A Poisson INAR(1) model with serially dependent innovations

Abstract

Motivated by a certain type of infinite-patch metapopulation model, we propose an extension to the popular Poisson INAR(1) model, where the innovations are assumed to be serially dependent in such a way that their mean is increased if the current population is large. We shall recognize that this new model forms a bridge between the Poisson INAR(1) model and the INARCH(1) model. We analyze the stochastic properties of the observations and innovations from an extended Poisson INAR(1) process, and we consider the problem of model identification and parameter estimation. A real-data example about iceberg counts shows how to benefit from the new model.

Proofs

1.1 Proof of Proposition 2

Since $(X_t)_{\mathbb N_0}$ is a homogeneous Markov chain with strictly positive 1-step transition probabilities, $(X_t)_{\mathbb N_0}$ is also irreducible and aperiodic. Furthermore, it is distributed as a BPI, so we can apply the theorem in Heathcote (1966). It remains to check two conditions:

• The mean related to the individual offspring, i.e., the mean of $Y$ according to our notations in Sect. 2, must be smaller than 1 (subcritical case). From Proposition 1, we obtain that $E[Y]=\rho +a$, so $E[Y]<1$ iff the condition $\rho +a<1$ holds.

• The immigration in (4) is described by the i.i.d. random variables $\nu _t\sim {\text {Po}}(b)$. Now consider the quantity $\sum _{n=1}^{\infty }\ {\textstyle \frac{1}{n}}\cdot P(\nu _t\ge n)$, i.e., the expectation $E[H(\nu _t)]$, where $H(n) := \sum _{k=1}^n \frac{1}{k}$ for $n\in \mathbb N$ and $H(0):=0$ denote the harmonic numbers. Since $H(\nu _t)\le \nu _t$, and since $E[\nu _t] = b<\infty$, also $E[H(\nu _t)]<\infty$ holds.

So the existence of a stationary marginal distribution follows, and together with the irreducibility and aperiodicity, it follows that $(X_t)_{\mathbb N_0}$ is an ergodic Markov chain (Feller 1968, p. 394). Note that this stationarity result could also be derived by applying Proposition 3.1 in Latour (1998), because binomial-Poisson thinning is an instance of the generalized thinning concept considered by Latour (1998).

With similar arguments as for Heathcote (1966), we now apply Theorem 1 in Pakes (1971). $(X_t)_{\mathbb N_0}$ is not only distributed as a subcritical BPI, the corresponding offspring also has a finite variance, i.e., $V[Y]<\infty$. Hence, $(X_t)_{\mathbb N_0}$ satisfies all conditions being required in Sect. 2 of Pakes (1971), so Theorem 1 in Pakes (1971) implies geometric ergodicity. From Theorem 1 of Nummelin and Tweedie (1978) and Theorem 3.7 in Bradley (2005), we conclude that a stationary extended Poisson INAR(1) process is also $\beta$-mixing (and thus $\alpha$-mixing) with exponentially decreasing weights.

1.2 Proofs of Sect. 3.1

Let $f(x) := a\cdot x+b$. Formula (5) about conditional mean and variance follows from

$$\begin{aligned} E[X_t\ |\ X_{t-1},\ldots ]&= \rho \cdot X_{t-1}+f(X_{t-1})=(\rho +a)\cdot X_{t-1} + b,\\ V[X_t\ |\ X_{t-1},\ldots ]&= \rho (1-\rho )\cdot X_{t-1}+f(X_{t-1})=\big (\rho (1-\rho )+a\big )\cdot X_{t-1} + b, \end{aligned}$$

respectively. For computing the variance, we used the conditional independence of $\rho \circ X_{t-1}$ and $\epsilon _t$, given $X_{t-1}$. From formula (5) and the stationarity assumption, we obtain the marginal mean and variance as given in formula (6) by solving

$$\begin{aligned} \mu _X=E[X_t]&= E\big [E[X_t|X_{t-1},\ldots ]\big ]=(\rho +a)\cdot \mu _X+ b,\\ \sigma _X^2=V[X_t]&= E\big [V[X_t|X_{t-1},\ldots ]\big ]+V\big [E[X_t|X_{t-1},\ldots ]\big ]\\&= \big (\rho (1-\rho )+a\big )\cdot \mu _X + b+(\rho +a)^2\cdot \sigma _X^2\\&= (1-\rho ^2)\cdot \mu _X+(\rho +a)^2\cdot \sigma _X^2. \end{aligned}$$

