Normality test in random coefficient autoregressive models

Abstract

In this paper, we consider the problem of testing for normality of the two unobservable random processes included in the first order random coefficient autoregressive models. To this end, we propose an information matrix based test and derive its limiting null distribution. We conduct simulations to evaluate the performance and characteristics of the introduced test, and provide a real data analysis.

Appendix

In this appendix, we provide some Lemmas and the proof of the main theorem.

Lemma 1

Under $H_0$, we have that for all $d\ge 1$,

$$\begin{aligned} \textrm{E}\sup _{\theta \in \Theta }\big |\partial _{\theta _i} l_t(\theta )\big |^d<\infty , \quad \textrm{E}\sup _{\theta \in \Theta }\big |\partial ^2_{\theta _i\theta _j} l_t(\theta )\big |^d<\infty , \quad \textrm{E}\sup _{\theta \in \Theta }\big |\partial ^3_{\theta _i\theta _j\theta _k} l_t(\theta )\big |^d <\infty . \end{aligned}$$

Proof

Letting $\gamma _t(\theta )=X_t -\phi X_{t-1}$ and $\delta ^2_t(\theta )=\sigma ^2+s^2X^2_{t-1}$, we can rewrite $l_t(\theta )$ as

$$\begin{aligned} l_t(\theta ) = -\frac{1}{2} \Big (\log \delta ^2_t(\theta )+ \frac{\gamma _t^2(\theta )}{\delta _t^2(\theta )}\Big ) \end{aligned}$$

and thus we have

$$\begin{aligned} \partial _\theta l_t(\theta )= & {} -\frac{1}{2}\frac{1}{\delta _t^2(\theta )} \left\{ \Big (1-\frac{\gamma ^2_t(\theta )}{\delta _t^2(\theta )}\Big )\partial _\theta \delta _t^2(\theta )+2\gamma _t(\theta )\partial _\theta \gamma _t (\theta )\right\} . \end{aligned}$$

(8)

Note that $\partial _\theta \gamma _t (\theta )=(-X_{t-1},0,0)'$, $\partial _\theta \delta _t^2(\theta )=(0,X^2_{t-1},1)'$, and $\gamma _t(\theta )=(\phi _0 -\phi +\xi _t)X_{t-1}+\eta _t$. Setting $c_0 =\min \big \{\inf _{\theta \in \Theta } s^2,\inf _{\theta \in \Theta }\sigma ^2 \big \}$ and denoting by K a general positive constant, which may have different values, then, since $\Theta$ is a bounded set by the compactness, we have that

$$\begin{aligned} \Big |\frac{1}{\delta _t^2(\theta )}\partial _{\theta _i} \delta _t^2(\theta )\big |\le & {} \frac{1}{\sigma ^2+s^2X^2_{t-1}} (X^2_{t-1}+1)\le \frac{1}{c_0}, \end{aligned}$$

(9)

$$\begin{aligned} \Big |\frac{\gamma ^2_t(\theta )}{\delta _t^4(\theta )} \partial _{\theta _i} \delta _t^2(\theta )\Big |\le & {} \frac{((\phi _0 -\phi +\xi _t)X_{t-1}+\eta _t)^2}{(\sigma ^2+s^2X^2_{t-1})^2} (X^2_{t-1}+1) \nonumber \\\le & {} \frac{1}{c_0} \frac{((\phi _0 -\phi +\xi _t)X_{t-1}+\eta _t)^2}{\sigma ^2+s^2X^2_{t-1}} \nonumber \\\le & {} K\big ( (\phi _0-\phi )^2 +\xi _t^2+\eta _t^2\big ) \le K \big ( 1 +\xi _t^2+\eta _t^2\big ) \end{aligned}$$

