Proofs for Poi-INAR(1) DGP 1.1 Linear approximation for squared residualsBefore turning to the specific case of a Poi-INAR(1) DGP, let us first consider the general case described in the beginning of Sect. 3 , i.e., Let \(M_t=E[X_t\ |\ X_{t-1},X_{t-2},\ldots ]\) and \(V_t=V[X_t\ |\ X_{t-1},\ldots ]\) depend on some parameter vector \({\varvec{\theta }}\) , then
$$\begin{aligned} \mathrm{MS}_R({\varvec{\theta }})\ =\ \frac{1}{n}\sum \limits _{t=1}^n\frac{\big (X_t-M_t({\varvec{\theta }})\big )^2}{V_t({\varvec{\theta }})}, \end{aligned}$$
where \(M_t({\varvec{\theta }})\) and \(V_t({\varvec{\theta }})\) denote the conditional mean and variance, respectively, as functions of \({\varvec{\theta }}\) . To derive a linear approximation for the statistic \(\mathrm{MS}_R(\hat{{\varvec{\theta }}})\) , we consider a first-order Taylor approximation of the function \(\mathrm{MS}_R({\varvec{\theta }})\) in \({\varvec{\theta }}\) . The first-order partial derivative in \(\theta _i\) equals
$$\begin{aligned} \tfrac{\partial }{\partial \theta _i}\,\mathrm{MS}_R\ =\ -\frac{1}{n}\sum \limits _{t=1}^n\frac{(X_t-M_t)^2}{V_t^2}\, \frac{\partial V_t}{\partial \theta _i}-\ \frac{2}{n}\sum \limits _{t=1}^n \frac{(X_t-M_t)}{V_t}\,\frac{\partial M_t}{\partial \theta _i}. \end{aligned}$$
The means of the summands are
$$\begin{aligned} \begin{array}{@{}rl} E\Big [\frac{(X_t-M_t)^2}{V_t^2}\, \frac{\partial V_t}{\partial \theta _i}\Big ]\ =&{} E\Big [\frac{\partial V_t}{\partial \theta _i}\,\frac{1}{V_t^2}\,E\big [(X_t-M_t)^2\ |\ X_{t-1},\ldots \big ]\Big ] \ =\ E\big [\frac{\partial V_t}{\partial \theta _i}\,\frac{1}{V_t}\big ],\\ E\Big [\frac{(X_t-M_t)}{V_t}\,\frac{\partial M_t}{\partial \theta _i}\Big ] \ =&{} E\Big [\frac{\partial M_t}{\partial \theta _i}\,\frac{1}{V_t}\,E[X_t-M_t\ |\ X_{t-1},\ldots ]\Big ]\ =\ 0, \end{array} \end{aligned}$$
where the first equation used that \(E\big [(X_t-M_t)^2\ |\ X_{t-1},\ldots \big ] = V_t\) . This is used to conclude that
$$\begin{aligned} \frac{1}{n}\sum \limits _{t=1}^n\left( \frac{(X_t-M_t)^2}{V_t^2}\, \frac{\partial V_t}{\partial \theta _i}\ -\ E\left[ \frac{\partial V_t}{\partial \theta _i}\,\frac{1}{V_t}\right] \right)= & {} o_P(1),\\ \frac{2}{n}\sum \limits _{t=1}^n \frac{(X_t-M_t)}{V_t}\,\frac{\partial M_t}{\partial \theta _i}= & {} o_P(1). \end{aligned}$$
Thus, the linear Taylor approximation
$$\begin{aligned} \mathrm{MS}_R(\hat{{\varvec{\theta }}})= & {} \textstyle \mathrm{MS}_R({\varvec{\theta }})\ +\ \sum \limits _{i=1}^m\, \tfrac{\partial }{\partial \theta _i}\,\mathrm{MS}_R({\varvec{\theta }})\,(\hat{\theta }_i-\theta _i) \ +\ o\big (\Vert \hat{{\varvec{\theta }}}-{\varvec{\theta }}\Vert \big )\\= & {} \textstyle \mathrm{MS}_R({\varvec{\theta }})\ -\ \sum \limits _{i=1}^m\, E\big [\tfrac{\partial V_t}{\partial \theta _i}\,\tfrac{1}{V_t}\big ]\,(\hat{\theta }_i-\theta _i) \ +\ o\big (\Vert \hat{{\varvec{\theta }}}-{\varvec{\theta }}\Vert \big )\\&\textstyle -\ \sum \limits _{i=1}^m\, \biggl (\frac{1}{n}\sum \limits _{t=1}^n\Big (\frac{(X_t-M_t)^2}{V_t^2}\, \frac{\partial V_t}{\partial \theta _i}\ -\ E\big [\frac{\partial V_t}{\partial \theta _i}\,\frac{1}{V_t}\big ]\Big )\biggl )\,(\hat{\theta }_i-\theta _i) \\&\textstyle -\ \sum \limits _{i=1}^m\, \Big (\frac{2}{n}\sum \limits _{t=1}^n \frac{(X_t-M_t)}{V_t}\,\frac{\partial M_t}{\partial \theta _i}\Big )\,(\hat{\theta }_i-\theta _i), \end{aligned}$$
together with Slutsky’s lemma, implies the linear approximation (5 ).
Now, consider the special case of a Poi-INAR(1) DGP, i.e., \({\varvec{\theta }}=(\beta ,\alpha )\) . According to (5 ), we require the partial derivatives of \(V_t=\beta +\alpha (1-\alpha )\,X_{t-1}\) , which equal \(\frac{\partial }{\partial \alpha }\,V_t=(1-2\alpha )\,X_{t-1}\) and \(\frac{\partial }{\partial \beta }\,V_t=1\) . Thus, it follows that
$$\begin{aligned} \textstyle E\big [\frac{\partial V_t}{\partial \alpha }\,\frac{1}{V_t}\big ]\ =\ (1-2\alpha )\,E\big [\frac{X_{t-1}}{V_t}\big ],\qquad E\big [\frac{\partial V_t}{\partial \beta }\,\frac{1}{V_t}\big ]\ =\ E\big [\frac{1}{V_t}\big ]. \end{aligned}$$
This implies the linear approximation (7 ). Note that \(E\big [\frac{1}{V_t}\big ]\) can be rewritten as \(\frac{1}{\beta }\big (1-\alpha (1-\alpha )\,E\big [\frac{ X_{t-1}}{V_t}\big ]\big )\) .
1.2 Joint distribution of residuals and momentsIn order to prepare the asymptotic results to be derived for Sect. 3 , we use a central limit theorem (CLT) for the four-dimensional vector-valued process \(({\varvec{Y}}_t)_{\mathbb {Z}}\) given by
$$\begin{aligned} {\varvec{Y}}_t\ :=\ \left( \begin{array}{c} Y_{t,0}\\ Y_{t,1}\\ Y_{t,2}\\ Y_{t,3}\\ \end{array}\right) \ :=\ \left( \begin{array}{c} \frac{(X_t-M_t)^2}{V_t} - 1\\ X_t - \mu \\ X_t^2 - \mu (0)\\ X_t X_{t-1} - \mu (1) \end{array}\right) \end{aligned}$$
(A.1)
with \(\mu (k) := E[X_t X_{t-k}]\) . Note that the Poi-INAR(1) process is \(\alpha \) -mixing with exponentially decreasing weights, and it has existing moments of any order (Weiß and Schweer 2016 ). So the CLT by Ibragimov (1962 ) can be applied to \(\frac{1}{\sqrt{T}} \sum _{t=1}^T {\varvec{Y}}_t\) . The following lemma summarizes the resulting asymptotics.
