Asymptotics of Alternative Interdependence Measures for Bivariate $\alpha -$Stable Autoregressive Model of Order 1

Abstract

In this paper, we concentrate on the two-dimensional $\alpha$-stable autoregressive model of order 1. For this model, we analyze the auto-dependence functions applied to describe the interdependence of the time series components considered separately as the one-dimensional processes. Since the classical second moment-based dependence measures are not defined in the $\alpha -$stable case, we examine here the auto-codifference and the auto-covariation functions. The main result of this paper is the derivation of the formulas which describe the asymptotics of the considered measures of interdependence. The theoretical results are supported by a simulation study.

Appendices

Appendix A:

Lemma 1

Let $\{\mathbf {X}(t)\}=\{X_1(t),X_2(t)\}$ with $t\in {\mathbb {Z}}$ be the bounded solution of Eq. (3) given by Eq. (5).

• 1. For two different eigenvalues of $\varTheta$ indicated as $\lambda _1$ and $\lambda _2$, $|\lambda _1|<1$ and $|\lambda _2|<1$, let us introduce the following notation

  $$\begin{aligned}\begin{aligned} A_1=A_1(s_1,s_2,a_1,a_2,\lambda _1,\lambda _2,j)&=\frac{\lambda _1^j(\lambda _2s_1-a_1s_1-a_2s_2)}{\lambda _2-\lambda _1},\\ B_1=B_1(s_1,s_2,a_1,a_2,\lambda _1,\lambda _2,j)&=\frac{\lambda _2^j(-\lambda _1s_1+a_1s_1+a_2s_2)}{\lambda _2-\lambda _1},\\ C_1=C_1(s_1,s_2,a_1,a_2,\lambda _1,\lambda _2,j)&=\frac{\lambda _1^j(-a_2s_2+\lambda _2s_1-a_1s_1)+\lambda _2^j(a_2s_2-\lambda _1s_1+a_1s_1)}{\lambda _2-\lambda _1}. \end{aligned}\end{aligned}$$

  Then, for $h \in {\mathbb {N}}_{0}$ we obtain that

  • (a) for $0<\alpha <2$

    $$\begin{aligned}&\mathrm {CD}(X_1(t),X_1(t-h))=\mathrm {CD}(X_1(t),X_1(t+h)) \nonumber \\&={\sum _{j=0}^{+\infty }\int _{S_2}\left( |\lambda _1^h{A_1}+\lambda _2^h{B_1}|^{\alpha }+|C_1|^{\alpha }-|C_1-(\lambda _1^h{A_1}+\lambda _2^h{B_1})|^\alpha \right) \varGamma (ds)}, \end{aligned}$$

    (13)

  • (b) for $1<\alpha <2$

    $$\begin{aligned} \mathrm {CV}&(X_1(t),X_1(t-h))=\sum _{j=0}^{+\infty }\int _{S_2}C_1^{\langle \alpha -1 \rangle }\left( \lambda _1^hA_1+\lambda _2^hB_1\right) \varGamma (ds), \end{aligned}$$

    (14)

    and

    $$\begin{aligned} \mathrm {CV}&(X_1(t),X_1(t+h))=\sum _{j=0}^{+\infty }\int _{S_2}C_1\left( \lambda _1^hA_1+\lambda _2^hB_1\right) ^{\langle \alpha -1 \rangle }\varGamma (ds). \end{aligned}$$

    (15)

• 2. For equal eigenvalues of $\varTheta$ indicated as $\lambda _1=\lambda _2=\lambda$, $|\lambda |<1$, let us introduce the following notation

  $$\begin{aligned}\begin{aligned} A_2=A_2(s_1,s_2,a_1,a_2,\lambda ,j)&=j\lambda ^{j-1}a_1s_1-j\lambda ^js_1+\lambda ^js_1+j\lambda ^{j-1}a_2s_2,\\ B_2=B_2(s_1,s_2,a_1,a_2,\lambda ,j)&=\lambda ^{j-1}a_1s_1-\lambda ^js_1+\lambda ^{j-1}a_2s_2,\\ C_2=C_2(s_1,s_2,a_1,a_2,\lambda ,j)&=j\lambda ^{j-1}a_1s_1+j\lambda ^{j-1}a_2s_2-j\lambda ^js_1+\lambda ^js_1. \end{aligned} \end{aligned}$$

  Then, for $h \in {\mathbb {N}}_{0}$ we obtain that

  • (a) for $0<\alpha <2$

    $$\begin{aligned}&\mathrm {CD}(X_1(t),X_1(t-h))=\mathrm {CD}(X_1(t),X_1(t+h)) \nonumber \\&={\sum _{j=0}^{+\infty }\int _{S_2}\left( |\lambda ^h{A_2}+h\lambda ^h{B_2}|^{\alpha }+|C_2|^{\alpha }-|C_2-(\lambda ^h{A_2}+h\lambda ^h{B_2})|^\alpha \right) \varGamma (ds)}, \end{aligned}$$

    (16)

  • (b) for $1<\alpha <2$

    $$\begin{aligned} \mathrm {CV}&(X_1(t),X_1(t-h))=\sum _{j=0}^{+\infty }\int _{S_2}C_2^{\langle \alpha -1 \rangle }\left( \lambda ^hA_2+h\lambda _2^hB_2\right) \varGamma (ds), \end{aligned}$$

    (17)

    and

    $$\begin{aligned} \mathrm {CV}&(X_1(t),X_1(t+h))=\sum _{j=0}^{+\infty }\int _{S_2}C_2\left( \lambda ^hA_2+h\lambda ^hB_2\right) ^{\langle \alpha -1 \rangle }\varGamma (ds). \end{aligned}$$

    (18)

Proof

The proof is analogous to the ones presented in the authors’ previous papers, see Grzesiek et al. (2019) for the codifference function and Grzesiek et al. (2020) for the covariation function. We also use the formulas given in Eqs. (7–8).

