CUSUM test for general nonlinear integer-valued GARCH models: comparison study

Abstract

This study considers the problem of testing a parameter change in general nonlinear integer-valued time series models where the conditional distribution of current observations is assumed to follow a one-parameter exponential family. We consider score-, (standardized) residual-, and estimate-based CUSUM tests and show that their limiting null distributions take the form of the functions of Brownian bridges. Based on the obtained results, we then conduct a comparison study of the performance of CUSUM tests through the use of Monte Carlo simulations. Our findings demonstrate that the standardized residual-based CUSUM test largely outperforms the others.

Appendix : Proofs of Theorems 1 and 2

Before proving Theorems 1 and 2, we prepare some lemmas. In what follows, we use notation $\eta _t^0 = \eta _t(\theta _0)$ and $\eta _t^n= \eta _t(\hat{\theta }_n)$.

Lemma 1

Suppose that conditions (A0), (A6), and (A10) hold. Then, we have

$$\begin{aligned} \sup _{\theta \in \Theta }|\tilde{X}_t(\theta ) - X_t (\theta )| \le V\rho ^t, \,\,\,\, \sup _{\theta \in \Theta }| \tilde{\eta }_t - \eta _t |\le V\rho ^t\,\,\,\,\, a.s. \end{aligned}$$

Proof

Note that

$$\begin{aligned} |\tilde{X}_t (\theta ) - X_t (\theta )|= & {} \left| f_{\theta }(\tilde{X}_{t-1}(\theta ), Y_{t-1}) -f_{\theta }( X_{t-1}(\theta ), Y_{t-1}) \right| \\\le & {} \omega _1 |\tilde{X}_{t-1}(\theta ) - X_{t-1} (\theta )| \le \omega _1^{t-1} |\tilde{X}_1 - X_1 (\theta )|. \end{aligned}$$

Then, using the mean value theorem and (A10), we have

$$\begin{aligned} | \tilde{\eta }_t - \eta _t |= & {} |B^{-1}(\tilde{X}_t(\theta )) - B^{-1}(X_t (\theta )) | \le \frac{\omega _1^{t-1}}{B^{\prime }(\eta _t^*)} |\tilde{X}_1 - X_1(\theta )|\\\le & {} \frac{\omega _1^{t-1}}{\underline{c}} |\tilde{X}_1 - X_1(\theta )|, \end{aligned}$$

where $\eta _t^* = B^{-1}(X_t^*)$ and $X_t^*$ is an intermediate point between $\tilde{X}_t(\theta )$ and $X_t(\theta )$. Hence, using (A6), we establish the lemma. $\square$

Lemma 2

Suppose that (A0)–(A13) hold. Then, under $H_0$, as $n\rightarrow \infty$,

$$\begin{aligned}&\mathrm {(i)}&\sup _{\theta \in \Theta } \left| \frac{1}{n} \sum _{t=1}^n \tilde{\ell }_t (\theta ) -\frac{1}{n} \sum _{t=1}^n \ell _t (\theta ) \right| \longrightarrow 0 \,\,\,\,\,a.s.;\\&\mathrm {(ii)}&\left\| \frac{1}{\sqrt{n}} \sum _{t=1}^n \frac{\partial \tilde{\ell }_t (\theta _0)}{\partial \theta } - \frac{1}{\sqrt{n}} \sum _{t=1}^n \frac{\partial \ell _t (\theta _0)}{\partial \theta } \right\| = o_P(1); \\&\mathrm {(iii)}&\sup _{\theta \in \Theta } \left\| \frac{\partial ^2 \tilde{\ell }_t (\theta )}{\partial \theta \partial \theta ^T } - \frac{\partial ^2 \ell _t (\theta )}{\partial \theta \partial \theta ^T } \right\| \longrightarrow 0 \,\,\,\,\, a.s.\,\,as\,\, t \rightarrow \infty ;\\&\mathrm {(iv)}&- \frac{1}{n} \frac{\partial ^2 \tilde{L}_n (\theta _n^*)}{\partial \theta \partial \theta ^T} \longrightarrow I(\theta _0)\,\,\,\,\,\,a.s., \end{aligned}$$

where $\theta _n^*$ is the intermediate point $\hat{\theta }_n$ and $\theta _0$.

