Asymptotic normality of the MLE in the level-effect ARCH model

Abstract

We establish consistency and asymptotic normality of the maximum likelihood estimator in the level-effect ARCH model of Chan et al. (J Financ 47(3):1209–1227, 1992). Furthermore, it is shown by simulations that the asymptotic properties also apply in finite samples.

Appendix

The analytical expressions for the first, second and third order derivatives of the quasi log likelihood function are given in a Supplementary Appendix available upon request from any of the authors. We provide now three important propositions that we need in order to prove Theorem 1. The proof technique for the MLE utilizes the classic Cramér type conditions for consistency and asymptotic normality (central limit theorem for the score, convergence of the Hessian and uniformly bounded third-order derivatives); see e.g. Lehmann (1999).

Proposition 1

Let $u_{jt}\left( \theta _{0}\right)$ be defined as in Theorem 1. Under Assumptions A and B, the joint distribution of the score functions evaluated at $\theta =\theta _{0}$ are asymptotically Gaussian,

$$\begin{aligned} \frac{1}{\sqrt{T}}\frac{\partial }{\partial \theta }L_{T}\left( \theta _{0}\right) \overset{d}{\rightarrow }N\left( 0,\Lambda \right) , \end{aligned}$$

where

$$\begin{aligned} \Lambda =\zeta \left[ \begin{array}{ccc} {\overline{m}}_{11} &{} \frac{1}{2}{\overline{m}}_{12} &{} \frac{1}{2}{\overline{m}} _{13} \\ \frac{1}{2}{\overline{m}}_{12} &{} \frac{1}{4}{\overline{m}}_{22} &{} \frac{1}{4} {\overline{m}}_{23} \\ \frac{1}{2}{\overline{m}}_{13} &{} \frac{1}{4}{\overline{m}}_{23} &{} \frac{1}{4} {\overline{m}}_{33} \end{array} \right] >0, \end{aligned}$$

and ${\overline{m}}_{ij}=E\left( u_{it}\left( \theta _{0}\right) u_{jt}\left( \theta _{0}\right) \right)$ for $i=1,2,3$ and $j=1,2,3$.

Proof of Proposition 1

For the proof of Proposition 1, we need first the following 2 Lemmas. $\square$

Lemma A

Let Assumptions A and B hold and define $u_{1t}\left( \theta _{0}\right) =\left( \ln \left| y_{t-1}\right| -\right.$ $\left. w_{0}\left( \frac{1}{w_{0}}-\frac{ 1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) \ln \left| y_{t-2}\right| \right) ,$

$u_{2t}\left( \theta _{0}\right) =\left( \frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right)$ and $u_{3t}\left( \theta _{0}\right) =\left( \frac{w_{0}}{\alpha _{0}}\right) \left( \frac{1}{w_{0}}-\frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right)$. Then $u_{it}\left( \theta _{0}\right)$ is a stationary and ergodic sequence. In addition $\frac{1}{T} \sum _{t=1}^{T}u_{it}\left( \theta _{0}\right) \overset{p}{\rightarrow } E\left( u_{it}\left( \theta _{0}\right) \right) \equiv {\overline{u}}_{i}$ and $\frac{1}{T}\sum _{t=1}^{T}u_{it}^{2}\left( \theta _{0}\right) \overset{p}{ \rightarrow }E\left( u_{it}^{2}\left( \theta _{0}\right) \right) \equiv {\overline{m}}_{ii}$ for $i=1,2,3.$

Proof of Lemma A

Define $I_{t}=\{y_{t},z_{t,}y_{t-1},z_{t-1,}y_{t-2},z_{t-2,}...\}.$ Note first that

$$\begin{aligned} \left| u_{1t}\left( \theta _{0}\right) \right| \le \left| \ln \left| y_{t-1}\right| \right| +w_{0}\left| \ln \left| y_{t-2}\right| \right| \left( \frac{1}{w_{0}}+\frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) \le \left| \ln \left| y_{t-1}\right| \right| +2\left| \ln \left| y_{t-2}\right| \right| , \end{aligned}$$

hence

$$\begin{aligned} E\left| u_{1t}\left( \theta _{0}\right) \right| \le 3E(\ln \left| y_{t}\right| )<\infty , \end{aligned}$$

where we have used assumptions A and B and where the last inequality follows from A3 where the first two moments of $\ln \left| y_{t}\right|$ are assumed to be bounded. Hence we can write

$$\begin{aligned} u_{1t}\left( \theta _{0}\right) \equiv g_{1}\left( y_{t-1},y_{t-2},\sigma _{t}^{2}\left( \theta _{0}\right) \right) , \end{aligned}$$

