Robust estimation for the one-parameter exponential family integer-valued GARCH(1,1) models based on a modified Tukey’s biweight function

Abstract

In this paper, we study a robust estimation method for observation-driven integer-valued time series models whose conditional distribution belongs to the one-parameter exponential family. Maximum likelihood estimator (MLE) is commonly used to estimate parameters, but it is highly affected by outliers. We resort to the Mallows’ quasi-likelihood estimator based on a modified Tukey’s biweight function as a robust estimator and establish its existence, uniqueness, consistency and asymptotic normality under some regularity conditions. Compared with MLE, simulation results illustrate the better performance of the new estimator. An application is performed on data for two real data sets, and a comparison with other existing robust estimators is also given.

Appendix

Partial derivatives. For the simplicity, in the proof below, we denote $\psi _c(u)$ by $\psi$. Now we give the first and second derivative of every components of $s_t(\theta )$. From Lemma 3, we have

$$\begin{aligned} s_{ti}=\left[ \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \dfrac{\partial \lambda _t}{\partial \theta _i}-E\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \dfrac{\partial \lambda _t}{\partial \theta _i}\bigg |{\mathcal {F}}_{t-1}\right) \right] , \end{aligned}$$

furthermore, from (2.1) and (2.2),

$$\begin{aligned} \frac{\partial \lambda _t}{\partial \alpha _0}=\frac{1-\beta _1^t}{1-\beta _1},~ \frac{\partial \lambda _t}{\partial \alpha _1}=\sum \limits _{d=0}^{t-1}\beta _1^dX_{t-1-d}, ~\hbox {and}~ \frac{\partial \lambda _t}{\partial \beta _1}=\sum \limits _{d=0}^{t-1}\beta _1^d\lambda _{t-1-d}. \end{aligned}$$

(B.1)

The first and second derivative of $s_{ti}(\theta )$ are $\dfrac{\partial s_{ti}(\theta )}{\partial \theta _j}$ and $\dfrac{\partial ^2 s_{ti}(\theta )}{\partial \theta _j\partial \theta _k}$, respectively, given by

$$\begin{aligned} \frac{\partial s_{ti}(\theta )}{\partial \theta _j}&\,=\,\frac{\partial }{\partial \theta _j}\left[ \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta _i}\right] \\&\quad -E\left[ \frac{\partial }{\partial \theta _j}\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta _i}\right) \bigg |{\mathcal {F}}_{t-1}\right] ,~j=1,2,3, \\ \frac{\partial ^2 s_{ti}(\theta )}{\partial \theta _j\partial \theta _k}&=\frac{\partial ^2}{\partial \theta _j\partial \theta _k}\left[ \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta _i}\right] \\&\quad -E\left[ \frac{\partial ^2}{\partial \theta _j\partial \theta _k}\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta _i}\right) \bigg |{\mathcal {F}}_{t-1}\right] ,~k=1,2,3, \end{aligned}$$

where

$$\begin{aligned} \frac{\partial }{\partial \theta _j}\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta _i}\right) \,=\,&(m_{1t}+m_{2t})\frac{\partial \lambda _t}{\partial \theta _j}\frac{\partial \lambda _t}{\partial \theta _i} +m_{3t}\frac{\partial ^2\lambda _t}{\partial \theta _i\partial \theta _j}\\ \frac{\partial ^2}{\partial \theta _j\partial \theta _k}\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}}\frac{\partial \lambda _t}{\partial \theta _i}\right) =&(m_{4t}+m_{5t}+m_{6t})\frac{\partial \lambda _t}{\partial \theta _k} \frac{\partial \lambda _t}{\partial \theta _j}\frac{\partial \lambda _t}{\partial \theta _i}\\&+(m_{1t}+m_{2t})\left( \frac{\partial ^2\lambda _t}{\partial \theta _i\partial \theta _k}\frac{\partial \lambda _t}{\partial \theta _j} +\frac{\partial ^2\lambda _t}{\partial \theta _j\partial \theta _k}\frac{\partial \lambda _t}{\partial \theta _i} +\frac{\partial ^2\lambda _t}{\partial \theta _i\partial \theta _j}\frac{\partial \lambda _t}{\partial \theta _k}\right) \\&+m_{3t}\frac{\partial ^3\lambda _t}{\partial \theta _i\partial \theta _j\partial \theta _k}, \end{aligned}$$

