Proof of Theorem 2.1

The formula ${\sigma }_t^2={\alpha }_0+\sum _{i=1}^r{\alpha }_i{\epsilon }_{t-i}^2+\sum _{j=1}^s{\beta }_j{\sigma }_{t-j}^2$ in model (1) can be rewritten in vector form as $$\begin{array}{}{\underset{\_}{\tilde{\sigma }}}_t^2={\underset{\_}{\tilde{c}}}_t+B{\underset{\_}{\tilde{\sigma }}}_{t-1}^2,\end{array}$$(10)

where $$\begin{array}{}{\underset{\_}{\tilde{\sigma }}}_t^2=\left(\begin{array}{c}{\sigma }_t^2\\ {\sigma }_{t-1}^2\\ ⋮\\ {\sigma }_{t-s+1}^2\end{array}\right),\underset{\_}{{\tilde{c}}_t}=\left(\begin{array}{c}{\alpha }_0+\sum _{i=1}^r{\alpha }_i{\epsilon }_{t-i}^2\\ 0\\ ⋮\\ 0\end{array}\right),B=\left(\begin{array}{cccc}{\beta }_1& {\beta }_2& \cdots & {\beta }_r\\ 1& 0& \cdots & 0\\ ⋮\\ 0& \cdots & 1& 0\end{array}\right).\end{array}$$(11)

Then following intermediate results can be used to prove Theorem 2.1.

1. lim_n→∞ sup_θ∈Θ|I_n (θ) − Ĩ_n(θ)| = 0 a.s.

2. ∃ t ∈ Z makes ${\sigma }_t^2$ (θ) = ${\sigma }_t^2$ (θ ₀) P_θ0 a.s. ⇒ θ = θ₀.

3. E_θ0 | l_t(θ₀)|<∞, and if θ ≠ θ₀, E_θ0l_t(θ) > E_θ0l_t(θ₀).

4. For θ ≠ θ₀, there is a neighbourhood V(θ) making lim inf_n→∞ inf_{θ^* ∈ V(θ)} Ĩ_n(θ^*)>E_θ0l₁(θ₀) a.s.

1. As Assumption 1, $$\begin{array}{}\underset{\theta \in \mathit{\Theta }}{sup}\rho (B)<1.\end{array}$$(12)

   Iterating (10) produces the following: $$\begin{array}{}{\underset{\_}{\sigma }}_t^2={\underset{\_}{c}}_t+B{\underset{\_}{c}}_{t-1}+B^2{\underset{\_}{c}}_{t-1}+\cdots +B^{t-1}{\underset{\_}{c}}_1+B^t{\underset{\_}{\sigma }}_0^2=\sum _{k=0}^{\mathrm{\infty }}{B}^k{\underset{\_}{c}}_{t-k}.\end{array}$$(13)

   It is supposed here that ${\underset{\_}{\tilde{\sigma }}}_t^2$ may be the vector obtained by replacing ${\sigma }_{t-i}^2\text{\hspace{0.17em}by\hspace{0.17em}}{\tilde{\sigma }}_{t-i}^2.$ Let c̃ be the vector obtained by replacing ${\epsilon }_0^2,\cdots ,{\epsilon }_{1-p}^2$ with the initial values (3) or (4). We have $$\begin{array}{}{\underset{\_}{\tilde{\sigma }}}_t^2={\underset{\_}{c}}_t+B{\underset{\_}{\tilde{c}}}_{t-1}+\cdots +B^{t-r-1}{\underset{\_}{\tilde{c}}}_r+1+B^{t-r}{\underset{\_}{\tilde{c}}}_r+\cdots +B^{t-1}{\underset{\_}{\tilde{c}}}_1+B^t{\underset{\_}{\tilde{\sigma }}}_0^2.\end{array}$$(14)

   Through (12)-(14), it is almost certain the following is true: $$\begin{array}{}\underset{\theta \in \mathit{\Theta }}{sup}\parallel {\underset{\_}{\sigma }}_t^2-{\underset{\_}{\tilde{\sigma }}}_t^2\parallel & =\underset{\theta \in \mathit{\Theta }}{sup}∥\left\{\sum _{k=1}^r{B}^{t-k}({\underset{\_}{c}}_k-{\underset{\_}{\tilde{c}}}_k)+B^t({\underset{\_}{\sigma }}_0^2-{\underset{\_}{\tilde{\sigma }}}_0^2)\right\}∥\\ & \le K{\rho }^t,\mathrm{\forall }t.\end{array}$$(15)

