Integer-valued Bilinear Model with Dependent Counting Series

Abstract

The present work proposes a new stationary integer-valued bilinear time series model with dependent counting series. The model will enable one to tackle the presence of some correlation between underlying events. The various probabilistic and statistical properties of the model are discussed, unknown parameters are estimated by several methods. Moreover, the performance of the estimation methods is illustrated through a simulation study and an empirical application to two data sets.

Appendices

Appendix A: Proof of Theorem 2.1

To investigate the stationarity of the precess, we follow a similar approach as that of Kim and Park (2008). Let $\{X_{t}^{(n)}, t{\in \mathbb {Z}}\}$ be a sequence of random variables as

$$ X_{t}^{(n)}=\left \{ \begin{array}{cl} 0 & n<0 \\ e_{t} & n=0 \\ a\diamond_{\theta} X_{t-1}^{(n-1)}+b\diamond_{\delta}X_{t-1}^{(n-1)}e_{t-1} +e_{t} & n>0. \end{array} \right. $$

(12)

where sequence $\{X_{s}^{(n)}\}$ is independent of e_t for s < t.

B1

The process $\{X_{t}^{(n)}, t{\in \mathbb {Z}}\}$ is strictly stationary for any $n \in \mathbb {N}$.

The strict stationarity of the process $\{X_{t}^{(n)}, t{\in \mathbb {Z}}\}$ will be deduced by showing that the two vectors $(X_{1}^{(n)},...,X_{k}^{(n)})^{T}$ and $(X_{1+h}^{(n)},...,X_{k+h}^{(n)})^{T}$ are identically distributed. It is clear that process $\{X_{t}^{(n)}, t{\in \mathbb {Z}}\}$ is strictly stationary for n = 0. Now suppose that the process $\{X_{t}^{(m)}, t{\in \mathbb {Z}}\}$ is strictly stationary for all 1 ≤ m ≤ n − 1. Hence for m = n we have

$$ \begin{pmatrix} X_{1}^{(n)} \\ {\vdots} \\ X_{k}^{(n)} \\ \end{pmatrix} = \begin{pmatrix} a\diamond_{\theta} & {\cdots} & 0\diamond_{\theta} \\ {\vdots} & {\ddots} & {\vdots} \\ 0\diamond_{\theta} & {\cdots} & a\diamond_{\theta} \\ & & \end{pmatrix} \begin{pmatrix} X_{0}^{(n-1)} \\ {\vdots} \\ X_{k-1}^{(n-1)} \\ \end{pmatrix} + \begin{pmatrix} b\diamond_{\delta} & {\cdots} & 0\diamond_{\delta} \\ {\vdots} & {\ddots} & {\vdots} \\ 0\diamond_{\delta} & {\cdots} & b\diamond_{\delta} \\ & & \end{pmatrix} \begin{pmatrix} X_{0}^{(n-1)} e_{0} \\ {\vdots} \\ X_{k-1}^{(n-1)} e_{k-1} \\ \end{pmatrix} + \begin{pmatrix} e_{1} \\ {\vdots} \\ e_{k} \\ \end{pmatrix} $$

and

$$ \begin{pmatrix} X_{1+h}^{(n)} \\ {\vdots} \\ X_{k+h}^{(n)} \\ \end{pmatrix} = \begin{pmatrix} a\diamond_{\theta} & {\cdots} & 0\diamond_{\theta} \\ {\vdots} & {\ddots} & {\vdots} \\ 0\diamond_{\theta} & {\cdots} & a\diamond_{\theta} \\ & & \end{pmatrix} \begin{pmatrix} X_{h}^{(n-1)} \\ {\vdots} \\ X_{k+h-1}^{(n-1)} \\ \end{pmatrix} + \begin{pmatrix} b\diamond_{\delta} & {\cdots} & 0\diamond_{\delta} \\ {\vdots} & {\ddots} & {\vdots} \\ 0\diamond_{\delta} & {\cdots} & b\diamond_{\delta} \\ & & \end{pmatrix} \begin{pmatrix} X_{h}^{(n-1)} e_{h} \\ {\vdots} \\ X_{k+h-1}^{(n-1)} e_{k+h-1} \\ \end{pmatrix} + \begin{pmatrix} e_{h+1} \\ {\vdots} \\ e_{k+h} \\ \end{pmatrix} $$

According to the induction hypothesis, the random vectors $(X_{1}^{(n)},...,X_{k}^{(n)})^{T}$ and $(X_{1+h}^{(n)},...,X_{k+h}^{(n)})^{T}$ are identically distributed.

