Mallows’ quasi-likelihood estimation for log-linear Poisson autoregressions

Abstract

We consider the problems of robust estimation and testing for a log-linear model with feedback for the analysis of count time series. We study inference for contaminated data with transient shifts, level shifts and additive outliers. It turns out that the case of additive outliers deserves special attention. We propose a robust method for estimating the regression coefficients in the presence of interventions. The resulting robust estimators are asymptotically normally distributed under some regularity conditions. A robust score type test statistic is also examined. The methodology is applied to real and simulated data.

Appendix

In the following, the symbol C denotes a constant which depends upon the context. Define also $d_M=\max (|d_L|,|d_U|)$, $a_M=\max (|a_L|,|a_U|)$ and $b_M=\max (|b_L|,|b_U|)$. In addition, when a quantity is evaluated at the true value of the parameter $\varvec{\theta }$, denoted it by $\varvec{\theta }_{0}$, then the notation will be simplified by dropping $\varvec{\theta }_{0}$. For instance, $m_{t} \equiv m_{t}(\varvec{\theta }_{0})$ and so on. The following two results are taken from Fokianos and Tjøstheim (2011) and are included for completeness.

Lemma 6.1

Assume model (2) and suppose that $|a|<1$. In addition, assume that when $b>0$ then $|a+b|<1$, and when $b<0$ then $|a||a+b|<1$. Then, the following conclusions hold:

• 1. The process $\{\nu _t^m,t \ge 0\}$ is a geometrically ergodic Markov chain with finite moments of order k, for an arbitrary k.

• 2. The process $\{(Y_t^m,U_t,\nu _t^m),t \ge 0\}$ is a $V_{(Y,U,\nu )}$-geometrically ergodic Markov chain with $V_{Y,U,\nu }(Y,U,\nu )=1+\log ^{2k}(1+Y)+\nu ^{2k}+U^{2k}$, k being a positive integer.

Lemma 6.2

Suppose that $(Y_t,\nu _t)$ and $(Y_t^m,\nu _t^m)$ are defined by (1) and (2) respectively. Assume that $|a+b|<1$, if a and b have the same sign, and $a^2+b^2<1$ if a and b have different signs. Then the following statements are true:

• 1. $E|\nu _t^m-\nu _t| \rightarrow 0$ and $|\nu _t^m-\nu _t|<\delta _{1,m}$ almost surely for m large.

• 2. $E(\nu _t^m-\nu _t)^2 \le \delta _{2,m}$,

• 3. $E|\lambda _t^m-\lambda _t| \le \delta _{3,m}$,

• 4. $E|Y_t^m-Y_t| \le \delta _{4,m}$,

• 5. $E(\lambda _t^m-\lambda _t)^2 \le \delta _{5,m}$,

• 6. $E(Y_t^m-Y_t)^2 \le \delta _{6,m}$,

where $\delta _{i,m} \rightarrow 0$ as $m \rightarrow \infty$ for $i=1,\ldots ,6$. Furthermore, almost surely, with m large enough

$$\begin{aligned} |\lambda _t^m-\lambda _t| \le \delta \quad \text {and} ~~ |Y_t^m-Y_t| \le \delta , \quad \text {for any} \, \delta >0. \end{aligned}$$

We will also need the following lemma whose proof is given below.

Lemma 6.3

Define the Pearson residuals for both perturbed and unperturbed models by

$$\begin{aligned} r_t^m=\frac{Y_t^m-e^{\nu _t^m}}{e^{{\nu _t^m}/2}}, ~~~ r_t=\frac{Y_t-e^{\nu _t}}{e^{{\nu _t}/2}} \end{aligned}$$

respectively. Suppose that $(Y_t,\nu _t)$ and $(Y_t^m,\nu _t^m)$ are defined by (1) and (2) respectively. Assume that $|a+b|<1$, if a and b have the same sign, and $a^2+b^2<1$ if a and b have different signs. Then,

• 1. $E|r_t^m-r_t| \rightarrow 0$,

• 2. $E(r_t^m-r_t)^2 \le \delta _{7;m}$,

where $\delta _{7,m} \rightarrow 0$ as $m \rightarrow \infty$. Furthermore, almost surely, with m large enough

$$\begin{aligned} |r_t^m-r_t| \le \delta ~~~ \text {for any} ~~ \delta >0. \end{aligned}$$

