Appendix
The following propositions are due to Lee and Song (2008) and Francq and Zakoïan (2004). Throughout this section, \(K>0\) and \(0<\rho <1\) denote universal constants and the norm of a matrix \(A=(a_{ij})\) is defined by \(\Vert A\Vert =\sum |a_{ij}|\).
Proposition 1
Under the same conditions in Theorem 3, we have
$$\begin{aligned} \max _{1\le k\le n}\frac{k}{\sqrt{n}}\Vert \varDelta _{1,k}\Vert =o_{\mathrm {P}}(1), \end{aligned}$$
where \(\varDelta _{1,k}= {\partial _\theta {\tilde{L}}_{k}(\theta _0)}-{\partial _\theta L_k(\theta _0)}\).
Proposition 2
Under the same conditions in Theorem 3, we have
$$\begin{aligned} \max _{1\le k\le n}\frac{k}{\sqrt{n}}\Vert \varDelta _{2,k}\Vert =o_{\mathrm {P}}(1), \end{aligned}$$
where \(\varDelta _{2,k}= \bigg ({\partial _{\theta \theta } {\tilde{L}}_k(\tilde{\theta }_{k})}-\mathcal {J}\bigg )({\hat{\theta }}_{k}-\theta _0)\).
Proposition 3
If \((\mathbf{A}0)\)–\((\mathbf{A}3)\) holds, then under \(H_0\),
$$\begin{aligned}&\sup _{\theta \in \varTheta }|\epsilon _t-{\tilde{\epsilon }}_t|\le K \rho ^t \ \ a.s.,\ \ \Big |\frac{1}{\sigma _t^2}-\frac{1}{{\tilde{\sigma }}_t^2}\Big |\le K\rho ^t \frac{S_{t-1}}{\sigma _t^2}, \ \ |{\tilde{\epsilon }}_t^2-\epsilon _t^2|\le K\rho ^t(|\epsilon _t|+1), \end{aligned}$$
where \(S_{t-1}=\sum _{k=1-q}^{t-1}(|\epsilon _k|+1)\). Further,
$$\begin{aligned} |{\tilde{\sigma }}_t^2-\sigma _t^2|\le & {} K\rho ^t, \ \ \Big |\frac{\partial {\tilde{\epsilon }}_t}{\partial \theta _i}-\frac{\partial \epsilon _t}{\partial \theta _i}\Big |\le K\rho ^t, \ \ \Big |\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _i}-\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |\nonumber \\\le & {} K\rho ^t \Big \{\Big |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |S_{t-1}+1 \Big \}. \end{aligned}$$
(7)
From the above, it can be obtained that
$$\begin{aligned} \Big |\frac{{\tilde{\epsilon }}_t^2}{{\tilde{\sigma }}_t^2}-\frac{\epsilon _t^2}{\sigma _t^2} \Big |\le K\rho ^t\Bigg (1+|\epsilon _t|+\frac{\epsilon _t^2}{\sigma _t^2}S_{t-1}\Bigg ) \end{aligned}$$
(8)
of which the proofs are omitted for brevity.
