SYMARMA: a new dynamic model for temporal data on conditional symmetric distribution

Abstract

Gaussian models of time series, ARMA, have been widely used in the literature. Benjamin et al. (J Am Stat Assoc 98:214–223, 2003) extended these models to the exponential family distributions. Also in that direction, Rocha and Cribari-Neto (Test 18:529–545, 2009) proposed a time series model for the class of beta distributions. In this paper, we develop an autoregressive and moving average symmetric model, named SYMARMA, which is a dynamic model for random variables belonging to the class of symmetric distributions including also a set of regressors. We discuss methods for parameter estimation, hypothesis testing and forecasting. In particular, we provide closed-form expressions for the score function and Fisher information matrix. Robust study is presented based on influence function. We conduct simulation studies to evaluate the consistency and asymptotic normality of the conditional maximum likelihood estimator for the model parameters. An application with real data is presented and discussed.

Appendices

Appendix 1: Proof of Theorem 1

Proof

Let $\Phi (B) = 1 - \phi _1B - \cdots - \phi _pB^p$ the autoregressive polynomial, $\Theta (B) = 1 + \theta _1B + \cdots + \theta _qB^q$ the moving averages polynomial and $B^ky_t=y_{t-k}$ the lag operator and $\Psi (B) =\sum \nolimits _{i=0}^{\infty }\psi _iB^i=\Theta (B)\Phi (B)^{-1}$, $\psi _0=1$ and assuming that $\Phi (B)$ is invertible. The SYMARMA model can be rewritten as

$$\begin{aligned} \Phi (B)(y_t-\mathbf{x}_{t}^{\top }\varvec{\beta }) = \Theta (B)r_t \end{aligned}$$

and, since $\Phi (B)$ is invertible,

$$\begin{aligned} y_t = \mathbf{x}_{t}^{\top }\varvec{\beta }+ \Psi (B)r_t, \end{aligned}$$

Therefore, assuming that $\Phi (B)$ is invertible, the marginal mean of $y_t$ of the SYMARMA model is given by

$$\begin{aligned} \mathrm {E}(y_t) = \mathbf{x}_t^{\top }\varvec{\beta }. \end{aligned}$$

$\square$

Appendix 2: Proof of Theorem 2

Proof

Let $Y_t = \mu _t + r_t$ where $r_t's$ are uncorrelated residuals with mean zero. We have

$$\begin{aligned} \mathrm {Var}(r_t)= & {} \mathrm {E}(r_t^2) = \mathrm {E}(\mathrm {E}(r_t^2|\mathrm {\mathcal{F}_{t-1}})) = \mathrm {E}(\mathrm {Var}(r_t|\mathrm {\mathcal{F}_{t-1}})) = \mathrm {E}(\mathrm {Var}(Y_t - \mu _t|\mathrm {\mathcal{F}_{t-1}}))\\= & {} \mathrm {E}(\mathrm {Var}(y_t|\mathrm {\mathcal{F}_{t-1}})) = \mathrm {E}(\xi \varphi ) = \xi \varphi . \end{aligned}$$

Note that $\mu _t$ given in (2) is $\mathcal{F}_{t-1}$-measurable. Therefore, the marginal variance of $Y_t$, $\mathrm {Var}(Y_t)$, is given by

$$\begin{aligned} \mathrm {Var}(Y_t)= & {} \mathrm {Var}(\mathbf{x}_{t}^{\top }\varvec{\beta }+ \Psi (B)r_t) = \mathrm {Var}(\Psi (B)r_t) = \mathrm {E}[(\Psi (B)r_t)^2] \\= & {} \sum \limits _{i=0}^\infty \sum \limits _{j=0}^\infty \psi _i\psi _j\mathrm {E}(r_{t-i}r_{t-j}) = \sum \limits _{i=0}^\infty \psi _i^2\mathrm {E}(r_{t-i}^2) = \sum \limits _{i=0}^\infty \psi _i^2\mathrm {Var}(r_{t-i})\\= & {} \xi \varphi \sum \limits _{i=0}^\infty \psi _i^2. \end{aligned}$$

