2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\Theta^{(N)}\right)}+\frac{\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\int_{0}^{N^{2\varepsilon}T}\mathds{1}\left(\bar{Q}^{(N)}_{3}(s)>0\right)ds\right)}{\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\Theta^{(N)}\right)}$
$\displaystyle\leq\frac{\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left({\color[rgb]{0,0,0}(\Theta^{(N)})^{2}}\right)}{N^{2\varepsilon}T\
\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\Theta^{(N)}\right)}+\frac{N^{2\varepsilon}T\,\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\sup_{s\leq
N^{2\varepsilon}T}\bar{Q}^{(N)}_{3}(s)>0\right)}{\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\Theta^{(N)}\right)}$
$\displaystyle\leq\frac{\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left({\color[rgb]{0,0,0}(\Theta^{(N)})^{2}}\right)}{N^{2\varepsilon}T\
\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\Theta^{(N)}\right)}+\frac{N^{2\varepsilon}T\,\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\sup_{s\leq
N^{2\varepsilon}T}Q^{(N)}_{2}(s)\geq N\right)}{\mathbb{E}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\left(\Theta^{(N)}\right)}.$

Now, applying Proposition 4.3 for the first term and Proposition 3.9 along
with Proposition 4.3 for the second term above, we conclude that there exists
a finite constant $c_{\circ}>0$ independent of $N,T$ such that

$\limsup_{N\rightarrow\infty}\
\pi(\bar{Q}^{(N)}_{3}(\infty)>0)\leq\frac{c_{\circ}}{T}.$

As $T>0$ is arbitrary, (4.32) follows.

∎

##  5 Proof of process-level limit.

In this section, we will analyze the process-level limit of the scaled
occupancy process, and in particular, prove Theorem 2.1. Throughout this
section, we consider fixed $\varepsilon\in(0,1/2)$. First, we claim that it is
enough to establish the following ‘deterministic initial state’ version of
Theorem 2.1.

######  Theorem 5.1.

Fix $\beta>0$ and $\varepsilon\in(0,\frac{1}{2})$ and assume that the initial
states are deterministic. Suppose there exists a subsequence
$(\hat{N})\subset\mathbb{N}$ such that, as $\hat{N}\rightarrow\infty$,
$X^{(\hat{N})}(0)\to x_{0}$ where $x_{0}\in(0,\infty)$ is a constant,
$Q_{3}^{(\hat{N})}(0)\to 0$, and
$\limsup_{\hat{N}\rightarrow\infty}\hat{N}^{-(\frac{1}{2}-\varepsilon)}I^{(\hat{N})}(0)<\infty$.
Then, for any fixed $T>0$, the scaled process $X^{(\hat{N})}$ converges weakly
as $\hat{N}\rightarrow\infty$ to the path-wise unique solution of the
following stochastic differential equation, uniformly on $[0,T]$:

$dX(t)=\Big{(}\frac{1}{X(t)}-\beta\Big{)}dt+\sqrt{2}dW(t),$ (5.1)

with $X(0)=x_{0}$, where $W=\big{(}W(t),t\geq 0\big{)}$ is the standard
Brownian motion.

Assuming Theorem 5.1, we now prove Theorem 2.1 below.

###### Proof.

Proof of Theorem 2.1. Note that, under the assumptions of Theorem 2.1, for any
subsequence of $(N)$, there exists a further subsequence $(\hat{N})$, such
that

$\big{(}X^{(\hat{N})}(0),\hat{N}^{-\frac{1}{2}+\varepsilon}I^{(\hat{N})}(0),\hat{N}^{-\frac{1}{2}-\varepsilon}Q^{(\hat{N})}_{2}(0),Q^{(\hat{N})}_{3}(0)\big{)}\
\xrightarrow{d}\ \big{(}X^{*},I^{*},Q^{*}_{2},0\big{)}\text{ as
}\hat{N}\rightarrow\infty$

where $X^{*}$ is as in Theorem 2.1 and $I^{*}$, $Q^{*}_{2}$ are finite random
variables. Therefore, by the Skorohod representation theorem, there exists a
probability space $(\Omega,\mathcal{F},\mathbb{P})$ such that this convergence
happens almost surely. Let $\hat{\Omega}\in\mathcal{F}$ be a subset of
$\Omega$ with $\mathbb{P}(\hat{\Omega})=1$ such that for all
$\omega\in\hat{\Omega}$,

$\big{(}X^{(\hat{N})}(0,\omega),\hat{N}^{-\frac{1}{2}+\varepsilon}I^{(\hat{N})}(0,\omega),\hat{N}^{-\frac{1}{2}-\varepsilon}Q^{(\hat{N})}_{2}(0,\omega),Q^{(\hat{N})}_{3}(0,\omega)\big{)}\to\big{(}X^{*}(\omega),I^{*}(\omega),Q^{*}_{2}(\omega),0\big{)}\text{
as }\hat{N}\rightarrow\infty$

and $X^{*}(\omega),I^{*}(\omega),Q^{*}_{2}(\omega)$ are finite. Therefore, in
the above probability space, the (sub-)sequence of initial states satisfies
the conditions stated in Theorem 5.1 for each fixed $\omega\in\hat{\Omega}$.
Hence, applying Theorem 5.1 conditionally on the initial configuration, we
deduce that for any $\omega\in\hat{\Omega}$ and any fixed $T>0$,
$X^{(\hat{N})}$ converges weakly to the path-wise unique solution $X$ of
(2.3), with $X(0)=X^{*}(\omega)$, uniformly on $[0,T]$ as
$\hat{N}\rightarrow\infty$. We thus conclude that $X^{(N)}$ converges weakly
to the solution of (2.3) uniformly on $[0,T]$ as $N\rightarrow\infty$.
∎

The rest of the section is devoted to the proof of Theorem 5.1. For notational
convenience, we will write $(N)$ for the subsequence $(\hat{N})$ in the
theorem. Also, observe from the initial conditions that there exist constants
$N_{0}\geq 1$ and $K_{1}>0$, such that $I^{(N)}(0)\leq
K_{1}N^{\frac{1}{2}-\varepsilon}$, and $Q^{(N)}_{3}(0)=0$ (and hence,
$\sum_{i\geq 3}Q^{(N)}_{i}=0$) for all $N\geq N_{0}$. Furthermore, since
$Q^{(N)}_{2}=N^{\frac{1}{2}+\varepsilon}X^{(N)}+I^{(N)}-\sum_{i\geq
3}Q^{(N)}_{i}$, we also have $K_{2}>0$ such that $Q^{(N)}_{2}(0)\leq
K_{2}N^{\frac{1}{2}+\varepsilon}$ for all $N\geq N_{0}$. Henceforth, we will
assume that $N$ is large enough to satisfy these bounds.

The main ingredient is to analyze the idle-server process $I^{(N)}$. We start
with the martingale representation of the process $X^{(N)}$.

#### Martingale representation.

Recall the representation (2.2) of $X^{(N)}(t)$, the arrival process
$A(N^{1+2\varepsilon}\lambda_{N}t)$, and the departure process
$D\left(\int_{0}^{N^{2\varepsilon}t}(N-I^{(N)}(s))ds\right)$. Let us introduce
the related filtrations
$\mathbf{F}=\big{\\{}\mathcal{F}_{N,t}:N\in\mathbb{N},t\in[0,\infty]\big{\\}}$
where

$\mathcal{F}_{N,t}:=\sigma\Big{(}S^{(N)}(0),A(N^{1+2\varepsilon}\lambda_{N}s),D\Big{(}\int_{0}^{N^{2\varepsilon}s}(N-I^{(N)}(u))du\Big{)},\,0\leq
s\leq t\Big{)},\quad t\geq 0,$ (5.2)

and $\mathcal{F}_{N,\infty}:=\sigma(\cup_{t\geq 0}\mathcal{F}_{N,t})$. We
write

$\displaystyle X^{(N)}(t)-X^{(N)}(0)$
$\displaystyle=N^{-\frac{1}{2}-\varepsilon}\left[A(N^{1+2\varepsilon}\lambda_{N}t)-D\Big{(}\int_{0}^{N^{2\varepsilon}t}(N-I^{(N)}(s))ds\Big{)}\right]$
$\displaystyle=\mathcal{M}_{a}^{(N)}(\lambda_{N}t)-\mathcal{M}_{d}^{(N)}\Big{(}t-\frac{1}{N^{1+2\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds\Big{)}$
(5.3)
$\displaystyle\quad+\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds$
(5.4) $\displaystyle\quad-\beta t+\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds$ (5.5)

where
$\mathcal{M}_{a}^{(N)}(t)=\frac{A(N^{1+2\varepsilon}t)-N^{1+2\varepsilon}t}{N^{\frac{1}{2}+\varepsilon}},\quad\mathcal{M}_{d}^{(N)}(t)=\frac{D(N^{1+2\varepsilon}t)-N^{1+2\varepsilon}t}{N^{\frac{1}{2}+\varepsilon}}.$
Note that $\mathcal{M}_{a}^{(N)}(t)$ and
$\mathcal{M}_{d}^{(N)}\Big{(}t-\frac{1}{N^{1+2\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds\Big{)}$
are martingales adapted to the filtration $\mathbf{F}$. We will proceed by
first showing in Proposition 5.5 that the integral in (5.3) converges to 0
uniformly on any (scaled) finite time interval. Using this, we will be able to
show that the difference of martingales in (5.3) convergence weakly to
$\sqrt{2}W$ as $N\to\infty$, where $W$ is the standard Brownian motion. The
next major challenge is to show that the difference of the two terms in (5.4)
converges to 0 as $N\to\infty$. This is achieved in Proposition 5.6. Finally,
a continuous mapping theorem-type result will complete the proof of Theorem
5.1.

In its core, the analysis of $I^{(N)}$ will be done by upper and lower
bounding it with suitable birth-and-death processes. Now, for any fixed $B>0$,
recall the stopping time $\tau^{(N)}_{2}(B)$ from (3.1) and the process
$\bar{I}^{(N)}_{B}$, and note that if
$Q^{(N)}_{2}(0)>BN^{\frac{1}{2}+\varepsilon}$, then for all
$t\leq\tau^{(N)}_{2}(B)$, $I^{(N)}(t)$ can be stochastically upper bounded by
$\bar{I}^{(N)}_{B}(t)$ with $\bar{I}^{(N)}_{B}(0)=I^{(N)}(0)$. As before,
throughout we assume $N$ to be large enough so that
$N>BN^{\frac{1}{2}+\varepsilon}>\beta N^{\frac{1}{2}-\varepsilon}$. We
emphasize that, unlike in Section 3, we will be interested in _small_ values
of $B$ for the process level limit and thus cannot directly apply the
estimates in Section 3 which deals with _large_ values of $B$.

######  Lemma 5.2.

There exist $N_{0},a,b>0$ depending only on $T$, $B$, and $\beta$, such that
for all $N\geq N_{0}$ and $\delta>0$,

$\mathbb{P}\Big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\
\big{|}\ \bar{I}^{(N)}_{B}(0)=0\Big{)}\leq ae^{-bN^{\frac{\delta}{2}}}.$

###### Proof.

Proof. The proof follows in three steps: first, we upper bound the tail
probability of the stationary distribution of $\bar{I}^{(N)}_{B}$. Next, we
upper bound the tail probability of $\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)$ when $\bar{I}^{(N)}_{B}(0)$ is a
random variable having the same distribution as the steady state of
$\bar{I}^{(N)}_{B}$, and finally, we consider $\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)$ when $\bar{I}^{(N)}_{B}(0)=0$.

######  Claim 5.3.

Let $\bar{I}^{(N)}_{B}(\infty)$ denote a random variable having stationary
distribution of $\bar{I}^{(N)}_{B}$. Then there exist constants
$N_{0}^{\prime}$, $a_{1}$ and $b_{1}$, that only depend on $B$ and $\beta$,
such that for all $N\geq N_{0}^{\prime}$,

$\mathbb{P}\big{(}\bar{I}^{(N)}_{B}(\infty)\geq
N^{\frac{1}{2}-\varepsilon+\delta}\big{)}\leq a_{1}e^{-b_{1}N^{\delta}}.$

_Proof._ Note that the stationary distribution of $\bar{I}^{(N)}_{B}$ is given
by $\mathbb{P}(\bar{I}^{(N)}_{B}(t)=k)=(1-\rho)\rho^{k}$ for $k\geq 0$, where
$\rho=\frac{N-BN^{\frac{1}{2}+\varepsilon}}{N-\beta
N^{\frac{1}{2}-\varepsilon}}$. Therefore, there exists $N_{0}^{\prime}>0$,
independent of $\delta$,  such that for all $N\geq N_{0}^{\prime}$,
$B/2\geq\beta N^{-2\varepsilon}$, and

$\begin{split}\mathbb{P}\big{(}\bar{I}^{(N)}_{B}(\infty)\geq
N^{\frac{1}{2}-\varepsilon+\delta}\big{)}&=\Big{(}\frac{N-BN^{\frac{1}{2}+\varepsilon}}{N-\beta
N^{\frac{1}{2}-\varepsilon}}\Big{)}^{N^{\frac{1}{2}-\varepsilon+\delta}}=\Big{(}1-\frac{BN^{\frac{1}{2}+\varepsilon}-\beta
N^{\frac{1}{2}-\varepsilon}}{N-\beta
N^{\frac{1}{2}-\varepsilon}}\Big{)}^{N^{\frac{1}{2}-\varepsilon+\delta}}\\\
&\leq\Big{(}1-\frac{B}{2N^{\frac{1}{2}-\varepsilon}}\Big{)}^{N^{\frac{1}{2}-\varepsilon+\delta}}\leq
a_{1}e^{-b_{1}N^{\delta}},\end{split}$ (5.6)

for appropriate constants $a_{1}$ and $b_{1}$ that depend only on $B$ and
$\beta$, where the last inequality is due to the observation that $\exists
n_{0}>0$ s.t. $\forall n\geq n_{0}$, $(1-1/n)^{n}\leq 2e^{-1}$, derived from
$\lim_{n\rightarrow\infty}(1-1/n)^{n}=e^{-1}$. This completes the proof of
Claim 5.3.

######  Claim 5.4.

Assume that $\\{\bar{I}^{(N)}_{B}(t),t\geq 0\\}$ is an equilibrium process.
Then there exist positive constants $N_{1},a_{2},$ and $b_{2}$ which only
depend on $T$, $B$ and $\beta$ such that for all $N\geq N_{1}$,

$\mathbb{P}\big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\big{)}\leq
a_{2}e^{-b_{2}N^{\frac{\delta}{2}}}.$

_Proof._ Let
$j=\lceil\frac{N^{2\varepsilon}T}{N^{-\varepsilon-\frac{1}{2}+\delta}}\rceil$
and consider the times $t_{i}=iN^{-\varepsilon-\frac{1}{2}+\delta}$,
$i=0,1,...,j-1$. Denote the number of positive increments in
$\bar{I}^{(N)}_{B}$ in a subinterval $[t_{i},t_{i+1})$ by $\zeta^{(N)}_{i}$.
Then $\zeta^{(N)}_{i}$ has a Poisson distribution with parameter
$N^{\frac{1}{2}-\varepsilon+\delta}-BN^{\delta}$. For any Poisson random
variable $\mathrm{Po}(\lambda)$ with parameter $\lambda$, we have (see [29,
Theorem 2.3(b)]) that for $0\leq\xi\leq 1$,
$\mathbb{P}\left(\mathrm{Po}(\lambda)-\lambda\geq\xi\lambda\right)\leq
e^{-\frac{3}{8}\xi^{2}\lambda}.$ Hence, for all $i=0,1,...,j-1$,
$\mathbb{P}\big{(}\zeta^{(N)}_{i}\geq\frac{3}{2}N^{\frac{1}{2}-\varepsilon+\delta}\big{)}\leq
e^{-\frac{3}{32}N^{\frac{1}{2}-\varepsilon+\delta}}.$ Take $N_{0}^{\prime}$ as
in Claim 5.3. Since we are considering the equilibrium process, due to Claim
5.3, we know for all $N\geq N_{0}^{\prime}$ and $t\geq 0$,
$\mathbb{P}\big{(}\bar{I}^{(N)}_{B}(t)\geq
N^{\frac{1}{2}-\varepsilon+\delta}\big{)}\leq a_{1}e^{-b_{1}N^{\delta}}.$ Now,
note that

$\Big{\\{}\sup_{t_{i}\leq t\leq
t_{i+1}}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\Big{\\}}\subseteq\Big{\\{}\bar{I}^{(N)}_{B}(t_{i})\geq
N^{\frac{1}{2}-\varepsilon+\delta}\Big{\\}}\cup\Big{\\{}\zeta^{(N)}_{i}\geq\frac{3}{2}N^{\frac{1}{2}-\varepsilon+\delta}\Big{\\}},$

for $i=0,2,...,j-1$, and we have,

$\displaystyle\mathbb{P}\big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\big{)}$
$\displaystyle\leq\sum_{i=0}^{j-1}\mathbb{P}\big{(}\bar{I}^{(N)}_{B}(t_{i})\geq
N^{\frac{1}{2}-\varepsilon+\delta}\big{)}+\sum_{i=0}^{j-1}\mathbb{P}\big{(}\zeta^{(N)}_{i}\geq\frac{3}{2}N^{\frac{1}{2}-\varepsilon+\delta}\big{)}$
$\displaystyle\leq\big{(}N^{\frac{1}{2}+3\varepsilon-\delta}T+1\big{)}\big{(}a_{1}e^{-b_{1}N^{\delta}}+e^{-\frac{3}{32}N^{\frac{1}{2}-\varepsilon+\delta}}\big{)}.$

Thus, there exist constants $a_{2},b_{2}>0$, and $N_{1}\geq N_{0}^{\prime}$,
which depend only on $B$, $\beta$, and $T$, such that for all $N\geq N_{1}$,
$\mathbb{P}\big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\big{)}\leq
a_{2}e^{-b_{2}N^{\frac{\delta}{2}}}.$ This completes the proof of Claim 5.4.