   Hence, $$\begin{array}{}\underset{\theta \in \mathit{\Theta }}{sup}|I_n(\theta )-{\tilde{I}}_n(\theta )|& \le n^{-1}{\displaystyle \sum _{t=1}^n\underset{\theta \in \mathit{\Theta }}{sup}\left\{\left|\frac{{\tilde{\sigma }}_t-{\sigma }_t}{{\sigma }_t{\tilde{\sigma }}_t}({\epsilon }_t-1)\right|+\frac{1}{2}\left|\log \left(1+\frac{{\sigma }_t^2-{\tilde{\sigma }}_t^2}{{\tilde{\sigma }}_t^2}\right)\right|\right\}}\\ & \le {\displaystyle n^{-1}\sum _{t=1}^n\underset{\theta \in \mathit{\Theta }}{sup}\left\{\sqrt{\left|\frac{{\tilde{\sigma }}_t^2-{\sigma }_t^2}{{\tilde{\sigma }}_t^2{\sigma }_t^2}({\epsilon }_t-1{)}^2\right|}+\frac{1}{2}\left|\log \left(1+\frac{{\sigma }_t^2-{\tilde{\sigma }}_t^2}{{\tilde{\sigma }}_t^2}\right)\right|\right\}}\\ & \le {\displaystyle \frac{\sqrt{K}}{n}\left\{\underset{\theta \in \mathit{\Theta }}{sup}\frac{1}{{\alpha }_0}\right\}\sum _{t=1}^n\sqrt{{\rho }^t({\epsilon }_t-1{)}^2}+\frac{K}{2n}\left\{\underset{\theta \in \mathit{\Theta }}{sup}\frac{1}{{\alpha }_0}\right\}\sum _{t=1}^n{\rho }^t.}\end{array}$$

   By the Markov inequality, the following equation can be determined: $$\begin{array}{}\sum _{t=1}^{\mathrm{\infty }}P({\rho }^t({\epsilon }_t-1{)}^2>ϵ)\le \sum _{t=1}^{\mathrm{\infty }}{\displaystyle \frac{E({\rho }^t({\epsilon }_t-1{)}^2{)}^m}{{ϵ}^m}<\mathrm{\infty }.}\end{array}$$

   From the Borel-Cantelli lemma, (i) is obtained.

2. It’s obvious that the result (ii) can be easily proved with Assumption 2 and Assumption 4.

3. Because of $\begin{array}{}{E}_{{\theta }_0}{l}_t^{-}(\theta )\le E_{{\theta }_0}{\ln }^{-}{\sigma }_t^2\le max\{0,-\ln \omega \}<\mathrm{\infty }.\end{array}$ It remains to be shown that E_θ0$l_t^{+}$(θ) < ∞. By Jenson inequality, $$\begin{array}{}{\displaystyle E_{{\theta }_0}\log {\sigma }_t({\theta }_0)=\frac{1}{2}{E}_{{\theta }_0}\log {\sigma }_t^2({\theta }_0)=\frac{1}{2}{E}_{{\theta }_0}\frac{1}{m}\log \{{\sigma }_t^2({\theta }_0){\}}^m\le \frac{1}{2m}\log E_{{\theta }_0}\{{\sigma }_t^2({\theta }_0){\}}^m<\mathrm{\infty },}\end{array}$$

   Thus, $$\begin{array}{}{\displaystyle E_{{\theta }_0}{l}_t({\theta }_0)=E_{{\theta }_0}\left\{\frac{1}{2}\log {\sigma }_t^2({\theta }_0)+\frac{|{\epsilon }_t-1|}{{\sigma }_t({\theta }_0)}\right\}=1+\frac{1}{2}{E}_{{\theta }_0}\log {\sigma }_t^2({\theta }_0)<\mathrm{\infty }.}\end{array}$$

   Therefore, $$\begin{array}{}{\displaystyle E_{{\theta }_0}{l}_t(\theta )-E_{{\theta }_0}{l}_t({\theta }_0)=E_{{\theta }_0}\ln \frac{{\sigma }_t(\theta )}{{\sigma }_t({\theta }_0)}+E_{{\theta }_0}\frac{{\sigma }_t({\theta }_0)}{{\sigma }_t(\theta )}-1.}\end{array}$$