Lemma 1
Let \((X_t)_{\mathbb {Z}}\) be a Poi-INAR(1) process with \(\mu =\frac{\beta }{1-\alpha }\) , \(\mu (k)=\mu (\alpha ^k+\mu )\) and
$$\begin{aligned} Y_{t,0}\ =\ \frac{(X_t-M_t)^2}{V_t} - 1 \ =\ \frac{(X_t-\alpha \, X_{t-1} - \beta )^2}{\alpha (1-\alpha )\, X_{t-1} + \beta } - 1. \end{aligned}$$
Then, \(\frac{1}{\sqrt{n}} \sum _{t=1}^n {\varvec{Y}}_t\) is asymptotically normally distributed with mean \({\varvec{0}}\) and covariance matrix \(\tilde{{\varvec{\Sigma }}} = (\tilde{\sigma }_{ij})\) , where
$$\begin{aligned} \begin{array}{@{}rl} \tilde{\sigma }_{00}\ =&{} 2+\frac{1}{\mu (1-\alpha )}-\frac{\alpha }{\mu }\, E\big [\frac{ X_{t-1}}{V_t}\big ]-6\alpha ^2(1-\alpha )^2\, E\big [\frac{ X_{t-1}}{V_t^2}\big ], \\ \tilde{\sigma }_{01}\ =&{} \frac{1}{1-\alpha }-2\alpha ^2\, E\big [\frac{ X_{t-1}}{V_t}\big ], \\ \tilde{\sigma }_{02}\ =&{} \frac{2(3\alpha +2)\mu }{1+\alpha } +\frac{1}{1-\alpha } +\frac{2\alpha ^2(-3+2\alpha )}{1+\alpha }\, E\big [\frac{ X_{t-1}}{V_t}\big ], \\ \tilde{\sigma }_{03}\ =&{} \frac{2\mu (1+2\alpha +2\alpha ^2)}{1+\alpha }+\frac{\alpha }{1-\alpha }+2\alpha \mu (1-\alpha )^2\, E\big [\frac{ X_{t-1}}{V_t}\big ]+\frac{2\alpha ^3(-3+2\alpha )}{1+\alpha }\, E\big [\frac{ X_{t-1}}{V_t}\big ]. \end{array} \end{aligned}$$
The expressions for \(\tilde{\sigma }_{11},\ldots ,\tilde{\sigma }_{33}\) can be found in Theorem 2.1 of Weiß and Schweer (2016 ).
1.2.1 Proof of Lemma 1
Note that the conditional variance and the conditional mean are related to each other in several ways, namely:
$$\begin{aligned} (M_t-\beta )(1-\alpha )+\beta= & {} V_t,\end{aligned}$$
(A.2)
$$\begin{aligned} M_t(1-\alpha )+\alpha \beta= & {} V_t,\end{aligned}$$
(A.3)
$$\begin{aligned} M_t-V_t= & {} \alpha ^2 X_{t-1},\end{aligned}$$
(A.4)
$$\begin{aligned} 1-\alpha (1-\alpha )\, E\big [\tfrac{ X_{t-1}}{V_t}\big ]= & {} \beta \, E\big [\tfrac{1}{V_t}\big ]. \end{aligned}$$
(A.5)
We first need to calculate some conditional moments. Obviously, we have \(E\left[ X_t^2\ |\ X_{t-1}\right] =V_t+M_t^2\) . Since for Poisson and binomial variates, factorial moments take a particularly simple form, we shall use the falling factorials \((x)_{(k)}=x\cdots (x-k+1)\) in the sequel. We have
$$\begin{aligned} E\left[ X_t^3\ |\ X_{t-1}\right]= & {} E[(\alpha \circ X_{t-1}+\epsilon _t)^3\ | \ X_{t-1}]\nonumber \\= & {} E[(\alpha \circ X_{t-1})^3\ | \ X_{t-1}]+3E[(\alpha \circ X_{t-1})^2\ | \ X_{t-1}]E[\epsilon _t]\nonumber \\&+\,3E[(\alpha \circ X_{t-1})\ | \ X_{t-1}]E[\epsilon _t^2]+E[\epsilon _t^3],\nonumber \\= & {} \alpha X_{t-1}+3\alpha ^2(X_{t-1})_{(2)}+\alpha ^3(X_{t-1})_{(3)} +3\alpha X_{t-1}(\beta ^2+\beta )\qquad \qquad \nonumber \\&+\,3(\alpha X_{t-1}+\alpha ^2(X_{t-1})_{(2)})\beta +(\beta +3\beta ^2+\beta ^3)\nonumber \\= & {} M_t^3+\alpha X_{t-1}+3\alpha ^2( X_{t-1})_{(2)}+\alpha ^3(-3 X_{t-1}^2+2 X_{t-1})\nonumber \\&+\,3(\alpha X_{t-1}-\alpha ^2 X_{t-1})\beta +3\alpha X_{t-1}\beta +(\beta +3\beta ^2)\nonumber \\= & {} M_t^3+3M_t V_t+\alpha X_{t-1}-3\alpha ^2 X_{t-1}+2\alpha ^3 X_{t-1}+\beta \nonumber \\= & {} M_t^3+3M_t\cdot V_t+(1-2\alpha )V_t+2\beta \alpha , \end{aligned}$$
(A.6)
where we used the moment formulae from Johnson et al. (2005 ), page 110, and the following simplification
$$\begin{aligned} M_t^3= & {} \beta ^3+3\beta ^2\alpha X_{t-1}+3\beta \alpha ^2 X_{t-1}^2+\alpha ^3 X_{t-1}^3,\\ M_t\cdot V_t= & {} \beta ^2+\alpha \beta (2-\alpha ) X_{t-1}+\alpha ^2(1-\alpha ) X_{t-1}^2. \end{aligned}$$
Analogously, we get for the fourth conditional moment
$$\begin{aligned} E[X_t^4\ | \ X_{t-1}]= & {} E[(\alpha \circ X_{t-1}+\epsilon _t)^4\ | \ X_{t-1}] \nonumber \\= & {} E[(\alpha \circ X_{t-1})^4\ | \ X_{t-1}]+4E[(\alpha \circ X_{t-1})^3\ | \ X_{t-1}]E[\epsilon _t]+E[\epsilon _t^4]\nonumber \\&+\,6E[(\alpha \circ X_{t-1})^2\ | \ X_{t-1}]E[\epsilon _t^2]+4E[(\alpha \circ X_{t-1})\ | \ X_{t-1}]E[\epsilon _t^3]\nonumber \\= & {} \alpha ^4( X_{t-1})_{(4)}+6\alpha ^3( X_{t-1})_{(3)}+7\alpha ^2( X_{t-1})_{(2)}+\alpha X_{t-1}\nonumber \\&+\,4(\alpha ^3( X_{t-1})_{(3)}+3\alpha ^2( X_{t-1})_{(2)}+\alpha X_{t-1})\cdot \beta \nonumber \\&+\,6(\alpha ^2( X_{t-1})_{(2)}+\alpha X_{t-1})\cdot (\beta +\beta ^2)\nonumber \\&+\,4\alpha X_{t-1}\cdot (\beta +3\beta ^2+\beta ^3)+(\beta ^4+6\beta ^3+7\beta ^2+\beta )\nonumber \\= & {} M_t^4+6M_t^2 V_t+M_t V_t(7-11\alpha )+V_t(1+11\alpha \beta )\nonumber \\&+\,2\alpha ^2(-3-3\alpha ^2+6\alpha +4\alpha \beta ) X_{t-1}. \end{aligned}$$
(A.7)
We are now in a position to compute the asymptotic covariances required for Lemma 1 . The CLT by Ibragimov (1962 ) implies that
$$\begin{aligned} \tilde{\sigma }_{00}= & {} E\big [(\tfrac{(X_{t}-M_{t})^2}{V_{t}}-1)^2\big ]\ +\ 2\sum \limits _{k=1}^\infty E\big [(\tfrac{(X_{t+k}-M_{t+k})^2}{V_{t+k}}-1) \cdot (\tfrac{(X_{t}-M_{t})^2}{V_{t}}-1)\big ]\\= & {} E\big [\tfrac{(X_t-M_t)^4}{V_t^2}\big ]-1\\&+\,2\sum \limits _{k=1}^\infty \left( E\Big [\tfrac{(X_{t}-M_{t})^2}{V_{t}}\cdot \tfrac{\overbrace{E[(X_{t+k}-M_{t+k})^2\ |\ X_{t+k-1}, \ldots ]}^{=V_{t+k}}}{V_{t+k}}\Big ]-1\right) \\= & {} \tfrac{1}{ V_t^2}\,E\big [E[(X_t-M_t)^4 \ | \ X_{t-1} ]\big ]-1\\= & {} 2+E\big [\tfrac{1}{V_t}\big ]-6\alpha ^2(1-\alpha )^2\,E\big [\tfrac{ X_{t-1}}{V_t^2}\big ]\\= & {} 2+\tfrac{1}{\beta }(1-\alpha (1-\alpha )E[\tfrac{ X_{t-1}}{V_t}])-6\alpha ^2(1-\alpha )^2\,E\big [\tfrac{ X_{t-1}}{V_t^2}\big ], \end{aligned}$$
where we used that \(E\left[ \frac{(X_t-M_t)^2}{V_t}\right] =1\) and
$$\begin{aligned}&E\Big [(X_t-M_t)^4 \ | \ X_{t-1} \Big ]\\&\quad =E[X_t^4\ | \ X_{t-1} ]-4E[X_t^3\ | \ X_{t-1} ]M_t+6E[X_t^2\ | \ X_{t-1} ]M_t^2-3M_t^4\\&\quad =M_t^4+6M_t^2 V_t+M_t V_t(7-11\alpha )+V_t(1+11\alpha \beta )\\&\qquad +\,2\alpha ^2(-3-3\alpha ^2+6\alpha +4\alpha \beta ) X_{t-1}\\&\qquad -\,4M_t(M_t^3+3M_t\cdot V_t+(1-2\alpha )V_t+2\beta \alpha )+6M_t^2(M_t^2+V_t^2)-3M_t^4\\&\quad =3(1-\alpha )M_tV_t+V_t(1+11\alpha \beta )-8\beta \alpha M_t+2\alpha ^2(-3-3\alpha ^2+6\alpha +4\alpha \beta ) X_{t-1}\\&\quad \overset{(A.4)}{=}3(1-\alpha )(\tfrac{V_t-\beta }{1-\alpha }+\beta )V_t+V_t(1+11\alpha \beta )-8\beta \alpha (\tfrac{V_t-\beta }{1-\alpha }+\beta )\\&\qquad +\,2(-3-3\alpha ^2+6\alpha +4\alpha \beta )((\tfrac{V_t-\beta }{1-\alpha }+\beta )-V_t)\\\\&\quad =3V_t^2+V_t+6\alpha (1-\alpha )(\beta -V_t)\\&\quad =3V_t^2+V_t-6\alpha ^2(1-\alpha )^2 X_{t-1}. \end{aligned}$$
Here, we used formulas (A.7 ) and (A.6 ) for the conditional moments.