B2

The sequence $\{X_{t}^{(n)}, t{\in \mathbb {Z}}\}$ belongs to the space $\pounds ^{2}=\{X|EX^{2}<\infty \}$.

Let $\mu _{n}=E(|X_{t}^{(n)}|)$. According to Eq. 12, we have

$$ \mu_{n}\leqslant |a|E|X_{t-1}^{(n-1)}|+|b|E|X_{t-1}^{(n-1)}e_{t-1}|+\mu . $$

(13)

As $E\left \vert X_{t}^{(n)}-e_{t}\right \vert =E|a\mathbf {\diamond _{\theta }} X_{t-1}^{(n-1)}+b\mathbf {\diamond _{\delta } }X_{t-1}^{(n-1)}e_{t-1}| \leqslant E|X_{t}^{(n)}|+\mu ,$ we have

$$ E|X_{t-1}^{(n-1)}e_{t-1}|=E\left\vert \left( a\mathbf{\diamond_{\theta}} X_{t-2}^{(n-2)}+b\mathbf{\diamond_{\delta} }X_{t-2}^{(n-2)}e_{t-2}+e_{t-1} \right) e_{t-1}\right\vert \leqslant \mu_{e}E|X_{t-1}^{(n-1)}|+\mu_{e^{2}}+\mu^{2}. $$

(14)

Substitution of Eqs. 14 into 13 gives

$$ \begin{array}{@{}rcl@{}} \mu_{n} &\leqslant &(|a|+|b|\mu )\mu_{n-1}+|b|(\mu_{e^{2}}+\mu^{2})+\mu \\ &\leqslant &(|a|+|b|\mu)^{n}\mu +[|b|(\mu_{e^{2}}+\mu^{2})+\mu ]{\sum}_{i=0}^{n-1}(|a|+|b|\mu )^{i}<\infty . \end{array} $$

Now we show that $E|X_{t}^{(n)}|^{2}\leqslant \infty$.

$$ \begin{array}{@{}rcl@{}} E|X_{t}^{(n)}|^{2} &\leqslant &E|a\mathbf{\diamond_{\theta}} X_{t-1}^{(n-1)}|^{2}+E|b\mathbf{\diamond_{\delta} } X_{t-1}^{(n-1)}e_{t-1}|^{2}+2E|(a\mathbf{\diamond_{\theta} } X_{t-1}^{(n-1)})(b\mathbf{\diamond_{\delta}}X_{t-1}^{(n-1)}e_{t-1})|+ \\ &&2E|(a\mathbf{\diamond_{\theta} }X_{t-1}^{(n-1)})e_{t}|+2E|(b\mathbf{ \diamond_{\delta}}X_{t-1}^{(n-1)}e_{t-1})e_{t}|+\mu_{e^{2}} \\ &\leqslant &|a \theta| E|X_{t-1}^{(n-1)}|^{2}+a(1-\theta) E|X_{t-1}^{(n-1)}|+|b \delta| E|X_{t-1}^{(n-1)}e_{t-1}|^{2}+b(1-\delta) E|X_{t-1}^{(n-1)}e_{t-1}|+ \\ &&2|ab|E|(X_{t-1}^{(n-1)})^{2}e_{t-1}|+2\mu |a|E|X_{t-1}^{(n-1)}|+2\mu |b|E|X_{t-1}^{(n-1)}e_{t-1}|+\mu_{e^{2}}. \end{array} $$

(15)