Proof of Lemma 6.3

We have that

$$\begin{aligned} |r_t^m-r_t|= & {} \left| \frac{Y_t^m-e^{\nu _t^m}}{e^{{\nu _t^m}/2}} - \frac{Y_t-e^{\nu _t}}{e^{{\nu _t}/2}} \right| = \left| \frac{Y_t^m}{e^{{\nu _t^m}/2}} - \frac{Y_t}{e^{{\nu _t}/2}} + \left( e^{{\nu _t}/2} - e^{{\nu _t^m}/2} \right) \right| \\\le & {} \left| \frac{Y_t^m e^{{\nu _t}/2} - Y_t e^{{\nu _t^m}/2} \pm Y_t^m e^{{\nu _t^m}/2}}{e^{{\nu _t^m}/2} e^{{\nu _t}/2}}\right| + \left| e^{{\nu _t}/2} - e^{{\nu _t^m}/2} \right| \\\le & {} \left| \frac{Y_t^m-Y_t}{e^{{\nu _t}/2}} \right| + \left| \frac{Y_t^m \left( e^{{\nu _t}/2} - e^{{\nu _t^m}/2} \right) }{e^{{\nu _t^m}/2} e^{{\nu _t}/2}} \right| + \left| e^{{\nu _t^m}/2}- e^{{\nu _t}/2} \right| \\\le & {} |Y_t^m-Y_t| + \left( |Y_t^m|+1 \right) \left| \lambda _{t}^{m} -\lambda _{t} \right| ^{1/2} \\< & {} \delta , \end{aligned}$$

for any $\delta >0$ almost surely and for m large enough by using the results of Lemma 6.2. The claims follow. $\square$

Proof of Lemma 2.1

We will show that

$$\begin{aligned} E \left( m_t^m(\varvec{\theta }) \left( m_t^m(\varvec{\theta }) \right) ^T \right) - E \left( m_t(\varvec{\theta }) \left( m_t(\varvec{\theta }) \right) ^T \right) \rightarrow 0 \end{aligned}$$

(9)

and

$$\begin{aligned} E \left( m_t^m(\varvec{\theta }) \right) E^{T} \left( m_t^m(\varvec{\theta }) \right) - E \left( m_t(\varvec{\theta }) \right) E^{T} \left( m_t(\varvec{\theta }) \right) \rightarrow 0, \end{aligned}$$

(10)

as $m \rightarrow \infty$. Consider first (9). Working along the lines of Fokianos and Tjøstheim (2011), we consider differences of the perturbed and non perturbed matrix along the diagonal individually for $\theta _i=d,a,b$. Then, we need to evaluate

$$\begin{aligned} E \left| (Z_t^m)^2 \left( \frac{\partial \nu _t^m}{\partial \theta _i} \right) ^2 - Z_t^2 \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2 \right| , \end{aligned}$$

with $Z_t=\psi _c(r_t)w_t e^{\nu _t/2}$ and similarly for $Z_t^m$. We have,

$$\begin{aligned}&E\left| \left( Z_t^m\right) ^2 \left( \frac{\partial \nu _t^m}{\partial \theta _i} \right) ^2 - Z_t^2 \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2 \right| \\&\quad = E \left| \left( Z_t^m\right) ^2 \left[ \left( \frac{\partial \nu _t^m}{\partial \theta _i} \right) ^2 - \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2 \right] + \left( \left( Z_t^m\right) ^2 - Z_t^2\right) \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2 \right| \\&\quad \le E \left| \left( Z_t^m\right) ^2 \left[ \left( \frac{\partial \nu _t^m}{\partial \theta _i} \right) ^2- \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2 \right] \right| + E\left| \left( \left( Z_t^m\right) ^2-Z_t^2 \right) \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2 \right| \\&\quad \le C E \left| \exp \left( \nu _{t}^{m}\right) \Bigl ( \left( \frac{\partial \nu _t^m}{\partial \theta _i} \right) -\left( \frac{\partial \nu _t}{\partial \theta _i} \right) \Bigr ) \Bigl ( \left( \frac{\partial \nu _t^m}{\partial \theta _i} \right) +\left( \frac{\partial \nu _t}{\partial \theta _i} \right) \Bigr ) \right| \\&\qquad +\, \sqrt{ E \Bigl ( \left( Z_t^{m}\right) ^{2}- Z_{t}^{2} \Bigr )^{2} E \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^{4} } \end{aligned}$$