where $g_{1}$ is a $I_{t}$-measurable function and where all arguments $y_{t-1},y_{t-2}$ and $\sigma _{t}^{2}\left( \theta _{0}\right)$ are stationary and ergodic as a consequence of Lemmas 1 and 2. This implies that $u_{1t}\left( \theta _{0}\right)$ is stationary and ergodic by Theorem 3.35 in White (1984). Consequently $\frac{1}{T}\sum _{t=1}^{T}u_{1t}\left( \theta _{0}\right) \overset{p}{\rightarrow }E\left( u_{1t}\left( \theta _{0}\right) \right)$ follows by the Ergodic Theorem. Similarly, it follows straightforwardly that $E\left| u_{2t}\left( \theta _{0}\right) \right| \le \left( \frac{1}{w_{0}}\right)$ and $E\left| u_{3t}\left( \theta _{0}\right) \right| \le \left( \frac{2}{\alpha _{0}} \right) .$ We can write $u_{2t}\left( \theta _{0}\right) \equiv g_{2}\left( \sigma _{t}^{2}\left( \theta _{0}\right) \right)$ and $u_{3t}\left( \theta _{0}\right) \equiv g_{3}\left( \sigma _{t}^{2}\left( \theta _{0}\right) \right)$ and as above conclude that $(u_{2t}\left( \theta _{0}\right) ,u_{3t}\left( \theta _{0}\right) )$ is stationary and ergodic, and hence $\frac{1}{T}\sum _{t=1}^{T}u_{it}\left( \theta _{0}\right) \overset{p}{ \rightarrow }E\left( u_{it}\left( \theta _{0}\right) \right)$ for $i=2,3$. Second, notice that

$$\begin{aligned} \left| u_{1t}^{2}\left( \theta _{0}\right) \right|= & {} |\ln ^{2}\left| y_{t-1}\right| -2w_{0}\left( \frac{1}{w_{0}}-\frac{1}{ \sigma _{t}^{2}\left( \theta _{0}\right) }\right) \ln \left| y_{t-2}\right| \ln \left| y_{t-1}\right| \\&\quad +\left( \ln \left| y_{t-2}\right| \right) ^{2}w_{0}^{2}\left( \frac{1}{w_{0}}-\frac{1}{ \sigma _{t}^{2}\left( \theta _{0}\right) }\right) ^{2}| \\\le & {} \ln ^{2}\left| y_{t-1}\right| +\frac{4w_{0}^{2}}{\sigma _{t}^{4}\left( \theta _{0}\right) }\ln ^{2}\left| y_{t-2}\right| +4 \frac{w_{0}}{\sigma _{t}^{2}\left( \theta _{0}\right) }\left| \ln \left| y_{t-1}\right| \ln \left| y_{t-2}\right| \right| \\\le & {} \ln ^{2}\left| y_{t-1}\right| +4\ln ^{2}\left| y_{t-2}\right| +4\left| \ln \left| y_{t-1}\right| \ln \left| y_{t-2}\right| \right| , \end{aligned}$$

such that

$$\begin{aligned} E\left| u_{1t}^{2}\left( \theta _{0}\right) \right| \le 5E\left( (\ln \left| y_{t}\right| )^{2}\right) +4E\left| \ln \left| y_{t}\right| \ln \left| y_{t-1}\right| \right| <\infty . \end{aligned}$$

On the right hand side of the first inequality we have used Lemmas 1 and 2 and the second inequality follows from A3 (existence of second order moments). In addition, $E\left| u_{2t}^{2}\left( \theta _{0}\right) \right| \le \left( \frac{1}{w_{0}^{2}}\right)$ and $E\left| u_{3t}^{2}\left( \theta _{0}\right) \right| \le \left( \frac{4}{\alpha _{0}^{2}}\right) .$ We can therefore conclude, by Theorem 3.35 in White (1984), that since $u_{it}\left( \theta _{0}\right)$ is stationary and ergodic then so is $u_{it}^{2}\left( \theta _{0}\right)$ for $i=1,2,3.$ Furthermore as $E|u_{it}^{2}\left( \theta _{0}\right) |$ is bounded then $\frac{1}{T}\sum _{t=1}^{T}u_{1t}^{2}\left( \theta _{0}\right) \overset{p}{ \rightarrow }E\left( u_{1t}^{2}\left( \theta _{0}\right) \right)$ for $i=1,2,3$ follows from the ergodicity theorem. This completes the proof of Lemma A. $\square$

Lemma B

Under Assumptions A and B, the marginal distributions of the score functions given by Eqs. (9)–(11) evaluated at $\theta =\theta _{0}$ are asymptotically Gaussian,

$$\begin{aligned} \frac{1}{\sqrt{T}}\frac{\partial }{\partial \gamma }L_{T}\left( \theta _{0}\right)= & {} \frac{-1}{\sqrt{T}}\sum _{t=1}^{T}\left( 1-z_{t}^{2}\right) u_{1t}\left( \theta _{0}\right) \overset{d}{\rightarrow }N\left( 0,\zeta {\overline{m}}_{11}\right) , \end{aligned}$$

(4)

$$\begin{aligned} \frac{1}{\sqrt{T}}\frac{\partial }{\partial w}L_{T}\left( \theta _{0}\right)= & {} \frac{-1}{\sqrt{T}}\sum _{t=1}^{T}\frac{1}{2}\left( 1-z_{t}^{2}\right) u_{2t}\left( \theta _{0}\right) \overset{d}{\rightarrow }N\left( 0,\zeta {\overline{m}}_{22}\right) , \end{aligned}$$

(5)

$$\begin{aligned} \frac{1}{\sqrt{T}}\frac{\partial }{\partial \alpha }L_{T}\left( \theta _{0}\right)= & {} \frac{-1}{\sqrt{T}}\sum _{t=1}^{T}\frac{1}{2}\left( 1-z_{t}^{2}\right) u_{3t}\left( \theta _{0}\right) \overset{d}{\rightarrow } N\left( 0,\zeta {\overline{m}}_{33}\right) , \end{aligned}$$

(6)

where ${\overline{m}}_{ii},$ $i=1,2,3$ and $\zeta$ are defined by Lemma A and A3 respectively.

