_{k=1}^n \rho_n^{k-n} u_{k} \nonumber\\ &\qquad +o_P\big(1+|l_n|\big) \end{align} $$

and

(5.35) $$ \begin{align} \rho_n^{-2n}\, \sigma_n^{-2} (\rho_n^2-1 )\,\sum_{k=1}^n y_{k-1}^2 =\Big( \sigma_n^{-1}\sum_{k=1}^n \rho_n^{-k} u_{k}+l_n \Big)^2 + o_P(1+|l_n|^2){.} \end{align} $$

Indeed, if $|l_n|\to \infty $ , it follows from (5.32)–(5.35) that

$$ \begin{align*} &\frac { l_n\,\rho_n^n}{\rho_n^2-1} \, \big(\widehat \rho_n- \rho_n\big) \nonumber\\ &= \frac {\frac {l_n^{-1}}{\rho_n^n\sigma_n^2}\, \sum_{k=1}^n y_{k-1} u_{k} -\frac 1n \, \frac {l_n^{-1}}{\rho_n^n\sigma_n^2}\, \sum_{k=1}^n y_{k-1}\,\sum_{k=1}^n u_{k}} {\frac {l_n^{-2}( \rho_n^2-1) }{\rho_n^{2n}\, \sigma_n^{2}}\, \,\sum_{k=1}^n y_{k-1}^2-\frac {l_n^{-2}( \rho_n^2-1) }{\rho_n^{2n}\, \sigma_n^{2}}\, \frac 1 n\,\big(\sum_{k=1}^n y_{k-1}\big)^2} \\ &=\frac { \sigma_n^{-1}\sum_{k=1}^n \rho_n^{k-n} u_{k} +o_P(1)}{1+o_P(1)}\to_D N(0, 1), \end{align*} $$

where we have used (5.5) in Lemma 5.2. Similarly, if $l_n\to l$ with $|l|<\infty $ , then

$$ \begin{align*} &\frac { \rho_n^n}{\rho_n^2-1} \, \big(\widehat \rho_n- \rho_n\big) \nonumber\\ &= \frac {\frac {1}{\rho_n^n\sigma_n^2}\, \sum_{k=1}^n y_{k-1} u_{k} -\frac 1n \, \frac {1}{\rho_n^n\sigma_n^2}\, \sum_{k=1}^n y_{k-1}\,\sum_{k=1}^n u_{k}} {\frac { \rho_n^2-1}{\rho_n^{2n}\, \sigma_n^{2}}\, \,\sum_{k=1}^n y_{k-1}^2-\frac { \rho_n^2-1 }{\rho_n^{2n}\, \sigma_n^{2}}\, \frac 1 n\,\big(\sum_{k=1}^n y_{k-1}\big)^2} \\ &=\frac {\Big( \sigma_n^{-1}\sum_{k=1}^n \rho_n^{-k} u_{k}+l \Big)\, \sigma_n^{-1}\sum_{k=1}^n \rho_n^{k-n} u_{k} +o_P(1)}{\Big( \sigma_n^{-1}\sum_{k=1}^n \rho_n^{-k} u_{k}+l \Big)^2+o_P(1)}\\ &\to_D X/(Y+l), \end{align*} $$

where X and Y are two independent standard normal variates, as required.

We next prove (5.34) and (5.35), starting with (5.34). It follows from (5.30) that

$$ \begin{align*} \sum_{k=1}^n y_{k-1} u_{k} &=\sum_{k=1}^n \tilde y_{k-1} u_{k} +\sum_{k=1}^n \tilde \alpha_{n, k-1} u_{k} \nonumber\\ &= \sum_{k=1}^n \tilde y_{k-1} u_{k} + \tau^{-1}\,k_n\,\alpha_n\,\Big[ \sum_{k=1}^n \rho_n^{k-1} u_{k}- \sum_{k=1}^n u_{k} \Big]. \end{align*} $$

Recall $ \tilde y_k =\rho _n \tilde y_{k-1}+ u_k$ . We have

(5.36) $$ \begin{align} \rho_n^{-n}\, \sigma_n^{-2}\sum_{k=1}^n \tilde y_{k-1} u_{k} =\Big[ \sigma_n^{-1} \sum_{i=0}^n \rho_n^{-i} u_{i} +l_n\Big]\, \sigma_n^{-1}\sum_{k=1}^n \rho_n^{k-n} u_{k} +o_P(1). \end{align} $$

See (5.14) and (5.17) in the proof of Theorem 2.1. This yields by using (5.28) and $(n/k_n)\rho _n^{-n}\to 0$ that

$$ \begin{align*} \rho_n^{-n}\, \sigma_n^{-2}\, \sum_{k=1}^n y_{k-1} u_{k} &= \rho_n^{-n}\, \sigma_n^{-2}\sum_{k=1}^n \tilde y_{k-1} u_{k} +\rho_n^{-n}\, \sigma_n^{-2} \tau^{-1}\,k_n\,\alpha_n\,\Big[ \sum_{k=1}^n \rho_n^{k-1} u_{k}- \sum_{k=1}^n u_{k} \Big] \nonumber\\ &= \rho_n^{-n}\, \sigma_n^{-2}\sum_{k=1}^n \tilde y_{k-1} u_{k} + l_n\sigma_n^{-1}\, \sum_{k=1}^n \rho_n^{k-n} u_{k} +o_P( 1+|l_n| )\nonumber\\ &=\Big[ \sigma_n^{-1} \sum_{i=0}^n \rho_n^{-i} u_{i} +l_n\Big]\, \sigma_n^{-1}\sum_{k=1}^n \rho_n^{k-n} u_{k} +o_P(1+|l_n|). \end{align*} $$

That is, (5.34) holds. To prove (5.35), we observe that $ y_k = \rho _n y_{k-1}+(u_k+\alpha _n).$ Squaring $y_k$ and summing over $k\in \{1, \ldots ,k\}$ yield

$$ \begin{align*} (\rho_n^2-1 )\, \sum_{k=1}^n y_{k-1}^2 &= y_n^2 - 2 \sum_{k=1}^n y_{k-1} (u_k+\alpha_n)- \sum_{k=1}^n (u_{k}+\alpha_n)^2 \\ &= y_n^2 - 2 \sum_{k=1}^n y_{k-1} u_k- 2\alpha_n \sum_{k=1}^n y_{k-1}- \sum_{k=1}^n (u_{k}+\alpha_n)^2. \end{align*} $$

In terms of (5.31), (5.34), and $\sum _{k=1}^n (u_{k}+\alpha _n)^2=O_P\big [n(1+\alpha _n^2)\big ]$ , simple calculation shows that

$$ \begin{align*} \rho_n^{-2n}\, \sigma_n^{-2} (\rho_n^2-1 )\,\sum_{k=1}^n y_{k-1}^2 &=\big( \frac { y_n}{\rho_n^n\sigma_n}\big)^2 + o_P(1+|l_n|^2)\\ &=\Big( \sigma_n^{-1}\sum_{k=1}^n \rho_n^{-k} u_{k}+l_n \Big)^2 + o_P(1+|l_n|^2), \end{align*} $$

as required in (5.35).

5.8 Proof of Theorem 3.2

This follows from the same arguments as in the proof of Theorem 3.1, by noting that the key mid-steps (5.28), (5.31), and (5.36) do not change if we replace ${u_k=\sum _{j=0}^\infty \psi _j\epsilon _{k-j}}$ by $u_k=\epsilon _k+z_{k-1}-z_k$ , as pointed out in the proof of Theorem 2.2.