On the $M_t/M_t/K_t+M_t$ queue in heavy traffic

Abstract

The focus of this paper is on the asymptotics of large-time numbers of customers in time-periodic Markovian many-server queues with customer abandonment in heavy traffic. Limit theorems are obtained for the periodic number-of-customers processes under the fluid and diffusion scalings. Other results concern limits for general time-dependent queues and for time-homogeneous queues in steady state.

Appendix

Lemma 5

Let $(F(t),\,t\in \mathbb R _+)$ be a function of locally bounded variation and $(f(t),\,t\in \mathbb R _+)$ be a locally bounded Lebesgue measurable function. If a locally integrable function $(y(t),\,t\in \mathbb R _+)$ is such that $y(t)\prec F(t)-\int _0^t f(s)y(s)\,ds$, then

$$\begin{aligned} y(t)\le e^{-\int _0^t f(s)\,ds}F(0)+e^{-\int _0^t f(s)\,ds}\int \limits _0^t e^{\int _0^s f(u)\,du}\,dF(s). \end{aligned}$$

Proof

Let $g(t)=F(t)-\int _0^t f(s)y(s)\,ds-y(t)$. The function $(g(t))$ is nondecreasing, $g(0)\ge 0$, and

$$\begin{aligned} y(t)= F(t)-g(t)-\int \limits _0^t f(s)y(s)\,ds. \end{aligned}$$

Hence,

$$\begin{aligned} y(t)&= e^{-\int _0^t f(s)\,ds}(F(0)-g(0))+e^{-\int _0^t f(s)\,ds}\int \limits _0^t e^{\int _0^s f(u)\,du}\,d(F(s)-g(s))\\&\le e^{-\int _0^t f(s)\,ds}F(0)+e^{-\int _0^t f(s)\,ds}\int \limits _0^t e^{\int _0^s f(u)\,du}\,dF(s). \end{aligned}$$

$\square$

The next lemma provides the bounds that have been used for the analysis of large-time behaviour. Let $T>0$ and $\sigma ^n_s=|\alpha ^n_s-\gamma ^n_s(q_s-\kappa _s)^+-\beta ^n_s(q_s\wedge \kappa _s)| +|(\theta ^n_s-\mu ^n_s)\delta ^n_s|.$ Let me recall that $q_t$ is defined by (2.6).

Lemma 6

• 1.

  • (a) If the functions $(\lambda _t,\,t\in \mathbb R _+), (\mu _t,\,t\in \mathbb R _+)$, and $(\theta _t,\,t\in \mathbb R _+)$ are $T$-periodic and $\int _0^T(\mu _s\wedge \theta _s)\,ds>0$ , then, for all $t\in \mathbb R _+$,

    $$\begin{aligned} q_t\le e^{-\lfloor t/T\rfloor \int _0^T (\mu _s\wedge \theta _s)\,ds}\, q_0 +\frac{ e^{\int _0^{T} (\mu _s\wedge \theta _s)\,ds}}{1-e^{-\int _0^{T} (\mu _s\wedge \theta _s)\,ds}}\int \limits _{0}^{T}\lambda _s\,ds. \end{aligned}$$

  • (b) If the functions $(\lambda ^n_t,\,t\in \mathbb R _+), (\mu ^n_t,\,t\in \mathbb R _+)$, and $(\theta ^n_t,\,t\in \mathbb R _+)$ are $T$-periodic and $\int \limits _0^T(\mu ^n_s\wedge \theta ^n_s)\,ds>0$ , then, for all $t\in \mathbb R _+$ and $V>0$,

    $$\begin{aligned} \mathbf E Q^n_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,&\le e^{-\lfloor t/T\rfloor \int _0^T (\mu ^n_s\wedge \theta ^n_s)\,ds} \, \mathbf E Q^n_0\,\mathbf 1 _{\{Q^n_0\le V\}}\,\\&+ \frac{ e^{\int _0^{T} (\mu ^n_s\wedge \theta ^n_s)\,ds}}{1-e^{-\int _0^{T} (\mu ^n_s\wedge \theta ^n_s)\,ds}}\int \limits _{0}^{T}\lambda ^n_s\,ds. \end{aligned}$$

    .