Finally, the proof for $\bar{I}^{(N)}_{B}(0)=0$ follows from Claim 5.4 by
observing that for any $k\geq 1$,

$\mathbb{P}\big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\
\big{|}\ \bar{I}^{(N)}_{B}(0)=0\big{)}\leq\mathbb{P}\big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\
\big{|}\ \bar{I}^{(N)}_{B}(0)=k\big{)}.$

This completes the proof of the lemma. ∎

######  Proposition 5.5.

Fix any $T>0$ and $0<\delta<\frac{1}{2}+\varepsilon$, and take $N_{0}$ as in
Lemma 5.2. For any $K_{1}>0$, there exist constants $N_{1},a,b>0$ that depend
only on $B$, $\beta$, and $T$, such that for all $N\geq N_{1},x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},y\geq BN^{\frac{1}{2}+\varepsilon}$,

$\sup_{\underline{z}}\mathbb{P}_{(x,y,\underline{z})}\Big{(}\sup_{0\leq
t\leq(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B)}I^{(N)}(t)\geq
5N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}\leq ae^{-bN^{\frac{\delta}{2}}},$
(5.7)

and consequently, for all $\tilde{\varepsilon}>0$,

$\lim_{N\rightarrow\infty}\sup_{\underline{z}}\mathbb{P}_{(x,y,\underline{z})}\Big{(}\frac{1}{N^{1+2\varepsilon}}\int_{0}^{(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B)}I^{(N)}(s)ds\geq\tilde{\varepsilon}\Big{)}=0.$
(5.8)

###### Proof.

Proof. Recall $N_{0}$ as in Lemma 5.2. Take $N_{1}\geq N_{0}$ such that
$\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\geq
K_{1}N^{\frac{1}{2}-\varepsilon}$ and consider $N\geq N_{1}$. Note that for
any $x\leq K_{1}N^{\frac{1}{2}-\varepsilon}$, $y\geq
BN^{\frac{1}{2}+\varepsilon}$ and $\underline{z}\in\mathbb{N}_{0}^{\infty}$
with $z_{1}\geq z_{2}\geq\dots$,

$\displaystyle\mathbb{P}_{(x,y,\underline{z})}\Big{(}\sup_{0\leq
t\leq(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B)}I^{(N)}(t)\geq
5N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\left(K_{1}N^{\frac{1}{2}-\varepsilon}+\bar{I}^{(N)}_{B}(t)\right)\geq
5N^{\frac{1}{2}-\varepsilon+\delta}\ \big{|}\ \bar{I}^{(N)}_{B}(0)=0\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sup_{0\leq t\leq
N^{2\varepsilon}T}\bar{I}^{(N)}_{B}(t)\geq\frac{5}{2}N^{\frac{1}{2}-\varepsilon+\delta}\
\big{|}\ \bar{I}^{(N)}_{B}(0)=0\Big{)}\leq ae^{-bN^{\frac{\delta}{2}}}.$

The first inequality follows from the fact that for $t\leq\tau^{(N)}_{2}(B)$,
the process $I^{(N)}(\cdot)$ starting from $x\leq
K_{1}N^{\frac{1}{2}-\varepsilon}$ is stochastically dominated by
$K_{1}N^{\frac{1}{2}-\varepsilon}+\bar{I}^{(N)}_{B}(\cdot)$ with
$\bar{I}^{(N)}_{B}(0)=0$. The last inequality follows from Lemma 5.2. Next,
for all $N\geq N_{1}$, $x\leq K_{1}N^{\frac{1}{2}-\varepsilon}$, $y\geq
BN^{\frac{1}{2}+\varepsilon}$, and feasible $\underline{z}$,

$\displaystyle\mathbb{P}_{(x,y,\underline{z})}\Big{(}\frac{1}{N^{1+2\varepsilon}}\int_{0}^{(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B)}I^{(N)}(s)ds\geq
5N^{-\frac{1}{2}-\varepsilon+\delta}T\Big{)}$ $\displaystyle\hskip
85.35826pt\quad\leq\mathbb{P}_{(x,y,\underline{z})}\Big{(}\sup_{0\leq
s\leq(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B)}I^{(N)}(s)\geq
5N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}\leq ae^{-bN^{\frac{\delta}{2}}},$

and thus, (5.8) holds for any $0<\delta<\frac{1}{2}+\varepsilon$. ∎

######  Proposition 5.6.

For any fixed $B>0$, recall $\tau^{(N)}_{2}(B)$ from (3.1). Under the
assumptions on the initial conditions stated in Theorem 5.1, the following
holds as $N\to\infty$:

$\sup_{0\leq t\leq
T\wedge(N^{-2\varepsilon}\tau^{(N)}_{2}(B))}\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds\Big{|}\xrightarrow{\
\mathbb{P}\ }0.$

The proof of Proposition 5.6 is given in Section 6.

######  Lemma 5.7.

The stochastic differential equation

$dX(t)=\Big{(}\frac{1}{X}-\beta\Big{)}dt+\sqrt{2}dW(t),$ (5.9)

with $X(0)>0$, has a path-wise unique strong solution. Also, if
$\tau_{\varepsilon}\coloneqq\inf\\{t>0:X(t)\leq\varepsilon\\}$,
$\varepsilon>0$, and $\tau:=\lim_{\varepsilon\rightarrow
0}\tau_{\varepsilon}$, then $\tau=\infty$ almost surely.

###### Proof.

Proof. For $\varepsilon>0$, the process $X(t)$, for $t\leq\tau_{\varepsilon}$
satisfies an SDE with Lipschitz coefficients. Such SDE are known to have path-
wise unique strong solutions (see [36, Theorem V.11.2]).

Next, to show that $\tau=\infty$ almost surely, consider the SDE
$d\hat{X}(t)=\frac{1}{\hat{X}(t)}+\sqrt{2}dW(t),$ and define the analogous
quantities $\hat{\tau}_{\varepsilon}$, $\varepsilon>0$, and $\hat{\tau}$ for
$\hat{X}$. Note that $\frac{\hat{X}(t)}{\sqrt{2}}$ is a Bessel process of
dimension 2. By Girsanov’s Theorem, we can add and remove the drift $\beta t$
to $\hat{X}$ with an exponential change of measure. Hence, the law of $X$ and
that of $\hat{X}$ are mutually absolutely continuous on compact time
intervals. For $n\geq 2$, the $n$-dimensional Bessel process is transient from
its starting point with probability one, i.e., the $n$-dimensional Bessel
process will be greater than 0 for all $t>0$ almost surely [21, §3.3.C]. Thus,
for any $a>0$, $\mathbb{P}(\tau\leq a)=\mathbb{P}(\hat{\tau}\leq a)=0$.
Therefore, $\tau=\infty$ almost surely. ∎

###### Proof.

Proof of Theorem 5.1. We will proceed as in the proof of [16, Theorem 1],
using Proposition 5.6 stated above in place of their Proposition EC.3. Recall
the martingale representation of $X^{(N)}$ in (5.3)–(5.5). First consider
(5.3). By the assumption on $\lambda_{N}$ and Proposition F.1, we have that,
as $N\rightarrow\infty$, $\lambda_{N}t\rightarrow t\text{ and
}t-\frac{1}{N^{1+2\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds\xrightarrow{\
\mathbb{P}\ }t,$ uniformly on the interval $[0,T]$. By the Martingale FCLT
[39] and the independence of $\mathcal{M}_{a}$ and $\mathcal{M}_{d}$, we have
that as $N\rightarrow\infty$,

$\displaystyle\Big{\\{}\mathcal{M}_{a}^{(N)}(\lambda_{N}t)-\mathcal{M}_{d}^{(N)}\Big{(}t-\frac{1}{N^{1+2\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds\Big{)}:t\geq
0\Big{\\}}\Rightarrow\Big{\\{}\sqrt{2}W(t):t\geq 0\Big{\\}},$

where $W$ is a standard Brownian motion. Next, we will consider (5.4). For any
fixed $B>0$, define $\hat{\tau}^{(N)}(B)\coloneqq\inf\\{t\geq 0:X^{(N)}(t)\leq
B\\}$ and $\hat{\tau}(B)\coloneqq\inf\\{t\geq 0:X(t)\leq B\\}$, where $X(t)$
is the unique strong solution to the S.D.E. (5.9) with initial value $X(0)>0$.
The claim below establishes a relation between $\hat{\tau}^{(N)}$ and
$\tau^{(N)}_{2}$.

######  Claim 5.8.

For any fixed $B>0$,

$\lim_{N\rightarrow\infty}\mathbb{P}\left(N^{2\varepsilon}\hat{\tau}^{(N)}(B)\wedge(N^{2\varepsilon}T)\leq\tau^{(N)}_{2}(B)\wedge(N^{2\varepsilon}T)\right)=1.$

###### Proof.

[Proof.] First, note that on the event $\big{\\{}\tau^{(N)}_{2}(B)\geq
N^{2\varepsilon}T\big{\\}}$, trivially,
$N^{2\varepsilon}\hat{\tau}^{(N)}(B)\wedge(N^{2\varepsilon}T)\leq\tau^{(N)}_{2}(B)\wedge(N^{2\varepsilon}T).$
Now, on the event
$\big{\\{}\tau^{(N)}_{2}(B)<N^{2\varepsilon}T\big{\\}}\cap\big{\\{}\sup_{0\leq
t\leq N^{2\varepsilon}T}Q^{(N)}_{3}(t)=0\big{\\}}$, we have that for $0\leq
t\leq N^{2\varepsilon}T$, $S^{(N)}(t)-N=Q^{(N)}_{2}(t)-I^{(N)}(t)\leq
Q^{(N)}_{2}(t),$ and thus, $S^{(N)}(\tau^{(N)}_{2}(B))-N\leq
Q^{(N)}_{2}(\tau^{(N)}_{2}(B))=\lfloor BN^{\frac{1}{2}+\varepsilon}\rfloor,$
which implies that $N^{2\varepsilon}\hat{\tau}^{(N)}(B)\leq\tau^{(N)}_{2}(B)$.
Hence, we have

$\displaystyle\mathbb{P}\left(N^{2\varepsilon}\hat{\tau}^{(N)}(B)\wedge(N^{2\varepsilon}T)\leq\tau^{(N)}_{2}(B)\wedge(N^{2\varepsilon}T)\right)\geq\mathbb{P}\Big{(}\sup_{0\leq
t\leq N^{2\varepsilon}T}Q^{(N)}_{3}(t)=0\Big{)}.$ (5.10)

By Proposition 3.9, the right hand side of (5.10) tends to 1 as
$N\rightarrow\infty$. This completes the proof of Claim 5.8. ∎

Using Claim 5.8 and Proposition 5.6, we can conclude

$\sup_{0\leq t\leq\hat{\tau}^{(N)}(B)\wedge
T}\left|\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds\right|\xrightarrow{\
\mathbb{P}\ }0\quad\text{as}\quad N\rightarrow\infty.$

Therefore, defining
$\delta^{(N)}(t):=\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds,\
t\geq 0,$ (when the integrals are well defined) we have that for any fixed
$B>0$, the process $\big{(}\delta^{(N)}(t\wedge\hat{\tau}^{(N)}(B)):t\geq
0\big{)}$ converges weakly to a process that is identically equal to 0, as
$N\rightarrow\infty$. Also, recall that $X^{(N)}(0)\xrightarrow{\ \mathbb{P}\
}X(0)$ where $X(0)$ is a positive constant. Thus, by the Skorohod
representation theorem, there exists a probability space
$(\Omega,\mathcal{F},\mathbb{P})$ such that, almost surely, the following
convergence holds

$\displaystyle\Big{(}X^{(N)}(0),\Big{\\{}\mathcal{M}_{a}^{(N)}(\lambda_{N}t)-\mathcal{M}_{d}^{(N)}\Big{(}t-\frac{1}{N^{1+2\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds\Big{)},$
$\displaystyle\hskip
199.16928pt\delta^{(N)}(t\wedge\hat{\tau}^{(N)}(K^{-1})\wedge\hat{\tau}(K^{-1})\big{)}:t\in[0,T],K\in\mathbb{N}\Big{\\}}\Big{)}$
$\displaystyle\xrightarrow{u.o.c.}\Big{(}X(0),\big{\\{}\sqrt{2}W(t),0\big{)}:t\in[0,T],K\in\mathbb{N}\big{\\}}\Big{)}\quad\text{as}\quad
N\rightarrow\infty,$ (5.11)

where ‘u.o.c.’ denotes convergence of the associated processes uniformly on
compact subsets of $[0,T]$, and the above random variables are seen as
$\mathbb{R}_{+}\times\left(\mathcal{D}\left([0,T]:\mathbb{R}^{2}\right)\right)^{\mathbb{N}}$
valued random variables. For $K\in\mathbb{N}$, define the event
$\Omega_{K}:=\\{\inf_{t\in[0,T]}X(t)\geq K^{-1}\\}.$ By Lemma 5.7,
$\lim_{K\rightarrow\infty}\mathbb{P}\left(\Omega_{K}\right)=1$. Let
$b^{(N)}(t):=\mathcal{M}_{a}^{(N)}(\lambda_{N}t)-\mathcal{M}_{d}^{(N)}\Big{(}t-\frac{1}{N^{1+2\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds\Big{)}+\delta^{(N)}(t)(\omega)$
and $b(t):=\sqrt{2}W(t)$. Define the following subsets of $\Omega$:

$\mathcal{S}^{(N)}_{\varepsilon,K}:=\left\\{\left|X^{(N)}(0)-X(0)\right|+\sup_{0\leq
s\leq
T\wedge\hat{\tau}^{(N)}(K^{-1})\wedge\hat{\tau}(K^{-1})}\left|b^{(N)}(s)-b(s)\right|<\varepsilon\right\\},\
\varepsilon>0,\,K\in\mathbb{N}.$

By (5.11), for any $\varepsilon>0,\,K\in\mathbb{N}$,
$\lim_{N\rightarrow\infty}\mathbb{P}\left(\mathcal{S}^{(N)}_{\varepsilon,K}\right)=1.$

Using the triangle inequality, we have that for all
$t\in[0,T\wedge\hat{\tau}^{(N)}(B)\wedge\hat{\tau}(B)]$,

$\left|X^{(N)}(t)-X(t)\right|\leq\left|X^{(N)}(0)-X(0)\right|+\left|b^{(N)}(t)-b(t)\right|+\int_{0}^{t}\left|\frac{1}{X^{(N)}(s)}-\frac{1}{X(s)}\right|ds.$