with

$$\begin{aligned}&m_{1t}\,=\,-\psi ^{'}(r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \left( \frac{\frac{1}{2}(X_t-\lambda _t)B^{''}(\eta _t)}{{B^{'}(\eta _t)}^\frac{5}{2}} +\frac{1}{\sqrt{B^{'}(\eta _t)}}\right) ,\nonumber \\&m_{2t}=-\psi (r_t)w_t\frac{\frac{1}{2}B^{''}(\eta _t)}{{B^{'}(\eta _t)}^2\sqrt{B^{'}(\eta _t)}},\nonumber \\&m_{3t}=\psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}},\nonumber \\&m_{4t}=\psi ^{''}(r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} {\left( \frac{\frac{1}{2}(X_t-\lambda _t)B^{''}(\eta _t)}{{B^{'}(\eta _t)}^\frac{5}{2}} +\frac{1}{\sqrt{B^{'}(\eta _t)}}\right) }^2,\nonumber \\&m_{5t}=-\psi ^{'}(r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \left( \frac{\frac{1}{2}(X_t-\lambda _t)B^{'''}(\eta _t)}{{B^{'}(\eta _t)}^\frac{7}{2}} -\frac{\frac{7}{4}(X_t-\lambda _t){B^{''}(\eta _t)}^2}{{B^{'}(\eta _t)}^\frac{9}{2}} -\frac{\frac{3}{2}B^{''}(\eta _t)}{{B^{'}(\eta _t)}^\frac{5}{2}}\right) ,\nonumber \\&m_{6t}=\psi (r_t)w_t \left( \frac{\frac{5}{4}{B^{''}(\eta _t)}^2\sqrt{B^{'}(\eta _t)}}{{B^{'}(\eta _t)}^5} -\frac{\frac{1}{2}B^{'''}(\eta _t)\sqrt{B^{'}(\eta _t)}}{{B^{'}(\eta _t)}^4}\right) ,\nonumber \\&\frac{\partial ^2\lambda _t}{\partial \alpha _0^2}=0,~~~ \frac{\partial ^2\lambda _t}{\partial \alpha _1^2}=0,~~~ \frac{\partial ^2\lambda _t}{\partial \alpha _0\partial \alpha _1}=0,~~~ \frac{\partial ^2\lambda _t}{\partial \alpha _0\partial \beta _1}=\frac{\partial \lambda _{t-1}}{\partial \alpha _0}+\beta _1\frac{\partial ^2\lambda _{t-1}}{\partial \alpha _0\partial \beta _1},\end{aligned}$$

(B.2)

$$\begin{aligned}&\frac{\partial ^2\lambda _t}{\partial \alpha _1\partial \beta _1}\,=\,\frac{\partial \lambda _{t-1}}{\partial \alpha _1}+\beta _1 \frac{\partial ^2\lambda _{t-1}}{\partial \alpha _1\partial \beta _1},~~~ \frac{\partial ^2\lambda _t}{\partial \beta _1^2}=\frac{\partial \lambda _{t-1}}{\partial \beta _1}+\beta _1\frac{\partial ^2\lambda _{t-1}}{\partial \beta _1^2},\end{aligned}$$

(B.3)

$$\begin{aligned}&\frac{\partial ^3\lambda _t}{\partial \alpha _0\partial \beta _1^2} \,=\,2\frac{\partial ^2\lambda _{t-1}}{\partial \alpha _0\partial \beta _1} +\beta _1\frac{\partial ^3\lambda _{t-1}}{\partial \alpha _0\partial \beta _1^2},~~~ \frac{\partial ^3\lambda _t}{\partial \alpha _1\partial \beta _1^2} =2\frac{\partial ^2\lambda _{t-1}}{\partial \alpha _1\partial \beta _1} +\beta _1\frac{\partial ^3\lambda _{t-1}}{\partial \alpha _1\partial \beta _1^2},\end{aligned}$$

(B.4)

$$\begin{aligned}&\frac{\partial ^3\lambda _t}{\partial \beta _1^3} =2\frac{\partial ^2\lambda _{t-1}}{\partial \beta _1^2} +\beta _1\frac{\partial ^3\lambda _{t-1}}{\partial \beta _1^3}. \end{aligned}$$

(B.5)