   For all x > 0, log x ≤ x − 1, where the equality is true if and only if x = 1. Thus, it is true that $$\begin{array}{}{\displaystyle E_{{\theta }_0}{l}_t(\theta )-E_{{\theta }_0}{l}_t({\theta }_0)\ge E_{{\theta }_0}\left\{\log \frac{{\sigma }_t(\theta )}{{\sigma }_t({\theta }_0)}+\log \frac{{\sigma }_t({\theta }_0)}{{\sigma }_t(\theta )}\right\}=0}\end{array}$$(16)

   It is noted that the equality in (16) holds if σ_t(θ₀) = σ_t(θ).

4. From result (i), $$\begin{array}{}\underset{n\to \mathrm{\infty }}{lim inf}\underset{{\theta }^{\ast }\in V_k(\theta )\cap \mathit{\Theta }}{inf}{\tilde{I}}_n({\theta }^{\ast })& \ge \underset{n\to \mathrm{\infty }}{lim inf}\underset{{\theta }^{\ast }\in V_k(\theta )\cap \mathit{\Theta }}{inf}{I}_n({\theta }^{\ast })-\underset{n\to \mathrm{\infty }}{lim sup}\underset{\theta \in \mathit{\Theta }}{sup}|{\tilde{I}}_n(\theta )-I_n({\theta }^{\ast })|\\ & \ge \underset{n\to \mathrm{\infty }}{lim inf}{\displaystyle \frac{1}{n}}\sum _{t=1}^n\underset{{\theta }^{\ast }\in V_k(\theta )\cap \mathit{\Theta }}{inf}{l}_t({\theta }^{\ast })\end{array}$$

   Based on the ergodic theorem, Beppo-Levi theorem and the formula (16), the result (iv) can be proved.

By compactness theory, the proof of Theorem 2.1 is finished.□

Proof of Theorem 2.2

Through a Taylor expansion at θ₀, $$\begin{array}{}0& =n^{-1/2}\sum _{t=1}^n{\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\theta }{l}_t({\hat{\theta }}_n)}\\ & =n^{-1/2}\sum _{t=1}^n{\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\theta }{l}_t({\theta }_0)+\left(\frac{1}{n}\sum _{t=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i{\theta }_j}{l}_t({\theta }_{ij}^{\ast })\right)\sqrt{n}({\hat{\theta }}_n-{\theta }_0),}\end{array}$$

which indicates that both $$\begin{array}{}{n}^{-1/2}\sum _{t=1}^n{\displaystyle \frac{\mathrm{\partial }}{\mathrm{\partial }\theta }{l}_t({\theta }_0)\Rightarrow N(0,(\frac{\tau -1}{4})J)}\end{array}$$(17)

and $$\begin{array}{}{n}^{-1}\sum _{t=1}^n{\displaystyle \frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i{\theta }_j}{l}_t({\theta }_{ij}^{\ast })\to J(i,j)inprobability.}\end{array}$$(18)

hold. The proof of Theorem 2.2 is divided into the following six conclusions.

1. E_θ0∥(∂l_t(θ₀)/∂θ)(∂l_t(θ₀)/∂θ^T)∥<∞, E_θ0∥∂²l_t(θ₀)/∂θ∂θ^T∥<∞.

2. J is nonsingular, var_θ0 {∂l_t(θ₀)/∂θ} = MJ.

3. There exists a neighborhood V(θ₀) of θ, with regard to i, j, k ∈ {1, ⋯, r + s + 1}, such that $$\begin{array}{}{E}_{{\theta }_0}{\displaystyle \underset{\theta \in V({\theta }_0)}{sup}\left|\frac{{\mathrm{\partial }}^3{l}_t(\theta )}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j\mathrm{\partial }{\theta }_k}\right|<\mathrm{\infty }.}\end{array}$$

4. ∥n^−1/2$\sum _{t=1}^n$ {∂l_t(θ₀)/∂θ − ∂l̃_t(θ₀)/∂θ} ∥ → 0 when n → ∞, sup_θ∈V(θ0) ∥ n⁻¹ $\sum _{t=1}^n$ {∂²l_t(θ)/∂θ∂θ^T − ∂²l̃_t(θ)/∂θ∂θ^T} ∥ → 0 in probability.

5. n^−1/2 $\sum _{t=1}^n$ ∂l_t(θ₀)/∂θ⇒ N(0, MJ).