The first term can become arbitrarily small because it can be shown (following the proof of Fokianos et al. (2009, Lemma 3.1)) that

$$\begin{aligned} \left| \frac{\partial \nu _t^m}{\partial \theta _i}- \frac{\partial \nu _t^m}{\partial \theta _i} \right| < \delta ,~~ E \left| \frac{\partial \nu _t^m}{\partial \theta _i}- \frac{\partial \nu _t^m}{\partial \theta _i} \right| < \delta ,~~ E \left( \frac{\partial \nu _t^m}{\partial \theta _i}- \frac{\partial \nu _t^m}{\partial \theta _i} \right) ^{2} < \delta . \end{aligned}$$

(11)

For the second term, note first that $E (\partial \nu _t / \partial \theta _i)^{4}$ is bounded by a finite constant for $i=1,2,3$ since

$$\begin{aligned} \frac{\partial \nu _t}{\partial d}\le & {} \frac{1}{(1-a_M)},~~ \frac{\partial \nu _t}{\partial a}\le \frac{c_0}{(1-a_M)}+b_M \sum \limits _{i=1}^{t-1} ia_M^{i-1} \log (1+Y_{t-i-1}),\nonumber \\ \frac{\partial \nu _t}{\partial b}\le & {} \sum \limits _{i=0}^{t-1} a_M^{i-1} \log (1+Y_{t-i-1}), \end{aligned}$$

(12)

by using Lemma 6.1. Furthermore

$$\begin{aligned} \left| \left( Z_t^{m}\right) ^{2}- Z_{t}^{2} \right|\le & {} \left| \psi ^2\left( r_t^m\right) e^{\nu _t^m} - \psi ^2(r_t) e^{\nu _t} \right| \\= & {} \left| \psi ^2\left( r_t^m\right) e^{\nu _t^m} - \psi ^2(r_t) e^{\nu _t} \pm \psi ^2(r_t) e^{\nu _t^m} \right| \\\le & {} \left| \psi ^2\left( r_t^m\right) - \psi ^2(r_t)\right| \lambda _{t}^{m} + \psi ^2(r_t)\left| e^{\nu _t^m} -e^{\nu _t}\right| \\\le & {} C \Bigl (\Bigl |r_t^m - r_t\Bigr | \lambda _{t}^{m} + \Bigl |\lambda _{t}^{m} - \lambda _t\Bigr | \Bigr ) < \delta \end{aligned}$$

where we have used the boundedness of the function $\psi (\cdot )$, the mean-value theorem and Lemmas 6.2 and 6.3. Hence (9) follows. To prove (10), consider

$$\begin{aligned}&\Bigl | E^{2} \Bigl ( Z_t^m \frac{\partial \nu _t^m}{\partial \theta _{i}} \Bigr )- E^{2} \Bigl ( Z_t \frac{\partial \nu _t}{\partial \theta _{i}} \Bigr ) \Bigr | \\&\quad \le E \left| Z_t^m \frac{\partial \nu _t^m}{\partial \theta _{i}} - Z_t \frac{\partial \nu _t}{\partial \theta _{i}} \right| E \left| Z_t^m \frac{\partial \nu _t^m}{\partial \theta _{i}} + Z_t \frac{\partial \nu _t}{\partial \theta _{i}} \right| \\&\quad \le C E \left| Z_t^m \frac{\partial \nu _t^m}{\partial \theta _{i}} - Z_t \frac{\partial \nu _t}{\partial \theta _{i}} \pm Z_t^m \frac{\partial \nu _t}{\partial \theta _{i}} \right| E \left| e^{\nu _t^m/2} \frac{\partial \nu _t^m}{\partial \theta _{i}} + e^{\nu _t/2} \frac{\partial \nu _t}{\partial \theta _{i}} \pm e^{\nu _t^m/2} \frac{\partial \nu _t}{\partial \theta _{i}} \right| \\&\quad = C E \left| Z_t^m \left( \frac{\partial \nu _t^m}{\partial \theta _{i}} -\frac{\partial \nu _t}{\partial \theta _{i}} \right) + \left( Z_t^m-Z_t \right) \frac{\partial \nu _t}{\partial \theta _{i}}\right| \\&\qquad \times \,E\left| e^{\nu _t^m/2}\left( \frac{\partial \nu _t^m}{\partial \theta _{i}} +\frac{\partial \nu _t}{\partial \theta _{i}} \right) +\left( e^{\nu _t^m/2}- e^{\nu _t/2}\right) \frac{\partial \nu _t}{\partial \theta _{i}}\right| . \end{aligned}$$

