Mean targeting estimator for the integer-valued GARCH(1, 1) model

Abstract

The integer-valued GARCH model is commonly used in modeling time series of counts. Maximum likelihood estimation (MLE) is used to estimate unknown parameters, but numerical results for MLE are sensitive to the choice of initial values, which also occurs in estimating the GARCH model. To alleviate this numerical difficulty, we propose an alternative to MLE and name it as mean targeting estimation (MTE), which is an analogue to variance targeting estimation used in the GARCH model. Consistency and asymptotic normality for MTE are established. Comparisons with the standard MLE are provided and the merits of the mean targeting method are discussed. In particular, it is shown that MTE can be superior to MLE for estimating parameters or prediction when the model is well specified and misspecified. We conduct numerical studies to confirm our theoretical findings and illustrate the practical utility of our proposals.

Appendix

Define

$$\begin{aligned} L_n(\phi )=\frac{1}{n}\sum _{t=1}^nl_t(\mu _0,\phi ),~~~l_t(\mu ,\phi )=l_t(\theta )=X_t\ln \lambda _t-\lambda _t. \end{aligned}$$

For $t\ge 1$, define

$$\begin{aligned} \tilde{l}_t=\tilde{l}_t(\theta )=X_t\ln \tilde{\lambda }_t-\tilde{\lambda }_t, \end{aligned}$$

where $\tilde{\lambda }_t=\tilde{\lambda }_t(\theta )=\mu (1-\alpha -\beta )+\alpha X_{t-1}+\beta \tilde{\lambda }_{t-1}$ with fixed initial values $X_0$ and $\tilde{\lambda }_0$. Then we know that $\lambda _{t,n}(\phi )=\tilde{\lambda }_t(\hat{\mu }_n,\phi )$.

In this appendix, the letters K and $\rho$ denote generic constants, whose values can vary but always satisfy $K>0$ and $0<\rho <1$. Note that from Lemma 1 in Ferland et al. (2006) we know that $E(X_t^k)$ is finite for all k, so there is no trouble about existences of moments as in the classic GARCH model. We use arguments similar to Francq and Zakoïan (2004) and Francq et al. (2011) to prove Theorem 1.

Proof of consistency in Theorem 1

The almost sure convergence of $\hat{\mu }_n$ to $\mu _0$ is a direct consequence of the ergodic theorem. To show the strong consistency of $\hat{\phi }_n$, it suffices to establish the following results:

• (i) $\displaystyle \lim _{n\rightarrow \infty }\sup _{\phi \in \Phi }|L_n(\phi )-L_n^*(\phi )|=0$ a.s.;

• (ii) If $\lambda _t(\mu _0,\phi )=\lambda _t(\mu _0,\phi _0)$ a.s., then $\phi =\phi _0$;

• (iii) If $\phi \ne \phi _0$, then $El_t(\mu _0,\phi )<El_t(\mu _0,\phi _0)$;

• (iv) Any $\phi \ne \phi _0$ has a neighbourhood $V(\phi )$ such that $\displaystyle \limsup _{n\rightarrow \infty } \sup _{\phi ^*\in V(\phi )}L_n^*(\phi ^*)<El_1(\mu _0,\phi _0)$ a.s.

We first show (i). Note that

$$\begin{aligned} \lambda _{t,n}(\phi )-\lambda _t(\mu _0,\phi )&=(1-\alpha -\beta )(\hat{\mu }_n-\mu _0)+\beta (\lambda _{t-1,n}(\phi )-\lambda _{t-1}(\mu _0,\phi ))\\&=(1-\alpha -\beta )\frac{1-\beta ^t}{1-\beta }(\hat{\mu }_n-\mu _0)+\beta ^t(\tilde{\lambda }_0-\lambda _0), \end{aligned}$$

then we have $\displaystyle \sup _{\phi \in \Phi }|\lambda _{t,n}(\phi )-\lambda _t(\mu _0,\phi )|\le K\rho ^t+o(1)$ a.s. since $\hat{\mu }_n$ converges to $\mu _0$ a.s. Note that K is a measurable function of $\{X_k, k<0\}$. For the almost sure consistency, the trajectory is fixed in a set of probability one and n tends to infinity. Thus, K can be considered as a constant, that is, K is almost surely invariant with n, see Francq et al. (2011) for similar arguments. Since $\ln x\le x-1$ for all $x>0$, we have