Proof of Lemma B

We will prove (4) in detail. The results in (5) and ( 6) hold by identical arguments. Define again $I_{t}= \{y_{t},z_{t,}y_{t-1},z_{t-1,}y_{t-2},z_{t-2,}...\}$ and recall from Result 1 that

$$\begin{aligned} s_{1t}\left( \theta _{0}\right) =-\left( 1-z_{t}^{2}\right) u_{1t}\left( \theta _{0}\right) . \end{aligned}$$

Consequently

$$\begin{aligned} E\left( s_{1t}|I_{t-1}\right)= & {} -E\left( \left( 1-z_{t}^{2}\right) u_{1t}\left( \theta _{0}\right) |I_{t-1}\right) =-E\left( \left( 1-z_{t}^{2}\right) \right) u_{1t}\left( \theta _{0}\right) \nonumber \\= & {} 0. \end{aligned}$$

(7)

Since $\{ s_{1t},I_{t}\}$ is an adapted stochastic sequence the result in (7) implies that $\{s_{1t},I_{t}\}$ is a martingale difference sequence according to Definition 3.75 in White (1984). Further, notice that

$$\begin{aligned} V_{1T}^{2}\left( \theta _{0}\right) =\sum _{t=1}^{T}E\left( s_{1t}^{2}\left( \theta _{0}\right) |I_{t-1}\right) =\sum _{t=1}^{T}E\left( \left( 1-z_{t}^{2}\right) ^{2}\right) u_{1t}^{2}\left( \theta _{0}\right) =\zeta \sum _{t=1}^{T}u_{1t}^{2}\left( \theta _{0}\right) . \end{aligned}$$

Hence,

$$\begin{aligned} E(V_{1T}^{2}\left( \theta _{0}\right) )=\zeta \sum _{t=1}^{T}E\left( u_{1t}^{2}\left( \theta _{0}\right) \right) =T\zeta {\overline{m}}_{11}. \end{aligned}$$

Furthermore, according to Lemma A we have that

$$\begin{aligned} \frac{1}{T}\sum _{t=1}^{T}u_{1t}^{2}\left( \theta _{0}\right) \overset{p}{ \rightarrow }{\overline{m}}_{11}, \end{aligned}$$

implying that

$$\begin{aligned} \frac{1}{T}V_{1T}^{2}\left( \theta _{0}\right) \overset{p}{\rightarrow } \zeta {\overline{m}}_{11}. \end{aligned}$$

From this we see that

$$\begin{aligned} \left( V_{1T}^{2}\left( \theta _{0}\right) \right) \left( E(V_{1T}^{2}\left( \theta _{0}\right) )\right) ^{-1}\overset{p}{\rightarrow }1. \end{aligned}$$

(8)

Importantly, the result given by equation (8) corresponds to Condition (1), p. 60 in Brown (1971).⁷

Finally, we need to prove that the Lindeberg type condition, which is Condition (2) in Brown (1971). In particular, we need to show that

$$\begin{aligned} \left( E(V_{1T}^{2}\left( \theta _{0}\right) )\right) ^{-1}\sum _{t=1}^{T}E\left( s_{1t}^{2}\left( \theta _{0}\right) 1\left\{ \left| s_{1t}\left( \theta _{0}\right) \right| >\epsilon \sqrt{ E(V_{1T}^{2}\left( \theta _{0}\right) )}\right\} \right) \overset{p}{ \rightarrow }0, \end{aligned}$$

for all $\epsilon >0.$ By inserting the expression for $s_{1t}^{2}$ and $E(V_{1T}^{2}\left( \theta _{0}\right) )$ we get

$$\begin{aligned}&\underset{T\rightarrow \infty }{\lim }\frac{1}{T\zeta {\overline{m}}_{11}} \sum _{t=1}^{T}E\left( s_{1t}^{2}\left( \theta _{0}\right) 1\left\{ \left| s_{1t}\left( \theta _{0}\right) \right|>\epsilon \sqrt{ T\zeta {\overline{m}}_{11}}\right\} \right) =\underset{T\rightarrow \infty }{\lim }\frac{1}{\zeta {\overline{m}}_{1}}\\&\quad E\left( \left( \left( 1-z_{t}^{2}\right) ^{2}u_{1t}^{2}\left( \theta _{0}\right) \right) 1\left\{ \left| \left( 1-z_{t}^{2}\right) ^{2}u_{1t}^{2}\left( \theta _{0}\right) \right| >\sqrt{T\zeta {\overline{m}}_{11}}\right\} \right) \rightarrow 0, \end{aligned}$$

for all $\zeta {\overline{m}}_{11}$ because, from Lemma A and A1, $u_{1t}^{2}\left( \theta _{0}\right)$ and $z_{t}^{2}$ have finite moments and are stationary and ergodic. Consequently, the Lindeberg condition holds.