  • (c) If the functions $(\lambda ^n_t,\,t\in \mathbb R _+), (\mu ^n_t,\,t\in \mathbb R _+), (\theta ^n_t,\,t\in \mathbb R _+), (\lambda _t,\,t\in \mathbb R _+), (\mu _t,\,t\in \mathbb R _+)$, and $(\theta _t,\,t\in \mathbb R _+)$ are $T$-periodic, $\int _0^T(\mu _s\wedge \theta _s)\,ds>0$ , and, for some $\epsilon >0, \int _0^{T}\left(2(\mu ^n_s\wedge \theta ^n_s)- \epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n})\right)\,ds>0$, then, for all $t\in \mathbb R _+$ and $V>0$,

    $$\begin{aligned}&\mathbf E (X^n_t)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\\&\quad \le e^{-\lfloor t/T\rfloor \int _0^T \left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)\,ds +\int _0^T |2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n})|\,ds}\\&\qquad \times \mathbf E (X^n_0)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\, + \frac{ e^{2\int _0^{T}|2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n})|\,ds}}{1-e^{-\int _0^{T}\left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n})\right)\,ds}}\\&\qquad \times \int \limits _0^{T}\left(\frac{1}{\epsilon }\,\sigma ^n_s +\frac{\lambda ^n_s}{n} \right. \left. + (\mu ^n_s\vee \theta ^n_s)\sup _{u\in \mathbb R _+}q_u+ \frac{1}{2\sqrt{n}}(\mu ^n_s\vee \theta ^n_s)\right)\,ds. \end{aligned}$$
• 2.

  • (a) If $\sup _{t\in \mathbb R _+}\lambda _t<\infty$ and $\inf _{t\in \mathbb R _+}( \mu _t\wedge \theta _t)>0$, then, for all $t\in \mathbb R _+$,

    $$\begin{aligned} q_t\le e^{-\int _0^t (\mu _s\wedge \theta _s)\,ds} q_0+ \frac{\sup _{s\in \mathbb R _+}\lambda _s}{\inf _{s\in \mathbb R _+} (\theta _s\wedge \mu _s)}. \end{aligned}$$

  • (b) If $\sup _{t\in \mathbb R _+}\lambda ^n_t<\infty$ and $\inf _{t\in \mathbb R _+}( \mu ^n_t\wedge \theta ^n_t)>0$, then, for all $t\in \mathbb R _+$ and $V>0$ ,

    $$\begin{aligned} \mathbf E Q^n_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,\le e^{-\int _0^t (\mu ^n_s\wedge \theta ^n_s)\,ds}\, \mathbf E Q^n_0\,\mathbf 1 _{\{Q^n_0\le V\}}\, +\frac{\sup _{s\in \mathbb R _+}\lambda ^n_s}{\inf _{s\in \mathbb R _+} (\theta ^n_s\wedge \mu ^n_s)}. \end{aligned}$$

  • (c) If $\sup _{t\in \mathbb R _+}\lambda _t<\infty , \inf _{t\in \mathbb R _+}( \mu _t\wedge \theta _t)>0, \sup _{t\in \mathbb R _+}\lambda ^n_t<\infty , \sup _{t\in \mathbb R _+}\theta ^n_t<\infty , \sup _{t\in \mathbb R _+}\mu ^n_t<\infty , \sup _{t\in \mathbb R _+}\sigma ^n_t<\infty$, and $\inf _{t\in \mathbb R _+}\left(2(\mu ^n_t\wedge \theta ^n_t)-\epsilon \,\sigma ^n_t\right.\left.- (\mu ^n_t\vee \theta ^n_t)/(2\sqrt{n})\right)>0$ for some $\epsilon >0$ , then, for all $t\in \mathbb R _+$ and $V>0$,

    $$\begin{aligned}&\mathbf E (X^n_t)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\\&\quad \le e^{-\int _0^t \left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)\,ds} \,\mathbf E (X^n_0)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\, \\&\qquad + \dfrac{\sup _{s\in \mathbb R _+}\left(\sigma ^n_s/\epsilon +\lambda ^n_s/n+ (\mu ^n_s\vee \theta ^n_s)q_s+ (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n})\right)}{\inf _{s\in \mathbb R _+}\left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \, \sigma ^n_s -(\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)}. \end{aligned}$$

