High-Order Self-excited Threshold Integer-Valued Autoregressive Model: Estimation and Testing

Abstract

Motivated by the need of modeling and inference for high-order integer-valued threshold time series models, this paper introduces a pth-order two-regime self-excited threshold integer-valued autoregressive (SETINAR(2, p)) model. Basic probabilistic and statistical properties of the model are discussed. The parameter estimation problem is addressed by means of conditional least squares and conditional maximum likelihood methods. The asymptotic properties of the estimators, including the threshold parameter, are obtained. A method to test the nonlinearity of the underlying process is provided. Some simulation studies are conducted to show the performances of the proposed methods. Finally, an application to the number of people suffering from meningococcal disease in Germany is provided.

8 Appendix

We first introduce some notations. These notations will be used in the following proofs. Let ${\varvec{A}}= {\left( a_{ij} \right) }_{m \times n}$ be an $m \times n$ matrix. We write ${\varvec{A}} \ge {\varvec{0}}$ if $a_{ij} \ge 0$ for all $1 \le i \le m$ and $1 \le j \le n$. Meanwhile, we write ${\varvec{A}} < \infty$ if $a_{ij} < \infty$ for all $1\le i \le m$ and $1 \le j \le n$. Let ${\varvec{X}}$ and ${\varvec{Y}}$ be two $n \times 1$ column vectors. We write ${\varvec{X}} \ge {\varvec{Y}}$ if ${\varvec{X}} - {\varvec{Y}} \ge {\varvec{0}}$.

Proof of Theorem 2.1

Let $L^2:= \{ X|E(X^2) < + \infty \}$. Define the scalar product operation on $L^2$ as $( X,Y) = E(XY)$, then $L^2$ is a Hilbert space.

Let us introduce a sequence $\{ X_t^{(n)}\}_{n \in \mathbb {Z}}$,

$$\begin{aligned} X_t^{(n)} = \left\{ \begin{matrix} 0, &{} n < 0, \\ Z_{t}, &{} n=0, \\ \phi _{t,1} \circ X_{t-1}^{(n-1)} + \phi _{t,2} \circ X_{t-2}^{(n-2)} + \cdots +\phi _{t,p} \circ X_{t-p}^{(n-p)} + Z_t, &{} n > 0, \\ \end{matrix} \right. \end{aligned}$$

(8.1)

where $\phi _{t,i}: = \alpha _{1,i} I_{t-d,1} + \alpha _{2,i}I_{t-d,2}$, $i = 1,2, \ldots , p$, $1 \le d \le p$, $I_{t-d,1}:= I\{ W_{t,d} \le r \}$, $I_{t-d,2}:= I\{ W_{t,d} > r \}$, $W_{t,d} \in L^2$, and $Cov( X_s^{(n)}, Z_t ) = 0$ for any n and $s < t$.

Step 1 Existence.

(A1) $X_t^{(n)} \in {L^2}$.

Let ${{\varvec{X}}_t^{(n)} = {( {X_t^{(n)},X_{t - 1}^{(n - 1)}, \ldots ,X_{t - p + 1}^{(n - p + 1)}} )}^{\textsf {T}}}$, ${\varvec{Z}}_t = ( Z_t,0$, $\ldots ,0 )^{\textsf {T}}$. Denote by ${\varvec{\varSigma }}_z = E({\varvec{Z}}_t {\varvec{Z}}_t^{\textsf {T}} )$, ${\varvec{\mu }} _{z} = {( {\lambda ,0, \ldots ,0} )^{\textsf {T}} }$, where $E({Z_t}) = \lambda$. Denote by ${\varvec{I}}_j=diag(I_{t - d,j},\ldots ,I_{t - d,j})_{p\times p}$ ($j=1,2$) a p-dimensional diagonal matrix, further denote

$$\begin{aligned} {\varvec{A}} _j = \left( \begin{matrix} {{\alpha _{j,1}}} &{} \cdots &{} {{\alpha _{j,p- 1}}} &{} {{\alpha _{j,p}}} \\ 1 &{} \cdots &{} 0 &{} 0 \\ \vdots &{} \ddots &{} \vdots &{} \vdots \\ 0 &{} \cdots &{} 1 &{} 0 \ \end{matrix} \right) _{p \times p}, \quad j = 1,2. \end{aligned}$$

Then, we have ${\varvec{X}}_t^{(n)} = ({\varvec{A}}_1 {\varvec{I}}_1 + {\varvec{A}}_2 {\varvec{I}}_2 ) \circ {\varvec{X}}_{t - 1}^{(n - 1)} + {\varvec{Z}}_t$. With this representation, we have