Observe that, for any $B>0$, the map $x\mapsto x^{-1}$ is Lipschitz on
$[B,\infty)$ with Lipschitz constant $B^{-2}$. Thus, for
$t\in[0,T\wedge\hat{\tau}^{(N)}(K^{-1}/2)\wedge\hat{\tau}(K^{-1}/2)]$,

$\displaystyle\sup_{0\leq s\leq t}\left|X^{(N)}(s)-X(s)\right|$
$\displaystyle\leq$ $\displaystyle\left|X^{(N)}(0)-X(0)\right|+\sup_{0\leq
s\leq t}\left|b^{(N)}(s)-b(s)\right|+4K^{2}\int_{0}^{t}\sup_{0\leq
s\leq\mu}\left|X^{(N)}(s)-X(s))\right|d\mu.$

Applying Gronwall’s inequality (see [33, Lemma 4.1]), we have on the set
$\mathcal{S}^{(N)}_{\varepsilon,2K}$, $\sup_{0\leq t\leq
T\wedge\hat{\tau}^{(N)}(K^{-1}/2)\wedge\hat{\tau}(K^{-1}/2)}|X^{(N)}(t)-X(t)|\leq\varepsilon
e^{4K^{2}T}.$ Set
$\varepsilon=\varepsilon_{K}=\frac{X(0)}{2}\wedge\frac{1}{4K}e^{-4K^{2}T}$.
Let $K_{0}:=\lceil 4/X(0)\rceil$. Take any $K\geq K_{0}$. The above bound
implies that, on the event $\mathcal{S}^{(N)}_{\varepsilon_{K},2K}$,
$X^{(N)}(s)>\frac{1}{2K}$ for all
$s\in[0,T\wedge\hat{\tau}^{(N)}(K^{-1}/2)\wedge\hat{\tau}(K^{-1}/2)]$.
Moreover, on $\Omega_{K}$, $\hat{\tau}(K^{-1}/2)\geq\hat{\tau}(K^{-1})\geq T$.
Hence, on $\Omega_{K}\cap\mathcal{S}^{(N)}_{\varepsilon_{K},2K}$,
$\hat{\tau}^{(N)}(K^{-1}/2)\wedge\hat{\tau}(K^{-1}/2)>T$. Thus, the above
bound gives on the event
$\Omega_{K}\cap\mathcal{S}^{(N)}_{\varepsilon_{K},2K}$, $\sup_{0\leq t\leq
T}|X^{(N)}(t)-X(t)|\leq\frac{1}{4K}.$ Therefore, for any $K\geq K_{0}$,
$\limsup_{N\rightarrow\infty}\mathbb{P}\left(\sup_{0\leq t\leq
T}|X^{(N)}(t)-X(t)|>\frac{1}{4K}\right)\leq\mathbb{P}\left(\Omega_{K}^{c}\right)+\limsup_{N\rightarrow\infty}\mathbb{P}\left(\left(\mathcal{S}^{(N)}_{\varepsilon_{K},2K}\right)^{c}\right)=\mathbb{P}\left(\Omega_{K}^{c}\right).$
On recalling $\lim_{K\rightarrow\infty}\mathbb{P}\left(\Omega_{K}\right)=1$,
we obtain $\sup_{0\leq t\leq T}|X^{(N)}(t)-X(t)|\xrightarrow{\ \mathbb{P}\ }0\
\text{ as }\ N\rightarrow\infty,$ proving the theorem. ∎

###### Proof.

Proof of Proposition 2.2. Write the SDE as

$\displaystyle dX(t)$
$\displaystyle=\Big{(}\frac{1}{X}-\beta\Big{)}dt+\sqrt{2}dW(t)=\sqrt{2}dW(t)-V^{\prime}(X)dt,$
(5.12)

where $V(X)$ is a function with derivative $V^{\prime}(X)=-\frac{1}{X}+\beta.$
The diffusion (5.12) is a Langevin diffusion and, for any $B>0$, it has an
invariant measure with density given by
$\frac{d\tilde{\pi}}{dx}=\exp\\{-V(x)\\},$ where
$V(x)=\int_{B}^{x}\Big{(}\beta-\frac{1}{u}\Big{)}du=\beta(x-B)+[\ln{B}-\ln{x}]=\beta(x-B)+\ln{\frac{B}{x}},\
x>0.$ Therefore, we have an invariant distribution $\pi(x)$ with density
$\frac{d\pi}{dx}=C\frac{x}{B}e^{-\beta(x-B)},$ where $C$ satisfies that
$\frac{1}{C}=\int_{0}^{\infty}\frac{x}{B}e^{-\beta(x-B)}dx=\frac{e^{\beta
B}}{\beta^{2}B}$. The computation of moments of $\pi$ is routine. This
completes the proof. ∎

##  6 Proof of Proposition 5.6.

For convenience, denote
$T(N,B)\coloneqq\tau^{(N)}_{2}(B)\wedge(N^{2\varepsilon}T)$. The key idea to
estimate the integral of $I^{(N)}$ is to consider its excursions in the
interval $[0,T(N,B)]$ and estimate the integral of $I^{(N)}$ within each
excursion. For that, define the following stopping times: $\xi^{(N)}_{0}=0$
and for $i\geq 0$,

$\begin{split}\xi^{(N)}_{2i+1}&=\inf\big{\\{}t\geq\xi^{(N)}_{2i}:I^{(N)}(t)=0\big{\\}},\\\
\xi^{(N)}_{2i+2}&=\inf\big{\\{}t\geq\xi^{(N)}_{2i+1}:I^{(N)}(t)>0\big{\\}}.\end{split}$
(6.1)

Thus, $(\xi_{1}^{(N)},\xi_{3}^{(N)},\xi_{5}^{(N)},\ldots)$ constitute
successive excursions of $I^{(N)}$, and $\xi_{2i+2}^{(N)}-\xi_{2i+1}^{(N)}$,
for $i\geq 0$ are the intervals of time $I^{(N)}$ spends at $0$. Also, let
$i^{*}_{N}:=\min\big{\\{}i\geq 0:\xi^{(N)}_{2i+3}\geq T(N,B)\big{\\}}$ be the
number of excursions in $[0,T(N,B)]$.

Recall that $I^{(N)}$ has instantaneous transition rates at time $t$ as
follows:

$\begin{split}&I^{(N)}(t)\nearrow I^{(N)}(t)+1\text{ at rate
}Q_{1}^{(N)}(t)-Q_{2}^{(N)}(t),\\\ &I^{(N)}(t)\searrow(I^{(N)}(t)-1)_{+}\text{
at rate }N-\beta N^{\frac{1}{2}-\varepsilon}.\end{split}$ (6.2)

As before, fix any $B>0$ and we will consider $N$ to be large enough so that
$N>BN^{\frac{1}{2}+\varepsilon}>\beta N^{\frac{1}{2}-\varepsilon}$. The main
challenge in estimating the integral $I^{(N)}$ is that the above rates are
state-dependent. That is why, our first step is to approximate, for each
excursion $[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$, the term
$Q_{1}^{(N)}(t)-Q_{2}^{(N)}(t)$ by
$Q_{1}^{(N)}(\xi^{(N)}_{2i+1})-Q_{2}^{(N)}(\xi^{(N)}_{2i+1})$, which is static
within that excursion. The approximation is justified in two steps: First, by
showing that the length of each excursion of $I^{(N)}$ is short with high
probability uniformly in $[0,T(N,B)]$ (Lemma 6.1) and second, by showing that
the fluctuation of $S^{(N)}$ within each of these excursions is also uniformly
small (Proposition 6.2). Finally, Lemma 6.5 and Lemma 6.6 show that the error
in using the above approximation for the integral of $I^{(N)}$ is
asymptotically negligible.

In the following, the set $\\{0,-1\\}$ is assumed to be the null set.

######  Lemma 6.1.

For any $0<\delta<\frac{1}{2}-\varepsilon$, there exist constants
$a_{1},b_{1},N_{1}>0$, such that for all $N\geq N_{1}$,

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\exists\
i\in\big{\\{}0,...,i^{*}_{N}-1\big{\\}}\text{ such that
}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+1}\geq 2N^{\delta-2\varepsilon}\Big{)}\leq
a_{1}e^{-b_{1}N^{\frac{\delta}{5}}}.$ (6.3)

The proof is given in Appendix E and we provide the high-level proof idea
here. We consider the excursion of $I^{(N)}$ under two mutually exclusive
events: $\sup_{t\in[0,T(N,B)]}Q^{(N)}_{2}(t)\geq
N^{\frac{1}{2}+\varepsilon+\delta}$ or not. By Proposition 3.9, we have
$\mathbb{P}(\sup_{t\in[0,T(N,B)]}Q^{(N)}_{2}(t)\geq
N^{\frac{1}{2}+\varepsilon+\delta})\xrightarrow{N\rightarrow\infty}0$.
Conditional on the event
$\\{\sup_{t\in[0,T(N,B)]}Q^{(N)}_{2}(t)<N^{\frac{1}{2}+\varepsilon+\delta}\\}$,
we can bound the excursion time of $I^{(N)}$ by coupling it with an
appropriate $M/M/1$ queue.

As mentioned above, the next proposition states that the fluctuation of
$S^{(N)}$ within each of these excursions is also uniformly small.

######  Proposition 6.2.

For any $0<\delta<\frac{1}{2}-\varepsilon$, there exist constants
$N_{2},a_{2},b_{2}>0$, such that for all $N\geq N_{2}$,

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
13N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}\leq
a_{2}e^{-b_{2}N^{\frac{\delta}{5}}}.$

Here also, the idea is to condition on the event
$\\{\sup_{t\in[0,T(N,B)]}Q^{(N)}_{2}(t)\leq
N^{\frac{1}{2}+\varepsilon+\delta}\\}$ and then use Lemma 6.1 to bound the
change in $Q^{(N)}_{2}$ in this ‘short’ interval
$[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3}]$ and then use Lemma 5.2 to bound the
maximum change in $I^{(N)}$. The proof is given in Appendix E

From now on, we will work under the assumptions stated in Theorem 5.1 for
fixed $T>0$ and
$\delta<(\varepsilon/2)\wedge\left(\frac{1}{2}-\varepsilon\right)$.
Proposition 6.2 and the properties of $I^{(N)}$ and $Q_{2}^{(N)}$ established
in Sections 3 and 4 hint that we can define the following high-probability
event, which is stated in Lemma 6.3:

$\begin{split}\bar{E}^{(N)}\coloneqq\Big{\\{}\sup_{0\leq t\leq
T(N,B)}I^{(N)}(t)&\leq 5N^{\frac{1}{2}-\varepsilon+\delta},\sup_{0\leq t\leq
T(N,B)}Q^{(N)}_{2}(t)\leq N^{\frac{1}{2}+\varepsilon+\delta},\\\
&\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi_{2i+1})\right|\leq
13N^{\frac{1}{2}-\varepsilon+\delta},\forall i\leq
i^{*}_{N}-1\Big{\\}}.\end{split}$ (6.4)

######  Lemma 6.3.

$\lim_{N\rightarrow\infty}\mathbb{P}(\bar{E}^{(N)})=1.$

###### Proof.

Proof. The result follows from Proposition 3.9, Proposition 5.5 and
Proposition 6.2. ∎

Note that on the event $\bar{E}^{(N)}$, the following equation holds:

$2N=S^{(N)}(t)+Q_{1}^{(N)}(t)-Q_{2}^{(N)}(t)+2I^{(N)}(t),\quad
t\in[0,T(N,B)].$ (6.5)

Hence, for $t\in[0,T(N,B)]$, the evolution of $I^{(N)}$ can be described as
follows:

$\displaystyle I^{(N)}(t)$ $\displaystyle\nearrow I^{(N)}(t)+1\text{ at rate
}2N-S^{(N)}(t)-2I^{(N)}(t),$ $\displaystyle I^{(N)}(t)$
$\displaystyle\searrow(I^{(N)}(t)-1)_{+}\text{ at rate }N-\beta
N^{\frac{1}{2}-\varepsilon}.$

Also, on the event $\bar{E}^{(N)}$, we have the following bounds on the rate
of increase of $I^{(N)}(t)$.

######  Lemma 6.4.

On the event $\bar{E}^{(N)}$, for each $i\in\\{0,...,i^{*}_{N}-1\\}$ and
$t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$, we have

$\lambda_{l,i}\leq 2N-S^{(N)}(t)-2I^{(N)}(t)\leq\lambda_{u,i},$

where
$\lambda^{(N)}_{l,i}=2N-S^{(N)}(\xi^{(N)}_{2i+1})-23N^{\frac{1}{2}-\varepsilon+\delta}$
and
$\lambda_{u,i}=2N-S^{(N)}(\xi^{(N)}_{2i+1})+13N^{\frac{1}{2}-\varepsilon+\delta}$.

Lemma 6.4 is easy to check (we omit its proof). Thus, Lemma 6.4 hints that on
$\bar{E}^{(N)}$, for each excursion interval
$[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$, we can construct two $M/M/1$ queues,
$I^{(N)}_{u,i}$ and $I^{(N)}_{l,i}$, on the same probability space as
$I^{(N)}$, and both starting from zero, to bound the process $I^{(N)}$ from
above and below respectively. For $t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$,
we define $I^{(N)}_{u,i}$ and $I^{(N)}_{l,i}$ with the following transition
rates:

$\displaystyle I^{(N)}_{u,i}(t)\nearrow I^{(N)}_{u,i}(t)+1$
$\displaystyle\text{ at rate }\lambda^{(N)}_{u,i},$ $\displaystyle
I^{(N)}_{u,i}(t)\searrow I^{(N)}_{u,i}(t)-1$ $\displaystyle\text{ at rate
}\mu^{(N)}_{u,i}=N-\beta N^{\frac{1}{2}-\varepsilon};$ $\displaystyle
I^{(N)}_{l,i}(t)\nearrow I^{(N)}_{l,i}(t)+1$ $\displaystyle\text{ at rate
}\lambda^{(N)}_{l,i},$ $\displaystyle I^{(N)}_{l,i}(t)\searrow
I^{(N)}_{l,i}(t)-1$ $\displaystyle\text{ at rate }\mu^{(N)}_{l,i}=N.$

Due to Lemma 6.4, we can naturally couple $I^{(N)}_{u,i}$ and $I^{(N)}_{l,i}$
with $I^{(N)}$ by setting
$I^{(N)}_{u,i}(\xi^{(N)}_{2i})=I^{(N)}_{l,i}(\xi^{(N)}_{2i+1})=I^{(N)}(\xi^{(N)}_{2i+1})=0$,
so that for $t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$, $I^{(N)}_{l,i}(t)\leq
I^{(N)}(t)\leq I^{(N)}_{u,i}(t)$. Note that since $\big{\\{}(I^{(N)}(t),t\geq
0)\big{\\}}_{N=1}^{\infty}$ are defined on the same probability space (see the
representation (5.2)),
$\big{\\{}(I_{u,i}^{(N)}(t),t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})),i\in\\{0,...,i^{*}_{N}-1\\}\big{\\}}_{N=1}^{\infty}$
and
$\big{\\{}(I_{l,i}^{(N)}(t),t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})),i\in\\{0,...,i^{*}_{N}-1\\}\big{\\}}_{N=1}^{\infty}$
are on the same probability space as well.

Having the above sample path-wise bounds, Lemma 6.5 below provides an
approximation of the integral of the upper and lower-bounding process at the
end of each excursion.