The above quantity can be made arbitrarily small because of finite moments of $\partial \nu _t / \partial \theta _{i}$, $\partial \nu _t^{m} / \partial \theta _{i}$, $\exp (\nu _{t}^{m})$, (11) and the fact that $E \left| \left( Z_t^m-Z_t \right) \ \partial \nu _t / \partial \theta _{i} \right| \rightarrow 0$, as $\rightarrow \infty$ which is proved following the previous arguments. $\square$

Proof of Lemma 2.2

The score function $S_n^m$ for the perturbed model is a martingale sequence, with $E(S_n^m||\mathcal{F}_{t-1}^m)=S_{n-1}^m$ at the true value $\varvec{\theta }=\varvec{\theta }_0$ and $\mathcal{F}_{t-1}^m$ denotes the $\sigma$-field generated by $\{Y_0^m,\ldots ,Y_{t-1}^m,\mathcal{U}_0,\ldots ,\mathcal{U}_{t-1}\}$. We will show that it is square integrable. Proving that $E||m_t^m|| ^2$ is finite for $\varvec{\theta }_0=d_0,a_0$ and $b_0$ guarantees an application of the strong law of large numbers for martingales (Chow 1967), which gives almost sure convergence to 0 of $S_n^m/n$ as $n \rightarrow \infty$. But

$$\begin{aligned} E\left\{ \Bigl ( \psi \left( r_t^m\right) w_t e^{\nu _t^m/2} \frac{\partial \nu _t^m}{\partial \theta _{i}} \Bigr )^{2} \right\} \le C \left( E| e^{\nu _t^m} |^2 \right) ^{1/2} \left( E \left( \frac{\partial \nu _t^m}{\partial \theta _{i}} \right) ^4 \right) ^{1/2} \end{aligned}$$

and this is finite because of Lemma 6.1 and (12). To show asymptotic normality of the perturbed score function $S_n^m$ we apply the CLT for martingales (Hall and Heyde 1980, Cor. 3.1). $(S_n^m)_{n \ge 1}$ is a zero mean, square integrable martingale sequence with $(s_t^m)_{t \ge \mathbb {N}}$ a martingale difference sequence. To prove the conditional Lindeberg’s condition note that

$$\begin{aligned} \frac{1}{n} \sum \limits _{t=1}^n E\left( \parallel s_t^m \parallel ^2 \mathbbm {1} \left( \parallel s_t^m \parallel > \sqrt{n} \delta \right) \parallel \mathcal{F}_{t-1}^m \right) \rightarrow 0, \end{aligned}$$

since $E||s_t^m||^4 < \infty$. In addition,

$$\begin{aligned} \frac{1}{n} \sum _{t=1}^n { Var\left( s_t^m \parallel \mathcal{F}_{t-1}^m\right) } \xrightarrow {p}&E \Bigl \{E\left[ \left( s_t^m\right) \left( s_t^m\right) ^T \parallel \mathcal{F}_{t-1}^m\right] \Bigr \}=W^{m}. \end{aligned}$$

This concludes the second result of the Lemma.