$$\begin{aligned} \begin{aligned} E{\varvec{X}}_t^{(n)}&\le {\varvec{A}}_{\max }E({\varvec{X}}_{t - 1}^{(n - 1)} ) + {\varvec{\mu }} _{z} \ \le {\varvec{A}}_{\max }^2 E({\varvec{X}}_{t - 2}^{(n - 2)}) + {\varvec{A}}_{\max } {\varvec{\mu }} _{z} + {\varvec{\mu }} _{z} \\&\le \cdots \le {\varvec{A}}_{\max }^n {\varvec{\mu }} _{z} + {\varvec{A}}_{\max }^{n - 1} {\varvec{\mu }} _{z} + \cdots + {\varvec{A}}_{\max } {\varvec{\mu }}_{z} + {\varvec{\mu }} _{z} \\&= ({\varvec{I}} - {\varvec{A}}_{\max }^{n + 1}){({\varvec{I}} - {\varvec{A}}_{\max })^{ - 1}}{\varvec{\mu }} _{z} < \infty . \\ E\left( {\varvec{X}}_t^{(n)}{\varvec{X}}{_t^{(n)}}^{\textsf {T}}\right)&= E\left[ {(({\varvec{A}}_1 {\varvec{I}}_1 + {\varvec{A}}_2 {\varvec{I}}_2) \circ {\varvec{X}}_{t - 1}^{(n - 1)} + {\varvec{Z}}_t)(({\varvec{A}}_1 {\varvec{I}}_1 + {\varvec{A}}_2 {\varvec{I}}_2) \circ {\varvec{X}}_{t - 1}^{(n - 1)} + {\varvec{Z}}_t)^{\textsf {T}}} \right] \\&\le E\left[ {({\varvec{A}}_{\max } \circ {\varvec{X}}_{t - 1}^{(n - 1)})({\varvec{A}}_{\max } \circ {\varvec{X}}_{t - 1}^{(n - 1)})^{\textsf {T}}} \right] + {\varvec{A}}_{\max }E({\varvec{X}}_{t - 1}^{(n - 1)}){\varvec{\mu }} _{z}^{\textsf {T}} \\&\quad + {\varvec{\mu }} _{z}E\left( {{\varvec{X}}_{t - 1}^{(n - 1)}}^{\textsf {T}}\right) {\varvec{A}}_{\max }^{\textsf {T}} + {\varvec{\varSigma }}_z = {\varvec{A}}_{\max }E\left( {{\varvec{X}}_{t - 1}^{(n - 1)} {{\varvec{X}}{_{t - 1}^{(n - 1)}}}^{\textsf {T}}} \right) {\varvec{A}}_{\max }^{\textsf {T}} \\&\quad + {\varvec{B}}_{n - 1} + {\varvec{A}}_{\max } E({\varvec{X}}_{t - 1}^{(n - 1)}){\varvec{\mu }} _{z}^{\textsf {T}} + {\varvec{\mu }} _{z} E\left( {{\varvec{X}}_{t - 1}^{(n - 1)}}^{\textsf {T}}\right) {\varvec{A}}_{\max }^{\textsf {T}} + {\varvec{\varSigma }}_z, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} {\varvec{B}}_{n - 1} = {\left[ \begin{matrix} {\sum \limits _{i = 1}^p {{\alpha _{\max ,i}}\left( {1 - {\alpha _{\max ,i}}} \right) E\left( {X_{t - i}^{(n - i)}} \right) } } &{} {\varvec{0}} \\ {\varvec{0}} &{} {\varvec{0}} \\ \end{matrix} \right] _{p \times p}}. \end{aligned}$$

Repeating the above recurrence n times, we can show that $E\left| {X_t^{(n)}}\right| ^2 < \infty$, which implies $X_t^{(n)} \in {L^2}$.

(A2) $X_t^{(n)}$ is a Cauchy sequence.

Let $U(n,t,k) = \left| {X_t^{(n)} - X_t^{(n - k)}} \right|$, $k = 1,2, \ldots$, note that $U(n,t,k) > 0$, we have

$$\begin{aligned} E U(n,t,k)&= E\left| {\sum _{i = 1}^p {\left( \sum _{j=1}^2\alpha _{j,i} I_{t - d,j}\right) \circ X_{t - i}^{(n - i)}} - \sum _{i = 1}^p {\left( \sum _{j=1}^2 \alpha _{j,i}{I_{t - d,j}} \right) \circ X_{t - i}^{(n - k - i)}} } \right| \\&\le E\sum _{i = 1}^p {\left| {\left( \sum _{j=1}^2 {\alpha _{j,i}}{I_{t - d,j}} \right) \circ X_{t - i}^{(n - i)} - \left( \sum _{j=1}^2 {{\alpha _{j,i}}{I_{t - d,j}} } \right) \circ X_{t - i}^{(n - k - i)}} \right| } \\&= E\sum _{i=1}^p{\left| \sum _{j=1}^2 {I_{t - d,j}\sum _{m=1}^{X_{t - i}^{(n-i)}} {U_{m,t}(\alpha _{j,i} )} -\sum _{j=1}^2 {I_{t - d,j}}\sum _{m = 1}^{X_{t - i}^{(n - k - i)}} {{U_{m,t}}\left( {{\alpha _{j,i}}} \right) } } \right| } \\&\le E\sum _{i=1}^p \left| \sum _{j=1}^2 I_{t-d,j}\sum _{m=1}^{\left| {X_{t-i}^{(n-i)}-X_{t-i}^{(n-k-i)}}\right| } {{U_{m,t}}\left( {{\alpha _{j,i}}} \right) } \right| \\&= E\sum _{i = 1}^p {\left| {\left( \sum _{j=1}^2 {{\alpha _{j,i}}{I_{t - d,j}} } \right) \circ \left| {X_{t - i}^{(n - i)} - X_{t - i}^{(n - k - i)}} \right| } \right| } \\&\le \sum _{i = 1}^p {{\alpha _{\max ,i}}E\left| {X_{t - i}^{(n - i)} - X_{t - i}^{(n - k - i)}} \right| } \\&= \sum _{i = 1}^p {{\alpha _{\max ,i}}E\left( {U(n - i,t - i,k)} \right) }. \\ \end{aligned}$$