$$\begin{aligned} \sup _{\phi \in \Phi }|L_n(\phi )-L_n^*(\phi )|&\le \frac{1}{n}\sum _{t=1}^n\sup _{\phi \in \Phi } \left\{ X_t\left| \ln \left( 1+\frac{\lambda _t(\mu _0,\phi )-\lambda _{t,n}(\phi )}{\lambda _{t,n}(\phi )}\right) \right| +|\lambda _{t,n}(\phi )-\lambda _t(\mu _0,\phi )|\right\} \\&\le \frac{K}{\hat{\mu }_n(1-\alpha _U-\beta _U)}\frac{1}{n}\sum _{t=1}^n\rho ^tX_t+K\frac{1}{n}\sum _{t=1}^n\rho ^t+o(1), \end{aligned}$$

then (i) holds using arguments similar to Francq and Zakoïan (2004, p. 616).

Note that using the parametrization

$$\begin{aligned} \lambda _t(\theta )=\mu (1-\alpha -\beta )\frac{1}{1-\beta }+\alpha \sum _{k=1}^\infty \beta ^{k-1}X_{t-k} \end{aligned}$$

and the technique in the proof of Theorem 3 of Davis and Liu (2016), it can be shown that if $\lambda _t(\theta )=\lambda _t(\theta _0)$ a.s., then $\theta =\theta _0$, which proves (ii).

Let $a^*$ be a constant. For any $x>0$, $f(x)\equiv a^*\ln x-x$ reaches its maximum at $x=a^*$. Note that $El_t(\theta )=E(X_t\ln \lambda _t(\theta )-\lambda _t(\theta ))=E(\lambda _t(\theta _0)\ln \lambda _t(\theta )-\lambda _t(\theta ))$. Thus $El_t(\theta )$ reaches its maximum at $\lambda _t(\theta )=\lambda _t(\theta _0)$, which is equivalent to $\theta =\theta _0$ in terms of (ii).

To show (iv), let $V_k(\phi )$ be the open ball with center $\phi$ and radius 1 / k. Using (i), the ergodicity of the process, the monotone convergence theorem and (iii), we obtain

$$\begin{aligned}&\limsup _{n\rightarrow \infty }\sup _{\phi ^*\in V_k(\phi )\cap \Phi }L_n^*(\phi ^*)\\&\quad \le \limsup _{n\rightarrow \infty }\sup _{\phi ^*\in V_k(\phi )\cap \Phi }L_n(\phi ^*)+\limsup _{n\rightarrow \infty }\sup _{\phi \in \Phi }|L_n(\phi )-L_n^*(\phi )|\\&\quad \le \limsup _{n\rightarrow \infty }\frac{1}{n}\sum _{t=1}^n\sup _{\phi ^*\in V_k(\phi )\cap \Phi }l_t(\mu _0,\phi ^*)\\&\quad =E\left( \sup _{\phi ^*\in V_k(\phi )\cap \Phi }l_1(\mu _0,\phi ^*)\right) \\&\quad <El_1(\mu _0,\phi _0) \end{aligned}$$

for k large enough, when $\phi \ne \phi _0$.$\square$

Proof of asymptotical normality in Theorem 1

Write $\phi =(\phi _1,\phi _2)^\top ,\theta =(\theta _1,\theta _2,\theta _3)^\top$. Noting that $l_{t,n}(\phi )=\tilde{l}_t(\hat{\mu }_n,\phi )$, we have

$$\begin{aligned} (0,0)^{\top }&=\frac{1}{\sqrt{n}}\sum _{t=1}^n\frac{\partial }{\partial \phi }l_{t,n}(\hat{\phi }_n) =\frac{1}{\sqrt{n}}\sum _{t=1}^n\frac{\partial }{\partial \phi }\tilde{l}_t(\hat{\theta }_n)\\&=\frac{1}{\sqrt{n}}\sum _{t=1}^n\frac{\partial }{\partial \phi }\tilde{l}_t(\hat{\theta }_0)+J_n\sqrt{n}(\hat{\phi }_n-\phi _0)+K_n\sqrt{n}(\hat{\mu }_n-\mu _0), \end{aligned}$$