Proof

(i) It suffices to show that

$$\begin{aligned} \sup _{\theta \in \Theta } |\tilde{\ell }_t(\theta ) - \ell _t(\theta ) | \rightarrow 0\ \ a.s.\ \ \text{ as }\,\,\, t \rightarrow \infty . \end{aligned}$$

(8)

Note that, by the mean value theorem, (A10), and Lemma 1,

$$\begin{aligned} | \tilde{\ell }_t(\theta ) - \ell _t(\theta ) |\le & {} | \tilde{\eta }_t - \eta _t| Y_t + |A(\tilde{\eta }_t) - A(\eta _t)|\\= & {} |B^{-1}(\tilde{X}_t(\theta )) - B^{-1}( X_t(\theta )) | Y_t\\&+ | A(B^{-1}(\tilde{X}_t(\theta )) ) - A(B^{-1}(X_t(\theta )) )| \\\le & {} \frac{Y_t + X_t^*}{B'(\eta _t^*)} |\tilde{X}_t(\theta ) - X_t(\theta )| \le \frac{Y_t + X_t^*}{\underline{c}} V\rho ^t \end{aligned}$$

for some intermediate points $X_t^*$ between $X_t(\theta )$ and $\tilde{X}_t(\theta )$ and $\eta _t^* = B^{-1}(X_t^*)$. Since

$$\begin{aligned} Y_t + X_t^* \le Y_t + X_t(\theta ) + |\tilde{X}_t(\theta ) - X_t(\theta )| \le Y_t + X_t(\theta ) + V\rho ^t, \end{aligned}$$

according to (A6), we can show that (8) holds.

(ii) Note that

$$\begin{aligned} \frac{\partial \ell _t(\theta )}{\partial \theta } = \left( Y_t - B(\eta _t) \right) \frac{\partial \eta _t}{\partial \theta } := U_t(\theta )\frac{\partial \eta _t}{\partial \theta }. \end{aligned}$$

Hence, we have

$$\begin{aligned}&\left\| \frac{1}{\sqrt{n}} \sum _{t=1}^n \frac{\partial \tilde{\ell }_t (\theta _0)}{\partial \theta } - \frac{1}{\sqrt{n}} \sum _{t=1}^n \frac{\partial \ell _t (\theta _0)}{\partial \theta } \right\| \nonumber \\&\quad \le \left\| \frac{1}{\sqrt{n}} \sum _{t=1}^n \tilde{U}_t(\theta _0 ) \left( \frac{\partial \tilde{\eta }_t^0}{\partial \theta } -\frac{\partial \eta _t^0}{\partial \theta }\right) \right\| + \left\| \frac{1}{\sqrt{n}} \sum _{t=1}^n \left( \tilde{U}_t(\theta _0)-U_t(\theta _0)\right) \frac{\partial \eta _t^0}{\partial \theta } \right\| .\qquad \end{aligned}$$

(9)

Using (A6), (A7), (A9)–(A11), Lemma 1, and the following facts: $|\tilde{U}_t (\theta )|\le |U_t(\theta )| + V\rho ^t$ and

$$\begin{aligned} \left\| \frac{\partial \tilde{\eta }_t^0}{\partial \theta } -\frac{\partial \eta _t^0}{\partial \theta } \right\|= & {} \left\| \frac{1}{B^{\prime }(\tilde{\eta }_t^0)} \frac{\partial \tilde{X}_t (\theta _0)}{\partial \theta }- \frac{1}{B^{\prime }(\eta _t^0)} \frac{\partial X_t (\theta _0)}{\partial \theta } \right\| \\\le & {} \frac{1}{B^{\prime }(\tilde{\eta }_t^0)} \left\| \frac{\partial \tilde{X}_t (\theta _0)}{\partial \theta }-\frac{\partial X_t (\theta _0)}{\partial \theta } \right\| + \left| \frac{1}{B^{\prime }(\tilde{\eta }_t^0)} - \frac{1}{B^{\prime }(\eta _t^0)} \right| \left\| \frac{\partial X_t (\theta _0)}{\partial \theta } \right\| , \end{aligned}$$

we can see that the first term of (9) is $o_P(1)$.

On the other hand, by Lemma 1, the second term of (9) is bounded by

$$\begin{aligned} \left\| \frac{1}{\sqrt{n}} \sum _{t=1}^n \left( \tilde{X}_t(\theta _0)-X_t(\theta _0)\right) \frac{\partial \eta _t^0}{\partial \theta } \right\| \le \frac{1}{\sqrt{n}} \sum _{t=1}^n \left\| \frac{\partial \eta _t^0}{\partial \theta } \right\| V\rho ^t, \end{aligned}$$

which is $o_P (1)$ due to (A7) and (A10).

