Validation tests for the innovation distribution in INAR time series models

Abstract

Goodness-of-fit tests are proposed for the innovation distribution in INAR models. The test statistics incorporate the joint probability generating function of the observations. Special emphasis is given to the INAR(1) model and particular instances of the procedures which involve innovations from the general family of Poisson stopped-sum distributions. A Monte Carlo power study of a bootstrap version of the test statistic is included as well as a real data example. Generalizations of the proposed methods are also discussed.

Appendix: Proofs

Proof of Equation (3.5)

From the definition below Eq. (3.4) we compute

$$\begin{aligned} D^2_T(u,v)&= u^2\left( \frac{\partial G_T(u,v)}{\partial u}\right) ^2+v^2\left( \frac{\partial G_T(u,v)}{\partial v}\right) ^2+\lambda ^2 (u-v)^2 G^2_T(u,v)\\&-2uv\frac{\partial G_T(u,v)}{\partial u}\frac{\partial G_T(u,v)}{\partial v}-2\lambda u(u-v)G_T(u,v)\frac{\partial G_T(u,v)}{\partial u}\\&+2\lambda v(u-v)G_T(u,v)\frac{\partial G_T(u,v)}{\partial v}:=\sum _{j=1}^6S_j(u,v) \end{aligned}$$

Then the test statistic in Eq. (3.4) may be written as

$$\begin{aligned} W_{T,a}=\sum _{j=1}^6 \mathcal{{I}}_j, \end{aligned}$$

(7.1)

where $\mathcal{{I}}_j=\int _0^1 \int _0^1 S_j(u,v) u^av^adudv, \, j=1,\ldots ,6$.

We shall illustrate the computation of

$$\begin{aligned} \mathcal{{I}}_3=\int \limits _0^1 \int \limits _0^1 S_3(u,v) u^a v^a dudv=\int \limits _0^1 \int \limits _0^1 \lambda ^2 (u-v)^2 G^2_T(u,v) u^a v^a dudv. \end{aligned}$$

Analogous are the computations of $\mathcal{{I}}_j, \ j\ne 3$. To this end we have

$$\begin{aligned} G^2_T(u,v)=\frac{1}{(T-1)^2}\sum _{t,s=2}^T u^{Y_t+Y_s} v^{Y_{t-1}+Y_{s-1}}, \end{aligned}$$

and integration term-by-term leads to

$$\begin{aligned} \mathcal{{I}}_3&= \frac{\lambda ^2}{(T-1)^2}\sum _{t,s=2}^T \int \limits _0^1 \int \limits _0^1 (u-v)^2 u^{Y_t+Y_s+a} v^{Y_{t-1}+Y_{s-1}+a}dudv\\&= \frac{\lambda ^2}{(T-1)^2}\sum _{t,s=2}^T \left( I_{t-1,s-1,a}I_{t,s,a+2}\!+\!I_{t-1,s-1,a+2}I_{t,s,a} \!-\!2I_{t-1,s-1,a+1}I_{t,s,a+1}\right) , \end{aligned}$$

where $I_{t,s,a}:=I(Y_t+Y_s+a)$ and

$$\begin{aligned} I(x)=\int \limits _0^1 u^x du=\frac{1}{1+x}, \ x>-1. \end{aligned}$$

So then by using the notation $r_{t,s}$ introduced in Eq. (3.5) we have

$$\begin{aligned} \mathcal{{I}}_3&= \frac{\lambda ^2}{(T-1)^2}\sum _{t,s=2}^T \frac{1}{(1+r_{t-1,s-1})(3+r_{t,s})} \\&+ \frac{1}{(3+r_{t-1,s-1})(1+r_{t,s})}- \frac{2}{(2+r_{t-1,s-1})(2+r_{t,s})}, \end{aligned}$$

which after some simple algebra is seen to coincide with the second term in Eq. (3.5). Similarly we obtain

$$\begin{aligned} \mathcal{{I}}_1&= \frac{1}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_t Y_s}{(1+r_{t,s})(1+r_{t-1,s-1})},\\ \mathcal{{I}}_2&= \frac{1}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_{t-1} Y_{s-1}}{(1+r_{t,s})(1+r_{t-1,s-1})},\\ \mathcal{{I}}_4&= -\frac{2}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_t Y_{s-1}}{(1+r_{t,s})(1+r_{t-1,s-1})},\\ \mathcal{{I}}_5&= \frac{-2\lambda }{(T-1)^2}\sum _{t,s=2}^T \frac{Y_t (r_{t,s}-r_{t-1,s-1})}{(1+r_{t,s})(2+r_{t,s}) (1+r_{t-1,s-1})(2+r_{t-1,s-1})}~~\hbox {and}\\ \mathcal{{I}}_6&= \frac{2\lambda }{(T-1)^2}\sum _{t,s=2}^T \frac{Y_{t-1}(r_{t,s}-r_{t-1,s-1})}{(1+r_{t,s}) (2+r_{t,s})(1+r_{t-1,s-1})(2+r_{t-1,s-1})}. \end{aligned}$$

