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Stationary distribution of a two-dimensional SRBM: geometric views and boundary measures

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Abstract

We present three sets of results for the stationary distribution of a two-dimensional semimartingale-reflecting Brownian motion (SRBM) that lives in the non-negative quadrant. The SRBM data can equivalently be specified by three geometric objects, an ellipse and two lines, in the two-dimensional Euclidean space. First, we revisit the variational problem (VP) associated with the SRBM. Building on Avram et al. (Queueing Syst. 37: 259–289, 2001), we show that the value of the VP at a point in the quadrant is equal to the optimal value of a linear function over a convex domain. Depending on the location of the point, the convex domain is either \(\mathcal{D}^{(1)}\) or \(\mathcal{D}^{(2)}\) or \(\mathcal{D}^{(1)}\cap \mathcal{D}^{(2)},\) where each \(\mathcal{D}^{(i)},\) \(i=1, 2,\) can easily be described by the three geometric objects. Our results provide a geometric interpretation for the value function of the VP and allow one to see geometrically when one edge of the quadrant has influence on the optimal path traveling from the origin to a destination point. Second, we provide a geometric condition that characterizes the existence of a product form stationary distribution. Third, we establish exact tail asymptotics of two boundary measures that are associated with the stationary distribution; a key step in our proof is to sharpen two asymptotic inversion lemmas in Dai and Miyazawa (Stoch. Syst. 1:146–208, 2011) which allow one to infer the exact tail asymptotic of a boundary measure from the singularity of its moment-generating function.

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Acknowledgments

This research was supported in part by NSF Grants CMMI-0825840, CMMI-1030589, CNS-1248117, and by Japan Society for the Promotion of Science under Grant No. 24310115.

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Correspondence to Masakiyo Miyazawa.

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On leave from H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology, Atlanta, USA.

Appendices

Appendix

A Proof of Lemma 6.1

We only prove the claim for \(n=1\) because the case for \(n \ge 2\) is iteratively obtained. Since \(f\) is ultimately nonincreasing, we can find \(x_{0} \ge 0\) such that \(f(x)\) is nonincreasing for \(x \ge x_{0}.\) By the assumption, we have, for any \(\epsilon > 0,\) there exists some \(x_{1} > x_{0}\) such that

$$\begin{aligned} (1-\epsilon ) b x^{\kappa } e^{-\alpha x} < T(f)(x) = \int _{x}^{\infty } f(u) du < (1+\epsilon ) b x^{\kappa } e^{-\alpha x}, \qquad \forall x \ge x_{1}.\qquad \quad \end{aligned}$$
(7.1)

Hence, for each \(\delta > 0\) and each \(x \ge x_{1} + \delta ,\)

$$\begin{aligned} \int _{x - \delta }^{x} f(u) du&< (1+\epsilon ) b (x-\delta )^{\kappa } e^{-\alpha (x-\delta )} - (1-\epsilon ) b x^{\kappa } e^{-\alpha x}\\&= \left(\left(\frac{x-\delta }{x}\right)^{\kappa } e^{\alpha \delta } - 1\right) b x^{\kappa } e^{-\alpha x} + \epsilon \left( \left(\frac{x-\delta }{x}\right)^{\kappa } e^{\alpha \delta } + 1 \right) b x^{\kappa } e^{-\alpha x} \end{aligned}$$

Since \(f(u)\) is nonincreasing in \(u,\) we have

$$\begin{aligned} \delta \limsup _{x \rightarrow \infty } \frac{f(x)}{x^{\kappa } e^{-\alpha x}} \le ( e^{\alpha \delta } - 1) b + \epsilon ( e^{\alpha \delta } + 1) b. \end{aligned}$$

Since \(\epsilon \) can be arbitrarily small, we obtain

$$\begin{aligned} \limsup _{x \rightarrow \infty } \frac{f(x)}{x^{\kappa } e^{-\alpha x}} \le \frac{1}{\delta } (e^{\alpha \delta } - 1) b. \end{aligned}$$

Letting \(\delta \) go to zero, we arrive at

$$\begin{aligned} \limsup _{x \rightarrow \infty } \frac{f(x)}{x^{\kappa } e^{-\alpha x}} \le \alpha b. \end{aligned}$$
(7.2)

Similarly, for each \(\delta > 0\) and each \(x \ge x_{1},\)

$$\begin{aligned} \int _{x}^{x+\delta } f(u) du&> (1-\epsilon ) b x^{\kappa } e^{-\alpha x} - (1+\epsilon ) b (x+\delta )^{\kappa } e^{-\alpha (x+\delta )}\\&= \left(1 \!-\! \left(\frac{x+\delta }{x}\right)^{\kappa } e^{-\alpha \delta } \right) b x^{\kappa } e^{-\alpha x} - \epsilon \left( 1 \!+\! \left(\frac{x+\delta }{x}\right)^{\kappa } e^{-\alpha \delta } \right) b x^{\kappa } e^{-\alpha x}, \end{aligned}$$

which yields

$$\begin{aligned} \liminf _{x \rightarrow \infty } \frac{f(x)}{x^{\kappa } e^{-\alpha x}} \ge \alpha b. \end{aligned}$$
(7.3)