The expression (7) for the pgf is shown analogously:

$$\begin{aligned} E[z^{X_t}]&= E\big [E[z^{\rho \circ X_{t-1}}|X_{t-1}]\cdot E[z^{\epsilon _t}\ |\ X_{t-1}]\big ]\\&= E\big [(1-\rho +\rho z)^{X_{t-1}}\cdot \exp {\big ((a\cdot X_{t-1}+b)(z-1)\big )}\big ]\\&= e^{b(z-1)}\cdot E\big [(1-\rho +\rho z)^{X_{t-1}}\cdot \big (e^{a(z-1)}\big )^{X_{t-1}}\big ]. \end{aligned}$$

Recursions (8) and (9) for higher-order moments follow from

$$\begin{aligned} E\left[ \left. X_t^n\right| X_{t-1},\ldots \right]&= E[(\rho \circ X_{t-1} + \epsilon _t)^n|X_{t-1}]\\&= \sum _{j=0}^n\ {n \atopwithdelims ()j}\cdot E[(\rho \circ X_{t-1})^j|X_{t-1}] \cdot E\left[ \left. \epsilon _t^{n-j}\right| X_{t-1}\right] . \end{aligned}$$

To compute the second and third moments of $X_t$, we first require the first few moments of the binomial and the Poisson distribution:

$$\begin{aligned} \mu _{1,{\text {B}}(n,p)}&= np,\quad \mu _{2,{\text {B}}(n,p)}=n^2 p^2+n p(1-p),\\ \mu _{3,{\text {B}}(n,p)}&= n^3 p^3+3n^2p^2 (1-p)+np(1-p)(1-2p),\\ \mu _{1,{\text {Po}}(\lambda )}&= \lambda ,\quad \mu _{2,{\text {Po}}(\lambda )}=\lambda (1+\lambda ),\quad \mu _{3,{\text {Po}}(\lambda )}=\lambda (1+3\lambda +\lambda ^2). \end{aligned}$$

Now we apply formula (9) to obtain

$$\begin{aligned} \mu _{2,X}&:= E\left[ X_t^2\right] \\&= E\big [\mu _{2,{\text {Po}}(a\cdot X_{t-1}+b)}\big ]\!+\!2\cdot E\big [\mu _{1,{\text {B}}(X_{t-1},\rho )}\cdot \mu _{1,{\text {Po}}(a\cdot X_{t-1}+b)}\big ]+E\big [\mu _{2,{\text {B}}(X_{t-1},\rho )}\big ]\\&= E\big [(a\cdot X_{t-1}+b)(1+a\cdot X_{t-1}+b)\big ]\\&\quad +\,2\cdot E\big [\rho X_{t-1}\cdot (a\cdot X_{t-1}+b)\big ]+ E\left[ X_{t-1}^2 \rho ^2+X_{t-1} \rho (1-\rho )\right] \\&= E\left[ b(1+b)+a(1+2b) X_{t-1}+a^2 X_{t-1}^2\right] \\&\quad +\,2\cdot E\left[ \rho a\cdot X_{t-1}^2+\rho b X_{t-1}\right] + E\left[ \rho ^2 X_{t-1}^2 + \rho (1-\rho ) X_{t-1}\right] \\&= \mu _X\cdot \big ((1+b)(1-(\rho +a))+(1+2b)(\rho +a)-\rho ^2\big )+ (\rho +a)^2\cdot \mu _{2,X}. \end{aligned}$$

This confirms formula (6) about the variance. Analogously,

$$\begin{aligned} \mu _{3,X}&:= E\left[ X_t^3\right] =\ E\big [\mu _{3,{\text {Po}}(a\cdot X_{t-1}+b)}\big ]+ E\big [\mu _{3,{\text {B}}(X_{t-1},\rho )}\big ]\\&\quad +\,3\cdot E\big [\mu _{1,{\text {B}}(X_{t-1},\rho )}\cdot \mu _{2,{\text {Po}}(a\cdot X_{t-1}+b)}\big ]+3\cdot E\big [\mu _{2,{\text {B}}(X_{t-1},\rho )}\cdot \mu _{1,{\text {Po}}(a\cdot X_{t-1}+b)}\big ]\\&= E\big [b(1+3 b+b^2) + a(1+6b+3b^2)\cdot X_{t-1} + 3 a^2(1+b)\cdot X_{t-1}^2+a^3\cdot X_{t-1}^3\big ]\\&\quad +\,E\big [\rho (1-\rho )(1-2\rho )\cdot X_{t-1} +3\rho ^2 (1-\rho ) \cdot X_{t-1}^2+\rho ^3\cdot X_{t-1}^3\big ]\\&\quad +\,3\cdot E\left[ \rho b(1+b)\cdot X_{t-1} + \rho a(1+2b)\cdot X_{t-1}^2 + \rho a^2\cdot X_{t-1}^3\right] \\&\quad +\,3\cdot E\left[ \rho (1-\rho ) b\cdot X_{t-1} + \big (\rho ^2 b + \rho (1-\rho ) a\big )\cdot X_{t-1}^2 + \rho ^2 a\cdot X_{t-1}^3\right] \\&= b(1+3 b+b^2) + \big ((\rho +a)(1+6b+3b^2)-3\rho ^2 (1+b)+2\rho ^3\big )\cdot \mu _{X}\\&\quad +\,3(\rho +a)\big ((\rho +a) (1+b)-\rho ^2\big )\cdot \mu _{2,X}+ (\rho +a)^3\cdot \mu _{3,X}, \end{aligned}$$