Proposition 4
Suppose that \((\mathbf{A}0)\)–\((\mathbf{A}3)\) hold. Then, for any neighborhood \(\mathcal {N}(\theta _0)\subset \varTheta \) whose elements have components \(\alpha _i\) and \(\beta _j\) bounded below from zero, under \(H_0\),
$$\begin{aligned} \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _t|^4<\infty , \ \ \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Big |\frac{\partial \epsilon _t}{\partial \theta _i} \Big |^4<\infty ,\ \ \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Bigg |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Bigg |^4<\infty , \end{aligned}$$
(9)
and
$$\begin{aligned} \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|S_{t-1}|^4\le K(t+q-1)^3, \end{aligned}$$
(10)
Moreover, if \((\mathbf{A}0')\), \((\mathbf{A}1)\)–\((\mathbf{A}3)\) hold,
$$\begin{aligned} \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _t|^6<\infty , \ \ \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Big |\frac{\partial \epsilon _t}{\partial \theta _i} \Big |^6<\infty ,\ \ \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Big |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |^6<\infty .\qquad \end{aligned}$$
(11)
Lemma 1
Under the same conditions in Theorem 3, we have
$$\begin{aligned}&\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\bigg \Vert \sum _{t=1}^k \partial _\theta {\tilde{l}}_t ({\hat{\theta }}_n) -\bigg (\sum _{t=1}^k \partial _\theta {l}_t (\theta _0)+k\mathcal {J}({\hat{\theta }}_n-\theta _0) \bigg )\bigg \Vert =o_\mathrm{P}(1). \end{aligned}$$
Proof
By the mean value theorem and Propositions 1 and 2, we get
$$\begin{aligned}&\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\bigg \Vert \sum _{t=1}^k \partial _\theta {\tilde{l}}_t ({\hat{\theta }}_n) -\bigg (\sum _{t=1}^k \partial _\theta {l}_t (\theta _0)+k\mathcal {J}({\hat{\theta }}_n-\theta _0) \bigg )\bigg \Vert \\&\quad \le \frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\bigg \Vert \sum _{t=1}^k \partial _\theta {\tilde{l}}_t ({\hat{\theta }}_n) -\bigg (\sum _{t=1}^k \partial _\theta {\tilde{l}}_t (\theta _0)+k\mathcal {J}({\hat{\theta }}_n-\theta _0) \bigg )\bigg \Vert \\&\qquad +\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\big \Vert \sum _{t=1}^{k}\partial _\theta {\tilde{l}}_t(\theta _0)-\sum _{t=1}^{k}\partial _\theta {l}_t(\theta _0)\big \Vert \\&\quad \le \max _{v_n \le k\le n}\frac{k}{\sqrt{n}}\big \Vert \big (\partial _{\theta \theta }{\tilde{L}}_k ({\tilde{\theta }}_n)-\mathcal {J}\big )({\hat{\theta }}_n-\theta _0) \big \Vert +\max _{1\le k\le n}\frac{k}{\sqrt{n}}\Vert \varDelta _{1,k}\Vert \\&\quad \le \max _{1\le k\le n}\frac{k}{\sqrt{n}}\Vert \varDelta _{2,k}\Vert +\max _{1\le k\le n}\frac{k}{\sqrt{n}}\Vert \varDelta _{1,k}\Vert =o_{\mathrm {P}}(1). \end{aligned}$$
This completes the proof. \(\square \)
Lemma 2
Under the same conditions in Theorem 3, we have
$$\begin{aligned} \frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Bigg \Vert \sum _{t=1}^{k}\partial _\theta {\tilde{l}}_t({\hat{\theta }}_n)-\left( \sum _{t=1}^{k}\partial _\theta l_t(\theta _0)-\frac{k}{n}\sum _{t=1}^{n}\partial _\theta l_t(\theta _0) \right) \Bigg \Vert =o_{\mathrm {P}}(1). \end{aligned}$$
Proof
Owing to Propositions 1 and 2 and Lemma 1,
