Invariant measures and error bounds for random walks in the quarter-plane based on sums of geometric terms

Abstract

We consider homogeneous random walks in the quarter-plane. The necessary conditions which characterize random walks of which the invariant measure is a sum of geometric terms are provided in Chen et al. (arXiv:1304.3316, 2013, Probab Eng Informational Sci 29(02):233–251, 2015). Based on these results, we first develop an algorithm to check whether the invariant measure of a given random walk is a sum of geometric terms. We also provide the explicit form of the invariant measure if it is a sum of geometric terms. Second, for random walks of which the invariant measure is not a sum of geometric terms, we provide an approximation scheme to obtain error bounds for the performance measures. Our results can be applied to the analysis of two-node queueing systems. We demonstrate this by applying our results to a tandem queue with server slow-down.

Appendices

Appendix 1: Proof of Theorem 2

In order to prove Theorem 2, we first present a lemma. The vertical branch points which will be used here are defined similarly to the horizontal branch points before. Since the algebraic curve Q has a unique connected component in $[0,\infty )^2$ (see [4, Lemma 7]), it has two vertical branch points in $[0,\infty )^2$, denoted by $(x_l, y_l)$, $(x_r, y_r)$ with $x_l \ge x_r$.

Lemma 4

Consider the measure m induced by the set $\varGamma$, which is the invariant measure of the random walk R. If we connect every two geometric terms with the same horizontal or vertical coordinates from the set $\varGamma$ with a line segment, then these line segments cannot form a cycle.

In order to prove Lemma 4, we define two types of partition of Q; see Fig. 10.

Definition 6

(Partition I of Q) The partition $\{Q_{00}, Q_{01}, Q_{10}, Q_{11}\}$ of Q is defined as follows: $Q_{00}$ is the part of Q connecting $(x_l, y_l)$ and $(x_b, y_b)$; $Q_{10}$ is the part of Q connecting $(x_b, y_b)$ and $(x_r, y_r)$; $Q_{01}$ is the part of Q connecting $(x_l, y_l)$ and $(x_t, y_t)$; $Q_{11}$ is the part of Q connecting $(x_r, y_r)$ and $(x_t, y_t)$.

Definition 7

(Partition II of Q) Let $\{Q_l, Q_c, Q_r\}$ denote a partition of Q, where

Fig. 10

Different partitions of Q. a Partition I of Q. b Partition II of Q.

Proof of Lemma 4

Denote the two pieces of $Q_c$ in Fig. 10b by $Q_c^t$ and $Q_c^b$, which satisfy $\tilde{y} > y$ if $(x,\tilde{y}) \in Q_c^t$ and $(x,y) \in Q_c^b$. Since the algebraic curve Q contains no singularity, because of [4, Theorem 12], $Q_l, Q_c$ and $Q_r$ are all non-empty.

In addition, we let $\{\varGamma _1,\dots ,\varGamma _K\}$ denote a partition of $\varGamma$, where the elements of $\varGamma _i$ are denoted by $\varGamma _i=\{(\rho _{i,1},\sigma _{i,1}),\dots ,(\rho _{i,L(i)},\sigma _{i,L(i)})\}$ and each $\varGamma _i$ satisfies

(18)

In addition, the partition $\{\varGamma _1,\dots ,\varGamma _K\}$ is maximal in the sense that no $\varGamma _i\cup \varGamma _j$, $i\ne j$, satisfies (18).

Assume that the line segments which connect every two geometric terms with the same horizontal or vertical coordinates from set $\varGamma$ form a cycle. Without loss of generality, we will have $\varGamma _1$, $\varGamma _2$, where $|\varGamma _1| > 1$ and $|\varGamma _2| > 1$, such that $\rho _{1,1} = \rho _{2,1}$ and $\rho _{1,1},\rho _{2,1} \in Q_r$. Moreover, either $\rho _{1,1}$ or $\rho _{2,1}$ must be on $Q_{11}$. However, $y_t \ge 1$ and $x_r \ge 1$, by [9, Lemma 2.3.8]. Also, using the fact that $Q_{11}$ is monotonic, by Lemma 9 from [4], we conclude that $Q_{11}$ is outside of U, which contradicts that m is a finite measure.

We are now ready to prove Theorem 2.

Proof of Theorem 2

First, it follows from [5, Theorem 1] that $\varGamma \subset Q \cap (0,1)^2$. Moreover, it follows from [5, Theorem 4] that $\varGamma$ must be a pairwise-coupled set.