For the next covariance, we need
$$\begin{aligned} E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_t\right]= & {} E\left[ \tfrac{1}{V_{t}}\, E\left[ (X_t-M_t)^2\ X_t\ |\ X_{t-1}\right] \right] \nonumber \\= & {} E\left[ \tfrac{1}{V_{t}} \big (E\left[ X_t^3\ |\ X_{t-1}\right] -2M_tE\left[ X_t^2\ |\ X_{t-1}\right] \right. \nonumber \\&\left. +M_t^2E\left[ X_t\ |\ X_{t-1}\right] \big )\right] \nonumber \\&\overset{(A.6)}{=}&E\left[ \tfrac{1}{V_{t}}(M_t^3+3M_tV_t\right. \nonumber \nonumber \\&\left. +(1-2\alpha )V_t+2\beta \alpha -2M_t(V_t+M_t^2)+M_t^3)\right] \qquad \nonumber \\= & {} E\left[ \tfrac{1}{V_{t}}(M_tV_t+(1-2\alpha )V_t+2\beta \alpha \right] \nonumber \\= & {} E[M_t]+1-2\alpha +2\beta \alpha E[\tfrac{1}{V_t}]\nonumber \\&\overset{(A.5)}{=}&\mu +1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
(A.8)
So it follows that
$$\begin{aligned} \tilde{\sigma }_{01}= & {} E\left[ (\tfrac{(X_t-M_t)^2}{V_{t}}-1)\cdot (X_t-\mu )\right] +\sum \limits _{k=1}^\infty E\left[ (\tfrac{(X_t-M_t)^2}{V_{t}}-1)\cdot ( X_{t+k}-\mu )\right] \\&+\,\sum \limits _{k=1}^\infty E\left[ (\tfrac{(X_{t+k}-M_{t+k})^2}{V_{t+k}}-1) \cdot (X_{t}-\mu )\right] \\= & {} E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_t\right] -\mu +\sum \limits _{k=1}^\infty \left( E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}\right] -\mu \right) \\&+\,\sum \limits _{k=1}^\infty \left( E\left[ \tfrac{(X_{t+k}-M_{t+k})^2}{V_{t+k}} \cdot X_{t}\right] -\mu \right) . \end{aligned}$$
Now, let us take a look at
$$\begin{aligned} E\left[ \tfrac{(X_{t+k}-M_{t+k})^2}{V_{t+k}} \cdot X_{t}\right] -\mu =E\left[ \tfrac{X_{t}}{V_{t+k}}\,E\left[ (X_{t+k}-M_{t+k})^2|\ X_{t+k-1},\ldots \right] \right] -\mu = 0, \end{aligned}$$
thus the last infinite sum vanishes. And the terms inside the first infinite sum can be calculated in the following way:
$$\begin{aligned}&E\Big [\tfrac{(X_t-M_t)^2}{V_{t}}\cdot \underbrace{E\left[ X_{t+k}|\ X_{t+k-1},\ldots \right] }_{\beta +\alpha X_{t+k-1}}\Big ]\nonumber \\&\quad =\ldots \ =\ \beta +\alpha \beta +\alpha ^2\beta +\ldots +\alpha ^{k-1}\beta +\alpha ^k E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_t\right] \nonumber \\&\quad \overset{(A.8)}{=}\tfrac{\beta (1-\alpha ^k)}{1-\alpha }+\alpha ^k(\mu +1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}])\nonumber \\&\quad =\mu +\alpha ^k(1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}]). \end{aligned}$$
(A.9)
Note that (A.9 ) also holds for \(k=0\) , see (A.8 ). So we get
$$\begin{aligned} \tilde{\sigma }_{01}= & {} \sum \limits _{k=0}^\infty (E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}\right] -\mu ) \ =\ \sum \limits _{k=0}^\infty \alpha ^k(1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}])\nonumber \\= & {} \frac{1}{1-\alpha }-2\alpha ^2\,E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
(A.10)
Now, let us look at
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_t^2\right] =E\left[ \tfrac{1}{V_{t}} E\left[ (X_t-M_t)^2\ X_t^2\ |\ X_{t-1}\right] \right] \nonumber \\&\quad =E\left[ \tfrac{1}{V_{t}}\, \big (E\left[ X_t^4\ |\ X_{t-1}\right] -2M_tE\left[ X_t^3\ |\ X_{t-1}\right] +M_t^2E\left[ X_t^2\ |\ X_{t-1}\right] \big )\right] \nonumber \\&\quad =E\Big [\tfrac{1}{V_{t}}\,\big (M_t^4+6M_t^2 V_t+M_t V_t(7-11\alpha )+V_t(1+11\alpha \beta )+M_t^2(V_t+M_t^2)\nonumber \\&\qquad +\,2\alpha ^2(-3-3\alpha ^2+6\alpha +4\alpha \beta ) X_{t-1}-2M_t(M_t^3+3M_t\cdot V_t+(1-2\alpha )V_t+2\beta \alpha )\big )\Big ]\nonumber \\&\quad =E[M_t^2]+E[M_t](5-7\alpha )+1+11\alpha (1-\alpha )\mu +2\alpha ^2(-3(1-\alpha )^2\nonumber \\&\qquad +\,4\alpha \mu (1-\alpha ))E[\tfrac{ X_{t-1}}{V_t}]-4\alpha \beta E[\tfrac{M_t}{V_t}]\nonumber \\&\quad =\alpha ^2\mu +\mu ^2+\mu (5-7\alpha )+1+11\alpha (1-\alpha )\mu \nonumber \\&\qquad +\,2\alpha ^2(1-\alpha )(-3+3\alpha +4\alpha \mu )E[\tfrac{ X_{t-1}}{V_t}]-4\alpha \mu (1-\alpha )(1+\alpha ^2 E[\tfrac{ X_{t-1}}{V_t}])\nonumber \\&\quad =(-6\alpha ^2+5+\mu )\mu +1+2\alpha ^2(1-\alpha )(-3+3\alpha +2\alpha \mu )E[\tfrac{ X_{t-1}}{V_t}], \end{aligned}$$
(A.11)
where we used (A.6 ) and (A.7 ) for the conditional moments, and \(M_t=V_t+\alpha ^2 X_{t-1}\) . This will be used for
$$\begin{aligned} \tilde{\sigma }_{02}= & {} \sum \limits _{k=0}^\infty E\left[ \left( \tfrac{(X_t-M_t)^2}{V_{t}}-1\right) \cdot ( X_{t+k}^2-\mu -\mu ^2)\right] \nonumber \\= & {} \sum \limits _{k=0}^\infty \left( E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}^2\right] -\mu -\mu ^2\right) . \end{aligned}$$
(A.12)
Now, let us take a look at the terms inside the infinite sum.