On the other hand, after some calculations we have

$$ E|(X_{t-1}^{(n-1)})^{2}e_{t-1}|\leqslant \mu E|X_{t-1}^{(n-1)}|^{2}+2(\mu^{2}+\mu_{e^{2}})E|X_{t-1}^{(n-1)}|+L_{1}(e) $$

and

$$ E|X_{t-1}^{(n-1)}e_{t-1}|^{2}\leqslant \mu_{e^{2}}E|X_{t-1}^{(n-1)}|^{2}+2(\mu \mu_{e^{2}}+\mu_{e^{3}})E|X_{t-1}^{(n-1)}|+L_{2}(e) $$

where $L_{1}(e)=\mu (3\mu _{e^{2}}+2\mu ^{2})+2\mu _{e^{2}}\mu +\mu _{e^{3}}$ and $L_{2}(e)=\mu _{e^{2}}(3\mu _{e^{2}}+2\mu ^{2})+2\mu _{e^{3}}\mu +\mu _{e^{4}}$. By substituting these equations and Eqs. 14 into 15, we have

$$ E|X_{t}^{(n)}|^{2}\leqslant K_{1}(e)E|X_{t-1}^{(n-1)}|^{2}+K_{2}(e)\mu_{n-1}+K_{3}(e) $$

(16)

where $K_{1}(e)=(|a \theta |+|b\delta |\mu _{e^{2}} )+2ab\mu$, $K_{2}(e)=|a(1-\theta ) |+|b(1-\delta ) |\mu +2\mu _{e^{2}}\mu +\mu _{e^{3}}+4|a||b|(\mu _{e^{2}}+\mu ^{2})+2\mu (|a|+|b|\mu )$ and $K_{3}(e)=|b(1-\delta ) |(\mu _{e^{2}}+\mu ^{2})+|b|^{2}L_{2}(e)+2|a||b|L_{1}(e)+2\mu |b|(\mu _{e^{2}}+{\mu _{e}^{2}})$.

Repeating (16), we have

$$ E|X_{t}^{(n)}|^{2}\leqslant {K_{1}^{n}}(e)\mu_{e^{2}}+K_{2}(e)\sum\limits_{i=0}^{n-1}{K_{1}^{n}}(e)\mu_{n-(i+1)}+K_{3}(e)\sum\limits_{i=0}^{n-1}{K_{1}^{i}}(e). $$

Therefore under strictly stationary condition, it can be concluded that $E|X_{t}^{(n)}|^{2}\leqslant \infty ,$ n ≥ 1. Hence $\{X_{t}^{(n)}\}\in \pounds ^{2}$.

B3

The sequence $\{X_{t}^{(n)}\}$ is Cauchy.

Let $\psi (t,n,m)=|X_{t}^{(n)}-X_{t}^{(n-m)}|,$ m = 1, 2,.... Using the definition of $\{X_{t}^{(n)}\}$, we have

$$ \begin{array}{@{}rcl@{}} E\psi (t,n,m) &\leq &E|a\mathbf{\diamond_{\theta} } (X_{t-1}^{(n)-1}-X_{t-1}^{(n-m-1)})|+E|b\mathbf{\diamond_{\delta}} (X_{t-1}^{(n-1)}-X_{t-1}^{(n-m-1)})e_{t-1}| \\ &\leq &(|a|+|b|\mu )E\psi (t-1,n-1,m) \\ &\leq &(|a|+|b|\mu )^{n}E\psi (t-n,0,m)=(|a|+|b|\mu )^{n}\mu \end{array} $$

As $n\rightarrow \infty$, under strictly stationary condition, Eψ(t,n,m) converges to 0. Similarly, we have

$$ \begin{array}{@{}rcl@{}} E\psi^{2}(t,n,m) &\leq &(|a \theta|+|b\delta|(\mu_{e^{2}})+2|a b \mu|)E\psi^{2}(t-1,n-1,m)+ (a+b \mu )E\psi (t-1,n-1,m) \\&& -(a \theta+b \delta \mu )E\psi (t-1,n-1,m) \\ &&{\vdots} \\ &\leq &(|a \theta|+|b\delta|(\mu_{e^{2}})+2|a b \mu|)^{n}E\psi^{2}(t-n,0,m)+ \\ &&(a (1-\theta)+b (1-\delta) \mu )\sum\limits_{i=1}^{n}(|a \theta|+|b\delta|(\mu_{e^{2}}) +2|a b \mu|)^{i-1}E\psi (t-i,n-i,m). \end{array} $$

As $n\rightarrow \infty$, under given condition, it easily can be seen that Eψ²(t,n,m) converges to 0. So $X_{t}^{\left (n\right ) }$ is a Cauchy sequence. Suppose $\lim _{n\rightarrow \infty }X_{t}^{(n)}=X_{t}$, then X_t ∈£².