from which we obtain an expression for $\mu _{3,X}$. After tedious algebra, we find an expression for the third central moment:

$$\begin{aligned} E[(X_t-\mu _X)^3]&= \mu _{3,X}-3\mu _{X}\mu _{2,X}+2\mu _{X}^3=\cdots \\&= \mu _{X}\cdot \frac{(1-\rho ^3)(1-\rho ^2+a \rho +2a^2)+3a \rho (1-\rho )}{\big (1-(a+\rho )^2\big )\big (1-(a+\rho )^3\big )}. \end{aligned}$$

Next, we derive an expression for the autocovariance function. For $k\ge 1$, we obtain by applying formula (5) and by conditioning that

$$\begin{aligned} Cov[X_t, X_{t-k}]=Cov\big [(\rho +a)\cdot X_{t-1}+b,\ X_{t-k}\big ]=\cdots =(\rho +a)^k\cdot \sigma _X^2. \end{aligned}$$

Finally, the proof of formula (11) is done by induction, with this formula being true for $k=1$ according to the expression for $E[X_t\ |\ X_{t-1},\ldots ]$ in formula (5).

1.3 Proofs of Sect. 3.2

Since $E[\epsilon _t | X_{t-1},\ldots ] = V[\epsilon _t | X_{t-1},\ldots ] = a\cdot X_{t-1}+b$, we obtain

$$\begin{aligned} E[\epsilon _t]&= E\big [E[\epsilon _t | X_{t-1},\ldots ]\big ] = a\cdot \mu _X+b = (1-\rho )\cdot \mu _X,\\ V[\epsilon _t]&= E\big [V[\epsilon _t | X_{t-1},\ldots ]\big ] + V\big [E[\epsilon _t\ |\ X_{t-1},\ldots ]\big ] = a\cdot \mu _X+b\ +\ a^2\cdot \sigma _X^2\\&= (1-\rho )\cdot \frac{1-(\rho +a)^2}{1-\rho ^2}\cdot \sigma _X^2 + a^2\cdot \sigma _X^2 = \frac{\sigma _X^2}{1+\rho }\cdot \big (1-\rho ^2-\rho a (2-a)\big ). \end{aligned}$$

In particular, this result shows that $V[\epsilon _t] = E[\epsilon _t] + a^2\cdot \sigma _X^2$, so the proof of formula (12) is complete. Similarly, we prove formula (13) via

$$\begin{aligned} E[z^{\epsilon _t}]&= E\big [E[z^{\epsilon _t}| X_{t-1},\ldots ]\big ]=E\big [\exp {\big ((a\cdot X_{t-1}+b)(z-1)\big )}\big ]\\&= e^{b(z-1)}\cdot p_X\big (e^{a(z-1)}\big ). \end{aligned}$$

Next, we consider the covariance between $X_t$ and $\epsilon _{t-j}$ with $j\in \mathbb N_0$. First, using the conditional independence between $\rho \circ X_{t-1}$ and $\epsilon _{t}$ given $X_{t-1}$, we have

$$\begin{aligned} Cov[X_t, \epsilon _{t}]&= Cov[\rho \circ X_{t-1}, \epsilon _{t}]+ Cov[\epsilon _{t},\epsilon _{t}]\\&= Cov[\rho \cdot X_{t-1},\ a\cdot X_{t-1}+b]+\sigma _{\epsilon }^2 =\rho a \cdot \sigma _{X}^2+\sigma _{\epsilon }^2. \end{aligned}$$

Secondly, for $j\ge 1$ and using formula (5),

$$\begin{aligned} Cov[X_t, \epsilon _{t-j}]&= Cov\big [(\rho +a)\cdot X_{t-1} + b,\ \epsilon _{t-j}\big ] \\&= \ldots \ =\ (\rho +a)^j\cdot Cov[X_{t-j}, \epsilon _{t-j}]\ =\ (\rho +a)^j\cdot \left( \rho a \cdot \sigma _{X}^2 + \sigma _{\epsilon }^2\right) . \end{aligned}$$