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}^2 \right] =E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot E\left[ X_{t+k}^2|\ X_{t+k-1},\ldots \right] \right] \nonumber \\&\quad = E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot (V_{t+k}+M_{t+k}^2)\right] \nonumber \\&\quad = E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot \big [(\beta +\alpha (1-\alpha ) X_{t+k-1})+(\beta +\alpha X_{t+k-1})^2\big ]\right] \nonumber \\&\quad =\beta (1+\beta )+\alpha (1-\alpha +2\beta ) E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}\right] +\alpha ^2E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}^2 \right] \nonumber \\&\quad \overset{A.9}{=}\beta (1+\beta )+\alpha (1-\alpha +2\beta ) (\mu +\alpha ^{k-1}(1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}]))\nonumber \\&\qquad +\,\alpha ^2E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}^2 \right] \nonumber \\&\quad =\mu (1-\alpha )(1+\mu (1-\alpha ))+\alpha (1-\alpha )(1+2\mu )\mu \nonumber \\&\qquad +\,\alpha ^k(1-\alpha )(1+2\mu )(1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}])+\alpha ^2E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}^2 \right] \nonumber \\&\quad =\mu (\mu +1)(1-\alpha ^2)+\alpha ^k(1-\alpha )(1+2\mu )(1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}])\nonumber \\&\qquad +\,\alpha ^2E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}^2 \right] . \end{aligned}$$
(A.13)
The above relationship can be viewed as a recurrence. Let us define \(g_k:= E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}^2 \right] \) with
$$\begin{aligned} g_0:=(-6\alpha ^2+5+\mu )\mu +1+2\alpha ^2(1-\alpha )(-3+3\alpha +2\alpha \mu )E\left[ \frac{ X_{t-1}}{V_t}\right] \end{aligned}$$
according to (A.11 ). Then, the first-order linear difference equation
$$\begin{aligned} g_k=\mu (\mu +1)(1-\alpha ^2)+\alpha ^k(1-\alpha )(1+2\mu )(1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}])+\alpha ^2\,g_{k-1} \end{aligned}$$
has the unique solution given by
$$\begin{aligned} g_k= & {} \textstyle g_0\cdot \prod _{i=0}^{k-1}\alpha ^2+\sum \limits _{j=0}^{k-1}\Big (\mu (\mu +1)(1-\alpha ^2)\nonumber \\&\textstyle +\,\alpha ^{j+1}(1-\alpha )(1+2\mu )\big (1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}]\big )\Big )\cdot \prod _{w =j+1}^{k-1}\alpha ^2\nonumber \\= & {} \textstyle g_0\cdot \alpha ^{2k}+\mu (\mu +1)(1-\alpha ^2)\sum \limits _{j=0}^{k-1}\alpha ^{2(k-1-j)}\nonumber \\&\textstyle +(1-\alpha )(1+2\mu )\big (1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}]\big )\sum \limits _{l=0}^{k-1} \alpha ^{l+1}\cdot \alpha ^{2(k-1-l)}\nonumber \\= & {} g_0\cdot \alpha ^{2k}+\mu (\mu +1)(1-\alpha ^2)\cdot \tfrac{1-\alpha ^{2k}}{1-\alpha ^2}\nonumber \\&+\,(1-\alpha )(1+2\mu )\big (1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}]\big )\cdot \alpha ^k\cdot \tfrac{1-\alpha ^k}{1-\alpha }\nonumber \\= & {} g_0\cdot \alpha ^{2k}+\mu (\mu +1)(1-\alpha ^{2k})\nonumber \\&+\,(1+2\mu )\big (1-2\alpha ^2(1-\alpha )\,E[\tfrac{ X_{t-1}}{V_t}]\big )(\alpha ^k-\alpha ^{2k}). \end{aligned}$$
(A.14)
If we insert (A.14 ) into Eq. (A.12 ), we get
$$\begin{aligned} \tilde{\sigma }_{02}= & {} \sum \limits _{k=0}^\infty (g_k-\mu -\mu ^2)\\= & {} \sum \limits _{k=0}^\infty \Big (g_0\cdot \alpha ^{2k}-\mu (\mu +1)\alpha ^{2k}\\&+\,(1+2\mu )\big (1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}]\big )(\alpha ^k-\alpha ^{2k})\Big )\\= & {} \frac{g_0}{1-\alpha ^2}-\frac{\mu (\mu +1)}{1-\alpha ^2}+\frac{(1+2\mu )\big (1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}]\big )\alpha }{1-\alpha ^2}. \end{aligned}$$
Together with \(g_0\) according to (A.11 ), it follows that
$$\begin{aligned} \tilde{\sigma }_{02}= & {} \frac{(-6\alpha ^2+5+\mu )\mu +1+2\alpha ^2(1-\alpha )(-3+3\alpha +2\alpha \mu )E[\tfrac{ X_{t-1}}{V_t}]}{1-\alpha ^2}\\&-\,\frac{\mu (\mu +1)}{1-\alpha ^2}+\frac{(1+2\mu )(1-2\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}])\alpha }{1-\alpha ^2}\\= & {} \frac{(-6\alpha ^2+2\alpha +4)\mu +1+\alpha +2\alpha ^2(1-\alpha )(-3+2\alpha )E[\tfrac{ X_{t-1}}{V_t}]}{1-\alpha ^2}\\= & {} \frac{2(3\alpha +2)\mu }{1+\alpha } +\frac{1}{1-\alpha } +\frac{2\alpha ^2(-3+2\alpha )}{1+\alpha }E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
Now, we have to calculate \(\tilde{\sigma }_{03}\) . For this purpose, we start with
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_t X_{t-1}\right] =E\left[ \tfrac{ X_{t-1}}{V_{t}}\, E\left[ (X_t-M_t)^2X_t \ | \ X_{t-1}\right] \right] \nonumber \\&\overset{(A.8)}{=}&E\left[ \tfrac{ X_{t-1}}{V_{t}}\cdot (M_t^3+3M_tV_t+(1-2\alpha )V_t+2\beta \alpha -2M_t(V_t+M_t^2)+M_t^3)\right] \nonumber \\&\quad =E[ X_{t-1} M_t]+2\beta \alpha \, E[\tfrac{ X_{t-1}}{V_t}]+(1-2\alpha )\mu \nonumber \\&\quad =(1-\alpha )\mu ^2+\alpha (\mu ^2+\mu )+2\beta \alpha \, E[\tfrac{ X_{t-1}}{V_t}]+(1-2\alpha )\mu \nonumber \\&\quad =\mu (\mu -\alpha +1)+2\mu (1-\alpha )\alpha \, E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
(A.15)
Then,
$$\begin{aligned} \tilde{\sigma }_{03}= & {} \sum \limits _{k=0}^\infty E\left[ (\tfrac{(X_t-M_t)^2}{V_{t}}-1)\cdot ( X_{t+k}X_{t+k-1}-\alpha \mu -\mu ^2) \right] \\&+\,\sum \limits _{k=1}^\infty E\left[ (\tfrac{(X_{t+k}-M_{t+k})^2}{V_{t+k}}-1)\cdot ( X_{t}X_{t-1}-\alpha \mu -\mu ^2) \right] ,\\ \end{aligned}$$
where
$$\begin{aligned}&E\left[ (\tfrac{(X_{t+k}-M_{t+k})^2}{V_{t+k}}-1)\cdot ( X_{t}X_{t-1}-\alpha \mu -\mu ^2) \right] \\&\quad =E\left[ \tfrac{X_{t}X_{t-1}}{V_{t+k}}\cdot E\left[ (X_{t+k}-M_{t+k})^2\ | \ X_{t+k-1}, \ldots \right] \right] -\alpha \mu -\mu ^2\\&\quad =E\left[ X_{t}X_{t-1} \right] -\alpha \mu -\mu ^2\ =\ 0. \end{aligned}$$
Furthermore,
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}X_{t+k-1} \right] -\alpha \mu -\mu ^2\\&\quad = E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}\cdot E\left[ X_{t+k}\ | \ X_{t+k-1},\ldots \right] \right] -\alpha \mu -\mu ^2\\&\quad = E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1} (\beta +\alpha X_{t+k-1})\right] -\alpha \mu -\mu ^2\\&\quad = \mu (1-\alpha )\, E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1} \right] +\alpha \, E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k-1}^2\right] -\alpha \mu -\mu ^2. \end{aligned}$$