Proof

Let me start with part 1(b). Since $Q^n_t\le Q^n_0+A^n_t$ by (2.1), I have that $\mathbf E Q^n_t \,\mathbf 1 _{\{Q^n_0\le V\}}\,\le V+\int _0^t\lambda ^n_s\,ds$. By Lemma 1, the processes $(M^{n,A}_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,,\,t\in \mathbb R _+), (M^{n,R}_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,,\,t\in \mathbb R _+)$, and $(M^{n,B}_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,,\,t\in \mathbb R _+)$ are $\mathbf F ^n$-locally square integrable martingales with respective predictable quadratic variation processes $(\,\mathbf 1 _{\{Q^n_0\le V\}}\,\int _0^t\lambda ^n_s\,ds\,,t\in \mathbb R _+), (\,\mathbf 1 _{\{Q^n_0\le V\}}\,\int _0^t\theta ^n_s\, (Q^n_s-K^n_s)^+\,ds,\,t\in \mathbb R _+)$, and $(\,\mathbf 1 _{\{Q^n_0\le V\}}\,\int _0^t\mu ^n_s\, (Q^n_s\wedge K^n_s)\,ds,\,t\in \mathbb R _+)$. Since the latter processes are of finite expectation, I obtain that $\mathbf E (M^{n,A}_t)^2\,\mathbf 1 _{\{Q^n_0\le V\}}\,<\infty , \mathbf E (M^{n,R}_t)^2\,\mathbf 1 _{\{Q^n_0\le V\}}\,<\infty$, and $\mathbf E (M^{n,B}_t)^2\,\mathbf 1 _{\{Q^n_0\le V\}}\,<\infty$. In particular, $(M^{n,A}_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,,\,t\in \mathbb R _+), (M^{n,R}_t\,\mathbf 1 _{\{Q^n_0\!\le \! V\}}\,,t\!\in \!\mathbb R _+)$, and $(M^{n,B}_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,,t\in \mathbb R _+)$ are martingales. By (2.3),

$$\begin{aligned} \mathbf E Q^n_t\,\mathbf 1 _{\{Q^n_0\!\le \! V\}}\,\!\prec \! \mathbf E Q^n_0\,\mathbf 1 _{\{Q^n_0\le V\}}\,\!+\! \int \limits _0^t\lambda ^n_s ds\!-\!\int \limits _0^t(\mu ^n_s\wedge \theta ^n_s)\, \mathbf E Q^n_s\,\mathbf 1 _{\{Q^n_0\le V\}}\,\,ds. \end{aligned}$$

(5.1)

By Lemma 5,

$$\begin{aligned}&\mathbf E Q^n_t\,\mathbf 1 _{\{Q^n_0\le V\}}\,\nonumber \\&\quad \le e^{-\int _0^t (\mu ^n_s\wedge \theta ^n_s)\,ds}\,\mathbf E Q^n_0\,\mathbf 1 _{\{Q^n_0\le V\}}\, \!+\!e^{-\int _0^t (\mu ^n_s\wedge \theta ^n_s)\,ds}\int \limits _0^t e^{\int _0^s (\mu ^n_u\wedge \theta ^n_u)\,du}\lambda ^n_s\,ds.\quad \end{aligned}$$

(5.2)