The third result of the Lemma is identical to Lemma 2.1 by Brockwell and Davis (1991, Prop. 6.4.9.). Consider now the last result of the Lemma.

$$\begin{aligned} \frac{1}{\sqrt{n}}\left( S_n^m-S_n\right)= & {} \frac{1}{\sqrt{n}} \sum \limits _{t=1}^n \left\{ s_t^m - s_t \right\} = \frac{1}{\sqrt{n}}\sum _{t=1}^{n} \left( W_{t}^{m}\frac{\partial \nu _{t}^{m}}{\partial \varvec{\theta } }-W_{t}\frac{\partial \nu _{t}}{\partial \varvec{\theta } }\right) \\= & {} \frac{1}{\sqrt{n}}\sum _{t=1}^{n} \left[ W_{t}^{m}\left( \frac{\partial \nu _{t}^{m}}{\partial \varvec{\theta } }-\frac{\partial \nu _{t}}{\partial \varvec{\theta }} \right) + \left( W_{t}^{m}-W_{t}\right) \frac{\partial \nu _{t}}{\partial \varvec{\theta }} \right] , \end{aligned}$$

where $W_{t}= Z_{t}-E[Z_{t} \mid \mathcal{F}_{t-1}]$ and similarly for the perturbed model. For the first summand in the above representation, we obtain that

$$\begin{aligned} {P}\left( \left\| \sum _{t=1}^{n}W_{t}^{m}\left( \frac{\partial \nu _{t}^{m}}{\partial \varvec{\theta } }-\frac{\partial \nu _{t}}{\partial \varvec{\theta } } \right) \right\| >\delta \sqrt{n}\right)\le & {} {P}\left( \epsilon _{m} \left| \sum _{t=1}^{n} W_{t}^{m}\right| >\delta \sqrt{n}\right) \\\le & {} \frac{\epsilon _{m}^{2}}{\delta ^{2}n}\sum _{t=1}^{n} \text{ E } \left| W_{t}^{m}\right| ^{2}\le C\epsilon _{m}^{2}\rightarrow 0, \end{aligned}$$

as $m \rightarrow \infty$, for some $\epsilon _{m}$. For the second summand, note that

$$\begin{aligned}&\Bigl |\Bigl (\psi \left( r_t^m\right) - E\left( \psi \left( r_t^m\right) || \mathcal{F}_{t-1}^m \right) \Bigr ) - \Bigl (\psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \Bigr ) \right] \Bigr |\nonumber \\&\quad \le C \Bigl (|r_t^m-r_t| + E\left( |r_t^m-r_t| || \mathcal{F}_{t-1}^m \right) \Bigr ) \end{aligned}$$

(13)

and therefore its expected value tends to 0 by Lemma 6.3. The fact that $E||\partial \nu _t / \partial \varvec{\theta }||^2< \infty$ yields the desired conclusion. $\square$

Proof of Lemma 2.3

Because $S_n(\varvec{\theta })=0$ is an unbiased estimating function, it holds that

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

where $l_t(\varvec{\theta })= \nu _{t}(\varvec{\theta })Y_{t}- \exp ( \nu _{t}(\varvec{\theta }))$, is the logarithm of the conditional probability of $Y_t||\mathcal{F}_{t-1}$ under the Poisson assumption. Then, the matrix $V_n(\varvec{\theta })$ is rewritten in the form

$$\begin{aligned} V_n(\varvec{\theta })=\frac{1}{n} \sum \limits _{t=1}^n E\left( s_t(\varvec{\theta }) \frac{\partial l_t(\varvec{\theta })}{\partial \varvec{\theta }} || \mathcal{F}_{t-1} \right) \end{aligned}$$

and the matrix $V_n^m(\varvec{\theta })$ for the perturbed model is defined analogously. We again examine the difference $s_t^m ({\partial l_t^m}/{\partial \theta _{i}}) - s_t ({\partial l_t}/{\partial \theta _i})$. Notice that

$$\begin{aligned} s_t \frac{\partial l_t}{\partial \theta _i}= & {} (m_t-E(m_t|| \mathcal{F}_{t-1}))(Y_t-e^{\nu _t}) \frac{\partial \nu _t}{\partial \theta _i}\\= & {} \left( \psi (r_t) w_t e^{\nu _t/2} \frac{\partial \nu _t}{\partial \theta _i}- E\left( \psi (r_t) w_t e^{\nu _t/2} \frac{\partial \nu _t}{\partial \theta _i} || \mathcal{F}_{t-1} \right) \right) \left( Y_t-e^{\nu _t} \right) \frac{\partial \nu _t}{\partial \theta _i}\\= & {} w_t e^{\nu _t} r_t \Bigl ( \psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \right) \Bigr ) \left( \frac{\partial \nu _t}{\partial \theta _i} \right) ^2. \end{aligned}$$