B4

The process {X_t} satisfies (1)

Since $X_{t}^{(n)}\rightarrow ^{\pounds ^{2}}X_{t}$, so

$$ E|a\mathbf{\diamond_{\theta} }X_{t-1}^{(n-1)}-a\mathbf{\diamond_{\theta} } X_{t-1}|^{2}=|a \theta| E|X_{t-1}^{(n-1)}-X_{t-1}|^{2}+a (1-\theta) E|X_{t-1}^{(n-1)}-X_{t-1}| $$

and

$$ E|b\mathbf{\diamond_{\delta} }(X_{t-1}^{(n-1)}-X_{t-1})e_{t-1}|^{2}=|b \delta| E|(X_{t-1}^{(n-1)}-X_{t-1})e_{t-1}|^{2}+b(1-\delta) E|(X_{t-1}^{(n-1)}-X_{t-1})e_{t-1}| $$

converge to zero. Hence process {X_t} satisfies (1).

B5

Uniqueness

Suppose that another solution $X_{t}^{\ast }$ of Eq. 1 exists. By Minkowski inequality, we have

$$ E^{1/2}(|X_{t}-X_{t}^{\ast }|^{2})\leq E^{1/2}(|X_{t}^{(n)}-X_{t}^{\ast }|^{2})+E^{1/2}(|X_{t}^{(n)}-X_{t}|^{2}), $$

so $X_{t}=X_{t}^{\ast }$ a.s.

B6

Strictly stationarity of the process {X_t}.

As the process $\{X_{t}^{(n)}\}$ is strictly stationary, so $(X_{0}^{(n)},...,X_{k}^{(n)})^{T}$ and $(X_{h}^{(n)},...,X_{k+h}^{(n)})^{T}$ have the same distribution for each n,h and k. Since $X_{t}^{(n)}\rightarrow ^{\pounds ^{2}}X_{t}$, it can be deduced $X_{t}^{(n)}\rightarrow ^{P}X_{t}$ and hence ${\sum }_{i=0}^{h}b_{i}X_{t+i}^{(n)}\rightarrow ^{P}{\sum }_{i=0}^{h}b_{i}X_{t+i}.$ Using Cramer-Wold device, it can be concluded that

$$ (X_{0}^{(n)},...,X_{k}^{(n)})\Rightarrow (X_{0},...,X_{k}) $$

and

$$ (X_{h}^{(n)},...,X_{k+h}^{(n)})\Rightarrow (X_{h},...,X_{k+h}). $$

Since $(X_{0}^{(n)},...,X_{k}^{(n)})$ and $(X_{h}^{(n)},...,X_{k+h}^{(n)})$ have the same distribution, hence their limits, (X₀,...,X_k) and (X_h,...,X_k+h) have the same distribution. This completes the proof.

Appendix B: Proof of Proposition 2.1

Using the stationarity of the process and the properties of the operator, E(X_t) will be obtained. To obtain $E({X^{2}_{t}})$, we have

$$ \begin{array}{@{}rcl@{}} E({X^{2}_{t}})=E(B_{t-1}^{2} + {e_{t}^{2}}+2B_{t-1}e_{t}), \end{array} $$

where B_t− 1 = a ◇_𝜃X_t− 1 + b ◇_δX_t− 1e_t− 1. Using the stationarity of the process and the properties of the operator, we have

$$ \begin{array}{@{}rcl@{}} E(B_{t-1})=E(X_{t}-e_{t})=E(X_{t})-\mu, \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} E(B_{t-1}^{2})&=&E(a\diamond_{\theta} X_{t-1})^{2} +E(b\diamond_{\delta}X_{t-1}e_{t-1})^{2}+2E((a\diamond_{\theta} X_{t-1})(b\diamond_{\delta}X_{t-1}e_{t-1})) \\ &=& a [\theta E(X_{t-1}^{2}) +(1-\theta)E(X_{t-1})]+b[\delta E(X_{t-1}^{2}e_{t-1}^{2}) +(1-\delta)E(X_{t-1}e_{t-1})]\\&&+2ab E(X_{t-1}^{2}e_{t-1}), \end{array} $$

