Parameter estimation and diagnostic tests for INMA(1) processes

Abstract

The INMA(1) model, an integer-valued counterpart to the usual moving-average model of order 1, gained recently importance for insurance applications. After a comprehensive discussion of stochastic properties of the INMA(1) model, we develop diagnostic tests regarding the marginal distribution (overdispersion, zero inflation) and the autocorrelation structure. We also derive formulae for correcting the bias of point estimators and for constructing joint confidence regions. These inferential approaches rely on asymptotic properties, the finite-sample performance of which is investigated with simulations. A real-data example illustrates the application of the novel diagnostic tools.

This is a preview of subscription content, log in to check access.

Fig. 1
Fig. 2
Fig. 3

References

  1. Al-Osh MA, Alzaid AA (1988) Integer-valued moving average (INMA) process. Stat Pap 29(1):281–300

    MathSciNet  Article  Google Scholar 

  2. Brännäs K, Hall A (2001) Estimation in integer-valued moving average models. Appl Stoch Models Bus Ind 17(3):277–291

    MathSciNet  Article  Google Scholar 

  3. Brännäs K, Quoreshi AMMS (2010) Integer-valued moving average modelling of the number of transactions in stocks. Appl Financ Econ 20(18):1429–1440

    Article  Google Scholar 

  4. Cossette H, Marceau E, Maume-Deschamps V (2010) Discrete-time risk models based on time series for count random variables. ASTIN Bull J IAA 40(1):123–150

    MathSciNet  Article  Google Scholar 

  5. Cossette H, Marceau E, Toureille F (2011) Risk models based on time series for count random variables. Insur Math Econ 48(1):19–28

    MathSciNet  Article  Google Scholar 

  6. Davis RA, Holan SH, Lund R, Ravishanker N (eds) (2016) Handbook of discrete-valued time series. Chapman & Hall, Boca Raton

    Google Scholar 

  7. Freeland RK (1998) Statistical analysis of discrete time series with applications to the analysis of workers compensation claims data. Ph.D. thesis, University of British Columbia, Canada. https://open.library.ubc.ca/cIRcle/collections/ubctheses/831/items/1.0088709

  8. Hall A (2001) Extremes of integer-valued moving averages models with regularly varying tails. Extremes 4(3):219–239

    MathSciNet  Article  Google Scholar 

  9. Hall A (2003) Extremes of integer-valued moving averages models with exponential type tails. Extremes 6(4):361–379

    MathSciNet  Article  Google Scholar 

  10. Hall A, Scotto MG, Cruz JP (2010) Extremes of integer-valued moving average sequences. Test 19(2):359–374

    MathSciNet  Article  Google Scholar 

  11. Hu X, Zhang L, Sun W (2018) Risk model based on the first-order integer-valued moving average process with compound Poisson distributed innovations. Scand Actuar J 5:412–425

    MathSciNet  Article  Google Scholar 

  12. Ibragimov I (1962) Some limit theorems for stationary processes. Theory Probab Appl 7(4):349–382

    MathSciNet  Article  Google Scholar 

  13. McKenzie E (1985) Some simple models for discrete variate time series. Water Resour Bull 21(4):645–650

    Article  Google Scholar 

  14. McKenzie E (1988) Some ARMA models for dependent sequences of Poisson counts. Adv Appl Probab 20(4):822–835

    MathSciNet  Article  Google Scholar 

  15. Romano JP, Thombs LA (1996) Inference for autocorrelations under weak assumptions. J Am Stat Assoc 91(434):590–600

    MathSciNet  Article  Google Scholar 

  16. Schweer S, Weiß CH (2014) Compound Poisson INAR(1) processes: stochastic properties and testing for overdispersion. Comput Stat Data Anal 77:267–284

    MathSciNet  Article  Google Scholar 

  17. Steutel FW, van Harn K (1979) Discrete analogues of self-decomposability and stability. Ann Probab 7(5):893–899

    MathSciNet  Article  Google Scholar 

  18. Weiß CH (2008) Serial dependence and regression of Poisson INARMA models. J Stat Plan Inference 138(10):2975–2990

    MathSciNet  Article  Google Scholar 

  19. Weiß CH (2018) An introduction to discrete-valued time series. Wiley, Chichester

    Google Scholar 

  20. Weiß CH, Homburg A, Puig P (2016) Testing for zero inflation and overdispersion in INAR(1) models. Stat Pap. https://doi.org/10.1007/s00362-016-0851-y

    Article  MATH  Google Scholar 

  21. Yu K, Zou H (2015) The combined Poisson INMA(q) models for time series of counts. J Appl Math. Article ID 457842

  22. Zhang L, Hu X, Duan B (2015) Optimal reinsurance under adjustment coefficient measure in a discrete risk model based on Poisson MA(1) process. Scand Actuar J 5:455–467

    MathSciNet  Article  Google Scholar 

  23. Zou H, Yu K (2014) First order threshold integer-valued moving average processes. Dyn Contin Discrete Impuls Sys Ser B 21(2–3):197–205

    MathSciNet  MATH  Google Scholar 

Download references

Acknowledgements

The authors thank the Associate Editor and the Reviewer for carefully reading the article and for their comments, which greatly improved the article.

Author information

Affiliations

Authors

Corresponding author

Correspondence to Christian H. Weiß.

Additional information

Publisher's Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Electronic supplementary material

Below is the link to the electronic supplementary material.

Supplementary material 1 (pdf 246 KB)

A Proofs

A Proofs

A.1 A central limit theorem

In order to prepare the asymptotic results to be derived for Sect. 3, we start with a central limit theorem (CLT) for the vector-valued process \(({\varvec{Y}}_t)_{\mathbb {Z}}\) of dimension \(K+3\) with some \(K\in \mathbb {N}\), which is defined by

$$\begin{aligned} {\varvec{Y}}_t\ :=\ \left( \begin{array}{c} Y_{t,0}\\ Y_{t,1}\\ Y_{t,2}\\ Y_{t,2+1}\\ \vdots \\ Y_{t,2+K}\\ \end{array}\right) \ :=\ \left( \begin{array}{c} {\mathbb {1}}_{\{X_t=0\}} - p_0\\ X_t - \mu \\ X_t^2 - \mu (0)\\ X_t X_{t-1} - \mu (1)\\ \vdots \\ X_t X_{t-K} - \mu (K) \end{array}\right) \end{aligned}$$
(A.1)

with \(\mu (k) := E[X_t X_{t-k}]\) and \(p_0 := P(X_t=0)\). More generally, we denote the joint mixed moments by

$$\begin{aligned} \begin{array}{rl} \mu (s_1,\ldots ,s_{r})\ :=&E[X_t\cdot X_{t+s_1}\cdots X_{t+s_{r}}] \end{array} \end{aligned}$$
(A.2)

with \(0\le s_1\le \cdots \le s_{r}\) and \(r\in \mathbb {N}\). Since the underlying INMA(1) process \((X_t)_{\mathbb {Z}}\) is 1-dependent, the derived process \(({\varvec{Y}}_t)_{\mathbb {Z}}\) is \((K+1)\)-dependent and hence \(\alpha \)-mixing with only finitely many non-zero weights. Therefore, with an analogous reasoning as in Schweer and Weiß (2014), we can apply the CLT by Ibragimov (1962) to \(\frac{1}{\sqrt{T}} \sum _{t=1}^T {\varvec{Y}}_t\), provided that moments of order \(>4\) exist.

Theorem 4

(CLT for INMA(1)) Let \((X_t)_{\mathbb {Z}}\) be an INMA(1) process according to (1), and assume that the underlying innovations \(\epsilon _t\) have existing moments of order \(>4\). Then

$$\begin{aligned} \begin{array}{l} \frac{1}{\sqrt{T}} \sum _{t=1}^T {\varvec{Y}}_t {\mathop {\longrightarrow }\limits ^{\mathcal {D}}} \text{ N }\left( \mathbf {0}, {\varvec{\varSigma }}\right) \quad \,\mathrm{with } \,{\varvec{\varSigma }}= \left( \sigma _{ij} \right) \, \mathrm{given \,by}\\ \sigma _{ij}\ =\ E\big [ Y_{0,i} Y_{0,j} \big ] + \sum _{r=1}^\infty \Big (E\big [Y_{0,i} Y_{r,j}\big ] + E\big [ Y_{r,i} Y_{0,j}\big ] \Big ). \end{array} \end{aligned}$$

Note that the infinite sums for computing \(\sigma _{ij}\) actually reduce to finite sums thanks to the \((K+1)\)-dependence of \(({\varvec{Y}}_t)_{\mathbb {Z}}\). Furthermore, the moment condition on the innovations is satisfied by any common count distribution; in particular, we have existing moments of any order for the Poisson INMA(1) process. Finally, the CLT in Theorem 4 is easily extended to arbitrary INMA(q) processes according to (2), where we just have \((K+\textit{q})\)-dependence for the corresponding process \(({\varvec{Y}}_t)_{\mathbb {Z}}\).

In the sequel, we have to compute some of the \(\sigma _{ij}\) according to Theorem 4 explicitly for proving the results from Sect. 3. For this purpose, we have to calculate some mixed moments (A.2), which is done in the subsequent appendix.

A.2 Some mixed moments for Poisson INMA(1) processes

In this appendix, we compute some mixed moments \(\mu (s_1,\ldots ,s_{r})\) of order \(r+1\le 4\) according to (A.2) for an underlying Poisson INMA(1) process. The computations behind the required moments are all done in a similar way; therefore, we exemplify these computations for one example, and we only provide the final results otherwise. If an equality makes explicit use of the Poisson assumption, then this is highlighted by writing \(\overset{\text{ Poi }}{=}\). But as pointed out in the end of this appendix, it is easily possible to extend our moment calculations to non-Poisson cases.

Lemma 1

Let \((X_t)_{\mathbb {Z}}\) be a Poisson INMA(1) process with \(\epsilon _t\sim \text{ Poi }({\uplambda })\), then

  1. (a)

    \(\mu (1)=\beta {\uplambda }+(1+\beta )^2{\uplambda }^2\),

  2. (b)

    \(\mu (1,1)=\beta {\uplambda }+ (1 + \beta ) (1 + 3 \beta ) {\uplambda }^2 + (1 + \beta )^3 {\uplambda }^3\),

  3. (c)

    \(\mu (1,2)=2 \beta (1 + \beta ) {\uplambda }^2 + (1 + \beta )^3 {\uplambda }^3\),

  4. (d)

    \( \mu (0,0,1)=\beta {\uplambda }+ (1 + \beta ) (1 + 7 \beta ) {\uplambda }^2 +3 (1 + \beta )^2 (1 + 2 \beta ) {\uplambda }^3 + (1 + \beta )^4 {\uplambda }^4\),

  5. (e)

    \(\mu (0,1,1)=\beta {\uplambda }+ (1+6\beta +7\beta ^2) {\uplambda }^2 + 2 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3 + (1 + \beta )^4 {\uplambda }^4\),

  6. (f)

    \( \mu (0,1,2)=2 \beta (1 + \beta ) {\uplambda }^2 + (1 + \beta )^2 (1 + 4 \beta ) {\uplambda }^3 + (1 +\beta )^4 {\uplambda }^4\),

  7. (g)

    \(\mu (1,1,2)=2 \beta (1 + 2 \beta ) {\uplambda }^2 + (1 + \beta )^2 (1+ 5 \beta ) {\uplambda }^3 + (1 + \beta )^4 {\uplambda }^4 \),

  8. (h)

    \( \mu (1,2,3)=\beta ^2 {\uplambda }^2 + 3 \beta (1 + \beta )^2 {\uplambda }^3 + (1 + \beta )^4 {\uplambda }^4\).