Lemma 2

Let $\{\mathbf {X}(t)\}=\{X_1(t),X_2(t)\}$ with $t\in {\mathbb {Z}}$ be the bounded solution of Eq. (3) given by Eq. (5).

• 1. For two different eigenvalues of $\varTheta$ indicated as $\lambda _1$ and $\lambda _2$, $|\lambda _1|<1$ and $|\lambda _2|<1$, let us introduce the following notation

  $$\begin{aligned} \begin{aligned} A_3=A_3(s_1,s_2,a_3,a_4,\lambda _1,\lambda _2,j)&=\frac{\lambda _1^j(\lambda _2s_2-a_3s_1-a_4s_2)}{\lambda _2-\lambda _1},\\ B_3=B_3(s_1,s_2,a_3,a_4,\lambda _1,\lambda _2,j)&=\frac{\lambda _2^j(-\lambda _1s_2+a_3s_1+a_4s_2)}{\lambda _2-\lambda _1},\\ C_3=C_3(s_1,s_2,a_3,a_4,\lambda _1,\lambda _2,j)&=\frac{\lambda _1^j(-a_3s_1+\lambda _2s_2-a_4s_2)+\lambda _2^j(a_3s_1-\lambda _1s_2+a_4s_2)}{\lambda _2-\lambda _1}.\\ \end{aligned}\end{aligned}$$

  Then, for $h \in {\mathbb {N}}_{0}$ we obtain that

  • (a) for $0<\alpha <2$

    $$\begin{aligned}&\mathrm {CD}(X_2(t),X_2(t-h))=\mathrm {CD}(X_2(t),X_2(t+h)) \nonumber \\&={\sum _{j=0}^{+\infty }\int _{S_2}\left( |\lambda _1^h{A_3}+\lambda _2^h{B_3}|^{\alpha }+|C_3|^{\alpha }-|C_3-(\lambda _1^h{A_3}+\lambda _2^h{B_3})|^\alpha \right) \varGamma (ds)}, \end{aligned}$$

    (19)

  • (b) for $1<\alpha <2$

    $$\begin{aligned} \mathrm {CV}&(X_2(t),X_2(t-h))=\sum _{j=0}^{+\infty }\int _{S_2}C_3^{\langle \alpha -1 \rangle }\left( \lambda _1^hA_3+\lambda _2^hB_3\right) \varGamma (ds), \end{aligned}$$

    (20)

    and

    $$\begin{aligned} \mathrm {CV}&(X_2(t),X_2(t+h))=\sum _{j=0}^{+\infty }\int _{S_2}C_3\left( \lambda _1^hA_3+\lambda _2^hB_3\right) ^{\langle \alpha -1 \rangle }\varGamma (ds). \end{aligned}$$

    (21)

• 2. For equal eigenvalues of $\varTheta$ indicated as $\lambda _1=\lambda _2=\lambda$, $|\lambda |<1$, let us introduce the following notation

  $$\begin{aligned}\begin{aligned} A_4=A_4(s_1,s_2,a_3,a_4,\lambda ,j)=&j\lambda ^{j-1}a_3s_1-j\lambda ^js_2+\lambda ^js_2+j\lambda ^{j-1}a_4s_2,\\ B_4=B_4(s_1,s_2,a_3,a_4,\lambda ,j)=&\lambda ^{j-1}a_3s_1-\lambda ^{j}s_2+\lambda ^{j-1}a_4s_2,\\ C_4=C_4(s_1,s_2,a_3,a_4,\lambda ,j)=&j\lambda ^{j-1}a_3s_1+j\lambda ^{j-1}a_4s_2-j\lambda ^{j}s_2+\lambda ^js_2. \end{aligned}\end{aligned}$$

  Then, for $h \in {\mathbb {N}}_{0}$ we obtain that

  • (a) for $0<\alpha <2$

    $$\begin{aligned}&\mathrm {CD}(X_2(t),X_2(t-h))=\mathrm {CD}(X_2(t),X_2(t+h)) \nonumber \\&={\sum _{j=0}^{+\infty }\int _{S_2}\left( |\lambda ^h{A_4}+h\lambda ^h{B_4}|^{\alpha }+|C_4|^{\alpha }-|C_4-(\lambda ^h{A_4}+h\lambda ^h{B_4})|^\alpha \right) \varGamma (ds)}, \end{aligned}$$

    (22)

  • (b) for $1<\alpha <2$

    $$\begin{aligned} \mathrm {CV}&(X_2(t),X_2(t-h))=\sum _{j=0}^{+\infty }\int _{S_2}C_4^{\langle \alpha -1 \rangle }\left( \lambda ^hA_4+h\lambda _2^hB_4\right) \varGamma (ds), \end{aligned}$$

    (23)

    and

    $$\begin{aligned} \mathrm {CV}&(X_2(t),X_2(t+h))=\sum _{j=0}^{+\infty }\int _{S_2}C_4\left( \lambda ^hA_4+h\lambda ^hB_4\right) ^{\langle \alpha -1 \rangle }\varGamma (ds). \end{aligned}$$

    (24)

Proof

The proof is analogous to the ones presented in the authors’ previous papers, see Grzesiek et al. (2019) for the codifference function and Grzesiek et al. (2020) for the covariation function. We also use the formulas given in Eqs. (7–8).

Appendix B:

Proof

• A) Let us consider the auto-codifference function of $\{X_1(t)\}$ given in Lemma 1.

  • For $\lambda _1 \ne \lambda _2$, and $|\lambda _1|<1$, $|\lambda _2|<1$, we examine the auto-codifference given in Eq. (13).