Thus, using that \(\alpha \mu +\mu ^2=\alpha (\mu +\mu ^2)+(1-\alpha )\mu ^2\) , we obtain
$$\begin{aligned} \tilde{\sigma }_{03}= & {} E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t}X_{t-1}\right] -\alpha \mu -\mu ^2\\&+\,\sum \limits _{k=0}^\infty \left( \mu (1-\alpha )\, E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k} \right] +\alpha \, E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}^2\right] -\alpha \mu -\mu ^2\right) \\= & {} E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t}X_{t-1}\right] -\alpha \mu -\mu ^2+(1-\alpha )\mu \, \sum \limits _{k=0}^\infty \left( E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k} \right] -\mu \right) \\&+\,\alpha \, \sum \limits _{k=0}^\infty \left( E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t+k}^2\right] -\mu -\mu ^2\right) \\= & {} E\left[ \tfrac{(X_t-M_t)^2}{V_{t}}\cdot X_{t}X_{t-1}\right] -\alpha \mu -\mu ^2+(1-\alpha )\mu \, \tilde{\sigma }_{01}+\alpha \, \tilde{\sigma }_{02}, \end{aligned}$$
where we used (A.10 ) and (A.12 ) in the last step. Plugging-in \(\tilde{\sigma }_{01}, \tilde{\sigma }_{02}\) from Lemma 1 as well as (A.15 ), it follows that
$$\begin{aligned} \tilde{\sigma }_{03}= & {} \mu (\mu -\alpha +1)+2\mu (1-\alpha )\alpha \, E[\tfrac{ X_{t-1}}{V_t}]-\alpha \mu -\mu ^2\\&+\,(1-\alpha )\mu \, \big (\frac{1}{1-\alpha }-2\alpha ^2\,E[\tfrac{ X_{t-1}}{V_t}]\big )\\&+\,\alpha \, \big (\tfrac{2(3\alpha +2)\mu }{1+\alpha } +\tfrac{1}{1-\alpha } +\tfrac{2\alpha ^2(-3+2\alpha )}{1+\alpha }E[\tfrac{ X_{t-1}}{V_t}]\big ) \\= & {} 2\mu (1-\alpha )+2\alpha (1-\alpha )^2\mu \, E[\tfrac{ X_{t-1}}{V_t}]+\tfrac{2\alpha (3\alpha +2)\mu }{1+\alpha } +\tfrac{\alpha }{1-\alpha } +\tfrac{2\alpha ^3(-3+2\alpha )}{1+\alpha }E[\tfrac{ X_{t-1}}{V_t}]\\= & {} \tfrac{2\mu }{1+\alpha }\,(1+2\alpha +2\alpha ^2)+\tfrac{\alpha }{1-\alpha }+2\alpha \mu (1-\alpha )^2\, E[\tfrac{ X_{t-1}}{V_t}]+\tfrac{2\alpha ^3(-3+2\alpha )}{1+\alpha }\,E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
So the proof of Lemma 1 is complete.
1.3 Proof of Theorem 1
We use the Delta method the same way as described in the proof of Corollary 1 in Weiß et al. (2017 ), page 599. Let \({\varvec{g}}:\mathbb R^4\rightarrow \mathbb R^3\) be defined as
$$\begin{aligned} {\varvec{g}}(y_1,\ y_2,\ y_3,\ y_4)\ =\ (y_1,\ y_2\frac{y_3-y_4}{y_3-y_2^2},\ \frac{y_4-y_2^2}{y_3-y_2^2})^{\top }, \end{aligned}$$
where
$$\begin{aligned} {\varvec{g}}(1,\mu ,\mu +\mu ^2,\alpha \mu +\mu ^2)\ =\ (1,\ \beta ,\alpha ). \end{aligned}$$
Then, the Jacobian of \({\varvec{g}}\) is given by
$$\begin{aligned} \mathbf J _{{\varvec{g}}}(y_1,\ y_2,\ y_3,\ y_4)\ =\ \left( \begin{array}{cccc} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad \frac{(y_3-y_4)(y_3+y_2^2)}{(y_3-y_2^2)^2} &{}\quad \frac{y_2(y_4-y_2^2)}{(y_3-y_2^2)^2} &{}\quad \frac{-y_2}{y_3-y_2^2} \\ 0 &{}\quad \frac{2y_2(y_4-y_3)}{(y_3-y_2^2)^2} &{}\quad \frac{y_2^2-y_4}{(y_3-y_2^2)^2} &{}\quad \frac{1}{y_3-y_2^2} \\ \end{array} \right) , \end{aligned}$$
such that
$$\begin{aligned} \mathbf D :=\mathbf J _{{\varvec{g}}}(1,\mu ,\mu +\mu ^2,\alpha \mu ^2+\mu ^2)=\left( \begin{array}{cccc} 1 &{}\quad 0 &{}\quad 0 &{}\quad 0\\ 0 &{}\quad (1-\alpha )(1+2\mu ) &{} \quad \alpha &{} -1 \\ 0 &{}\quad -2(1-\alpha ) &{}\quad -\frac{\alpha }{\mu } &{}\quad \frac{1}{\mu } \\ \end{array} \right) . \end{aligned}$$
We need to calculate the covariance matrix \({\varvec{\Sigma }}=\mathbf D \tilde{\varvec{\Sigma }}\mathbf D ^{\top }\) with \(\tilde{\varvec{\Sigma }}\) from the previous Lemma 1 . The components \(\sigma _{22},\sigma _{23},\sigma _{33}\) are provided by Weiß and Schweer (2016 ), page 127. Furthermore, it obviously holds that \(\sigma _{11}=\tilde{\sigma }_{00}\) . Then,
$$\begin{aligned} \sigma _{12}= & {} \left( 0,\ (1-\alpha )(1+2\mu ),\ \alpha ,\ -1\right) \cdot \left( \tilde{\sigma }_{00},\ \tilde{\sigma }_{01},\ \tilde{\sigma }_{02},\ \tilde{\sigma }_{03} \right) ^{\top }\\= & {} 1+2\mu -2\alpha ^2(1-\alpha )(1+2\mu )E[\tfrac{ X_{t-1}}{V_t}]+\tfrac{2\mu (3\alpha ^2+2\alpha )}{1+\alpha }\\&-\,\tfrac{2\mu (1+2\alpha +2\alpha ^2)}{1+\alpha }-2\alpha \mu (1-\alpha )^2 E[\tfrac{ X_{t-1}}{V_t}]\\= & {} 1+2\mu -\tfrac{2\mu (1-\alpha ^2)}{1+\alpha }-2(1-\alpha )\alpha (\alpha +\mu \alpha +\mu )E[\tfrac{ X_{t-1}}{V_t}]\\= & {} 1+2\alpha \mu -2\alpha (1-\alpha )(\alpha +\mu \alpha +\mu )E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
Finally,
$$\begin{aligned} \sigma _{13}= & {} \left( 0,\ -2(1-\alpha ),\ -\tfrac{\alpha }{\mu },\ \tfrac{1}{\mu } \right) \cdot \left( \tilde{\sigma }_{00},\ \tilde{\sigma }_{01},\ \tilde{\sigma }_{02},\ \tilde{\sigma }_{03} \right) ^{\top }\\= & {} -2+4\alpha ^2(1-\alpha )E[\tfrac{ X_{t-1}}{V_t}]-\tfrac{2(3\alpha ^2+2\alpha )}{1+\alpha }+\tfrac{2(1+2\alpha +2\alpha ^2)}{1+\alpha }+2\alpha (1-\alpha )^2 E[\tfrac{ X_{t-1}}{V_t}]\\= & {} -2\alpha +2\alpha (1-\alpha ^2) E[\tfrac{ X_{t-1}}{V_t}]. \end{aligned}$$
1.4 Proof of Theorem 2
Using Theorem 1 and the Cramer–Wold device with
$$\begin{aligned} \textstyle {\varvec{l}}\ =\ \Big (1\ , -\frac{1}{\mu (1-\alpha )}(1-\alpha (1-\alpha )E[\frac{ X_{t-1}}{V_t}]),\ -(1-2\alpha )\,E[\frac{X_{t-1}}{V_{t}}]\Big )\in \mathbb R^3, \end{aligned}$$
it follows for the linear approximation (7 ) that
$$\begin{aligned} \textstyle \sqrt{n}\Big (\mathrm{MS}_R(\beta ,\alpha )-1-(\hat{\alpha }-\alpha )(1-2\alpha )E[\frac{X_{t-1}}{V_{t}}]-(\hat{\beta }-\beta )\frac{1}{\beta }(1-\alpha (1-\alpha )E[\frac{ X_{t-1}}{V_t}])\ \Big ) \end{aligned}$$
converges to the normal distribution with mean 0 and variance \(\sigma _{\mathrm{MS}_R}^2={\varvec{l}}{\varvec{\Sigma }}{\varvec{l}}^{\top }\) , where \({\varvec{\Sigma }}\) is the covariance matrix from Theorem 1 . After tedious calculations, the variance simplifies to
$$\begin{aligned} \sigma _{\mathrm{MS}_R}^2= & {} \frac{3-5\alpha }{1-\alpha }-6 (1-\alpha )^2 \alpha ^2 E[\tfrac{X_{t-1}}{V_t^2}]+\big (\tfrac{\alpha (2 \alpha (\mu +2)+8 \mu -1)}{\mu }-2\big )E[\tfrac{X_{t-1}}{V_t}]\\&+\,\tfrac{(1-\alpha )}{\mu }\left( 5 \alpha ^3 \mu -\alpha ^2 (\mu +3)-5 \alpha \mu +\alpha +\mu \right) E[\tfrac{X_{t-1}}{V_t}]^2. \end{aligned}$$
The proof is complete.