(iii) Note that

$$\begin{aligned} \frac{\partial ^2 \ell _t(\theta )}{\partial \theta \partial \theta ^T} = -B^{\prime }(\eta _t) \frac{\partial \eta _t}{\partial \theta }\frac{\partial \eta _t}{\partial \theta ^T} + U_t(\theta )\frac{\partial ^2 \eta _t}{\partial \theta \partial \theta ^T}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \left\| \frac{\partial ^2 \tilde{\ell }_t(\theta )}{\partial \theta \partial \theta ^T} - \frac{\partial ^2 \ell _t(\theta )}{\partial \theta \partial \theta ^T} \right\|\le & {} \left\| B^{\prime }(\tilde{\eta }_t) \left( \frac{\partial \tilde{\eta }_t}{\partial \theta } - \frac{\partial \eta _t}{\partial \theta } \right) \frac{\partial \tilde{\eta }_t}{\partial \theta ^T} \right\| + \left\| B^{\prime }(\tilde{\eta }_t) \frac{\partial \eta _t}{\partial \theta } \left( \frac{\partial \tilde{\eta }_t}{\partial \theta ^T} - \frac{\partial \eta _t}{\partial \theta ^T} \right) \right\| \nonumber \\&+ \left\| \left\{ B^{\prime }(\tilde{\eta }_t)-B^{\prime }(\eta _t) \right\} \frac{\partial \eta _t}{\partial \theta } \frac{\partial \eta _t}{\partial \theta ^T} \right\| + \left\| \tilde{U}_t(\theta ) \left( \frac{\partial ^2 \tilde{\eta }_t}{\partial \theta \partial \theta ^T} - \frac{\partial ^2 \eta _t}{\partial \theta \partial \theta ^T} \right) \right\| \nonumber \\&+ \left\| \left\{ \tilde{X}_t (\theta )- X_t (\theta ) \right\} \frac{\partial ^2 \eta _t}{\partial \theta \partial \theta ^T} \right\| . \end{aligned}$$

(10)

Since $B^{\prime }(\eta _t)\partial \eta _t / \partial \theta = \partial X_t(\theta )/\partial \theta$, the first and second terms of the RHS of (10) converge to 0 a.s. as $t \rightarrow \infty$ because of (A7), (A9), and (A10). On the other hand, the fourth and fifth terms converge to 0 a.s. as $t \rightarrow \infty$ owing to (A6), (A7), (A9), and Lemma 1. Due to (A11), we have

$$\begin{aligned} \left\| \left\{ B^{\prime }(\tilde{\eta }_t)-B^{\prime }(\eta _t) \right\} \frac{\partial \eta _t}{\partial \theta } \frac{\partial \eta _t}{\partial \theta ^T} \right\| \le \left\| \frac{1}{{B^{\prime }(\eta _t)^2}} \frac{\partial X_t(\theta )}{\partial \theta } \frac{\partial X_t(\theta )}{\partial \theta ^T} \right\| \cdot V\rho ^t. \end{aligned}$$

Henceforth, the third term converges to 0 a.s. as $t \rightarrow \infty$ owing to (A7) and (A10).

(iv) This can be proven similarly to the proof of Proposition 5 of Lee et al. (2016). $\square$

Lemma 3

Suppose that conditions (A0)–(A13) hold. Then, under $H_0$, as $n\rightarrow \infty$,

$$\begin{aligned} \frac{1}{\sqrt{n}} \max _{1 \le k \le n} \left\| \tilde{S}_k(\hat{\theta }_n) - \left\{ \tilde{S}_k (\theta _0) - \frac{k}{n} \tilde{S}_n(\theta _0) \right\} \right\| = o_P(1), \end{aligned}$$

where $\tilde{S}_k (\theta ) = \sum _{t=1}^k \partial \tilde{\ell }_t (\theta )/\partial \theta .$

Proof

As $\hat{\theta }_n$ is the CMLE of $\theta _0$, it suffices to show that

$$\begin{aligned} \frac{1}{\sqrt{n}} \max _{1 \le k \le n} \left\| \tilde{S}_k(\hat{\theta }_n) - \tilde{S}_k (\theta _0) - \frac{k}{n} \left\{ \tilde{S}_n(\hat{\theta }_n) - \tilde{S}_n(\theta _0) \right\} \right\| = o_P(1). \end{aligned}$$