$$\begin{aligned}&\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Bigg \Vert \sum _{t=1}^{k}\partial _\theta {\tilde{l}}_t({\hat{\theta }}_n)-\left( \sum _{t=1}^{k}\partial _\theta l_t(\theta _0)-\frac{k}{n}\sum _{t=1}^{n}\partial _\theta l_t(\theta _0) \right) \Bigg \Vert \\&\quad \le \frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Big \Vert \sum _{t=1}^k \partial _\theta {\tilde{l}}_t ({\hat{\theta }}_n) -\Big (\sum _{t=1}^k \partial _\theta {l}_t (\theta _0)+k\mathcal {J}({\hat{\theta }}_n-\theta _0) \Big )\Big \Vert \\&\qquad +\,\max _{v_n \le k\le n}\frac{k}{\sqrt{n}}\Big \Vert \frac{1}{n}\sum _{t=1}^{n}\partial _\theta l_t(\theta _0)+\mathcal {J}({\hat{\theta }}_n-\theta _0)\Big \Vert \\&\quad \le o_{\mathrm {P}}(1)+\max _{1\le k\le n}\frac{k}{\sqrt{n}}\Vert \varDelta _{1,k}+\varDelta _{2,k}\Vert =o_{\mathrm {P}}(1). \end{aligned}$$
\(\square \)
Lemma 3
Suppose that \(\mathbf{(A0)}-\mathbf{(A3)}\) hold. Then, we have that as \(n\rightarrow \infty \),
$$\begin{aligned}&\frac{1}{n}\sum _{t=1}^{n}\frac{\partial _\theta \tilde{\sigma }_t^2}{{\tilde{\sigma }}_t^2}\frac{\partial _{\theta ^T}\tilde{\sigma }_t^2}{{\tilde{\sigma }}_t^2}({\hat{\theta }}_n){\mathop {\longrightarrow }\limits ^{a.s.}}\mathbb {E}\frac{\partial _\theta \sigma _t^2}{\sigma _t^2}\frac{\partial _{\theta ^T}\sigma _t^2}{\sigma _t^2}(\theta _0), \\&\frac{1}{n}\sum _{t=1}^{n}\frac{\partial _\theta \tilde{\sigma }_t^2}{{\tilde{\sigma }}_t^2}\frac{\partial _{\theta ^T}\tilde{\epsilon }_t}{{\tilde{\sigma }}_t}({\hat{\theta }}_n){\mathop {\longrightarrow }\limits ^{a.s.}}\mathbb {E}\frac{\partial _\theta \sigma _t^2}{\sigma _t^2}\frac{\partial _{\theta ^T}\epsilon _t}{\sigma _t}(\theta _0), \\&\frac{1}{n}\sum _{t=1}^{n}\frac{\partial _\theta \tilde{\epsilon }_t}{{\tilde{\sigma }}_t}\frac{\partial _{\theta ^T}\tilde{\epsilon }_t}{{\tilde{\sigma }}_t}({\hat{\theta }}_n){\mathop {\longrightarrow }\limits ^{a.s.}}\mathbb {E}\frac{\partial _\theta \epsilon _t}{\sigma _t}\frac{\partial _{\theta ^T}\epsilon _t}{\sigma _t}(\theta _0), \end{aligned}$$
and
$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n}\frac{{\tilde{\epsilon }}_t^4}{{\tilde{\sigma }}_t^4}{\mathop {\longrightarrow }\limits ^{a.s.}}\mathbb {E}\eta _t^4. \end{aligned}$$
Proof
Due to the strong consistency of \({\hat{\theta }}_n\), it suffices to prove that for \(i=1,2,3,4\),
$$\begin{aligned} \Big \Vert \frac{1}{n}\sum _{t=1}^{n}(\tilde{P}_{i,t}-P_{i,t})\Big \Vert \ \ \text {and} \ \ \Big \Vert \frac{1}{n}\sum _{t=1}^{n}P_{i,t}-\mathbb {E}P_{i,t}\Big \Vert \end{aligned}$$
converge to 0 a.s. uniformly over any neighborhood \(\mathcal {N}(\theta _0)\), where
$$\begin{aligned} P_{1,t}=\frac{\partial _\theta \sigma _t^2}{\sigma _t^2}\frac{\partial _{\theta ^T}\sigma _t^2}{\sigma _t^2}, \ \ P_{2,t}=\frac{\partial _\theta \sigma _t^2}{\sigma _t^2}\frac{\partial _{\theta ^T}\epsilon _t}{\sigma _t}, \ \ P_{3,t}=\frac{\partial _\theta \epsilon _t}{\sigma _t}\frac{\partial _{\theta ^T}\epsilon _t}{\sigma _t}, \ \ \text {and} \ \ P_{4,t}=\frac{\epsilon _t^4}{\sigma _t^4}, \end{aligned}$$
and \(\tilde{P}_{i,t}\), \(i=1,2,3,4\), are similarly defined with \(\sigma _t\), \(\epsilon _t\), \(\partial _\theta \sigma _t\) and \(\partial _\theta \epsilon _t\) replaced by \({\tilde{\sigma }}_t\), \({\tilde{\epsilon }}_t\), \(\partial _\theta {\tilde{\sigma }}_t\) and \(\partial _\theta {\tilde{\epsilon }}_t\), respectively.
Using Hölder’s inequality and the fact that \(\sigma _t^2\ge \inf _{\theta \in \varTheta }w\) and (9), we can see that \(\mathbb {E}[\sup _{\theta \in \mathcal {N}(\theta _0)}|P_{i,t}|]<\infty \), \(i=1,2,3,4\). Hence, \(\sup _{\theta \in \mathcal {N}(\theta _0)}\Big \Vert \frac{1}{n}\sum _{t=1}^{n}P_{i,t}-\mathbb {E}P_{i,t}\Big \Vert =o(1)\) a.s. for \(i=1,2,3,4\), owing to Theorem 2.7 of Straumann and Mikosch (2006).