where

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{2}}{e_{t}^{2}})&=&E(B_{t-1}^{2}) E({e_{t}^{2}}) +E({e_{t}^{4}})+2E({e_{t}^{3}}) E(B_{t-1}) \\ E({X_{t}^{2}}e_{t})&=&E(B_{t-1}^{2}) E(e_{t}) +E({e_{t}^{3}})+2E({e_{t}^{2}}) E(B_{t-1}) \\ E(X_{t}e_{t})&=&E(B_{t-1}) E(e_{t}) +E({e_{t}^{2}}). \end{array} $$

After some simplifications, we have

$$ \begin{array}{@{}rcl@{}} E(B_{t-1}^{2})=\frac{a \theta E(X_{t-1}^{2}) + C}{1-[b\delta E({e_{t}^{2}}) +2ab\mu]}, \end{array} $$

where $C =a(1-\theta )E(X_{t-1})+b\delta (E({e_{t}^{4}})+2E({e_{t}^{3}}) E(B_{t-1}))+b(1-\delta )(E(B_{t-1}) E(e_{t}) +E({e_{t}^{2}})) + 2ab(E({e_{t}^{3}})+2E({e_{t}^{2}}) E(B_{t-1}) )$. Hence substitution of $E(B_{t-1}^{2})$ into $E({X_{t}^{2}})$, $E({X_{t}^{2}})$ can be obtained.

Appendix C: Proof of Proposition 2.2

For the autocovariance function of order one, we have

$$ E(X_{t}X_{t+1})=aE({X_{t}^{2}})+bE({X_{t}^{2}}e_{t})+E(X_{t})E(e_{t+1}), $$

where $E({X_{t}^{2}}e_{t})$ and $E(B_{t-1}^{2})$ are obtained as

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{2}}e_{t}) &=&E({e_{t}^{3}})+E(B_{t-1}^{2})E(e_{t})+2E(B_{t-1})E({e_{t}^{2}}) \\ E(B_{t-1}^{2}) &=&\frac{a \theta E(X_{t-1}^{2}) + C}{1-[b\delta E({e_{t}^{2}}) +2ab\mu]}. \end{array} $$

Substitution $E(B_{t-1}^{2})$ into the second term of $E({X_{t}^{2}}e_{t})\$, after some calculations we have

$$ E({X_{t}^{2}}e_{t})=E({X_{t}^{2}})E(e_{t})+E({e_{t}^{3}})+2E({e_{t}^{2}})E(B_{t-1})-E(e_{t})(2\mu E(B_{t-1})+E({e_{t}^{2}})), $$

and hence

$$ \begin{array}{@{}rcl@{}} E(X_{t}X_{t+1})&=&(a+b\mu)E({X_{t}^{2}})+b[E({e_{t}^{3}})+2E({e_{t}^{2}})E(B_{t-1}) -\mu(2\mu E(B_{t-1})-E({e_{t}^{2}}))]+E(X_{t})\mu \\ &=&(a+b\mu )E({X_{t}^{2}})+\mu_{X}\mu +b[2\sigma^{2}(\mu_{X}-\mu )+E({e_{t}^{3}})-\mu (\mu^{2}+ \sigma^{2})]. \end{array} $$

Also

$$ E(X_{t}X_{t+2})=aE(X_{t}X_{t+1})+bE(X_{t}X_{t+1}e_{t+1})+E(X_{t})E(e_{t+2}). $$

After some tedious computations, we obtain E(X_tX_t+ 1e_t+ 1) = σ²E(X_t) + μE(X_tX_t+ 1). By substituting E(X_tX_t+ 1e_t+ 1) in E(X_tX_t+ 2), we have

$$ E(X_{t}X_{t+2})=(a+b\mu )E(X_{t}X_{t+1})+b\sigma^{2}\mu_{X}+\mu \mu_{X}. $$

On the other hand, Eq. 2 implies that $b\sigma ^{2}\mu _{X}+\mu \mu _{X}+[(a+b\mu )-1]{\mu _{X}^{2}}=0$. Hence