6. n⁻¹ $\sum _{t=1}^n$ ∂²l_t(${\theta }_{ij}^{\ast }$)/∂θ_i∂θ_j → J(i, j) a.s.

1. Because of l_t(θ) = ln $\sqrt{{\sigma }_t^2}+|{\epsilon }_t-1|/\sqrt{{\sigma }_t^2},$ it is true that $$\begin{array}{}{\displaystyle \frac{\mathrm{\partial }{l}_t}{\mathrm{\partial }\theta }}& {\displaystyle =\frac{1}{2{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }+\left(-\frac{|{\epsilon }_t-1|}{2}\frac{1}{({\sigma }_t^2{)}^{3/2}}\right)\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }}\\ & {\displaystyle =\frac{1}{2}\left(1-\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right)\left(\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }\right),}\end{array}$$(19) $$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial }}^2{l}_t}{\mathrm{\partial }\theta \mathrm{\partial }{\theta }^T}=\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }\theta \mathrm{\partial }{\theta }^T}\left(\frac{1}{2{\sigma }_t^2}-\frac{|{\epsilon }_t-1|}{2}\frac{1}{\sqrt{{\sigma }_t^6}}\right)+\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }^T}\left(\frac{3|{\epsilon }_t-1|}{4}\frac{1}{\sqrt{{\sigma }_t^{10}}}-\frac{1}{2}\frac{1}{{\sigma }_t^4}\right)}\\ {\displaystyle =\frac{1}{2}\left(\frac{1}{{\sigma }_t^2}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }\theta \mathrm{\partial }{\theta }^T}\right)\left(1-\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right)+\frac{1}{2}\left(\frac{1}{{\sigma }_t^4}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }^T}\right)\left(\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}-1\right).}\end{array}$$(20)

   For θ = θ₀, we have $$\begin{array}{c}{\displaystyle E_{{\theta }_0}∥\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }({\theta }_0)∥<\mathrm{\infty },E_{{\theta }_0}∥\frac{1}{{\sigma }_t^2}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }\theta \mathrm{\partial }{\theta }^T}({\theta }_0)∥<\mathrm{\infty },}\\ {\displaystyle E_{{\theta }_0}∥\frac{1}{{\sigma }_t^4}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }^T}({\theta }_0)∥<\mathrm{\infty }.}\end{array}$$

   The proof of (i) is finished.

2. From (i), it have $$\begin{array}{}{\displaystyle E_{{\theta }_0}\left\{\frac{\mathrm{\partial }{l}_t({\theta }_0)}{\mathrm{\partial }\theta }\right\}=E_{{\theta }_0}\left(\frac{1}{2}-\frac{|{\epsilon }_t-1|}{2\sqrt{{\sigma }_t^2}}\right)E_{{\theta }_0}\left\{\frac{1}{{\sigma }_t^2({\theta }_0)}\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }\theta }\right\}=0.}\end{array}$$

   Since (20), (i) and (9), the following must also be determined. $$\begin{array}{}{\displaystyle va{r}_{{\theta }_0}\left\{\frac{\mathrm{\partial }{l}_t({\theta }_0)}{\mathrm{\partial }\theta }\right\}}& ={\displaystyle E_{{\theta }_0}\left\{\frac{\mathrm{\partial }{l}_t({\theta }_0)}{\mathrm{\partial }\theta }\frac{\mathrm{\partial }{l}_t({\theta }_0)}{\mathrm{\partial }{\theta }^{{}^{\prime }}}\right\}}\\ & {\displaystyle =E\left\{\frac{(1-\frac{|{\epsilon }_t-1|}{{\sigma }_t}{)}^2}{4}\right\}{E}_{{\theta }_0}\left\{\frac{1}{{\sigma }_t^4({\theta }_0)}\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }\theta }\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }{\theta }^{{}^{\prime }}}\right\}}\\ & ={\displaystyle \left(\frac{\tau -1}{4}\right)J.}\end{array}$$(21)

   This shows that J is non-singular. So we establish the conclusion of (ii).