(11)

By Taylor’s theorem, we have

$$\begin{aligned} \tilde{S}_k (\hat{\theta }_n) = \tilde{S}_k(\theta _0) + \sum _{t=1}^k \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} (\hat{\theta }_n - \theta _0 ), \end{aligned}$$

where $\theta _n^*$ is an intermediate point between $\theta _0$ and $\hat{\theta }_n$. Thus, we have

$$\begin{aligned}&\frac{1}{\sqrt{n}} \max _{1 \le k \le n} \left\| \tilde{S}_k(\hat{\theta }_n) - \tilde{S}_k (\theta _0) - \frac{k}{n} \left\{ \tilde{S}_n(\hat{\theta }_n) - \tilde{S}_n(\theta _0) \right\} \right\| \\&\quad \le \frac{1}{\sqrt{n}} \max _{1 \le k \le n} \left\| \sum _{t=1}^k \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} (\hat{\theta }_n - \theta _0 ) - \frac{k}{n} \sum _{t=1}^n \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} (\hat{\theta }_n - \theta _0 ) \right\| \\&\quad \le \max _{1 \le k \le n} \frac{k}{n} \left\| \frac{1}{k} \sum _{t=1}^k \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} + I(\theta _0) \right\| \cdot \sqrt{n} \Vert \hat{\theta }_n - \theta _0 \Vert \\&\qquad + \max _{1 \le k \le n} \frac{k}{n} \left\| \frac{1}{n} \sum _{t=1}^n \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T}+ I(\theta _0) \right\| \cdot \sqrt{n} \Vert \hat{\theta }_n - \theta _0 \Vert := I_n + II_n. \end{aligned}$$

Note that, by Proposition 1 and (iv) in Lemma 2,

$$\begin{aligned} II_n\le \left\| \frac{1}{n} \sum _{t=1}^n \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T}+ I(\theta _0) \right\| \cdot \sqrt{n} \Vert \hat{\theta }_n - \theta _0 \Vert = o_p(1)\cdot O_P(1)= o_P(1). \end{aligned}$$

Meanwhile, due to (A7), for some $0<\gamma <1/2$,

$$\begin{aligned}&\max _{1 \le k \le n^{\gamma }} \frac{k}{n} \left\| \frac{1}{k}\sum _{t=1}^k \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} + I(\theta _0) \right\| \\&\quad \le \frac{n^{\gamma }}{n} \sum _{t=1}^{n^{\gamma }} \left\| \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} \right\| + \frac{n^{\gamma }}{n} \Vert I(\theta _0)\Vert = o_P(1)+o(1)=o_P(1). \end{aligned}$$

Furthermore, since

$$\begin{aligned} \max _{n^{\gamma } < k \le n} \left\| \frac{1}{k}\sum _{t=1}^k \frac{\partial ^2 \tilde{\ell }_t (\theta _n^*)}{\partial \theta \partial \theta ^T} + I(\theta _0) \right\| \rightarrow 0\,\,\,\,a.s. \end{aligned}$$

owing to (iv) in Lemma 2, we can show that $I_n = o_P(1)$. This asserts (11), and the lemma is validated. $\square$

Proof of Theorem 1

Since $\{\partial \ell _t (\theta _0)/\partial \theta , {\mathcal {F}}_t \}$ forms a sequence of stationary ergodic martingale differences, using a functional central limit theorem, we have

$$\begin{aligned} I(\theta _0)^{-1/2} \frac{1}{\sqrt{n}} S_{[ns]} (\theta _0) {\mathop {\longrightarrow }\limits ^{w}} {\mathbf B}_{d} (s), \end{aligned}$$

where $S_k (\theta ) = \sum _{t=1}^k {\partial \ell _t (\theta )}/{\partial \theta }$ and $\{\mathbf B_d (s), 0<s<1\}$ is a d-dimensional standard Brownian motion. Further, from (ii) in Lemma 2, we have

$$\begin{aligned} I(\theta _0)^{-1/2} \frac{1}{\sqrt{n}} \tilde{S}_{[ns]} (\theta _0) {\mathop {\longrightarrow }\limits ^{w}} {\mathbf B}_{d} (s). \end{aligned}$$

Then, using Lemma 3, we obtain

$$\begin{aligned} \hat{I}_n^{-1/2} \frac{1}{\sqrt{n}} \tilde{S}_{[ns]} (\hat{\theta }_n) {\mathop {\longrightarrow }\limits ^{w}} {\mathbf B}_{d}^{\circ } (s). \end{aligned}$$

This establishes the theorem. $\square$

Proof of Theorem 2

We consider $T_n^{\mathrm{res},2}$ only. (The proof of $T_n^{\mathrm{res},1} {\mathop {\longrightarrow }\limits ^{w}} \sup _{0\le s \le 1} | \mathbf {B}_1^{\circ } (s) |$ is similar to that of Kang and Lee (2014).)