(10)

and for any $n\in {\mathbb {N}}$,

$$\begin{aligned} \Big | \frac{\gamma _t(\theta )}{\delta _t^2(\theta )} \partial _{\theta _i} \gamma _t(\theta )\Big |^{2n}\le & {} \Big |\frac{(\phi _0 -\phi +\xi _t)X_{t-1}+\eta _t}{\sigma ^2+s^2X^2_{t-1}}X_{t-1}\Big |^{2n}\nonumber \\\le & {} 2^{2n-1}\frac{(\phi _0 -\phi +\xi _t)^{2n}X^{2n}_{t-1}+\eta ^{2n}_t}{(\sigma ^2+s^2X^2_{t-1})^{2n}}X^{2n}_{t-1}\nonumber \\\le & {} 2^{2n-1} \Big \{ \frac{2^{2n-1}}{c_0^{2n}}\big ((\phi _0 -\phi )^{2n} +\xi _t^{2n}\big ) +\frac{1}{c_0^{2n}}\eta _t^{2n} \Big \}\nonumber \\\le & {} \left( \frac{2}{c_0}\right) ^{2n} \Big \{ 2^{2n}\big (K^{2n} +\xi _t^{2n}\big ) +\eta _t^{2n} \Big \}. \end{aligned}$$

(11)

Therefore, it follows that

$$\begin{aligned} \big | \partial _{\theta _i} l_t(\theta )\big |^{2n} \le K^n \big ( 1+\xi _t^{2n}+\eta _t^{2n}+\xi _t^{4n}+\eta _t^{4n}\big ). \end{aligned}$$

(12)

Since $\partial ^2_{\theta _i\theta _j} \gamma _t(\theta )=0$ and $\partial ^2_{\theta _i\theta _j} \delta ^2_t(\theta )=0$, we have by simple calculations that

$$\begin{aligned} \partial ^2_{\theta _i\theta _j} l_t(\theta )&= - \frac{\partial _{\theta _i} l_t(\theta )}{\delta _t^2(\theta )}\partial _{\theta _j} \delta _t^2(\theta ) + \left\{ \frac{\gamma _t(\theta )}{\delta _t^2(\theta )}\partial _{\theta _j} \gamma _t(\theta ) -\frac{1}{2}\frac{\gamma ^2_t(\theta )}{\delta _t^4(\theta )}\partial _{\theta _j} \delta _t^2(\theta )\right\} \frac{1}{\delta _t^2(\theta )} \partial _{\theta _i} \delta _t^2(\theta )\nonumber \\&\quad - \frac{1}{\delta _t^2(\theta )} \partial _{\theta _i}\gamma _t(\theta )\partial _{\theta _j} \gamma _t (\theta ). \end{aligned}$$

(13)

Using (9)–(12) and $|\partial _{\theta _i}\gamma _t(\theta )\partial _{\theta _j} \gamma _t (\theta ) /\sigma _t^2(\theta )| \le K$, one can see that

$$\begin{aligned} \big | \partial ^2_{\theta _i\theta _j} l_t(\theta ) \big |^{2n} \le K^n \big ( 1+\xi _t^{2n}+\eta _t^{2n}+\xi _t^{4n}+\eta _t^{4n}\big ). \end{aligned}$$

Similarly to the above, we can also obtain

$$\begin{aligned} \big | \partial ^3_{\theta _i\theta _j\theta _k} l_t(\theta ) \big |^{2n} \le K^n \big ( 1+\xi _t^{2n}+\eta _t^{2n}+\xi _t^{4n}+\eta _t^{4n}\big ). \end{aligned}$$

Since every moment of $\xi _t$ and $\eta _t$ exists under $H_0$, we thus have that

$$\begin{aligned} \textrm{E}\sup _{\theta \in \Theta }\big |\partial _{\theta _i} l_t(\theta )\big |^{2n}<\infty , \quad \textrm{E}\sup _{\theta \in \Theta }\big |\partial ^2_{\theta _i\theta _j} l_t(\theta )\big |^{2n}<\infty , \quad \textrm{E}\sup _{\theta \in \Theta }\big |\partial ^3_{\theta _i\theta _j\theta _k} l_t(\theta )\big |^{2n} <\infty . \end{aligned}$$

Therefore, the lemma follows from Lyapunov’s inequality. $\square$

Lemma 2

Under $H_0$, it holds that

$$\begin{aligned} \textrm{E}\big [ \partial _{\theta }l_t(\theta _{0})\partial _{\theta '}l_t(\theta _{0})\big ]= -\textrm{E}\big [ \partial ^{2}_{\theta \theta '} l_t(\theta _{0})\big ]. \end{aligned}$$