######  Lemma 6.5.

The following hold:

$\sup_{j\in\\{0,...,i^{*}_{N}-1\\}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds\xrightarrow{\
\mathbb{P}\ }0,$
$\inf_{j\in\\{0,...,i^{*}_{N}-1\\}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{l,i}(s)-\frac{\lambda^{(N)}_{l,i}}{\mu^{(N)}_{l,i}-\lambda^{(N)}_{l,i}}\Big{)}ds\xrightarrow{\
\mathbb{P}\ }0.$

The proof is given in Appendix E. Here we provide the proof scheme of the
first convergence since the second one is similar. The idea is similar to the
one used in [16], with one technical difference: The term
$\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds$
can be approximated by
$Z^{(N)}_{j}\coloneqq\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{u,i}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds$
plus
$\int^{\xi^{(N)}_{u,i}}_{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds$
where $\xi^{(N)}_{u,i}$ is the time that $I^{(N)}_{u,i}$ hits 0. Although
$\mathbb{E}(Z^{(N)}_{j})=0$, we cannot use the LLN argument (as used in [16])
since the order of $i^{*}_{N}$ is higher than the scaling
$N^{\frac{1}{2}+\varepsilon}$ with high probability. Hence, we estimate the
second moment of
$\int_{\xi^{(N)}_{2i-1}}^{\xi^{(N)}_{u,i-1}}I^{(N)}_{u,i-1}(s)ds$ by Lemma E.3
and use Doob’s $L^{2}$ inequality to get
$\sup_{j}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}Z^{(N)}_{j}\xrightarrow{\
\mathbb{P}\ }0$. For the second term, $|\xi^{(N)}_{u,i}-\xi^{(N)}_{2i+3}|$ can
be bounded due to the known expectation of excursion time of an $M/M/1$ queue,
and we use Markov’s inequality to get the desired convergence.

Next, Lemma 6.6 connects the term that is used in the integrand as an
approximation in Lemma 6.5 to the integral of certain functional of
$S^{(N)}(t)$, which will later be shown to be close to the integral of
$I^{(N)}(t)$.

######  Lemma 6.6.

The following hold as $N\to\infty$:

$\sup_{j\in\\{0,...,i^{*}_{N}-1\\}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\Big{|}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}\frac{\lambda^{(N)}_{l,i}}{\mu^{(N)}_{l,i}-\lambda^{(N)}_{l,i}}-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\Big{|}\xrightarrow{\
\mathbb{P}\ }0,$ (6.6)
$\sup_{j\in\\{0,...,i^{*}_{N}-1\\}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\Big{|}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\Big{|}\xrightarrow{\
\mathbb{P}\ }0.$ (6.7)

The key step in proving Lemma 6.6 is to condition on the event
$\bar{E}^{(N)}$, where
$\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi_{2i+1})\right|\leq
13N^{\frac{1}{2}-\varepsilon+\delta}$ and to use the fact that
$\frac{2N-x}{x-N}$ (which is the form of two terms of the integrand) is
Lipschitz continuous on the interval
$[N+\frac{1}{2}BN^{\frac{1}{2}-\varepsilon+\delta},\infty)$ with Lipschitz
constant $\frac{4}{B^{2}}N^{-2\varepsilon}$. The proof is given in Appendix E.

Note that both Lemma 6.5 and Lemma 6.6 concerns approximation of respective
integrals where the upper limits are excursion end points. Lemma 6.7 shows
that it is sufficient to consider the latter case, since the contribution of
the integrals within a single excursion is asymptotically negligible.

######  Lemma 6.7.

Under the assumptions on the initial state as given in Theorem 5.1, the
following holds as $N\to\infty$:

$\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+3}}\left[\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{\xi^{(N)}_{2i+1}}^{t}I^{(N)}(s)ds+\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{\xi^{(N)}_{2i+1}}^{t}\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}ds\right]\xrightarrow{\
\mathbb{P}\ }0.$ (6.8)

The proof of Lemma 6.7 is given in Appendix E. We now have all the ingredients
to prove Proposition 5.6.

###### Proof.

Proof of Proposition 5.6. Again, we will consider events within
$\bar{E}^{(N)}$. Observe that for any $t\leq T(N,B)$, either
$t\leq\xi^{(N)}_{1}$, or $t\in[\xi^{(N)}_{2j+1},\xi^{(N)}_{2j+3}]$ for some
$0\leq j\leq i^{*}_{N}$, or $t\in[\xi^{(N)}_{2i^{*}_{N}+1},T(N,B)]$. Hence,

$\begin{split}&\sup_{0\leq t\leq
T(N,B)}\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{t}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\Big{|}\\\
\leq&\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{\xi^{(N)}_{1}}\left|I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\right|ds\\\
&+\sup_{j\in\\{0,...,i^{*}_{N}-1\\}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\left|\int_{\xi^{(N)}_{1}}^{\xi^{(N)}_{2j+3}}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\right|\\\
&+\sup_{j\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2j+1}\leq
t\leq\xi^{(N)}_{2j+3}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\left|\int_{\xi^{(N)}_{2j+1}}^{t}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\right|\\\
&+\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{\xi^{(N)}_{2i^{*}_{N}+1}}^{T(N,B)}\left|I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\right|ds.\end{split}$
(6.9)

For $t\in[0,T(N,B)]$, $I^{(N)}(t)$ can be upper bounded by
$\bar{I}^{(N)}_{B}(t)$ as defined in Section 3. Then, there exist constants
$C,C^{\prime}$ such that

$\begin{split}\mathbb{P}\big{(}\xi^{(N)}_{1}\geq
N^{\delta-2\varepsilon}\big{)}\leq\mathbb{P}(\inf\\{t>0:\bar{I}^{(N)}_{B}(t)=0\\}\geq
N^{\delta-2\varepsilon}\ \big{|}\ \bar{I}^{(N)}_{B}(0)=\lfloor
K_{1}N^{\frac{1}{2}-\varepsilon}\rfloor)\leq
Ce^{-C^{\prime}N^{\delta}},\end{split}$ (6.10)

where the first inequality is due to the assumption $I^{(N)}(0)\leq
K_{1}N^{\frac{1}{2}-\varepsilon}$ and the second inequality is due to Lemma
B.1 and Markov’s inequality. Also, on the event $\bar{E}^{(N)}$, we have

$\sup_{0\leq t\leq T(N,B)}I^{(N)}(t)\leq
5N^{\frac{1}{2}-\varepsilon+\delta}\quad\text{and}\quad\sup_{0\leq t\leq
T(N,B)}\frac{2N-S^{(N)}(t)}{S^{(N)}(t)-N}\leq\frac{2N^{\frac{1}{2}-\varepsilon}}{B}.$
(6.11)

Hence, on
$\big{\\{}\xi^{(N)}_{1}<N^{\delta-2\varepsilon}\big{\\}}\cap\bar{E}^{(N)}$, we
have

$\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{\xi^{(N)}_{1}}\Big{|}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{|}ds\leq
CN^{-4\varepsilon+2\delta}\to 0,\quad\text{as}\quad N\to\infty.$

With
$\lim_{N\rightarrow\infty}\mathbb{P}\big{(}\xi^{(N)}_{1}<N^{\delta-2\varepsilon}\big{)}=1$
due to (6.10) and $\lim_{N\to\infty}\mathbb{P}(\bar{E}^{(N)})=1$ due to Lemma
6.3, we have

$\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{\xi^{(N)}_{1}}\Big{|}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{|}ds\xrightarrow{\
\mathbb{P}\ }0,$ (6.12)

as $N\to\infty$. Now, consider the second term in the right hand side of
(6.9). For $j\in\\{0,...,i^{*}_{N}-1\\}$,

$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{l,i}(s)-\frac{\lambda^{(N)}_{l,i}}{\mu^{(N)}_{l,i}-\lambda^{(N)}_{l,i}}\Big{)}ds+\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}\frac{\lambda^{(N)}_{l,i}}{\mu^{(N)}_{l,i}-\lambda^{(N)}_{l,i}}-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds$
$\displaystyle\leq$
$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{\xi^{(N)}_{1}}^{\xi^{(N)}_{2j+3}}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds$
$\displaystyle\leq$
$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds+\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds.$

Thus, by Lemma 6.6 and Lemma 6.5,

$\sup_{j\in\\{0,...,i^{*}_{N}-1\\}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\left|\int_{\xi^{(N)}_{1}}^{\xi^{(N)}_{2j+3}}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\right|\xrightarrow{\
\mathbb{P}\ }0\text{ as }N\rightarrow\infty.$ (6.13)

By Lemma 6.7, we have

$\sup_{j\in\\{0,...,i^{*}-1\\}}\sup_{\xi^{(N)}_{2j+1}\leq
t\leq\xi^{(N)}_{2j+3}}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\left|\int_{\xi^{(N)}_{2j+1}}^{t}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\right|\xrightarrow{\
\mathbb{P}\ }0.$ (6.14)

To estimate the last term in the bound (6.9), note that using the same
argument used to derive (E) and (E.6),
$\mathbb{P}\left(T(N,B)-\xi^{(N)}_{2i^{*}_{N}+1}\geq
N^{\delta-2\varepsilon}\right)\leq Ce^{-C^{\prime}N^{\delta}}.$ From this and
(6.11), we conclude

$\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{\xi^{(N)}_{2i^{*}_{N}+1}}^{T(N,B)}\left|I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\right|ds\xrightarrow{\
\mathbb{P}\ }0.$ (6.15)

Using (6.12), (6.13), (6.14) and (6.15) in (6.9), we have

$\sup_{0\leq t\leq
T(N,B)}\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{t}\Big{(}I^{(N)}(s)-\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}\Big{)}ds\Big{|}\xrightarrow{\
\mathbb{P}\ }0.$ (6.16)

For any $t$ such that $N^{2\varepsilon}t\leq T(N,B)$, by the triangle
inequality,

$\begin{split}&\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds\Big{|}\\\
\leq&\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}ds\Big{|}\\\
&\hskip
56.9055pt+\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds\Big{|}.\end{split}$
(6.17)

Note that

$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}ds=$
$\displaystyle\frac{1}{N^{\frac{1}{2}-\varepsilon}}\int_{0}^{t}\frac{2N-S^{(N)}(N^{2\varepsilon}s)}{S^{(N)}(N^{2\varepsilon}s)-N}ds$
$\displaystyle=$
$\displaystyle\frac{1}{N^{\frac{1}{2}-\varepsilon}}\int_{0}^{t}\frac{N^{\frac{1}{2}-\varepsilon}+\frac{N-S^{(N)}(N^{2\varepsilon}s)}{N^{\frac{1}{2}+\varepsilon}}}{\frac{S^{(N)}(N^{2\varepsilon}s)-N}{N^{\frac{1}{2}+\varepsilon}}}ds=-\frac{t}{N^{\frac{1}{2}-\varepsilon}}+\int_{0}^{t}\frac{1}{X^{(N)}}ds,$

which implies that for all $t$ such that $N^{2\varepsilon}t\leq T(N,B)$,

$\displaystyle\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds\Big{|}$
$\displaystyle\hskip
56.9055pt\leq\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}\frac{2N-S^{(N)}(s)}{S^{(N)}(s)-N}ds\Big{|}+\frac{t}{N^{\frac{1}{2}-\varepsilon}}.$

This, combined with (6.16), yields that as $N\to\infty$, $\sup_{0\leq t\leq
T\wedge(N^{-2\varepsilon}\tau^{(N)}_{2}(B))}\Big{|}\frac{1}{N^{\frac{1}{2}+\varepsilon}}\int_{0}^{N^{2\varepsilon}t}I^{(N)}(s)ds-\int_{0}^{t}\frac{1}{X^{(N)}(s)}ds\Big{|}\xrightarrow{\
\mathbb{P}\ }0,$ which concludes the proof of Proposition 5.6. ∎

##  Appendix A Useful concentration estimates.

The following elementary lemma gives a supremum bound on the path of a
compensated Poisson process.

######  Lemma A.1.

Suppose $A(\cdot)$ is a Poisson process with unit rate. For any $T>0$ and
$x\in[0,4T]$,

$\mathbb{P}\big{(}\sup_{s\in[0,T]}\left|A(s)-s\right|\geq x\big{)}\leq
2e^{-\frac{x^{2}}{8T}}.$

###### Proof.

Proof. By [24, Theorem 5, Chapter 4, Page 351], for any $x\geq 0$,
$\mathbb{P}\big{(}\sup_{s\in[0,T]}\big{(}A(s)-s\big{)}\geq
x\big{)}\leq\exp\big{\\{}-\sup_{\lambda>0}\big{(}\lambda
x-2T(e^{\lambda}-1-\lambda)\big{)}\big{\\}}.$ Since
$e^{\lambda}-1-\lambda\leq\lambda^{2}$, for all $\lambda\in[0,1],$ we have for
any $x\in[0,4T]$, $\sup_{\lambda>0}\big{(}\lambda
x-2T(e^{\lambda}-1-\lambda)\big{)}\geq\sup_{\lambda\in[0,1]}\big{(}\lambda
x-2T\lambda^{2}\big{)}=\frac{x^{2}}{8T}.$ We can perform the same calculation
with $-(A(s)-s)$ in place of $(A(s)-s)$. This proves the lemma. ∎

The next lemma gives a concentration bound on sums of random variables with
stretched exponential tails. It is used multiple times in this article, and is
of independent interest.

######  Lemma A.2.

Suppose $\\{\Phi^{(r)}_{i}:i\geq 1,r\in R\\}$ are independent nonnegative
random variables adapted to a filtration $\big{\\{}\mathcal{F}^{(r)}_{i}:i\geq
1,r\in R\big{\\}}$, where $R$ is an arbitrary indexing set. Let
$\mathcal{F}^{(r)}_{0}\subseteq\mathcal{F}^{(r)}_{1}$ be an arbitrary
$\sigma$-field. Suppose there exist deterministic constants $c,c_{1}>0$,
$\theta\in(0,1)$ not dependent on $r$ such that for any $x\geq 0$ and $i\geq
1$, $\mathbb{P}\big{(}\Phi^{(r)}_{i}\geq x|\mathcal{F}^{(r)}_{i-1}\big{)}\leq
ce^{-c_{1}x^{\theta}},x\geq 0.$ Define
$\mu:=\int_{0}^{\infty}ce^{-c_{1}x^{\theta}}dx\geq\sup_{r\in
R}\mathbb{E}\big{(}\Phi^{(r)}_{i}|\mathcal{F}^{(r)}_{i-1}\big{)},\forall i\geq
1$. Then there exist $c_{2},c_{3}>0$, not depending on $r$, such that for any
$a\geq 4\mu$,

$\sup_{r\in R}\mathbb{P}\big{(}\sum_{i=1}^{n}\Phi^{(r)}_{i}\geq
an|\mathcal{F}^{(r)}_{0}\big{)}\leq
c_{2}\left(1+\frac{n^{\frac{1}{2+\theta}}}{a^{\frac{\theta}{2+\theta}}}\right)\exp\big{\\{}-c_{3}(a^{2}n)^{\frac{\theta}{2+\theta}}\big{\\}},\quad
n\geq 1.$

###### Proof.

Proof. Let
$\tilde{\Phi}^{(r)}_{i}:=\Phi^{(r)}_{i}\mathds{1}[\Phi^{(r)}_{i}\leq d]$ for
some $d>0$ that does not depend on $r$, to be chosen later. Then by Azuma’s
inequality,

$\begin{split}\sup_{r\in
R}\mathbb{P}\big{(}\sum_{i=1}^{n}\tilde{\Phi}^{(r)}_{i}\geq\frac{an}{2}|\mathcal{F}^{(r)}_{0}\big{)}\leq&\sup_{r\in
R}\mathbb{P}\big{(}\sum_{i=1}^{n}(\tilde{\Phi}^{(r)}_{i}-\mathbb{E}(\tilde{\Phi}^{(r)}_{i}|\mathcal{F}_{i-1}))\geq\frac{an}{2}-n\mu|\mathcal{F}^{(r)}_{0}\big{)}\\\
\leq&\sup_{r\in
R}\mathbb{P}\big{(}\sum_{i=1}^{n}(\tilde{\Phi}^{(r)}_{i}-\mathbb{E}(\tilde{\Phi}^{(r)}_{i}|\mathcal{F}_{i-1}))\geq\frac{an}{4}|\mathcal{F}^{(r)}_{0}\big{)}\\\
\leq&\exp\big{\\{}-\frac{a^{2}n}{32d^{2}}\big{\\}}.\end{split}$ (A.1)

Moreover, using Markov’s inequality and the union bound

$\sup_{r\in
R}\mathbb{P}\big{(}\sum_{i=1}^{n}(\Phi^{(r)}_{i}-\tilde{\Phi}^{(r)}_{i})\geq\frac{an}{2}|\mathcal{F}^{(r)}_{0}\big{)}\leq\frac{2}{a}\left[dce^{-c_{1}d^{\theta}}+\int_{d}^{\infty}ce^{-c_{1}x^{\theta}}dx\right]\leq\frac{\tilde{c}de^{-\tilde{c}^{\prime}d^{\theta}}}{a},$
(A.2)

where $\tilde{c}$ and $\tilde{c}^{\prime}$ are positive constants not
depending on $d$. Using (A.1) and (A.2) with
$d=(a^{2}n)^{\frac{1}{2+\theta}}$, the lemma follows upon using

$\sup_{r\in R}\mathbb{P}\big{(}\sum_{i=1}^{n}\Phi^{(r)}_{i}\geq
an|\mathcal{F}^{(r)}_{0}\big{)}\leq\sup_{r\in
R}\mathbb{P}\big{(}\sum_{i=1}^{n}\tilde{\Phi}^{(r)}_{i}\geq\frac{an}{2}|\mathcal{F}^{(r)}_{0}\big{)}+\sup_{r\in
R}\mathbb{P}\big{(}\sum_{i=1}^{n}(\Phi^{(r)}_{i}-\tilde{\Phi}^{(r)}_{i})\geq\frac{an}{2}|\mathcal{F}^{(r)}_{0}\big{)},$

and choosing appropriate constants $c_{2}$ and $c_{3}$. ∎

##  Appendix B Integral and supremum of the upper-bounding birth-death process

The goal of this appendix is to prove Lemma 3.1. Define the following stopping
time and expectation notation: For any $x\geq 0$ and $\eta>0$,

$\bar{\tau}^{(N)}_{B}(x):=\inf\big{\\{}t\geq 0:\bar{I}^{(N)}_{B}(t)=\lfloor
xN^{\frac{1}{2}-\varepsilon}\rfloor\big{\\}}\quad\text{and}\quad\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}\cdot\big{)}:=\mathbb{E}\big{(}\cdot|\bar{I}_{B}^{(N)}(0)=\lfloor\eta
N^{\frac{1}{2}-\varepsilon}\rfloor\big{)}.$

The next lemma is used to control the time taken by $\bar{I}_{B}^{(N)}$ to hit
a level $\lfloor yN^{\frac{1}{2}-\varepsilon}\rfloor$ starting from a larger
point $\lfloor\eta N^{\frac{1}{2}-\varepsilon}\rfloor$ with $\eta>y$.