Note that owing to (7),
$$\begin{aligned} \Big |\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _i}\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _j}-\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _j}\Big |\le & {} \Big |\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _i}-\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |\Big |\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _j}-\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _j}\Big |\\&+\,\Big |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |\Big |\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _j}-\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _j}\Big |\\&+\,\Big |\frac{1}{{\tilde{\sigma }}_t^2}\frac{\partial {\tilde{\sigma }}_t^2}{\partial \theta _i}-\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |\Big |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _j}\Big |\\\le & {} K\rho ^t \Bigg \{\Big |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _i}\Big |S_{t-1}+1 \Bigg \}\Bigg \{\Big |\frac{1}{\sigma _t^2}\frac{\partial \sigma _t^2}{\partial \theta _j }\Big |S_{t-1}\\&+\,1 \Bigg \}\\:= & {} K\rho ^t Q_{t,i,j}. \end{aligned}$$
Using (9), (10), Markov’s inequality, and Hölder’s inequality, we have that for each \(\epsilon >0\),
$$\begin{aligned} \sum _{t=1}^{\infty }\mathrm {P}\Big ( \rho ^t\sup _{\theta \in \mathcal {N}(\theta _0)}Q_{t,i,j}>\epsilon \Big )\le & {} \frac{1}{{\epsilon }}\sum _{t=1}^{\infty }\rho ^{t}\mathbb {E}\Big |\sup _{\theta \in \mathcal {N}(\theta _0)}Q_{t,i,j}\Big |\\\le & {} K\sum _{t=1}^{\infty }(t+q-1)^{3/2}\rho ^t<\infty , \end{aligned}$$
which implies \(\rho ^t\sup _{\theta \in \mathcal {N}(\theta _0)}Q_{t,i,j}=o(1)\) a.s.. Therefore, \(\sup _{\theta \in \mathcal {N}(\theta _0)}\Big \Vert \frac{1}{n}\sum _{t=1}^{n}(\tilde{P}_{1,t}-P_{1,t})\Big \Vert =o(1)\) a.s.. Since the cases of \(i=2,3,4\) can be similarly handled, the lemma is established. \(\square \)
Lemma 4
Let \(R_{i,t}\), \(i=1,2\), be those in (6) and suppose that the conditions in Theorem 4, we have
$$\begin{aligned} \frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Big |\sum _{t=1}^{k}R_{i,t}-\frac{k}{n}\sum _{t=1}^{n}R_{i,t} \Big |=o_{\mathrm {P}}(1). \end{aligned}$$
Proof
First, we deal with \(R_{1,t}\). For any \(\delta >0\), neighborhood \(N(\theta _0)\) satisfying (9), we have that
$$\begin{aligned} \mathrm {P}\Big ( \frac{1}{\sqrt{n}}\sum _{t=1}^{n}\Big |\Big (\frac{{\tilde{\epsilon }}_t^2}{{\tilde{\sigma }}_t^2}-\frac{\epsilon _t^2}{\sigma _t^2}\Big )({\hat{\theta }}_n)\Big |>\delta \Big )\le & {} \mathrm {P}\Big ({\hat{\theta }}_n\in \varTheta /N(\theta _0)\Big )\\&+\,\mathrm {P}\Big (\frac{1}{\sqrt{n}}\sum _{t=1}^{n}\sup _{\theta \in \mathcal {N}(\theta _0)}\Big |\frac{{\tilde{\epsilon }}_t^2}{{\tilde{\sigma }}_t^2}-\frac{\epsilon _t^2}{\sigma _t^2}\Big |>\delta \Big ) \end{aligned}$$
and by (8), (9) and the fact \(\sup _{\theta \in \mathcal {N}(\theta _0)}\frac{1}{\sigma _t^2(\theta )}\le \sup _{\theta \in \mathcal {N}(\theta _0)}\frac{1}{\omega }<\infty \) and Hölder’s inequality
$$\begin{aligned}&\mathbb {E}\Big (\frac{1}{\sqrt{n}}\sum _{t=1}^{n}\sup _{\theta \in \mathcal {N}(\theta _0)}\Big |\frac{{\tilde{\epsilon }}_t^2}{{\tilde{\sigma }}_t^2}-\frac{\epsilon _t^2}{\sigma _t^2}\Big |\Big )\\&\quad \le \frac{1}{\sqrt{n}}\sum _{t=1}^{n}K\rho ^t\cdot \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Bigg (1+|\epsilon _t|+\frac{\epsilon _t^2}{\sigma _t^2}S_{t-1}\Bigg )\\&\quad \le \frac{1}{\sqrt{n}}\sum _{t=1}^{n}K\rho ^t\bigg (1+\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _t|+\Big (\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _t|^4\Big )^{1/2}\Big (\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|S_{t-1}|^2\Big )^{1/2}\bigg )\\&\quad \le \Big (1+\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _1| \Big )\frac{1}{\sqrt{n}}\sum _{t=1}^{n}K\rho ^t+\Big (\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _1|^4\Big )^{1/2}\frac{1}{\sqrt{n}}\sum _{t=1}^{n}K\rho ^t(t{+}q{-}1)\\&\qquad \longrightarrow 0, \ \ n\rightarrow \infty \end{aligned}$$
which imply with Markov’s inequality and the fact \({\hat{\theta }}_n\rightarrow \theta _0\) a.s.