Proof

From (8), we have

$$\begin{aligned} \partial _\theta l_t(\theta ) \partial _{\theta '} l_t(\theta )&=\frac{1}{4}\frac{1}{\delta _t^4(\theta )} \left\{ \Big (1-\frac{\gamma ^2_t(\theta )}{\delta _t^2(\theta )}\Big )^2\partial _\theta \delta _t^2(\theta )\partial _{\theta '} \delta _t^2(\theta ) +4\gamma ^2_t(\theta )\partial _\theta \gamma _t (\theta )\partial _{\theta '} \gamma _t (\theta )\right. \\&\quad \left. +2\gamma _t(\theta )\Big (1-\frac{\gamma ^2_t(\theta )}{\delta _t^2(\theta )}\Big )\big (\partial _\theta \gamma _t (\theta )\partial _{\theta '}\delta _t^2(\theta )+\partial _{\theta '} \gamma _t (\theta )\partial _{\theta }\delta _t^2(\theta )\big )\right\} . \end{aligned}$$

Note that $\gamma _t(\theta _0)=\xi _t X_{t-1} +\eta _t$. Then, since $\xi _t \sim N(0,s_0^2)$ and $\eta _t \sim N(0,\sigma _0^2)$ under $H_0$, one can see that

$$\begin{aligned} \textrm{E}\big [\gamma _t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ]=0,\ \textrm{E}\big [\gamma ^2_t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ]=\delta _t^2(\theta _0),\ \textrm{E}\big [\gamma ^3_t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ]=0,\ \textrm{E}\big [\gamma ^4_t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ]=3\delta _t^4(\theta _0). \end{aligned}$$

(14)

Since $\partial _\theta \gamma _t (\theta )$ and $\partial _\theta \delta _t^2(\theta )$ are measurable w.r.t. ${\mathcal {F}}_{t-1}$, we have by (14) that

$$\begin{aligned} \textrm{E}\big [\partial _\theta l_t(\theta _0) \partial _{\theta '} l_t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ] =\frac{1}{2}\frac{1}{\delta _t^4(\theta _0)} \left\{ \partial _\theta \delta _t^2(\theta _0)\partial _{\theta '} \delta _t^2(\theta _0) +2\delta ^2_t(\theta _0)\partial _\theta \gamma _t (\theta _0)\partial _{\theta '} \gamma _t (\theta _0)\right\} . \end{aligned}$$

Similarly, it can be readily shown that $\textrm{E}\big [\partial _\theta l_t(\theta _0) \big |{\mathcal {F}}_{t-1}\big ]=0$. Using this and (14), we have from (13) that

$$\begin{aligned} \textrm{E}\big [\partial ^2_{\theta \theta '} l_t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ]= & {} -\frac{1}{2}\frac{1}{\delta _t^4(\theta _0)} \Big \{ \partial _\theta \delta _t^2(\theta _0)\partial _{\theta '} \delta _t^2(\theta _0) +2\delta ^2_t(\theta _0)\partial _\theta \gamma _t (\theta _0)\partial _{\theta '} \gamma _t (\theta _0)\Big \} \nonumber \\= & {} - \textrm{E}\big [\partial _\theta l_t(\theta _0) \partial _{\theta '} l_t(\theta _0) \big | {\mathcal {F}}_{t-1}\big ], \end{aligned}$$

(15)

which asserts the lemma. $\square$

Lemma 3

Under $H_0$, we have

$$\begin{aligned} \frac{1}{n} \sum _{t=1}^n \partial ^2_{\theta \theta '} l_t(\theta _n^*)=\textrm{E}\big [\partial ^2_{\theta \theta '} l_t(\theta _0)\big ]+o(1)\quad a.s. \end{aligned}$$

(16)

and

$$\begin{aligned} \frac{1}{n} \sum _{t=1}^n \nabla d(X_t;\theta ^*_n) =E \big [\nabla d(X_t;\theta _0)\big ]+o(1)\quad a.s., \end{aligned}$$

(17)

where $\theta _n^*$ is any point on the line segment between ${\hat{\theta }}_n$ and $\theta _0$.