######  Lemma B.1.

There exists $B_{0}\geq 1,N_{0}\geq 1$ such that for any $\eta>0$, $B\geq
B_{0}$, $N\geq N_{0}$, and $y\in[0,\eta)$,

$\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\left(\exp\left\\{\frac{BN^{2\varepsilon}}{8}\,\bar{\tau}^{(N)}_{B}(y)\right\\}\right)\leq
e^{\eta-y}.$

###### Proof.

Proof. Take any $B>2\beta$. Let us denote
$\hat{I}^{(N)}_{B}(t)=\frac{\bar{I}^{(N)}_{B}(t)}{N^{\frac{1}{2}-\varepsilon}}$
and $\sigma^{(N)}=\bar{\tau}^{(N)}_{B}(y)$. For $\theta>0$ to be chosen later,
define
$Z^{(N)}_{B}(t):=\exp\big{\\{}\hat{I}^{(N)}_{B}(t)+\frac{\theta}{2}t\big{\\}}.$
For $t<\bar{\tau}_{B}^{(N)}(0)$,

$\displaystyle\mathcal{L}Z^{(N)}_{B}(t)$
$\displaystyle=\frac{\theta}{2}Z^{(N)}_{B}(t)+e^{\frac{\theta}{2}t}\Big{[}\big{(}e^{\hat{I}^{(N)}_{B}(t)+\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-e^{\hat{I}^{(N)}_{B}(t)}\big{)}\big{(}N-BN^{\frac{1}{2}+\varepsilon}\big{)}$
$\displaystyle\hskip
156.49014pt+\big{(}e^{\hat{I}^{(N)}_{B}(t)-\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-e^{\hat{I}^{(N)}_{B}(t)}\big{)}\big{(}N-\beta
N^{\frac{1}{2}-\varepsilon}\big{)}\Big{]}$
$\displaystyle=\frac{\theta}{2}Z^{(N)}_{B}(t)+Z^{(N)}_{B}(t)N(e^{\frac{1}{N^{\frac{1}{2}-\varepsilon}}}+e^{-\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-2)$
$\displaystyle\hskip
85.35826pt+Z^{(N)}_{B}(t)\big{[}-\big{(}e^{\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-1\big{)}BN^{\frac{1}{2}+\varepsilon}-\big{(}e^{-\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-1\big{)}\beta
N^{\frac{1}{2}-\varepsilon}\big{]},$

where $\mathcal{L}(\cdot)$ is the infinitesimal generator of the associated
continuous time Markov process. Fix two constants $c,a>0$, such that for all
$x\in(-a,a)$, $e^{x}+e^{-x}-2\leq cx^{2},$ and note that for all
$x\in\mathbb{R}$, $e^{x}-1\geq x.$ Now, let $N_{0}\geq 1$ be such that
$N^{-\frac{1}{2}+\varepsilon}<a$ for all $N\geq N_{0}$. Also, as $B>2\beta$,
$\frac{1}{2}BN^{2\varepsilon}>\beta$ for all $N\geq 1$. Then for all $N\geq
N_{0}$,

$\displaystyle\mathcal{L}Z^{(N)}_{B}(t\wedge\sigma^{(N)})$
$\displaystyle\leq\frac{\theta}{2}Z^{(N)}_{B}(t\wedge\sigma^{(N)})+cN^{2\varepsilon}Z^{(N)}_{B}(t\wedge\sigma^{(N)})$
$\displaystyle\quad+Z^{(N)}_{B}(t\wedge\sigma^{(N)})\big{[}-\big{(}e^{\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-1\big{)}BN^{\frac{1}{2}+\varepsilon}-\big{(}e^{-\frac{1}{N^{\frac{1}{2}-\varepsilon}}}-1\big{)}\beta
N^{\frac{1}{2}-\varepsilon}\big{]}$
$\displaystyle\leq\frac{\theta}{2}Z^{(N)}_{B}(t\wedge\sigma^{(N)})+cN^{2\varepsilon}Z^{(N)}_{B}(t\wedge\sigma^{(N)})-\frac{BN^{\frac{1}{2}+\varepsilon}}{N^{\frac{1}{2}-\varepsilon}}Z^{(N)}_{B}(t\wedge\sigma^{(N)})+\beta
Z^{(N)}_{B}(t\wedge\sigma^{(N)})$
$\displaystyle\leq\frac{\theta}{2}Z^{(N)}_{B}(t\wedge\sigma^{(N)})+cN^{2\varepsilon}Z^{(N)}_{B}(t\wedge\sigma^{(N)})-\frac{BN^{2\varepsilon}}{2}Z^{(N)}_{B}(t\wedge\sigma^{(N)}).$

Let $B_{0}\geq 1+2\beta$ be such that $c-\frac{B}{2}\leq-\frac{B}{4}$,
$\forall B\geq B_{0}$. Take $\theta=\frac{B}{4}N^{2\varepsilon}$. Then
$\mathcal{L}Z^{(N)}_{B}(t\wedge\sigma^{(N)})\leq 0$, $\forall t\geq 0$,
implying for all $B\geq B_{0}$,

$\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}Z^{(N)}_{B}(t\wedge\sigma^{(N)})\big{)}\leq\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}Z^{(N)}_{B}(0)\big{)},\quad\forall t\geq
0.$

By Fatou’s lemma and the observation
$\lim_{t\rightarrow\infty}Z^{(N)}_{B}(t\wedge\sigma^{(N)})=Z^{(N)}_{B}(\sigma^{(N)})$
a.s., we get

$\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}Z^{(N)}_{B}(\sigma^{(N)})\big{)}\leq\liminf_{t\rightarrow\infty}\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}Z^{(N)}_{B}(t\wedge\sigma^{(N)})\big{)}\leq\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}Z^{(N)}_{B}(0)\big{)}\leq e^{\eta},$

implying that $\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}e^{\frac{\theta}{2}\bar{\tau}^{(N)}_{B}(y)}\big{)}\leq
e^{\eta-y}.$ This completes the proof of the lemma. ∎

Fix $\eta=1\wedge(\beta/8)$, and recall the notations introduced below the
statement of Lemma B.2, such as
$\bar{\sigma}^{(N)}_{i},\bar{\xi}^{(N)}_{i},\bar{u}^{(N)}_{i},$ and
$\bar{K}^{(N)}_{t}$, and the following upper bound on the integral of
$\bar{I}_{B}^{(N)}$:

$\int_{0}^{N^{2\varepsilon}t}\bar{I}_{B}^{(N)}(s)ds\leq\eta
N^{\frac{1}{2}+\varepsilon}t+\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\leq\frac{\beta}{8}N^{\frac{1}{2}+\varepsilon}t+\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}.$

In the next few lemmas, probability bounds on the integral are obtained by
analyzing the sum in the above upper bound. Lemma B.2 furnishes control on
$\bar{u}^{(N)}_{1}\bar{\xi}^{(N)}_{1}$ which stochastically dominates the
integral of $\bar{I}^{(N)}_{B}$ over the excursion interval
$[\bar{\sigma}^{(N)}_{2i-1},\bar{\sigma}^{(N)}_{2i}]$. Lemma B.3 bounds the
number of excursions $\bar{K}^{(N)}_{t}$ of $\bar{I}_{B}^{(N)}$ on the time
interval $[0,N^{2\varepsilon}t]$ for large enough $t$.

######  Lemma B.2.

Take $B_{0}$ as in Lemma B.1. There exist positive constants $\bar{c}$,
$c_{1}$, $c_{2}$, such that for any $B\geq B_{0}$, $\exists\tilde{N}_{B}>0$
such that for all $N\geq\tilde{N}_{B}$,

  1. (i)

$\mathbb{P}\big{(}\bar{u}^{(N)}_{1}\bar{\xi}^{(N)}_{1}\geq\frac{xN^{\frac{1}{2}-3\varepsilon}}{B^{\frac{3}{2}}}\big{)}\leq
c_{1}e^{-c_{2}\sqrt{x}},\quad\forall x\geq 0$;

  2. (ii)

$\mathbb{E}\big{(}\bar{u}^{(N)}_{1}\bar{\xi}^{(N)}_{1}\big{)}\leq\frac{\bar{c}N^{\frac{1}{2}-3\varepsilon}}{B^{\frac{3}{2}}}$.

###### Proof.

Proof. Note that

$\mathbb{P}\big{(}\bar{u}^{(N)}_{1}\bar{\xi}^{(N)}_{1}\geq\frac{xN^{\frac{1}{2}-3\varepsilon}}{B^{\frac{3}{2}}}\big{)}\leq\mathbb{P}\big{(}\bar{\xi}^{(N)}_{1}\geq\frac{\sqrt{x}}{B}N^{-2\varepsilon}\big{)}+\mathbb{P}\big{(}\bar{u}^{(N)}_{1}\geq\frac{\sqrt{x}}{\sqrt{B}}N^{\frac{1}{2}-\varepsilon},\bar{\xi}^{(N)}_{1}<\frac{\sqrt{x}}{B}N^{-2\varepsilon}\big{)}.$
(B.1)

By Lemma B.1 and Markov’s inequality, for $B\geq B_{0}$, $N\geq N_{0}$, $x\geq
0$,

$\mathbb{P}\big{(}\bar{\xi}^{(N)}_{1}\geq\frac{\sqrt{x}}{B}N^{-2\varepsilon}\big{)}\leq
e^{-\frac{\theta\sqrt{x}}{2BN^{2\varepsilon}}}\mathbb{E}_{\eta
N^{\frac{1}{2}-\varepsilon}}\big{(}e^{\frac{\theta}{2}\bar{\tau}^{(N)}_{B}(\frac{\eta}{2})}\big{)}\leq
e^{\eta}e^{-\sqrt{x}/8},$ (B.2)

where $\theta=BN^{2\varepsilon}/4$ as in Lemma B.1. Now, for
$s\in[0,\sigma^{(N)}_{2}-\sigma^{(N)}_{1})$,

$\displaystyle\bar{I}_{B}^{(N)}\big{(}\sigma^{(N)}_{1}+s\big{)}$
$\displaystyle=\lfloor\eta
N^{\frac{1}{2}-\varepsilon}\rfloor+A_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-A_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}$ $\displaystyle=\lfloor\eta
N^{\frac{1}{2}-\varepsilon}\rfloor+\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}-BN^{\frac{1}{2}+\varepsilon}s+\beta
N^{\frac{1}{2}-\varepsilon}s,$

where $A_{1}$, $A_{2}$ are i.i.d. Poisson processes with unit rate and
$\hat{A}_{i}(s)=A_{i}(s)-s$, $i=1,2$. Therefore, choosing $\tilde{N}_{B}\geq
N_{0}$ such that
$\frac{\sqrt{x}}{2\sqrt{B}}N^{\frac{1}{2}-\varepsilon}\leq\frac{4\sqrt{x}}{B}N^{1-2\varepsilon}$
for all $N\geq\tilde{N}_{B}$, we obtain for any $N\geq\tilde{N}_{B}$, $x\geq
0$,

$\begin{split}&\mathbb{P}\big{(}\bar{u}^{(N)}_{1}\geq\frac{\sqrt{x}}{\sqrt{B}}N^{\frac{1}{2}-\varepsilon},\bar{\xi}^{(N)}_{1}<\frac{\sqrt{x}}{B}N^{-2\varepsilon}\big{)}\\\
&=\mathbb{P}\big{(}\sup_{s\in[0,\sigma^{(N)}_{2}-\sigma^{(N)}_{1})}\bar{I}_{B}^{(N)}(\sigma^{(N)}_{1}+s)\geq\lfloor\eta
N^{\frac{1}{2}-\varepsilon}\rfloor+\frac{\sqrt{x}}{\sqrt{B}}N^{\frac{1}{2}-\varepsilon},\
\sigma^{(N)}_{2}-\sigma^{(N)}_{1}<\frac{\sqrt{x}}{B}N^{-2\varepsilon}\big{)}\\\
&\leq\mathbb{P}\big{(}\sup_{s\in[0,\sqrt{x}/(BN^{2\varepsilon}))}\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}-BN^{\frac{1}{2}+\varepsilon}s+\beta
N^{\frac{1}{2}-\varepsilon}s\geq\frac{\sqrt{x}}{\sqrt{B}}N^{\frac{1}{2}-\varepsilon}\big{)}\\\
&\leq\mathbb{P}\big{(}\sup_{s\in[0,\sqrt{x}/(BN^{2\varepsilon}))}\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}\geq\frac{\sqrt{x}}{\sqrt{B}}N^{\frac{1}{2}-\varepsilon}\big{)}\\\
&\leq
2\mathbb{P}\big{(}\sup_{s\in[0,\sqrt{x}/(BN^{2\varepsilon}))}|\hat{A}_{1}\big{(}Ns\big{)}|\geq\frac{\sqrt{x}}{2\sqrt{B}}N^{\frac{1}{2}-\varepsilon}\big{)}\leq
4\exp\big{\\{}-\frac{\sqrt{x}}{32}\big{\\}},\end{split}$ (B.3)

where the second inequality is due to $BN^{\frac{1}{2}+\varepsilon}>\beta
N^{\frac{1}{2}-\varepsilon}$ that we assumed while defining
$\bar{I}^{(N)}_{B}$, and the last inequality is due to Lemma A.1. Plugging
(B.2) and (B.3) into (B.1), we get part (i). Part (ii) follows directly from
part (i). ∎

######  Lemma B.3.

There exist $\tilde{c},t_{0}>0$ such that for all $B\geq 1$,
$a\geq\frac{256B}{\eta^{2}}$, $N\geq 1$, and $t\geq t_{0}$,

$\mathbb{P}\big{(}\bar{K}^{(N)}_{t}\geq
aN^{4\varepsilon}t\big{)}\leq\exp\\{-\tilde{c}aN^{4\varepsilon}t\\}.$