where

$$\begin{aligned} J_n= & {} \left( \frac{1}{n}\sum _{t=1}^n\frac{\partial ^2}{\partial \phi _i\partial \phi _j}\tilde{l}_t(\hat{\theta }_{i+1,j+1}^*)\right) _{2\times 2},\\ K_n= & {} \left( \frac{1}{n}\sum _{t=1}^n\frac{\partial ^2}{\partial \mu \partial \phi _1}\tilde{l}_t(\hat{\theta }_{12}^*), \frac{1}{n}\sum _{t=1}^n\frac{\partial ^2}{\partial \mu \partial \phi _2}\tilde{l}_t(\hat{\theta }_{13}^*)\right) ^\top \end{aligned}$$

with $\Vert \hat{\theta }_{ij}^*-\theta _0\Vert \le \Vert \hat{\theta }-\theta _0\Vert ,i=1,2,j=1,2,3.$

We will show that

• (i) $\displaystyle E\sup _{\theta \in \Theta }\left\| \frac{\partial l_t(\theta )}{\partial \theta } \frac{\partial l_t(\theta )}{\partial \theta ^\top }\right\|<\infty , \quad E\sup _{\theta \in \Theta }\left\| \frac{\partial ^2l_t(\theta )}{\partial \theta \partial \theta ^\top }\right\|<\infty , \quad E\sup _{\theta \in \Theta }\left| \frac{\partial ^3l_t(\theta )}{\partial \theta _i\partial \theta _j\partial \theta _k}\right| <\infty ,i,j,k=1,2,3$;

• (ii) $\displaystyle A\equiv \mathrm{Var}\left( \frac{\partial l_t(\theta _0)}{\partial \theta }\right) =E\left( \frac{1}{\lambda _t(\theta _0)} \frac{\partial \lambda _t(\theta _0)}{\partial \theta }\frac{\partial \lambda _t(\theta _0)}{\partial \theta ^\top }\right)$ is nonsingular;

• (iii) There exists a neighbourhood $\mathcal {V}(\theta _0)$ of $\theta _0$ such that $\displaystyle \left\| \frac{1}{\sqrt{n}}\sum _{t=1}^n\left\{ \frac{\partial l_t(\theta _0)}{\partial \theta } -\frac{\partial \tilde{l}_t(\theta _0)}{\partial \theta }\right\} \right\| \rightarrow 0$ and $\displaystyle \sup _{\theta \in \mathcal {V}(\theta _0)}\left\| \frac{1}{n}\sum _{t=1}^n\left\{ \frac{\partial ^2l_t(\theta )}{\partial \theta \partial \theta ^\top } -\frac{\partial ^2\tilde{l}_t(\theta )}{\partial \theta \partial \theta ^\top }\right\} \right\| \rightarrow 0$ in probability when $n\rightarrow \infty$;

• (iv) $\displaystyle \frac{1}{n}\sum _{t=1}^n\frac{\partial ^2l_t(\theta _{ij}^*)}{\partial \theta _i\partial \theta _j}\rightarrow A(i,j)$ a.s.;

• (v) $\displaystyle \left( \begin{array}{c}\sqrt{n}(\hat{\mu }_n-\mu _0)\\ \displaystyle \frac{1}{\sqrt{n}} \sum _{t=1}^n\frac{\partial l_t(\theta _0)}{\partial \phi }\end{array}\right) {\mathop {\longrightarrow }\limits ^{d}}\mathcal {N}\left( 0,\left( \begin{array}{cc}b&{}0\\ 0&{}J\end{array}\right) \right)$.