3. It is shown in (20) that l_t(θ) is differentiated. Then $$\begin{array}{}{\displaystyle \frac{{\mathrm{\partial }}^3{l}_t(\theta )}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j\mathrm{\partial }{\theta }_k}}& ={\displaystyle \frac{1}{2}\left(1-\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right)\left(\frac{1}{{\sigma }_t^2}\frac{{\mathrm{\partial }}^3{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j\mathrm{\partial }{\theta }_k}\right)}\\ & +{\displaystyle \left(1-\frac{15}{8}\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right)\left(\frac{1}{{\sigma }_t^6}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_k}\right)}\\ & +{\displaystyle \frac{1}{2}\frac{1}{{\sigma }_t^4}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_k}\left(\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}-1\right)}\\ & +{\displaystyle \frac{1}{2}\frac{1}{{\sigma }_t^4}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_k}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}\left(\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}-1\right)}\\ & +{\displaystyle \frac{1}{2}\frac{1}{{\sigma }_t^4}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_j\mathrm{\partial }{\theta }_k}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}\left(\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}-1\right).}\end{array}$$

   Since $$\begin{array}{rl}{\displaystyle E_{{\theta }_0}\underset{\theta \in V({\theta }_0)}{sup}\frac{|{\epsilon }_t-1|}{{\sigma }_t}<\mathrm{\infty },}& {\displaystyle E_{{\theta }_0}\underset{\theta \in V({\theta }_0)}{sup}\left|\frac{1}{{\sigma }_t^2}\frac{{\mathrm{\partial }}^3{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j\mathrm{\partial }{\theta }_k}\right|<\mathrm{\infty },}\\ {\displaystyle E_{{\theta }_0}\underset{\theta \in V({\theta }_0)}{sup}|\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}|<\mathrm{\infty },}& {\displaystyle E_{{\theta }_0}\underset{\theta \in V({\theta }_0)}{sup}\left|\frac{1}{{\sigma }_t^4}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}\right|<\mathrm{\infty },}\end{array}$$

   we have $$\begin{array}{}{\displaystyle E_{{\theta }_0}\underset{\theta \in V({\theta }_0)}{sup}\left|\frac{{\mathrm{\partial }}^3{l}_t(\theta )}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j\mathrm{\partial }{\theta }_k}\right|<\mathrm{\infty },}\end{array}$$

   which shows that the conclusion of (iii) is true.

4. It follows from (3), (4), (13) and (14) that $$\begin{array}{}{\displaystyle \underset{\theta \in \mathit{\Theta }}{sup}∥\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }-\frac{\mathrm{\partial }{\tilde{\sigma }}_t^2}{\mathrm{\partial }\theta }∥<K{\rho }^t,\underset{\theta \in \mathit{\Theta }}{sup}∥\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }\theta \mathrm{\partial }{\theta }^T}-\frac{{\mathrm{\partial }}^2{\tilde{\sigma }}_t^2}{\mathrm{\partial }\theta \mathrm{\partial }{\theta }^T}∥<K{\rho }^t,\mathrm{\forall }t,}\end{array}$$(22)

   which yields $$\begin{array}{}{\displaystyle \left|\frac{1}{{\sigma }_t^2}-\frac{1}{{\tilde{\sigma }}_t^2}\right|=\left|\frac{{\tilde{\sigma }}_t^2-{\sigma }_t^2}{{\sigma }_t^2{\tilde{\sigma }}_t^2}\right|\le \frac{K{\rho }^t}{{\sigma }_t^2},\frac{{\sigma }_t^2}{{\tilde{\sigma }}_t^2}\le 1+K{\rho }^t.}\end{array}$$(23)

   Because of $$\begin{array}{}{\displaystyle \frac{\mathrm{\partial }{l}_t(\theta )}{\mathrm{\partial }\theta }=\frac{1}{2}\left(1-\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right)\left(\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }\theta }\right),\frac{\mathrm{\partial }{\tilde{l}}_t(\theta )}{\mathrm{\partial }\theta }=\frac{1}{2}(1-\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}})(\frac{1}{{\tilde{\sigma }}_t^2}\frac{\mathrm{\partial }{\tilde{\sigma }}_t^2}{\mathrm{\partial }\theta }),}\end{array}$$