In addition, we consider the vector

$$\begin{aligned} {\varvec{U}}_{t,k}^{(n)} = {\left( {U(n,t,k),U(n - 1,t - 1,k), \ldots ,U(n - p + 1,t - p + 1,k)} \right) ^{\textsf {T}}}. \end{aligned}$$

It is easy to see that $E{\varvec{U}}_{t,k}^{(n)} \le {\varvec{A}}_{\max }E{\varvec{U}}_{t - 1,k}^{(n - 1)} \le \cdots \le {\varvec{A}}_{\max }^n E{\varvec{U}}_{t - n,k}^{(0)} = {\varvec{A}}_{\max }^n {\varvec{\mu }} _{z}.$ Since the eigenvalues of ${\varvec{A}}_{\max }$ are all in the unit circle, we can draw a conclusion

$$\begin{aligned} \mathop {\lim }\limits _{n \rightarrow \infty } {\varvec{U}}_{t,k}^{(n)} = 0. \end{aligned}$$

In a similar way, we have

$$\begin{aligned}&E\left( {{\varvec{U}}_{t,k}^{(n)} {\varvec{U}}{{_{t,k}^{(n)}}^{\textsf {T}}}} \right) \le {\varvec{A}}_{\max }E\left( {{\varvec{U}}_{t - 1,k}^{(n - 1)} {\varvec{U}}{{_{t - 1,k}^{(n - 1)}}^{\textsf {T}}}} \right) {\varvec{A}}_{\max }^{\textsf {T}} {\left( \begin{matrix} {\sum \limits _{i = 1}^p {\beta _i EU(n - i,t - i,k)} } &{} {\varvec{0}} \\ {\varvec{0}} &{} {\varvec{0}} \\ \end{matrix} \right) } \\&\quad \le {\varvec{A}}_{\max }E\left( {{\varvec{U}}_{t - 1,k}^{(n - 1)}{\varvec{U}}{{_{t - 1,k}^{(n - 1)}}^{\textsf {T}}}} \right) {\varvec{A}}_{\max }^{\textsf {T}} + \textrm{diag}\left( {{\varvec{B}}{\varvec{A}}_{\max }^{n - 1} {\varvec{\mu }} _{z}} \right) \\&\quad \le {\varvec{A}}_{\max }\left( {{\varvec{A}}_{\max } E\left( {{\varvec{U}}_{t - 2,k}^{(n - 2)}{\varvec{U}}{{_{t - 2,k}^{(n - 2)}}}^{\textsf {T}}} \right) {\varvec{A}}_{\max }^{\textsf {T}} + \textrm{diag}\left( {{\varvec{B}}{\varvec{A}}_{\max }^{n - 2}{\varvec{\mu }} _{z}} \right) } \right) {\varvec{A}}_{\max }^{\textsf {T}} \\&\quad + \textrm{diag}\left( {{\varvec{B}}{\varvec{A}}_{\max }^{n - 1}{\varvec{\mu }} _{z}} \right) \le \cdots \le \sum \limits _{j = 0}^{n - 1} {{\varvec{A}}_{\max }^j \textrm{diag}\left( {{\varvec{B}}{\varvec{A}}^{(n - j - 1)}{\varvec{\mu }} _{z}} \right) {{\left( {\varvec{A}}_{\max }^j \right) }^{\textsf {T}} }} \\&\quad + {\varvec{A}}_{\max }^n \left[ \begin{matrix} {\lambda + {\lambda ^2}} &{} {\varvec{0}} \\ {\varvec{0}} &{} {\varvec{0}} \\ \end{matrix} \right] {\left( {\varvec{A}}_{\max }^n \right) ^{\textsf {T}} }, \end{aligned}$$

where $\beta _i={\alpha _{\max ,i}}\left( {1 - {\alpha _{\max ,i}}} \right)$ and

$$\begin{aligned} {\varvec{B}} = {\left( \begin{matrix} \beta _1 &{} \cdots &{} \beta _p \\ 0 &{} \cdots &{} 0 \\ \vdots &{} \ddots &{} \vdots \\ 0 &{} \cdots &{} 0 \\ \end{matrix} \right) _{p \times p}}. \end{aligned}$$

Since the eigenvalues of ${\varvec{A}}_{\max }$ are all in the unit circle, the second term of the above formula converges to 0. The upper bound of the first term is

$$\begin{aligned} \mathop {\max }\limits _{1 \le j \le p} \left\{ {{\alpha _{\max ,j}}\left( {1 - {\alpha _{\max ,j}}} \right) } \right\} \lambda S(n), \end{aligned}$$

where $S( n) = \sum _{j = 0}^{n - 1} {{\varvec{1}}^{\textsf {T}}{\varvec{A}}_{\max }^{n - j - 1}{\varvec{1}}{\varvec{A}}_{\max }^j ({\varvec{A}}_{\max }^j)^{\textsf {T}}}$, ${\varvec{1}} = {\left( {1, \ldots ,1} \right) ^{\textsf {T}} }$. According to Lemma 3.1 in Latour [13], the sequence S(n) converges to 0. Therefore, $\{X_t^{(n)}\}$ is a Cauchy sequence. Thus, there exists a random variable $X_t$ in $L^2$ such that ${X_t} = \mathop {\lim }\limits _{n \rightarrow \infty } X_t^{(n)}$.