As already noted, $\mathcal{{I}}_3$ coincides with the second term in Eq. (3.5). Also it is easy to see that the sum $\mathcal{{I}}_1+\mathcal{{I}}_2+\mathcal{{I}}_4$ gives the first term in Eq. (3.5), and $\mathcal{{I}}_5+\mathcal{{I}}_6$ gives the third term in Eq. (3.5). Therefore by Eq. (7.1) the proof is complete.

Proof of Equation (4.2)

From Eq. (4.1) it follows that $g_\varepsilon (U_{p^k})=e^{\lambda [\exp \{\varphi p^k(u-1)\}-1]}$, which when inserted in Eq. (2.2) yields

$$\begin{aligned} g_Y(u)&= \prod _{k=0}^\infty e^{\lambda [\exp \{\varphi p^k(u-1)\}-1]}= \prod _{k=0}^\infty \exp \left\{ \lambda \sum _{l=1}^\infty \frac{[\varphi p^k(u-1)]^l}{l!}\right\} \nonumber \\&= \prod _{k=0}^\infty \prod _{l=1}^\infty \exp \left\{ \frac{\lambda \varphi ^l (p^k)^l(u-1)^l}{l!}\right\} = \prod _{l=1}^\infty \prod _{k=0}^\infty \exp \left\{ \frac{\lambda \varphi ^l (p^k)^l(u-1)^l}{l!}\right\} \nonumber \\&= \prod _{l=1}^\infty \exp \left\{ \frac{\lambda \varphi ^l (u-1)^l}{l!}\sum _{k=0}^\infty (p^l)^k \right\} = \prod _{l=1}^\infty \exp \left\{ \frac{\lambda \varphi ^l (u-1)^l}{l!} \frac{1}{1-p^l} \right\} . \nonumber \\ \end{aligned}$$

(7.2)

Proof of Equation (4.5)

From Eq. (4.4) it follows that

$$\begin{aligned} g_\varepsilon (U_{p^k})=\exp \left\{ \lambda \sum _{l=1}^\nu \nu _l \varpi ^l (p^k)^l(u-1)^l\right\} , \end{aligned}$$

which when inserted in (2.2) yields,

$$\begin{aligned} g_Y(u)&= \prod _{k=0}^\infty \exp \left\{ \lambda \sum _{l=1}^\nu \nu _l \varpi ^l (p^k)^l(u-1)^l\right\} \\&= \exp \left\{ \lambda \sum _{l=1}^\nu \nu _l \varpi ^l (u-1)^l\sum _{k=0}^\infty (p^k)^l\right\} = \exp \left\{ \lambda \sum _{l=1}^\nu \nu _l \varpi ^l (u-1)^l\frac{1}{1-p^l}\right\} . \end{aligned}$$

Proof of Equations (4.10) and (4.11)

From Eq. (4.3) we have

$$\begin{aligned} \frac{\partial G_\vartheta (u,v)}{\partial u}&= \lambda \left( \frac{\partial h(u)}{\partial u} +\frac{\partial \widetilde{h}(u,v)}{\partial u}\right) G_\vartheta (u,v)~~\text{ and } \\ \frac{\partial G_\vartheta (u,v)}{\partial v}&= \lambda \frac{\partial \widetilde{h}(u,v)}{\partial v}G_\vartheta (u,v). \end{aligned}$$

Substituting these equations in the left-hand side of Eq. (4.10) yields

$$\begin{aligned} \lambda \left\{ U_p\left[ \left\{ \frac{\partial h(u)}{\partial u} +\frac{\partial \widetilde{h}(u,v)}{\partial u}\right\} -\varphi e^{\varphi (u-1)}\right] -pv \frac{\partial \widetilde{h}(u,v)}{\partial v}\right\} G_\vartheta (u,v). \end{aligned}$$

(7.3)

Clearly

$$\begin{aligned} \frac{\partial h(u)}{\partial u}=\varphi e^{\varphi (u-1)}. \end{aligned}$$

(7.4)