If $vT\le t<(v+1)T$, where $v\in \mathbb Z _+$, then by $T$-periodicity,

$$\begin{aligned}&e^{-\int _0^t (\mu ^n_s\wedge \theta ^n_s)\,ds}\int \limits _0^t e^{\int _0^s (\mu ^n_u\wedge \theta ^n_u)\,du}\lambda ^n_s\,ds\\&\quad \le e^{-\int _0^{vT} (\mu ^n_s\wedge \theta ^n_s)\,ds}\sum _{i=1}^{v+1} e^{\int _0^{iT} (\mu ^n_u\wedge \theta ^n_u)\,du}\int \limits _{(i-1)T}^{iT}\lambda ^n_s\,ds\\&\quad =e^{-v\int _0^{T} (\mu ^n_s\wedge \theta ^n_s)\,ds}\sum _{i=1}^{v+1} e^{i\int _0^{T} (\mu ^n_u\wedge \theta ^n_u)\,du}\int \limits _{0}^{T}\lambda ^n_s\,ds\\&\quad =e^{-(v-1)\int _0^{T} (\mu ^n_s\wedge \theta ^n_s)\,ds} \frac{ e^{(v+1)\int _0^{T} (\mu ^n_u\wedge \theta ^n_u)\,du}-1}{e^{\int _0^{T} (\mu ^n_u\wedge \theta ^n_u)\,du}-1}\int \limits _{0}^{T}\lambda ^n_s\,ds\\&\quad \le \frac{ e^{2\int _0^{T} (\mu ^n_u\wedge \theta ^n_u)\,du}}{e^{\int _0^{T} (\mu ^n_u\wedge \theta ^n_u)\,du}-1}\int \limits _{0}^{T}\lambda ^n_s\,ds. \end{aligned}$$

In addition, $e^{-\int _0^t (\mu ^n_s\wedge \theta ^n_s)\,ds}\le e^{-\lfloor t/T\rfloor \int _0^T (\mu ^n_s\wedge \theta ^n_s)\,ds}$. Part 1(b) has been proved.

Part 1(a) follows by a similar argument if one observes that by (2.6),

$$\begin{aligned} q_t\prec q_0+\int \limits _0^t\lambda _s\,ds -\int \limits _0^t(\theta _s\wedge \mu _s)q_s\,ds\, \end{aligned}$$

so that, by Lemma 5,

$$\begin{aligned} q_t\le e^{-\int _0^t (\theta _s\wedge \mu _s)\,ds}q_0+ e^{-\int _0^t (\theta _s\wedge \mu _s)\,ds} \int \limits _0^te^{\int _0^s (\theta _u\wedge \mu _u)\,du}\,\lambda _s\,ds. \end{aligned}$$

(5.3)

In order to prove part 1(c), let me note that by (2.13),

$$\begin{aligned} (X^n_t)^2&= (X^n_0)^2+ 2\int \limits _0^t X^n_{s-}\,dX^n_s+\sum _{0<s\le t}\left( \Delta X^n_s\right) ^2= (X^n_0)^2+ 2\int \limits _0^t\alpha ^n_s X^n_{s}\,ds\\&\quad -2\int \limits _0^t\theta ^n_sX^n_{s}\left(\left(X^n_s-\delta ^n_s+ \sqrt{n}( q_s-\kappa _s)\right)^+ -\sqrt{n}(q_s-\kappa _s)^+\right)\,ds\\&\quad -2\int \limits _0^t\gamma ^n_s(q_s-\kappa _s)^+X^n_s\,ds\\&\quad -2 \int \limits _0^t\mu ^n_sX^n_{s}\left(\left(X^n_s+ \sqrt{n} q_s\right)\wedge (\delta ^n_s+ \sqrt{n}\kappa _s) - \sqrt{n}\,(q_s\wedge \kappa _s)\right)\,ds\\&\quad -2\int \limits _0^t\beta ^n_s(q_s\wedge \kappa _s)X^n_s\,ds\\&\quad +\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,A}_s- \frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,R}_s -\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,B}_s\\&\quad +\frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,A}_s\right) ^2+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,R}_s\right) ^2+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,B}_s\right) ^2. \end{aligned}$$

On noting that

$$\begin{aligned}&\theta ^n_sX^n_{s}\left(\left(X^n_s-\delta ^n_s+ \sqrt{n}( q_s-\kappa _s)\right)^+ -\sqrt{n}(q_s-\kappa _s)^+\right)\\&\quad +\mu ^n_sX^n_{s}\left(\left(X^n_s+ \sqrt{n} q_s\right)\wedge (\delta ^n_s \right. \left.\quad + \sqrt{n}\kappa _s) - \sqrt{n}\,(q_s\wedge \kappa _s)\right)\\&\quad \ge \left( \mu ^n_s\wedge \theta ^n_s\right) \left( X^n_s\right) ^2- \left| {\left( \theta ^n_s-\mu ^n_s\right) \delta ^n_sX^n_s}\right| \,, \end{aligned}$$

