A new estimation for INAR(1) process with Poisson distribution

Abstract

The first-order Poisson autoregressive model may be suitable in situations where the time series data are non-negative integer valued. In this article, we propose a new parameter estimator based on empirical likelihood. Our results show that it can lead to efficient estimators by making effective use of auxiliary information. As a by-product, a test statistic is given, testing the randomness of the parameter. The simulation values show that the proposed test statistic works well. We have applied the suggested method to a real count series.

Appendix

Proof of Theorem 1 The proof follows essentially the lines of the classical proof of Owen (2001) for the i.i.d. case, which may be applied when the following three conditions are satisfied. Let $S({{\varvec{\theta }}})=\frac{1}{n}\sum _{t=1}^n {\varvec{m}}_t({{\varvec{\theta }}})m_t({{\varvec{\theta }}})'.$

(C1) $S({{\varvec{\theta }}}_0) \rightarrow {\varvec{\Sigma }}$ in probability and ${\varvec{\Sigma }}$ positive definite.

(C2) $m_n^\star =\max _{1\le t\le n}\Vert {\varvec{m}}_t({{\varvec{\theta }}}_0)\Vert =o_p(n^{1/2})$.

(C3) $\frac{1}{n}\sum _{t=1}^n \Vert {\varvec{m}}_t({{\varvec{\theta }}}_0)\Vert ^3=o_p(n^{1/2}),$ where ${{\varvec{\theta }}}_0$ is the true value of the parameter.

For the first condition, let $\mathcal F_n=\sigma (X_0,X_1,\ldots ,X_n), A_n=\sum _{t=1}^n m_{1t}(\theta _0)=\sum _{t=1}^n(X_t-\phi _0 X_{t-1}-\lambda _0),A_0=0,$ then,

$$\begin{aligned} E(A_n|\mathcal F_{n-1})=A_{n-1}+E(X_n-\phi _0 X_{n-1}-\lambda _0)=A_{n-1}, \end{aligned}$$

so, $\{A_n, \mathcal F_n, n\ge 0\}$ is a martingale. For $E|X_t|^2<\infty ,$ $\{(X_t-\phi _0 X_{t-1}-\lambda _0)^2, n\ge 1\}$ is square integrable. By the strictly stationary and ergodic theorem,

$$\begin{aligned} \displaystyle \frac{\;1\;}{\;n\;}\sum \limits _{t=1}^n(X_t-\phi _0 X_{t-1}-\lambda _0)^2 {\mathop {\longrightarrow }\limits ^{a.s}} E(X_1-\phi _0 X_0-\lambda _0)^2. \end{aligned}$$

Similarly, we can proof that $B_n=\sum _{t=1}^nm_{2t}({{\varvec{\theta }}}_0), C_n=\sum _{t=1}^n m_{3t}(\theta _0)$ are martingale. Then the first condition holds. To check the second condition, we note that $E\{{\varvec{m}}_t({{\varvec{\theta }}}_0)/{\varvec{m}}_t({{\varvec{\theta }}}_0)\}<\infty ,$ then, $\sum _{n=1}^\infty P({\varvec{m}}_t({{\varvec{\theta }}}_0)'$ ${\varvec{m}}_t({{\varvec{\theta }}}_0)>n)<\infty .$ By the Borel-Cantelli lemma, $\Vert {\varvec{m}}_t({{\varvec{\theta }}}_0)\Vert >\sqrt{n}$ happens with probability 1 only for finitely many n, since $\{X_t\}$ is a strictly stationary process. This implies that there are only finitely many n for which $m_n^\star >\sqrt{n}.$ Similarly, for any $\varepsilon >0,$ there are only finitely many n for which $m_n^\star >\varepsilon \sqrt{n},$ hence $\lim \sup _{n\rightarrow \infty }m_n^\star /\sqrt{n}\le \varepsilon$ holds with probability 1, so $m_n^\star =o_p(\sqrt{n}).$ From the (C1) and (C2), we have $\frac{\;1\;}{\;n\;}\sum _{t=1}^n \Vert {\varvec{m}}_t({{\varvec{\theta }}}_0)\Vert ^3\le m_n^\star \frac{\;1\;}{\;n\;}\sum _{t=1}^n {\varvec{m}}_t({{\varvec{\theta }}}_0){\varvec{m}}_t({{\varvec{\theta }}}_0)' = o_p(\sqrt{n})O_p(1)=o_p(n^{1/2}).$ See Zhang et al. (2011a) and Zhao and Yu (2016) also.