$$\begin{aligned} \frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\sum _{t=1}^{k}\Big |\Big (\frac{{\tilde{\epsilon }}_t^2}{{\tilde{\sigma }}_t^2}-\frac{\epsilon _t^2}{\sigma _t^2}\Big )({\hat{\theta }}_n)\Big |\le \frac{1}{\sqrt{n}}\sum _{t=1}^{n}\Big |\Big (\frac{{\tilde{\epsilon }}_t^2}{{\tilde{\sigma }}_t^2}-\frac{\epsilon _t^2}{\sigma _t^2}\Big )({\hat{\theta }}_n)\Big |=o_{\mathrm {P}}(1). \end{aligned}$$
(12)
Next, we deal with \(R_{2,t}\). By the mean value theorem, we have
$$\begin{aligned} \Big (\frac{\epsilon _t^2}{\sigma _t^2}({\hat{\theta }}_n)-\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0)\Big )= & {} ({\hat{\theta }}_n-\theta _0)^T\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )({\tilde{\theta }}_n)\nonumber \\= & {} ({\hat{\theta }}_n-\theta _0)^T\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\nonumber \\&+\,({\hat{\theta }}_n-\theta _0)^T\Bigg \{\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )({\tilde{\theta }}_n)\nonumber \\&-\,\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0)\Bigg \} \end{aligned}$$
where \({\tilde{\theta }}_n\) is an appropriate intermediate point between \({\hat{\theta }}_n\) and \(\theta _0\). By using (9), one can readily check that \(\mathbb {E}\Big \Vert \Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\Big \Vert <\infty \) and, \(\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )\) is stationary and ergodic. Thus the invariance principle for stationary processes, we have
$$\begin{aligned}&\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Bigg \Vert \sum _{t=1}^{k}\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0) -\frac{k}{n}\sum _{t=1}^{n}\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0)\Bigg \Vert \\&\quad =\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\bigg \Vert \sum _{t=1}^{k}\Bigg \{\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0) -\mathbb {E}\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0)\Bigg \}\\&\qquad -\,\frac{k}{n}\sum _{t=1}^{n}\Bigg \{\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0) -\mathbb {E}\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0)\Bigg \}\bigg \Vert \\&\quad =O_{\mathrm {P}}(1) \end{aligned}$$
which implies with the fact \({\hat{\theta }}_n\rightarrow \theta _0\) a.s.,
$$\begin{aligned}&\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\bigg \Vert \sum _{t=1}^{k}({\hat{\theta }}_n-\theta _0)^T\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\nonumber \\&\qquad -\frac{k}{n}\sum _{t=1}^{n}({\hat{\theta }}_n-\theta _0)^T\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\bigg \Vert \nonumber \\&\quad =o_{\mathrm {P}}(1). \end{aligned}$$
(13)
Since \(\mathbb {E}\Big \Vert \Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\Big \Vert <\infty \) and \(\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )\) is stationary and ergodic, the convergence of \({\tilde{\theta }}_n\) to \(\theta _0\), almost surely, implies