First, we will prove (i). Note that

$$\begin{aligned} \frac{\partial l_t(\theta )}{\partial \theta }&=\left( \frac{X_t}{\lambda _t(\theta )}-1\right) \frac{\partial \lambda _t(\theta )}{\partial \theta }, \quad \frac{\partial ^2l_t(\theta )}{\partial \theta \partial \theta ^\top }=\left( \frac{X_t}{\lambda _t(\theta )}-1\right) \frac{\partial ^2\lambda _t(\theta )}{\partial \theta \partial \theta ^\top }\\&\quad -\,\frac{X_t}{\lambda _t^2(\theta )}\frac{\partial \lambda _t(\theta )}{\partial \theta }\frac{\partial \lambda _t(\theta )}{\partial \theta ^\top },\\ \frac{\partial ^3l_t(\theta )}{\partial \theta _i\partial \theta _j\partial \theta _k}&= \frac{-X_t}{\lambda _t^2(\theta )}\left( \frac{\partial ^2\lambda _t(\theta )}{\partial \theta _i\partial \theta _j} \frac{\partial \lambda _t(\theta )}{\partial \theta _k} +\frac{\partial ^2\lambda _t(\theta )}{\partial \theta _i\partial \theta _k}\frac{\partial \lambda _t(\theta )}{\partial \theta _j} +\frac{\partial ^2\lambda _t(\theta )}{\partial \theta _j\partial \theta _k}\frac{\partial \lambda _t(\theta )}{\partial \theta _i}\right) \\&\quad +\,\frac{2X_t}{\lambda _t^3(\theta )}\frac{\partial \lambda _t(\theta )}{\partial \theta _i}\frac{\partial \lambda _t(\theta )}{\partial \theta _j} \frac{\partial \lambda _t(\theta )}{\partial \theta _k} +\left( \frac{X_t}{\lambda _t(\theta )}-1\right) \frac{\partial ^3\lambda _t(\theta )}{\partial \theta _i\partial \theta _j\partial \theta _k}. \end{aligned}$$

It can be shown that

$$\begin{aligned} \frac{X_t}{\lambda _t^2(\theta )}\le \frac{X_t}{\omega _L^2},~~~\frac{X_t}{\lambda _t^3(\theta )}\le \frac{X_t}{\omega _L^3}, \quad \left| \frac{X_t}{\lambda _t(\theta )}-1\right| \le \frac{X_t}{\omega _L}+1, \end{aligned}$$

where $\omega _L=\mu _L(1-\alpha _U-\beta _U)>0$. The most involved derivatives are those with respect to $\theta _3=\beta$. Using arguments similar to the proof of Lemma 3.4 in Fokianos et al. (2009) and the fact that $E(X_t^k)$ is finite for all k we obtain (i).

If A is singular, there exists $x=(x_1,x_2,x_3)^\top \ne 0$ such that $x^\top \dfrac{\partial \lambda _t(\theta _0)}{\partial \theta }=0$. Since

$$\begin{aligned} \frac{\partial \lambda _t(\theta )}{\partial \theta }=\frac{\partial \mu (1-\alpha -\beta )}{\partial \theta } +X_{t-1}\frac{\partial \alpha }{\partial \theta }+\lambda _{t-1}(\theta _0)\frac{\partial \beta }{\partial \theta } +\beta \frac{\partial \lambda _{t-1}(\theta )}{\partial \theta }, \end{aligned}$$

the strict stationary of $\lambda _t(\theta _0)$ implies $x_1(1-\alpha _0-\beta _0)-\mu _0(x_2+x_3)+x_2X_{t-1}+x_3\lambda _{t-1}(\theta _0)=0$. This means that $x_2X_{t-1}$ is a function of $\{X_{t-i},i>1\}$, which is impossible unless $x_2=0$. We have $x_3=0$ because $\lambda _{t-1}(\theta )$ is not almost surely constant, then we know $x_1=0$. From this contradiction we know that (ii) holds.