I obtain that, for $\epsilon >0$,

$$\begin{aligned} \left( X^n_t\right) ^2&\prec \left( X^n_0\right) ^2+ 2\int \limits _0^t\sigma ^n_s\left| { X^n_{s}}\right| \,ds -2\int \limits _0^t\left( \mu ^n_s\wedge \theta ^n_s\right) \left( X^n_{s}\right) ^2\,ds\\&+\,\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,A}_s- \frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,R}_s -\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,B}_s\\&+\,\frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,A}_s\right) ^2+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,R}_s\right) ^2 + \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,B}_s\right) ^2\\&\prec \left( X^n_0\right) ^2+ \frac{1}{\epsilon }\,\int \limits _0^t\sigma ^n_s\,ds -\int \limits _0^t\left(2\left( \mu ^n_s\wedge \theta ^n_s\right) -\epsilon \sigma ^n_s\right) \left( X^n_{s}\right) ^2\,ds\\&+\,\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,A}_s- \frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,R}_s -\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\,dM^{n,B}_s\\&+\,\frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,A}_s\right) ^2+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,R}_s\right) ^2+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,B}_s\right) ^2. \end{aligned}$$

Hence, for $V>0$,

$$\begin{aligned} \left( X^n_t\right) ^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,&\prec \left( X^n_0\right) ^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,+ \frac{1}{\epsilon }\int \limits _0^t\sigma ^n_s\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\,ds\nonumber \\&-\int \limits _0^t\left(2\left( \mu ^n_s\wedge \theta ^n_s\right) -\epsilon \,\sigma ^n_s\right)(X^n_{s})^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\,ds\nonumber \\&+\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\, dM^{n,A,V}_s- \frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\, \,dM^{n,R,V}_s\nonumber \\&-\frac{2}{\sqrt{n}}\,\int \limits _0^t X^n_{s-}\, \,dM^{n,B,V}_s +\frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,A,V}_s\right) ^2\nonumber \\&+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,R,V}_s\right) ^2+ \frac{1}{n}\,\sum _{0<s\le t}\left( \Delta M^{n,B,V}_s\right) ^2\,\, \end{aligned}$$

(5.4)

where $M^{n,i,V}_s=M^{n,i}_s\,\mathbf 1 _{\{|X^n_0|\le V\}}\,$, for $i=A,R,B$.

By Lemma 1, the processes $M^{n,i,V}=(M^{n,i,V}_t,\,t\in \mathbb R _+)$ are $\mathbf F ^n$-locally square integrable martingales with predictable quadratic variation processes $\langle M^{n,i,V}\rangle =\langle M^{n,i}\rangle \,\mathbf 1 _{\{|X^n_0|\le V\}}\,$. Since $Q^n_t\le Q^ n_0+A^n_t$ by (2.1) and $\mathbf E (A^n_t)^2=\int _0^t\lambda ^n_s\,ds+\left(\int _0^t\lambda ^n_s\,ds\right)^2<\infty$, I have that $\mathbf E (Q^n_t)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,<\infty$. Hence, by (2.4a), (2.4b), (2.4c), $\mathbf E \langle M^{n,i,V}\rangle _t<\infty$, which implies that $\mathbf E \left(\sup _{s\le t}\left( M^{n,i,V}_s\right) ^2\right)<\infty$, that the $M^{n,i,V}$ are $\mathbf F ^n$-martingales, and that $\mathbf E ( M^{n,i,V}_t)^2=\mathbf E \langle M^{n,i,V}\rangle _t$. Consequently, the processes $\left(\int _0^{t} X^n_{s-}\, \,dM^{n,i,V}_s,\,t\in \mathbb R _+\right)$ are $\mathbf F ^n$-martingales. Since the $M^{n,i,V}$ are purely discontinuous locally square integrable martingales by being of locally bounded variation, $\mathbf E \sum _{0<s\le t}(\Delta M^{n,i,V}_s)^2= \mathbf E \langle M^{n,i,V}\rangle _{t}$.