Proof

By Lemma 1, we have $\textrm{E}\sup _{\theta \in \Theta }\Vert \partial ^2_{\theta \theta '} l_t(\theta ) -\partial ^2_{\theta \theta '} l_t(\theta _0)\Vert <\infty$, where $\Vert \cdot \Vert$ is any norm for matrices. Then, for any $\epsilon >0$, using the continuity of $\partial ^2_{\theta \theta '} l_t(\theta )$ and the dominated convergence theorem, we can take a positive constant $r_\epsilon$ such that

$$\begin{aligned} \textrm{E}\sup _{\theta \in N_\epsilon (\theta _0)}\Vert \partial ^2_{\theta \theta '} l_t(\theta ) -\partial ^2_{\theta \theta '} l_t(\theta _0)\Vert <\frac{\epsilon }{2}, \end{aligned}$$

where $N_\epsilon (\theta _0)$ is the neighborhood of $\theta _0$ with radius $r_\epsilon$. Hence, it follows from the strong consistency of ${\hat{\theta }}_n$ and the ergodic theorem that, for sufficiently large n,

$$\begin{aligned}{} & {} \Big \Vert \frac{1}{n} \sum _{t=1}^n \partial ^2_{\theta \theta '} l_t(\theta _n^*)-\textrm{E}\big [\partial ^2_{\theta \theta '} l_t(\theta _0)\big ]\Big \Vert \\{} & {} \quad \le \Big \Vert \frac{1}{n} \sum _{t=1}^n \partial ^2_{\theta \theta '} l_t(\theta _n^*)- \frac{1}{n} \sum _{t=1}^n \partial ^2_{\theta \theta '} l_t(\theta _0)\Big \Vert +\Big \Vert \frac{1}{n} \sum _{t=1}^n \partial ^2_{\theta \theta '} l_t(\theta _0)-\textrm{E}\big [\partial ^2_{\theta \theta '} l_t(\theta _0)\big ]\Big \Vert \\{} & {} \quad \le \frac{1}{n} \sum _{t=1}^n \sup _{\theta \in N_\epsilon (\theta _0)} \big \Vert \partial ^2_{\theta \theta '} l_t(\theta )- \ \partial ^2_{\theta \theta '} l_t(\theta _0)\big \Vert +\Big \Vert \frac{1}{n} \sum _{t=1}^n \partial ^2_{\theta \theta '} l_t(\theta _0)-\textrm{E}\big [\partial ^2_{\theta \theta '} l_t(\theta _0)\big ]\Big \Vert <\epsilon \quad a.s., \end{aligned}$$

which asserts (16).

Since $\textrm{E}\sup _{ \theta \in \Theta }\Vert \nabla d(X_t;\theta )\Vert$ is also finite by Lemma 1 and $\nabla d(X_t;\theta )$ is continuous in $\theta$, one can show the second on in the same manner as the above. This completes the proof. $\square$

Proof of Theorem 2

Note from (15) that $\left\{ (d(X_t;\theta _0), {\mathcal {F}}_t)\right\}$ is a martingale difference. Then, by the CLT for the martingale differences, we have

$$\begin{aligned} D_n(\theta _0):=\frac{1}{\sqrt{n}} \sum _{t=1}^n d(X_t;\theta _0) {\mathop {\longrightarrow }\limits ^{d}} N_q (\textbf{0}, \Sigma _0), \end{aligned}$$

where $\Sigma _0= \textrm{cov}(d(X;\theta _0))$ exists by Lemma 1. By Taylor’s theorem, we can write that

$$\begin{aligned} D_n({\hat{\theta }}_n) =D_n(\theta _0)+ \frac{1}{\sqrt{n}}\nabla D_n({\tilde{\theta }}_n)\sqrt{n}({\hat{\theta }}_n -\theta _0), \end{aligned}$$