Note that $\lambda _t(\theta )-\tilde{\lambda }_t(\theta )=\beta ^t(\lambda _0-\tilde{\lambda }_0)$, choosing $\mathcal {V}(\theta _0)$ such that $\phi \in \Phi$ for all $\theta \in \mathcal {V}(\theta _0)$, then we have

$$\begin{aligned}&\sup _{\theta \in \mathcal {V}(\theta _0)}|\lambda _t(\theta )-\tilde{\lambda }_t(\theta )|\le K\rho ^t, \quad \sup _{\theta \in \mathcal {V}(\theta _0)}\left| \frac{\partial \lambda _t(\theta )}{\partial \theta } -\frac{\partial \tilde{\lambda }_t(\theta )}{\partial \theta }\right| \le K\rho ^t, \quad \\&\sup _{\theta \in \mathcal {V}(\theta _0)}\left| \dfrac{1}{\lambda _t(\theta )}-\dfrac{1}{\tilde{\lambda }_t(\theta )}\right| \le K\rho ^t. \end{aligned}$$

Thus we obtain

$$\begin{aligned}&\left\| \frac{\partial l_t(\theta )}{\partial \theta }-\frac{\partial \tilde{l}_t(\theta )}{\partial \theta }\right\| \\&\quad =\left\| \left( \frac{X_t}{\lambda _t(\theta )}-1\right) \frac{\partial \lambda _t(\theta )}{\partial \theta } -\left( \frac{X_t}{\tilde{\lambda }_t(\theta )}-1\right) \frac{\partial \tilde{\lambda }_t(\theta )}{\partial \theta }\right\| \\&\quad =\left\| \left( \frac{X_t}{\lambda _t(\theta )}-\frac{X_t}{\tilde{\lambda }_t(\theta )}\right) \frac{\partial \lambda _t(\theta )}{\partial \theta } -\left( \frac{X_t}{\tilde{\lambda }_t(\theta )}-1\right) \left( \frac{\partial \tilde{\lambda }_t(\theta )}{\partial \theta } -\frac{\partial \lambda _t(\theta )}{\partial \theta }\right) \right\| \\&\quad \le K\rho ^t(1+X_t)\left( 1+\frac{\partial \lambda _t(\theta )}{\partial \theta }\right) , \end{aligned}$$

$$\begin{aligned} \sup _{\theta \in \mathcal {V}(\theta _0)}\left\| \frac{1}{\sqrt{n}}\sum _{t=1}^n\left\{ \frac{\partial l_t(\theta )}{\partial \theta } -\frac{\partial \tilde{l}_t(\theta )}{\partial \theta }\right\} \right\| \le \frac{K}{\sqrt{n}}\sum _{t=1}^n\rho ^t(1+X_t)\sup _{\theta \in \mathcal {V}(\theta _0)}\left( 1+\frac{\partial \lambda _t(\theta )}{\partial \theta }\right) , \end{aligned}$$

the right side converges in probability to 0, which can be proved using arguments similar to Francq and Zakoïan (2004, p. 616) and this technique has been used in the proof of consistency of this theorem. Using the same arguments and replacing first derivatives with second derivatives, (iii) is shown.

Similar to the proof of (4.36) in Francq and Zakoïan (2004), (iv) follows from the Taylor expansion of $\displaystyle \frac{1}{n}\sum _{t=1}^n\frac{\partial ^2l_t(\theta _{ij}^*)}{\partial \theta \partial \theta ^\top }$ around $\theta _0$, the convergence of $\theta _{ij}^*$ to $\theta _0$, the ergodic theorem and (i).

Note that $\lambda _t(\theta _0)=\mu _0(1-\alpha _0-\beta _0)+\alpha _0X_{t-1}+\beta _0\lambda _{t-1}(\theta _0)$, then we have

$$\begin{aligned} \frac{1}{n}\sum _{t=1}^n\lambda _t(\theta _0)&=\mu _0(1-\alpha _0-\beta _0)+\alpha _0\frac{1}{n}\sum _{t=1}^nX_{t-1}+\beta _0 \frac{1}{n}\sum _{t=1}^n\lambda _{t-1}(\theta _0)\\&=\mu _0(1-\alpha _0-\beta _0)+\alpha _0\hat{\mu }_n+\beta _0\frac{1}{n}\sum _{t=1}^n\lambda _t(\theta _0)+o_p(n^{-{1\over 2}}),\\ \frac{1}{n}\sum _{t=1}^n\lambda _t(\theta _0)&=\frac{\mu _0(1-\alpha _0-\beta _0)}{1-\beta _0}+\frac{\alpha _0}{1-\beta _0}\hat{\mu }_n+o_p(n^{-{1\over 2}}),\\ \hat{\mu }_n&=\frac{\mu _0(1-\alpha _0-\beta _0)}{1-\beta _0} +\frac{1}{n}\sum _{t=1}^n(X_t-\lambda _t(\theta _0))+\frac{\alpha _0}{1-\beta _0}\hat{\mu }_n+o_p(n^{-{1\over 2}}), \end{aligned}$$