According to Theorem 2, p. 60, in Brown (1971) we can therefore conclude that

$$\begin{aligned} \frac{1}{\sqrt{T\zeta {\overline{m}}_{11}}}\sum _{t=1}^{T}s_{1t}\left( \theta _{0}\right) \overset{d}{\rightarrow }N(0,1), \end{aligned}$$

which completes the proof. $\square$

Along the same lines

$$\begin{aligned} \frac{1}{T}\sum _{t=1}^{T}E\left( s_{2t}^{2}\mid I_{t-1}\right)= & {} \frac{1}{T} \sum _{t=1}^{T}\frac{\zeta }{4}\frac{1}{\left( w_{0}+\alpha _{0}\left( \frac{ y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }\overset{p}{\longrightarrow }\frac{\zeta }{4w_{0}^{2}}>0, \\ \frac{1}{T}\sum _{t=1}^{T}E\left( s_{3t}^{2}\mid I_{t-1}\right)= & {} \frac{1}{T} \sum _{t=1}^{T}\frac{\zeta }{4}\frac{\left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}}{\left( w_{0}+\alpha _{0}\left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }\overset{p}{\longrightarrow }\frac{\zeta }{4\alpha _{0}^{2}}>0. \end{aligned}$$

and

$$\begin{aligned}&\frac{1}{T}\sum _{t=1}^{T}E\left( s_{2t}^{2}1\left\{ \left| s_{2t}\right|>\sqrt{T}\delta \right\} \right) \\&\quad \le E\left( \left( \frac{ \left( 1-z_{t}^{2}\right) ^{2}}{4w_{0}^{2}}\right) 1\left\{ \left| \frac{ \left( 1-z_{t}^{2}\right) }{2w_{0}}\right|>\sqrt{T}\delta \right\} \right) \rightarrow 0, \\&\frac{1}{T}\sum _{t=1}^{T}E\left( s_{3t}^{2}1\left\{ \left| s_{3t}\right|>\sqrt{T}\delta \right\} \right) \\&\quad \le E\left( \left( \frac{ \left( 1-z_{t}^{2}\right) ^{2}}{4\alpha _{0}^{2}}\right) 1\left\{ \left| \frac{\left( 1-z_{t}^{2}\right) }{2\alpha _{0}}\right| >\sqrt{T}\delta \right\} \right) \rightarrow 0, \end{aligned}$$

for some $\delta >0$ and as T tends to $\infty$ . $\square$

Proof of Proposition 1

In order to fully characterize the asymptotic distribution we need to determine the off-diagonal elements of the variance covariance matrix of the score vectors given by $\Lambda .$ In particular, because $u_{1t}\left( \theta _{0}\right) ,u_{2t}\left( \theta _{0}\right)$ and $u_{3t}\left( \theta _{0}\right)$ are all stationary and ergodic with finite first moments (from Lemma A) it follows straightforwardly that

$$\begin{aligned} \frac{1}{T}\sum _{t=1}^{T}s_{1t}\left( \theta _{0}\right) s_{2t}\left( \theta _{0}\right)= & {} \frac{1}{T}\sum _{t=1}^{T}\left( 1-z_{t}^{2}\right) ^{2}u_{1t}\left( \theta _{0}\right) u_{2t}\left( \theta _{0}\right) \overset{ p}{\rightarrow }\frac{1}{2}\zeta {\overline{m}}_{12}, \\ \frac{1}{T}\sum _{t=1}^{T}s_{1t}\left( \theta _{0}\right) s_{3t}\left( \theta _{0}\right)= & {} \frac{1}{T}\sum _{t=1}^{T}\frac{1}{2}\left( 1-z_{t}^{2}\right) ^{2}u_{1t}\left( \theta _{0}\right) u_{3t}\left( \theta _{0}\right) \overset{ p}{\rightarrow }\frac{1}{2}\zeta {\overline{m}}_{13}, \\ \frac{1}{T}\sum _{t=1}^{T}s_{2t}\left( \theta _{0}\right) s_{3t}\left( \theta _{0}\right)= & {} \frac{1}{T}\sum _{t=1}^{T}\frac{1}{4}\left( 1-z_{t}^{2}\right) ^{2}u_{2t}\left( \theta _{0}\right) u_{3t}\left( \theta _{0}\right) \overset{ p}{\rightarrow }\frac{1}{4}\zeta {\overline{m}}_{23}. \end{aligned}$$

Since all the elements in the score vector are asymptotically normal (see Lemma B), the result follows directly from application of the Cramer-Wold device, see for example Proposition 5.1 in White (1984), which completes the proof. $\square$