(A3) $\left\{ {X_t}\right\}$ satisfies the SETINAR(2, p) process.

Let $W_{t,d}$ be $X_{t-d}$ ($1 \le d \le p$) and substitute it into (8.1). Then, we have

$$\begin{aligned} \begin{aligned}&E{\left| {\left( {{\alpha _{1,i}}{I_{t - d,1}} + {\alpha _{2,i}}{I_{t - d,2}}} \right) \circ X_{t - i}^{(n - i)} - \left( {{\alpha _{1,i}}{I_{t - d,1}} + {\alpha _{2,i}}{I_{t - d,2}}} \right) \circ {X_{t - i}}} \right| ^2} \\&\quad = E \left| \sum _{j=1}^2 I_{t - d,j}\sum _{m = 1}^{X_{t - i}^{(n - i)}} U_{m,t} ( {\alpha _{j,i}}) - \sum _{j=1}^2 I_{t - d,j}\sum _{m = 1}^{X_{t - i} } U_{m,t} ( {\alpha _{j,i}}) \right| ^2 \\&\quad \le E{\left| \sum _{j=1}^2 {{I_{t - d,j}}\sum _{m = 1}^{\left| {X_{t - i}^{(n - i)} - {X_{t - i}}}\right| } {{U_{m,t}}( {{\alpha _{j,i}}} )} } \right| ^2} \\&\quad \le {\alpha _{\max ,i}}\left( {1 - {\alpha _{\max ,i}}} \right) E\left| {X_{t - i}^{(n - i)} - {X_{t - i}}} \right| + \alpha _{\max ,i}^2 E{\left| {X_{t - i}^{(n - i)} - {X_{t - i}}} \right| ^2},\quad i = 1, \ldots ,p. \end{aligned} \end{aligned}$$

$X_t^{(n)}\buildrel {{L_2}} \over \longrightarrow {X_t}$, where $\buildrel {{L_2}} \over \longrightarrow$ denotes convergence in ${L^2}$. Thus,

$$\begin{aligned} E{\left| {\left( {{\alpha _{1,i}}{I_{t - d,1}} + {\alpha _{2,i}}{I_{t - d,2}}} \right) \circ X_{t - i}^{(n - i)} - \left( {{\alpha _{1,i}}{I_{t - d,1}} + {\alpha _{2,i}}{I_{t - d,2}}} \right) \circ {X_{t - i}}} \right| ^2} \rightarrow 0, ~ n \rightarrow \infty . \end{aligned}$$

Note that $X_t^{(n)}$ converges to $X_t$ implying $\mathop {\lim }\limits _{n \rightarrow \infty } \textrm{Cov}( {X_s^{(n)},{Z_t}} ) = \textrm{Cov}\left( {{X_s},{Z_t}} \right)$. $\textrm{Cov}\left( {X_s^{(n)},{Z_t}} \right) = 0$ implies $\textrm{Cov}\left( {{X_s},{Z_t}} \right) = 0$ for $s < t$. Therefore, the process $\left\{ {X_t} \right\}$ satisfies the equation of SETINAR(2, p) process.

Step 2: Uniqueness.

Suppose there is another process $\left\{ {X_t^ * } \right\}$ such that $X_t^{(n)}\buildrel {{L_2}} \over \longrightarrow {X_t^ * }$. According to the Hölder inequality, we have

$$\begin{aligned} \begin{aligned} E\left| {{X_t} - X_t^*} \right| \le {\left( {E{{\left| {{X_t} - X_t^{(n)}} \right| }^2}} \right) ^{{1 \over 2}}}{\left( {E{{\left| {X_t^ * - X_t^{(n)}} \right| }^2}} \right) ^{{1 \over 2}}} \rightarrow 0. \\ \end{aligned} \end{aligned}$$

Therefore, $E\left| {{X_t} - X_t^*} \right| =0$, which implies ${X_t} = X_t^*$ a.s.

Step 3 Strict stationarity.

Since $X_t^{(n)} = 0$ for $n < 0$, recursively solving (8.1), regardless of t, there exists a function ${F}\left( \cdot \right)$ such that

$$\begin{aligned} \begin{aligned} X_t^{(n)}&= {F}\left( {\varvec{\alpha }}_1,{\varvec{\alpha }}_2,{Z_t},{Z_{t - 1}}, \ldots ,{Z_{t - n}} \right) \\&\mathop = \limits ^d {F}\left( {{\varvec{\alpha }}_1,{\varvec{\alpha }}_2,{Z_n},{Z_{n - 1}}, \ldots ,{Z_0}} \right) , \end{aligned} \end{aligned}$$