$$\begin{aligned}&\max _{v_n \le k\le n}\Bigg \Vert \frac{1}{k}\sum _{t=1}^{k} \Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )({\tilde{\theta }}_n)-\frac{1}{k}\sum _{t=1}^{k}\Bigg ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Bigg )(\theta _0)\Bigg \Vert \\&\quad =o(1),\ \ a.s., \end{aligned}$$
which implies with the fact \(\sqrt{n}({\hat{\theta }}_n-\theta _0)=O_{\mathrm {P}}(1)\) (cf. Theorem 2),
$$\begin{aligned}&\frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Big \Vert \sum _{t=1}^{k}({\hat{\theta }}_n-\theta _0)^T\Big \{ \Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )({\tilde{\theta }}_n)\nonumber \\&\qquad -\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\Big \}\Big \Vert \nonumber \\&\quad \le \Vert \sqrt{n}({\hat{\theta }}_n-\theta _0)\Vert \max _{v_n \le k\le n}\Big \Vert \frac{1}{k}\sum _{t=1}^{k}\Big \{ \Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )({\tilde{\theta }}_n)\nonumber \\&\qquad -\Big ( \frac{2\epsilon _t\partial _\theta \epsilon _t}{\sigma _t^2}-\frac{\epsilon _t^2\partial _\theta \sigma _t^2}{\sigma _t^4}\Big )(\theta _0)\Big \}\Big \Vert \nonumber \\&\quad =o_{\mathrm {P}}(1). \end{aligned}$$
(14)
Thus, combining (13) and (14), we obtain
$$\begin{aligned} \frac{1}{\sqrt{n}}\max _{v_n \le k\le n}\Big |\sum _{t=1}^{k}R_{2,t}-\frac{k}{n}\sum _{t=1}^{n}R_{2,t} \Big |=o_{\mathrm {P}}(1). \end{aligned}$$
This together with (12) validates the lemma. \(\square \)
Lemma 5
Under the same conditions in Theorem 4, we have
$$\begin{aligned} {\hat{\tau }}_n^2=\frac{1}{n}\sum _{t=1}^{n}{\hat{\eta }}_t^4-\Bigg (\frac{1}{n}\sum _{t=1}^{n}{\hat{\eta }}_t^2\Bigg )^2{\mathop {\longrightarrow }\limits ^{\mathrm {P}}}\mathrm {Var}(\eta _1^2). \end{aligned}$$
Proof
To show that \(\frac{1}{n}\sum _{t=1}^n {\hat{\eta }}_t^2{\mathop {\longrightarrow }\limits ^{\mathrm {P}}}\mathbb {E}(\eta _1^2)\), we verity that
$$\begin{aligned} \Big |\frac{1}{n}\sum _{t=1}^{n}{\hat{\eta }}_t^2-\frac{1}{n}\sum _{t=1}^{n}\eta _t^2\Big |\le \Big |\frac{1}{n}\sum _{t=1}^{n}R_{1,t}\Big |+\Big |\frac{1}{n}\sum _{t=1}^{n}R_{2,t}\Big |=o_{\mathrm {P}}(1). \end{aligned}$$
(15)
Since \(\mathbb {E}\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0)=\mathbb {E}\eta _t^2<\infty \) and \(\frac{\epsilon _t^2}{\sigma _t^2}(\theta )\) is stationary and ergodic and \({\hat{\theta }}_n\rightarrow \theta _0\) a.s., we have
$$\begin{aligned} \Big |\frac{1}{n}\sum _{t=1}^{n}R_{2,t}\Big |\le & {} \Big |\frac{1}{n}\sum _{t=1}^{n}\frac{\epsilon _t^2}{\sigma _t^2}({\hat{\theta }}_n)-\frac{1}{n}\sum _{t=1}^{n}\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0)\Big |\longrightarrow 0 \ \ a.s. \end{aligned}$$
(16)
Further, \(\frac{1}{\sqrt{n}}\sum _{t=1}^{n}|R_{1,t}|=o_{\mathrm {P}}(1)\) as we have in proof of Lemma 4. Hence, combining this with (16), we can obtain (15), which in turn implies \(\frac{1}{n}\sum _{t=1}^n {\hat{\eta }}_t^2{\mathop {\longrightarrow }\limits ^{\mathrm {P}}}\mathbb {E}(\eta _1^2)\).