    • I) Let us assume that $|\lambda _1|>|\lambda _2|$. In this case, one can show that

      $$\begin{aligned} \begin{aligned}&\lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-(\lambda _1^hA_1+\lambda _2^hB_1)\right| ^\alpha }{\lambda _1^h}\varGamma (ds)\\{\mathop {=}\limits ^{(\star )}}&\sum _{j=0}^{+\infty }\lim _{h \rightarrow +\infty }\int _{S_2}\frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-(\lambda _1^hA_1+\lambda _2^hB_1)\right| ^\alpha }{\lambda _1^h}\varGamma (ds)\\{\mathop {=}\limits ^{(\star \star )}}&\sum _{j=0}^{+\infty }\int _{S_2}\lim _{h \rightarrow +\infty }\frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-(\lambda _1^hA_1+\lambda _2^hB_1)\right| ^\alpha }{\lambda _1^h}\varGamma (ds). \end{aligned} \end{aligned}$$

      (25)

      From the dominated convergence theorem Weir (1973), let us notice that $(\star )$ holds if the sum over $j \in {\mathbb {N}}_{0}$ in Eq. (25) converges uniformly. Now, from the below inequalities

      $$\begin{aligned} \begin{aligned} \left| |a|^\alpha +|b|^\alpha -|a+b|^\alpha \right|&\le (\alpha +1)|a|^\alpha +\alpha |a||b|^{\alpha -1},\\ \left| a+b\right| ^\alpha&\le 2^{\alpha -1}\left( |a|^\alpha +|b|^\alpha \right) , \end{aligned} \end{aligned}$$

      (26)

      satisfied for $a,b \in {\mathrm{I{\!}\mathrm R}}$ and $1<\alpha <2$, see Maejima and Yamamoto (2003), for all $j \in {\mathbb {N}}_{0}$ we can show that

      $$\begin{aligned}&\left| \int _{S_2}\frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-(\lambda _1^hA_1+\lambda _2^hB_1)\right| ^\alpha }{\lambda _1^h}\varGamma (ds)\right| \\&\qquad \qquad \le \,2^{\alpha -1}(\alpha +1)\left( \int _{S_2}\left| A_1\right| ^\alpha \varGamma (ds)+\int _{S_2}\left| B_1\right| ^{\alpha }\varGamma (ds)\right) \\&\qquad \qquad \quad \,\,\,\,+\, \alpha \left( \int _{S_2}\left| A_1\right| |C_1|^{\alpha -1}\varGamma (ds)+\int _{S_2}\left| B_1\right| |C_1|^{\alpha -1}\varGamma (ds)\right) =M_j, \end{aligned}$$

      which means that each component of the infinite sum is bounded by an expression which does not depend on h. As a consequence, the sum over $j \in {\mathbb {N}}_{0}$ given in Eq. (25) converges uniformly if the sum of $M_j$ over $j \in {\mathbb {N}}_{0}$ is finite, i.e. if the below conditions are satisfied

      $$\begin{aligned} \begin{aligned}&\sum _{j=0}^{+\infty }\int _{S_2}\left| A_1\right| ^\alpha \varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| A_1\right| |C_1|^{\alpha -1}\varGamma (ds)<+\infty ,\\&\sum _{j=0}^{+\infty }\int _{S_2}\left| B_1\right| ^{\alpha }\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| B_1\right| |C_1|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned} \end{aligned}$$

      (27)

      which is true (see Remark 1 in Appendix D). Now, to prove that $(\star \star )$ in Eq. (25) holds we again apply the dominated convergence theorem. From the inequalities in Eq. (26), for the integrand given in Eq. (25) we have that

      $$\begin{aligned}&\left| \frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-(\lambda _1^hA_1+\lambda _2^hB_1)\right| ^\alpha }{\lambda _1^h}\right| \\&\qquad \le \,2^{\alpha -1}(\alpha +1)\left( \left| A_1\right| ^\alpha +\left| B_1\right| ^{\alpha }\right) +\alpha \left( \left| A_1\right| |C_1|^{\alpha -1}+\left| B_1\right| |C_1|^{\alpha -1}\right) . \end{aligned}$$

      for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$. Let us notice that the dominating function does not depend on h. Moreover, it is integrable if for all $j \in {\mathbb {N}}_0$ the below conditions are true

      $$\begin{aligned} \begin{aligned}&\int _{S_2}\left| A_1\right| ^\alpha \varGamma (ds)<+\infty , \quad \int _{S_2}\left| A_1\right| |C_1|^{\alpha -1}\varGamma (ds)<+\infty ,\\&\int _{S_2}\left| B_1\right| ^{\alpha }\varGamma (ds)<+\infty , \quad \int _{S_2}\left| B_1\right| |C_1|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned} \end{aligned}$$

      which is satisfied (see Remark 1 in Appendix D). Then, let us notice that for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$ and $h \rightarrow +\infty$ we have

      $$\begin{aligned}&\left| \lambda _1^h{A_1}+\lambda _2^h{B_1}\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-\left( \lambda _1^h{A_1}+\lambda _2^h{B_1}\right) \right| ^\alpha \\&=\left| \lambda _1^h\left( {A_1}+\left( {\lambda _2}/{\lambda _1}\right) ^h{B_1}\right) \right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-\lambda _1^h\left( {A_1}+\left( \lambda _2/\lambda _1\right) ^h{B_1}\right) \right| ^\alpha \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \sim \left| \lambda _1^h\right| ^{\alpha }\left| A_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-\lambda _1^hA_1\right| ^\alpha \end{aligned}$$

      and the following limit holds

      $$\begin{aligned} \lim _{x \rightarrow 0}\frac{|dx|^\alpha +|c|^\alpha -|c-dx|^\alpha }{x}=\alpha \, dc^{\langle \alpha -1\rangle } \quad \mathrm {for\ 1<\alpha <2,\ d,c \in {\mathrm{I{\!}\mathrm R}}}. \end{aligned}$$

      (28)

      Moreover, we have $\lambda _1^h \rightarrow 0$ for $h \rightarrow +\infty$. Finally we can write that

      $$\begin{aligned}&\lim _{h \rightarrow +\infty }\frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-\left( \lambda _1^hA_1+\lambda _2^hB_1\right) \right| ^\alpha }{\lambda _1^h} \nonumber \\&\qquad \quad =\lim _{h \rightarrow +\infty }\frac{\left| \lambda _1^hA_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-\lambda _1^hA_1\right| ^\alpha }{\lambda _1^h}=\alpha \, A_1C_1^{\langle \alpha -1\rangle } \end{aligned}$$