Proof

We show the detailed derivation for part (b) only, a similar procedure is used for all other mixed moments. In the first step, we expand the argument of the expectation operator (utilizing the independence of the innovations):

$$\begin{aligned} \mu (1,1)= & {} E[X_tX_{t+1}^2]\ =\ E\big [(\epsilon _t+\beta \circ \epsilon _{t-1})\cdot (\epsilon _{t+1}+\beta \circ \epsilon _t)^2\big ]\\= & {} E[\epsilon _t\,\epsilon _{t+1}^2] + E\big [\epsilon _t\cdot 2\epsilon _{t+1}\,(\beta \circ \epsilon _t)\big ] \\&+ E\big [\epsilon _t\,(\beta \circ \epsilon _t)^2\big ] + E\big [(\beta \circ \epsilon _{t-1})\,\epsilon _{t+1}^2\big ]\\&+\, E\big [(\beta \circ \epsilon _{t-1})\cdot 2\epsilon _{t+1}\,(\beta \circ \epsilon _t)\big ] + E\big [(\beta \circ \epsilon _{t-1})\,(\beta \circ \epsilon _t)^2\big ]\\= & {} E[\epsilon _t]\,E[\epsilon _{t+1}^2] + 2\,E[\epsilon _{t+1}]\,E\big [\epsilon _t\,(\beta \circ \epsilon _t)\big ] + E\big [\epsilon _t\,(\beta \circ \epsilon _t)^2\big ] \\&+ E[\beta \circ \epsilon _{t-1}]\,E[\epsilon _{t+1}^2]\\&+\, 2\,E[\epsilon _{t+1}]\,E[\beta \circ \epsilon _{t-1}]\, E[\beta \circ \epsilon _t] \\&+ E[\beta \circ \epsilon _{t-1}]\,E\big [(\beta \circ \epsilon _t)^2\big ]. \end{aligned}$$

In the second step, we make use of the innovations’ stationarity, and moments related to binomial thinnings are solved via conditioning on \(\epsilon _t\):

$$\begin{aligned} \mu (1,1)= & {} E[\epsilon _t]\,E[\epsilon _{t}^2] + 2\,E[\epsilon _{t}]\,E\big [\epsilon _t\,(\beta \circ \epsilon _t)\big ] + E\big [\epsilon _t\,(\beta \circ \epsilon _t)^2\big ]\\&+\, E[\beta \circ \epsilon _{t}]\,E[\epsilon _{t}^2] + 2\,E[\epsilon _{t}]\,E[\beta \circ \epsilon _{t}]^2 + E[\beta \circ \epsilon _{t}]\,E\big [(\beta \circ \epsilon _t)^2\big ]\\= & {} E[\epsilon _t]\,E[\epsilon _{t}^2] + 2\,E[\epsilon _{t}]\,E\big [\epsilon _t\,(\epsilon _t\,\beta )\big ] + E\big [\epsilon _t\,\big (\epsilon _t\,\beta (1-\beta ) + \epsilon _t^2\,\beta ^2\big )\big ]\\&+\, E[\epsilon _{t}\,\beta ]\,E[\epsilon _{t}^2] + 2\,E[\epsilon _{t}]\,E[\epsilon _{t}\,\beta ]^2 + E[\epsilon _{t}\,\beta ]\,E\big [\big (\epsilon _t\,\beta (1-\beta ) + \epsilon _t^2\,\beta ^2\big )\big ]\\= & {} (1+3\beta + \beta ^3)\,E[\epsilon _t]\,E[\epsilon _{t}^2] + \beta (1-\beta )\,E\big [\epsilon _t^2\big ] + \beta ^2\,E\big [\epsilon _t^3\big ]\\&+\, 2\,\beta ^2\,E[\epsilon _{t}]^3 + \beta ^2(1-\beta )\,E[\epsilon _{t}]^2. \end{aligned}$$

Only in the last step, we make use of the Poisson assumption:

$$\begin{aligned} \mu (1,1)&\overset{\text{ Poi }}{=} (1+3\beta + \beta ^3)\,{\uplambda }^{2}(1+{\uplambda }) + \beta (1-\beta )\,{\uplambda }(1+{\uplambda }) + \beta ^2\,{\uplambda }(1+3 {\uplambda }+{\uplambda }^2)\\&+\, 2\,\beta ^2\,{\uplambda }^3 + \beta ^2(1-\beta )\,{\uplambda }^2\\&= \beta {\uplambda }+ (1 + \beta ) (1 + 3 \beta ) {\uplambda }^2 + (1 + \beta )^3 {\uplambda }^3, \end{aligned}$$

which concludes the proof. \(\square \)

Note that the derivation is easily adopted to any other distributional assumption concerning \(\epsilon _t\), we just have to use the respective moments of \(\epsilon _t\) in the last step of the derivation.

Furthermore, the calculation of the mixed moments remains essentially the same if we extend the INMA(1) to the INMA(q) model from (2). The major difficulty will be to calculate the mixed moments of the form \(E\big [(\beta _{i_1} \circ _{t+i_1} \epsilon _t)^{k_1}\cdots ( \beta _{i_j}\circ _{t+i_j}\epsilon _{t})^{k_j}\big ]\). Since there are different INMA(q) models of the same model order \(\textit{q}\), based on the specification of the conditional distribution of \((\beta _0 \circ _{t} \epsilon _t, \beta _1\circ _{t+1}\epsilon _{t},\ldots , \beta _{\textit{q}}\circ _{t+\textit{q}}\epsilon _{t})\) given \(\epsilon _{t}\) (Weiß 2008), also see Sect. 1, the calculation of \(E\big [(\beta _{i_1} \circ _{t+i_1} \epsilon _t)^{k_1}\cdots ( \beta _{i_j}\circ _{t+i_j}\epsilon _{t})^{k_j}\big ]\) depends on the specific type of selected INMA(q) model.

A.3 Proofs of Section 3.1

The results derived in this and the subsequent appendices refer to the case of a Poisson INMA(1) process. In this special case, we can make use of the time reversibility (TR) of \((X_t)_{\mathbb {Z}}\) (Al-Osh and Alzaid 1988), also see Sect. 2. This will help us to simplify some of the computations, which is highlighted below accordingly. Generally, it is also possible to extend our derivations to non-Poisson cases, but then some of the simplifications below are not valid anymore. This is exemplified by Supplement S.2, where we derive the asymptotic distribution of \(\hat{I}_{\text{ disp }}\) for general INMA(1) processes.

For deriving the joint asymptotic distribution of the statistics \(\hat{I}_{\text{ disp }}\) from (9) and \(\hat{I}_{\text{ zero }}\) from (10) referring to the marginal distribution, it suffices to concentrate on the components

$$\begin{aligned} \big (Y_{t,0}, Y_{t,1}, Y_{t,2}\big )^{\top } \ =\ \big ({\mathbb {1}}_{\{X_t=0\}} - p_0,\ X_t - \mu ,\ X_t^2 - \mu (0)\big )^{\top } \end{aligned}$$

of the vector \({\varvec{Y}}_t\) from (A.1). So we need to calculate the entries \(\sigma _{ij}\) with \(0\le i\le j\le 2\) from the covariance matrix \({\varvec{\varSigma }}\) in Theorem 4. Here and in the sequel, if an equality makes explicit use of the Poisson assumption, then this is highlighted by writing \(\overset{\text{ Poi }}{=}\), and the use of time reversibility by \(\overset{\text{ TR }}{=}\).

$$\begin{aligned} \sigma _{0,0}&= \textstyle E\left[ (\mathbb {1}_{\{X_t=0\}}-p_0)^2\right] +2\sum \limits _{r=1}^\infty \underbrace{E\left[ (\mathbb {1}_{\{X_{t+r}=0\}}-p_0)(\mathbb {1}_{\{X_t=0\}}-p_0)\right] }_{=0\text { for }r>2}\\&= -3p_0^2+E[\mathbb {1}_{\{X_{t}=0\}}^2]+2E[\mathbb {1}_{\{X_{t+1}=0\}}\mathbb {1}_{\{X_t=0\}}]\\= & {} -3p_0^2+E[\mathbb {1}_{\{X_{t}=0\}}]+2\, E[{\mathbb {1}}_{\{\beta \circ \epsilon _t+\epsilon _{t+1}=0\}}{\mathbb {1}}_{\{\beta \circ \epsilon _{t-1}+\epsilon _t=0\}}]\\&= -3p_0^2+p_0+2\, E[{\mathbb {1}}_{\{\epsilon _{t+1}=0\}}{\mathbb {1}}_{\{\beta \circ \epsilon _t=0\}}{\mathbb {1}}_{\{\epsilon _t=0\}}{\mathbb {1}}_{\{\beta \circ \epsilon _{t-1}=0\}}]\\&= -3p_0^2+p_0+2\, \underbrace{E[{\mathbb {1}}_{\{\epsilon _{t+1}=0\}}]}_{=P(\epsilon _{t+1}=0)}\, \underbrace{E[{\mathbb {1}}_{\{\beta \circ \epsilon _t=0\}}{\mathbb {1}}_{\{\epsilon _t=0\}}]}_{=E[{\mathbb {1}}_{\{\epsilon _t=0\}}]=P(\epsilon _{t}=0)}\, \underbrace{E[{\mathbb {1}}_{\{\beta \circ \epsilon _{t-1}=0\}}]}_{=E[(1-\beta )^{\epsilon _{t-1}}]}\\&\overset{\text{ Poi }}{=} -3p_0^2+p_0+2\hbox {e}^{-{\uplambda }}\cdot \hbox {e}^{-{\uplambda }}\cdot \hbox {e}^{-{\uplambda }\beta }\ =\ p_0(-3p_0+1+2 \hbox {e}^{-{\uplambda }}). \end{aligned}$$