$\square$

Appendix 3: Proof of Theorem 3

Proof

By Theorem 1 and 2 we have

$$\begin{aligned} \mathrm {E}(Y_t) = x_t^\top \mathbf{\beta } \ \ \mathrm {e} \ \ \ \mathrm {Var}(Y_t) = \xi \varphi \sum _{i=0}^{\infty }\psi _i^2.\quad \mathrm{And} \end{aligned}$$

$$\begin{aligned} \mathrm {Cov}(Y_t,Y_{t-k})= & {} \mathrm {Cov}(\mathbf{x}_{t}^{\top }\varvec{\beta }+ \sum _{i=0}^{\infty }\psi _ir_{t-i}, \mathbf{x}_{t-k}^{\top }\varvec{\beta }+ \sum _{i=0}^{\infty }\psi _ir_{t-k-i}) \\= & {} \mathrm {Cov}(\psi _0r_t + \psi _1r_{t-1} + \cdots , \psi _0r_{t-k} + \psi _1r_{t-k-1} + \cdots )\\= & {} \mathrm {Var}(r_t)\sum _{i=0}^{\infty }\psi _i\psi _{i+k} = \xi \varphi \sum _{i=0}^{\infty }\psi _i\psi _{i+k} \end{aligned}$$

where $\Psi _0=1$.

$$\begin{aligned} \mathrm {Corr}(Y_t,Y_{t-k})= & {} \dfrac{\mathrm {Cov}(Y_t,Y_{t-k})}{\sqrt{\mathrm {Cov}(Y_t,Y_{t})\mathrm {Cov}(Y_{t-k},Y_{t-k})}} = \dfrac{\xi \varphi \sum _{i=0}^{\infty }\psi _i\Psi _{i+k}}{\xi \varphi \sum _{i=0}^{\infty }\psi _i^2} \\= & {} \dfrac{\sum _{i=0}^{\infty }\psi _i\psi _{i+k}}{\sum _{i=0}^{\infty }\psi _i^2}. \end{aligned}$$

$\square$

Expected conditional Fisher information matrix

The elements of the expected conditional Fisher information matrix, $\mathbf{K}$, are obtained from expression

$$\begin{aligned} \mathbf{K}_{\omega _r\omega _s} = -\mathrm {E}\left[ \dfrac{\partial ^2 \ell ({\varvec{\delta }},\varphi )}{\partial \omega _r\partial \omega _s}\left| \mathcal{F}_{t-1}\right. \right] = \mathrm {E}\left[ \dfrac{\partial \ell ({\varvec{\delta }},\varphi )}{\partial \omega _r}\dfrac{\partial \ell ({\varvec{\delta }},\varphi )}{\partial \omega _s}\left| \mathcal{F}_{t-1}\right. \right] , \end{aligned}$$

where $\omega _r$ and $\omega _s$ are model parameters and $\ell$ is the logarithm of the conditional likelihood function.

Under suitable regularity conditions

$$\begin{aligned} \mathrm {E}\left( \dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \mu _t}\left| \right. \mathcal{F}_{t-1}\right)= & {} \mathrm {E}\left( \dfrac{\partial \mathrm {log}f(y_t|\mathcal{F}_{t-1})}{\partial \mu _t}\right) \nonumber \\= & {} \displaystyle \int _{-\infty }^{\infty }\dfrac{\partial \mathrm {log}f(y_t|\mathcal{F}_{t-1})}{\partial \mu _t}f(y_t|\mathcal{F}_{t-1})d\mu _t\nonumber \\= & {} \displaystyle \int _{-\infty }^{\infty }\left( \dfrac{1}{f(y_t|\mathcal{F}_{t-1})}\dfrac{\partial f(y_t|\mathcal{F}_{t-1})}{\partial \mu _t}\right) f(y_t|\mathcal{F}_{t-1})d\mu _t\nonumber \\= & {} \displaystyle \int _{-\infty }^{\infty }\dfrac{\partial f(y_t|\mathcal{F}_{t-1})}{\partial \mu _t}d\mu _t = \dfrac{\partial }{\partial \mu _t}\displaystyle \int _{-\infty }^{\infty }f(y_t|\mathcal{F}_{t-1})d\mu _t = 0.\nonumber \\ \end{aligned}$$