Now, to show that \(\frac{1}{n}\sum _{t=1}^{n}{\hat{\eta }}_t^4{\mathop {\longrightarrow }\limits ^{\mathrm {P}}}\mathbb {E}\eta _1^4\), we first verify that
$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n}({\hat{\eta }}_t^2-\eta _t^2)^2\le \frac{2}{n}\Bigg (\sum _{t=1}^{n}|R_{1,t}|^2+\sum _{t=1}^{n}|R_{2,t}|^2\Bigg )=o_{\mathrm {P}}(1). \end{aligned}$$
(17)
For \(|R_{1,t}|^2\), note that by Hölder inequality and (11)
$$\begin{aligned} \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\frac{\epsilon _t^4}{\sigma _t^4}S_{t-1}^2\le & {} K \Big (\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _t|^6\Big )^{2/3}\Big (\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|S_{t-1}|^6\Big )^{1/3}\\\le & {} K \Big (\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}|\epsilon _1|^6\Big )^{2/3}(t+q-1)^2 \end{aligned}$$
Thus, similar to (12), we can have easily
$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n}|R_{1,t}|^2=o_{\mathrm {P}}(1). \end{aligned}$$
(18)
By using (11), one can readily check that \(\mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Big |\frac{\epsilon _t^4}{\sigma _t^4}\Big |<\infty \), and thus, by the ergodic theorem
$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n}\sup _{\theta \in \mathcal {N}(\theta _0)}\Bigg |\frac{\epsilon _t^2}{\sigma _t^2}(\theta )-\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0) \Bigg |^2\longrightarrow \mathbb {E}\sup _{\theta \in \mathcal {N}(\theta _0)}\Bigg |\frac{\epsilon _t^2}{\sigma _t^2}(\theta )-\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0) \Bigg |^2 \ \ a.s. \end{aligned}$$
Hence, for any \(\delta >0\), there exist a neighborhood \(N_r(\theta _0)=\{\theta :\Vert \theta - \theta _0\Vert <r(\delta )\}\) and \(N_1>1\) such that
$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n}\sup _{\theta \in N_r(\theta _0)}\Bigg |\frac{\epsilon _t^2}{\sigma _t^2}(\theta )-\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0) \Bigg |^2\le \delta \ \ a.s. \ \ \text {for all} \ \ n\ge N_1. \end{aligned}$$
Further, since \({\hat{\theta }}_n\rightarrow \theta _0\) a.s., there exists \(N_2\ge 1\) such that \(\Vert {\hat{\theta }}_n-\theta _0\Vert <r(\delta )\), almost surely, for \(n\ge N_2\). Thus for any \(\delta >0\), all \(n\ge \max \{N_1,N_2\}\), almost surely
$$\begin{aligned} \frac{1}{n}\sum _{t=1}^{n}|R_{2,t}|^2\le \frac{1}{n}\sum _{t=1}^{n}\sup _{\theta \in N_r(\theta _0)}\Bigg |\frac{\epsilon _t^2}{\sigma _t^2}(\theta )-\frac{\epsilon _t^2}{\sigma _t^2}(\theta _0) \Bigg |^2\le \delta . \end{aligned}$$
(19)
Thus, combining (18) and (19), we obtain (17). Therefore
$$\begin{aligned} \Big |\frac{1}{n}\sum _{t=1}^{n}{\hat{\eta }}_t^4-\frac{1}{n}\sum _{t=1}^{n}\eta _t^4\Big |\le & {} \Big (\frac{1}{n}\sum _{t=1}^{n}({\hat{\eta }}_t^2-\eta _t^2)^2 \Big )^{1/2}\Big (\frac{1}{n}\sum _{t=1}^{n}({\hat{\eta }}_t^2+\eta _t^2)^2 \Big )^{1/2}\\\le & {} \Big (\frac{1}{n}\sum _{t=1}^{n}({\hat{\eta }}_t^2-\eta _t^2)^2 \Big )^{1/2}\Big (\frac{2}{n}\sum _{t=1}^{n}({\hat{\eta }}_t^2-\eta _t^2)^2+\frac{8}{n}\sum _{t=1}^{n}\eta _t^4 \Big )^{1/2}\\= & {} o_{\mathrm {P}}(1)(o_{\mathrm {P}}(1)+O_{\mathrm {P}}(1))=o_{\mathrm {P}}(1), \end{aligned}$$
and thus, \(\frac{1}{n}\sum _{t=1}^{n}{\hat{\eta }}_t^4{\mathop {\longrightarrow }\limits ^{\mathrm {P}}}\mathbb {E}\eta _1^4\). Hence, the lemma is validated. \(\square \)