      (29)

      for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$. Finally, from Eq. (25) we finally have that

      $$\begin{aligned}&\lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{\left| \lambda _1^hA_1+\lambda _2^hB_1\right| ^{\alpha }+\left| C_1\right| ^{\alpha }-\left| C_1-(\lambda _1^hA_1+\lambda _2^hB_1)\right| ^\alpha }{\lambda _1^h}\varGamma (ds) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad =\alpha D_1, \end{aligned}$$

      (30)

      which is equivalent to

      $$\begin{aligned}&\mathrm{{CD}}(X_1(t),X_1(t-h))=\mathrm{{CD}}(X_1(t),X_1(t+h)) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha D_1\,\lambda _1^h\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (31)

      where

      $$\begin{aligned} D_1:=\sum _{j=0}^{+\infty }\int _{S_2} A_1C_1^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (32)

    • II) Let us assume that $|\lambda _1|<|\lambda _2|$. Proceeding similarly as above we have

      $$\begin{aligned}&\mathrm{{CD}}(X_1(t),X_1(t-h))=\mathrm{{CD}}(X_1(t),X_1(t+h))\nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha D_2\,\lambda _2^h\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (33)

      where

      $$\begin{aligned} D_2:=\sum _{j=0}^{+\infty }\int _{S_2}B_1C_1^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (34)

    • III) Let us assume that $\lambda _1=-\lambda _2$ and even h. Similarly as above, we obtain

      $$\begin{aligned}&\mathrm{{CD}}(X_1(t),X_1(t-h))=\mathrm{{CD}}(X_1(t),X_1(t+h)) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha (D_1+D_2)\,\lambda _1^h\ \ \text { for}\ h \rightarrow +\infty , \end{aligned}$$

      (35)

      where

      $$\begin{aligned} D_1+D_2=\sum _{j=0}^{+\infty }\int _{S_2}(A_1+B_1)C_1^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

    • IV) Let us assume that $\lambda _1=-\lambda _2$ and h is odd. Similarly as above, we have

      $$\begin{aligned}&\mathrm{{CD}}(X_1(t),X_1(t-h))=\mathrm{{CD}}(X_1(t),X_1(t+h)) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha (D_1-D_2)\,\lambda _1^h\ \text { for}\ h \rightarrow +\infty , \end{aligned}$$

      (36)

      where

      $$\begin{aligned} D_1-D_2=\sum _{j=0}^{+\infty }\int _{S_2}(A_1-B_1)C_1^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

  • For $\lambda _1=\lambda _2=\lambda$, and $|\lambda |<1$, we examine the auto-codifference function given in Eq. (16). Similarly to previous case, we show that

    $$\begin{aligned} \begin{aligned}&\lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{\left| \lambda ^hA_2+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-(\lambda ^hA_2+h\lambda ^hB_2)\right| ^\alpha }{h\lambda ^h}\varGamma (ds)\\ {}&{\mathop {=}\limits ^{(\star )}}\sum _{j=0}^{+\infty }\lim _{h \rightarrow +\infty }\int _{S_2}\frac{\left| \lambda ^hA_2+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-(\lambda ^hA_2+h\lambda ^hB_2)\right| ^\alpha }{h\lambda ^h}\varGamma (ds)\\ {}&{\mathop {=}\limits ^{(\star \star )}}\sum _{j=0}^{+\infty }\int _{S_2}\lim _{h \rightarrow +\infty }\frac{\left| \lambda ^hA_2+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-(\lambda ^hA_2+h\lambda ^hB_2)\right| ^\alpha }{h\lambda ^h}\varGamma (ds). \end{aligned} \end{aligned}$$

    (37)

    From the dominated convergence theorem, to justify $(\star )$ one has to prove that the sum over $j \in {\mathbb {N}}_0$ in Eq. (37) converges uniformly. From the inequalities in Eq. (26) we have that

    $$\begin{aligned}&\left| \int _{S_2}\frac{\left| \lambda ^hA_2+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-(\lambda ^hA_2+h\lambda ^hB_2)\right| ^\alpha }{h\lambda ^h}\varGamma (ds)\right| \\&\qquad \qquad \le \,2^{\alpha -1}(\alpha +1)M\left( \int _{S_2}\left| A_2\right| ^\alpha \varGamma (ds)+\int _{S_2}\left| B_2\right| ^{\alpha }\varGamma (ds)\right) \\&\qquad \qquad \qquad +\alpha \left( \int _{S_2}\left| A_2\right| |C_2|^{\alpha -1}\varGamma (ds)+\int _{S_2}\left| B_2\right| |C_2|^{\alpha -1}\varGamma (ds)\right) =N_j, \end{aligned}$$

    for all $j \in {\mathbb {N}}_0$, where M is the boundary of $\{h\lambda ^h\}$ over ${h\in {\mathbb {N}}_{0}}$. Let us notice that $N_j$ does not depend on h. Consequently, the sum over ${j\in {\mathbb {N}}_{0}}$ in Eq. (37) converges uniformly if the sum of $N_j$ over $j \in {\mathbb {N}}_0$ converges, i.e. if the below sums are finite

    $$\begin{aligned} \begin{aligned}&\sum _{j=0}^{+\infty }\int _{S_2}\left| A_2\right| ^\alpha \varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| A_2\right| |C_2|^{\alpha -1}\varGamma (ds)<+\infty ,\\&\sum _{j=0}^{+\infty }\int _{S_2}\left| B_2\right| ^{\alpha }\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| B_2\right| |C_2|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned}\end{aligned}$$