Hence, (7.2) and (7.3) conclude \(f(x) \sim \alpha b \, x^{\kappa } e^{-\alpha x}.\)

B Asymptotic inversion lemmas

In this appendix, we state Lemmas C.1 and C.2 in [5]. Actually, we here present a refined version of Lemma C.1, whereas Lemma C.2 is unchanged. This refinement is closer to Theorems 35.1 of Doetsch [6]. The reason for modifying Lemma C.1 is clear in the proof of Lemma 6.2 in Sect. 6. For two functions \(h_1\) and \(h_2,\) we say \(h_1=o(h_2)\) as \(x\rightarrow \infty \) if

$$\begin{aligned} \lim _{x\rightarrow \infty } h_1(x)/h_2(x)=0. \end{aligned}$$

Lemma 6.4

Let \(g\) be the moment-generating function in (6.5) of a non-negative, continuous, and integrable function \(f.\) Assume the following conditions are satisfied for some integer \(m \ge 1:\)

(B1a) there is a complex variable function

$$\begin{aligned} g_{0}(z) = \sum _{j=1}^{m} \sum _{\ell =1}^{k_{j}} \frac{c^{(j)}_{\ell }}{(\alpha _{j} - z)^{\ell }} \end{aligned}$$

for some positive integers \(k_{j}\) and some positive numbers \(\beta , \alpha _{j}, c^{(j)}_{\ell }\) for \(\ell =1, \ldots , k_j\) and \(j= 1,\ldots m\) with \(0 < \alpha _{1} < \ldots < \alpha _{m-1} < \alpha _{m} < \beta \) such that \(g(z) - g_{0}(z)\) is analytic for \(\mathfrak R z < \beta ,\)

(B1b) \(g(z)\) uniformly converges to \(0\) as \(z \rightarrow \infty \) for \(0 \le \mathfrak R z \le \beta ,\)

(B1c) for some \(T > 0,\) \(\int _{-\infty }^{+\infty } e^{- i y x} g(\beta +iy) dy\) uniformly converges for \(x > T.\) Then,

$$\begin{aligned} f(x) = \sum _{j=1}^{m} \sum _{\ell =1}^{k_{j}} \frac{c^{(j)}_{\ell }}{\Gamma (\ell )} x^{\ell -1} e^{-\alpha _{j} x} + o(e^{-\beta x}), \qquad x \rightarrow \infty . \end{aligned}$$
(7.4)

Remark 6.1

Conditions (B1b) and (B1c) are satisfied if, for some constants \(a, b, \delta > 0,\)

$$\begin{aligned} |g(z)| < \frac{a}{|z|^{1+\delta }}, \quad \text{ for} \mathfrak R z \in [0, \beta ] \text{ and} |\mathfrak I z| > b. \end{aligned}$$
(7.5)

Lemma 6.5

Let \(f\) and \(g\) be two functions in Lemma 6.4. Assume that the following two conditions hold for some \(\alpha > 0\) and some \(\delta \in [0, \frac{\pi }{2}):\)

(B2a) \(g(z)\) is analytic on \(\mathcal{G}_{\delta }(\alpha ) \equiv \{z \in \mathbb C ; z \ne \alpha , |\arg (z-\alpha )| > \delta \},\) where for \(z\in \mathbb C \setminus \{0\},\) \(\arg (z)\) is the angle of \(z,\)

(B2b) \(g(z) \rightarrow 0\) as \(|z| \rightarrow \infty \) for \(z \in \mathcal{G}_{\delta }(\alpha ),\)

(B2c) for some complex number \(d,\) some \(\lambda \in \mathbb R \) and \(c_{0} \in \mathbb R ,\)

$$\begin{aligned} \lim \limits _{\small {\begin{array}{l} z \rightarrow \alpha \\ z\in \mathcal{G}_{\delta }(\alpha )\end{array}}} (\alpha - z)^{\lambda } \left(g(z) - d \right) = c_{0}. \end{aligned}$$
(7.6)

Then

$$\begin{aligned} f(x)=\frac{c_{0}}{\Gamma (\lambda )} x^{\lambda -1} e^{-\alpha x}(1+o(1)) . \end{aligned}$$
(7.7)

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Dai, J.G., Miyazawa, M. Stationary distribution of a two-dimensional SRBM: geometric views and boundary measures. Queueing Syst 74, 181–217 (2013). https://doi.org/10.1007/s11134-012-9339-1

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