On taking expectations in (5.4),

$$\begin{aligned} \mathbf E (X^n_{t})^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,&\prec \mathbf E (X^n_{0})^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\, + \frac{1}{\epsilon }\int \limits _0^t\sigma ^n_s\,ds +\frac{1}{n}\,\left(\mathbf E \left\langle M^{n,A}\right\rangle _{t}\,\mathbf 1 _{\{|X^n_0|\le V\}}\, \right.\\&\left. +\mathbf E \left\langle M^{n,R}\right\rangle _{t}\,\mathbf 1 _{\{|X^n_0|\le V\}}\, +\mathbf E \left\langle M^{n,B}\right\rangle _{t}\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\right)\\&- \int \limits _0^t\left(2\left( \mu ^n_s\wedge \theta ^n_s\right) -\epsilon \,\sigma ^n_s\right) \mathbf E (X^n_{s})^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\,ds. \end{aligned}$$

By (2.4a), (2.4b), (2.4c), and (2.12),

$$\begin{aligned}&\frac{1}{n}\,\left(\mathbf{E }\left\langle M^{n,A}\right\rangle _{t}\,\mathbf 1 _{\{|X^n_0|\le V\}}\, +\mathbf{E }\left\langle M^{n,R}\right\rangle _{t}\,\mathbf 1 _{\{|X^n_0|\le V\}}\, +\mathbf{E }\left\langle M^{n,B}\right\rangle _{t}\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\right)\\&\quad \prec \int \limits _0^t\frac{\lambda ^n_s}{n}\,ds +\int \limits _0^{t}\left( \mu ^n_s\vee \theta ^n_s\right) \frac{\mathbf{E }Q^n_s\,\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\,}{n} ds\\&\quad \prec \int \limits _0^t\frac{\lambda ^n_s}{n}\,ds +\int \limits _0^{t}\left( \mu ^n_s\vee \theta ^n_s\right) \left(\frac{\mathbf{E }X^n_s\,\mathbf 1 _{\{|X^n_0|\le V\}}\, }{\sqrt{n}}+q_s\right)\,ds\\&\quad \prec \int \limits _0^t\frac{\lambda ^n_s}{n}\,ds+ \int \limits _0^{t}\left( \mu ^n_s\vee \theta ^n_s\right) \left(q_s+ \frac{1}{2\sqrt{n}}\right)\,ds\\&\qquad +\frac{1}{2\sqrt{n}}\int \limits _0^{t}\left( \mu ^n_s\vee \theta ^n_s\right) \,\mathbf{E }(X^n_s)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\, \,ds . \end{aligned}$$

Thus, for $t\in \mathbb R _+$,

$$\begin{aligned}&\mathbf E (X^n_{t})^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\\&\quad \prec \mathbf{E }(X^n_0)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,+ \frac{1}{\epsilon }\int \limits _0^t\sigma ^n_s\,ds +\int \limits _0^t\frac{\lambda ^n_s}{n}\,ds\\&\quad \quad + \int \limits _0^{t}\left( \mu ^n_s\vee \theta ^n_s\right) \left(q_s+ \frac{1}{2\sqrt{n}}\right)\,ds\\&\quad \quad - \int \limits _0^t\left(2\left( \mu ^n_s\wedge \theta ^n_s\right) -\epsilon \,\sigma ^n_s- \frac{1}{2\sqrt{n}}\left( \mu ^n_s\vee \theta ^n_s\right) \right)\mathbf{E }(X^n_{s})^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\,ds. \end{aligned}$$

By Lemma 5,

$$\begin{aligned}&\mathbf E (X^n_t)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\nonumber \\&\quad \le e^{-\int _0^t \left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)\,ds} \mathbf E (X^n_0)^2\,\mathbf 1 _{\{|X^n_0|\le V\}}\,\nonumber \\&\quad \quad +\,e^{-\int _0^t \left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)\,ds}\int \limits _0^te^{\int _0^s\left(2(\mu ^n_u\wedge \theta ^n_u)-\epsilon \,\sigma ^n_u- (\mu ^n_u\vee \theta ^n_u)/(2\sqrt{n})\right)\,du}\nonumber \\&\quad \quad \times \left(\frac{\sigma ^n_s}{\epsilon }\, +\frac{\lambda ^n_s}{n}+ (\mu ^n_s\vee \theta ^n_s)\left(q_s+ \frac{1}{2\sqrt{n}}\right)\right)\,ds. \end{aligned}$$