$$ \gamma_{X}(2)=(a+b\mu )\gamma_{X}(1). $$

So by induction, we can conclude that

$$ \gamma_{X}(k)=(a+b\mu )\gamma_{X}(k-1)=(a+b\mu )^{k-1}\gamma_{X}(1). $$

Appendix D: Proof of Proposition 2.3

The one step ahead conditional expectation can be obtained directly. Since

$$ \begin{array}{@{}rcl@{}} E[X_{t+1}e_{t+1}|t] &=&E(e_{t+1}^{2})+ E(e_{t+1})E((a\diamond_{\theta} X_{t}+b\diamond_{\delta} X_{t}e_{t})|t) \\ &=&E(e_{t+1}^{2})+\mu (a+be_{t})X_{t}, \end{array} $$

we have

$$ \begin{array}{@{}rcl@{}} E(X_{t+2}|t) &=& E(a\diamond_{\theta} X_{t+1}+b\diamond_{\delta} X_{t+1}e_{t+1} +e_{t+2}|t) \\ &=& a E(X_{t+1}|t)+ b E (X_{t+1}e_{t+1}|t)+E(e_{t+2}) \\ &=& a E(X_{t+1}|t)+ b(\mu^{2} +\sigma^{2}+\mu (a+be_{t})X_{t})+\mu \\ &=& a E(X_{t+1}|t)+b\sigma^{2} +\mu +b\mu(\mu+ (a+be_{t})X_{t}). \end{array} $$

Using the one step ahead conditional expectation E[X_t+ 1|t] = μ + (a + be_t)X_t, we can be conclude that

$$ \begin{array}{@{}rcl@{}} E(X_{t+2}|t)=(a+b\mu)E[X_{t+1}|t]+b\sigma^{2} +\mu, \end{array} $$

and

$$ E(X_{t+3}|t)=(a+b\mu)E[X_{t+2}|t]+b\sigma^{2} +\mu. $$

Hence by induction, we conclude

$$ \begin{array}{@{}rcl@{}} E(X_{t+k}|t) &=&(a+b\mu)E(X_{t+k-1}|t)+b\sigma^{2} +\mu \\ &=&(a+b\mu)^{2}E(X_{t+k-2}|t)+(b\sigma^{2} +\mu)(1+(a+b\mu)) \\ &=&(a+b\mu)^{k-1}E(X_{t+1}|t)+(b\sigma^{2} +\mu)\sum\limits_{i=0}^{k-2}(a+b\mu)^{i}. \end{array} $$

Now we obtain an expression for $E(X_{t+k}^{2}|t).$ One can easily find $E(X_{t+1}^{2}|t)$. Also

$$ \begin{array}{@{}rcl@{}} E(X_{t+2}^{2}|t) &=& E([B_{t+1} + e_{t+2}]^{2}|t) \\ &=& E(B_{t+1}^{2}|t) +E(e_{t+2}^{2})+2E(B_{t+1}|t)E(e_{t+2}) \\ &=& a\theta E(X_{t+1}^{2}|t)+a(1-\theta)E(X_{t+1}|t)+b\delta E(X_{t+1}^{2}e_{t+1}^{2}|t)+b(1-\delta)E(X_{t+1}e_{t+1}|t) \\ &&+2abE(X_{t+1}^{2}e_{t+1}|t)+ E(e_{t+2}^{2}|t)+2\mu E(B_{t+1}|t), \end{array} $$

where $E(X_{t+1}^{2}e_{t+1}|t)$ and $E(X_{t+1}^{2}e_{t+1}^{2}|t)$ are obtained as

$$ E(X_{t+1}^{2}e_{t+1}|t)=E(e_{t+1})E({B_{t}^{2}}|t) +2E(e_{t+1}^{2})E(B_{t}|t)+E(e_{t+1}^{3}), $$

$$ E(X_{t+1}^{2}e_{t+1}^{2}|t)=E(e_{t+1}^{2})E({B_{t}^{2}}|t) +2E(e_{t+1}^{3})E(B_{t}|t)+E(e_{t+1}^{4}). $$