Next,

$$\begin{aligned} \sigma _{1,1}= & {} \textstyle E[(X_t-\mu )^2]+2\sum \limits _{r=1}^\infty \underbrace{E[(X_t-\mu )(X_{t+r}-\mu )]}_{=0\text { for }r>2}\\= & {} \mu (0)-\mu ^2+2\big (\mu (1) -\mu ^2\big )\ \overset{\text{ Poi }}{=}\ {\uplambda }(1+\beta )+2\beta {\uplambda }\ =\ {\uplambda }(1+3\beta ), \end{aligned}$$

where we used Lemma 1 (a).

$$\begin{aligned} \sigma _{2,2}&=\textstyle E\big [\big (X_t^2-\mu (0)\big )^2\big ]+2\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t}^2-\mu (0)\big )\big (X_{t+r}^2-\mu (0)\big )\big ]}_{=0\text { for }r>1}\\&= E[X_t^4]-\mu (0)^2+2\,E[X_t^2X_{t+1}^2]-2\mu (0)^2\\&= \mu (0,0,0)+2\mu (0,1,1)-3\mu (0)^2\\&\overset{\text{ Poi }}{=} ({\uplambda }+{\uplambda }\beta )+7({\uplambda }+{\uplambda }\beta )^2+6({\uplambda }+{\uplambda }\beta )^3+({\uplambda }+{\uplambda }\beta )^4\\&+\,2\big (\beta {\uplambda }+ (1 + 6 \beta + 7 \beta ^2) {\uplambda }^2 + 2 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3 + (1 + \beta )^4 {\uplambda }^4\big )\\&-\,3\big ({\uplambda }+{\uplambda }\beta +({\uplambda }+{\uplambda }\beta )^2\big )^2\\&= (1+3 \beta ) {\uplambda }+2(3+10 \beta +9\beta ^2)\, {\uplambda }^2+4 (1+\beta )^2 (1+3 \beta ) {\uplambda }^3, \end{aligned}$$

where we used Lemma 1 (e).

$$\begin{aligned} \sigma _{1,2}&= E\big [(X_t-\mu )\big (X_t^2-\mu (0)\big )\big ]\\&\textstyle +\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_t-\mu )\big (X_{t+r}^2-\mu (0)\big )\big ]}_{=0\text { for }r>1}+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}-\mu )\big (X_t^2-\mu (0)\big )\big ]}_{=0\text { for }r>1}\\&= \mu (0,0)-\mu \,\mu (0)+ E[X_t\,X_{t+1}^2]+E[X_{t+1}\,X_t^2]-2\mu \mu (0)\\&\overset{\text{ TR }}{=} \mu (0,0)-\mu \,\mu (0)+ 2\big (\mu (1,1)-\mu \,\mu (0)\big )\\&\overset{\text{ Poi }}{=} {\uplambda }(1+\beta )+2({\uplambda }+{\uplambda }\beta )^2\ +\ 2\big (\beta {\uplambda }+2\beta (1+\beta ){\uplambda }^2\big )\\&= (1 + 3 \beta ) {\uplambda }+ 2 (1 +\beta )(1+3 \beta ) {\uplambda }^2, \end{aligned}$$

where we used Lemma 1 (b).

$$\begin{aligned} \sigma _{0,1}&= E\left[ ({\mathbb {1}}_{\{X_t=0\}}-p_0)(X_t-\mu )\right] \\&\textstyle +\,\sum \limits _{r=1}^\infty \underbrace{E\left[ ({\mathbb {1}}_{\{X_t=0\}}-p_0)(X_{t+r}-\mu )\right] }_{=0\text { for }r>1}+\sum \limits _{r=1}^\infty \underbrace{E\left[ ({\mathbb {1}}_{\{X_{t+r}=0\}}-p_0)(X_t-\mu )\right] }_{=0\text { for }r>1}\\&= -3p_0\mu +\underbrace{E[{\mathbb {1}}_{\{X_t=0\}}X_t]}_{=0}+E[{\mathbb {1}}_{\{X_t=0\}}X_{t+1}]+E[{\mathbb {1}}_{\{X_{t+1}=0\}}X_t]\\&\overset{\text{ TR }}{=} -3p_0\mu +2E[{\mathbb {1}}_{\{\beta \circ \epsilon _{t-1}=0\}}]\,E\big [{\mathbb {1}}_{\{\epsilon _t=0\}}(\beta \circ \epsilon _t+\epsilon _{t+1})\big ]\\&= -3p_0\mu +2E[(1-\beta )^{\epsilon _{t-1}}]\,\Big (\underbrace{E\big [{\mathbb {1}}_{\{\epsilon _t=0\}}(\beta \circ \epsilon _t)\big ]}_{=0}+E[{\mathbb {1}}_{\{\epsilon _t=0\}}]\,E[\epsilon _{t+1}]\Big )\\&\overset{\text{ Poi }}{=} -3p_0\mu +2\hbox {e}^{-{\uplambda }\beta }\cdot \hbox {e}^{-{\uplambda }}{\uplambda }\ =\ p_0(-3\mu +2{\uplambda }) \ =\ -p_0\,{\uplambda }(1+3\beta ). \end{aligned}$$

Analogously,

$$\begin{aligned} \sigma _{0,2}&= E\left[ ({\mathbb {1}}_{\{X_t=0\}}-p_0)(X_t^2-\mu (0))\right] \\&\textstyle +\,\sum \limits _{r=1}^\infty \underbrace{E\left[ ({\mathbb {1}}_{\{X_t=0\}}-p_0)(X_{t+r}^2-\mu (0))\right] }_{=0\text { for }r>1}+\sum \limits _{r=1}^\infty \underbrace{E\left[ ({\mathbb {1}}_{\{X_{t+r}=0\}}-p_0)(X_t^2-\mu (0))\right] }_{=0\text { for }r>1}\\&= -3p_0\mu (0)+E[{\mathbb {1}}_{\{X_t=0\}}X_t^2]+E[{\mathbb {1}}_{\{X_t=0\}}X_{t+1}^2]+E[{\mathbb {1}}_{\{X_{t+1}=0\}}X_t^2]\\&\overset{\text{ TR }}{=} -3p_0\mu (0)+2\,E[{\mathbb {1}}_{\{\beta \circ \epsilon _{t-1}=0\}}]E[{\mathbb {1}}_{\{\epsilon _t=0\}}X_{t+1}^2]\\&= -3p_0\mu (0)+2\,E[(1-\beta )^{\epsilon _{t-1}}]\,\underbrace{E[{\mathbb {1}}_{\{\epsilon _t=0\}}(\beta \circ \epsilon _t+\epsilon _{t+1})^2]}_{=E[{\mathbb {1}}_{\{\epsilon _t=0\}}\epsilon _{t+1}^2]=E[{\mathbb {1}}_{\{\epsilon _t=0\}}]\, E[\epsilon _{t+1}^2]}\\&\overset{\text{ Poi }}{=} -3p_0\,\big ({\uplambda }(1+\beta )+({\uplambda }+{\uplambda }\beta )^2\big )\ +\ 2\hbox {e}^{-{\uplambda }\beta }\cdot \hbox {e}^{-{\uplambda }}({\uplambda }+{\uplambda }^2)\\&= -p_0\,\big ((1 + 3 \beta ) {\uplambda }+(1 + 6 \beta + 3 \beta ^2) {\uplambda }^2\big ). \end{aligned}$$

After having computed the required \(\sigma _{ij}\), we proceed as in Schweer and Weiß (2014) and Weiß et al. (2016) and compute the asymptotic distribution and bias of \(\hat{I}_{\text{ disp }}\) and \(\hat{I}_{\text{ zero }}\), respectively, based on appropriate Taylor expansions. For this purpose, let us define the following functions \(g_i: \mathbb {R}^3\rightarrow \mathbb {R}\), \(i=1,2\), by

$$\begin{aligned} g_1(y_0,y_1,y_2)\ :=\ \frac{y_2-y_1^2}{y_1},\quad g_2(y_0,y_1,y_2)\ :=\ y_0\,\exp {y_1}, \end{aligned}$$

since \(g_1\big (p_0,\mu ,\mu (0)\big )=I_{\text{ disp }}\) and \(g_2\big (p_0,\mu ,\mu (0)\big )=I_{\text{ zero }}\). Note that \(\big (p_0,\mu ,\mu (0)\big ) = \big (\hbox {e}^{-\mu },\mu ,\mu (1+\mu )\big )\) in the case of the Poisson INMA(1) model. First, we need to compute the Jacobian \({\mathbf{J }}_{{\varvec{g}}}(y_0,y_1,y_2)\) of \({\varvec{g}}:=(g_1,g_2)^{\top }\) as well as the Hessians \({\mathbf{H }}_{g_i}(y_0,y_1,y_2)\) of \(g_i\) for \(i=1,2\).

$$\begin{aligned} {\mathbf{J }}_{{\varvec{g}}}(y_0,y_1,y_2)= & {} \left( \begin{array}{lll} 0 &{}\quad -1-\frac{y_2}{y_1^2} &{}\quad \frac{1}{y_1}\\ \hbox {e}^{y_1} &{}\quad \hbox {e}^{y_1}y_0 &{}\quad 0 \end{array} \right) , \\ {\mathbf{H }}_{g_1}(y_0,y_1,y_2)= & {} \left( \begin{array}{lll} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \frac{2 y_2}{y_1^3} &{}\quad \frac{-1}{y_1^2} \\ 0 &{}\quad \frac{-1}{y_1^2} &{}\quad 0 \\ \end{array} \right) ,\qquad {\mathbf{H }}_{g_2}(y_0,y_1,y_2) = \left( \begin{array}{lll} 0 &{}\quad \hbox {e}^{y_1} &{}\quad 0 \\ \hbox {e}^{y_1} &{}\quad \hbox {e}^{y_1} y_0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \\ \end{array} \right) . \end{aligned}$$

Next, we evaluate the Jacobian in \((\hbox {e}^{-\mu },\ \mu ,\ \mu (1+\mu ))\) and obtain \({\mathbf{D }}:={\mathbf{J }}_{{\varvec{g}}}\big (\hbox {e}^{-\mu },\ \mu ,\ \mu (1+\mu )\big )\) as

$$\begin{aligned} {\mathbf{D }}=\left( \begin{array}{lll} 0 &{}\quad -2-\frac{1}{\mu } &{}\quad \frac{1}{\mu }\\ \hbox {e}^{\mu } &{}\quad 1 &{}\quad 0 \end{array} \right) . \end{aligned}$$

The application of the Delta method implies that, asymptotically, \((\hat{I}_{\text{ disp }},\hat{I}_{\text{ zero }})^{\top }\) are jointly normally distributed with covariance matrix \({\mathbf{D }}\, \big (\sigma _{ij}\big )_{0\le i,j\le 2}\, {\mathbf{D }}^{\top }\) given as

$$\begin{aligned} \left( \begin{array}{ll} \frac{2\,(1 + 2 \beta + 3 \beta ^2)}{(1+\beta )^2} &{}\quad \frac{{\uplambda }\,(1 + 2\beta + 3 \beta ^2) }{1+\beta }\\ \frac{{\uplambda }\,(1 + 2\beta + 3 \beta ^2) }{1+\beta } &{}\quad 2 \hbox {e}^{\beta {\uplambda }} + \hbox {e}^{(1 + \beta ) {\uplambda }} - (1 + 3 \beta ) {\uplambda }-3 \end{array} \right) , \end{aligned}$$

where we used the relationship \(\mu ={\uplambda }(1+\beta )\).