###### Proof.

Proof.Let
$H^{(N)}_{i}=\mathds{1}\big{[}\bar{\sigma}^{(N)}_{2i+1}-\bar{\sigma}^{(N)}_{2i}>\frac{\eta^{2}}{64B}N^{-2\varepsilon}\big{]}$,
$i\geq 1$. Recall $A_{i}$, $i=1,2$ as defined in the proof of Lemma B.2. Also,
define the one-dimensional Skorohod map $\Psi$ as follows: for any function
$x:\mathbb{R}\to\mathbb{R}$ having càdlàg paths, $\Psi[x](t):=x(t)+\sup_{0\leq
s\leq t}\max\\{-x(s),0\\},\,t\geq 0.$ Also write

$x^{(N)}(s):=\lfloor\frac{\eta}{2}N^{\frac{1}{2}-\varepsilon}\rfloor+\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}-BN^{\frac{1}{2}+\varepsilon}s+\beta
N^{\frac{1}{2}-\varepsilon}s,\,s\geq 0.$

Then, note that

$\displaystyle\mathbb{P}\Big{(}H^{(N)}_{i}=0\Big{)}=\mathbb{P}\Big{(}\sup_{s\in[0,\bar{\sigma}^{(N)}_{2i+1}-\bar{\sigma}^{(N)}_{2i}]}\bar{I}_{B}^{(N)}\big{(}\sigma^{(N)}_{2i}+s\big{)}\geq\eta
N^{\frac{1}{2}-\varepsilon},\bar{\sigma}^{(N)}_{2i+1}-\bar{\sigma}^{(N)}_{2i}\leq\frac{\eta^{2}}{32}N^{-2\varepsilon}\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sup_{s\in[0,\frac{\eta^{2}}{64B}N^{-2\varepsilon}]}\Psi\Big{[}x^{(N)}\Big{]}(s)\geq\eta
N^{\frac{1}{2}-\varepsilon}\Big{)}\leq\mathbb{P}\Big{(}\sup_{s\in[0,\frac{\eta^{2}}{64B}N^{-2\varepsilon}]}\Big{|}x^{(N)}(s)\Big{|}\geq\frac{\eta}{2}N^{\frac{1}{2}-\varepsilon}\Big{)},$
(B.4)

where, in the last step, we used the fact that
$\sup_{s\in[0,T]}\Psi[x^{(N)}](s)\leq 2\sup_{s\in[0,T]}|x^{(N)}(s)|$ for any
$T\geq 0$. Now, since $BN^{\frac{1}{2}+\varepsilon}>\beta
N^{\frac{1}{2}-\varepsilon}$ as before and $\eta\leq 1$, we have

$\displaystyle\sup_{s\in[0,\frac{\eta^{2}}{64B}N^{-2\varepsilon}]}\Big{|}x^{(N)}(s)\Big{|}$
$\displaystyle\leq\sup_{s\in[0,\frac{\eta^{2}}{64B}N^{-2\varepsilon}]}\big{|}\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}\big{|}+\frac{\eta}{2}N^{\frac{1}{2}-\varepsilon}+\frac{\eta^{2}}{64}N^{\frac{1}{2}-\varepsilon}$
$\displaystyle\leq\sup_{s\in[0,\frac{\eta^{2}}{64B}N^{-2\varepsilon}]}\big{|}\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}\big{|}+\frac{3\eta}{4}N^{\frac{1}{2}-\varepsilon}.$

Using this in (B), we obtain

$\displaystyle\mathbb{P}\Big{(}H^{(N)}_{i}=0\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sup_{s\in[0,\frac{\eta^{2}}{64B}N^{-2\varepsilon}]}\big{|}\hat{A}_{1}\big{(}(N-BN^{\frac{1}{2}+\varepsilon})s\big{)}-\hat{A}_{2}\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})s\big{)}\big{|}\geq\frac{\eta}{4}N^{\frac{1}{2}-\varepsilon}\Big{)}$
$\displaystyle\leq\frac{2(\eta^{2}/64B)N^{1-2\varepsilon}}{(\eta^{2}/16)N^{1-2\varepsilon}}\leq\frac{1}{2},$

where the last inequality follows from Doob’s $L^{2}$-maximal inequality and
the assumption $B\geq 1$. Then, for $a\geq\frac{256B}{\eta^{2}}$ we can write
for $t\geq t_{0}$ sufficiently large,

$\displaystyle\mathbb{P}\Big{(}\bar{K}^{(N)}_{t}\geq
aN^{4\varepsilon}t\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\left\lfloor{atN^{4\varepsilon}}\right\rfloor}(\bar{\sigma}^{(N)}_{2i+1}-\bar{\sigma}^{(N)}_{2i})<N^{2\varepsilon}t\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\left\lfloor{atN^{4\varepsilon}}\right\rfloor}H^{(N)}_{i}<\frac{64B}{\eta^{2}}N^{4\varepsilon}t\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\left\lfloor{atN^{4\varepsilon}}\right\rfloor}(H^{(N)}_{i}-\mathbb{E}(H^{(N)}_{i}))<\frac{64B}{\eta^{2}}N^{4\varepsilon}t-\frac{1}{2}\left\lfloor{atN^{4\varepsilon}}\right\rfloor\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\left\lfloor{atN^{4\varepsilon}}\right\rfloor}(H^{(N)}_{i}-\mathbb{E}(H^{(N)}_{i}))<-\frac{1}{8}atN^{4\varepsilon}\Big{)}$
$\displaystyle\leq\exp\\{-\frac{\tilde{c}a^{2}N^{8\varepsilon}t^{2}}{aN^{4\varepsilon}t}\\}=\exp\\{-\tilde{c}aN^{4\varepsilon}t\\}$

for some $\tilde{c}>0$, where the last inequality above follows from Azuma’s
inequality. ∎

The following lemma gives bounds on a random sum that appears in (3.2). It
will be used to bound the integral
$\int_{0}^{N^{2\varepsilon}t}\bar{I}_{B}^{(N)}(s)ds$.

######  Lemma B.4.

Take $B_{0}$ as in Lemma B.1, $\tilde{N}_{B}$ for $B\geq B_{0}$ as in Lemma
B.2, and $t_{0}$ as in Lemma B.3. There exist constants
$c,c^{*},c^{\prime}>0$, such that the following holds for all $B\geq B_{0}$,
$N\geq\tilde{N}_{B}$, $t\geq t_{0}$,

$\mathbb{P}\big{(}\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\geq\frac{c^{*}}{\eta^{2}B^{\frac{1}{2}}}N^{\frac{1}{2}+\varepsilon}t\big{)}\leq
c\exp\\{-c^{\prime}B^{\frac{1}{5}}N^{\frac{4\varepsilon}{5}}t^{\frac{1}{5}}\\}.$

###### Proof.

Proof. Write

$\mathbb{P}\big{(}\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\geq\frac{c^{*}}{\eta^{2}B^{\frac{3}{2}}}N^{\frac{1}{2}+\varepsilon}t\big{)}\leq\mathbb{P}\big{(}\bar{K}^{(N)}_{t}\geq
a^{\prime}N^{4\varepsilon}t\big{)}+\mathbb{P}\big{(}\sum_{i=1}^{\lfloor
a^{\prime}N^{4\varepsilon}t\rfloor}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\geq\frac{c^{*}}{\eta^{2}B^{\frac{3}{2}}}N^{\frac{1}{2}+\varepsilon}t\big{)}.$
(B.5)

For the first term, we take $a^{\prime}=\frac{256B}{\eta^{2}}$, and thus,
Lemma B.3 yields

$\mathbb{P}\big{(}\bar{K}^{(N)}_{t}\geq
a^{\prime}N^{4\varepsilon}t\big{)}\leq\exp\\{-\tilde{c}a^{\prime}N^{4\varepsilon}t\\}.$
(B.6)

For the second term, define
$\Phi_{i}^{(N)}:=\frac{B^{3/2}}{N^{\frac{1}{2}-3\varepsilon}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i},\quad
i\geq 1,$ and a sequence of filtrations
$\mathcal{F}_{N,i}:=\sigma\\{\Phi_{j}^{(N)}:j\leq i\\}$, $i\geq 1$. Then, as
$\\{\Phi_{j}^{(N)}:i\in\mathbb{N}\\}$ are i.i.d., from Lemma B.2 we know for
$B\geq B_{0}$ and $N\geq\tilde{N}_{B}$, $\mathbb{P}\big{(}\Phi_{i}^{(N)}\geq
x|\mathcal{F}_{N,i-1}\big{)}\leq c_{1}e^{-c_{2}\sqrt{x}},\quad\forall x\geq
0.$ We will use Lemma A.2 for the random variables $\\{\Phi_{i}^{(N)}:i\geq
1,N\in\mathbb{N}\\}$. Note that $\mu$ in the lemma takes the form
$\mu=\int_{0}^{\infty}c_{1}e^{-c_{2}\sqrt{x}}dx$ in our case. Write
$c^{*}=4\mu\times 256$. Applying Lemma A.2 with $\theta=1/2$, $n=\lfloor
a^{\prime}N^{4\varepsilon}t\rfloor$, and $a=4\mu$, we get for some constants
$c^{\prime}_{1},c^{\prime}_{2}>0$,

$\begin{split}\mathbb{P}\big{(}\sum_{i=1}^{\lfloor
a^{\prime}N^{4\varepsilon}t\rfloor}&\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\geq\frac{c^{*}}{\eta^{2}B^{\frac{1}{2}}}N^{\frac{1}{2}+\varepsilon}t|\mathcal{F}_{N,0}\big{)}=\mathbb{P}\big{(}\sum_{i=1}^{\lfloor
a^{\prime}N^{4\varepsilon}t\rfloor}\frac{N^{\frac{1}{2}-3\varepsilon}}{B^{\frac{3}{2}}}\Phi_{i}^{(N)}\geq\frac{c^{*}}{\eta^{2}B^{\frac{1}{2}}}N^{\frac{1}{2}+\varepsilon}t|\mathcal{F}_{N,0}\big{)}\\\
&=\mathbb{P}\Big{(}\sum_{i=1}^{\lfloor
a^{\prime}N^{4\varepsilon}t\rfloor}\Phi_{i}^{(N)}\geq\frac{c^{*}B}{\eta^{2}}N^{4\varepsilon}t|\mathcal{F}_{N,0}\Big{)}\leq
c^{\prime}_{1}\exp\Big{\\{}-c^{\prime}_{2}\Big{[}\Big{(}4\mu\Big{)}^{2}a^{\prime}N^{4\varepsilon}t\Big{]}^{1/5}\Big{\\}}.\end{split}$
(B.7)

Plugging the bounds from (B.6) and (B.7) into (B.5) completes the proof of the
lemma for appropriately chosen constants $c$ and $c^{\prime}$ dependent on
$\beta$ but not $B$. ∎

We now have all the required ingredients to prove Lemma 3.1.

###### Proof.

Proof of Lemma 3.1. (i) Recall $\eta=1\wedge(\beta/8)$ and

$\int_{0}^{N^{2\varepsilon}t}\bar{I}_{B}^{(N)}(s)ds\leq\eta
N^{\frac{1}{2}+\varepsilon}t+\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\leq\frac{\beta}{8}N^{\frac{1}{2}+\varepsilon}t+\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}.$
(B.8)

Now, choose $B_{1}\geq B_{0}$ (from Lemma B.1) such that
$\frac{c^{*}}{\eta^{2}B_{1}^{\frac{1}{2}}}\leq\frac{\beta}{8}$, where $c^{*}$
is the constant from Lemma B.4. By Lemma B.4 and (B.8), there exist constants
$c_{1}$ and $c_{2}$, such that for all $B\geq B_{1}$, $t\geq t_{0}$,
$N\geq\tilde{N}_{B}$,

$\displaystyle\mathbb{P}\Big{(}\int_{0}^{N^{2\varepsilon}t}\bar{I}_{B}^{(N)}(s)ds\geq\frac{\beta}{4}N^{\frac{1}{2}+\varepsilon}t\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\geq\frac{\beta}{8}N^{\frac{1}{2}+\varepsilon}t\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\bar{K}^{(N)}_{t}}\bar{u}^{(N)}_{i}\bar{\xi}^{(N)}_{i}\geq\frac{c^{*}}{\eta^{2}B^{\frac{1}{2}}}N^{\frac{1}{2}+\varepsilon}t\Big{)}\leq
c_{1}\exp\\{-c_{2}B^{\frac{1}{5}}N^{\frac{4\varepsilon}{5}}t^{\frac{1}{5}}\\}.$
(B.9)

For any $t\geq t_{0}$, define $s_{k}=kt,k\geq 1$. Then, for any $k\geq 1$,

$\displaystyle\mathbb{P}\Big{(}\int_{0}^{N^{2\varepsilon}s}\bar{I}_{B}^{(N)}(u)du\geq\frac{\beta
s}{2}N^{\frac{1}{2}+\varepsilon}\text{ for some }s\in[s_{k},s_{k+1})\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\int_{0}^{N^{2\varepsilon}s_{k+1}}\bar{I}_{B}^{(N)}(u)du\geq\frac{\beta
s_{k+1}}{4}N^{\frac{1}{2}+\varepsilon}\Big{)}\leq
c_{1}\exp\\{-c_{2}B^{\frac{1}{5}}N^{\frac{4\varepsilon}{5}}s_{k+1}^{\frac{1}{5}}\\}.$

Therefore, i follows on summing over $k$ and applying the union bound.

(ii) Write $a=\frac{256B}{\eta^{2}}$, and note that

$\mathbb{P}\Big{(}\sup_{s\leq
t}\bar{I}_{B}^{(N)}(N^{2\varepsilon}s)\geq\frac{\beta}{4}N^{\frac{1}{2}-\varepsilon}t\Big{)}\leq\mathbb{P}\Big{(}\bar{K}^{(N)}_{t}\geq
aN^{4\varepsilon}t\Big{)}+\mathbb{P}\Big{(}\sup_{1\leq i\leq
aN^{4\varepsilon}t}\bar{u}^{(N)}_{i}\geq\frac{\beta}{4}N^{\frac{1}{2}-\varepsilon}t-\eta
N^{\frac{1}{2}-\varepsilon}\Big{)}.$ (B.10)

By Lemma B.3, for $t\geq t_{0}$,

$\mathbb{P}\big{(}\bar{K}^{(N)}_{t}\geq
aN^{4\varepsilon}t\big{)}\leq\exp\\{-\tilde{c}aN^{4\varepsilon}t\\}.$ (B.11)

Recall the upper bounds from Equations (B.2) and (B.3), and note that for
$B\geq B_{0}$, and large enough $N\geq\tilde{N}_{B}$,

$\displaystyle\mathbb{P}\Big{(}\sup_{1\leq i\leq
aN^{4\varepsilon}t}\bar{u}^{(N)}_{i}\geq\frac{\beta}{4}N^{\frac{1}{2}-\varepsilon}t-\eta
N^{\frac{1}{2}-\varepsilon}\Big{)}$ $\displaystyle\leq$
$\displaystyle\mathbb{P}\Big{(}\sup_{1\leq i\leq
aN^{4\varepsilon}t}\bar{u}^{(N)}_{i}\geq\frac{\beta}{8}N^{\frac{1}{2}-\varepsilon}t\Big{)}$
$\displaystyle\leq$ $\displaystyle
aN^{4\varepsilon}t\cdot\Big{[}\mathbb{P}\big{(}\bar{u}^{(N)}_{1}\geq\frac{\beta}{8}N^{\frac{1}{2}-\varepsilon}t,\bar{\xi}^{(N)}_{1}<\frac{\beta
tN^{-2\varepsilon}}{8\sqrt{B}}\big{)}+\mathbb{P}\big{(}\bar{\xi}^{(N)}_{1}\geq\frac{\beta
tN^{-2\varepsilon}}{8\sqrt{B}}\big{)}\Big{]}$ $\displaystyle\leq$
$\displaystyle aN^{4\varepsilon}t\cdot
c_{0}\exp\\{-c^{\prime}_{0}\beta\sqrt{B}t\\}\text{ for some constants
}c_{0},c^{\prime}_{0}>0,$ (B.12)

where the last inequality is due to (B.2) and (B.3). Plugging (B.11) and (B)
into (B.10) and choosing appropriate constants
$\tilde{c}_{1},\tilde{c}_{2},\tilde{c}_{3}$ complete the proof. ∎

##  Appendix C Proofs from Section 3 (sample path analysis).

Here, we give proofs of some lemmas that were used to prove the main results
in Section 3.