   it is true that $$\begin{array}{}{\displaystyle \left|\frac{\mathrm{\partial }{l}_t({\theta }_0)}{\mathrm{\partial }\theta }-\frac{\mathrm{\partial }{\tilde{l}}_t({\theta }_0)}{\mathrm{\partial }\theta }\right|}& ={\displaystyle \frac{1}{2}|\left\{\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}-\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right\}\left\{\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}\right\}}\\ & +{\displaystyle \left\{1-\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}\right\}\left\{\frac{1}{{\sigma }_t^2}-\frac{1}{{\tilde{\sigma }}_t^2}\right\}\left\{\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}\right\}}\\ & +{\displaystyle \left\{1-\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}\right\}\left\{\frac{1}{{\tilde{\sigma }}_t^2}\right\}\left\{\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}-\frac{\mathrm{\partial }{\tilde{\sigma }}_t^2}{\mathrm{\partial }{\theta }_i}\right\}|({\theta }_0)}\\ & \le {\displaystyle \frac{1}{2}K{\rho }^t(1+\frac{|{\epsilon }_t-1|}{{\sigma }_t})\left|1+\left\{\frac{1}{{\sigma }_t^2({\theta }_0)}\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }{\theta }_i}\right\}\right|.}\end{array}$$

   Thus, $$\begin{array}{}{\displaystyle \left|n^{-1/2}\sum _{t=1}^n\left\{\frac{\mathrm{\partial }{l}_t({\theta }_0)}{\mathrm{\partial }\theta }-\frac{\mathrm{\partial }{\tilde{l}}_t({\theta }_0)}{\mathrm{\partial }\theta }\right\}\right|}\\ {\displaystyle \le \frac{K}{2}{n}^{-1/2}\sum _{t=1}^n{\rho }^t(1+\frac{|{\epsilon }_t-1|}{{\sigma }_t})\left\{1+\frac{1}{{\sigma }_t^2({\theta }_0)}\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }{\theta }_i}\right\}.}\end{array}$$

   Similarly to the proof (i), according to the Markov inequality and the independent relationship between η_t and ${\sigma }_t^2$(θ₀), for all ε>0 we have $$\begin{array}{}{\displaystyle P\left(n^{-1/2}\sum _{t=1}^n{\rho }^t(1+\frac{|{\epsilon }_t-1|}{{\sigma }_t})\left\{1+\frac{1}{{\sigma }_t^2({\theta }_0)}\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }\theta }\right\}>\epsilon \right)}\\ \le {\displaystyle \frac{2}{\epsilon }\left(1+E_{{\theta }_0}\left|\frac{1}{{\sigma }_t^2({\theta }_0)}\frac{\mathrm{\partial }{\sigma }_t^2({\theta }_0)}{\mathrm{\partial }\theta }\right|\right)n^{-1/2}\sum _{t=1}^n{\rho }^t\to 0.}\end{array}$$(24)

   Thus, the first part of (iv) was obtained. Due to (20), (22), and (23), we have $$\begin{array}{}{\displaystyle \underset{\theta \in V({\theta }_0)}{sup}\left|n^{-1}\sum _{t=1}^n\left\{\frac{{\mathrm{\partial }}^2{l}_t(\theta )}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}-\frac{{\mathrm{\partial }}^2{\tilde{l}}_t(\theta )}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}\right\}\right|}\\ {\displaystyle \le \frac{n^{-1}}{2}\sum _{t=1}^n\underset{\theta \in V({\theta }_0)}{sup}|\left\{\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}-\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right\}\left\{\frac{1}{{\sigma }_t^2}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}\right\}}\\ {\displaystyle +\left\{1-\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}\right\}\left\{\left(\frac{1}{{\sigma }_t^2}-\frac{1}{{\tilde{\sigma }}_t^2}\right)\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}+\frac{1}{{\tilde{\sigma }}_t^2}\left(\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}-\frac{{\mathrm{\partial }}^2{\tilde{\sigma }}_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}\right)\right\}}\\ {\displaystyle +\left\{\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}-\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}\right\}\left\{\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}\right\}\left\{\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}\right\}}\\ {\displaystyle +\left\{\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}-1\right\}\left\{\left(\frac{1}{{\sigma }_t^2}-\frac{1}{{\tilde{\sigma }}_t^2}\right)\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}+\frac{1}{{\tilde{\sigma }}_t^2}\left(\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}-\frac{\mathrm{\partial }{\tilde{\sigma }}_t^2}{\mathrm{\partial }{\theta }_i}\right)\right\}\left\{\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}\right\}}\\ {\displaystyle +\left\{\frac{3}{2}\frac{|{\epsilon }_t-1|}{\sqrt{{\tilde{\sigma }}_t^2}}-1\right\}\left\{\frac{1}{{\tilde{\sigma }}_t^2}\frac{\mathrm{\partial }{\tilde{\sigma }}_t^2}{\mathrm{\partial }{\theta }_i}\right\}\left\{\left(\frac{1}{{\sigma }_t^2}-\frac{1}{{\tilde{\sigma }}_t^2}\right)\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}+\frac{1}{{\tilde{\sigma }}_t^2}\left(\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}-\frac{\mathrm{\partial }{\tilde{\sigma }}_t^2}{\mathrm{\partial }{\theta }_j}\right)\right\}|}\\ {\displaystyle \le \frac{1}{2}K{n}^{-1}\sum _{t=1}^n{\rho }^t{N}_t.}\end{array}$$