To obtain the asymptotic bias corrections, we first evaluate the Hessians \({\mathbf{H }}_{g_i}\) for \(i=1,2\) in \((\hbox {e}^{-\mu },\ \mu ,\ \mu (1+\mu ))\), leading to the matrices \({\mathbf{H }}_{i}={\mathbf{H }}_{g_i}\big (\hbox {e}^{-\mu },\ \mu ,\ \mu (1+\mu )\big )\) given by

$$\begin{aligned} {\mathbf{H }}_1= \left( \begin{array}{lll} 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad \frac{2 (\mu +1)}{\mu ^2} &{}\quad -\frac{1}{\mu ^2} \\ 0 &{}\quad -\frac{1}{\mu ^2} &{}\quad 0 \end{array} \right) ,\qquad {\mathbf{H }}_2=\left( \begin{array}{lll} 0 &{}\quad \hbox {e}^{\mu } &{}\quad 0 \\ \hbox {e}^{\mu } &{}\quad 1 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 \end{array} \right) . \end{aligned}$$

Then, abbreviating \({\varvec{Z}}_T := \frac{1}{\sqrt{T}} \sum _{t=1}^T (Y_{t,0},Y_{t,1},Y_{t,2})^{\top }\), we compute the bias correction for the dispersion index and zero index as

$$\begin{aligned}&T\,E\big [\hat{I}_{\text{ disp }}-I_{\text{ disp }}\big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{1}\,{\varvec{Z}}_T\big ]\ \approx \ -3+\frac{2}{1+\beta },\\&T\,E\big [\hat{I}_{\text{ zero }}-I_{\text{ zero }}\big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{2}\,{\varvec{Z}}_T\big ]\ \approx \ -\frac{(1+3\beta ){\uplambda }}{2}. \end{aligned}$$

This concludes the proof of Theorem 1.

A.4 Proofs of Section 3.2

For deriving the joint asymptotic distribution of \(\big (\hat{\rho }(1),\ldots ,\hat{\rho }(K)\big )^{\top }\), we concentrate on the components

$$\begin{aligned}&\big (Y_{t,1}, Y_{t,2+0},Y_{t,2+1},\ldots ,Y_{t,2+K}\big )^{\top } \\&\quad = \big (X_t - \mu ,\ X_t^2 - \mu (0),\ X_t X_{t-1} - \mu (1),\ \ldots ,\ X_t X_{t-K} - \mu (K)\big )^{\top } \end{aligned}$$

of the vector \({\varvec{Y}}_t\) from (A.1). So we need to calculate the entries \(\sigma _{ij}\) with \(1\le i\le j\le 2+K\) from the covariance matrix \({\varvec{\varSigma }}\) in Theorem 4, where some of them we have already calculated in the “Appendix A.3”, namely \(\sigma _{1,1},\ \sigma _{1,2},\ \sigma _{2,2}\). We now compute the remaining ones, using the mixed moments provided by Lemma 1:

$$\begin{aligned} \sigma _{2+1,2+1}&= \textstyle E\big [\big (X_tX_{t-1}-\mu (1)\big )^2\big ]+2\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_tX_{t-1}-\mu (1)\big )\big (X_{t+r}X_{t-1+r}-\mu (1)\big )\big ]}_{=0\text { for }r>2}\\&= -5\,\mu (1)^2+E[X_{t-1}^2X_t^2]+2\,E[X_{t-1}X_t^2X_{t+1}]+2\,E[X_{t-1}X_tX_{t+1}X_{t+2}]\\&= -5\,\mu (1)^2+\mu (0,1,1)+2\,\mu (1,1,2)+2\,\mu (1,2,3)\\&\overset{\text{ Poi }}{=} \beta {\uplambda }+ (1 + 10 \beta + 12 \beta ^2) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3, \end{aligned}$$

where we used Lemma 1 (a), (e), (g) and (h).

$$\begin{aligned} \sigma _{2+1,2+2}&= \textstyle E\big [\big (X_tX_{t-1}-\mu (1)\big )(X_tX_{t-2}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_tX_{t-1}-\mu (1)\big )(X_{t+r}X_{t+r-2}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t+r}X_{t-1+r}-\mu (1)\big )(X_tX_{t-2}-\mu ^2)\big ]}_{=0\text { for }r>2}\\&= -6\,\mu (1)\,\mu ^2+E[X_{t-1}X_t^2X_{t-2}]+E[X_tX_{t-1}^2X_{t+1}]+E[X_t^2X_{t-1}]\,E[X_{t+2}]\\&+\,E[X_{t-1}X_{t+1}]\,E[X_{t+3}]+E[X_{t+1}X_{t}^2]\,E[X_{t-2}]+E[X_{t+2}X_{t+1}X_{t}]\,E[X_{t-2}]\\&\overset{\text{ TR }}{=} -6\,\mu (1)\,\mu ^2+2\,E[X_{t-1}^2X_tX_{t+1}]+2\,E[X_{t-1}X_t^2]\,E[X_{t+2}]\\&+2\,E[X_{t-1}X_tX_{t+1}]\,E[X_{t+3}]\\&= -6\,\mu (1)\,\mu ^2+2\,\mu (0,1,2)+2\,\mu (1,1)\,\mu +2\,\mu (1,2)\,\mu \\&\overset{\text{ Poi }}{=} 6 \beta (1 + \beta ) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3, \end{aligned}$$

where we first used the time reversibility and then the Poissonity of the considered INMA(1) process. Similarly,

$$\begin{aligned} \sigma _{2+1,2+3}&= \textstyle E\big [\big (X_tX_{t-1}-\mu (1)\big )(X_tX_{t-3}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t+r}X_{t-1+r}-\mu (1)\big )(X_tX_{t-3}-\mu ^2)\big ]}_{=0\text { for }r>2}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_tX_{t-1}-\mu (1)\big )(X_{t+r}X_{t+r-3}-\mu ^2)\big ]}_{=0\text { for }r>4}\\&= -7\mu (1)\mu ^2+E[X_t^2X_{t-1}]\,E[X_{t-3}]+E[X_t^2X_{t+1}]\,E[X_{t-3}]\\&+\,E[X_tX_{t+1}X_{t+2}]E[X_{t-3}]+E[X_tX_{t-1}X_{t+1}X_{t-2}]+E[X_tX_{t-1}^2]\,E[X_{t+2}]\\&+\,E[X_t^2X_{t-1}]\,E[X_{t+3}]+E[X_tX_{t-1}X_{t+1}]\,E[X_{t+4}]\\&\overset{\text{ TR }}{=} -7\,\mu (1)\,\mu ^2+4\,E[X_{t-1}X_t^2]\,E[X_{t-3}]+2\,E[X_tX_{t+1}X_{t+2}]\,E[X_{t-3}]\\&+E[X_{t-2}X_{t-1}X_tX_{t+1}]\\&= -7\,\mu (1)\,\mu ^2+4\,\mu (1,1)\,\mu +2\,\mu (1,2)\,\mu +\mu (1,2,3)\\&\overset{\text{ Poi }}{=} \beta (4 + 5 \beta ) {\uplambda }^2 +4 (1 + \beta )^2 (1 +3 \beta ) {\uplambda }^3. \end{aligned}$$

For arbitrary \(k>3\), we obtain

$$\begin{aligned} \sigma _{2+1,2+k}&= \textstyle E\big [\big (X_tX_{t-1}-\mu (1)\big )(X_tX_{t-k}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t+r}X_{t-1+r}-\mu (1)\big )(X_tX_{t-k}-\mu ^2)\big ]}_{=0\text { for }r>2}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_tX_{t-1}-\mu (1)\big )(X_{t+r}X_{t+r-k}-\mu ^2)\big ]}_{=0\text { for }r>k+1}\\&= -8\,\mu (1)\,\mu ^2+E[X_t^2X_{t-1}]\,E[X_{t-k}]+E[X_{t+1}X_{t}^2]\,E[X_{t-k}]\\&+E[X_{t+2}X_{t+1}X_{t}]\,E[X_{t-k}]\\&+\,E[X_tX_{t-1}X_{t+1}]\,E[X_{t-k+1}]+E[X_tX_{t-1}X_{t-2}]\,E[X_{t+k-2}]\\&+E[X_tX_{t-1}^2]\,E[X_{t+k-1}]\\&+\,E[X_t^2X_{t-1}]\,E[X_{t+k}]+E[X_tX_{t-1}X_{t+1}]\,E[X_{t+k+1}]\\&\overset{\text{ TR }}{=} -8\,\mu (1)\,\mu ^2+4\,E[X_t^2X_{t-1}]\,E[X_{t-k}]+4\,E[X_tX_{t+1}X_{t+2}]\,E[X_{t-k}]\\&= -8\,\mu (1)\,\mu ^2+4\,\mu (1,1)\,\mu +4\,\mu (1,2)\,\mu \\&\overset{\text{ Poi }}{=} 4\beta (1 + \beta ) {\uplambda }^2 +4 (1 + \beta )^2 (1 +3 \beta ) {\uplambda }^3. \end{aligned}$$

Next, we consider lag 2.

$$\begin{aligned} \sigma _{2+2,2+2}&= \textstyle E\big [(X_tX_{t-2}-\mu ^2)^2\big ]+2\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-2}-\mu ^2)(X_{t+r}X_{t-2+r}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&= -7\,\mu ^4+E[X_t^2]\,E[X_{t-2}^2]+2\,E[X_tX_{t-2}X_{t+1}X_{t-1}]\\&+2\,E[X_{t+2}]\,E[X_t^2]\,E[X_{t-2}]\\&+\,2\,E[X_{t+3}]\,E[X_{t+1}X_t]\,E[X_{t-2}]\\&= -7\,\mu ^4+\mu (0)^2+2\,\mu (1,2,3)+2\,\mu ^2\,\mu (0)+2\,\mu ^2\,\mu (1)\\&\overset{\text{ Poi }}{=} (1 + 2 \beta + 3 \beta ^2) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3, \end{aligned}$$

where we used Lemma 1 (a) and (h).