Substituting the above equations in $E(X_{t+2}^{2}|t)$ and after some calculations, we have

$$ \begin{array}{@{}rcl@{}} E(X_{t+2}^{2}|t) &=& (a\theta + b\delta E(e_{t+1}^{2}) + 2ab \mu )E(X_{t+1}^{2}|t)+2abE(e_{t+1}^{3}) +b(1-\delta)E(X_{t+1}e_{t+1}|t)\\ &&+a(1-\theta)E(X_{t+1}|t) +b\delta(2E(e_{t+1}^{3})E(B_{t}|t) +E(e_{t+1}^{4}) )+2a b2E(e_{t+1}^{2})E(B_{t}|t) \\&&+(b\delta E(e_{t+1}^{2}) +2ab\mu)[-(2\mu E(B_{t}|t) +E(e_{t+1}^{2})) ] +2\mu E(B_{t+1}|t) +E(e_{t+2}^{2}). \end{array} $$

Also, we can find that

$$ \begin{array}{@{}rcl@{}} E(X_{t+3}^{2}|t) &=& (a\theta + b\delta E(e_{t+2}^{2}) + 2ab \mu )E(X_{t+2}^{2}|t)+b(1-\delta)E(X_{t+2}e_{t+2}|t) +2\mu E(B_{t+2}|t) \\ &&+a(1-\theta)E(X_{t+2}|t) +b\delta(2E(e_{t+2}^{3})E(B_{t+1}|t) +E(e_{t+2}^{4}) )+4a bE(e_{t+2}^{2})E(B_{t+1}|t) \\&&+2abE(e_{t+2}^{3}) +(b\delta E(e_{t+2}^{2}) +2ab\mu)[-(2\mu E(B_{t+1}|t) +E(e_{t+2}^{2}))]+E(e_{t+3}^{2}). \end{array} $$

So by induction and properties of the i.i.d. process $\{e_{t}, t{\in \mathbb {Z}}\}$, we can conclude that

$$ \begin{array}{@{}rcl@{}} E(X_{t+k}^{2}|t) &=&(a\theta + b\delta E({e_{t}^{2}}) + 2ab \mu )E(X_{t+k-1}^{2}|t)+2abE({e_{t}^{3}}) +b(1-\delta)E(X_{t+k-1}e_{t+k-1}|t) \\ &&+a(1-\theta)E(X_{t+k-1}|t) +b\delta(2E({e_{t}^{3}})E(B_{t+k-2}|t) +E({e_{t}^{4}}) )+4a bE({e_{t}^{2}})E(B_{t+k-2}|t) \\&&+(b\delta E(e_{t+k-1}^{2}) +2ab\mu)[-(2\mu E(B_{t+k-2}|t) +E(e_{t+k-1}^{2})) ]+2\mu E(B_{t+k-1}|t) +E(e_{t+k}^{2}), \end{array} $$

where E(B_t+k−i|t) = E(X_t+k−i+ 1 − μ|t) for i = 0, 1,...,k.

Appendix E

Some expectations which are used in the Yule Walker method are calculated in this part.

Let B_t− 1 = a ◇_𝜃X_t− 1 + b ◇_δX_t− 1e_t− 1. So B_t− 1 = X_t − e_t and we have

$$ \begin{array}{@{}rcl@{}} E(X_{t}e_{t}) &=&E((B_{t-1}+e_{t})e_{t})=E(B_{t-1})E(e_{t})+E({e_{t}^{2}}) =(E(X_{t})-\mu )\mu+E({e_{t}^{2}}) \\ &=&\mu_{X}\mu+\sigma^{2}, \end{array} $$

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{2}}e_{t}) &=&E((B_{t-1}+e_{t})^{2}e_{t}) \\ &=&E(B_{t-1}^{2})E(e_{t})+E({e_{t}^{3}})-2E(B_{t-1})E({e_{t}^{2}}) \\ &=&[E({X_{t}^{2}})+\mu(-2E(X_{t})+\mu)-\sigma^{2}]\mu +E({e_{t}^{3}})-2[\mu_{X}- \mu](\sigma^{2}+\mu^{2}), \end{array} $$

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{2}}{e_{t}^{2}})&=&E((B_{t-1}+e_{t})^{2}{e_{t}^{2}}) \\ &=&E(B_{t-1}^{2}) E({e_{t}^{2}}) +E({e_{t}^{4}})+2E({e_{t}^{3}}) E(B_{t-1}),\\ &=&[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\sigma^{2}](\sigma^{2}+\mu^{2})+E({e_{t}^{4}})+2E({e_{t}^{3}})[\mu_{X}-\mu], \end{array} $$