where $\mathop = \limits ^d$ stands for equal in distribution. Therefore, the distribution of $X_t^{(n)}$ depends only on n, not on t. Then, for any n and T, $( {X_0^{(n)},X_1^{(n)},\ldots ,X_T^{(n)}} )$ and $( {X_k^{(n)},X_{k + 1}^{(n)}, \ldots ,X_{k + T}^{(n)}} )$ are identically distributed for any k. Since $X_t^{(n)}\buildrel {{L_2}} \over \longrightarrow {X_t}$, then $X_t^{(n)}\buildrel {{P}} \over \longrightarrow {X_t}$. Thus, for any real number ${a_i}$’s, we have $\sum \nolimits _{i = 0}^T {{a_i}X_{t + i}^{(n)}} \buildrel P \over \longrightarrow \sum \nolimits _{i = 0}^T {{a_i}{X_{t + i}}}$. By the Cramer–Wold device, we have

$$\begin{aligned} \begin{aligned}&\left( {X_0^{(n)},X_1^{(n)}, \ldots ,X_T^{(n)}} \right) \buildrel L \over \longrightarrow \left( {{X_0},{X_1}, \ldots ,{X_T}} \right) , \\&\left( {X_k^{(n)},X_{k + 1}^{(n)}, \ldots ,X_{k + T}^{(n)}} \right) \buildrel L \over \longrightarrow \left( {{X_k},{X_{k + 1}}, \ldots ,{X_{k + T}}} \right) . \\ \end{aligned} \end{aligned}$$

Since $\left( {X_0^{(n)},X_1^{(n)}, \ldots ,X_T^{(n)}} \right) \mathop = \limits ^d \left( {X_k^{(n)},X_{k + 1}^{(n)}, \ldots ,X_{k + T}^{(n)}} \right)$ for every k, $\left( {{X_0},{X_1}, \ldots ,{X_T}} \right)$ and $( {X_k},{X_{k + 1}}, \ldots$, ${X_{k + T}})$ have the same distribution, which implies that $\left\{ {X_t} \right\}$ is a strictly stationary process.

Step 4 Ergodicity.

Let $\left\{ {Y(t)} \right\}$ be all the counting series of ${\phi _{t,1}} \circ {X_{t - 1}} + {\phi _{t,2}} \circ {X_{t - 2}} + \cdots + {\phi _{t,p}} \circ {X_{t - p}}$ in (2.1), then $\left\{ {Y(t)} \right\}$ is a time series. Let $\sigma (X)$ be a $\sigma$-field generated by the random variable X. According to (2.1), we obtain $\sigma ({X_t},{X_{t + 1}}, \ldots ) \subset \sigma ({Z_t},Y(t),{Z_{t + 1}},Y(t + 1), \ldots )$ and $\bigcap \limits _{t = 1}^{ + \infty } {\sigma ({X_t},{X_{t + 1}}, \ldots )} \subset \bigcap \limits _{t = 1}^{ + \infty } {\sigma ({Z_t},Y(t),{Z_{t + 1}},Y(t + 1), \ldots )}.$ Because $\left\{ {\left( {{Z_t},Y(t)} \right) } \right\}$ is an independent sequence, Kolmogorov’s 0–1 law implies that for any event $A \in \bigcap \limits _{t = 1}^{ + \infty } {\sigma ({Z_t},Y(t),{Z_{t + 1}},Y(t + 1), \ldots )}$, we have $P(A) = 0$ or $P(A) = 1$, so the tail event of $\sigma$-field of $\left\{ {X_t} \right\}$ only contains the set with measure of 0 or 1. We know from Wang [23] that $\left\{ {X_t} \right\}$ is ergodic. $\square$

Proof of Theorem 3.2

Let $h_t ({\varvec{\phi }})= -B_t ({\varvec{\phi }})$. The proof will be done in three steps.

Step 1: We show that $E(B_t ({\varvec{\phi }}))$ is continuous in ${\varvec{\phi }}$, so that $E (h_t ({\varvec{\phi }}))$ is also continuous in ${\varvec{\phi }}$. For any ${\varvec{\phi }} \in \Theta \times \mathbb {N}_{0}$, let $V _\eta ( {\varvec{\phi }} )= B ( {{\varvec{\phi }}, \eta } )$ be an open ball centered at ${\varvec{\phi }}$ with radius $\eta$ ($\eta < 1$). We go on to show the following property:

$$\begin{aligned} E\left( \sup \limits _{{\varvec{\phi }}^{'} \in V _\eta ( {\varvec{\phi }} ) }\left| B_t ({\varvec{\phi }})- B_t ({\varvec{\phi }}^{'})\right| \right) \rightarrow 0, \quad as \quad \eta \rightarrow 0. \end{aligned}$$

To see this, observe that

$$\begin{aligned} \left| B_t ({\varvec{\phi }})- B_t ({\varvec{\phi }}^{'})\right| =&\left| \left( X_t - \sum _{i=1}^p\sum _{j=1}^2 \alpha _{j,i}I_{t - d,j}X_{t-i}-\lambda \right) ^2\right. \\&- \left. \left( X_t - \sum _{i=1}^p\sum _{j=1}^2 \alpha _{j,i}^{'}I_{t - d,j}X_{t-i} - \lambda ^{'} \right) ^2 \right| \\ =&\left| \left( 2 X_t - \sum _{i=1}^p {\sum _{j=1}^2 { \left( {\alpha _{j,i}}+{\alpha _{j,i}^{'}}\right) {I_{t - d,j}}{X_{t-i}}}}- \lambda -\lambda ^{'} \right) \right| \\&\times \left| \sum _{i=1}^p {\sum _{j=1}^2 { \left( {\alpha _{j,i}^{'}}-{\alpha _{j,i}}\right) {I_{t - d,j}}{X_{t-i}}}}+\lambda ^{'}-\lambda \right| \\ \le&\left( 2 X_t + 2 \sum _{i=1}^p {{X_{t-i}}}+ 2 \lambda _{\max } \right) \times \eta \left( \sum \limits _{i=1}^p {{X_{t-i}}} +1\right) \\ \le&2\eta \left( \sum _{i=0}^p {{X_{t-i}}} + C \right) ^2, \end{aligned}$$