Also

$$\begin{aligned} \frac{\partial \widetilde{h}(u,v)}{\partial v}= \frac{\partial }{\partial v} \sum _{\ell =1}^\infty \frac{(\varphi (vU_p-1))^\ell }{\ell !} \frac{1}{1-p^\ell }=\varphi U_p S(u,v) \end{aligned}$$

(7.5)

where $S(u,v)=\sum \nolimits _{\ell =1}^\infty \frac{(\varphi (vU_p-1))^{\ell }}{\ell !} \frac{1}{1-p^{\ell +1}}$ and likewise

$$\begin{aligned} \frac{\partial \widetilde{h}(u,v)}{\partial u}=\varphi v p S(u,v). \end{aligned}$$

(7.6)

(Notice that $S(u,v)<\infty$ by the ratio test for convergence of series). Then by substitution of Eqs. (7.4), (7.5) and (7.6) in Eq. (7.3), the proof is complete. The proof of Eq. (4.11) is completely analogous and it is therefore omitted.

Proof of Equation (4.12)

First replace the theoretical PGF and its derivatives by the corresponding empirical quantities in the definition below Eq. (4.10), and compute (use the notation $U_p=1+p(u-1)$),

$$\begin{aligned} D^2_T(u,v)&= U^2_p\left( \frac{\partial G_T(u,v)}{\partial u}\right) ^2+\lambda ^2 \varphi ^2 U^2_p e^{2\varphi (u-1)} G^2_T(u,v)\\&-2\lambda \varphi U^2_p e^{\varphi (u-1)}\frac{\partial G_T(u,v)}{\partial u} G_T(u,v) \\&+p^2 v^2 \left( \frac{\partial G_T(u,v)}{\partial v}\right) ^2 -2pU_p v \frac{\partial G_T(u,v)}{\partial u}\frac{\partial G_T(u,v)}{\partial v}\\&+2\lambda \varphi p U_p v e^{\varphi (u-1)} \frac{\partial G_T(u,v)}{\partial v} G_T(u,v) :=\sum _{j=1}^6 S_j(u,v). \end{aligned}$$

Then the test statistic may be written as in Eq. (7.1). The corresponding integrals can be calculated analogously as in the proof of Eq. (3.5) above as

$$\begin{aligned} \mathcal{{I}}_1&= \frac{1}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_t Y_sJ^{(2)}(0,r_{t,s}-2)}{(1+r_{t-1,s-1})},\\ \mathcal{{I}}_2&= \frac{\lambda ^2 \varphi ^2 e^{-2\varphi }}{(T-1)^2}\sum _{t,s=2}^T \frac{ J^{(2)}(2\varphi ,r_{t,s})}{(1+r_{t-1,s-1})},\\ \mathcal{{I}}_3&= -\frac{2\lambda \varphi e^{-\varphi }}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_t J^{(2)}(\varphi ,r_{t,s}-1)}{(1+r_{t-1,s-1})},\\ \mathcal{{I}}_4&= \frac{p^2}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_{t-1} Y_{s-1}}{(1+r_{t,s})(1+r_{t-1,s-1})},\\ \mathcal{{I}}_5&= -\frac{2p}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_{t} Y_{s-1}J^{(1)}(0,r_{t,s}-1)}{(1+r_{t-1,s-1})}~~\text{ and }\\ \mathcal{{I}}_6&= \frac{2\lambda \varphi p e^{-\varphi }}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_{t-1} J^{(1)}(\varphi ,r_{t,s})}{(1+r_{t-1,s-1})}. \end{aligned}$$

Collecting now the terms in Eq. (7.1) and comparing with the terms in Eq. (4.12), the proof is complete.

Proof of Equation (4.13)

In an analogous manner (see the proof of Eq. (4.12) above), from the definition below Eq. (4.11) compute

$$\begin{aligned} D^2_T(u,v)&= U^2_p\left( \frac{\partial G_T(u,v)}{\partial u}\right) ^2+\lambda ^2 \nu ^2 \varpi ^2 U^2_p U^{2(\nu -1)}_\varpi G^2_T(u,v)\\&-2\lambda \nu \varpi U^2_p U^{\nu -1}_\varpi \frac{\partial G_T(u,v)}{\partial u} G_T(u,v)\\&+p^2 v^2 \left( \frac{\partial G_T(u,v)}{\partial v}\right) ^2 -2pU_p v \frac{\partial G_T(u,v)}{\partial u}\frac{\partial G_T(u,v)}{\partial v} \\&+2\lambda p \nu \varpi U_p v U^{\nu -1}_\varpi \frac{\partial G_T(u,v)}{\partial v} G_T(u,v) := \sum _{j=1}^6 S_j(u,v). \end{aligned}$$

