Exact asymptotic formulae of the stationary distribution of a discrete-time two-dimensional QBD process

Abstract

We consider a discrete-time two-dimensional process $\{(X_{1,n},X_{2,n})\}$ on $\mathbb {Z}_+^2$ with a supplemental process $\{J_n\}$ on a finite set, where the individual processes $\{X_{1,n}\}$ and $\{X_{2,n}\}$ are both skip-free. We assume that the joint process $\{\varvec{Y}_n\}=\{(X_{1,n},X_{2,n},J_n)\}$ is Markovian and that the transition probabilities of the two-dimensional process $\{(X_{1,n},X_{2,n})\}$ are modulated depending on the state of the supplemental process $\{J_n\}$. This modulation is space homogeneous except for the boundaries of $\mathbb {Z}_+^2$. We call this process a discrete-time two-dimensional quasi-birth-and-death process. Under several conditions, we obtain the exact asymptotic formulae of the stationary distribution in the coordinate directions.

Proof of Lemma 3.1

Recall that, for $r>0$, we denote by $\Delta _r$, $\bar{\Delta }_r$ and $\partial \Delta _r$ the open disk, closed disk and circle of center 0 and radius r on the complex plane, respectively. The generating function $\varvec{\varphi }_+(z,w)$ is the function of two variables defined by the power series

$$\begin{aligned} \varvec{\varphi }_+(z,w)=\sum _{z=1}^\infty \sum _{w=1}^\infty \varvec{\nu }_{i,j} z^i w^j, \end{aligned}$$

(A.1)

which is also the Taylor series for the function $\varvec{\varphi }_+(z,w)$ at $(z,w)=(0,0)$. Since $\varvec{\varphi }_+(1,1)<\infty$, power series (A.1) converges absolutely on $\bar{\Delta }_1\times \bar{\Delta }_1$ and $\varvec{\varphi }_+(z,w)$ is analytic in $\Delta _1\times \Delta _1$. The lemma asserts that, for every $(p,q)\in \mathcal {D}_0$, the power series (A.1) converges absolutely on $\bar{\Delta }_p\times \bar{\Delta }_q$, and to prove it, it suffices to show that, for every $(p,q)\in \mathcal {D}_0$, the power series (A.1) converges absolutely at $(z,w)=(p,q)$, since any coefficient in the power series is nonnegative. Define the vector function $\varvec{h}(z,w)$ by $\varvec{h}(z,w) = \varvec{f}(z,w)/g(z,w)$, where

$$\begin{aligned} \varvec{f}(z,w)&= z^{s_0} w^{s_0} \bigl ( \varvec{\varphi }_1(z) (C_1(z,w)-I) + \varvec{\varphi }_2(w) (C_2(z,w)-I) \\&\quad + \varvec{\nu }_{0,0} (C_0(z,w)-I) \bigr )\,\mathrm{adj}(I-C(z,w)), \\ g(z,w)&=z^{s_0} w^{s_0}\det (I-C(z,w)). \end{aligned}$$

The function g(z, w) is analytic in $\mathbb {C}^2$ and, by Proposition 3.1, each element of $\varvec{f}(z,w)$ is analytic in $\Delta _{r_1}\times \Delta _{r_2}$. For $(z,w)\in \bar{\Delta }_1\times \bar{\Delta }_1$, since the power series (A.1) is absolutely convergent, Eq. (3.3) holds and we have the $\varvec{\varphi }_+(z,w)=\varvec{h}(z,w)$. Hence, by the identity theorem, we see that $\varvec{h}(z,w)$ is an analytic extension of $\varvec{\varphi }_+(z,w)$. Hereafter, we denote the analytic extension by the same notation $\varvec{\varphi }_+(z,w)$. The proof of the lemma is given as follows.

Fig. 4

Convergence domain of $\varvec{\varphi }_+(e^{s_1},e^{s_2})$

Proof of Lemma 3.1

Under Assumption 2.3, either $a_1^{\{1,2\}}$ or $a_2^{\{1,2\}}$ is negative. Here, we assume $a_1^{\{1,2\}}<0$; the proof for the case where $a_2^{\{1,2\}}<0$ is analogous. Since $\chi _{z_1}(1,1)=a_1^{\{1,2\}}<0$, where $\chi _{z_1}(z_1,z_2)=(\partial /\partial \,z_1)\,\chi (z_1,z_2)$, we have the open interval $(1,\min \{\bar{\zeta _1}(1),r_1\})\subset \mathcal {D}_0$ (see Fig. 4). Let (p, q) be an arbitrary point in $\mathcal {D}_0$, and consider a path on $\mathcal {D}_0\cup \{(1,1)\}$ connecting different points $(p_0,q_0)=(1,1)$, $(p_1,q_0)$, $(p_1,q)$ and (p, q) by lines in this order, where we assume $1<p_1<\min \{\bar{\zeta _1}(1),r_1\}$ (see Fig. 4). This is always possible because $\mathcal {D}_0$ is given by