where $\lambda _{\max }=\max (\lambda ,\lambda ^{'})$, $C=\max (\lambda _{\max },1)$. Then,

$$\begin{aligned} E\left( \sup \limits _{{\varvec{\phi }}^{'} \in V _\eta ( {\varvec{\phi }} ) }\left| B_t ({\varvec{\phi }})- B_t ({\varvec{\phi }}^{'})\right| \right) \le 2\eta E \left( \sum _{i=0}^p {{X_{t-i}}} + C \right) ^2 \rightarrow 0, \quad \eta \rightarrow 0. \end{aligned}$$

Step 2 We prove that $E_{{\varvec{\phi }}_0}\left[ h_t ({\varvec{\phi }})- h_t ({\varvec{\phi }}_0)\right] < 0$, or equivalently, $E_{{\varvec{\phi }}_0}\left[ B_t ({\varvec{\phi }})-B_t ({\varvec{\phi }}_0)\right] > 0$ for any ${\varvec{\phi }}\ne {\varvec{\phi }}_0$. Denote by $\nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}):=g( {{\varvec{\phi }}_0,X_{t-1},\ldots ,X_{t-p}})- g( {{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}})$. With the fact that the true value of ${\varvec{\phi }}$ is ${\varvec{\phi }}_0$, we have

$$\begin{aligned} E_{{\varvec{\phi }}_0}[ B_t ({\varvec{\phi }})] =&E_{{\varvec{\phi }}_0} [ \left( X_t - g\left( {{\varvec{\phi }}, {X_{t - 1}},\ldots ,X_{t-p}} \right) \right) ^2 ] \\ =&E_{{\varvec{\phi }}_0} [ \left( X_t - g( {{\varvec{\phi }}_0,X_{t-1},\ldots ,X_{t-p}})+ \nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}) \right) ^2 ] \\ =&E_{{\varvec{\phi }}_0} [ ( X_t - g( {{\varvec{\phi }}_0, {X_{t - 1}},\ldots ,X_{t-p}} ))^2 +( \nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}) )^2 \\&+ 2( X_t - g( {{\varvec{\phi }}_0, {X_{t - 1}},\ldots ,X_{t-p}} ))( \nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}))] \\ =&E_{{\varvec{\phi }}_0} [B_t ({\varvec{\omega }}_0)]+ \mathbb {I}+\mathbb{I}\mathbb{I}, \end{aligned}$$

where $\mathbb {I}= E_{{\varvec{\phi }}_0} ( \nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}))^2 > 0 ~ (\text{ by } \text{ Assumption } \text{(A2) })$, and

$$\begin{aligned} \mathbb{I}\mathbb{I}&=2E_{{\varvec{\phi }}_0}[( X_t - g( {{\varvec{\phi }}_0, X_{t - 1},\ldots ,X_{t-p}} ))( \nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}))] \\&=2E_{{\varvec{\phi }}_0}[ E_{{\varvec{\phi }}_0}[\left( X_t -g( {{\varvec{\phi }}_0, X_{t - 1},\ldots ,X_{t-p}} ) \right) \\&( \nabla g({\varvec{\phi }}_0,{\varvec{\phi }},X_{t-1},\ldots ,X_{t-p}))|X_{t - 1},\ldots ,X_{t-p}] ] \\&= 2 E_{{\varvec{\phi }}_0} (0)=0. \end{aligned}$$

Thus, we have $E_{{\varvec{\phi }}_0}[B_t ({\varvec{\phi }})- B_t ({\varvec{\phi }}_0)] > 0$.

Step 3: We prove that $\hat{{\varvec{\phi }}}_{CLS} \in V$ holds for arbitrary (small) open neighborhood V of ${{\varvec{\phi }}}_{0}$, which means $\hat{{\varvec{\phi }}}_{CLS}$ is a consistent estimator of ${\varvec{\phi }}_0$. This proof will be conducted in the similar arguments given in Yang et al. [28]. Therefore, we omit the details. $\square$

Proof of Theorems 3.4 and 3.5

As discussed in Franke and Seligmann [7], we only have to check that the conditions (C1–C6) hold, which implies the regularity conditions of Theorems 2.1 and 2.2 in Billingsley [2] hold.