thus we obtain

$$\begin{aligned} \hat{\mu }_n=\mu _0+\frac{1-\beta _0}{1-\alpha _0-\beta _0}\frac{1}{n}\sum _{t=1}^n(X_t-\lambda _t(\theta _0))+o_p(n^{-{1\over 2}}). \end{aligned}$$

Based the above results we have the representation

$$\begin{aligned} \left( \begin{array}{c}\sqrt{n}(\hat{\mu }_n-\mu _0)\\ \displaystyle \frac{1}{\sqrt{n}}\sum _{t=1}^n\frac{\partial l_t(\theta _0)}{\partial \phi }\end{array}\right) =\frac{1}{\sqrt{n}}\sum _{t=1}^n(X_t-\lambda _t(\theta _0))Z_t+o_p(1), \quad Z_t=\left( \begin{array}{c}\displaystyle \frac{1-\beta _0}{1-\alpha _0-\beta _0}\\ \displaystyle \frac{1}{\lambda _t(\theta _0)} \frac{\partial \lambda _t(\theta _0)}{\partial \phi }\end{array}\right) . \end{aligned}$$

Note that $E((X_t-\lambda _t(\theta _0))Z_t|\mathcal {F}_{t-1})=0$. Moreover,

$$\begin{aligned}&\frac{\lambda _t(\theta _0)}{\partial \alpha }=X_{t-1}-\mu _0+\beta _0\frac{\lambda _{t-1}(\theta _0)}{\partial \alpha } =\sum _{i=0}^\infty \beta _0^i(X_{t-1-i}-\mu _0), \quad \\&\quad \frac{\lambda _t(\theta _0)}{\partial \beta }=\sum _{i=0}^\infty \beta _0^i(\lambda _{t-1-i}-\mu _0), \end{aligned}$$

thus $E\displaystyle \left( \frac{\lambda _t(\theta _0)}{\partial \phi }\right) =0$. If follows that Var $((X_t-\lambda _t(\theta _0))Z_t)=\left( \begin{array}{cc}b&{}0\\ 0&{}J\end{array}\right)$. Using the martingale difference central limit theorem and the Cramer-Wold device, we obtain the asymptotic normality in (v).

Now we will complete the proof of the theorem. From (ii) to (iv) if follows that the matrix $J_n$ is a.s. invertible for sufficiently large n, then

$$\begin{aligned} \sqrt{n}(\hat{\phi }_n-\phi _0)=-J_n^{-1}\left\{ \frac{1}{\sqrt{n}}\sum _{t=1}^n\frac{\partial }{\partial \phi }\tilde{l}_t(\theta _0) +K_n\sqrt{n}(\hat{\mu }_n-\mu _0)\right\} . \end{aligned}$$

Using (iii), we obtain that

$$\begin{aligned} \sqrt{n}\left( \begin{array}{c}\hat{\mu }_n-\mu _0\\ \hat{\phi }_n-\phi _0\end{array}\right) =\left( \begin{array}{cc} 1&{}0\\ -J_n^{-1}K_n&{}-J_n^{-1}\end{array}\right) \left( \begin{array}{c}\sqrt{n}(\hat{\mu }_n-\mu _0)\\ \displaystyle \frac{1}{\sqrt{n}} \sum _{t=1}^n\frac{\partial l_t(\theta _0)}{\partial \phi }\end{array}\right) +o_p(1). \end{aligned}$$