(18)

where $\nabla D_n$ is the Jacobian matrix of $D_n$ and ${\tilde{\theta }}_n$ is a point on the line segment between ${\hat{\theta }}_n$ and $\theta _0$. Also, using Taylor’s theorem again and the fact that $\sum _{t=1}^n\partial _{\theta }l_t({\hat{\theta }}_n)=0$, we have

$$\begin{aligned} \sum _{t=1}^n\partial _{\theta }l_t({\hat{\theta }}_n)=\sum _{t=1}^n\partial _{\theta }l_t(\theta _0)+\sum _{t=1}^n\partial ^2_{\theta \theta '}l_t(\theta ^*_n)(\hat{\theta }_n-\theta _0)=0, \end{aligned}$$

where $\theta ^*_n$ lies between ${\hat{\theta }}_n$ and $\theta _0$, and thus we can write that

$$\begin{aligned} \sqrt{n}(\hat{\theta }_n-\theta _0)=-{\mathcal {J}}^{-1} \frac{1}{\sqrt{n}} \sum _{t=1}^n\partial _{\theta }l_t(\theta _0)-{\mathcal {J}}^{-1}\Big (\frac{1}{n}\sum _{t=1}^n\partial ^2_{\theta \theta '}l_t(\theta ^*_n)-{\mathcal {J}}\Big ) \sqrt{n}(\hat{\theta }_n-\theta _0). \end{aligned}$$

(19)

Since $\{(\partial _\theta l_t(\theta _0),{\mathcal {F}}_t)\}$ is also martingale difference, we have $\frac{1}{\sqrt{n}}\sum _{t=1}^n\partial _{\theta }l_t(\theta _0)=O_P(1)$, which together with (16) implies $\sqrt{n}(\hat{\theta }_n-\theta _0)=O_P(1)$. By this and (16) again, the second term on the right hand side of (19) is converges to zero, and thus we have

$$\begin{aligned} \sqrt{n}(\hat{\theta }_n-\theta _0)=-{\mathcal {J}}^{-1} \frac{1}{\sqrt{n}} \sum _{t=1}^n \partial _\theta l_t(\theta _0) +o_P(1). \end{aligned}$$

(20)

Now, set ${\mathcal {K}}=\textrm{E}[\nabla d(X_t;\theta _0)]$. From (20) and (17), one can see that

$$\begin{aligned}{} & {} \frac{1}{\sqrt{n}}\nabla D_n({\tilde{\theta }}_n)\sqrt{n} (\hat{\theta }_{n}-\theta _0)\\{} & {} \quad =-{\mathcal {K}}{\mathcal {J}}^{-1}\frac{1}{\sqrt{n}} \sum _{t=1}^n \partial _\theta l_t(\theta _0) -\Big (\frac{1}{\sqrt{n}}\nabla D_n({\tilde{\theta }}_n) - {\mathcal {K}}\Big ) {\mathcal {J}}^{-1}\frac{1}{\sqrt{n}}\sum _{t=1}^n \partial _\theta l_t(\theta _0)+o_P(1)\\{} & {} \quad =-{\mathcal {K}}{\mathcal {J}}^{-1}\frac{1}{\sqrt{n}} \sum _{t=1}^n \partial _\theta l_t(\theta _0)+o_P(1), \end{aligned}$$

and thus, it follows from (18) that

$$\begin{aligned} D_n({\hat{\theta }}_n)= & {} D_n(\theta _0)-{\mathcal {K}}{\mathcal {J}}^{-1}\frac{1}{\sqrt{n}} \sum _{t=1}^n \partial _\theta l_t(\theta _0)+o_P(1)\\= & {} \frac{1}{\sqrt{n}}\sum _{t=1}^n \big ( d(X_t;\theta _0) -{\mathcal {K}}{\mathcal {J}}^{-1}\partial _\theta l_t(\theta _0)\big )+o_P(1). \end{aligned}$$

Since $\left\{ (d(X_t;\theta _0), {\mathcal {F}}_t)\right\}$ and $\left\{ (\partial _\theta l_t(\theta _0), {\mathcal {F}}_{t})\right\}$ are martingale differences, $\left\{ (d(X_t;\theta _0) -{\mathcal {K}}{\mathcal {J}}^{-1}\partial _\theta l_t(\theta _0), {\mathcal {F}}_{t})\right\}$ also becomes a martingale difference. Hence, applying the CLT for martingale differences to the sequence, we have

$$\begin{aligned} D_n({\hat{\theta }}_n) {\mathop {\longrightarrow }\limits ^{d}} N_q ( \textbf{0}, \Sigma ), \end{aligned}$$

where $\Sigma = \textrm{cov}\left( d(X_t;\theta _0)- {\mathcal {K}}{\mathcal {J}}^{-1}\partial _\theta l(X_t;\theta _0) \right)$ and it exists by Lemma 1. This completes the proof. $\square$