(C1) The set $\left\{ {m:~P\left( {{Z_t} = m} \right) = f(m,\lambda ) = {{{\lambda ^m}} \over {m!}}{e^{ - \lambda }} > 0} \right\}$ does not depend on $\lambda$;

(C2) $E\left[ {Z_t^3} \right] = {\lambda ^3} + 3{\lambda ^2} + \lambda < \infty$;

(C3) $P\left( {{Z_t} = m} \right)$ is three times continuously differentiable with respect to $\lambda$;

(C4) For any $\lambda ^{'} \in B$, where B is an open subset of $\mathbb {R}_+$, there exists a neighborhood U of $\lambda ^{'}$ such that:

1. ${\sum _{k = 0}^\infty {\mathop {\sup }_{\lambda \in U} f(k,\lambda )} < \infty }$,

2. ${\sum _{k = 0}^\infty {\mathop {\sup }_{\lambda \in U} \left| {{{\partial f(k,\lambda )} \over {\partial \lambda }}} \right| } < \infty }$,

3. ${\sum _{k = 0}^\infty {\mathop {\sup }_{\lambda \in U} \left| {{{{\partial ^2}f(k,\lambda )} \over {\partial {\lambda ^2}}}} \right| } < \infty }$;

(C5) For any $\lambda ^{'}\in B$, there exists a neighborhood U of $\lambda ^{'}$ and the sequences ${\psi _1}(n) = \textrm{const}1 \cdot n$, ${\psi _{11}}(n) = \textrm{const}2 \cdot {n^2}$ and ${\psi _{111}}(n) = \textrm{const}3 \cdot {n^3}$, with suitable constants const1, const2, const3 and $n \ge 0$ such that $\forall \lambda \in U$ and $\forall m \le n$, with $P\left( {{Z_t}} \right)$ nonvanishing $f(m,\lambda )$,

• (i) $\left| {{{\partial f(m,\lambda )} \over {\partial \lambda }}} \right| \le {\psi _1}(n)f(m,\lambda )$, $\left| {{{{\partial ^2}f(m,\lambda )} \over {\partial {\lambda ^2}}}} \right| \le {\psi _{11}}(n)f(m,\lambda )$, $\left| {{{{\partial ^3}f(m,\lambda )} \over {\partial {\lambda ^3}}}} \right| \le {\psi _{111}}(n)f(m,\lambda )$,

and with respect to the stationary distribution of the process $\{ {X_t} \}$,

• (ii) $E[ {\psi _1^3( {{X_1}} )}] < \infty$, $E[ {{X_1}{\psi _{11}}( {{X_2}} )}] < \infty$, $E[ {{\psi _1}( {{X_1}} ){\psi _{11}}( {{X_2}} )}] < \infty$, $E[ {{\psi _{111}}( {{X_1}} )}] < \infty$.

(C6) Let ${\varvec{I}}\left( {\varvec{\theta }} \right) = {\left( {{\sigma _{ij}}} \right) _{\left( {2p + 1} \right) \times \left( {2p + 1} \right) }}$ denote the Fisher information matrix, i.e.,

$$\begin{aligned}&\sigma _{ij}= E\left( {\partial \over \partial \theta _i} \log {P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})}{\partial \over \partial \theta _j}\log {P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})}\right) ,\\&\quad ~i,j\in \{1,\ldots ,2p+1\}, \end{aligned}$$

where $P(X_t,\ldots ,X_{t-p};{\varvec{\theta }}):=P(X_t =x_t|X_{t-1}=x_{t-1}, \ldots ,X_{t-p}=x_{t - p})$ denotes the transition probability given in (2.2), $\theta _i$ denotes the ith component of ${\varvec{\theta }}$, and ${\varvec{I}}( {\varvec{\theta }} )$ is nonsingular.

Franke and Seligmann [7] have proved that the above conditions (C1)–(C4) all hold. Therefore, for the SETINAR(2, p) model it is only necessary to verify conditions (C5) and (C6) also hold. It is easy to verify that statement (i) in (C5) holds trivially by the properties of Poisson distribution. Condition (C2) implies $E X_t^3<\infty$ with respect to the stationary distribution of $\{ {X_t} \}$. Therefore, statement (ii) in (C5) also holds.

In the following, we prove condition (C6) holds. Then, we need to check the following statements are all true.

(S1) $E{\left| {{\partial \over {\partial {\alpha _{j,i}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| ^2} < \infty$, $j = 1,2$; $\quad i = 1,2, \ldots p$;

(S2) $E{\left| {{\partial \over {\partial \lambda }}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| ^2} < \infty$;

(S3) $E\left| {{\partial \over {\partial {\alpha _{j,i}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }}){\partial \over {\partial \lambda }}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| < \infty$, $j = 1,2$; $i = 1,2, \ldots p$;

(S4) $E\left| {{\partial \over {\partial {\alpha _{j,i}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }}){\partial \over {\partial {\alpha _{j,k}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| < \infty$, $j = 1,2$, $i,k \in \{1,\ldots p\}$;

(S5) $E\left| {{\partial \over {\partial {\alpha _{1,i}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }}){\partial \over {\partial {\alpha _{2,k}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| < \infty$, $i,k \in \{1,\ldots p\}$.