Proof of Theorem 2 The proof is similar to that of Theorem 6.2 of Zhang et al. (2011a) and that of Theorem 1 of Qin and Lawless (1994). Let

$$\begin{aligned} Q_{1n}({{\varvec{\theta }}},{\varvec{\beta }})=\frac{1}{n}\sum \limits _{t=1}^n \frac{{\varvec{m}}_t({{\varvec{\theta }}})}{1+{\varvec{\beta }}'{\varvec{m}}_t({{\varvec{\theta }}})},\quad Q_{2n}({{\varvec{\theta }}},{\varvec{\beta }})=\frac{1}{n}\sum \limits _{t=1}^n \frac{{\varvec{\beta }}({{\varvec{\theta }}})}{1+{\varvec{\beta }}'{\varvec{m}}_t({{\varvec{\theta }}})}\left( \frac{\partial {\varvec{m}}_t({{\varvec{\theta }}})}{\partial {{\varvec{\theta }}}}\right) '. \end{aligned}$$

Expanding $Q_{1n}(\check{{\varvec{\theta }}}, \check{{\varvec{\beta }}}),Q_{2n}(\check{{\varvec{\theta }}},\check{{\varvec{\beta }}})$ at $({{\varvec{\theta }}}_0,0),$ we have

$$\begin{aligned} 0= & {} Q_{1n}(\check{{\varvec{\theta }}},\check{{\varvec{\beta }}})=Q_{1n}({{\varvec{\theta }}}_0,0)+\frac{\partial Q_{1n}({{\varvec{\theta }}}_0,0)}{\partial {{\varvec{\theta }}}}(\check{{\varvec{\theta }}}-{{\varvec{\theta }}}_0)+\frac{\partial Q_{1n}({{\varvec{\theta }}}_0,0)}{\partial {\varvec{\beta }}'}(\check{{\varvec{\beta }}}-0)+o_p(\delta _n), \\ 0= & {} Q_{2n}(\check{{\varvec{\theta }}},\check{{\varvec{\beta }}})=Q_{2n}({{\varvec{\theta }}}_0,0)+\frac{\partial Q_{2n}({{\varvec{\theta }}}_0,0)}{\partial {{\varvec{\theta }}}}(\check{{\varvec{\theta }}}-{{\varvec{\theta }}}_0)+\frac{\partial Q_{2n}({{\varvec{\theta }}}_0,0)}{\partial {\varvec{\beta }}'}(\check{{\varvec{\beta }}}-0)+o_p(\delta _n), \end{aligned}$$

where $\delta _n=\Vert \check{{\varvec{\theta }}}-{{\varvec{\theta }}}_0\Vert +\Vert \check{{\varvec{\beta }}}\Vert =O_p(n^{-1/2}).$ So, we have

$$\begin{aligned} S_n\left( \begin{array}{c} \check{{\varvec{\beta }}}\\ \check{{\varvec{\theta }}}-{{\varvec{\theta }}}_0 \end{array}\right) =\left( \begin{array}{c} -Q_{1n}({{\varvec{\theta }}}_0,0)+o_p(\delta _n)\\ o_p(\delta _n) \end{array}\right) . \end{aligned}$$