    (38)

    which is true (see Remark 1 in Appendix D). As the second step, to prove that $(\star \star )$ in Eq. (37) holds we again use the dominated convergence theorem. Namely, from the inequalities given in Eq. (26), for the integrand in Eq. (37) we obtain that

    $$\begin{aligned}&\left| \frac{\left| \lambda ^hA_2+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-(\lambda ^hA_2+h\lambda ^hB_2)\right| ^\alpha }{h\lambda ^h}\right| \nonumber \\&\, \le \,2^{\alpha -1}(\alpha +1)M\left( \left| A_2\right| ^\alpha +\left| B_2\right| ^{\alpha }\right) +\alpha \left( \left| A_2\right| |C_2|^{\alpha -1}+\left| B_2\right| |C_2|^{\alpha -1}\right) , \end{aligned}$$

    (39)

    for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$. Let us notice that the dominating function does not depend on h. Moreover, it is integrable if for all $j \in {\mathbb {N}}_{0}$ the below integrals are finite

    $$\begin{aligned} \begin{aligned}&\int _{S_2}\left| A_2\right| ^\alpha \varGamma (ds)<+\infty , \quad \int _{S_2}\left| A_2\right| |C_2|^{\alpha -1}\varGamma (ds)<+\infty ,\\&\int _{S_2}\left| B_2\right| ^{\alpha }\varGamma (ds)<+\infty , \quad \int _{S_2}\left| B_2\right| |C_2|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned}\end{aligned}$$

    which is true (see Remark 1 in Appendix D). Then, let us notice that for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$ and $h \rightarrow \infty$ we have

    $$\begin{aligned}&\left| \lambda ^h{A_2}+h\lambda ^h{B_2}\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-\left( \lambda ^h{A_2}+h\lambda ^h{B_2}\right) \right| ^\alpha \\&\qquad \,\, =\left| h\lambda ^h\left( {A_2/h}+{B_2}\right) \right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-h\lambda ^h\left( {A_2/h}+{B_2}\right) \right| ^\alpha \\&\qquad \qquad \qquad \qquad \qquad \qquad \,\,\, \sim \left| h\lambda ^h\right| ^{\alpha }\left| B_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-h\lambda ^hB_2\right| ^\alpha . \end{aligned}$$

    From the limit in Eq. (28) and since $h\lambda ^h \rightarrow 0$ for $h \rightarrow +\infty$, we have that

    $$\begin{aligned}&\lim _{h \rightarrow +\infty }\frac{\left| \lambda ^hA+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-\left( \lambda ^hA+h\lambda ^hB_2\right) \right| ^\alpha }{h\lambda ^h}\\&\qquad \qquad \qquad \quad =\lim _{h \rightarrow +\infty }\frac{\left| h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-h\lambda ^hB_2\right| ^\alpha }{h\lambda ^h}=\alpha \, B_2C_2^{\langle \alpha -1\rangle } \end{aligned}$$

    for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$. Finally, from Eq. (37) we have

    $$\begin{aligned}&\lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{\left| \lambda ^hA_2+h\lambda ^hB_2\right| ^{\alpha }+\left| C_2\right| ^{\alpha }-\left| C_2-(\lambda ^hA_2+h\lambda ^hB_2)\right| ^\alpha }{h\lambda ^h}\varGamma (ds) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =\alpha D_3, \end{aligned}$$

    (40)

    which is equivalent to the fact that

    $$\begin{aligned} \mathrm{{CD}}(X_1(t),X_1(t-h))=\mathrm{{CD}}(X_1(t),X_1(t+h)) \sim \alpha D_3\,h\lambda ^h \ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

    (41)

    where

    $$\begin{aligned} D_3:=\sum _{j=0}^{+\infty }\int _{S_2} B_2C_2^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

    (42)

• B) Let us consider the auto-codifference function of $\{X_2(t)\}$ given in Lemma 2.

  • For $\lambda _1 \ne \lambda _2$, and $|\lambda _1|<1$, $|\lambda _2|<1$ we examine the auto-codifference function given in Eq. (19).

    • I) Let us assume that $|\lambda _1|>|\lambda _2|$. Proceeding as in A) leads to

      $$\begin{aligned}&\mathrm{{CD}}(X_2(t),X_2(t-h))=\mathrm{{CD}}(X_2(t),X_2(t+h))\nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha E_1\,\lambda _1^h\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (43)

      where

      $$\begin{aligned} E_1:=\sum _{j=0}^{+\infty }\int _{S_2} A_3C_3^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (44)

      We mention here, similarly as in A), to use the dominated convergence theorem we need to guarantee that

      $$\begin{aligned} \begin{aligned}&\sum _{j=0}^{+\infty }\int _{S_2}\left| A_3\right| ^\alpha \varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| A_3\right| |C_3|^{\alpha -1}\varGamma (ds)<+\infty ,\\&\sum _{j=0}^{+\infty }\int _{S_2}\left| B_3\right| ^{\alpha }\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| B_3\right| |C_3|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned}\end{aligned}$$

      (45)

      that is true (see Remark 1 in Appendix D).

    • II) Let us assume that $|\lambda _1|<|\lambda _2|$. Again, proceeding as in A) we obtain

      $$\begin{aligned}&\mathrm{{CD}}(X_2(t),X_2(t-h))=\mathrm{{CD}}(X_2(t),X_2(t+h)) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha E_2\,\lambda _2^h\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (46)

      where

      $$\begin{aligned} E_2:=\sum _{j=0}^{+\infty }\int _{S_2} B_3C_3^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (47)

    • III) Let us assume that $\lambda _1=-\lambda _2$ and even h. Proceeding as in A) we have

      $$\begin{aligned}&\mathrm{{CD}}(X_2(t),X_2(t-h))=\mathrm{{CD}}(X_2(t),X_2(t+h)) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha (E_1+E_2)\,\lambda _1^h\ \text { for}\ h \rightarrow +\infty , \end{aligned}$$

      (48)

      where

      $$\begin{aligned} E_1+E_2=\sum _{j=0}^{+\infty }\int _{S_2} (A_3+B_3)C_3^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