Proposition 2

Let $u_{jt}\left( \theta _{0}\right)$ be defined as in Theorem 1. Under Assumptions A and B, the observed information evaluated at $\theta =\theta _{0}$ converges in probability, i.e.,

$$\begin{aligned} -\frac{1}{T}\frac{\partial ^{2}}{\partial \theta \partial \theta ^{\prime }} L_{T}\left( \theta _{0}\right) \overset{p}{\rightarrow }\Omega , \end{aligned}$$

where

$$\begin{aligned} \Omega =\left[ \begin{array}{ccc} 2{\overline{m}}_{11} &{} {\overline{m}}_{12} &{} {\overline{m}}_{13} \\ {\overline{m}}_{12} &{} \frac{1}{2}{\overline{m}}_{22} &{} \frac{1}{2}{\overline{m}} _{23} \\ {\overline{m}}_{13} &{} \frac{1}{2}{\overline{m}}_{23} &{} \frac{1}{2}{\overline{m}} _{33} \end{array} \right] >0, \end{aligned}$$

and ${\overline{m}}_{ij}=E\left( u_{it}\left( \theta _{0}\right) u_{jt}\left( \theta _{0}\right) \right)$ for $i=1,2,3$ and $j=1,2,3$.

Proof of Proposition 2

Recall from Result 2 (see Supplementary Appendix) that

$$\begin{aligned}&-\frac{1}{T}\frac{\partial ^{2}}{\partial \gamma ^{2}}L_{T}\left( \theta _{0}\right) =2\frac{1}{T}\sum _{t=1}^{T}z_{t}^{2}u_{1t}^{2}\left( \theta _{0}\right) -\frac{2}{T}\sum _{t=1}^{T}\left( 1-z_{t}^{2}\right) \nonumber \\&\quad \left( \ln \left| y_{t-2}\right| \right) ^{2}w_{0}^{2}\left( \frac{1}{w_{0}}- \frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) ^{2} \\&\qquad +\frac{2}{T}\sum _{t=1}^{T}\left( 1-z_{t}^{2}\right) w_{0}\left( \frac{1}{ w_{0}}-\frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) \left( \ln \left| y_{t-2}\right| \right) ^{2}. \end{aligned}$$

Since $z_{t}^{2}$ and $u_{1t}^{2}\left( \theta _{0}\right)$ are independent, the first term on the right hand side converges to $2{\overline{m}}_{11}$ by Lemma A. Furthermore, since $\left( \ln \left| y_{t-2}\right| \right) ^{2}w_{0}^{2}\left( \frac{1}{w_{0}}-\frac{1}{ \sigma _{t}^{2}\left( \theta _{0}\right) }\right) ^{2}$ and $w_{0}\left( \frac{1}{w_{0}}-\frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) \left( \ln \left| y_{t-2}\right| \right) ^{2}$ have bounded moments, they are ergodic and stationary and since $E\left( 1-z_{t}^{2}\right) =0,$ it follows from the ergodic theorem that the last term on the right hand side converges in probability to zero. Therefore, the result follows. Using identical arguments we find

$$\begin{aligned} -\frac{1}{T}\frac{\partial ^{2}}{\partial w^{2}}L_{T}\left( \theta _{0}\right)= & {} -\frac{1}{2}\frac{1}{T}\sum _{t=1}^{T}\left( 1-2z_{t}^{2}\right) u_{2t}^{2}\left( \theta _{0}\right) \overset{p}{\rightarrow }\frac{1}{2} {\overline{m}}_{22},-\frac{1}{T}\frac{\partial ^{2}}{\partial \alpha ^{2}} L_{T}\left( \theta _{0}\right) \\= & {} -\frac{1}{2}\frac{1}{T}\sum _{t=1}^{T}\left( 1-2z_{t}^{2}\right) u_{3t}^{2}\left( \theta _{0}\right) \overset{p}{\rightarrow }\frac{1}{2} {\overline{m}}_{33},-\frac{1}{T}\frac{\partial ^{2}}{\partial \gamma \partial w }L_{T}\left( \theta _{0}\right) \\= & {} \frac{1}{T}\sum _{t=1}^{T}z_{t}^{2}u_{1t}\left( \theta _{0}\right) u_{2t}\left( \theta _{0}\right) \\&+\,\frac{1}{T}\sum _{t=1}^{T}\left( 1-z_{t}^{2}\right) \left( \left( \ln \left| y_{t-2}\right| \right) \left( \frac{w_{0}}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) \left( \frac{1}{w_{0}}-\frac{1}{\sigma _{t}^{2}\left( \theta _{0}\right) }\right) \right) \overset{p}{\rightarrow }{\overline{m}}_{12}, \\&-\,\frac{1}{T}\frac{\partial ^{2}}{\partial \gamma \partial \alpha } L_{T}\left( \theta _{0}\right) =\frac{1}{T}\sum _{t=1}^{T}z_{t}^{2}u_{1t} \left( \theta _{0}\right) u_{3t}\left( \theta _{0}\right) \\&-\,\frac{1}{T} \sum _{t=1}^{T}\left( 1-z_{t}^{2}\right) w_{0}\left( \ln \left| y_{t-2}\right| \right) \left( \frac{w_{0}}{\alpha _{0}}\right) \left( \frac{1}{w_{0}}-\frac{1}{_{\sigma _{t}^{2}\left( \theta _{0}\right) }} \right) ^{2} \\&\overset{p}{\rightarrow }{\overline{m}}_{13},-\frac{1}{T}\frac{\partial ^{2}L_{T}\left( \theta _{0}\right) }{\partial w\partial \alpha }=-\frac{1}{2} \frac{1}{T}\sum _{t=1}^{T}\left( 1-2z_{t}^{2}\right) u_{2t}\left( \theta _{0}\right) u_{3t}\left( \theta _{0}\right) \overset{p}{\rightarrow }\frac{1 }{2}{\overline{m}}_{23}. \end{aligned}$$