Next, without loss of generality, we only consider derivatives with respect to $\beta _1$. Under the assumptions of (A5)–(A7) and according to (B.1)–(B.5), we have

$$\begin{aligned}&|m_{1t}|\le \psi ^{'}(r_t)w_t \left( \frac{\frac{1}{2}(X_t+\lambda _t)K_2}{K_1} +\frac{1}{K_1}\right) ,\\&|m_{2t}|\le \frac{1}{2}\psi (r_t)w_t\frac{K_2}{\sqrt{K_1}},\\&|m_{3t}|\le \psi (r_t)w_t\frac{1}{\sqrt{K_1}},\\&|m_{4t}|\le \psi ^{''}(r_t)w_t\frac{1}{\sqrt{K_1}} {\left( \frac{\frac{1}{2}(X_t+\lambda _t)K_2}{\sqrt{K_1}} +\frac{1}{\sqrt{K_1}}\right) }^2,\nonumber \\&|m_{5t}|\le \psi ^{'}(r_t)w_t \left( \frac{\frac{1}{2}(X_t+\lambda _t)K_3}{K_1} +\frac{\frac{7}{4}(X_t+\lambda _t)K_2^2}{K_1} +\frac{\frac{3}{2}K_2}{K_1}\right) ,\\&|m_{6t}|\le \psi (r_t)w_t \left( \frac{\frac{5}{4}K_2^2}{\sqrt{K_1}} +\frac{\frac{1}{2}K_3}{\sqrt{K_1}}\right) ,\\&\lambda _t\le \mu _{0t}:=\alpha _{1u}\sum \limits _{d=1}^{t-1}\beta _{1u}^dX_{t-1-d}+c_0,~\hbox {where}~c_0=\frac{\alpha _{0u}}{1-\beta _{1u}}+\lambda _0,\\&\frac{\partial \lambda _t}{\partial \beta _1}\le \mu _{1t} :=\alpha _{1u}\sum \limits _{d=1}^{t-1}d\beta _{1u}^dX_{t-1-d}+c_1,~\hbox {where}~c_1=\frac{c_0}{1-\beta _{1u}},\\&\frac{\partial ^2\lambda _t}{\partial \beta _1^2}\le \mu _{2t} :=\alpha _{1u}\sum \limits _{d=1}^{t-2}d(d+1)\beta _{1u}^{d-1}X_{t-2-d}+c_2,~\hbox {where}~c_2=\frac{2c_0}{(1-\beta _{1u})^2},\\&\frac{\partial ^3\lambda _t}{\partial \beta _1^3}\le \mu _{3t} :=\alpha _{1u}\sum \limits _{d=1}^{t-2}d(d+1)(d+2)\beta _{1u}^{d-1}X_{t-3-d}+c_3,~\hbox {where}~c_3=\frac{6c_0}{(1-\beta _{1u})^3}, \end{aligned}$$

$$\begin{aligned} \Bigg |\frac{\partial s_{ti}(\theta )}{\partial \beta _1}\Bigg |&\le \left( |m_{1t}-E(m_{1t}|{\mathcal {F}}_{t-1})| +|m_{2t}-E(m_{2t}|{\mathcal {F}}_{t-1})|\right) \mu _{1t}^2 \nonumber \\&\quad +|m_{3t}-E(m_{3t}|{\mathcal {F}}_{t-1})|\mu _{2t}\equiv {\tilde{m}}_t,\end{aligned}$$

(B.6)

$$\begin{aligned} \Bigg |\frac{\partial ^2 s_{ti}(\theta )}{\partial \beta _1^2}\Bigg |&\le \left( |m_{4t}-E(m_{4t}|{\mathcal {F}}_{t-1})| +|m_{5t}-E(m_{5t}|{\mathcal {F}}_{t-1})|+|m_{6t}-E(m_{6t}|{\mathcal {F}}_{t-1})|\right) \mu _{1t}^3\nonumber \\&\quad +3\left( |m_{1t}-E(m_{1t}|{\mathcal {F}}_{t-1})| +|m_{2t}-E(m_{2t}|{\mathcal {F}}_{t-1})|\right) \mu _{1t}\mu _{2t}\nonumber \\&\quad +|m_{3t}-E(m_{3t}|{\mathcal {F}}_{t-1})|\mu _{3t}\equiv m_t, \end{aligned}$$

(B.7)

$E{\tilde{m}}_t<\infty$ and $Em_t<\infty$ hold by considering each term in ${\tilde{m}}_t$ and $m_t$ separately. For instance, $E(m_{1t}-E(m_{1t}|{\mathcal {F}}_{t-1}))^2\le Em_{1t}^2<\infty$ and $E\mu _{1t}^4<\infty$ as function $\psi$ is bounded and $EX_t^6$ is finite, thus $E|m_{1t}-E(m_{1t}|{\mathcal {F}}_{t-1})|\mu _{1t}^2<\infty$. In addition, the highest moment of $X_t$ contained in $m_t$ is 6 and $EX_t^6$ is finite, thus $Em_t<\infty$.