Then,

$$\begin{aligned} E\left| s_t^m \frac{\partial l_t^m}{\partial \theta _{i}} - s_t \frac{\partial l_t}{\partial \theta _{i}} \right|&\le E\left| e^{\nu _t^m} r_t^m \left( \frac{\partial \nu _t^m}{\partial \theta _{i}} \right) ^2 \Bigl (\left[ \psi (r_t^m) - E\left( \psi (r_t^m) || \mathcal{F}_{t-1}^m \right) \right] \right. \\&\quad \left. - \left[ \psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \right) \right] \Bigr ) \right| \\&\quad + E\left| \Bigl ( e^{\nu _t^m} r_t^m \left( \frac{\partial \nu _t^m}{\partial \theta _{i}} \right) ^2 - e^{\nu _t} r_t \left( \frac{\partial \nu _t}{\partial \theta _{i}} \right) ^2 \Bigr ) \left( \psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \right) \right) \right| . \end{aligned}$$

For the first summand, (13) shows that it tends to zero. We work similarly for the second summand to obtain the desired result. To show that $V_n$ is positive definite it is sufficient to show that $z^T \left( {\partial \nu _t}/{\partial \varvec{\theta }} \right) \left( {\partial \nu _t}/{\partial \varvec{\theta }} \right) ^T z >0$ for any non-zero three dimensional real vector z. If $z^{\prime } {\partial \nu _{t}}/{\partial \varvec{\theta }}=0$, then we obtain that $z^{\prime } (1, \nu _{t-1}, \log (Y_{t-1}+1))^{\prime }=0$. But if the last equation holds, then $z =0$ because $\nu _{t}$ is expressed as a past function of $\log (Y_{t}+1)$ and $Y_{t}$ is non-zero for some t. The same reasoning holds for $V_{n}^{m}$. $\square$

Proof of Lemma 2.4

The first assertion of the Lemma holds by using a LLN. For the second, the Hessian matrix $H_n$ can be represented as

$$\begin{aligned} H_n=\frac{1}{n} \sum \limits _{t=1}^n s_t \frac{\partial l_t}{\partial \varvec{\theta }} = \frac{1}{n} \sum \limits _{t=1}^n \left\{ w_t e^{\nu _t} r_t \left[ \psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \right) \right] \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) ^T \right\} . \end{aligned}$$

The matrix $H_n^m$ for the perturbed model is defined analogously. Examining the difference $H_n^m-H_n$, we obtain that

$$\begin{aligned} H_n^m-H_n= & {} \frac{1}{n} \sum \limits _{t=1}^n w_t \left\{ e^{\nu _t^m} r_t^m \left[ \psi (r_t^m) - E\left( \psi (r_t^m) || \mathcal{F}_{t-1}^m \right) \right] \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) ^T\right. \\&- \left. e^{\nu _t} r_t \left[ \psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \right) \right] \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) ^T \right\} \\= & {} \frac{1}{n} \sum \limits _{t=1}^n w_t \left\{ e^{\nu _t^m} r_t^m \left[ \psi (r_t^m) - E\left( \psi (r_t^m) || \mathcal{F}_{t-1}^m \right) \right] \right. \\&\left. \times \, \left[ \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) ^T - \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) ^T \right] \right. \\&\left. +\,\left( e^{\nu _t^m} r_t^m \left[ \psi (r_t^m) - E\left( \psi (r_t^m) || \mathcal{F}_{t-1}^m \right) \right] - e^{\nu _t} r_t \left[ \psi (r_t) - E\left( \psi (r_t) || \mathcal{F}_{t-1} \right) \right] \right) \right. \\&\left. \times \,\left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) ^T \right\} . \end{aligned}$$

The second term in the above representation tends to zero as $m \rightarrow \infty$ because of the previous Lemma and the fact that $E \left\| \left( {\partial \nu _{t}}/{\partial \varvec{\theta }} \right) \left( {\partial \nu _t}/{\partial \varvec{\theta }} \right) ^T \right\| < \infty$.