We proceed now to show that $\Lambda$ is positive definite. $\Lambda$ will be positive definite if for any non-zero column vector z with entries a, b and c, we show that $z^{T}\Lambda z>0$. In our case

$$\begin{aligned} z^{T}\Lambda z= & {} \zeta \left( \begin{array}{c} aE\left( u_{1t}^{2}\right) +\frac{b}{2}E\left( u_{1t}u_{2t}\right) +\frac{c}{ 2}E\left( u_{1t}u_{3t}\right) \\ \frac{a}{2}E\left( u_{1t}u_{2t}\right) +\frac{b}{4}E\left( u_{2t}^{2}\right) +\frac{c}{4}E\left( u_{2t}u_{3t}\right) \\ \frac{a}{2}E\left( u_{1t}u_{3t}\right) +\frac{b}{4}E\left( u_{2t}u_{3t}\right) +\frac{c}{4}E\left( u_{3t}^{2}\right) \end{array} \right) ^{T}\left( \begin{array}{c} a \\ b \\ c \end{array} \right) \\= & {} \zeta \Bigg [ a^{2}E\left( u_{1t}^{2}\right) +\frac{ab}{2}E\left( u_{1t}u_{2t}\right) +\frac{ac}{2}E\left( u_{1t}u_{3t}\right) +\frac{ab}{2} E\left( u_{1t}u_{2t}\right) \\&+\,\frac{b^{2}}{4}E\left( u_{2t}^{2}\right) +\frac{ bc}{4}E\left( u_{2t}u_{3t}\right) \\&+\,\frac{ac}{2}E\left( u_{1t}u_{3t}\right) +\frac{bc}{4}E\left( u_{2t}u_{3t}\right) +\frac{c^{2}}{4}E\left( u_{3t}^{2}\right) \Bigg ] \\= & {} \zeta \Bigg [ a^{2}E\left( u_{1t}^{2}\right) +\frac{b^{2}}{4}E\left( u_{2t}^{2}\right) +\frac{c^{2}}{4}E\left( u_{3t}^{2}\right) \\&+\, abE\left( u_{1t}u_{2t}\right) +acE\left( u_{1t}u_{3t}\right) +\frac{bc}{2}E\left( u_{2t}u_{3t}\right) \Bigg ] \end{aligned}$$

where we have written $u_{it}\left( \theta _{0}\right) =u_{it}$ for simplicity reasons. Since $\zeta ,$ by Assumption A1, is always positive and larger than zero, and from Lemma A we have that

$$\begin{aligned} u_{3t}=\left( \frac{w_{0}}{\alpha _{0}}\right) \left( \frac{1}{w_{0}} -u_{2t}\right) , \end{aligned}$$