Then the test statistic may be written as in Eq. (7.1). Now clearly $\mathcal{{I}}_1, \mathcal{{I}}_4$ and $\mathcal{{I}}_5$ coincide with the corresponding integrals in the proof of Eq. (4.12) above. Also, by straightforward algebra

$$\begin{aligned} \mathcal{{I}}_2&= \frac{\lambda ^2 \varpi ^2 \nu ^2}{(T-1)^2}\sum _{t,s=2}^T \frac{ K^{(2)}(2\nu -2,r_{t,s})}{(1+r_{t-1,s-1})},\\ \mathcal{{I}}_3&= -\frac{2\lambda \nu \varpi }{(T-1)^2}\sum _{t,s=2}^T \frac{Y_t K^{(2)}(\nu -1,r_{t,s}-1)}{(1+r_{t-1,s-1})},~~ \text{ and }\\ \mathcal{{I}}_6&= \frac{2\lambda \nu \varpi p}{(T-1)^2}\sum _{t,s=2}^T \frac{Y_{t-1} K^{(2)}(\nu -1,r_{t,s})}{(1+r_{t-1,s-1})}, \end{aligned}$$

and the proof is complete by a further appeal to Eq. (7.1) and by comparison with Eq. (4.13).

Proof of Equation (4.14)

From Eq. (2.3) we have

$$\begin{aligned} \frac{\partial G(u,v)}{\partial u}\!=\!\frac{\partial g_\varepsilon (u)}{\partial u} g_Y(vU_p)\!+\!g_\varepsilon (u) \frac{\partial g_Y(vU_p)}{\partial u}~~\text{ and }~~ \frac{\partial G(u,v)}{\partial v}\!=\!g_\varepsilon (u) \frac{\partial g_Y(vU_p)}{\partial v}. \end{aligned}$$

Substituting these equations in the left-hand side of Eq. (4.14) yields

$$\begin{aligned}&\left[ 1\!+\!\varrho (1-u)\right] \left[ U_p\left\{ \frac{\partial g_\varepsilon (u)}{\partial u} g_Y(vU_p)\!+\!g_\varepsilon (u) \frac{\partial g_Y(vU_p)}{\partial u}\right\} \!-\!pvg_\varepsilon (u)\frac{\partial g_Y(vU_p)}{\partial v}\right] \nonumber \\&\quad -\varphi \varrho U_p G_\vartheta (u,v). \end{aligned}$$

(7.7)

Clearly the PGF of the NB distribution in Eq. (4.7) satisfies

$$\begin{aligned} \frac{\partial g_\varepsilon (u)}{\partial u}=\frac{\varphi \varrho }{1+\varrho (1-u)}g_\varepsilon (u). \end{aligned}$$

(7.8)

Also from Eq. (4.8) write

$$\begin{aligned} g_Y(vU_p)=\prod _{k=0}^\infty g_k(u,v), \quad g_k(u,v)=\frac{1}{[1+\varrho p^k(1-vU_p)]^{\varphi }}, \end{aligned}$$

and compute

$$\begin{aligned} \frac{\partial g_Y(vU_p)}{\partial v}&= \frac{\partial }{\partial v} \prod _{k=0}^\infty g_k(u,v) = \sum _{k=0}^\infty \frac{\partial }{\partial v} g_k(u,v)\prod _{j \ne k} g_j(u,v) \nonumber \\&= \sum _{k=0}^\infty \frac{\varphi \varrho U_p p^k}{[1+\varrho p^k(1-vU_p)]} g_k(u,v)\prod _{j \ne k} g_j(u,v)\nonumber \\&= \varphi \varrho U_p \sum _{k=0}^\infty \frac{p^k}{1+\varrho p^k(1-vU_p)} \prod _{k=0}^\infty g_k(u,v)\nonumber \\&= \varphi \varrho U_p S(u,v) g_Y(vU_p), \end{aligned}$$

(7.9)

where $S(u,v)=\sum _{k=0}^\infty (\varrho (1-vU_p)+p^{-k})^{-1}$. (Notice that $S(u,v)<\infty$, since $\varrho (1-vU_p)>0$ and therefore $S(u,v)<\sum _{k=0}^\infty p^{k}<\infty ,\,0<p<1$). Likewise we can show that

$$\begin{aligned} \frac{\partial g_Y(vU_p)}{\partial u}=\varrho \varphi p v S(u,v)g_Y(vU_p), \end{aligned}$$

(7.10)

and by substitution of Eqs. (7.8), (7.9) and (7.10) in Eq. (7.7), the proof is complete.