We write

$$\begin{aligned} \hat{\epsilon }_{t,2} - \epsilon _{t,2}= & {} \frac{Y_t - \hat{X}_t}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{Y_t - X_t(\theta _0)}{\sqrt{B^{\prime }(\eta _t^0)}} \\= & {} (X_t(\theta _0)-\hat{X}_t) \left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t^0)}} \right) \\&+ \,\epsilon _{t,1}\left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t ^0)}} \right) + \frac{1}{\sqrt{B^{\prime }(\eta _t^0)}}(X_t(\theta _0) - \hat{X}_t))\\:= & {} R_{t,1} + R_{t,2}+R_{t,3}. \end{aligned}$$

Noting (3), it suffices to show that

$$\begin{aligned} \max _{1 \le k \le n} \frac{1}{\sqrt{n}} \left| \sum _{t=1}^k (\hat{\epsilon }_{t,2} - \epsilon _{t,2} ) - \frac{k}{n} \sum _{t=1}^n (\hat{\epsilon }_{t,2} - \epsilon _{t,2}) \right| = o_P(1), \end{aligned}$$

(12)

i.e., for $i=1,2,3$,

$$\begin{aligned} \max _{1 \le k \le n} \frac{1}{\sqrt{n}} \left| \sum _{t=1}^k R_{t,i} - \frac{k}{n} \sum _{t=1}^nR_{t,i} \right| = o_P(1). \end{aligned}$$

(13)

Firstly, we express

$$\begin{aligned} \max _{1 \le k \le n} \frac{1}{\sqrt{n}} \left| \sum _{t=1}^k R_{t,1} - \frac{k}{n} \sum _{t=1}^n R_{t,1} \right| \le \frac{2}{\sqrt{n}} \sum _{t=1}^n | R_{t,1} | \le I_{n,1} + I_{n,2}+I_{n,3}, \end{aligned}$$

where

$$\begin{aligned}&I_{n,1} = \frac{2}{\sqrt{n}} \sum _{t=1}^n \left| \left( X_t(\theta _0)-X_t(\hat{\theta }_n)\right) \left( \frac{1}{\sqrt{B^{\prime }(\eta _t^0)}} - \frac{1}{\sqrt{B^{\prime }( \eta _t^n)}} \right) \right| ,\\&I_{n,2} = \frac{2}{\sqrt{n}} \sum _{t=1}^n \left| \left( X_t(\theta _0)-X_t(\hat{\theta }_n)\right) \left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }( \eta _t^n)}} \right) \right| ,\\&I_{n,3} = \frac{2}{\sqrt{n}} \sum _{t=1}^n \left| \left( X_t(\hat{\theta }_n)-\hat{X}_t\right) \left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }( \eta _t^0)}} \right) \right| . \end{aligned}$$

Using the mean value theorem with intermediate points $\theta _{n,1}^*$ and $\theta _{n,2}^*$ between $\hat{\theta }_n$ and $\theta _0$, we have

$$\begin{aligned} I_{n,1}= & {} \frac{1}{\sqrt{n}} \sum _{t=1}^n \left| (\hat{\theta }_n - \theta _0)^T \frac{\partial X_t (\theta _{n,1}^*)}{\partial \theta }\cdot \frac{B^{\prime \prime }(\eta _t(\theta _{n,2}^*))}{B^{\prime }(\eta _t(\theta _{n,2}^*))^{3/2}} \frac{1}{B^{\prime }(\eta _t(\theta _{n,2}^*))} (\hat{\theta }_n - \theta _0)^T \frac{\partial X_t (\theta _{n,2}^*)}{\partial \theta } \right| \\\le & {} n \Vert \hat{\theta }_n - \theta _0 \Vert ^2 \frac{1}{\underline{c}} \frac{1}{n\sqrt{n}} \sum _{t=1}^n \left| \sup _{\theta \in \Theta }\frac{B^{\prime \prime }(\eta _t)}{B^{\prime }(\eta _t)^{3/2}}\right| \cdot \left\| \sup _{\theta \in \Theta } \frac{\partial X_t (\theta )}{\partial \theta }\right\| ^2=O_p(1) \cdot o_P(1) =o_P(1), \end{aligned}$$