$$\begin{aligned} \mathcal {D}_0=\{(e^{s_1},e^{s_2}): (s_1,s_2)\in \ \text{ the } \text{ interior } \text{ of } \bar{\Gamma },\ s_1<\log r_1,\ s_2<\log r_2\} \end{aligned}$$

and $\bar{\Gamma }$ is a convex set.

First, we consider a vector function of one variable given by $\varvec{\varphi }_+(z,1)$. The Taylor series for $\varvec{\varphi }_+(z,1)$ at $z=0$ is given by the power series (A.1), where w is set at 1, and $\varvec{\varphi }_+(z,1)$ is identical to $\varvec{h}(z,1)$ on a domain where $\varvec{h}(z,1)$ is well defined. Let $\varepsilon$ be a sufficiently small positive number. Since g(z, 1) and each element of $\varvec{f}(z,1)$ are analytic as a function of one variable in $\Delta _{p_1+\varepsilon }$, $\varvec{h}(z,1)$ is meromorphic as a function of one variable in that domain. We have, for $z\in \Delta _{p_1+\varepsilon }{\setminus }\bar{\Delta }_1$, $\mathrm{spr}(C(z,1))\le \mathrm{spr}(C(|z|,1))<1$, and, by Proposition 4.1 in Sect. 4, we have, for $z\in \partial \Delta _{1}{\setminus }\{1\}$, $\mathrm{spr}(C(z,1))<\mathrm{spr}(C(|z|,1))=1$. Hence, $g(z,1)\ne 0$ for any $z\in \Delta _{p_1+\varepsilon }{\setminus }(\Delta _1\cup \{1\})$, and each element of $\varvec{h}(z,1)$ is analytic on $\Delta _{p_1+\varepsilon }{\setminus }(\Delta _1\cup \{1\})$. Furthermore, we have $\varvec{\varphi }_+(1,1)=\varvec{h}(1,1)<\infty$, and this implies that the point $z=1$ is not a pole of any element of $\varvec{h}(z,1)$; hence, it is a removable singularity. From this and the fact that $\varvec{\varphi }_+(z,1)$ is analytic in $\Delta _{1}$, we see that $\varvec{\varphi }_+(z,1)$ is analytic in $\Delta _{p_1+\varepsilon }$. This implies that the radius of convergence of the Taylor series for $\varvec{\varphi }_+(z,1)$ at $z=0$ is greater than $p_1$, and the power series (A.1) converges at $(z,w)=(p_1,1)$.

Next, we consider a vector function of one variable given by the $\varvec{\varphi }_+(p_1,w)$. By the fact obtained above, the Taylor series for $\varvec{\varphi }_+(p_1,w)$ at $w=0$ is given by the power series (A.1), where z is set at $p_1$, and $\varvec{\varphi }_+(p_1,w)$ is analytic as a function of one variable in $\Delta _{1}$. Furthermore, we know that $\varvec{\varphi }_+(p_1,w)$ is identical to $\varvec{h}(p_1,w)$ on a domain where $\varvec{h}(p_1,w)$ is well defined. If $q\le 1$, then it is obvious that the power series (A.1) converges at $(z,w)=(p_1,q)$. Therefore, we assume $q>1$. Let $\varepsilon$ be a sufficiently small positive number. For $w\in \Delta _{q+\varepsilon }{\setminus }\bar{\Delta }_{1-\varepsilon }$, we have $\mathrm{spr}(C(p_1,w))\le \mathrm{spr}(C(p_1,|w|))<1$. Hence, for the same reason used in the case of $\varvec{h}(z,1)$, we see that $\varvec{h}(p_1,w)$ is analytic in $\Delta _{q+\varepsilon }{\setminus }\bar{\Delta }_{1-\varepsilon }$, and this implies that $\varvec{\varphi }_+(p_1,w)$ is analytic in $\Delta _{q+\varepsilon }$. Hence, the radius of convergence of the Taylor series for $\varvec{\varphi }_+(p_1,w)$ at $w=0$ is greater than q, and the power series (A.1) converges at $(z,w)=(p_1,q)$. Applying a similar procedure to the vector function $\varvec{\varphi }_+(z,q)$, we see that the power series (A.1) converges at $(z,w)=(p,q)$, and this completes the proof. $\square$