(5.5)

In analogy with the earlier argument, if $vT\le t<(v+1)T$, where $v\in \mathbb Z _+$, recalling that $\sup _{u\in \mathbb R _+}q_u<\infty$ by part 1(a),

$$\begin{aligned}&e^{-\int _0^t \left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)\,ds}\int \limits _0^te^{\int _0^s\left(2(\mu ^n_u\wedge \theta ^n_u)-\epsilon \,\sigma ^n_u- (\mu ^n_u\vee \theta ^n_u)/(2\sqrt{n})\right)\,du}\\&\quad \quad \times \left(\frac{\,\sigma ^n_s}{\epsilon } +\frac{\lambda ^n_s}{n} + (\mu ^n_s\vee \theta ^n_s)\left(q_s+ \frac{1}{2\sqrt{n}}\right)\right)\,ds\\&\quad \le e^{-v\int _0^{T} \left(2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n}) \right)\,ds+2\int _0^{T} |2(\mu ^n_s\wedge \theta ^n_s)-\epsilon \,\sigma ^n_s- (\mu ^n_s\vee \theta ^n_s)/(2\sqrt{n})|\,ds}\\&\quad \quad \times \sum _{i=0}^{v}e^{i\int \limits _0^{T}\left(2(\mu ^n_u\wedge \theta ^n_u)-\epsilon \,\sigma ^n_u- (\mu ^n_u\vee \theta ^n_u)/(2\sqrt{n})\right)\,du}\int \limits _0^{T}\left(\frac{1}{\epsilon }\,\sigma ^n_s +\frac{\lambda ^n_s}{n}+ (\mu ^n_s\vee \theta ^n_s) \right.\\&\quad \quad \times \left.\left(\sup _{u\in \mathbb R _+}q_u+ \frac{1}{2\sqrt{n}}\right)\right)\,ds\\&\quad \le \frac{ e^{ 2\int _0^{T}|2(\mu ^n_u\wedge \theta ^n_u)-\epsilon \,\sigma ^n_u- (\mu ^n_u\vee \theta ^n_u)/(2\sqrt{n})|\,du}}{1-e^{-\int _0^{T}\left(2(\mu ^n_u\wedge \theta ^n_u)-\epsilon \,\sigma ^n_u- (\mu ^n_u\vee \theta ^n_u)/(2\sqrt{n})\right)\,du}} \int \limits _0^{T}\left(\frac{1}{\epsilon }\,\sigma ^n_s +\frac{\lambda ^n_s}{n}+ (\mu ^n_s\vee \theta ^n_s)\right.\\&\quad \quad \times \left.\left(\sup _{u\in \mathbb R _+}q_u+ \frac{1}{2\sqrt{n}}\right)\right)\,ds\,, \end{aligned}$$

where the last inequality uses the fact that $\int _0^{T}\left(2(\mu ^n_u\wedge \theta ^n_u)-\epsilon \,\sigma ^n_u- (\mu ^n_u\vee \theta ^n_u)/\right.\left.(2\sqrt{n})\right)\,du>0$. The latter expression furnishes the required bound. Part 1 has been proved.

The assertions of part 2 also follow from the respective inequalities (5.2), (5.3), and (5.5). For instance part 2(b) is obtained by applying the bound

$$\begin{aligned} \int \limits _0^t e^{\int _0^s (\theta ^n_u\wedge \mu ^n_u)\,du}\,\lambda ^n_s\,ds\le \frac{\sup _{t\in \mathbb R _+}\lambda ^n_t}{\inf _{t\in \mathbb R _+} (\theta ^n_t\wedge \mu ^n_t)}\,\left(e^{\int _0^t (\theta ^n_s\wedge \mu ^n_s)\,ds}-1\right). \end{aligned}$$

$\square$