So through conditioning on $X_{t-1},\ldots , \epsilon _{t-1},\ldots$, we obtain for $j\ge 1$ that

$$\begin{aligned} Cov[\epsilon _t,\epsilon _{t-j}]=Cov[a\cdot X_{t-1}+b,\epsilon _{t-j}]=a\cdot (\rho +a)^{j-1}\cdot \left( \rho a\cdot \sigma _{X}^2 + \sigma _{\epsilon }^2\right) . \end{aligned}$$

1.4 Proofs of Sect. 4.1

The score functions (17) to (19) can be derived in analogy to Sect. 3 in Freeland and McCabe (2004). Since

$$\begin{aligned} \frac{\partial }{\partial \theta _i}\,\ell ({\varvec{\theta }})=\sum _{t=1}^T\ \frac{\frac{\partial }{\partial \theta _i}\,P(X_t|X_{t-1})}{P(X_t|X_{t-1})}, \end{aligned}$$

it is essential to discuss the first-order partial derivatives of the transition probabilities $P_{{\varvec{\theta }}}(k|l)$ according to (3). Let us start with the derivative in $\rho$. Since (Freeland and McCabe 2004, p. 716)

$$\begin{aligned}&\frac{\partial }{\partial \rho }\,\left( {l \atopwithdelims ()j}\,\rho ^j\,(1-\rho )^{l-j}\right) \\&\quad =\frac{l}{1-\rho }\,\left( {l-1\atopwithdelims ()j-1}\,\rho ^{j-1}\,(1-\rho )^{l-j}-{l \atopwithdelims ()j}\,\rho ^j\,(1-\rho )^{l-j}\right) , \end{aligned}$$

and since

$$\begin{aligned} e^{-(a\,l+b)}\,\frac{(a\,l+b)^{k-j}}{(k-j)!}=e^{-\big (a\,(l-1)+a+b\big )}\,\frac{\big (a\,(l-1)+a+b\big )^{(k-1)-(j-1)}}{\big ((k-1)-(j-1)\big )!}, \end{aligned}$$

we can express

$$\begin{aligned} \frac{\partial }{\partial \rho }\,P_{\rho ,a,b}(k|l)=\frac{l}{1-\rho }\,\big (P_{\rho ,a,a+b}(k-1|l-1)-P_{\rho ,a,b}(k|l)\big ). \end{aligned}$$

So (17) follows. Differentiating with respect to $a$, we have

$$\begin{aligned}&\frac{\partial }{\partial a}\,\left( e^{-(a\,l+b)}\,(a\,l+b)^{k-j}\right) \\&\quad =l\,\left( (k-j)\,e^{-(a\,l+b)}\,(a\,l+b)^{k-1-j}-e^{-(a\,l+b)}\,(a\,l+b)^{k-j}\right) . \end{aligned}$$

So (18) follows, since

$$\begin{aligned} \frac{\partial }{\partial a}\,P_{\rho ,a,b}(k|l)=l\,\big (P_{\rho ,a,b}(k-1|l)\ -\ P_{\rho ,a,b}(k|l)\big ). \end{aligned}$$

Finally, we have

$$\begin{aligned} \frac{\partial }{\partial b}\,\left( e^{-(a\,l+b)}\,(a\,l+b)^{k-j}\right) =(k-j)\,e^{-(a\,l+b)}\,(a\,l+b)^{k-1-j}-e^{-(a\,l+b)}\,(a\,l+b)^{k-j}, \end{aligned}$$

and hence

$$\begin{aligned} \frac{\partial }{\partial b}\,P_{\rho ,a,b}(k|l)=P_{\rho ,a,b}(k-1|l)-P_{\rho ,a,b}(k|l), \end{aligned}$$

in analogy to Proposition 2 in Freeland and McCabe (2004). This completes the proof of (19).