Lemma 4

Under $H_0$,

$$\begin{aligned} \textrm{E}\big [ \nabla d(X_t;\theta _0)\big ]= - \textrm{E}\big [ d(X_t;\theta _0) \partial _{\theta '} l_t(\theta _0)\big ] \end{aligned}$$

Proof

Denote the process generated from the RCA(1) model with the parameter $\theta$ by $\{X_{\theta ,t}\}$. Then, by the same argument as in the proof of Lemma 2, we have

$$\begin{aligned} \textrm{E}\big [ \partial _{\theta _i} l_\theta (X_{\theta ,t})\,\partial _{\theta _j} l_\theta (X_{\theta ,t})|{\mathcal {F}}_{\theta ,t-1}\big ]= -\textrm{E}\big [ \partial ^2_{\theta _i\theta _j} l_\theta (X_{\theta ,t})|{\mathcal {F}}_{\theta ,t-1}\big ], \end{aligned}$$

(21)

where ${\mathcal {F}}_{\theta ,t-1}=\sigma (X_{\theta ,k}:\, k\le t-1)$ and

$$\begin{aligned} l_\theta (X_{\theta ,t}) = -\frac{1}{2} \log (\sigma ^2+s^2 X_{\theta , t-1}^2)-\frac{1}{2} \frac{(X_{\theta ,t }-\phi X_{\theta , t-1})^2}{\sigma ^2+s^2 X_{\theta , t-1}^2}. \end{aligned}$$

Rewriting (21) in the integral form, we have

$$\begin{aligned} \int \partial _{\theta _i} l_\theta (x)\,\partial _{\theta _j} l_\theta (x)\,g_\theta (x)dx= -\int \partial ^2_{\theta _i \theta _j} l_\theta (x)\,g_\theta (x)dx, \end{aligned}$$

where $g(x;\theta )$ is the conditional pdf of $X_{\theta ,t}$ given ${\mathcal {F}}_{\theta ,t-1}$. Since $X_{\theta ,t}|{\mathcal {F}}_{\theta ,t-1}\sim N( \phi X_{\theta ,t-1}, \sigma ^2+s^2 X_{\theta ,t-1})$ under $H_0$, we have $\partial _\theta \log g(x;\theta ) = \partial _\theta l_\theta (x)$. Hence, by differentiating the both sides w.r.t. $\theta _l$, we obtain

$$\begin{aligned}&\int \big \{ \partial ^2_{\theta _i\theta _l} l_\theta (x)\,\partial _{\theta _j} l_\theta (x)+ \partial _{\theta _i} l_\theta (x)\,\partial ^2_{\theta _j\theta _l} l_\theta (x) + \partial _{\theta _i} l_\theta (x)\,\partial _{\theta _j} l_\theta (x)\, \partial _{\theta _l} l_\theta (x)\big \} g_\theta (x)dx\\&\quad =- \int \big \{ \partial ^3_{\theta _i \theta _j \theta _l} l_\theta (x)+\partial ^2_{\theta _i\theta _j} l_\theta (x)\, \partial _{\theta _l}l_\theta (x)\big \}g_\theta (x) dx. \end{aligned}$$

Noting $l_{\theta _0}(X_{\theta _0,t})=l_t(\theta _0)$ and using the above, one can see that

$$\begin{aligned} \textrm{E}\big [ \partial _{\theta _l}d_k(X_t;\theta _0) |{\mathcal {F}}_{t-1}\big ]= & {} \textrm{E}\big [ \partial ^3_{\theta _{i_k} \theta _{j_k} \theta _l}l_t(\theta _0)+ \partial ^2_{\theta _{i_k} \theta _l}l_t(\theta _0)\partial _{\theta _{j_k}}l_t(\theta _0)+\partial _{\theta _{i_k}}l_t(\theta _0)\partial ^2_{\theta _{j_k} \theta _l}l_t(\theta _0)|{\mathcal {F}}_{t-1}\big ]\\= & {} - \textrm{E}\big [\big \{\partial ^2_{\theta _{i_k} \theta _{j_k}}l_t(\theta _0)+\partial _{\theta _{i_k}}l_t(\theta _0)\partial _{\theta _{j_k}}l_t(\theta _0)\big \}\partial _{\theta _l}l_t(\theta _0)|{\mathcal {F}}_{t-1}\big ]\\= & {} -\textrm{E}\big [ d_k(X_t;\theta _0) \partial _{\theta _l}l_t(\theta _0)|{\mathcal {F}}_{t-1}\big ], \end{aligned}$$