(10)

From some algebraic manipulations, we also obtain that

$$\begin{aligned} \dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \mu _t} = -\dfrac{2}{\sqrt{\varphi }}W_g(u_t)z_t, \end{aligned}$$

with $z_t = \sqrt{u_t} = (y_t-\mu _t)/\sqrt{\varphi }$. Therefore, using the results in (10), we have

$$\begin{aligned} \mathrm {E}\left( W_g(u_t)z_t|\mathcal{F}_{t-1}\right) = 0. \end{aligned}$$

(11)

Furthermore, the expressions

$$\begin{aligned} \dfrac{\partial \mu _t}{\partial \beta _l} = x_{tl} - \sum \limits _{i=1}^{p}\phi _ix_{(t-i)l}, \qquad \ \dfrac{\partial \mu _t}{\partial \phi _i} = y_{t-i} - \mathbf{x}^\top _{t-i}\varvec{\beta }\qquad \ \mathrm {and} \qquad \ \dfrac{\partial \mu _t}{\partial \theta _j} = y_{t-j} - \mu _{t-j} \end{aligned}$$

(12)

are measurable with respect to $\mathcal{F}_{t-1}$.

The $\mathbf{K}_{{\varvec{\delta }}{\varvec{\delta }}}$ matrix elements

$$\begin{aligned}&\mathrm {E}\left( \dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \delta _i}\dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \delta _j}|\mathcal{F}_{t-1}\right) \\&\quad = \mathrm {E}\left[ \left( \dfrac{-2W_g(u_t)}{\sqrt{\varphi }}\dfrac{\partial \mu _t}{\partial \delta _i}z_t\right) \left( \dfrac{-2W_g(u_t)}{\sqrt{\varphi }}\dfrac{\partial \mu _t}{\partial \delta _j}z_t\right) |\mathcal{F}_{t-1}\right] \\&\quad = \dfrac{4}{\varphi }\mathrm {E}\left[ W^2_g(u_t)z_t^2\dfrac{\partial \mu _t}{\partial \delta _i}\dfrac{\partial \mu _t}{\partial \delta _j} |\mathcal{F}_{t-1}\right] \\&\quad = \dfrac{4}{\varphi }\mathrm {E}\left[ W^2_g(u_t)z_t^2|\mathcal{F}_{t-1}\right] \dfrac{\partial \mu _t}{\partial \delta _i}\dfrac{\partial \mu _t}{\partial \delta _j}\\&\quad = \dfrac{4}{\varphi }d_{g}\dfrac{\partial \mu _t}{\partial \delta _i}\dfrac{\partial \mu _t}{\partial \delta _j}, \end{aligned}$$

with $d_{g} = \mathrm {E}\left[ W^2_g(u_t)z_t^2|\mathcal{F}_{t-1}\right]$. So, $d_{g} = \mathrm {E}\left[ W^2_g(U^2)U^2|\mathcal{F}_{t-1}\right]$ with $U \sim S(0,1,g)$.

From the results presented in (12) one can easily find the expressions for the elements of $\mathbf{K}_{{\varvec{\delta }}{\varvec{\delta }}}$.