1.5 Proofs of Sect. 5.1

To compute the derivative $\frac{\partial }{\partial \tau }\,\ell \big ((1-\tau )\xi ,\ \tau \xi ,\ b\big )$, we proceed in analogy to the proof of formulae (17) and (18), but by also applying the chain rule. Since $\frac{\partial \rho }{\partial \tau } = -\xi$ and $\frac{\partial a}{\partial \tau } = \xi$, we obtain

$$\begin{aligned}&\frac{\partial }{\partial \tau }\,P_{(1-\tau )\xi ,\tau \xi ,b}(k|l)\\&\quad =\frac{-\xi \,l}{1-(1-\tau )\xi }\,\Big (P_{(1-\tau )\xi ,\tau \xi ,\tau \xi +b}(k-1|l-1) -P_{(1-\tau )\xi ,\tau \xi ,b}(k|l)\Big )\\&\qquad +\,\xi \,l\left( P_{(1-\tau )\xi ,\tau \xi ,b}(k-1|l)- P_{(1-\tau )\xi ,\tau \xi ,b}(k|l)\right) \ \ \\&\quad =\xi \,l\,\left( \frac{(1-\tau )\xi }{1-(1-\tau )\xi }\,P_{(1-\tau )\xi ,\tau \xi ,b}(k|l) +P_{(1-\tau )\xi ,\tau \xi ,b}(k-1|l)\right. \\&\left. \qquad -\,\frac{1}{1-(1-\tau )\xi }\,P_{(1-\tau )\xi ,\tau \xi ,\tau \xi +b}(k-1|l-1)\right) . \end{aligned}$$

Hence, it follows that

$$\begin{aligned}&\left. \frac{\frac{\partial }{\partial \tau }\,P_{(1-\tau )\xi ,\tau \xi ,b}(k|l)}{P_{(1-\tau )\xi ,\tau \xi ,b}(k|l)}\right| _{\tau =0}\\&\quad =\xi \,l\,\left( \frac{\xi }{1-\xi }+\frac{P_{\xi ,0,b}(k-1|l)}{P_{\xi ,0,b}(k|l)}-\frac{1}{1-\xi }\,\frac{P_{\xi ,0,b}(k-1|l-1)}{P_{\xi ,0,b}(k|l)}\right) , \end{aligned}$$

from which we obtain the formula for the score function $S_{\tau ;T}(\xi ,b)$.

For the sake of readability, let us now use the abbreviation $g(X_t,X_{t-1})$ for the summands of $S_{\tau ;\, T}(\xi ,b)$. Under the null of a Poisson INAR(1) model, it follows from Proposition 2 in Freeland and McCabe (2004) that

$$\begin{aligned} \frac{P_{\xi ,0,b}(X_t-1|X_{t-1})}{P_{\xi ,0,b}(X_t|X_{t-1})}=\frac{E[\epsilon _t|X_t,X_{t-1},\ldots ]}{b} \end{aligned}$$

and

$$\begin{aligned} \xi \,X_{t-1}\,\frac{P_{\xi ,0,b}(X_t-1|X_{t-1}-1)}{P_{\xi ,0,b}(X_t|X_{t-1})} =E[\xi \circ X_{t-1}|X_t,X_{t-1},\ldots ]. \end{aligned}$$

Hence, the conditional mean of $g(X_t,X_{t-1})$ becomes

$$\begin{aligned}&E\big [g(X_t,X_{t-1})\ |\ X_{t-1}, \ldots \big ]\\&\quad =\frac{\xi ^2\,X_{t-1}\,}{1-\xi }+\xi \,X_{t-1}\,\frac{E[\epsilon _t |X_{t-1},\ldots ]}{b}-\frac{1}{1-\xi }\,E[\xi \circ X_{t-1}|X_{t-1},\ldots ]\\&\quad =\frac{\xi ^2\,X_{t-1}\,}{1-\xi }+\xi \,X_{t-1}- \frac{\xi \,X_{t-1}}{1-\xi }=0. \end{aligned}$$

As a result,

$$\begin{aligned} E\big [S_{\tau ; T}(\xi ,b)\big ]=\sum _{t=1}^T\,E\Big [E\big [g(X_t,X_{t-1})|X_{t-1},\ldots \big ]\Big ]=0. \end{aligned}$$

Finally, the process of $g(X_t,X_{t-1})$ is $\alpha$-mixing with exponentially decreasing weights, see Proposition 2, such that Theorem 1.7 of Ibragimov (1962) is applicable, which gives the stated asymptotic normality with $\sigma ^2=E\big [g(X_1,X_{0})^2\big ]+ 2\,\sum _{k=1}^{\infty }\ E\big [g(X_1,X_{0})\,g(X_{k+1},X_{k})\big ]$. Using the above result $E\big [g(X_t,X_{t-1})| X_{t-1}, \ldots \big ]=0$, it follows that

$$\begin{aligned} E\big [g(X_1,X_{0})\,g(X_{k+1},X_{k})\big ]=E\Big [g(X_1,X_{0})\,E\big [g(X_{k+1},X_{k})|X_{k-1},\ldots \big ]\Big ]=0. \end{aligned}$$

So the proof of Proposition 3 is complete.