   As a consequence, $$\begin{array}{}{\displaystyle N_t=\underset{\theta \in V({\theta }_0)}{sup}\left\{1+\frac{|{\epsilon }_t-1|}{\sqrt{{\sigma }_t^2}}\right\}\left\{1+\frac{1}{{\sigma }_t^2}\frac{{\mathrm{\partial }}^2{\sigma }_t^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}+\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_i}\frac{1}{{\sigma }_t^2}\frac{\mathrm{\partial }{\sigma }_t^2}{\mathrm{\partial }{\theta }_j}\right\}.}\end{array}$$

   By (iii) and Holder inequality, N_t was integrable for some neighbourhood V(θ₀). So, it follows from the Markov inequality that the second part of (iv) is true.

5. The proof of (v) is easily obtained from the central limit theorem for martingale difference. Suppose that 𝓕_t is σ-domain generated from varibles {ε_t−i, i ≥ 0}. As E_θ0(∂l_t(θ₀)/∂θ|𝓕_t) = 0, var_θ0(∂ l_t(θ₀)/∂θ) exists. From Assumption 3 and (ii), 0 < τ − 1 < ∞ and J is non-singular. Hence, the matrix var_θ0(∂l_t(θ₀)/∂θ) is non-degenerate. Thus, λ ∈ R^p+q+1, {λ^T(∂ /∂θ)l_t(θ₀), 𝓕_t is a martingale difference sequence. From the central limit theorem and the Wold-Cramer device, the asymptotic normality result (v) is established.

6. To prove (vi), we firstly prove that the second-order derivatives of l_t(θ) exists. For all i, j, $$\begin{array}{}{\displaystyle n^{-1}\sum _{t=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}{l}_t({\theta }_{ij}^{\ast })=}& {\displaystyle n^{-1}\sum _{t=1}^n\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}{l}_t({\theta }_0)+n^{-1}\sum _{t=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{\theta }^T}\{\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}}\\ & l_t({\tilde{\theta }}_{ij})\}({\theta }_{ij}^{\ast }-{\theta }_0),\end{array}$$(25)

   Here, θ̃_ij locates between ${\theta }_{ij}^{\ast }$ and θ₀. As θ̃_ij almost certainly converges to θ₀, it follows from the ergodic theorem and (iii) that $$\begin{array}{}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}sup∥n^{-1}\sum _{t=1}^n\frac{\mathrm{\partial }}{\mathrm{\partial }{\theta }^T}\left\{\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_{{}_i}\mathrm{\partial }{\theta }_j}{l}_t({\tilde{\theta }}_{ij})\right\}∥}\\ \le {\displaystyle \underset{n\to \mathrm{\infty }}{lim}sup{n}^{-1}\sum _{t=1}^n\underset{\theta \in V({\theta }_0)}{sup}∥\frac{\mathrm{\partial }}{\mathrm{\partial }{\theta }^T}\left\{\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}{l}_t(\theta )\right\}∥}\\ {\displaystyle =E_{{\theta }_0}\underset{\theta \in V({\theta }_0)}{sup}∥\frac{\mathrm{\partial }}{\mathrm{\partial }{\theta }^T}\left\{\frac{{\mathrm{\partial }}^2}{\mathrm{\partial }{\theta }_i\mathrm{\partial }{\theta }_j}{l}_t(\theta )\right\}∥<\mathrm{\infty }.}\end{array}$$

   Since ∥${\theta }_{ij}^{\ast }$ − θ₀∥ → 0 a.s., the second term on the right-hand side of (25) converges to 0 with probability 1. The first term on the right-hand side of (25) is also proved by the ergodic theorem. As a result, the conclusion of (vi) is obtained immediately.