The $\mathbf{K}_{\varphi \varphi }$ matrix elements

$$\begin{aligned} \mathrm {E}\left( \dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \varphi }\dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \varphi }|\mathcal{F}_{t-1}\right)= & {} \mathrm {E}\left[ \left( -\dfrac{1}{2\varphi } - \dfrac{W_g(u_t)}{\varphi }u_t\right) \left( -\dfrac{1}{2\varphi } - \dfrac{W_g(u_t)}{\varphi }u_t\right) |\mathcal{F}_{t-1}\right] \\= & {} \mathrm {E}\left[ \dfrac{1}{4\varphi ^2} + \dfrac{W_g(u_t)u_t}{\varphi ^2} + \dfrac{W^2_g(u_t)u_t^2}{\varphi ^2} |\mathcal{F}_{t-1}\right] \\= & {} \dfrac{1}{4\varphi ^2} + \dfrac{1}{\varphi ^2}\mathrm {E}\left[ W_g(u_t)u_t|\mathcal{F}_{t-1}\right] + \dfrac{1}{\varphi ^2}\mathrm {E}\left[ W^2_g(u_t)u_t^2|\mathcal{F}_{t-1}\right] \\= & {} \dfrac{1}{4\varphi ^2} + \dfrac{1}{\varphi ^2}\left( -\dfrac{1}{2}\right) + \dfrac{1}{\varphi ^2}f_{g}\\= & {} \dfrac{1}{\varphi ^2}f_{g} - \dfrac{1}{4\varphi ^2} = \dfrac{1}{4\varphi ^2}\left( 4f_{g}-1\right) , \end{aligned}$$

with $f_{g} = \mathrm {E}\left[ W^2_g(u_t)u_t^2|\mathcal{F}_{t-1}\right]$. So, $f_{g} = \mathrm {E}\left[ W^2_g(U^2)U^4|\mathcal{F}_{t-1}\right]$ with $U \sim S(0,1,g)$. From Fang et al. (1990) (p. 94). we have $\mathrm {E}\left[ W_g(u_t)u_t|\mathcal{F}_{t-1}\right] = -1/2$. Therefore,

$$\begin{aligned} \mathbf{K}_{\varphi \varphi }= & {} \sum \limits _{t=m+1}^n\dfrac{1}{4\varphi ^2}\left( 4f_{g}-1\right) = \dfrac{(n-m)}{4\varphi ^2}\left( 4f_{g}-1\right) . \end{aligned}$$

$\mathbf{K}_{{\varvec{\delta }}\varphi }$ matrix

$$\begin{aligned}&\mathrm {E}\left( \dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \delta _i}\dfrac{\partial \ell _t({\varvec{\delta }},\varphi )}{\partial \varphi }|\mathcal{F}_{t-1}\right) \\&\quad = \mathrm {E}\left[ \left( \dfrac{-2W_g(u_t)}{\sqrt{\varphi }}\dfrac{\partial \mu _t}{\partial \delta _i}z_t\right) \left( -\dfrac{1}{2\varphi } - \dfrac{W_g(u_t)}{\varphi }u_t\right) |\mathcal{F}_{t-1}\right] \\&\quad = \mathrm {E}\left[ \dfrac{W_g(u_t)}{\varphi \sqrt{\varphi }}z_t|\mathcal{F}_{t-1}\right] \dfrac{\partial \mu _t}{\partial \delta _i} + \mathrm {E}\left[ \dfrac{2W^2_g(u_t)}{\varphi \sqrt{\varphi }}z_tu_t|\mathcal{F}_{t-1}\right] \dfrac{\partial \mu _t}{\partial \delta _i}\\&\quad = \dfrac{1}{\varphi \sqrt{\varphi }}\left\{ \mathrm {E}\left[ W_g(u_t)z_t|\mathcal{F}_{t-1}\right] + 2\mathrm {E}\left[ W^2_g(u_t)z_tu_t|\mathcal{F}_{t-1}\right] \right\} \dfrac{\partial \mu _t}{\partial \delta _i}\\&\quad = 0. \end{aligned}$$

From Fang et al. (1990) (p. 94) we have $\mathrm {E}\left[ W^2_g(u_t)z_tu_t|\mathcal{F}_{t-1}\right] = 0$ and in addition, $\mathrm {E}\left[ W_g(u_t)z_t|\mathcal{F}_{t-1}\right] =0$ because (11).