    • iV) Let us assume that $\lambda _1=-\lambda _2$ and odd h. Proceeding as in A) leads to

      $$\begin{aligned}&\mathrm{{CD}}(X_2(t),X_2(t-h))=\mathrm{{CD}}(X_2(t),X_2(t+h)) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \sim \alpha (E_1-E_2)\,\lambda _1^h\ \text { for}\ h \rightarrow +\infty , \end{aligned}$$

      (49)

      where

      $$\begin{aligned} E_1-E_2=\sum _{j=0}^{+\infty }\int _{S_2} (A_3-B_3)C_3^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty .\end{aligned}$$

  • For $\lambda _1=\lambda _2=\lambda$, and $|\lambda |<1$, we examine the auto-codifference function given in Eq. (22). Proceeding as in A) leads to

    $$\begin{aligned} \mathrm{{CD}}(X_2(t),X_2(t+h))=\mathrm{{CD}}(X_2(t),X_2(t+h)) \sim \alpha E_3\,h\lambda ^h \ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

    (50)

    where

    $$\begin{aligned} E_3:=\sum _{j=0}^{+\infty }\int _{S_2} B_4C_4^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

    (51)

    As in A), to use the dominated convergence theorem we need to guarantee that

    $$\begin{aligned} \begin{aligned}&\sum _{j=0}^{+\infty }\int _{S_2}\left| A_4\right| ^\alpha \varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| A_4\right| |C_4|^{\alpha -1}\varGamma (ds)<+\infty ,\\&\sum _{j=0}^{+\infty }\int _{S_2}\left| B_4\right| ^{\alpha }\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| B_4\right| |C_4|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned} \end{aligned}$$

    (52)

    that is true (see Remark 1 in Appendix D).

Appendix C:

Proof

• A) Let us consider the auto-covariation function of the time series $\{X_1(t)\}$ given in Lemma 1.

  • (a) At first, we examine the function $\mathrm{CV}(X_1(t),X_1(t-h))$.

    • Let us assume that $\lambda _1 \ne \lambda _2$, and $|\lambda _1|<1$, $|\lambda _2|<1$. The auto-covariation function given in Eq. (14) can be written as

      $$\begin{aligned} \mathrm{{CV}}(X_1(t),X_1(t-h))=\lambda _1^hD_1+\lambda _2^hD_2, \end{aligned}$$

      (53)

      which directly leads to the formulas given in Theorem 2. The constants $D_1$ and $D_2$ are specified in Eqs. (32) and (34), respectively.

    • Let us assume that $\lambda _1=\lambda _2=\lambda$, and $|\lambda |<1$. The auto-covariation function given in Eq. (17) can be written as

      $$\begin{aligned} \mathrm{{CV}}(X_1(t),X_1(t-h))=\lambda ^hD_{*}+h\lambda ^hD_3, \end{aligned}$$

      (54)

      which directly leads to the formula given in Theorem 2. The constant $D_3$ is given in Eq. (42) and

      $$\begin{aligned} D_{*}:={\sum _{j=0}^{+\infty }\int _{S_2}A_2C_2^{\langle \alpha -1 \rangle }\varGamma (ds)}. \end{aligned}$$

  • (b) Now, we examine the function $\mathrm{CV}(X_1(t),X_1(t+h))$.

  • For $\lambda _1 \ne \lambda _2$, and $|\lambda _1|<1$, $|\lambda _2|<1$, we consider the auto-covariation function given in Eq. (15).

    • I) Let us assume that $|\lambda _1|>|\lambda _2|$. In this case, one can show that

      $$\begin{aligned} \begin{aligned}&\lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)\\{\mathop {=}\limits ^{(\star )}}&\sum _{j=0}^{+\infty }\lim _{h \rightarrow +\infty }\int _{S_2}\frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)\\{\mathop {=}\limits ^{(\star \star )}}&\sum _{j=0}^{+\infty }\int _{S_2}\lim _{h \rightarrow +\infty }\frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds), \end{aligned}\end{aligned}$$

      (55)

      From the dominated convergence theorem, $(\star )$ holds if the sum over $j \in {\mathbb {N}}_{0}$ in Eq. (55) converges uniformly. From the following inequality

      $$\begin{aligned} \left| a+b\right| ^{\alpha -1}\le |a|^{\alpha -1}+|b|^{\alpha -1} \end{aligned}$$

      (56)

      satisfied for $a,b \in {\mathrm{I{\!}\mathrm R}}$, $1<\alpha <2$, for all $j \in {\mathbb {N}}_0$ we have

      $$\begin{aligned}&\left| \int _{S_2}\frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)\right| \nonumber \\&\qquad \qquad \,\, \le \int _{S_2}|C_1|\left| A_1\right| ^{\alpha -1}\varGamma (ds)+\int _{S_2}|C_1|\left| B_1\right| ^{\alpha -1}\varGamma (ds)=K_j. \end{aligned}$$

      (57)

      Let us notice that $K_j$ does not depend on h. Consequently, the sum over $j \in {\mathbb {N}}_{0}$ in Eq. (55) converges uniformly if the sum of $K_j$ over $j \in {\mathbb {N}}_0$ converges, which leads to the below conditions

      $$\begin{aligned} \sum _{j=0}^{+\infty }\int _{S_2}\left| C_1\right| |A_1|^{\alpha -1}\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| C_1\right| |B_1|^{\alpha -1}\varGamma (ds)<+\infty \end{aligned}$$

      (58)

      that are true, see Remark 1 in Appendix D. In the next step, we prove that $(\star \star )$ in Eq. (55) holds by the use of the dominated convergence theorem. From the inequality in Eq. (56) we have that

      $$\begin{aligned} \left| \frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }}\right| \le |C_1||A_1|^{\alpha -1}+|C_1||B_1|^{\alpha -1}. \end{aligned}$$

      for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$. Let us notice that the dominating function does not depend on h and is integrable if for all $j \in {\mathbb {N}}_0$ we have that

      $$\begin{aligned} \begin{aligned} \int _{S_2}\left| C_1||A_1\right| ^{\alpha -1}\varGamma (ds)<+\infty , \quad \int _{S_2}\left| C_1||B_1\right| ^{\alpha -1}\varGamma (ds)<+\infty ,\\ \end{aligned} \end{aligned}$$

      which is true, see Remark 1 in Appendix D. Then, let us notice that for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$ the following limit holds

      $$\begin{aligned} \lim _{h \rightarrow +\infty }\frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }} =C_1A_1^{\langle \alpha -1 \rangle } \end{aligned}$$

      and finally, from Eq. (55) we have

      $$\begin{aligned} \lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{C_1\left( A_1\lambda _1^h+B_1\lambda _2^h\right) ^{\langle \alpha -1 \rangle }}{\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)= D_4. \end{aligned}$$