Proof of Lemma 1

Part (1). From Eq. (2.3), we know that $s_t(\theta )$ is a martingale difference sequence as $E(s_t(\theta )|{\mathcal {F}}_{t-1})=0$ and $S_n(\theta )$ is a martingale sequence with $E(S_n(\theta )|{\mathcal {F}}_{n-1}) =S_{n-1}(\theta )$. In the following, it is sufficient to show $s_t(\theta )$ is square integrable, that is $E\Vert s_t(\theta )\Vert ^2<\infty$. According to the strong law of large numbers for martingales (Chow 1967), we have

$$\begin{aligned} \frac{1}{n}S_n(\theta )\xrightarrow {a.s.}0,~n\rightarrow \infty , \end{aligned}$$

for that

$$\begin{aligned} \Vert s_t(\theta )\Vert ^2&=\left\| \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta }\right\| ^2+\left\| E\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta }\bigg |{\mathcal {F}}_{t-1}\right) \right\| ^2\nonumber \\&~~~-2\psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}}E\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \bigg |{\mathcal {F}}_{t-1}\right) \left\| \frac{\partial \lambda _t}{\partial \theta }\right\| ^2\nonumber \\&\leqslant 2\left( \left\| \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta }\right\| ^2+\left\| E\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta }\bigg |{\mathcal {F}}_{t-1}\right) \right\| ^2\right) , \end{aligned}$$

then

$$\begin{aligned} E\Vert s_t(\theta )\Vert ^2&\le C_1^2E\left\| \frac{\partial \lambda _t}{\partial \theta }\right\| ^2, \end{aligned}$$

where $C_1$ is a constant and depends on function $\psi$ and $\frac{1}{\sqrt{B^{'}(\eta _t)}}$, as $EX_t^2<\infty$, $E\lambda _t^2<\infty$ and according to (B.1), it is easy to know that $E\left( \dfrac{\partial \lambda _t}{\partial \theta _i}\right) ^2,~i=1,2,3$ are finite, then $E\Vert s_t(\theta )\Vert ^2<\infty$.

Part (2).

$$\begin{aligned} \Vert s_t(\theta )\Vert ^4&=(\Vert s_t(\theta )\Vert ^2)^2 \le 8 \\&\quad \times \left( \left\| \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta }\right\| ^4+\left\| E\left( \psi (r_t)w_t\frac{1}{\sqrt{B^{'}(\eta _t)}} \frac{\partial \lambda _t}{\partial \theta }\bigg |{\mathcal {F}}_{t-1}\right) \right\| ^4\right) ,\\ E\Vert s_t(\theta )\Vert ^4&\le 16C_1^4E\left\| \frac{\partial \lambda _t}{\partial \theta }\right\| ^4. \end{aligned}$$

As $\left\| \dfrac{\partial \lambda _t}{\partial \theta }\right\| ^4$ is the function of $X_t$ and $\lambda _t$, $EX_t^6$ is finite, then $E\Vert s_t(\theta )\Vert ^4<\infty$. Note that

$$\begin{aligned} \frac{1}{n}\sum \limits _{t=1}^nE\left[ \Vert s_t(\theta )\Vert ^2I(\Vert s_t(\theta )\Vert >\sqrt{n}\varepsilon )|{\mathcal {F}}_{t-1}\right] \le \frac{1}{n^2\varepsilon ^2}\sum \limits _{t=1}^nE\left[ \Vert s_t(\theta )\Vert ^4|{\mathcal {F}}_{t-1}\right] \rightarrow 0, \end{aligned}$$

this result gives the conditional Lindeberg’s condition. In addition,

$$\begin{aligned} \frac{1}{n}\sum \limits _{t=1}^n\hbox {Var}(s_t(\theta )|{\mathcal {F}}_{t-1})\xrightarrow {P}E\left\{ E\left[ s_t(\theta )(s_t(\theta ))^\top \bigg |{\mathcal {F}}_{t-1}\right] \right\} =W. \end{aligned}$$

Applying the CLT for martingales (Taniguchi and Kakizawa 2000, Theorem 1.3.13), then

$$\begin{aligned} \frac{1}{\sqrt{n}}S_n(\theta )=\frac{1}{\sqrt{n}}\sum \limits _{t=1}^ns_t(\theta )\xrightarrow {d}N(0,W). \end{aligned}$$