###### Proof.

Proof of Lemma 3.8. Fix $N^{\frac{1}{2}-\varepsilon}\geq x\geq x_{B}$, where
$x_{B}\geq 2B$ will be chosen later. Define the stopping time
$\hat{\sigma}^{(N)}_{i}\coloneqq\inf\\{s\geq\sigma^{(N)}_{2i}:Q^{(N)}_{2}(s)\geq
xN^{\frac{1}{2}+\varepsilon}\text{ or }Q^{(N)}_{2}(s)\leq\lfloor
BN^{\frac{1}{2}+\varepsilon}\rfloor\\},\ i\geq 0.$ Note that, as $x\leq
N^{\frac{1}{2}-\varepsilon}$, when $Q^{(N)}_{3}(\sigma^{(N)}_{2i})=0$,
$Q^{(N)}_{3}(t)=0$ for all $t\in[\sigma^{(N)}_{2i},\hat{\sigma}^{(N)}_{i}]$,
and in that case, we have $S^{(N)}(t)=N-I^{(N)}(t)+Q^{(N)}_{2}(t)$. Therefore,
we obtain for $t\in[0,\hat{\sigma}^{(N)}_{i}-\sigma^{(N)}_{2i}]$,

$\displaystyle
Q^{(N)}_{2}(t+\sigma^{(N)}_{2i})-I^{(N)}(t+\sigma^{(N)}_{2i})-(Q^{(N)}_{2}(\sigma^{(N)}_{2i})-I^{(N)}(\sigma^{(N)}_{2i}))=S^{(N)}(t+\sigma^{(N)}_{2i})-S^{(N)}(\sigma^{(N)}_{2i})$
$\displaystyle=\left[A\left((N-\beta
N^{\frac{1}{2}-\varepsilon})(t+\sigma^{(N)}_{2i})\right)-A\left((N-\beta
N^{\frac{1}{2}-\varepsilon})\sigma^{(N)}_{2i}\right)\right]-D\left(\int_{\sigma^{(N)}_{2i}}^{\sigma^{(N)}_{2i}+t}(N-I^{(N)}(s))ds\right)$
$\displaystyle=\left[\hat{A}\left((N-\beta
N^{\frac{1}{2}-\varepsilon})(t+\sigma^{(N)}_{2i})\right)-\hat{A}\left((N-\beta
N^{\frac{1}{2}-\varepsilon})\sigma^{(N)}_{2i}\right)\right]-\hat{D}\left(\int_{\sigma^{(N)}_{2i}}^{\sigma^{(N)}_{2i}+t}(N-I^{(N)}(s))ds\right)$
$\displaystyle\qquad+\int_{\sigma^{(N)}_{2i}}^{\sigma^{(N)}_{2i}+t}I^{(N)}(s)ds-\beta
N^{\frac{1}{2}-\varepsilon}t,$

where $\hat{A}(s)=A(s)-s$ and $\hat{D}(s)=D(s)-s$. Using the above and the
strong Markov property at time $\sigma^{(N)}_{2i}$, we can write

$\begin{split}&\mathbb{P}\Big{(}\sup_{s\in[\sigma^{(N)}_{2i},\sigma^{(N)}_{2i+1}]}Q^{(N)}_{2}(s)\geq
xN^{\frac{1}{2}+\varepsilon}\ \Big{|}\
Q^{(N)}_{3}(\sigma^{(N)}_{2i})=0\Big{)}\\\
&\leq\mathbb{P}\left(\hat{\sigma}^{(N)}_{i}-\sigma^{(N)}_{2i}\geq
N^{2\varepsilon}t\ \Big{|}\ Q^{(N)}_{3}(\sigma^{(N)}_{2i})=0\right)\\\
&\qquad+\mathbb{P}\Big{(}\sup_{s\in[\sigma^{(N)}_{2i},\sigma^{(N)}_{2i}+(N^{2\varepsilon}t)\wedge(\hat{\sigma}^{(N)}_{i}-\sigma^{(N)}_{2i})]}(Q^{(N)}_{2}(s)-I^{(N)}(s))\geq
xN^{\frac{1}{2}+\varepsilon}\ \Big{|}\
Q^{(N)}_{3}(\sigma^{(N)}_{2i})=0\Big{)}\\\ &\leq\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\big{(}\tau^{(N)}_{2}(B)\geq
N^{2\varepsilon}t\big{)}\\\ &\qquad+\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\Big{(}\sup_{s\in[0,N^{2\varepsilon}t\wedge\tau_{2}^{(N)}(B)]}\left(2BN^{\frac{1}{2}+\varepsilon}+\hat{\mathcal{M}}(s)+\int_{0}^{s}I^{(N)}(u)du-\beta
N^{\frac{1}{2}-\varepsilon}s\right)\geq
xN^{\frac{1}{2}+\varepsilon}\Big{)},\end{split}$ (C.1)

where $\hat{\mathcal{M}}(s):=\hat{A}\left((N-\beta
N^{\frac{1}{2}-\varepsilon})s\right)-\hat{D}\left(\int_{0}^{s}(N-I^{(N)}(u))du\right)$,
and the first equality above is due to the fact that $Q^{(N)}_{2}$ can
increase only when $I^{(N)}=0$. Now, for the first term in the above bound,
observe that by Proposition 3.4, for $B\geq B_{1}$, $N\geq\tilde{N}_{B}$, and
$t\geq t_{0}\vee\frac{8B}{\beta}$, we have

$\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\big{(}\tau^{(N)}_{2}(B)\geq
N^{2\varepsilon}t\big{)}\leq
4e^{-c^{\prime}_{0}t}+c^{\prime}_{1}\exp\\{-c^{\prime}_{2}B^{\frac{1}{5}}N^{\frac{4\varepsilon}{5}}t^{\frac{1}{5}}\\}+c^{\prime}_{3}N^{4\varepsilon}t\exp\\{-c^{\prime}_{4}\sqrt{B}N^{2\varepsilon}t\\}.$
(C.2)

For the second term on the right side of (C.1),

$\displaystyle\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\Big{(}\sup_{s\in[0,N^{2\varepsilon}t\wedge\tau_{2}^{(N)}(B)]}\left(2BN^{\frac{1}{2}+\varepsilon}+\hat{\mathcal{M}}(s)+\int_{0}^{s}I^{(N)}(u)du-\beta
N^{\frac{1}{2}-\varepsilon}s\right)\geq xN^{\frac{1}{2}+\varepsilon}\Big{)}$
$\displaystyle\qquad\leq\mathbb{P}\Big{(}\int_{0}^{N^{2\varepsilon}t}\bar{I}_{B}^{(N)}(u)du>\frac{\beta}{2}N^{\frac{1}{2}+\varepsilon}t\Big{)}+\mathbb{P}\Big{(}\sup_{s\in[0,N^{2\varepsilon}t]}\hat{\mathcal{M}}(s)\geq(x-\frac{\beta
t}{2}-2B)N^{\frac{1}{2}+\varepsilon}\Big{)}.$

Thus, using Lemma 3.1 (i) for the first term and Lemma A.1 for the second term
in the above bound, we have that for $N\geq\tilde{N}_{B}$, $t\geq t_{0}$ and
$x-\frac{\beta t}{2}-2B\geq 0$,

$\displaystyle\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\Big{(}\sup_{s\in[0,N^{2\varepsilon}t]}\left(2BN^{\frac{1}{2}+\varepsilon}+\hat{\mathcal{M}}(s)+\int_{0}^{s}I^{(N)}(u)du-\beta
N^{\frac{1}{2}-\varepsilon}s\right)\geq xN^{\frac{1}{2}+\varepsilon}\Big{)}$
$\displaystyle\leq$ $\displaystyle
c_{1}\exp\\{-c_{2}B^{\frac{1}{5}}N^{\frac{4\varepsilon}{5}}t^{\frac{1}{5}}\\}+2\exp\\{-c_{3}(x-\beta
t/2-2B)^{2}/t\\},$ (C.3)

for positive constants $c_{1},c_{2},c_{3}$ not depending on $x,t,N$.

Thus, taking $t=\frac{x}{\beta}$, for $x\geq x_{B}:=(t_{0}\beta\vee 8B)$, and
putting the upper bounds in (C.2) and (C) into (C.1), the result holds for
appropriately chosen constants. ∎

##  Appendix D Proofs from Section 4 (steady state analysis).

We start by proving the following lemma that bounds the number of toggles of
$Q^{(N)}_{2}$ for $\bar{Q}^{(N)}_{3}$ to fall below level
$\frac{B}{4\beta}N^{2\varepsilon}$, which will be used to prove Lemma 4.1.
Define $\bar{K}^{*}\coloneqq\inf\\{k\geq
1:\bar{Q}^{(N)}_{3}(\sigma^{(N)}_{2k})\leq\frac{B}{4\beta}N^{2\varepsilon}\\}$.

######  Lemma D.1.

There exist $N_{0},c^{*}_{1},c^{*}_{2}>0$, such that for all $N\geq
N_{0},k\geq 1$,

$\sup_{0\leq\underline{z}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\big{(}\bar{K}^{*}\geq
1+kN^{\frac{1}{2}-\varepsilon}\big{)}\leq
c^{*}_{1}\exp\big{\\{}-c^{*}_{2}N^{(\frac{1}{2}-\varepsilon)/11}\big{\\}}\exp\Big{\\{}-c^{*}_{2}\big{(}k/N^{\frac{1}{2}-\varepsilon}\big{)}^{\frac{1}{11}}\Big{\\}}.$

###### Proof.

Proof.Define
$\bar{Z}^{(N)}_{i}\coloneqq\bar{Q}^{(N)}_{3}(\sigma^{(N)}_{2i+2})-\bar{Q}^{(N)}_{3}(\sigma^{(N)}_{2i}),i\geq
0$. Note that

$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\big{(}\bar{K}^{*}\geq
1+kN^{\frac{1}{2}-\varepsilon}\big{)}$ $\displaystyle\leq$
$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\big{(}\sum_{j=0}^{i}\bar{Z}^{(N)}_{j}>\frac{B}{4\beta}N^{2\varepsilon}-\bar{Q}^{(N)}_{3}(0),\forall\,0\leq
i\leq\lfloor kN^{\frac{1}{2}-\varepsilon}\rfloor-1\big{)}.$ (D.1)

To estimate (D), define
$\Phi^{(N)}_{i}\coloneqq\mathds{1}\big{[}\bar{Z}^{(N)}_{i}\leq-\big{(}\frac{B}{4\beta}N^{2\varepsilon}\wedge\bar{Q}^{(N)}_{3}(\sigma^{(N)}_{2i})\big{)}\big{]}$,
$i\geq 0$. By Lemma 3.3 (ii), for sufficiently large $N$,

$\inf_{\underline{z}}\mathbb{E}_{(0,2BN^{\frac{1}{2}+\varepsilon},\underline{z})}\big{(}\Phi^{(N)}_{i}|\mathcal{F}_{\sigma^{(N)}_{2i}}\big{)}\geq\frac{1}{2},\quad
i\geq 0.$ (D.2)

Using Azuma-Hoeffding inequality, we get for any $k\geq 1$,

$\begin{split}&\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}\Phi^{(N)}_{j}\leq\frac{kN^{\frac{1}{2}-\varepsilon}}{3}\Big{)}\\\
\leq&\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}\big{(}\Phi^{(N)}_{j}-\mathbb{E}\big{(}\Phi^{(N)}_{j}\
\big{|}\
\mathcal{F}_{\sigma^{(N)}_{2j}}\big{)}\big{)}\leq-\frac{kN^{\frac{1}{2}-\varepsilon}}{6}\Big{)}\leq
e^{-ckN^{\frac{1}{2}-\varepsilon}},\end{split}$ (D.3)

for some $c>0$, where in the first inequality, we have used (D.2). Therefore,
for $k\geq 1$,

$\begin{split}\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}&\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\big{(}\sum_{j=0}^{i}\bar{Z}^{(N)}_{j}>\frac{B}{4\beta}N^{2\varepsilon}-\bar{Q}^{(N)}_{3}(0),\forall\,0\leq
i\leq\lfloor kN^{\frac{1}{2}-\varepsilon}\rfloor-1\big{)}\\\
\leq&\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}-\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}\Big{(}\frac{B}{4\beta}N^{2\varepsilon}\Big{)}\Phi^{(N)}_{j}>0\Big{)}\\\
\leq&\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}\Phi^{(N)}_{j}\leq\frac{kN^{\frac{1}{2}-\varepsilon}}{3}\Big{)}\\\
&\hskip
113.81102pt+\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}\geq\frac{B}{12\beta}N^{\frac{1}{2}+\varepsilon}k\Big{)}\\\
\leq&e^{-ckN^{\frac{1}{2}-\varepsilon}}+\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}\geq\frac{B}{12\beta}N^{\frac{1}{2}+\varepsilon}k\Big{)},\end{split}$
(D.4)

where the last inequality uses (D.3). Note that, by Lemma 3.3 (i), for any
$j\geq 1,x>0$, and sufficient large $N$,

$\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}\Big{(}\frac{[\bar{Z}^{(N)}_{j}]_{+}}{N^{\frac{1}{2}+\varepsilon}}\geq
x\ \big{|}\ \mathcal{F}_{\sigma^{(N)}_{2j}}\Big{)}\leq
c_{1}e^{-c_{2}N^{(\frac{1}{2}-\varepsilon)/5}}e^{-c_{3}x^{1/5}},$ (D.5)

where $c_{1},c_{2},c_{3}>0$ are constants. Choose $N_{1}$ sufficiently large
such that (D.2) and (D.5) hold, and for $N\geq N_{1}$,

$\int_{0}^{\infty}c_{1}e^{-c_{2}N^{(\frac{1}{2}-\varepsilon)/5}}e^{-c_{3}x^{1/5}}dx\leq\frac{B}{4\times
12\beta N^{\frac{1}{2}-\varepsilon}}.$

Take any $N\geq N_{1}$, and $k\geq 1$. We will use Lemma A.2 with
$\theta=\frac{1}{5},\,r=\underline{z},\,R=\\{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}\\},\,a=\frac{B}{12\beta
N^{\frac{1}{2}-\varepsilon}},\,n=\lfloor kN^{\frac{1}{2}-\varepsilon}\rfloor,$
and
$\Phi^{(r)}_{j}=\frac{[\bar{Z}^{(N)}_{j}]_{+}}{N^{\frac{1}{2}+\varepsilon}},$
with starting configuration
$(I^{(N)}(0),Q_{2}^{(N)}(0),\bar{Q}^{(N)}_{3}(0))=(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})$, and associated
filtration $\big{\\{}\mathcal{F}^{(r)}_{j}:j\geq 1,r\in R\big{\\}}$ being the
natural filtration generated by the above random variables. We get

$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}$
$\displaystyle\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}\geq\frac{B}{12\beta}N^{\frac{1}{2}+\varepsilon}k\Big{)}$
$\displaystyle\leq\sup_{r\in
R}\mathbb{P}\Big{(}\sum_{j=0}^{n}\Phi^{(r)}_{j}\geq an\Big{)}\leq
c^{\prime}_{1}\Big{(}1+\frac{\big{(}\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor\big{)}^{5/11}}{\big{(}\frac{B}{12\beta
N^{\frac{1}{2}-\varepsilon}}\big{)}^{\frac{1}{11}}}\Big{)}\exp\Big{[}-c^{\prime}_{2}\Big{(}\frac{k}{N^{\frac{1}{2}-\varepsilon}}\Big{)}^{1/11}\Big{]}$
(D.6) $\displaystyle\leq
c_{1}^{\prime\prime}N^{\frac{1}{2}-\varepsilon}\exp\Big{[}-c^{\prime\prime}_{2}\Big{(}\frac{k}{N^{\frac{1}{2}-\varepsilon}}\Big{)}^{1/11}\Big{]}.$

Moreover, for any $k\geq 1$, from (D.5),

$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}$
$\displaystyle\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}\geq\frac{B}{12\beta}N^{\frac{1}{2}+\varepsilon}k\Big{)}$
$\displaystyle\leq
kN^{\frac{1}{2}-\varepsilon}\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}\Big{(}[\bar{Z}^{(N)}_{0}]_{+}\geq\frac{B}{12\beta}N^{2\varepsilon}\Big{)}\leq
kN^{\frac{1}{2}-\varepsilon}c_{1}\exp\big{\\{}-c_{2}N^{(\frac{1}{2}-\varepsilon)/5}\big{\\}}.$
(D.7)