$$\begin{aligned} \sigma _{2+2,2+3}&= \textstyle E\big [(X_tX_{t-2}-\mu ^2)(X_tX_{t-3}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}X_{t-2+r}-\mu ^2)(X_tX_{t-3}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-2}-\mu ^2)(X_{t+r}X_{t+r-3}-\mu ^2)\big ]}_{=0\text { for }r>4}\\&= -8\,\mu ^4+E[X_{t}^2]\,E[X_{t-2}X_{t-3}]+E[X_{t+1}X_{t-1}X_{t}]\,E[X_{t-3}]\\&+E[X_{t+2}]\,E[X_{t}^2]\,E[X_{t-3}]\\&+\,E[X_{t+3}]\,E[X_{t+1}X_{t}]\,E[X_{t-3}]+E[X_{t}X_{t+1}]\,E[X_{t-2}^2]\\&+E[X_{t}X_{t-2}X_{t-1}]\,E[X_{t+2}]\\&+\,E[X_{t}^2]\,E[X_{t-2}]\,E[X_{t+3}]+E[X_{t}X_{t+1}]\,E[X_{t-2}]\,E[X_{t+4}]\\&\overset{\text{ TR }}{=} -8\,\mu ^4+2\,E[X_{t}^2]\,E[X_{t-2}X_{t-3}]+2\,E[X_{t+1}X_{t-1}X_{t}]\,E[X_{t-3}]\\&+2\,E[X_{t}^2]\,E[X_{t-2}]^2\\&+\,2\,E[X_{t}X_{t+1}]\,E[X_{t-2}]^2\\&= -8\,\mu ^4+2\,\mu (0)\,\mu (1)+2\,\mu ^2\,\mu (1)+2\,\mu (1,2)\,\mu +2\,\mu (0)\,\mu ^2\\&\overset{\text{ Poi }}{=} 2 \beta (1 + \beta ) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3. \end{aligned}$$

Analogously,

$$\begin{aligned} \sigma _{2+2,2+4}&= \textstyle E\big [(X_tX_{t-2}-\mu ^2)(X_tX_{t-4}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}X_{t-2+r}-\mu ^2)(X_tX_{t-4}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-2}-\mu ^2)(X_{t+r}X_{t+r-4}-\mu ^2)\big ]}_{=0\text { for }r>5}\\&= -9\,\mu ^4+E[X_{t}^2]\,E[X_{t-2}]\,E[X_{t-4}]+E[X_{t+1}X_{t-1}X_{t}]\,E[X_{t-4}]\\&+E[X_{t+2}]\,E[X_{t}^2]\,E[X_{t-4}]\\&+\,E[X_{t+3}]\,E[X_{t+1}X_{t}]\,E[X_{t-4}]+E[X_{t}X_{t+1}]\,E[X_{t-2}X_{t-3}]\\&+E[X_{t}]\,E[X_{t-2}^2]\,E[X_{t+2}]\\&+\,E[X_{t}X_{t-2}X_{t-1}]\,E[X_{t+3}]+E[X_{t}^2]\,E[X_{t-2}]\,E[X_{t+4}]\\&+E[X_{t-2}]\,E[X_{t}X_{t+1}]\,E[X_{t+5}]\\&\overset{\text{ TR }}{=} -9\,\mu ^4+4\,E[X_{t}^2]\,E[X_{t-2}]^2+2\,E[X_{t-1}X_{t}X_{t+1}]\,E[X_{t-4}]\\&+\,2\,E[X_{t}X_{t+1}]\,E[X_{t+3}]^2+E[X_{t}X_{t+1}]^2\\&= -9\,\mu ^4+4\,\mu (0)\,\mu ^2+2\,\mu (1,2)\,\mu +2\,\mu (1)\,\mu ^2+\mu (1)^2\\&\overset{\text{ Poi }}{=} \beta ^2 {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3. \end{aligned}$$

For arbitrary \(k>4\), we obtain

$$\begin{aligned} \sigma _{2+2,2+k}&= \textstyle E\big [(X_tX_{t-2}-\mu ^2)(X_tX_{t-k}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}X_{t-2+r}-\mu ^2)(X_tX_{t-k}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-2}-\mu ^2)(X_{t+r}X_{t+r-k}-\mu ^2)\big ]}_{=0\text { for }r>k+1}\\&= -10\,\mu ^4+E[X_{t}^2]\,E[X_{t-2}]\,E[X_{t-k}]+E[X_{t+1}X_{t-1}X_{t}]\,E[X_{t-k}]\\&+E[X_{t+2}]\,E[X_{t}^2]\,E[X_{t-k}]\\&+\,E[X_{t+3}]\,E[X_{t+1}X_{t}]\,E[X_{t-k}]+E[X_{t-2}]\,E[X_{t}X_{t+1}]\,E[X_{t-k+1}]\\&+\,E[X_{t}]\,E[X_{t-2}X_{t-3}]\,E[X_{t+k-3}]+E[X_{t}]\,E[X_{t-2}^2]\,E[X_{t+k-2}]\\&+E[X_{t}X_{t-2}X_{t-1}]\,E[X_{t+k-1}]\\&+\,E[X_{t}^2]\,E[X_{t-2}]\,E[X_{t+k}]+E[X_{t-2}]\,E[X_{t}X_{t+1}]\,E[X_{t+k+1}]\\&\overset{\text{ TR }}{=} -10\,\mu ^4+4\,E[X_{t}^2]\,E[X_{t-2}]^2+2\,E[X_{t-1}X_{t}X_{t+1}]\,E[X_{t-k}]\\&+4\,E[X_{t}X_{t+1}]\,E[X_{t+3}]^2\\&= -10\,\mu ^4+4\,\mu (0)\,\mu ^2+2\,\mu (1,2)\,\mu +4\,\mu (1)\,\mu ^2\\&\overset{\text{ Poi }}{=} 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3. \end{aligned}$$

Finally, we consider lags \(l\ge 3\).

$$\begin{aligned} \sigma _{2+l,2+l}&= \textstyle E\big [(X_tX_{t-l}-\mu ^2)^2\big ]+2\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-l}-\mu ^2)(X_{t+r}X_{t+r-l}-\mu ^2)\big ]}_{=0\text { for }r>l+1}\\&= -9\,\mu ^4+E[X_t^2]\,E[X_{t-l}^2]+2\,E[X_tX_{t+1}]\,E[X_{t-l}X_{t+1-l}]\\&+2\,E[X_{t-l}]\,E[X_{t}X_{t-1}]\,E[X_{t+l-1}]\\&+\,2\,E[X_{t-l}]\,E[X_{t}^2]\,E[X_{t+l}]+2\,E[X_{t-l}]\,E[X_{t}X_{t+1}]\,E[X_{t+l+1}]\\&= -9\,\mu ^4+\mu (0)^2+2\,\mu (1)^2+4\,\mu (1)\,\mu ^2+2\,\mu (0)\,\mu ^2 \\&\overset{\text{ Poi }}{=} (1 + 2 \beta + 3 \beta ^2) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3, \end{aligned}$$

where we used Lemma 1 (a). \(\sigma _{2+l,2+l}\) remains the same as in the case for \(l=2\). Next, we continue with \(k=l+1\), i.e., the difference between the lags is 1.

$$\begin{aligned} \sigma _{2+l,2+k}&= \textstyle E\big [(X_tX_{t-l}-\mu ^2)(X_tX_{t-l-1}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}X_{t-l+r}-\mu ^2)(X_tX_{t-l-1}-\mu ^2)\big ]}_{=0\text { for }r>l+1}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-l}-\mu ^2)(X_{t+r}X_{t+r-l-1}-\mu ^2)\big ]}_{=0\text { for }r>l+2}\\&= -10\,\mu ^4+E[X_{t}^2]\,E[X_{t-l}X_{t-l-1}]+E[X_{t-l+1}]\,E[X_{t+1}X_t]\,E[X_{t-l-1}]\\&+\,E[X_{t+l-1}]\,E[X_{t-1}X_t]\,E[X_{t-l-1}]+E[X_{t+l}]\,E[X_{t}^2]\,E[X_{t-l-1}]\\&+\,E[X_{t+l+1}]\,E[X_{t+1}X_t]\,E[X_{t-l-1}]+E[X_{t}X_{t+1}]\,E[X_{t-l}^2]\\&+E[X_{t}]\,E[X_{t+2}]\,E[X_{t-l}X_{t-l+1}]\\&+\,E[X_{t-l}]\,E[X_{t}X_{t-1}]\,E[X_{t+l}]+E[X_{t}^2]\,E[X_{t-l}]\,E[X_{t+l+1}]\\&+E[X_{t-l}]\,E[X_{t}X_{t+1}]\,E[X_{t+l+2}]\\&\overset{\text{ TR }}{=} -10\,\mu ^4+2\,E[X_{t}^2]\,E[X_{t-1}X_{t}]+6\,E[X_tX_{t+1}]\,E[X_{t-l+1}]^2\\&+2\,E[X_t^2]\,E[X_{t-l+1}]^2\\&= -10\,\mu ^4+2\,\mu (0)\,\mu (1)+6\,\mu (1)\,\mu ^2+2\,\mu (0)\,\mu ^2\\&\overset{\text{ Poi }}{=} 2 \beta (1 + \beta ) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3. \end{aligned}$$

For \(l\ge 3\) and \(k=l+2\), i.e., the difference between the lags is 2, we get

$$\begin{aligned} \sigma _{2+l,2+k}&= \textstyle E\big [(X_tX_{t-l}-\mu ^2)(X_tX_{t-l-2}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}X_{t-l+r}-\mu ^2)(X_tX_{t-l-2}-\mu ^2)\big ]}_{=0\text { for }r>l+1}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-l}-\mu ^2)(X_{t+r}X_{t+r-l-2}-\mu ^2)\big ]}_{=0\text { for }r>l+3}\\&= -11\,\mu ^4+E[X_{t-l}]\,E[X_t^2]\,E[X_{t-l-2}]+E[X_{t-l+1}]\,E[X_{t+1}X_t]\,E[X_{t-l-2}]\\&+\,E[X_{t+l-1}]\,E[X_{t-1}X_t]\,E[X_{t-l-2}]+E[X_{t+l}]\,E[X_{t}^2]\,E[X_{t-l-2}]\\&+\,E[X_{t+l+1}]\,E[X_{t+1}X_t]\,E[X_{t-l-2}]\\&+\,E[X_{t}X_{t+1}]\,E[X_{t-l}X_{t-l-1}]+E[X_{t}]\,E[X_{t+2}]\,E[X_{t-l}^2]\\&+E[X_{t}]\,E[X_{t+3}]\,E[X_{t-l}X_{t-l+1}]\\&+\,E[X_{t-l}]\,E[X_{t}X_{t-1}]\,E[X_{t+l+1}]+E[X_{t-l}]\,E[X_{t}^2]\,E[X_{t+l+2}]\\&+\,E[X_{t-l}]\,E[X_{t}X_{t+1}]\,E[X_{t+l+3}]\\&\overset{\text{ TR }}{=} -11\,\mu ^4+4\,E[X_{t}^2]\,E[X_{t}]^2+6\,E[X_tX_{t+1}]\,E[X_{t}]^2+E[X_tX_{t+1}]^2\\&= -11\,\mu ^4+4\,\mu (0)\,\mu ^2+6\,\mu (1)\,\mu ^2+\mu (1)^2\\&\overset{\text{ Poi }}{=} \beta ^2 {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3. \end{aligned}$$