where we have used Proposition 1 and (A7), (A10), and (A12). Since $\hat{\eta }_t$ can be represented as $\tilde{\eta }_t(\hat{\theta }_n)=B^{-1}(\tilde{X}_t(\hat{\theta }_n))$ with $\tilde{X}_1 (\hat{\theta }_n)=\hat{X}_1$, we have

$$\begin{aligned} I_{n,2}\le & {} \frac{1}{\sqrt{n}} \frac{1}{\underline{c}^{3/2}}\sum _{t=1}^n \left| \left( X_t(\theta _0)-X_t(\hat{\theta }_n)\right) \left( B'(\eta _t^n) - B'(\hat{\eta }_t) \right) \right| \\\le & {} \sqrt{n} \Vert \hat{\theta }_n - \theta _0 \Vert \frac{V}{n\underline{c}^{3/2}} \sum _{t=1}^n \rho ^t \left\| \frac{\partial X_t (\theta _{n,1}^*)}{\partial \theta } \right\| =O_p(1) \cdot o_P(1) =o_P(1) \end{aligned}$$

with intermediate point $\theta _{n,1}^*$ between $\hat{\theta }_n$ and $\theta _0$, due to (A7), (A10), and (A11). Furthermore, note that $|\hat{X}_t - X_t (\hat{\theta }_n)| \le V\rho ^t$ a.s. since owing to (A0),

$$\begin{aligned} |\hat{X}_t - X_t (\hat{\theta }_n)|= & {} \left| f_{\hat{\theta }_n}(\hat{X}_{t-1}, Y_{t-1}) -f_{\hat{\theta }_n}( X_{t-1}(\hat{\theta }_n), Y_{t-1}) \right| \\\le & {} \omega _1|\hat{X}_{t-1} - X_{t-1} (\hat{\theta }_n)| \le \omega _1^{t-1} |\hat{X}_1 - X_1 (\hat{\theta }_n)|. \end{aligned}$$

Then, by using this and (A10),

$$\begin{aligned} I_{n,3}\le & {} \frac{2}{\sqrt{n}} \sum _{t=1}^n V\rho ^t \left\| \sup _{\theta \in \Theta } \frac{2}{\sqrt{B^{\prime }(\eta _t)}}\right\| \le \frac{4V}{\sqrt{\underline{c}}\sqrt{n}} \sum _{t=1}^n \rho ^t =o_P(1). \end{aligned}$$

Thus, (13) holds for $i=1$.

Secondly, we express

$$\begin{aligned} \max _{1 \le k \le n} \frac{1}{\sqrt{n}} \left| \sum _{t=1}^k R_{t,2} - \frac{k}{n} \sum _{t=1}^n R_{t,2} \right| \le \frac{2}{\sqrt{n}} \sum _{t=1}^n | R_{t,2} | \le II_{n,1} + II_{n,2}, \end{aligned}$$

where

$$\begin{aligned} II_{n,1}= & {} \max _{1 \le k \le n} \frac{1}{\sqrt{n}} \left| \sum _{t=1}^k \epsilon _{t,1} \left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t^n)}} \right) \right. \\&\left. - \frac{k}{n} \sum _{t=1}^n \epsilon _{t,1} \left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t^n)}} \right) \right| ,\\ II_{n,2}= & {} \max _{1 \le k \le n} \frac{1}{\sqrt{n}} \left| \sum _{t=1}^k \epsilon _{t,1} \left( \frac{1}{\sqrt{B^{\prime }(\eta _t^n)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t^0)}} \right) \right. \\&\left. - \frac{k}{n} \sum _{t=1}^n \epsilon _{t,1} \left( \frac{1}{\sqrt{B^{\prime }(\eta _t^n)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t^0)}} \right) \right| . \end{aligned}$$