If the measure induced by $\varGamma = \{(\rho , \sigma )\}$ is the invariant measure of the random walk R, then $Q(\rho , \sigma ) = 0, H(\rho , \sigma ) = 0$ and $V(\rho , \sigma ) = 0$. Hence, we have $\varGamma \subset Q \cap (0,1)^ 2$ and $(\rho , \sigma ) \in H_{\mathrm {set}} \cap V_{\mathrm {set}}$.

When $1< |\varGamma | < \infty$, from Lemma 4 the pairwise-coupled set $\varGamma$ cannot form a cycle. Hence, there must be two geometric terms which do not share the horizontal or vertical coordinate with other geometric terms from the set $\varGamma$. We denote these two geometric terms by $(\rho _1, \sigma _1), (\rho _2, \sigma _2)$.

It follows from Lemma 3 that the measure induced by any two geometric terms from the set $\varGamma$ which have the same horizontal coordinates must satisfy the horizontal balance equation.

Without loss of generality, we assume that $(\rho _1, \sigma _1), (\rho _2, \sigma _2) \in H_{\mathrm {set}}$. Thus, for $k = 1,2$, we have

$$\begin{aligned} {} B^h(\rho _k, \sigma _k) = \sum _{s=-1}^1 \big (\rho _k^{1-s} h_s+\rho _k^{1-s}\sigma _k p_{s,-1}\big ) - \rho _k = 0. \end{aligned}$$

(19)

Hence, for $k = 1,2$, there exists no $(\rho , \sigma ) \in \varGamma \backslash (\rho _k, \sigma _k)$ such that $\rho = \rho _k$. Otherwise, the balance for $(\rho _k, \sigma _k)$ and $(\rho , \sigma )$ cannot be satisfied. Moreover, because $\varGamma$ is a pairwise-coupled set, there exists $(\rho , \sigma ) \in \varGamma \backslash (\rho _k, \sigma _k)$ such that $\sigma \!=\! \sigma _k$. Similar results hold when $(\rho _1, \sigma _1), (\rho _2, \sigma _2)\! \in \! V_{\mathrm {set}}$ or when $(\rho _1, \sigma _1) \!\in \! H_{\mathrm {set}}$ and $(\rho _2, \sigma _2) \!\in \! V_{\mathrm {set}}$.

It can be readily verified that if $(\rho _1, \sigma _1) \in H_{\mathrm {set}}$ and $(\rho _2, \sigma _2) \in V_{\mathrm {set}}$, then $|\varGamma | = 2k+1$, where $k = 1,2,3, \ldots$. Otherwise, we have $|\varGamma | = 2k$, where $k = 1,2,3, \ldots$.

Finally, if such pairs $(\rho _1, \sigma _1), (\rho _2, \sigma _2)$ are not unique, then, by carefully choosing the coefficients, we find 2 signed measures to make all balance equations satisfied. However, this contradicts the uniqueness of the invariant measure, which completes the proof.

Appendix 2: Proof of Theorem 3

Proof of Theorem 3

Similar to the proof of Lemma 4, we find $\{\varGamma _1,\dots ,\varGamma _K\}$ which are defined in (18).

First, we prove $L(i)<\infty$ by demonstrating that

$$\begin{aligned} |\varGamma _i\cap Q_l| \le 1,\quad \left| \varGamma _i\cap Q_c\right| <\infty , \quad \left| \varGamma _i\cap Q_r\right| \le 1. \end{aligned}$$

Suppose that $\left| \varGamma _i\cap Q_r\right| \ge 2$. Then there exist $(\rho , \sigma )$ and $(\tilde{\rho }, \tilde{\sigma })$ on $Q_{11}$ or $Q_{10}$ satisfying $\tilde{\sigma } = \sigma$. This contradicts [4, Lemma 9], which indicates the monotonicity of $Q_{11}$ and $Q_{10}$.

Therefore, $\left| \varGamma _i\cap Q_r\right| \le 1$. Similarly, we show $\left| \varGamma _i\cap Q_l\right| \le 1$.