then, we need to show if the following term is strictly positive

$$\begin{aligned}&a^{2}E\left( u_{1t}^{2}\right) +\frac{b^{2}}{4}E\left( u_{2t}^{2}\right) +\left( \frac{cw_{0}}{2\alpha _{0}}\right) ^{2}E\left( \left( \frac{1}{w_{0}} -u_{2t}\right) ^{2}\right) \\&\qquad +abE\left( u_{1t}u_{2t}\right) +\frac{acw_{0}}{ \alpha _{0}}E\left( \frac{u_{1t}}{w_{0}}-u_{1t}u_{2t}\right) \\&\qquad +\frac{bcw_{0}}{2\alpha _{0}}E\left( \frac{u_{2t}}{w_{0}}-u_{2t}^{2}\right) =a^{2}E\left( u_{1t}^{2}\right) +\frac{1}{4}\left( b-\frac{cw_{0}}{\alpha _{0}}\right) ^{2}\\&E\left( u_{2t}^{2}\right) +\left( \frac{c}{2\alpha _{0}} \right) ^{2}+\frac{ac}{\alpha _{0}}E\left( u_{1t}\right) \\&\qquad +\frac{c}{2\alpha _{0}}\left( b-\frac{cw_{0}}{\alpha _{0}}\right) E\left( u_{2t}\right) +a\left( b-\frac{cw_{0}}{\alpha _{0}}\right) E\left( u_{1t}u_{2t}\right) \\&\quad =a^{2}E\left( u_{1t}^{2}\right) +\frac{1}{4}\left( b-\frac{cw_{0}}{\alpha _{0}}\right) ^{2}E\left( u_{2t}^{2}\right) +a\left( b-\frac{cw_{0}}{\alpha _{0}}\right) E\left( u_{1t}u_{2t}\right) \\&\qquad +\frac{c}{\alpha _{0}}\left[ aE\left( u_{1t}\right) +\frac{1}{2}\left( b- \frac{cw_{0}}{\alpha _{0}}\right) E\left( u_{2t}\right) \right] +\left( \frac{c}{2\alpha _{0}}\right) ^{2} \\&\quad =E\left[ a^{2}u_{1t}^{2}+\frac{1}{4}\left( b-\frac{cw_{0}}{\alpha _{0}} \right) ^{2}u_{2t}^{2}+a\left( b-\frac{cw_{0}}{\alpha _{0}}\right) u_{1t}u_{2t}\right. \\&\qquad \left. +\frac{c}{\alpha _{0}}\left[ au_{1t}+\frac{1}{2}\left( b-\frac{ cw_{0}}{\alpha _{0}}\right) u_{2t}\right] +\left( \frac{c}{2\alpha _{0}} \right) ^{2}\right] \\&\quad =E\left[ \left( au_{1t} +\frac{1}{2}\left( b-\frac{cw_{0}}{\alpha _{0}} \right) u_{2t}\right) ^{2}\right. \\&\left. \qquad +\frac{c}{\alpha _{0}}\left[ au_{1t} +\frac{1}{2} \left( b-\frac{cw_{0}}{\alpha _{0}}\right) u_{2t}\right] +\left( \frac{c}{ 2\alpha _{0}}\right) ^{2}\right] \\&\quad =E\left( au_{1t}+\frac{1}{2}\left( b-\frac{cw_{0}}{2\alpha _{0}}\right) u_{2t}+\frac{c}{2\alpha _{0}}\right) ^{2}>0. \end{aligned}$$

Finally notice that since $\Omega =2\Lambda \zeta ^{-1}=\Lambda ,$ then $\Omega >0.$ This completes the proof of Proposition 2. $\square$

Proposition 3

Define the lower and upper values for each parameter in $\theta _{0}$ as $\gamma _{L}<\gamma _{0}<\gamma _{U},w_{L}<w_{0}<w_{U},$ and $\alpha _{L}<\alpha _{0}<\alpha _{U},$ respectively and the neighborhood $N\left( \theta _{0}\right)$ around $\theta _{0}$ as

$$\begin{aligned} N\left( \theta _{0}\right) =\left\{ \theta \backslash \gamma _{L}\le \gamma \le \gamma _{U},w_{L}\le w\le w_{U},\text { and }\alpha _{L}\le \alpha \le \alpha _{U}\right\} . \end{aligned}$$

Under Assumptions A and B, there exists a neighborhood $N\left( \theta _{0}\right)$ for which for $i,j,k=1,2,3$

$$\begin{aligned} \sup _{\theta \in N\left( \theta _{0}\right) }\left| \frac{1}{T}\frac{ \partial ^{3}}{\partial \theta _{i}\partial \theta _{j}\partial \theta _{k}} L_{T}\left( \theta \right) \right| \le \frac{1}{T} \sum _{t=1}^{T}w_{ijkt}, \end{aligned}$$

where $w_{ijkt}$ is stationary. Furthermore $\frac{1}{T} \sum _{t=1}^{T}w_{ijkt}\overset{a.s.}{\longrightarrow }E\left( w_{ijkt}\right) <\infty$ for $\forall ijk.$

Proof of Proposition 3

Let us start from the components of $\left| \frac{1}{T}\frac{\partial ^{3}}{\partial \gamma ^{3}}L_{T}\left( \theta \right) \right|$ defined in Result 3 (see Supplementary Appendix). Part I (which is also defined in Result 3) can be written as

$$\begin{aligned}&\left| \frac{4}{T}\sum _{t=1}^{T}\frac{\alpha \left( \frac{y_{t-1}^{*} }{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\left( \ln \left( \left| y_{t-2}\right| \right) \right) ^{3}}{\left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }\left( 1-\frac{\left( y_{t}^{*}\right) ^{2}}{ y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }\right) \right| \end{aligned}$$