where

$$\begin{aligned} S_n=\left( \begin{array}{cc} \frac{\partial Q_{1n}}{\partial {\varvec{\beta }}'} &{}\frac{\partial Q_{1n}}{\partial {{\varvec{\theta }}}}\\ \frac{\partial Q_{2n}}{\partial {\varvec{\beta }}'} &{}0 \end{array}\right) _{({{\varvec{\theta }}}_0,0)} \rightarrow \left( \begin{array}{cc} -E(m_{t}m_{t}') &{}E\left( \frac{\partial {\varvec{m}}_t}{\partial {{\varvec{\theta }}}}\right) \\ E\left( \frac{\partial {\varvec{m}}_t}{\partial {{\varvec{\theta }}}}\right) ' &{}0 \end{array}\right) = \left( \begin{array}{cc} -{\varvec{\Sigma }}&{}{\varvec{\Omega }}\\ {\varvec{\Omega }}'&{}0 \end{array}\right) . \end{aligned}$$

since $\{X_t\}$ is strictly stationary and ergodic and note that $\sqrt{n} Q_{1n}({{\varvec{\theta }}}_0,0)=\frac{1}{\sqrt{n}}\sum _{t=1}^n m_{t}({{\varvec{\theta }}}_0){\mathop {\longrightarrow }\limits ^{d}} N(0,{\varvec{\Sigma }}).$ Then, we have

$$\begin{aligned} \sqrt{n}(\check{{\varvec{\theta }}}-{{\varvec{\theta }}}_0)=({\varvec{\Omega }}'{\varvec{\Sigma }}^{-1}{\varvec{\Omega }})^{-1}{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1}\sqrt{n} Q_{1n}({{\varvec{\theta }}}_0,0)+o_p(1). \end{aligned}$$

The proof of the theorem is completed.

Proof of Theorem 3: The proof is similar to that of Theorem 2 and Corollary 4 of Qin and Lawless (1994) and will be only sketched here.

Note that

$$\begin{aligned} \ell (\check{{\varvec{\theta }}},\check{{\varvec{\beta }}}) =\sum \limits _{t=1}^n\log [1+\check{{\varvec{\beta }}}{\varvec{m}}_t(\check{{{\varvec{\theta }}}})] =\frac{n}{2}\left[ \frac{1}{n}\sum \limits _{t=1}^n {{\varvec{m}}_t({{\varvec{\theta }}}_0)'}\right] {\varvec{A}} \left[ \frac{1}{n}\sum \limits _{t=1}^n {{\varvec{m}}_t({{\varvec{\theta }}}_0)}\right] +o_p(1), \end{aligned}$$

where ${\varvec{A}}={\varvec{\Sigma }}^{-1}[{\varvec{I}}-{\varvec{\Omega }}[{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1}{\varvec{\Omega }}]^{-1}{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1}].$ Thus,

$$\begin{aligned} 2\ell (\check{{\varvec{\theta }}})&=[{\varvec{\Sigma }}^{-1/2}\sqrt{n} Q_{1n}({{\varvec{\theta }}}_0,0)]' [{\varvec{I}}-{\varvec{\Sigma }}^{-1/2}{\varvec{\Omega }}[{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1}{\varvec{\Omega }}]^{-1}{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1/2}] [{\varvec{\Sigma }}^{-1/2}\\&\quad \sqrt{n} Q_{1n}({{\varvec{\theta }}}_0,0)]+o_p(1), \end{aligned}$$

Note that $({\varvec{\Sigma }})^{-1/2}{\sqrt{n}}Q_{1n}({{\varvec{\theta }}}_0,0)$ converges to a standard multivariate normal distribution and the matrix ${\varvec{I}}-{\varvec{\Sigma }}^{-1/2}{\varvec{\Omega }}[{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1}{\varvec{\Omega }}]^{-1}{\varvec{\Omega }}'{\varvec{\Sigma }}^{-1/2}$ is symmetric and idempotent, with trace equal to 1. Hence the test statistic $2\ell (\check{{{\varvec{\theta }}}})$ converges to $\chi ^2(1).$