For the first term in the representation of $H_n^m-H_n$, we obtain the following

$$\begin{aligned}&P\left( \left\| \sum \limits _{t=1}^n e^{\nu _t^m} r_t^m \left[ \psi (r_t^m) - E\left( \psi (r_t^m) || \mathcal{F}_{t-1}^m \right) \right] \right. \right. \\&\left. \left. \qquad \times \, \left[ \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) ^T -\left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) ^T\right] \right\| \ge \epsilon n \right) \\&\quad \le \frac{1}{\epsilon n } \sum \limits _{t=1}^n E \left\| e^{\nu _t^m} r_t^m \left[ \psi (r_t^m) - E\left( \psi (r_t^m) || \mathcal{F}_{t-1}^m \right) \right] \right. \\&\left. \qquad \times \,\left[ \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t^m}{\partial \varvec{\theta }} \right) ^T -\left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) \left( \frac{\partial \nu _t}{\partial \varvec{\theta }} \right) ^T\right] \right\| \\&\quad \rightarrow 0. \end{aligned}$$

$\square$

Proof of Lemma 2.5

Recall that the components of the MQLE score are given by

$$\begin{aligned} s_{ti}(\varvec{\theta })= & {} m_{ti}(\varvec{\theta })-E({m}_{ti}(\varvec{\theta }) || \mathcal{F}_{t-1}), ~~~\text {where}~~~m_{ti}(\varvec{\theta })\\= & {} \psi (r_t(\varvec{\theta }))w_te^{\nu _t(\varvec{\theta })/2} \frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _i}, ~~~i=1,2,3. \end{aligned}$$

The second derivative of the i-th component of the MQLE score $\partial ^2 s_{ti}(\varvec{\theta })/\partial \theta _k \partial \theta _j$ is given by

$$\begin{aligned} \frac{\partial ^2 s_{ti}(\varvec{\theta })}{\partial \theta _k \partial \theta _j}=\frac{\partial ^2 m_{ti}(\varvec{\theta })}{ \partial \theta _k \partial \theta _j}-E\left( \frac{\partial ^2 m_{ti}(\theta )}{\partial \theta _k \partial \theta _j} || \mathcal{F}_{t-1}\right) , \end{aligned}$$

where

$$\begin{aligned} \frac{\partial ^2 m_{ti}(\varvec{\theta })}{\partial \theta _k \partial \theta _j}= & {} \xi _{1t}(\varvec{\theta }) \frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _k}\frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _j}\frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _i}\\&+\,\xi _{2t}(\varvec{\theta }) \left\{ \frac{\partial ^2 \nu _t(\varvec{\theta })}{\partial \theta _k \partial \theta _j}\frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _i}+\frac{\partial ^2 \nu _t(\varvec{\theta })}{\partial \theta _k \partial \theta _i}\frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _j}+\frac{\partial ^2 \nu _t(\varvec{\theta })}{\partial \theta _j \partial \theta _i}\frac{\partial \nu _t(\varvec{\theta })}{\partial \theta _k} \right\} \\&+\,\xi _{3t}(\varvec{\theta }) \frac{\partial ^3 \nu _t(\varvec{\theta })}{\partial \theta _k \partial \theta _j \partial \theta _i}, \end{aligned}$$

with

$$\begin{aligned} \xi _{1t}(\varvec{\theta })= & {} -\frac{w_{t}}{2} \left\{ \psi '(r_t(\varvec{\theta }))e^{\nu _t(\varvec{\theta })}-\frac{1}{2} \psi ''(r_t(\varvec{\theta }))(Y_t+e^{\nu _t(\varvec{\theta })})(Y_t e^{-\nu _t(\varvec{\theta })/2} + e^{\nu _t(\varvec{\theta })/2}) \right. \\&\left. +\, \frac{1}{2} \psi '(r_t(\varvec{\theta })) (Y_t+e^{\nu _t(\varvec{\theta })})-\frac{1}{2}\psi (r_t(\varvec{\theta })) e^{\nu _t(\varvec{\theta })/2} \right\} ,\\ \xi _{2t} (\varvec{\theta })= & {} - \frac{w_{t}}{2} \left\{ (Y_t+e^{\nu _t(\varvec{\theta })})\psi '(r_t(\varvec{\theta }))- \psi (r_t(\varvec{\theta })) e^{\nu _t(\varvec{\theta })/2} \right\} ,\\ \xi _{3t} (\varvec{\theta })= & {} w_{t} \psi (r_t(\varvec{\theta })) e^{\nu _t(\varvec{\theta })/2}, \end{aligned}$$