$$\begin{aligned}&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\frac{\left( \left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }} \right) ^{2}\right) -w\right) \left| \ln \left( \left| y_{t-2}\right| \right) \right| ^{3}}{\left( w+\alpha \left( \frac{ y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) } \right. \\&\qquad \times \, \left. \left( \frac{\left( y_{t}^{*}\right) ^{2}}{y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }+1\right) \right| \\&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\left( \frac{\left( y_{t}^{*}\right) ^{2}}{y_{t-1}^{2\gamma } \left( w+\alpha \left( \frac{y_{t-1}^{*} }{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| \\&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\left( \left( \frac{ y_{t-1}^{2\gamma _{0}}\left( w_{0}+\alpha _{0}\left( \frac{y_{t-1}^{*}}{ \left| y_{t-2}\right| ^{\gamma _{0}}}\right) ^{2}\right) }{ y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }\right) z_{t}^{2}+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| \\&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\left( \left( \frac{w_{0}}{w} y_{t-1}^{2\left( \gamma _{0}-\gamma \right) }+\frac{\alpha _{0}}{\alpha } \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| }\right) ^{2\left( \gamma _{0}-\gamma \right) }\right) z_{t}^{2}+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| \\&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\left( \left\{ \frac{w_{U}}{w_{L}} \Lambda _{t-1}+\frac{\alpha _{U}}{\alpha _{L}}\Lambda _{t-2}\right\} z_{t}^{2}+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| \\&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\left( \left\{ \frac{w_{U}}{w_{L}} \Lambda _{t-1}+\frac{\alpha _{U}}{\alpha _{L}}\Lambda _{t-2}\right\} z_{t}^{2}+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| \\&\quad \le \left| \frac{4}{T}\sum _{t=1}^{T}\left( \left\{ \frac{w_{U}}{w_{L}} \Lambda _{t-1}+\frac{\alpha _{U}}{\alpha _{L}}\Lambda _{t-2}\right\} z_{t}^{2}+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| , \end{aligned}$$

where we can define the lower bound for all t,  $y_{L}\le \left| y_{t-1}\right| ,$ $y_{L}\le \left| y_{t-2}\right| ,$ $\Lambda _{t-1}=\max \left\{ y_{L}^{2\left| \gamma _{U}-\gamma _{L}\right| },y_{t-1}^{2\left| \gamma _{U}-\gamma _{L}\right| }\right\}$, $\Lambda _{t-2}=\max \left\{ 1,\left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| }\right) ^{2\left| \gamma _{U}-\gamma _{L}\right| }\right\}$ and the result follows by setting $2\left| \gamma _{U}-\gamma _{L}\right| =\varphi$, Assumptions A and B and the law of large numbers (see Jensen and Rahbek (2004a), Lemma 5). Part II requires also assumption A3 since

$$\begin{aligned}&\left| \frac{8}{T}\sum _{t=1}^{T}\frac{\alpha ^{3}\left( \frac{ y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{6}\left( \ln \left( \left| y_{t-2}\right| \right) \right) ^{3}}{\left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) ^{3}}\left( 1-\frac{3\left( y_{t}^{*}\right) ^{2}}{y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*} }{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }\right) \right| \\&\quad \le \left| \frac{8}{T}\sum _{t=1}^{T}\left( \frac{3\left( y_{t}^{*}\right) ^{2}}{y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*} }{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }-1\right) \left| \ln \left( \left| y_{t-2}\right| \right) \right| ^{3}\right| \\&\quad \le \left| \frac{8}{T}\sum _{t=1}^{T}\left( 3\left\{ \frac{w_{U}}{w_{L} }\Lambda _{t-1}+\frac{\alpha _{U}}{\alpha _{L}}\Lambda _{t-2}\right\} z_{t}^{2}+1\right) \left| \ln \left( \left| y_{t-1}\right| \right) \right| ^{3}\right| . \end{aligned}$$

Parts III, IV, V and VI follow the same argument.

Along the same lines for $\left| \frac{1}{T}\frac{\partial ^{3}}{ \partial \alpha ^{3}}L_{T}\left( \theta \right) \right|$

$$\begin{aligned} \left| \frac{1}{T}\frac{\partial ^{3}}{\partial \alpha ^{3}}L_{T}\left( \theta \right) \right|= & {} \left| \frac{1}{T}\sum _{t=1}^{T}\left( 3 \frac{\left( y_{t}^{*}\right) ^{2}}{y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*}}{\left| y_{t-2}\right| ^{\gamma }} \right) ^{2}\right) }-1\right) \frac{\left( \frac{y_{t-1}^{*}}{ y_{t-2}^{\gamma }}\right) ^{6}}{\left( w+\alpha \left( \frac{y_{t-1}^{*} }{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) ^{3}} \right| \\\le & {} \left| \frac{1}{T}\sum _{t=1}^{T}\left( 3\frac{\left( y_{t}^{*}\right) ^{2}}{y_{t-1}^{2\gamma }\left( w+\alpha \left( \frac{y_{t-1}^{*} }{\left| y_{t-2}\right| ^{\gamma }}\right) ^{2}\right) }-1\right) \right| \frac{1}{\alpha _{L}^{3}} \\\le & {} \frac{1}{T}\sum _{t=1}^{T}\left( 3\left\{ \frac{w_{U}}{w_{L}}\Lambda _{t-1}+\frac{\alpha _{U}}{\alpha _{L}}\Lambda _{t-2}\right\} z_{t}^{2}+1\right) \frac{1}{\alpha _{L}^{3}}. \end{aligned}$$

The rest of the cases follow directly using the same argument. This completes the proof of Proposition 3. $\square$

Proof of Theorem 1

Given the conditions provided by Propositions 1–3, Theorem 1 follows from Lumsdaine (1996, pp. 593–595, Theorem 3), the ergodic theorem and Lemma 1, p. 1206 in Jensen and Rahbek (2004b). $\square$