For \(l\ge 3\) and \(k>l+2\), i.e., the difference between the lags is greater than 2, we obtain

$$\begin{aligned} \sigma _{2+l,2+k}&= \textstyle E\big [(X_tX_{t-l}-\mu ^2)(X_tX_{t-k}-\mu ^2)\big ]\\&+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}X_{t-l+r}-\mu ^2)(X_tX_{t-k}-\mu ^2)\big ]}_{=0\text { for }r>l+1}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_tX_{t-l}-\mu ^2)(X_{t+r}X_{t+r-k}-\mu ^2)\big ]}_{=0\text { for }r>k+1}\\&= -12\,\mu ^4+E[X_{t}^2]\,E[X_{t-l}]\,E[X_{t-k}]+E[X_{t-l+1}]\,E[X_{t+1}X_t]\,E[X_{t-k}]\\&+\,E[X_{t+l-1}]\,E[X_{t-1}X_t]\,E[X_{t-k}]+E[X_{t+l}]\,E[X_{t}^2]\,E[X_{t-k}]\\&+\,E[X_{t+l+1}]\,E[X_{t+1}X_t]\,E[X_{t-k}]\\&+\,E[X_{t-l}]\,E[X_{t}X_{t+1}]\,E[X_{t+1-k}]+E[X_{t}]\,E[X_{t-l}X_{t-l-1}]\,E[X_{t+k-l-1}]\\&+\,E[X_{t}]\,E[X_{t-l}^2]\,E[X_{t+k-l}]+E[X_{t}]\,E[X_{t-l}X_{t-l+1}]\,E[X_{t+k-l+1}]\\&+\,E[X_{t-l}]\,E[X_{t}X_{t-1}]\,E[X_{t+k-1}]+E[X_{t-l}]\,E[X_{t}^2]\,E[X_{t+k}]\\&+\,E[X_{t-l}]\,E[X_{t}X_{t+1}]\,E[X_{t+k+1}]\\&\overset{\text{ TR }}{=} -12\,\mu ^4+4\,E[X_{t}^2]\,E[X_{t}]^2+8\,E[X_{t+1}X_t]\,E[X_{t}]^2\\&= -12\,\mu ^4+4\,\mu (0)\,\mu ^2+8\,\mu (1)\,\mu ^2\\&\overset{\text{ Poi }}{=} 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3. \end{aligned}$$

Next, we consider the cross-covariances between \(X_s\) and \(X_tX_{t-k}\).

$$\begin{aligned} \sigma _{1,2+1}&= \textstyle E\big [(X_t-\mu )\big (X_tX_{t-1}-\mu (1)\big )\big ]+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_t-\mu )\big (X_{t+r}X_{t+r-1}-\mu (1)\big )\big ]}_{=0\text { for }r>2}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}-\mu )\big (X_tX_{t-1}-\mu (1)\big )\big ]}_{=0\text { for }r>1}\\&= -4\,\mu \, \mu (1)+E[X_t^2X_{t-1}]+E[X_t^2X_{t+1}]+E[X_tX_{t+2}X_{t+1}]\\&+E[X_{t+1}X_tX_{t-1}]\\&\overset{\text{ TR }}{=} -4\,\mu \, \mu (1)+2\,E[X_{t-1}X_t^2]+2\,E[X_tX_{t+1}X_{t+2}]\\&= -4\,\mu \, \mu (1)+2\,\mu (1,1)+2\,\mu (1,2)\\&\overset{\text{ Poi }}{=} 2 \beta {\uplambda }+ 2(1+\beta )(1+3\beta ){\uplambda }^2, \end{aligned}$$

where we used Lemma 1 (a)–(c). Analogously,

$$\begin{aligned} \sigma _{1,2+2}&= \textstyle E\big [(X_t-\mu )(X_tX_{t-2}-\mu ^2)\big ]+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_t-\mu )(X_{t+r}X_{t+r-2}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\left[ (X_{t+r}-\mu )(X_tX_{t-2}-\mu ^2)\right] }_{=0\text { for }r>1}\\&= -5\,\mu ^3 + E[X_t^2]\,E[X_{t-2}]+E[X_{t-1}X_tX_{t+1}]+E[X_{t+2}]\,E[X_t^2]\\&+\,E[X_tX_{t+1}]\,E[X_{t+3}]+E[X_{t-2}]\,E[X_tX_{t+1}]\\&= -5\,\mu ^3+2\,\mu (0)\,\mu +2\,\mu (1)\,\mu +\mu (1,2)\\&\overset{\text{ Poi }}{=} 2(1 + \beta )(1+3\beta ) {\uplambda }^2, \end{aligned}$$

where we used the Lemma 1 (a) and (c). For lag \(k\ge 3\), we get

$$\begin{aligned} \sigma _{1,2+k}= & {} \textstyle E\big [(X_t-\mu )(X_tX_{t-k}-\mu ^2)\big ]+\sum \limits _{r=1}^\infty \underbrace{E\big [(X_{t+r}-\mu )(X_tX_{t-k}-\mu ^2)\big ]}_{=0\text { for }r>1}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [(X_t-\mu )(X_{t+r}X_{t+r-k}-\mu ^2)\big ]}_{=0\text { for }r>k+1}\\= & {} -6\,\mu ^3+E[X_t^2]\,E[X_{t-k}]+E[X_{t+1}X_t]\,E[X_{t-k}]+E[X_tX_{t+1}]\,E[X_{t+1-k}]\\&+\,E[X_{t+k-1}]\,E[X_tX_{t-1}]+E[X_t^2]\,E[X_{t+k}]+E[X_{t+k+1}]\,E[X_tX_{t+1}]\\= & {} -6\,\mu ^3+2\,\mu (0)\,\mu +4\,\mu (1)\,\mu \ \overset{\text{ Poi }}{=}\ 2(1 + \beta )(1+3\beta ) {\uplambda }^2, \end{aligned}$$

where we used the Lemma 1 (a). It remains the same as in the case for lag \(k=2\). Finally, we consider the cross-covariances between \(X_s^2\) and \(X_tX_{t-k}\).

$$\begin{aligned} \sigma _{2,2+1}&= \textstyle E\big [\big (X_t^2-\mu (0)\big )\big (X_tX_{t-1}-\mu (1)\big )\big ]+\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_t^2-\mu (0)\big )\big (X_{t+r}X_{t+r-1}-\mu (1)\big )\big ]}_{=0\text { for }r>2}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t+r}^2-\mu (0)\big )\big (X_tX_{t-1}-\mu (1)\big )\big ]}_{=0\text { for }r>1}\\&= -4\,\mu (0)\, \mu (1)+E[X_t^3X_{t-1}]+E[X_t^3X_{t+1}]+E[X_t^2X_{t+1}X_{t+2}]+E[X_{t-1}X_t X_{t+1}^2]\\&\overset{\text{ TR }}{=} -4\,\mu (0)\,\mu (1)+2\,E[X_t^3X_{t+1}]+2\,E[X_t^2X_{t+1}X_{t+2}]\\&= -4\,\mu (0)\,\mu (1)+2\,\mu (0,0,1)+2\,\mu (0,1,2)\\&\overset{\text{ Poi }}{=} 2 \beta {\uplambda }+ 2 (1 + \beta )(1 + 7 \beta ) {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3, \end{aligned}$$

where we used Lemma 1 (a), (d) and (f).

$$\begin{aligned} \quad \sigma _{2,2+2}&= \textstyle E\big [\big (X_t^2-\mu (0)\big )(X_tX_{t-2}-\mu ^2)\big ]+\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_t^2-\mu (0)\big )(X_{t+r}X_{t+r-2}-\mu ^2)\big ]}_{=0\text { for }r>3}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t+r}^2-\mu (0)\big )(X_tX_{t-2}-\mu ^2)\big ]}_{=0\text { for }r>1}\\&= -5\,\mu (0)\,\mu ^2+E[X_t^3]\,E[X_{t-2}]+E[X_t^2X_{t+1}X_{t-1}]+E[X_t^3]\,E[X_{t+2}]\\&+\,E[X_t^2X_{t+1}]\,E[X_{t+3}]+E[X_{t+1}^2X_t]\,E[X_{t-2}]\\&\overset{\text{ TR }}{=} -5\,\mu (0)\, \mu ^2+2\,E[X_t^3]\,E[X_{t-2}]+E[X_{t-1}X_t^2X_{t+1}]+2\,E[X_tX_{t+1}^2]\,E[X_{t+3}] \\&= -5\,\mu (0)\,\mu ^2+2\,\mu (0,0)\,\mu +\mu (1,1,2)+2\,\mu (1,1)\,\mu \\&\overset{\text{ Poi }}{=} 2 (1 + 2 \beta )^2 {\uplambda }^2 + 4 (1 + \beta )^2 (1 + 3 \beta ) {\uplambda }^3, \end{aligned}$$

where we used the Lemma 1 (b) and (g).

$$\begin{aligned} \sigma _{2,2+k}&= \textstyle E\big [\big (X_t^2-\mu (0)\big )(X_tX_{t-k}-\mu ^2)\big ]+\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_{t+r}^2-\mu (0)\big )(X_tX_{t-k}-\mu ^2)\big ]}_{=0\text { for }r>1}\\&+\,\sum \limits _{r=1}^\infty \underbrace{E\big [\big (X_t^2-\mu (0)\big )(X_{t+r}X_{t+r-k}-\mu ^2)\big ]}_{=0\text { for }r>k+1}\\&= -6\,\mu ^2\,\mu (0)+E[X_t^3]\,E[X_{t-k}]+E[X_{t+1}^2X_t]\,E[X_{t-k}]\\&+E[X_t^2X_{t+1}]\,E[X_{t+1-k}]\\&+\,E[X_t^2X_{t-1}]\,E[X_{t+k-1}]+E[X_t^3]\,E[X_{t+k}]+E[X_t^2X_{t+1}]\,E[X_{t+k+1}]\\&\overset{\text{ TR }}{=} -6\,\mu ^2\,\mu (0)+2\,E[X_t^3]\,\mu +4\,E[X_tX_{t+1}^2]\,\mu \\&= -6\,\mu ^2\,\mu (0)+2\,\mu (0,0)\,\mu +4\,\mu (1,1)\,\mu \\&\overset{\text{ Poi }}{=} 2(1 + \beta )(1+3\beta ) {\uplambda }^2 + 4 (1 + \beta ) ^2(1+3\beta ) {\uplambda }^3, \end{aligned}$$

where we used the Lemma 1 (b). So \(\sigma _{2,2+k}\) for \(k\ge 3\) is slightly different from lag \(k=2\).