Without loss of generality, we only consider derivatives with respect to a. For the derivatives with respect to d and b we use identical arguments. For the derivatives of $\nu _t$ with respect to the parameter a we obtain the following bounds

$$\begin{aligned}&\nu _t \le \mu _{0t}:=b_M \sum \limits _{j=1}^{t-1} a_M^{j} \log (1+Y_{t-j-1})+c_0,~~~\text {where}~~~c_0=d_M/(1-a_M)+\nu _0,\\&\frac{\partial \nu _t}{\partial a} \le \mu _{1t}:=b_M \sum \limits _{j=1}^{t-1} ja_M^{j-1} \log (1+Y_{t-j-1})+c_1,~~~\text {where}~~~c_1=c_0/(1-a_M),\\&\frac{\partial ^2 \nu _t}{\partial a^2} \le \mu _{2t}:=b_M \sum \limits _{j=1}^{t-2} j(j+1)a_M^{j-1} \log (1+Y_{t-j-2})+c_2,~~~\text {where}~~~c_2=2c_0/(1-a_M)^2,\\&\frac{\partial ^3 \nu _t}{\partial a^3} \le \mu _{3t}:=b_M \sum \limits _{j=1}^{t-3} j(j+1)(j+2)a_M^{j-1} \log (1+Y_{t-j-3})+c_3,~~~\text {where}~~~c_3=6c_0/(1-a_M)^3. \end{aligned}$$

With $\theta _i=\theta _j=\theta _k=a$,

$$\begin{aligned}&\Bigl |\frac{\partial ^2 m_{ti}(\varvec{\theta })}{\partial \theta _k \partial \theta _j}-E(\frac{\partial ^2 m_{ti}(\varvec{\theta })}{\partial \theta _k \partial \theta _j}||\mathcal{F}_{t-1})\Bigr |\nonumber \\&\quad < C \Bigl \{ \left| \xi _{1t}(\varvec{\theta })- E(\xi _{1t}(\varvec{\theta })|| \mathcal{F}_{t-1}) \right| \mu _{1t}^3+\left| \xi _{2t}(\varvec{\theta })- E(\xi _{2t}(\varvec{\theta }) || \mathcal{F}_{t-1})\right| \mu _{1t}\mu _{2t}\nonumber \\&\qquad +\,\left| \xi _{3t}(\varvec{\theta })-E(\xi _{3t}(\varvec{\theta })||\mathcal{F}_{t-1}) \right| \mu _{3t} \Bigr \}\nonumber \\&\quad \equiv \tilde{m}_t. \end{aligned}$$

(14)

By defining $\tilde{M}_{n}^{m}$ analogously and working as before the result of the Lemma follows. $\square$

Proof of Theorem 3.1

The first assertion of the Theorem follows from arguments given in Francq and Zakoïan (2010, Prop. 8.3). For the second assertion of the Theorem, we consider the difference

$$\begin{aligned} ST_n^m(\tilde{\varvec{\theta }}_n)-ST_n(\tilde{\varvec{\theta }}_n)=\left[ S_n^{m(2)}(\tilde{\varvec{\theta }}_n)\right] ^2 \frac{\tilde{\sigma }-\tilde{\sigma ^m}}{\tilde{\sigma ^m} \tilde{\sigma }} + \frac{\left[ S_n^{m(2)}(\tilde{\varvec{\theta }}_n)\right] ^2-\left[ S_n^{(2)}(\tilde{\varvec{\theta }}_n)\right] ^2}{\tilde{\sigma }} \end{aligned}$$

The above representation is composed of the following differences, $W_{22}^m-W_{22}$, $W_{12}^m-W_{12}$, $W_{21}^m-W_{21}$, $W_{11}^m-W_{11}$, ${V_{11}^m}^{-1}V_{12}^m-{V_{11}}^{-1}V_{12}$, $V_{21}^m{V_{11}^m}^{-1}-V_{21}V_{11}^{-1}$ and ${V_{11}^m}^{-1}V_{12}^m-V_{11}^{-1}V_{12}$ which all converge to zero as results of Lemmas 2.1 and 2.3. $\square$