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{3}}e_{t})&=&E((B_{t-1}+e_{t})^{3}e_{t}) \\ &=&E(B_{t-1}^{3}) E(e_{t}) +E({e_{t}^{4}})+3E({e_{t}^{3}}) E(B_{t-1}) +3E({e_{t}^{2}}) E(B_{t-1}^{2}),\\ &=&[E({X_{t}^{3}})-E({e_{t}^{3}})-3[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\sigma^{2}]\mu-3(\mu_{X}-\mu)(\sigma^{2}+\mu^{2})]\mu\\&&+E({e_{t}^{4}})+3E({e_{t}^{3}})[\mu_{X}-\mu]+3E({e_{t}^{2}})[E({X_{t}^{2}})+\mu(-2\mu_{X})+\mu)-\sigma^{2}], \end{array} $$

and

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{3}}{e_{t}^{2}})&=&E((B_{t-1}+e_{t})^{3}{e_{t}^{2}}) \\ &=&E(B_{t-1}^{3}) E({e_{t}^{2}}) +E({e_{t}^{5}})+3E({e_{t}^{4}}) E(B_{t-1})+3E({e_{t}^{3}}) E(B_{t-1}^{2}),\\ &=&[E({X_{t}^{3}})-E({e_{t}^{3}})-3[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\sigma^{2}]\mu-3(\mu_{X}-\mu)(\sigma^{2}+\mu^{2})](\sigma^{2}+\mu^{2})\\&&+E({e_{t}^{5}})+3E({e_{t}^{4}})[\mu_{X}-\mu] +3E({e_{t}^{3}})[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\sigma^{2}]. \end{array} $$

It is noted that under P(μ) distribution of innovations {e_t}, we have

$$ \begin{array}{@{}rcl@{}} E({X_{t}^{3}}{e_{t}^{2}})&=&[E({X_{t}^{3}})-(\mu^{2}+\mu (1+\mu)^{2})-3[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\mu]\mu -3(\mu_{X}-\mu) \\&&(\mu+\mu^{2})](\mu^{2}+\mu)+(3\mu^{3}+3\mu^{2}(1+\mu)^{2}+\mu(1+\mu)^{4}+2\mu^{2}(1+\mu)(2+\mu) \\&&+\mu^{2}(2+\mu)^{2})+3(\mu^{4}+6\mu^{3}+7\mu^{2}+\mu)(\mu_{X}-\mu)+3(\mu^{3}+3\mu^{2}+\mu )[E({X_{t}^{2}}) \\&&+\mu(-2\mu_{X}+\mu)-\mu], \\ && \\ E({X_{t}^{3}}e_{t})&=&[E({X_{t}^{3}})-(\mu^{2}+\mu (1+\mu)^{2})-3[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\mu]\mu \\&& -3(\mu_{X}-\mu)(\mu+\mu^{2})]\mu+(2\mu^{2} (1 + \mu) + \mu (1 + \mu)^{3} + \mu^{2} (2 + \mu)) \\&&+3(\mu^{2} + \mu (1 + \mu)^{2})[\mu_{X}-\mu]+3(\mu^{2}+\mu)[E({X_{t}^{2}})+\mu(-2\mu_{X})+\mu)-\mu], \\ && \\ E({X_{t}^{2}}{e_{t}^{2}})&=&[E({X_{t}^{2}})+\mu(-2\mu_{X}+\mu)-\mu](\mu^{2}+\mu)+2(\mu^{3}+3\mu^{2}+ \mu)(\mu_{X}-\mu) \\&&+(\mu^{4}+6\mu^{3}+7\mu^{2}+\mu), \\ && \\ E({X_{t}^{2}}e_{t}) &=&[E({X_{t}^{2}})+\mu (-2\mu_{X}+\mu)-\mu] \mu+(\mu^{3}+3\mu^{2}+\mu)+2(\mu^{2}+\mu )(\mu_{X}-\mu), \\ && \\ E(X_{t}e_{t})&=&\mu_{X}\mu+\mu. \end{array} $$

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