Similarly to the case of $I_{n,2}$, we have

$$\begin{aligned} II_{n,1}\le & {} \frac{2}{\sqrt{n}} \sum _{t=1}^n \left| \epsilon _{t,1} \left( \frac{1}{\sqrt{B^{\prime }(\hat{\eta }_t)}} - \frac{1}{\sqrt{B^{\prime }(\eta _t^n)}} \right) \right| \\\le & {} \frac{1}{\sqrt{n}} \frac{1}{\underline{c}^{3/2}} \sum _{t=1}^n \left| \epsilon _{t,1} (B'(\hat{\eta }_t) - B'(\eta _t^n)) \right| \le \frac{V}{\sqrt{n}} \frac{1}{\underline{c}^{3/2}} \sum _{t=1}^n | \epsilon _{t,1}|\rho ^t = o_P(1). \end{aligned}$$

Using Taylor’s theorem, we have

$$\begin{aligned} B^{\prime }(\eta _t^n)^{-1/2} = B^{\prime }(\eta _t^0)^{-1/2} - \frac{1}{2} Z_t(\theta _0)(\hat{\theta }_n - \theta _0)^T \frac{\partial \eta _t^0}{\partial \theta } -\frac{1}{2} (\hat{\theta }_n - \theta _0)^T \left( \zeta _t(\theta _n^*) - \zeta _t(\theta _0) \right) , \end{aligned}$$

where $\theta _n^*$ is an intermediate point between $\hat{\theta }_n$ and $\theta _0$, and $\zeta _t(\theta )=B''(\eta _t)B'(\eta _t)^{-3/2} \frac{\partial \eta _t}{\partial \theta }$, so that

$$\begin{aligned} II_{n,2}\le & {} II_{n,2}^{\prime } + II_{n,2}^{\prime \prime }, \end{aligned}$$

where

$$\begin{aligned} II_{n,2}^{\prime }= & {} \sqrt{n} \Vert \hat{\theta }_n - \theta _0\Vert \max _{1 \le k \le n} \frac{k}{n} \left| \frac{1}{k} \sum _{t=1}^k \epsilon _{t,1} \zeta _t(\theta _0) - \frac{1}{n}\sum _{t=1}^n \epsilon _{t,1} \zeta _t(\theta _0) \right| ,\\ II_{n,2}^{\prime \prime }= & {} \sqrt{n} \Vert \hat{\theta }_n - \theta _0\Vert \frac{2}{n}\sum _{t=1}^n |\epsilon _{t,1}|\left\| \zeta _t(\theta _n^*) - \zeta _t(\theta _0) \right\| . \end{aligned}$$

Since $\{\epsilon _{t,1} \zeta _t(\theta _0)\}$ is ergodic, $\sqrt{n} \Vert \hat{\theta }_n - \theta _0\Vert = O_P(1)$, and $E|\epsilon _{t,1}| \Vert \zeta _t(\theta _0)\Vert < \infty$ owing to (A6), (A7), (A10), and (A12), we have $II_{n,2}^{\prime }=o_P(1)$. Moreover, because

$$\begin{aligned} II_{n,2}^{\prime \prime } \le \sqrt{n} \Vert \hat{\theta }_n - \theta _0\Vert \frac{2}{n}\sum _{t=1}^n |\epsilon _{t,1}| \sup _{\Vert \theta -\theta _0\Vert \le \Vert \hat{\theta }_n - \theta _0\Vert } \left\| \zeta _t(\theta ) - \zeta _t(\theta _0) \right\| \end{aligned}$$

and $E \sup _{\theta \in \Theta } \Vert \zeta _t(\theta )\Vert ^2 < \infty$ owing to (A7), (A10), and (A12), using (A3), (A6), and the dominated convergence theorem, we can have $II_{n,2}^{\prime \prime }=o_P(1)$ (cf., Proposition 5 in Lee et al. (2016)). Thus, (13) holds for $i=2$.

Finally, using Taylor’s theorem, we have

$$\begin{aligned} X_t (\hat{\theta }_n) = X_t(\theta _0) + (\hat{\theta }_n - \theta _0)^T \frac{\partial X_t(\theta _0)}{\partial \theta } + (\hat{\theta }_n - \theta _0)^T \left( \frac{\partial X_t(\theta _n^*)}{\partial \theta } - \frac{\partial X_t(\theta _0)}{\partial \theta } \right) \end{aligned}$$

for some $\theta _n^*$ between $\hat{\theta }_n$ and $\theta _0$. Then, similarly to the case of $II_{n,2}$, we can show that (13) holds for $i=3$. Hence, (12) is verified. $\square$