      (59)

      which is equivalent to the fact that

      $$\begin{aligned} \mathrm{CV}(X_1(t),X_1(t+h)) \sim D_4\,\left( \lambda _1^h\right) ^{\langle \alpha -1 \rangle }\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (60)

      where

      $$\begin{aligned} D_4:=\sum _{j=0}^{+\infty }\int _{S_2} C_1A_1^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (61)

    • II) Let us assume that $|\lambda _1|<|\lambda _2|$. Proceeding similary as above, we obtain

      $$\begin{aligned} \mathrm{CV}(X_1(t),X_1(t+h))\sim D_5\,\left( \lambda _2^h\right) ^{\langle \alpha -1\rangle }\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (62)

      where

      $$\begin{aligned} D_5:=\sum _{j=0}^{+\infty }\int _{S_2} C_1B_1^{\langle \alpha -1\rangle }\varGamma (ds) < +\infty . \end{aligned}$$

      (63)

    • III) Let us assume that $\lambda _1=-\lambda _2$ and even h. In this case, we have the exact formula

      $$\begin{aligned} \mathrm{CV}(X_1(t),X_1(t+h)) = D_{7}\,\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }, \end{aligned}$$

      (64)

      where

      $$\begin{aligned} D_{7}:=\sum _{j=0}^{+\infty }\int _{S_2}C_1(A_1+B_1)^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty .\end{aligned}$$

      (65)

    • IV) Let us assume that $\lambda _1=-\lambda _2$ and odd h. In this case, as in the case given above, we have the exact formula

      $$\begin{aligned} \mathrm{CV}(X_1(t),X_1(t+h)) = D_{11}\,\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }, \end{aligned}$$

      (66)

      where

      $$\begin{aligned} D_{8}:=\sum _{j=0}^{+\infty }\int _{S_2}C_1(A_1-B_1)^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (67)

  • For $\lambda _1=\lambda _2=\lambda$, and $|\lambda |<1$, we consider the auto-covariation function given in Eq. (18). Similarly to the previous case, one can show that

    $$\begin{aligned} \begin{aligned}&\lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)\\{\mathop {=}\limits ^{(\star )}}&\sum _{j=0}^{+\infty }\lim _{h \rightarrow +\infty }\int _{S_2}\frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)\\{\mathop {=}\limits ^{(\star \star )}}&\sum _{j=0}^{+\infty }\int _{S_2}\lim _{h \rightarrow +\infty }\frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds). \end{aligned}\end{aligned}$$

    (68)

    From the uniform convergence theorem, to justify $(\star )$ the sum over $j \in {\mathbb {N}}_{0}$ in Eq. (68) has to converge uniformly. From the inequality in Eq. (56) for all $j \in {\mathbb {N}}_0$ we can show that

    $$\begin{aligned}&\left| \int _{S_2}\frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)\right| \\&\qquad \qquad \qquad \qquad \le \int _{S_2}|C_2|\left| A_2\right| ^{\alpha -1}\varGamma (ds)+\int _{S_2}|C_2|\left| B_2\right| ^{\alpha -1}\varGamma (ds)=L_j, \end{aligned}$$

    where $L_j$ does not depend on h. Consequently, the sum over $j \in {\mathbb {N}}_{0}$ in Eq. (68) converges uniformly if the sum of $L_j$ over $j \in {\mathbb {N}}_0$ converges, that leads to the below conditions

    $$\begin{aligned} \sum _{j=0}^{+\infty }\int _{S_2}\left| C_2\right| |A_2|^{\alpha -1}\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| C_2\right| |B_2|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned}$$

    (69)

    which are true (see Remark 1 in Appendix D). Now, to justify $(\star \star )$ in Eq. (68) we again use the dominated convergence theorem. From the inequality in Eq. (56) we have

    $$\begin{aligned} \left| \frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}\right| \le |C_2||A_2|^{\alpha -1}+|C_2||B_2|^{\alpha -1}, \end{aligned}$$

    for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$. Let us notice that the integrand in Eq. (68) is dominated a function that does not depend on h. Moreover, this function is integrable if for all $j \in {\mathbb {N}}$ the integrals given below are finite

    $$\begin{aligned} \begin{aligned} \int _{S_2}\left| C_2||A_2\right| ^{\alpha -1}\varGamma (ds)<+\infty , \quad \int _{S_2}\left| C_2||B_2\right| ^{\alpha -1}\varGamma (ds)<+\infty ,\\ \end{aligned} \end{aligned}$$

    which is true (see Remark 1 in Appendix D). Then, let us notice that for a fixed $\mathbf {s}=(s_1,s_2)\in S_2$ the following limit holds

    $$\begin{aligned} \lim _{h \rightarrow +\infty }\frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}=C_2B_2^{\langle \alpha -1 \rangle } \end{aligned}$$

    and finally from Eq. (68) we obtain

    $$\begin{aligned} \lim _{h \rightarrow +\infty }\sum _{j=0}^{+\infty }\int _{S_2}\frac{C_2\left( A_2\lambda ^h+B_2h\lambda ^h\right) ^{\langle \alpha -1 \rangle }}{\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle }}\varGamma (ds)= D_6, \end{aligned}$$

    (70)

    which is equivalent to the fact that

    $$\begin{aligned} \mathrm{{CV}}(X_1(t),X_1(t+h))\sim D_6\,\left( h\lambda ^h\right) ^{\langle \alpha -1 \rangle }\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

    (71)

    where

    $$\begin{aligned} D_6:=\sum _{j=0}^{+\infty }\int _{S_2} C_2B_2^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

    (72)

• B) Let us consider the auto-covariation function of the time series $\{X_2(t)\}$ given in Lemma 2.