We shall first prove statement (S1). Recall that

$$\begin{aligned}&P(X_t,\ldots ,X_{t-p};{\varvec{\theta }}) =P\left( {{X_t} = {x_t}|{X_{t - 1}} = {x_{t - 1}}, \ldots ,{X_{t - p}} = {x_{t - p}}} \right) \\&= \sum _{{i_1} = 0}^{\min ( {{x_{t - 1}},{x_t}} )} {\sum _{{i_2} = 0}^{\min ( {{x_{t - 2}},{x_t} - {i_1}} )} { \ldots \sum _{{i_p} = 0}^{\min \left( {{x_{t - p}},{x_t} - \sum _{s = 1}^{p - 1} {{i_s}} } \right) } {\left( \begin{matrix} {{x_{t - 1}}} \\ {{i_1}} \\ \end{matrix} \right) \left( \begin{matrix} {{x_{t - 2}}} \\ {{i_2}} \\ \end{matrix} \right) \cdots \left( \begin{matrix} {{x_{t - p}}} \\ {{i_p}} \\ \end{matrix} \right) } } } \nonumber \\&\quad \times \prod _{k=1}^p \left( \sum _{j=1}^2 I_{t-d,j}\alpha _{j,k}^{i_k}(1-\alpha _{j,k})^{x_{t-k}-i_k} \right) \times {f_z}\left( {{x_t} - \sum _{s = 1}^p {{i_s}} } \right) \nonumber \\&= \sum _{{i_1} = 0}^{\min ( {{x_{t - 1}},{x_t}} )} {\sum _{{i_2} = 0}^{\min ( {{x_{t - 2}},{x_t} - {i_1}} )} { \cdots \sum _{{i_p} = 0}^{\min \left( {{x_{t - p}},{x_t} - \sum _{s = 1}^{p - 1} {{i_s}} } \right) } {\left( \begin{matrix} {{x_{t - 1}}} \\ {{i_1}} \\ \end{matrix} \right) \left( \begin{matrix} {{x_{t - 2}}} \\ {{i_2}} \\ \end{matrix} \right) \cdots \left( \begin{matrix} {{x_{t - p}}} \\ {{i_p}} \\ \end{matrix} \right) } } } \nonumber \\&\quad \times \left( \sum _{j=1}^2 I_{t-d,j} h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} ) \right) \times {f_z}\left( {{x_t} - \sum _{s = 1}^p {{i_s}} } \right) , \nonumber \end{aligned}$$

(8.2)

where $h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} ) = \prod _{k = 1}^p {\alpha _{j,k}^{{i_k}}{{\left( {1 - {\alpha _{j,k}}} \right) }^{{x_{t - k}} - {i_k}}}}$, ${\varvec{i}}=(i_1,\ldots ,i_p)$, $j = 1,2$. Thus, for $j = 1,2$ and $k=1,2,\ldots ,p$, we have

$$\begin{aligned} {{\partial h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} )} \over {\partial {\alpha _{j,k}}}} = \left( { {{{{i_k}} \over {{\alpha _{j,k}}}}} - {{{{x_{t - k}} - {i_k}} \over {1 - {\alpha _{j,k}}}}} } \right) h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} ), \end{aligned}$$

yielding

$$\begin{aligned}{} & {} - {x_{t-k} \over 1 - \alpha _{j,k}} h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} ) \le {{\partial h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} )} \over {\partial {\alpha _{j,k}}}} \\{} & {} \quad \le {x_{t - k} \over \alpha _{j,k}} h( {{\varvec{\alpha }}_{j},{x_{t - 1},\ldots ,x_{t-p}},{{\varvec{i}}}} ). \end{aligned}$$

Together with (8.2), we have

$$\begin{aligned} - {{\sum _{i = 1}^p {{X_{t - i}}} } \over {1 - \max _{1 \le k \le p} \left\{ {{\alpha _{\max ,k}}} \right\} }} \le {{\partial \log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \over {\partial {\alpha _{j,k}}}} \le {{\sum _{i= 1}^p {{X_{t - i}}} } \over {\min _{1 \le k \le p}\left\{ {{\alpha _{\min ,k}}} \right\} }}, \end{aligned}$$

(8.3)

where $\alpha _{\max ,k}:=\max \{\alpha _{1,k},\alpha _{2,k}\}$, and $\alpha _{\min ,k}:=\min \{\alpha _{1,k},\alpha _{2,k}\}$, $k=1,2,\ldots ,p$. Therefore, we have

$$\begin{aligned} E{\left| {{\partial \over {\partial {\alpha _{j,k}}}}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| ^2}< C \cdot EX_1^2 < \infty , \end{aligned}$$

for some suitable constant C.

Next, we will prove statement (S2). By (C5), we know that $\left| {{{\partial f(m,\lambda )} \over {\partial \lambda }}} \right| \le {\psi _1}(n)f(m,\lambda )$ holds for $m \le n$, which means that

$$\begin{aligned} \left| {\partial {f_z}\left( {{x_t} - \sum \limits _{s = 1}^p {{i_s}} } \right) } \bigg / {\partial \lambda }\right| \le \psi _1(x_t){f_z}\left( {{x_t} - \sum \limits _{s = 1}^p {{i_s}} } \right) . \end{aligned}$$

(8.4)

This implies ${\left| {{\partial \over {\partial \lambda }}\log P(X_t,\ldots ,X_{t-p};{\varvec{\theta }})} \right| ^2}< E\psi _1^2\left( {{X_1}} \right) < \infty .$

Lastly, by (8.3), (8.4) and (C5), we can conclude that statements (S3), (S4) and (S5) hold. Therefore, the Fisher information matrix ${\varvec{I}}\left( {\varvec{\theta }} \right)$ is well defined. Finally, some elementary but tedious calculations show that (C6) is satisfied too. $\square$