Proofs for Poi-INARCH(1) DGP 1.1 Linear approximation for squared residualsWe proceed in the same way as in Appendix A.1 . To apply the linear approximation (5 ) to the special case of a Poi-INARCH(1) DGP, we require the partial derivatives of \(V_t=M_t=\beta +\alpha \,X_{t-1}\) , which equal \(\frac{\partial }{\partial \alpha }\,V_t=X_{t-1}\) and \(\frac{\partial }{\partial \beta }\,V_t=1\) . Thus, it follows that
$$\begin{aligned} \textstyle E\big [\frac{\partial V_t}{\partial \alpha }\,\frac{1}{V_t}\big ]\ =\ E\big [\frac{X_{t-1}}{M_t}\big ],\qquad E\big [\frac{\partial V_t}{\partial \beta }\,\frac{1}{V_t}\big ]\ =\ E\big [\frac{1}{M_t}\big ]. \end{aligned}$$
This implies the linear Taylor approximation (10 ). Note that \(E\big [\frac{1}{M_t}\big ]\) can be rewritten as \(\frac{1}{\beta }\big (1-\alpha \,E\big [\frac{ X_{t-1}}{M_t}\big ]\big )\) .
1.2 Joint distribution of residuals and momentsThe Poi-INARCH(1) process satisfies the same mixing and moment conditions as stated for the Poi-INAR(1) process in Appendix A.2 . So again, we can use a CLT in the same way as described in Appendix A.2 for the vectors from Eq. (A.1 ). Note that in the INARCH(1) case, \(V_t\) equals \(M_t\) .
Lemma 2
Let \((X_t)_{\mathbb {Z}}\) be a Poi-INARCH(1) process with \(\mu =\frac{\beta }{1-\alpha }\) , \(\mu (k)=\frac{\mu \,\alpha ^k}{1-\alpha ^2}+\mu ^2\) and
$$\begin{aligned} Y_{t,0}\ =\ \frac{(X_t-M_t)^2}{M_t} - 1 \ =\ \frac{(X_t-\alpha \, X_{t-1} - \beta )^2}{\alpha \, X_{t-1} + \beta } - 1. \end{aligned}$$
Then, \(\frac{1}{\sqrt{n}} \sum _{t=1}^n {\varvec{Y}}_t\) is asymptotically normally distributed with mean \({\varvec{0}}\) and covariance matrix \(\tilde{{\varvec{\Sigma }}} = (\tilde{\sigma }_{ij})\) , where
$$\begin{aligned} \begin{array}{@{}rl@{\qquad }rl} \tilde{\sigma }_{00}\ =&{} 2+E[\frac{1}{M_t}], &{} \tilde{\sigma }_{01}\ =&{} \frac{1}{1-\alpha }, \\ \tilde{\sigma }_{02}\ =&{} \frac{1+4\beta +2\alpha \beta }{(1-\alpha )^2(1+\alpha )}, &{} \tilde{\sigma }_{03}\ =&{} \frac{\alpha +2\beta +4\alpha \beta }{(1-\alpha )^2 (1+\alpha )}. \end{array} \end{aligned}$$
The expressions for \(\tilde{\sigma }_{11},\ldots ,\tilde{\sigma }_{33}\) can be found in Theorem 2.2 of Weiß and Schweer (2016 ).
1.2.1 Proof of Lemma 2
The proof is done in complete analogy to the proof of Lemma 1 in Appendix A.2 . For computing conditional moments, we use that the conditional distribution of \(X_t\) given \(X_{t-1}\) is \(Poi (M_t)\) , so we use the formulae for Poisson moments in Johnson et al. (2005 ). Like in Appendix A.2 , using that \(E\big [\tfrac{(X_t-M_t)^2}{M_t}\big ]=1\) , we get
$$\begin{aligned} \tilde{\sigma }_{00}= & {} E\big [\tfrac{(X_t-M_t)^4}{M_t^2}\big ]-1 \ =\ E\big [\tfrac{1}{M_t^2}\,E[(X_t-M_t)^4 \ | \ X_{t-1} ]\big ]-1\\= & {} E\big [\tfrac{1}{M_t^2}\,E[X_t^4-4X_t^3M_t+6X_t^2M_t^2-4X_tM_t^3+M_t^4\ |\ X_{t-1}\ldots ]\big ]-1\\= & {} E\big [\tfrac{1}{M_t^2}\,\big (M_t+7M_t^2+6M_t^3+M_t^4-4(M_t+3M_t^2+M_t^3)M_t\\&+\,6(M_t+M_t^2)M_t^2-3M_t^4\big )\big ]-1\\= & {} E\big [\tfrac{1}{M_t^2}\,(M_t+3M_t^2)\big ]-1 \ =\ 2+E[\frac{1}{M_t}]. \end{aligned}$$
To compute \(\tilde{\sigma }_{01}\) , we first need
$$\begin{aligned} E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_t\right]= & {} E\left[ \tfrac{1}{M_t} E\left[ (X_t-M_t)^2\ X_t\ |\ X_{t-1}\right] \right] \nonumber \\= & {} E\left[ \tfrac{1}{M_t}(M_t+3M_t^2+M_t^3-2M_t^2-2M_t^3+M_t^3)\right] \nonumber \\= & {} 1+E[M_t]\ =\ 1+\mu . \end{aligned}$$
(B.1)
Then, like in Appendix A.2 , it follows that
$$\begin{aligned} \tilde{\sigma }_{01} \ =\ \sum \limits _{k=0}^\infty \left( E\left[ \tfrac{(X_t-M_t)^2}{M_{t}}\cdot X_{t+k}\right] -\mu \right) . \end{aligned}$$
(B.2)
The terms inside the infinite sum can be calculated as
$$\begin{aligned}&E\Big [\tfrac{(X_t-M_t)^2}{M_t}\cdot \underbrace{E\left[ X_{t+k}|\ X_{t+k-1},\ldots \right] }_{\beta +\alpha X_{t+k-1}}\Big ]-\mu \nonumber \\&\quad = \cdots \ =\ \beta +\alpha \beta +\alpha ^2\beta +\cdots +\alpha ^{k-1}\beta +\alpha ^k\, E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_t\right] -\mu \nonumber \\&\overset{(B.1)}{=}&\tfrac{\beta (1-\alpha ^k)}{1-\alpha }+\alpha ^k(1+\mu )-\mu =\alpha ^k+\tfrac{\beta }{1-\alpha }-\mu \ =\ \alpha ^k. \end{aligned}$$
(B.3)
So we get
$$\begin{aligned} \tilde{\sigma }_{01} \ =\ \sum \limits _{k=0}^\infty \left( E\left[ \tfrac{(X_t-M_t)^2}{M_{t}}\cdot X_{t+k}\right] -\mu \right) \ \overset{(B.3)}{=}\ \sum \limits _{k=0}^\infty \alpha ^k \ =\ \frac{1}{1-\alpha }. \end{aligned}$$
(B.4)
Now, let us look at
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_t^2\right] \ =\ E\left[ \tfrac{1}{M_t}\, E\left[ (X_t-M_t)^2\, X_t^2\ |\ X_{t-1}\right] \right] \nonumber \\&\quad =E\left[ \tfrac{1}{M_t}\,\big (M_t+7M_t^2+6M_t^3+M_t^4-2M_t(M_t+3M_t^2+M_t^3)+M_t^2(M_t+M_t^2)\big )\right] \nonumber \\&\quad =1+5\,E[M_t]+E[M_t^2] = 1+5\mu +\alpha ^2\,V[X_{t-1}]+\mu ^2 \nonumber \\&\quad = 1+5\mu +\mu ^2+\alpha ^2\,\tfrac{\mu }{1-\alpha ^2}. \end{aligned}$$
(B.5)
This will be used for
$$\begin{aligned} \tilde{\sigma }_{02} \ =\ \sum \limits _{k=0}^\infty (E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k}^2\right] -\tfrac{\mu }{1-\alpha ^2}-\mu ^2). \end{aligned}$$
(B.6)
We can rewrite the terms inside the infinite sum as
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k}^2 \right] \ =\ E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot E\left[ X_{t+k}^2|\ X_{t+k-1},\ldots \right] \right] \nonumber \\&\quad = E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot (M_{t+k}+M_{t+k}^2)\right] \nonumber \\&\quad = E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot \big [(\beta +\alpha X_{t+k-1})+(\beta +\alpha X_{t+k-1})^2\big ]\right] \nonumber \\&\quad = \beta (1+\beta )+\alpha (1+2\beta )\, E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k-1}\right] +\alpha ^2\,E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k-1}^2 \right] \nonumber \\&\quad \overset{(B.3)}{=} \mu (1-\alpha )\,(1+\beta )+\alpha (1+2\beta )\, (\mu +\alpha ^{k-1})+\alpha ^2\,E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k-1}^2 \right] \nonumber \\&\quad = \mu \,(1+\beta +\alpha \beta )+(1+2\beta )\,\alpha ^k+\alpha ^2\,E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k-1}^2 \right] . \end{aligned}$$