  • (a) At first, let us examine the function $\mathrm{CV}(X_2(t),X_2(t-h))$.

    • Let us assume that $\lambda _1 \ne \lambda _2$, and $|\lambda _1|<1$, $|\lambda _2|<1$. The auto-covariation function given in Eq. (20) can be written as

      $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t-h))=\lambda _1^hE_1+\lambda _2^hE_2, \end{aligned}$$

      (73)

      which directly leads to the formulas given in Theorem 2. The constants $E_1$ and $E_2$ are specified in Eqs. (44) and (47), respectively.

    • Let us assume that $\lambda _1=\lambda _2=\lambda$, and $|\lambda |<1$. The auto-covariation function given in Eq. (23) can be written as

      $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t-h))=\lambda ^hE_{*}+h\lambda ^hE_3, \end{aligned}$$

      (74)

      which directly leads to the formulas given in Theorem 2. The constant $E_3$ is given in Eq. (51) and

      $$\begin{aligned} E_{*}:={\sum _{j=0}^{+\infty }\int _{S_2}A_4C_4^{\langle \alpha -1 \rangle }\varGamma (ds)}. \end{aligned}$$

  • (b) Now, we examine the function $\mathrm{CV}(X_2(t),X_2(t+h))$.

    • For $\lambda _1 \ne \lambda _2$, and $|\lambda _1|<1$, $|\lambda _2|<1$, we consider the auto-covariation function given in Eq. (21).

      • I) Let us assume that $|\lambda _1|>|\lambda _2|$. Proceeding as in A) we obtain

        $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t+h))\sim E_4\left( \lambda _1^h\right) ^{\langle \alpha -1 \rangle } \quad \text {for} \quad h \rightarrow \infty , \end{aligned}$$

        (75)

        where

        $$\begin{aligned} E_4:=\sum _{j=0}^{\infty }\int _{S_2}C_3A_3^{\langle \alpha -1 \rangle }\varGamma (ds)<+\infty . \end{aligned}$$

        (76)

        We mention here that similarly as in A), to use the dominated convergence theorem we need to guarantee that

        $$\begin{aligned} \begin{aligned} \sum _{j=0}^{+\infty }\int _{S_2}\left| C_3\right| |A_3|^{\alpha -1}\varGamma (ds)<+\infty ,\\ \sum _{j=0}^{+\infty }\int _{S_2}\left| C_3\right| |B_3|^{\alpha -1}\varGamma (ds)<+\infty . \end{aligned} \end{aligned}$$

        (77)

        which is true (see Remark 1 in Appendix D).

      • II) Let us assume that $|\lambda _1|<|\lambda _2|$. Again, proceeding as in A) we obtain

        $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t+h))\sim E_5\,\left( \lambda _2^h\right) ^{\langle \alpha -1 \rangle }\ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

        (78)

        where

        $$\begin{aligned} E_5:=\sum _{j=0}^{+\infty }\int _{S_2} C_3B_3^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

        (79)

      • III) Let us assume that $\lambda _1=-\lambda _2$ and h is even. In this case, we have the exact formula

        $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t+h))= E_7\,\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }\ \text { for}\ h \rightarrow +\infty , \end{aligned}$$

        (80)

        where

        $$\begin{aligned} E_7:=\sum _{j=0}^{+\infty }\int _{S_2} C_3(A_3+B_3)^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      • IV) Let us assume that $\lambda _1=-\lambda _2$ and h is odd. Similarly as above, we have the exact formula

        $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t+h))= E_8\left( \lambda _1^h\right) ^{\langle \alpha -1\rangle }\ \text { for}\ h \rightarrow +\infty , \end{aligned}$$

        (81)

        where

        $$\begin{aligned} E_8:=\sum _{j=0}^{+\infty }\int _{S_2} C_3(A_3-B_3)^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty .\end{aligned}$$

    • For $\lambda _1=\lambda _2=\lambda$, and $|\lambda |<1$, we consider the auto-covariation function given in Eq. (24). Proceeding as in A) we obtain

      $$\begin{aligned} \mathrm{{CV}}(X_2(t),X_2(t+h)) \sim E_6\,\left( h\lambda ^h\right) ^{\langle \alpha -1\rangle } \ \ \text { for}\ \ h \rightarrow +\infty , \end{aligned}$$

      (82)

      where

      $$\begin{aligned} E_6:=\sum _{j=0}^{+\infty }\int _{S_2} C_4B_4^{\langle \alpha -1\rangle }\varGamma (ds)<+\infty . \end{aligned}$$

      (83)

      As in A), to use the dominated convergence theorem we need to guarantee that

      $$\begin{aligned} \begin{aligned} \sum _{j=0}^{+\infty }\int _{S_2}\left| C_4\right| |A_4|^{\alpha -1}\varGamma (ds)<+\infty , \quad \sum _{j=0}^{+\infty }\int _{S_2}\left| C_4\right| |B_4|^{\alpha -1}\varGamma (ds)<+\infty , \end{aligned} \end{aligned}$$

      (84)

      which is true (see Remark 1 in Appendix D).

Appendix D

Remark 1

Let us notice that the constants $A_i,B_i,C_i$ for $i=1,2,3,4$ can be upper-bounded by $M\,\mathrm{max}(|\lambda _1|,|\lambda _2|)^j$ or by $M\,j\mathrm{max}(|\lambda _1|,|\lambda _2|)^j$ with the constant M independent of j. Since $\mathrm{max}(|\lambda _1|,|\lambda _2|)<1$ and the measure $\varGamma (\cdot )$ is finite, the conditions given in Eqs. (27), (38), (45), (52), (58), (69), (77) and (84) are always satisfied.