$\square$

Proof of Lemma 2

Because $S_n(\theta )=0$ is an unbiased estimating function, then

$$\begin{aligned} -E\left[ \frac{\partial }{\partial {\theta }^\top }s_t(\theta )\bigg |{\mathcal {F}}_{t-1}\right] =E\left[ s_t(\theta )\frac{\partial l_t(\theta )}{\partial {\theta }^\top }\bigg |{\mathcal {F}}_{t-1}\right] , \end{aligned}$$

for that

$$\begin{aligned} 0=\frac{\partial E(s_t(\theta )|{\mathcal {F}}_{t-1})}{\partial {\theta }^\top }&=\int \frac{\partial (s_t(\theta )\hbox {exp}\{\eta _t X_t-A(\eta _t)\}h(X_t))}{\partial {\theta }^\top }d_x\nonumber \\&=\int \frac{\partial s_t(\theta )}{\partial {\theta }^\top }\hbox {exp}\{\eta _t X_t-A(\eta _t)\}h(X_t)d_x+\\&\int s_t(\theta )\frac{\partial (\hbox {exp}\{\eta _t X_t-A(\eta _t)\}h(X_t))}{\partial {\theta }^\top }d_x \nonumber \\&=E\left[ \frac{\partial }{\partial {\theta }^\top }s_t(\theta )\bigg |{\mathcal {F}}_{t-1}\right] +\int s_t(\theta )\frac{\partial l_t(\theta )}{\partial {\theta }^\top }\hbox {exp}\{\eta _t X_t-A(\eta _t)\}h(X_t)d_x\nonumber \\&=E\left[ \frac{\partial }{\partial {\theta }^\top }s_t(\theta )\bigg |{\mathcal {F}}_{t-1}\right] +E\left[ s_t(\theta )\frac{\partial l_t(\theta )}{\partial {\theta }^\top }\bigg |{\mathcal {F}}_{t-1}\right] . \end{aligned}$$

Let

$$\begin{aligned} V_t=E\left[ s_t(\theta )\frac{\partial l_t(\theta )}{\partial {\theta }^\top }\bigg |{\mathcal {F}}_{t-1}\right] = M_{11}, \end{aligned}$$

(B.8)

where

$$\begin{aligned} M_{11}=\frac{\partial \lambda _t}{\partial \theta }E\left[ \psi (r_t)(X_t-\lambda _t)|{\mathcal {F}}_{t-1}\right] \left( \frac{1}{\sqrt{B^{'}(\eta _t)}}\right) ^3w_t\frac{\partial \lambda _t}{\partial {\theta }^\top }. \end{aligned}$$

To prove V is a positive definite matrix is equivalent to show $V_t$ is a positive definite matrix. Obviously, $V_t$ is a $3\times 3$ positive definite matrix. This is because for any non-zero three dimensional real vector z, $z^\top M_{11}z>0$ is equivalent to $z^\top \dfrac{\partial \lambda _t}{\partial {\theta }} \dfrac{\partial \lambda _t}{\partial {\theta }^\top }z>0$, if $z^\top \dfrac{\partial \lambda _t}{\partial {\theta }}=0$, from (A8), we derive $z=0$. Thus, $V_t$ is a positive definite matrix. Let $H_n(\theta )=-\dfrac{1}{n}\dfrac{\partial }{\partial {\theta }^\top }S_n(\theta )$, according to (B.6), $E\left\| \dfrac{\partial }{\partial {\theta }^\top }s_t(\theta )\right\| <\infty$. Using the LLN of Jensen and Rahbek (2007), it is easy to obtain

$$\begin{aligned} H_n(\theta )=-\frac{1}{n}\dfrac{\partial }{\partial {\theta }^\top }S_n(\theta )=-\frac{1}{n}\sum \limits _{t=1}^n\dfrac{\partial }{\partial {\theta }^\top }s_t(\theta )\xrightarrow {a.s.} -E\left[ \dfrac{\partial }{\partial {\theta }^\top }s_t(\theta )\right] =V. \end{aligned}$$

$\square$

Proof of Lemma 3

From (B.7), we have

$$\begin{aligned} \max _{i,j,k=1,2,3}\sup \limits _{\theta \in \Theta }\left| \frac{1}{n}\sum \limits _{t=1}^n\frac{\partial ^2 s_{ti}}{\partial \theta _j\partial \theta _k}\right| \le M_n: =\frac{1}{n}\sum \limits _{t=1}^nm_t, \end{aligned}$$

with $Em_t<\infty$, then $M_n\xrightarrow {a.s.} M$ and $M=Em_t$ hold by the LLN in Jensen and Rahbek (2007). Using Lemmas 1–3, we can claim Theorem 1 holds. $\square$