(B.7)
This relationship can be viewed as a recurrence, where \(1+\beta +\alpha \beta =1+\mu (1-\alpha ^2)\) . Defining \(g_k= E\left[ \frac{(X_t-M_t)^2}{M_t}\cdot X_{t+k}^2 \right] \) with
$$\begin{aligned} g_0\ =\ E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_t^2\right] \ =\ 1+5\mu +\mu ^2+\alpha ^2\,\tfrac{\mu }{1-\alpha ^2} \end{aligned}$$
according to (B.5 ), we have the first-order linear difference equation
$$\begin{aligned} g_k\ =\ \mu \,(1+\mu (1-\alpha ^2))+(1+2\beta )\,\alpha ^k\ +\ \alpha ^2\,g_{k-1}. \end{aligned}$$
It has the unique solution
$$\begin{aligned} g_k= & {} \textstyle g_0\cdot \prod \limits _{i=0}^{k-1}\alpha ^2+\sum \limits _{j=0}^{k-1}[\mu \,(1+\mu (1-\alpha ^2))+(1+2\beta )\,\alpha ^{j+1}]\cdot \prod \limits _{w =j+1}^{k-1}\alpha ^2\nonumber \\= & {} \textstyle g_0\cdot \alpha ^{2k}+\mu \,(1+\mu (1-\alpha ^2))\sum \limits _{j=0}^{k-1}\alpha ^{2(k-1-j)}+(1+2\beta )\,\sum \limits _{l=0}^{k-1} \alpha ^{l+1}\cdot \alpha ^{2(k-1-l)}\nonumber \\= & {} g_0\cdot \alpha ^{2k}+\mu \,(1+\mu (1-\alpha ^2))\cdot \tfrac{1-\alpha ^{2k}}{1-\alpha ^2}+(1+2\beta )\,\alpha ^k\cdot \tfrac{1-\alpha ^k}{1-\alpha }\nonumber \\= & {} g_0\cdot \alpha ^{2k}+\big (\tfrac{\mu }{1-\alpha ^2}+\mu ^2\big )\,(1-\alpha ^{2k})+\tfrac{1+2\beta }{1-\alpha }\cdot (\alpha ^k-\alpha ^{2k}), \end{aligned}$$
(B.8)
which also holds for \(k=0\) . Thus,
$$\begin{aligned} \tilde{\sigma }_{02}= & {} \sum \limits _{k=0}^\infty (g_k-\tfrac{\mu }{1-\alpha ^2}-\mu ^2)\nonumber \\= & {} \sum \limits _{k=0}^\infty \Big (g_0\cdot \alpha ^{2k} +\big (\tfrac{\mu }{1-\alpha ^2}+\mu ^2\big )\,(1-\alpha ^{2k}) +\tfrac{1+2\beta }{1-\alpha }\cdot (\alpha ^k-\alpha ^{2k})-\tfrac{\mu }{1-\alpha ^2}-\mu ^2\Big ). \end{aligned}$$
All the terms without the power k can be added together:
$$\begin{aligned} \big (\tfrac{\mu }{1-\alpha ^2}+\mu ^2\big )-\tfrac{\mu }{1-\alpha ^2}-\mu ^2\ =\ 0. \end{aligned}$$
So we can further calculate
$$\begin{aligned} \tilde{\sigma }_{02}= & {} \sum \limits _{k=0}^\infty \big (g_0\cdot \alpha ^{2k}-\big (\tfrac{\mu }{1-\alpha ^2}+\mu ^2\big )\,\alpha ^{2k}+\tfrac{1+2\beta }{1-\alpha }\,(\alpha ^k-\alpha ^{2k})\big )\\&\overset{(B.5)}{=}&\big (1+5\mu +\mu ^2+\alpha ^2\,\tfrac{\mu }{1-\alpha ^2}-\tfrac{\mu }{1-\alpha ^2}-\mu ^2-\tfrac{1+2\beta }{1-\alpha }\big )\,\sum \limits _{k=0}^\infty \alpha ^{2k} \ +\ \tfrac{1+2\beta }{1-\alpha }\,\sum \limits _{k=0}^\infty \alpha ^k\\= & {} \big (1+4\mu -\tfrac{1+2\beta }{1-\alpha }\big )\,\tfrac{1}{1-\alpha ^2} \ +\ \tfrac{1+2\beta }{1-\alpha }\,\tfrac{1}{1-\alpha } \ =\ \big (2\mu -\tfrac{\alpha }{1-\alpha }\big )\,\tfrac{1}{1-\alpha ^2} \ +\ \tfrac{1+2\beta }{(1-\alpha )^2}\\= & {} \tfrac{2\beta -\alpha }{(1-\alpha )^2(1+\alpha )} \ +\ \tfrac{(1+2\beta )(1+\alpha )}{(1-\alpha )^2(1+\alpha )} \ =\ \tfrac{1+4\beta +2\alpha \beta }{(1-\alpha )^2(1+\alpha )}. \end{aligned}$$
So it remains to calculate \(\tilde{\sigma }_{03}\) . First,
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_t X_{t-1}\right] =E\left[ \tfrac{ X_{t-1}}{M_t}\,E\left[ (X_t-M_t)^2X_t \ | \ X_{t-1}\right] \right] \nonumber \\&\quad = E\left[ \tfrac{ X_{t-1}}{M_t}\cdot (M_t+3M_t^2+M_t^3-2M_t(M_t+M_t^2)+M_t^3)\right] \nonumber \\&\quad = \mu +E\left[ X_{t-1} M_t\right] \ =\ \mu (1+\beta )+\alpha \,\mu (0) \nonumber \\&\quad = \mu (1+\beta )+\alpha \,\big (\tfrac{\mu }{1-\alpha ^2}+\mu ^2\big ) \ =\ \mu (1+\mu )+\tfrac{\alpha \,\mu }{1-\alpha ^2}. \end{aligned}$$
(B.9)
Then, like in Appendix A.2 , we have
$$\begin{aligned} \tilde{\sigma }_{03} \ =\ \sum \limits _{k=0}^\infty (E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k}X_{t+k-1} \right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2). \end{aligned}$$
Here,
$$\begin{aligned}&E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k}X_{t+k-1} \right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2\\&\quad =E\left[ \tfrac{(X_t-M_t)^2}{M_t}\,X_{t+k-1} \cdot E\left[ X_{t+k}\ | \ X_{t+k-1}, \ldots \right] \right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2\\&\quad =E\left[ \tfrac{(X_t-M_t)^2}{M_t}\,X_{t+k-1} \cdot (\beta +\alpha X_{t+k-1})\right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2\\&\quad =\beta \, E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k-1} \right] +\alpha \, E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k-1}^2\right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2. \end{aligned}$$
Thus, it follows that
$$\begin{aligned} \tilde{\sigma }_{03}= & {} E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t}X_{t-1} \right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2\\&+\sum \limits _{k=0}^\infty \Big (\beta \, E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k} \right] +\alpha \, E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t+k}^2\right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2\Big ). \end{aligned}$$
Comparing with (B.2 ) and (B.6 ), and using that
$$\begin{aligned} \beta \,\mu \ +\ \alpha \,\big (\tfrac{\mu }{1-\alpha ^2}+\mu ^2\big ) \ =\ \tfrac{\alpha \,\mu }{1-\alpha ^2}+\mu ^2, \end{aligned}$$
it follows that
$$\begin{aligned} \tilde{\sigma }_{03} \ =\ E\left[ \tfrac{(X_t-M_t)^2}{M_t}\cdot X_{t}X_{t-1} \right] -\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2 \ +\ \beta \,\tilde{\sigma }_{01} \ +\ \alpha \,\tilde{\sigma }_{02}. \end{aligned}$$
Plugging-in \(\tilde{\sigma }_{01},\tilde{\sigma }_{02}\) from Lemma 2 as well as (B.9 ), we obtain
$$\begin{aligned} \tilde{\sigma }_{03}= & {} \mu (1+\mu )+\tfrac{\alpha \,\mu }{1-\alpha ^2}-\tfrac{\alpha \,\mu }{1-\alpha ^2}-\mu ^2 \ +\ \beta \,\tfrac{1}{1-\alpha } \ +\ \alpha \,\tfrac{1+4\beta +2\alpha \beta }{(1-\alpha )^2(1+\alpha )}\\= & {} 2\mu \ +\ \alpha \,\tfrac{1+4\beta +2\alpha \beta }{(1-\alpha )^2(1+\alpha )} \ =\ \tfrac{2\beta (1-\alpha ^2)}{(1-\alpha )^2(1+\alpha )}\ +\ \tfrac{\alpha +4\alpha \beta +2\alpha ^2\beta }{(1-\alpha )^2(1+\alpha )} \ =\ \tfrac{\alpha +2\beta +4\alpha \beta }{(1-\alpha )^2 (1+\alpha )}. \end{aligned}$$
This completes the proof of Lemma 2 .