In terms of (iv), (v) and Slutsky’s lemma, $\sqrt{n}(\hat{\theta }-\theta _0)$ is asymptotically $\mathcal {N}(0,\Sigma )$ distributed, with

$$\begin{aligned} \Sigma =\left( \begin{array}{cc} 1&{}0\\ -J^{-1}K&{}-J^{-1}\end{array}\right) \left( \begin{array}{cc} b&{}0\\ 0&{}J\end{array}\right) \left( \begin{array}{cc}1&{}-K^\top J^{-1}\\ 0&{}-J^{-1}\end{array}\right) . \end{aligned}$$

$\square$

Proof of Corollary 2

Let

$$\begin{aligned} S_t=(X_t-\lambda _t(\theta _0))\left( \frac{1}{\lambda _t(\theta _0)}\frac{\partial \lambda _t(\theta _0)}{\partial \mu }, \frac{1}{\lambda _t(\theta _0)}\frac{\partial \lambda _t(\theta _0)}{\partial \phi },\frac{1-\beta _0}{1-\alpha _0-\beta _0}\right) ^\top , \end{aligned}$$

then we observe that

$$\begin{aligned} E(S_tS_t^\top )=\left( \begin{array}{ccc} d&{}K^\top &{}1\\ K&{}J&{}0\\ 1&{}0&{}b\end{array}\right) . \end{aligned}$$

First, we have

$$\begin{aligned} \displaystyle \Sigma ^*=\left( \begin{array}{cc}d&{}K^\top \\ K&{}J\end{array}\right) ^{-1}= \left( \begin{array}{cc}a&{}-aK^\top J^{-1}\\ -aJ^{-1}K&{}J^{-1}+aJ^{-1}KK^\top J^{-1}\end{array}\right) , \end{aligned}$$

thus $\Sigma -\Sigma ^*=(b-a)CC^\top$.

Second, observe that $\Sigma ^*=[E(GS_tS_t^\top G^\top )]^{-1},\Sigma =E(HS_tS_t^\top H^\top )$, where

$$\begin{aligned} G=(I_3~~0)_{3\times 4},~~~H=\left( \begin{array}{ccc}0&{}0&{}1\\ 0&{}J^{-1}&{}-J^{-1}K\end{array}\right) _{3\times 4} \end{aligned}$$

with $I_k$ denoting the identity matrix of size k. Note that $GE(S_tS_t^\top )H^\top =I_3$. Let $D_t=\Sigma ^*GS_t-HS_t$, then we have

$$\begin{aligned} E(D_tD_t^\top )=\Sigma ^*+\Sigma -\Sigma ^*GE(S_tS_t^\top )H^\top -HE(S_tS_t^\top )G^\top \Sigma ^*=\Sigma -\Sigma ^*, \end{aligned}$$

which shows that $\Sigma -\Sigma ^*$ is positive semidefinite. It can be seen that the matrix $E(D_tD_t^\top ) =(\Sigma ^*G-H)E(S_tS_t^\top )(\Sigma ^*G-H)^\top$ is not positive because

$$\begin{aligned} \Sigma ^*G-H=\left( \begin{array}{ccc}a&{}\quad -aK^\top J^{-1}&{}\quad -1\\ -aJ^{-1}K&{}\quad aJ^{-1}KK^\top J^{-1}&{}\quad J^{-1}K\end{array}\right) =(aCC^\top ~-C) \end{aligned}$$

is of rank 1.

Proof of Corollary 3

The first convergence in distribution is a direct consequence of Theorem 1 and of the delta method. In view of Corollary 3 and the delta method, we have $\sqrt{n}(\varphi (\tilde{\theta }_n)-\varphi (\theta _0)){\mathop {\longrightarrow }\limits ^{d}}\mathcal {N}(0,s^{*2})$, where

$$\begin{aligned} s^{*2}=\frac{\partial \varphi (\theta _0)}{\partial \theta ^\top }\Sigma ^*\frac{\partial \varphi (\theta _0)}{\partial \theta ^\top } =s^2-(b-a)\frac{\partial \varphi (\theta _0)}{\partial \theta ^\top }CC^\top \frac{\partial \varphi (\theta _0)}{\partial \theta ^\top }=s^2. \end{aligned}$$