After having computed the required \(\sigma _{ij}\), we proceed as in “Appendix A.3” before, and compute the asymptotic distribution and bias of \(\big (\hat{\rho }(1),\ldots ,\hat{\rho }(K)\big )^{\top }\) based on appropriate Taylor expansions. For this purpose, let us define the following functions \(g_k: \mathbb {R}^{2+K}\rightarrow \mathbb {R}\), \(k=1,\ldots ,K\), by

$$\begin{aligned} g_k(y_1,\ \ldots ,\ y_{2+K})\ :=\ \frac{y_{2+k}-y_1^2}{y_{2}-y_1^2}, \end{aligned}$$

since \(g_k\big (\mu ,\mu (0),\ldots , \mu (K)\big )=\rho (k)\). Note that \(\big (\mu ,\mu (0),\ldots ,\mu (K)\big ) = \big (\mu ,\mu (1+\mu ),\mu (\frac{\beta }{1+\beta }+\mu ),\mu ^2,\ldots ,\mu ^2\big )\) in the case of the Poisson INMA(1) model. First, we need to compute the Jacobian \({\mathbf{J }}_{{\varvec{g}}}(y_1,\ldots ,y_{2+K})\) of \({\varvec{g}}:=(g_1,\ldots ,g_K)^{\top }\) as well as the Hessians \({\mathbf{H }}_{g_k}(y_1,\ldots ,y_{2+K})\) of \(g_k\) for \(k=1,\ldots ,K\):

$$\begin{aligned} {\mathbf{J }}_{{\varvec{g}}}(y_1,\ldots ,y_{2+K})=\left( \begin{array}{llllll} \frac{2 y_1 (y_3-y_2)}{(y_2-y_1^2)^2} &{} \quad \frac{y_1^2-y_3}{(y_2-y_1^2)^2} &{} \quad \frac{1}{y_2-y_1^2} &{}\quad 0 &{}\quad \ldots &{}\quad 0\\ \vdots &{}\quad \vdots &{}\quad 0 &{}\quad \frac{1}{y_2-y_1^2} &{}\quad &{}\quad \vdots \\ &{}\quad &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad 0\\ \frac{2 y_1 (y_{2+K}-y_2)}{(y_2-y_1^2)^2} &{}\quad \frac{y_1^2-y_{2+K}}{(y_2-y_1^2)^2} &{}\quad 0 &{}\quad \ldots &{}\quad 0 &{}\quad \frac{1}{y_2-y_1^2}\\ \end{array} \right) , \end{aligned}$$

and

$$\begin{aligned} \frac{\partial ^2 g_k}{\partial y_1^2}= & {} \frac{2 (3 y_1^2 + y_2) (y_{k+2}-y_2)}{(y_2-y_1^2)^3},\qquad \frac{\partial ^2 g_k}{\partial y_2\ \partial y_1}=\frac{2 y_1 (y_1^2 + y_2 - 2 y_{k+2})}{(y_2-y_1^2)^3}, \\ \frac{\partial ^2 g_k}{\partial y_2^2}= & {} \frac{2 ( y_{k+2}-y_1^2 )}{(y_2-y_1^2)^3},\quad \frac{\partial ^2 g_k}{\partial y_1\ \partial y_{k+2}}=\frac{2y_1}{(y_2-y_1^2)^2},\quad \frac{\partial ^2 g_k}{\partial y_2\ \partial y_{k+2}}=\frac{-1}{(y_2-y_1^2)^2}. \end{aligned}$$

Note that all other derivatives equal zero.

Next, we evaluate the Jacobian in \(\big (\mu ,\mu (0),\ldots ,\mu (K)\big )\) and obtain \({\mathbf{D }}:={\mathbf{J }}_{{\varvec{g}}}\big (\mu ,\mu (0),\ldots ,\mu (K)\big )\) as

$$\begin{aligned} {\mathbf{D }}=\left( \begin{array}{llllll} \frac{-2}{1+\beta } &{}\quad \frac{-\beta }{\mu \,(1+\beta )} &{}\quad \frac{1}{\mu } &{}\quad 0 &{}\quad \ldots &{}\quad 0\\ -2 &{}\quad 0 &{}\quad 0 &{}\quad \frac{1}{\mu } &{}\quad &{}\quad \vdots \\ \vdots &{}\quad \vdots &{}\quad \vdots &{}\quad &{}\quad \ddots &{}\quad 0\\ -2 &{}\quad 0 &{}\quad 0 &{}\quad \ldots &{}\quad 0 &{}\quad \frac{1}{\mu }\\ \end{array} \right) . \end{aligned}$$

The application of the Delta method implies that asymptotically, also see Romano and Thombs (1996), \(\big (\hat{\rho }(1),\ldots ,\hat{\rho }(K)\big )^{\top }\) are jointly normally distributed with covariance matrix \({\mathbf{D }}\, \big (\sigma _{ij}\big )_{1\le i,j\le 2+K}\, {\mathbf{D }}^{\top }\) given as

$$\begin{aligned} \left( \begin{array}{lllll} \frac{{\uplambda }-2\beta ^3{\uplambda }+2\beta ^4{\uplambda }+\beta ^2(-1+3{\uplambda })+\beta (1+4{\uplambda })}{(1+\beta )^4{\uplambda }} &{}\quad \frac{2\beta (1+2\beta )}{(1+\beta )^3} &{}\quad \frac{\beta ^2}{(1+\beta )^2} &{}\quad &{}\quad 0 \\ * &{}\quad \frac{1+2\beta +3\beta ^2}{(1+\beta )^2} &{}\quad \frac{2\beta }{1+\beta } &{}\quad \ddots &{}\quad \\ * &{}\quad *&{}\quad \ddots &{}\quad \ddots &{}\quad \frac{\beta ^2}{(1+\beta )^2}\\ &{}\quad &{}\quad &{}\quad &{}\quad \frac{2\beta }{1+\beta } \\ 0 &{}\quad &{}\quad &{}\quad &{}\quad \frac{1+2\beta +3\beta ^2}{(1+\beta )^2} \\ \end{array} \right) . \end{aligned}$$

To obtain the asymptotic bias corrections, we first evaluate the Hessians \({\mathbf{H }}_{g_k}\) for \(k=1,\ldots ,K\) in \(\big (\mu ,\mu (0),\ldots ,\mu (K)\big )\), leading to

$$\begin{aligned}&\frac{\partial ^2 g_1}{\partial y_1^2} = -\frac{2+8\mu }{\mu (1+\beta )} ,\qquad \frac{\partial ^2 g_1}{\partial y_2\ \partial y_1}=\frac{2(1-\beta )}{(1+\beta )\mu },\qquad \frac{\partial ^2 g_1}{\partial y_2^2}=\frac{2\beta }{(1+\beta )\mu ^2}, \\&\frac{\partial ^2 g_1}{\partial y_1\ \partial y_{3}}=\frac{2}{\mu },\qquad \frac{\partial ^2 g_1}{\partial y_2\ \partial y_{3}}=-\frac{1}{\mu ^2}. \end{aligned}$$

For \(k\ge 2\), we have

$$\begin{aligned}&\frac{\partial ^2 g_k}{\partial y_1^2}=-8-\frac{2}{\mu } ,\qquad \frac{\partial ^2 g_k}{\partial y_2\ \partial y_1}=\frac{2}{\mu },\qquad \frac{\partial ^2 g_k}{\partial y_2^2}=0, \\&\frac{\partial ^2 g_k}{\partial y_1\ \partial y_{k+2}}=\frac{2}{\mu },\qquad \frac{\partial ^2 g_k}{\partial y_2\ \partial y_{k+2}}=-\frac{1}{\mu ^2}, \end{aligned}$$

all other derivatives are equal to zero.

Then, abbreviating \({\varvec{Z}}_T := \frac{1}{\sqrt{T}} \sum _{t=1}^T (Y_{t,1},\ldots ,Y_{t,2+K})^{\top }\), we compute the bias correction for the kth-order autocorrelation as

$$\begin{aligned}&T\,E\big [\hat{\rho }(1)-\rho (1)\big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{g_1}\,{\varvec{Z}}_T\big ]\ \approx \ \frac{-\beta + \beta ^2 - (1 + 6 \beta + 7 \beta ^2 - 2 \beta ^3) {\uplambda }}{(1 + \beta )^3 {\uplambda }}, \\&T\,E\big [\hat{\rho }(2)-\rho (2)\big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{g_2}\,{\varvec{Z}}_T\big ]\ \approx \ -\frac{1+4\beta +5\beta ^2}{(1+\beta )^2}, \end{aligned}$$

and for \(k>2\),

$$\begin{aligned} T\,E\big [\hat{\rho }(k)-\rho (k)\big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{g_k}\,{\varvec{Z}}_T\big ]\ \approx \ -3+\frac{2}{1+\beta }. \end{aligned}$$

A.5 Proofs of Section 3.3

For deriving the joint asymptotic distribution of \(({\hat{\mu }},{\hat{\rho }})^{\top }\) and \(({\hat{{\uplambda }}},{\hat{\beta }})^{\top }\), respectively, it suffices to consider the components

$$\begin{aligned} \big (Y_{t,1}, Y_{t,2},Y_{t,3}\big )^{\top } \ =\ \big (X_t - \mu ,\ X_t^2 - \mu (0),\ X_t X_{t-1} - \mu (1)\big )^{\top } \end{aligned}$$

of the vector \({\varvec{Y}}_t\) from (A.1). The corresponding entries \(\sigma _{ij}\) with \(1\le i\le j\le 3\) from the covariance matrix \({\varvec{\varSigma }}\) in Theorem 4 have already been calculated in “Appendices A.3 and A.4” before. We now compute the asymptotic distribution and bias of \(({\hat{\mu }},{\hat{\rho }})^{\top }\) and \(({\hat{{\uplambda }}},{\hat{\beta }})^{\top }\), respectively, based on appropriate Taylor expansions.