1.3 Proof of Theorem 3
The proof follows the same steps as in Appendix A.3 , and the required matrix \(\mathbf D \) can be directly taken from page 599 in Weiß et al. (2017 ):
$$\begin{aligned} \mathbf D =\left( \begin{array}{cccc} 1 &{} 0 &{} 0 &{} 0\\ 0 &{} (1-\alpha )\big (1+2\mu (1-\alpha ^2)\big )&{} \alpha (1-\alpha ^2) &{} -(1-\alpha ^2) \\ 0 &{} -2(1-\alpha )^2(1+\alpha ) &{} -(1-\alpha ^2)\,\frac{\alpha }{\mu } &{} (1-\alpha ^2)\,\frac{1}{\mu } \\ \end{array} \right) . \end{aligned}$$
Then, we calculate the covariance matrix \({\varvec{\Sigma }}=\mathbf D \tilde{\varvec{\Sigma }}\mathbf D ^{\top }\) with \(\tilde{\varvec{\Sigma }}\) from the previous Lemma 2 . The components \(\sigma _{22},\sigma _{23},\sigma _{33}\) are provided by Weiß and Schweer (2016 ), page 129, or by Weiß et al. (2017 ), page 599. Furthermore, it obviously holds that \(\sigma _{11}=\tilde{\sigma }_{00}\) . Then,
$$\begin{aligned} \sigma _{12}= & {} \left( 0,\ (1-\alpha )\big (1+2\mu (1-\alpha ^2)\big ),\ \alpha (1-\alpha ^2),\ -(1-\alpha ^2) \right) \cdot \left( \tilde{\sigma }_{00},\ \tilde{\sigma }_{01},\ \tilde{\sigma }_{02},\ \tilde{\sigma }_{03} \right) ^{\top }\\= & {} 1+2\mu (1-\alpha ^2)+\frac{\alpha (1+4\beta +2\alpha \beta )}{1-\alpha }-\frac{\alpha +2\beta +4\alpha \beta }{1-\alpha }\\= & {} 1+2\mu (1-\alpha ^2)+\frac{-2\beta +2\alpha ^2\beta }{1-\alpha }\ =\ 1. \end{aligned}$$
Finally,
$$\begin{aligned} \sigma _{13}= & {} \left( 0,\ -2(1-\alpha )^2(1+\alpha ),\ -(1-\alpha ^2)\,\tfrac{\alpha }{\mu },\ (1-\alpha ^2)\,\tfrac{1}{\mu } \right) \cdot \left( \tilde{\sigma }_{00},\ \tilde{\sigma }_{01},\ \tilde{\sigma }_{02},\ \tilde{\sigma }_{03} \right) ^{\top }\\= & {} -2(1-\alpha ^2)-\frac{1+4\beta +2\alpha \beta }{1-\alpha }\,\frac{\alpha }{\mu }+\frac{\alpha +2\beta +4\alpha \beta }{1-\alpha }\,\frac{1}{\mu }\\= & {} -2(1-\alpha ^2)+2(-\alpha ^2+1)\ =\ 0. \end{aligned}$$
1.4 Proof of Theorem 4
Using Theorem 3 and the Cramer–Wold device with
$$\begin{aligned} \textstyle {\varvec{l}}\ =\ \Big (1,\ -\frac{1}{\mu (1-\alpha )}\, \big (1-\alpha \,E\big [\frac{X_{t-1}}{M_t}\big ]\big ),\ -E[\frac{X_{t-1}}{M_t}] \Big )\in \mathbb {R}^3, \end{aligned}$$
it follows for the linear approximation (10 ) that
$$\begin{aligned} \textstyle \sqrt{n} \Big (\mathrm{MS}_R(\beta ,\alpha ) - E\big [\frac{X_{t-1}}{M_t}\big ]\,(\hat{\alpha }-\alpha ) - \frac{1}{\beta }\, \Big (1-\alpha \,E\big [\frac{X_{t-1}}{M_t}\big ]\Big )\,(\hat{\beta }-\beta )\ -1\Big ) \end{aligned}$$
converges to the normal distribution with mean 0 and variance \(\sigma _{\mathrm{MS}_R}^2={\varvec{l}}{\varvec{\Sigma }}{\varvec{l}}^{\top }\) , where \({\varvec{\Sigma }}\) is the covariance matrix from Theorem 3 . After tedious calculations, the variance simplifies to
$$\begin{aligned} \sigma _{\mathrm{MS}_R}^2= & {} \frac{\alpha ^4 (\mu +2)+\alpha ^3 (1-3 \mu )-\alpha \mu +3 \mu }{(1-\alpha )(1-\alpha ^3)\, \mu }\\&+\frac{(1+\alpha ) \left( \alpha ^3 (2 \mu -3)+\alpha ^2-\alpha -2 \mu \right) }{(1-\alpha )(1-\alpha ^3)\, \mu }\,E[\tfrac{X_{t-1}}{M_t}]\\&+\,\frac{\alpha ^4 (1-\mu ) +\alpha ^3(1-\mu )+\alpha \mu +\alpha +\mu }{(1-\alpha )(1-\alpha ^3)\, \mu }\,E[\tfrac{X_{t-1}}{M_t}]^2. \end{aligned}$$
Tables See Tables 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 and 9 .
Table 1 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) Table 2 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) (if \(I=1\) ) or NB-INAR(1) (if \(I>1\) ) Table 3 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) (if \(I=1\) ) or ZIP-INAR(1) (if \(I>1\) ) Table 4 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) (if \(I=1\) ) or Good-INAR(1) (if \(I<1\) ) Table 5 Hypothetical model: Poi-INARCH(1); DGP: Poi-INARCH(1) Table 6 Hypothetical model: Poi-INARCH(1); DGP: Poi-INARCH(1) (if \(\theta =1\) ) or NB-INARCH(1) (if \(\theta >1\) ) Table 7 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) (if \(I=1\) ) or NB-INAR(1) (if \(I>1\) ) Table 8 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) Table 9 Hypothetical model: Poi-INAR(1); DGP: Poi-INAR(1) (if \(I=1\) ) or NB-INAR(1) (if \(I>1\) ) 