Next, we prove that $\sigma _{i,j+2}\le \sigma _{i,j}-\min (D_1, D_2)$, where

$$\begin{aligned} D_1 = \frac{\varDelta _y(x_b)}{\sum _{s = -1} ^{1} p_{s,-1} x_t^{1 - s}}, D_2 = \frac{\varDelta _y(x_t)}{\sum _{s = -1} ^{1} p_{s,-1} x_t^{1 - s}}, \end{aligned}$$

for three consecutive elements in $\left| \varGamma _i\cap Q_c\right|$, $(\rho _{i,j}, \sigma _{i,j})$, $(\rho _{i+1, j+1}, \sigma _{i+1, j+1})$ and $(\rho _{i+2, j+2}, \sigma _{i+2, j+2})$ satisfying

Note that $\varDelta _y(x) > 0$ and $\varDelta _y(x)$ has at most one stationary point where the derivative is 0 for $x \in (x_b, x_t)$, because $\varDelta _y(x)$ is continuous over x and $\varDelta _y(x) = 0$ has 4 real solutions due to [4, Lemma 4]. We obtain that $\varDelta _y(x) \ge \min (\varDelta _y(x_b), \varDelta _y(x_t))$. Moreover, it can be readily verified that $\sum _{s = -1}^{1} p_{s,-1} x^{1 - s}$ is monotonically increasing in x for $x \in (x_b, x_t)$. Therefore, we have

$$\begin{aligned} {} \frac{\varDelta _y(x)}{\sum _{s = -1}^{1} p_{s,-1} x^{1 - s}} \ge \min \left( \frac{\varDelta _y(x_b)}{\sum _{s = -1}^{1} p_{s,-1} x_t^{1 - s}}, \frac{\varDelta _y(x_t)}{\sum _{s = -1}^{1} p_{s,-1} x_t^{1 - s}}\right) , \end{aligned}$$

(20)

for $x\in (x_b, x_t)$.

Notice that the left-hand side of Eq. (20) is the distance between two intersections of Q and a vertical line, i.e., $\frac{\varDelta _y(a)}{\sum _{s = -1}^{1} p_{s,-1} a^{1 - s}}$ is the distance between two intersections of Q and the line $x = a$. Therefore, we conclude that $\sigma _{i,j+2}\le \sigma _{i,j}-\min (D_1, D_2)$, where

$$\begin{aligned} D_1 = \frac{\varDelta _y(x_b)}{\sum _{s = -1} ^{1} p_{s,-1} x_t^{1 - s}}, D_2 = \frac{\varDelta _y(x_t)}{\sum _{s = -1} ^{1} p_{s,-1} x_t^{1 - s}}. \end{aligned}$$

Next, we show that if $K > 2$, then there exists $(\rho , \sigma ) \in \varGamma$ such that $\rho > 1$ or $\sigma > 1$. Without loss of generality, we assume $K=3$. Observe that $\{\varGamma _1,\varGamma _2,\varGamma _3\}$ forms a pairwise-coupled set. Using, from the above, that $|\varGamma _i|<\infty$ for $i=1,2,3$, we must have $\rho _{1,L(1)} = \rho _{2,L(2)}$ with $\rho _{1, L(1)}, \rho _{2, L(2)} \in Q_l$ and $\rho _{2, 1} = \rho _{3,1}$ with $\rho _{2, 1},\rho _{3,1} \in Q_r$ after a proper ordering of $\{\varGamma _1,\varGamma _2,\varGamma _3\}$. Moreover, either $\rho _{2,1}$ or $\rho _{3,1}$ must be on $Q_{11}$. However, $y_t \ge 1$ and $x_r \ge 1$ due to [9, Lemma 2.3.8]. Also, using the fact that $Q_{11}$ is monotonic, due to Lemma 9 from [4], we conclude that $Q_{11}$ is outside of U. Hence, there exists $(\rho , \sigma ) \in \varGamma$ such that $\rho > 1$ or $\sigma > 1$.

When $K \le 2$, we know that the distance between two intersections of Q and a vertical line $x = a$, where $a \in (x_b, x_t)$, is at least $\min (D_1, D_2)$.

Therefore, we conclude that if

$$\begin{aligned} |\varGamma | > M(R) = \frac{6}{\min (D_1, D_2)} + 4, \end{aligned}$$

then there exists $(\rho , \sigma ) \in \varGamma$ such that $\rho > 1$ or $\sigma > 1$. We know from [4, Theorem 12] that the algebraic curve Q can only have an accumulation point at the origin when $p_{1,0} + p_{1,1} + p_{0,1} = 0$. Hence, when $p_{1,0} + p_{1,1} + p_{0,1} \ne 0$, we have $x_b > 0$ and $x_t > 0$. This means that $D_1 > 0$ and $D_2> 0$. Therefore, we have $M(R) < \infty$.