Let us begin with \(({\hat{{\uplambda }}},{\hat{\beta }})^{\top }\), where we define the following functions \(g_i: \mathbb {R}^3\rightarrow \mathbb {R}\), \(i=1,2\), by

$$\begin{aligned} g_1(y_1,y_2,y_3)\ :=\ y_1\,\Big (1-\frac{y_3-y_1^2}{y_2-y_1^2}\Big ),\qquad g_2(y_1,y_2,y_3)\ :=\ \Big (1-\frac{y_3-y_1^2}{y_2-y_1^2}\Big )^{-1}-1, \end{aligned}$$

since \(g_1\big (\mu ,\mu (0),\mu (1)\big )={\uplambda }\) and \(g_2\big (\mu ,\mu (0),\mu (1)\big )=\beta \). Note that \(\big (\mu ,\mu (0),\mu (1)\big ) = \big (\mu ,\mu (1+\mu ),\mu (\frac{\beta }{1+\beta }+\mu )\big )\) in the case of the Poisson INMA(1) model. First, we need to compute the Jacobian \({\mathbf{J }}_{{\varvec{g}}}(y_1,y_2,y_3)\) of \({\varvec{g}}:=(g_1,g_2)^{\top }\) as well as the Hessians \({\mathbf{H }}_{g_i}(y_1,y_2,y_3)\) of \(g_i\) for \(i=1,2\):

$$\begin{aligned} {\mathbf{J }}_{{\varvec{g}}}(y_1,y_2,y_3)= & {} \left( \begin{array}{lll} \frac{(y_1^2+y_2) (y_2-y_3)}{(y_1^2-y_2)^2} &{}\quad \frac{y_1(y_3-y_1^2)}{(y_1^2-y_2)^2} &{}\quad \frac{y_1}{y_1^2-y_2} \\ \frac{-2 y_1}{y_2-y_3} &{}\quad \frac{y_1^2-y_3}{(y_2-y_3)^2} &{}\quad \frac{y_2-y_1^2}{(y_2-y_3)^2} \\ \end{array} \right) , \\ {\mathbf{H }}_{g_1}(y_1,y_2,y_3)= & {} \left( \begin{array}{lll} \frac{2 y_1 (y_1^2+3 y_2) (y_2-y_3)}{(y_2-y_1^2)^3} &{}\quad \frac{y_1^4+3 (y_2-y_3) y_1^2-y_2y_3}{(y_1^2-y_2)^3} &{}\quad -\frac{y_1^2+y_2}{(y_2-y_1^2)^2} \\ \frac{y_1^4+3 (y_2-y_3) y_1^2-y_2 y_3}{(y_1^2-y_2)^3} &{}\quad \frac{2 y_1 (y_1^2-y_3)}{(y_2-y_1^2)^3} &{}\quad \frac{y_1}{(y_2-y_1^2)^2} \\ -\frac{y_1^2+y_2}{(y_2-y_1^2)^2} &{}\quad \frac{y_1}{(y_2-y_1^2)^2} &{}\quad 0\\ \end{array} \right) ,\\ {\mathbf{H }}_{g_2}(y_1,y_2,y_3)= & {} \left( \begin{array}{lll} \frac{2}{y_3-y_2} &{}\quad \frac{2 y_1}{(y_3-y_2)^2} &{}\quad \frac{-2 y_1}{(y_3-y_2)^2} \\ \frac{2 y_1}{(y_3-y_2)^2} &{}\quad \frac{2 (y_1^2-y_3)}{(y_3-y_2)^3} &{}\quad \frac{y_2+y_3-2y_1^2}{(y_3-y_2)^3} \\ \frac{-2 y_1}{(y_3-y_2)^2} &{}\quad \frac{y_2+y_3-2 y_1^2}{(y_3-y_2)^3} &{}\quad \frac{2(y_1^2-y_2)}{(y_3-y_2)^3}\\ \end{array} \right) . \end{aligned}$$

Next, we evaluate the Jacobian in \(\big (\mu ,\mu (0),\mu (1)\big )\) with \(\mu ={\uplambda }(1+\beta )\) and obtain \({\mathbf{D }}:={\mathbf{J }}_{{\varvec{g}}}\big (\mu ,\mu (0),\mu (1)\big )\) as

$$\begin{aligned} {\mathbf{D }}=\left( \begin{array}{lll} 2 {\uplambda }+\frac{1}{1+\beta } &{}\quad \frac{\beta }{1+\beta } &{}\quad -1 \\ -2 (1+\beta ) &{}\quad -\frac{\beta }{{\uplambda }} &{}\quad \frac{1+\beta }{{\uplambda }} \\ \end{array} \right) . \end{aligned}$$

The application of the Delta method implies that, asymptotically, \(({\hat{{\uplambda }}},{\hat{\beta }})^{\top }\) are jointly normally distributed with covariance matrix \({\mathbf{D }}\, \big (\sigma _{ij}\big )_{1\le i,j\le 3}\, {\mathbf{D }}^{\top }\) given as

$$\begin{aligned} \left( \begin{array}{ll} {\uplambda }+\frac{(1+4 \beta +3 \beta ^2-2 \beta ^3+2 \beta ^4) {\uplambda }^2}{(1+\beta )^2} &{}\quad -\frac{(1+4 \beta +3 \beta ^2-2 \beta ^3+2 \beta ^4) {\uplambda }}{1+\beta } \\ -\frac{(1+4 \beta +3 \beta ^2-2 \beta ^3+2 \beta ^4) {\uplambda }}{1+\beta } &{}\quad \frac{\beta (1-\beta ) + (1+4 \beta +3 \beta ^2-2 \beta ^3+2 \beta ^4) {\uplambda }}{{\uplambda }} \\ \end{array} \right) . \end{aligned}$$

To obtain the asymptotic bias corrections, we first evaluate the Hessians \({\mathbf{H }}_{g_i}\) for \(i=1,2\) in \(\big (\mu ,\mu (0),\mu (1)\big )\), leading to the matrices \({\mathbf{H }}_i\) given by

$$\begin{aligned} {\mathbf{H }}_1= & {} \left( \begin{array}{lll} 8 {\uplambda }+\frac{6}{1+\beta } &{}\quad \frac{\beta -2 {\uplambda }(1-\beta ^2)}{(1+\beta )^2 {\uplambda }} &{}\quad -2-\frac{1}{(1+\beta ){\uplambda }}\\ \frac{\beta -2 {\uplambda }(1-\beta ^2)}{(1+\beta )^2 {\uplambda }} &{}\quad \frac{-2 \beta }{(1+\beta )^2 {\uplambda }} &{}\quad \frac{1}{(1+\beta ){\uplambda }}\\ -2-\frac{1}{(1+\beta ){\uplambda }} &{}\quad \frac{1}{(1+\beta ){\uplambda }} &{}\quad 0 \\ \end{array} \right) , \\ {\mathbf{H }}_2= & {} \left( \begin{array}{lll} -\frac{2}{{\uplambda }} &{}\quad \frac{2 (1+\beta )}{{\uplambda }} &{}\quad -\frac{2 (1+\beta )}{{\uplambda }} \\ \frac{2 (1+\beta )}{{\uplambda }} &{}\quad \frac{2 \beta }{{\uplambda }^2} &{}\quad -\frac{1+2\beta }{{\uplambda }^2} \\ -\frac{2 (1+\beta )}{{\uplambda }} &{}\quad -\frac{1+2\beta }{{\uplambda }^2} &{}\quad \frac{2 (1+\beta )}{{\uplambda }^2} \\ \end{array} \right) . \end{aligned}$$

Then, abbreviating \({\varvec{Z}}_T := \frac{1}{\sqrt{T}} \sum _{t=1}^T (Y_{t,0},Y_{t,1},Y_{t,2})^{\top }\), we compute the bias corrections as

$$\begin{aligned}&T\,E\big [{\hat{{\uplambda }}}-{\uplambda }\big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{1}\,{\varvec{Z}}_T\big ]\ \approx \ \frac{(1+6 \beta +7 \beta ^2-2 \beta ^3) {\uplambda }}{(1+\beta )^2}, \\&T\,E\big [{\hat{\beta }}-\beta \big ]\ \approx \ E\big [\tfrac{1}{2}\,{\varvec{Z}}_T^{\top }\,{\mathbf{H }}_{2}\,{\varvec{Z}}_T\big ]\ \approx \ -2 \beta \,(1+\beta -\beta ^2). \end{aligned}$$

Finally, we consider the estimators \(({\hat{\mu }},{\hat{\rho }})^{\top }\). Note that \({\hat{\mu }}=\bar{X}\) is exactly unbiased, while the bias correction for \({\hat{\rho }}={\hat{\rho }}(1)\) was already computed in “Appendix A.4”. Actually, we also already computed the asymptotic variances of these quantities, only there asymptotic cross-covariance is missing. So define the functions \(g_i: \mathbb {R}^3\rightarrow \mathbb {R}\), \(i=1,2\), by

$$\begin{aligned} g_1(y_1,y_2,y_3)\ :=\ y_1,\quad g_2(y_1,y_2,y_3)\ :=\ \frac{y_3-y_1^2}{y_2-y_1^2}, \end{aligned}$$

with \(g_1\big (\mu ,\mu (0),\mu (1)\big )=\mu \) and \(g_2\big (\mu ,\mu (0),\mu (1)\big )=\rho \). Then, the Jacobian \({\mathbf{J }}_{{\varvec{g}}}(y_1,y_2,y_3)\) of \({\varvec{g}}:=(g_1,g_2)^{\top }\) equals

$$\begin{aligned} {\mathbf{J }}_{{\varvec{g}}}(y_1,y_2,y_3)=\left( \begin{array}{lll} 1 &{}\quad 0 &{}\quad 0 \\ \frac{2 y_1 (y_3-y_2)}{(y_2-y_1^2)^2} &{}\quad \frac{y_1^2-y_3}{(y_2-y_1^2)^2} &{}\quad \frac{1}{y_2-y_1^2} \\ \end{array} \right) , \end{aligned}$$

which leads to \({\mathbf{D }}:={\mathbf{J }}_{{\varvec{g}}}\big (\mu ,\mu (0),\mu (1)\big )\) as

$$\begin{aligned} {\mathbf{D }}=\left( \begin{array}{lll} 1 &{}\quad 0 &{}\quad 0 \\ -2 (1-\rho ) &{}\quad -\frac{\rho }{\mu } &{}\quad \frac{1}{\mu } \\ \end{array} \right) . \end{aligned}$$

By applying the Delta method in analogy to the above derivations, we obtain asymptotic normality for \(({\hat{\mu }},{\hat{\rho }})^{\top }\) with covariance matrix \({\mathbf{D }}\, \big (\sigma _{ij}\big )_{1\le i,j\le 3}\, {\mathbf{D }}^{\top }\) given as:

$$\begin{aligned} \left( \begin{array}{ll} \mu (1+2 \rho ) &{}\quad \rho (1-2 \rho ) \\ \rho (1-2 \rho ) &{}\quad \frac{\mu (1-3 \rho ^2+4 \rho ^4)+\rho (1-\rho ) (1-2 \rho )}{\mu } \\ \end{array} \right) . \end{aligned}$$

Rights and permissions

Reprints and Permissions

About this article

Verify currency and authenticity via CrossMark

Cite this article

Aleksandrov, B., Weiß, C.H. Parameter estimation and diagnostic tests for INMA(1) processes. TEST 29, 196–232 (2020). https://doi.org/10.1007/s11749-019-00653-7

Download citation

Keywords

  • Count time series
  • INMA(1) model
  • Diagnostic tests
  • Overdispersion
  • Autocorrelation structure
  • Confidence regions

Mathematics Subject Classification

  • 60G10
  • 62F03
  • 62F05
  • 62M10