which yields the lemma. $\square$

Lemma 5

Under $H_0$, $V(\theta _0)$ corresponding to $d(X_t;\theta _0) = \left( d_1(X_t;\theta _0), d_2(X_t;\theta _0), d_3(X_t;\theta _0) \right) ^{'}$ is nonsingular, where $d_k(X_t;\theta _0)=\partial ^2_{\theta _k \theta _k} l(X_t;\theta _0) + \partial _{\theta _k} l(X_t;\theta _0) \partial _{\theta _k} l(X_t;\theta _0)$.

Proof

We follow the scheme used in Lemma 6 in (Aue et al., 2006). Assume that $\tau ^{'} V(\theta _0) \tau =0$ for some $\tau =(\tau _1,\tau _2,\tau _3)^{'}$, where $\tau _1, \tau _2$, and $\tau _3$ are not all zero. Then, we have by (7) that

$$\begin{aligned} \tau ^{'} V(\theta _0) \tau= & {} \textrm{E}\left[ (\tau ^{'} d(X_t;\theta _0))^2 \right] +\tau '\textrm{E}\left[ d(X_t;\theta _0) \partial _{\theta '} l_t(\theta _0) \right] {\mathcal {I}}^{-1} \textrm{E}\left[ \partial _\theta l_t(\theta _0) d(X_t;\theta _0)'\right] \tau . \end{aligned}$$

Since ${\mathcal {I}}^{-1}$ is a non-negative definite matrix, the second term in the right hand side of the above equality is greater than or equal to zero as well as the first term, and thus two terms should be zero as $\tau ^{'} V(\theta _0) \tau =0$. We just deal with the first term to prove the lemma. By some algebraic calculations, we can have that

$$\begin{aligned}&\tau ^{'} d(X_t;\theta _0) \\&\quad =\frac{3}{4} \big (\tau _2X_{t-1}^4+\tau _3\big ) \frac{\gamma _t^4(\theta _0)}{\delta _t^8(\theta _0)} +\Big (\tau _1X_{t-1}^2 -2 \frac{\tau _2X_{t-1}^4+\tau _3 }{\delta _t^2(\theta _0)}\Big )\frac{\gamma _t^2(\theta _0)}{\delta _t^4(\theta _0)} +\frac{3}{4} \frac{\tau _2X_{t-1}^4+\tau _3 }{\delta _t^4(\theta _0)}-\tau _1 \frac{X_{t-1}^2}{\delta _t^2(\theta _0)}\\&\quad =0 \end{aligned}$$

Following the similar steps in Lemma 6 in (Aue et al., 2006), it can be shown that $P\left( \tau _2X_0^4+\tau _3\ne 0\right) =1$. Hence, the fourth degree equation for $\gamma _t(\theta _0)$ has at most four solutions and thus we can express that

$$\begin{aligned} \textrm{P}\Big ( \gamma _t(\theta _0) \in \{C_1, C_2, C_3, C_4\} \Big )=1, \end{aligned}$$

where $C_1$ - $C_4$ are functions of $X_{t-1}$, implying that

$$\begin{aligned} \textrm{P}\Big ( \xi _t x +\eta _t \in \{C_1(x), C_2(x), C_3(x), C_4(x)\} \Big )=1\quad \text{ for }\ \textrm{P}_{X_0}(x)- a.s. \end{aligned}$$

This contradicts the assumption that $\xi _t$ and $\eta _t$ are independent normal random variables. $\tau =(0,0,0)'$ is therefore the only solution to $\tau ^{'} V(\theta _0) \tau =0$. This completes the proof. $\square$