Hence, (D) and (D) imply

$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}$
$\displaystyle\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}\geq\frac{B}{12\beta}N^{\frac{1}{2}+\varepsilon}k\Big{)}$
$\displaystyle\leq$
$\displaystyle\min\Big{\\{}c_{1}^{\prime\prime}N^{\frac{1}{2}-\varepsilon}\exp\Big{[}-c^{\prime\prime}_{2}\Big{(}\frac{k}{N^{\frac{1}{2}-\varepsilon}}\Big{)}^{1/11}\Big{]},\,kN^{\frac{1}{2}-\varepsilon}c_{1}\exp\big{\\{}-c_{2}N^{(\frac{1}{2}-\varepsilon)/5}\big{\\}}\Big{\\}}.$
(D.8)

Note that, there exists a constant $N_{2}$, such that for all $N\geq N_{2}$
and $k\leq N^{1-2\varepsilon}$,

$\displaystyle
kN^{\frac{1}{2}-\varepsilon}c_{1}\exp\big{[}-c_{2}N^{(\frac{1}{2}-\varepsilon)/5}\big{]}=$
$\displaystyle\Big{[}kN^{\frac{1}{2}-\varepsilon}c_{1}\exp\big{\\{}-\frac{c_{2}}{2}N^{(\frac{1}{2}-\varepsilon)/5}\big{\\}}\Big{]}\exp\big{[}-\frac{c_{2}}{2}N^{(\frac{1}{2}-\varepsilon)/5}\big{]}$
$\displaystyle\leq$
$\displaystyle\exp\big{[}-\frac{c_{2}}{4}k^{\frac{1}{10}}\big{]}\exp\big{[}-\frac{c_{2}}{2}N^{(\frac{1}{2}-\varepsilon)/5})\big{]},$

and for $k\geq N^{1-2\varepsilon}$,

$N^{\frac{1}{2}-\varepsilon}\exp\Big{[}-c^{\prime\prime}_{2}\Big{(}\frac{k}{N^{\frac{1}{2}-\varepsilon}}\Big{)}^{1/11}\Big{]}\leq
N^{\frac{1}{2}-\varepsilon}\exp\Big{[}-\frac{c^{\prime\prime}_{2}}{2}N^{(\frac{1}{2}-\varepsilon)/11}\Big{]}\exp\Big{[}-\frac{c^{\prime\prime}_{2}}{2}\Big{(}\frac{k}{N^{\frac{1}{2}-\varepsilon}}\Big{)}^{1/11}\Big{]}.$

Hence, there exist constants $\tilde{c},\tilde{c}^{\prime}>0$, such that for
all $N\geq N_{1}\vee N_{2}$ and $k\geq 1$,

$\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\Big{(}\sum_{j=0}^{\lfloor
kN^{\frac{1}{2}-\varepsilon}\rfloor-1}[\bar{Z}^{(N)}_{j}]_{+}\geq\frac{B}{12\beta}N^{\frac{1}{2}+\varepsilon}k\Big{)}\\\
\leq\tilde{c}\exp\big{\\{}-\tilde{c}^{\prime}N^{(\frac{1}{2}-\varepsilon)/11}\big{\\}}\exp\Big{\\{}-\tilde{c}^{\prime}\big{(}k/N^{\frac{1}{2}-\varepsilon}\big{)}^{\frac{1}{11}}\Big{\\}}.$
(D.9)

Finally, using (D) and plugging (D.9) into (D.4), we have

$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\big{(}\bar{K}^{*}\geq
1+kN^{\frac{1}{2}-\varepsilon}\big{)}$ $\displaystyle\leq$
$\displaystyle\sup_{\underline{z}:\,\sum_{i}z_{i}\leq\frac{B}{4\beta}N^{2\varepsilon}}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{z})}\big{(}\sum_{j=0}^{i}\bar{Z}^{(N)}_{j}>\frac{B}{4\beta}N^{2\varepsilon}-\bar{Q}^{(N)}_{3}(0),\forall\,0\leq
i\leq\lfloor kN^{\frac{1}{2}-\varepsilon}\rfloor-1\big{)}$ $\displaystyle\leq$
$\displaystyle
c^{*}_{1}\exp\big{\\{}-c^{*}_{2}N^{(\frac{1}{2}-\varepsilon)/11}\big{\\}}\exp\Big{\\{}-c^{*}_{2}\big{(}k/N^{\frac{1}{2}-\varepsilon}\big{)}^{\frac{1}{11}}\Big{\\}},$

where $c^{*}_{1}$ and $c^{*}_{2}$ are appropriate constants. ∎

###### Proof.

Proof of Lemma 4.1. Define $\bar{K}^{*}_{0}\coloneqq 0$ and for $j\geq 0$,
$\bar{K}^{*}_{j+1}\coloneqq\inf\big{\\{}l\geq\bar{K}^{*}_{j}+1:\bar{Q}^{(N)}_{3}(\sigma^{(N)}_{2l})\leq\frac{B}{4\beta}N^{2\varepsilon}\big{\\}}.$
Define
$\chi^{(N)}_{j}\coloneqq\mathds{1}\big{[}\bar{Q}^{(N)}_{3}(\sigma^{(N)}_{2\bar{K}_{j}^{*}+2})=0\big{]},\
j\geq 0$. By Lemma 3.3 (ii),

$\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\big{(}\chi^{(N)}_{j}=1|\mathcal{F}_{\sigma^{(N)}_{2\bar{K}^{*}_{j}}}\big{)}\geq\frac{1}{2},\quad\forall
j\geq 0.$ (D.10)

Thus, there exist constants $c_{2}^{*},N_{0}>0$, such that for $k\geq 1,N\geq
N_{0}$,

$\displaystyle\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\big{(}\bar{K}^{(N)}\geq
1+k+k^{2}N^{\frac{1}{2}-\varepsilon}\big{)}$ $\displaystyle\leq$
$\displaystyle\sum_{j=1}^{k}\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\big{(}\bar{K}^{*}_{j+1}-\bar{K}^{*}_{j}\geq
1+kN^{\frac{1}{2}-\varepsilon}\big{)}$ $\displaystyle+\mathbb{P}_{(0,\,\lfloor
2BN^{\frac{1}{2}+\varepsilon}\rfloor,\,\underline{0})}\big{(}\chi^{(N)}_{j}=0,\forall
0\leq j\leq k-1\big{)}$ $\displaystyle\leq$ $\displaystyle
kc^{*}_{1}\exp\big{\\{}-c^{*}_{2}N^{(\frac{1}{2}-\varepsilon)/11}\big{\\}}\exp\Big{\\{}-c^{*}_{2}\big{(}k/N^{\frac{1}{2}-\varepsilon}\big{)}^{\frac{1}{11}}\Big{\\}}+2^{-k},$

where the last inequality comes from Lemma D.1 and the inequality in (D.10). ∎

##  Appendix E Auxiliary results for the proof of Proposition 5.6.

In this appendix, we will prove the auxiliary results for the proof of
Proposition 5.6. Let us start with the proof of Lemma 6.1. The following
technical lemma from [16] will be used in the proof of Lemma 6.1.

######  Lemma E.1 ([16, Lemma EC.18]).

Let $Q$ be the length of an M/M/1 queue with arrival rate $\alpha$ and service
rate $\mu$, with $\mu>\alpha$. Also, let $\tilde{T}$ be the length of the
renewal cycle from the queue being of length 1 to the queue being empty. Then,
for $t\geq 0$,

$\mathbb{P}\left(\tilde{T}\geq
t\right)\leq\sqrt{\frac{\mu}{\alpha}}e^{-(\sqrt{\mu}-\sqrt{\alpha})^{2}t}.$

###### Proof.

Proof of Lemma 6.1. Note that for any $i\geq 0$,

$\begin{split}&\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}\Big{(}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+1}\geq
2N^{\delta-2\varepsilon},i<i^{*}_{N}\Big{)}\\\
\leq&\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{0\leq
t\leq T(N,B)}Q^{(N)}_{2}(t)\geq N^{\frac{1}{2}+\varepsilon+\delta}\Big{)}\\\
&\qquad+\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1}\geq
N^{\delta-2\varepsilon},\sup_{0\leq t\leq
T(N,B)}Q^{(N)}_{2}(t)<N^{\frac{1}{2}+\varepsilon+\delta},i<i^{*}_{N}\Big{)}\\\
&\qquad+\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+2}\geq
N^{\delta-2\varepsilon},\sup_{0\leq t\leq
T(N,B)}Q^{(N)}_{2}(t)<N^{\frac{1}{2}+\varepsilon+\delta},i<i^{*}_{N}\Big{)}.\end{split}$
(E.1)

For the first term of the right hand side of (E.1), take any $\tilde{K}\geq
K_{2}$ large enough such that the bound in Proposition 3.9 holds with $2B$
there replaced by $\tilde{K}$ for large enough $N$. Also assume that $N$ is
large enough such that $N^{\delta}>x^{\prime}_{\tilde{K}/2}$, where the latter
constant appears in Proposition 3.9. For such large enough $N$, we obtain
constants $c^{\prime}_{1}$ and $c^{\prime}_{2}$ such that

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{0\leq
t\leq T(N,B)}Q^{(N)}_{2}(t)\geq N^{\frac{1}{2}+\varepsilon+\delta}\Big{)}$
$\displaystyle\leq$
$\displaystyle\mathbb{P}_{(0,\lfloor\tilde{K}N^{\frac{1}{2}+\varepsilon}\rfloor,\underline{0})}\Big{(}\sup_{0\leq
t\leq T(N,B)}Q^{(N)}_{2}(t)\geq N^{\frac{1}{2}+\varepsilon+\delta}\Big{)}$
$\displaystyle\leq$
$\displaystyle\exp\\{-\bar{c}^{\prime}TN^{\delta}\\}+TN^{\delta}\big{(}\bar{c}_{1}\exp\big{\\{}-\bar{c}_{2}N^{(\frac{1}{2}-\varepsilon)/5}\big{\\}}+c^{*}_{1}\exp\\{-c^{*}_{2}N^{\delta}\\}+c^{*}_{3}\exp\\{-c^{*}_{4}N^{\frac{4\varepsilon}{5}}N^{\frac{\delta}{5}}\\}\big{)}$
$\displaystyle\leq$ $\displaystyle
c^{\prime}_{1}e^{-c^{\prime}_{2}N^{\delta/5}},$ (E.2)

where the second inequality is from Proposition 3.9. For the second term in
(E.1), there exist constants $c^{\prime}_{3}$ and $c^{\prime}_{4}$ such that
for large enough $N$ and any $i\geq 0$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1}\geq
N^{\delta-2\varepsilon},\sup_{0\leq t\leq
T(N,B)}Q^{(N)}_{2}(t)<N^{\frac{1}{2}+\varepsilon+\delta},i<i^{*}_{N}\Big{)}$
$\displaystyle\leq$
$\displaystyle\mathbb{P}\Big{(}\mathrm{Exp}(N-N^{\frac{1}{2}+\varepsilon+\delta})\geq
N^{\delta-2\varepsilon}\Big{)}\leq
c^{\prime}_{3}e^{-c^{\prime}_{4}N^{\delta+1-2\varepsilon}},$ (E.3)

where $\mathrm{Exp}(N-N^{\frac{1}{2}+\varepsilon+\delta})$ is an exponential
random variable with rate $N-N^{\frac{1}{2}+\varepsilon+\delta}$. The first
inequality in (E) is due to the fact that the rate of increase of $I^{(N)}(t)$
when $I^{(N)}(t)=0$ is $N-Q^{(N)}_{2}$. Now, consider the last term in the
right hand side of (E.1). Recall the evolution of the process $I^{(N)}(t)$ as
in (6.2). For $t\leq\tau_{2}^{(N)}(B)$, $Q^{(N)}_{1}(t)-Q^{(N)}_{2}(t)\leq
N-BN^{\frac{1}{2}+\varepsilon}$, and thus, there is a natural coupling so that
the process $I^{(N)}$ is upper bounded by an M/M/1 queue, $Q$, with arrival
rate $\alpha:=N-B^{\frac{1}{2}+\varepsilon}$ and service rate $\mu:=N-\beta
N^{\frac{1}{2}-\varepsilon}$. Therefore, for any $i\geq 0$,

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}\big{(}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+2}\geq
t,i<i^{*}_{N}\big{)}\leq\mathbb{P}\big{(}\tilde{T}\geq t\big{)},$ (E.4)

where $\tilde{T}$ for the process $Q$ is as defined in Lemma E.1. Now, note
that there exists $N_{0}>0$ such that for all $N\geq N_{0}$,

$\begin{split}\sqrt{\frac{\mu}{\alpha}}&=\sqrt{\frac{N-\beta
N^{\frac{1}{2}-\varepsilon}}{N-B^{\frac{1}{2}+\varepsilon}}}\leq 2\\\
\big{(}\sqrt{\mu}-\sqrt{\alpha}\big{)}^{2}&=\big{(}\sqrt{N-\beta
N^{\frac{1}{2}-\varepsilon}}-\sqrt{N-BN^{\frac{1}{2}+\varepsilon}}\big{)}^{2}\\\
&=\big{(}N-\beta
N^{\frac{1}{2}-\varepsilon}\big{)}\Big{(}1-\sqrt{1-\frac{B-\beta
N^{-2\varepsilon}}{N^{\frac{1}{2}-\varepsilon}-\beta
N^{-2\varepsilon}}}\Big{)}^{2}\geq\frac{B^{2}}{32}N^{2\varepsilon},\end{split}$
(E.5)

where the last inequality is due to the facts that
$1-\sqrt{1-x}\geq\frac{x}{2}$, for $x\in(0,1)$, and for all $N\geq N_{0}$,
$N-\beta N^{\frac{1}{2}-\varepsilon}\geq\frac{N}{2}$ and $\frac{B-\beta
N^{-2\varepsilon}}{N^{\frac{1}{2}-\varepsilon}-\beta
N^{-2\varepsilon}}\geq\frac{B}{2}N^{\varepsilon-\frac{1}{2}}$. Thus, plugging
(E.5) into Lemma E.1, we get for any $t\geq 0$, $i\geq 0$,

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}\big{(}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+2}\geq
t,i<i^{*}_{N}\big{)}\leq\mathbb{P}\big{(}\tilde{T}\geq t\big{)}\leq
2e^{-\frac{B^{2}}{32}N^{2\varepsilon}t}.$ (E.6)

Plugging (E), (E), and (E.6), into (E.1), we obtain constants $c^{\prime}_{5}$
and $c^{\prime}_{6}$ such that for any $i\geq 0$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}\Big{(}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+1}\geq
2N^{\delta-2\varepsilon},i<i^{*}_{N}\Big{)}$ $\displaystyle\leq$
$\displaystyle
c^{\prime}_{1}e^{-c^{\prime}_{2}N^{\delta/5}}+c^{\prime}_{3}e^{-c^{\prime}_{4}N^{\delta+1-2\varepsilon}}+2e^{-\frac{B^{2}}{32}N^{\delta}}\leq
c^{\prime}_{5}e^{-c^{\prime}_{6}N^{\delta/5}}.$ (E.7)

To prove (6.3), note that the value of $i^{*}_{N}$ is upper bounded by the
number of increments in $I^{(N)}$ in the interval
$[0,\tau^{(N)}_{2}(B)\wedge(N^{2\varepsilon}T)]$, which is stochastically
upper bounded by a Poisson random variable with parameter
$N^{1+2\varepsilon}T$. Hence, using standard concentration inequality for
Poisson random variables (see [29, Theorem 2.3(b)]), we have

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}\Big{(}i^{*}_{N}\geq\frac{3}{2}N^{1+2\varepsilon}T\Big{)}\leq\mathbb{P}\Big{(}\mathrm{Po}(N^{1+2\varepsilon}T)\geq\frac{3}{2}N^{1+2\varepsilon}T\Big{)}\leq
e^{-\frac{3}{32}N^{1+2\varepsilon}T},$ (E.8)

and therefore, using (E), there exists $N_{1}\geq N_{0}$, depending on $T$,
such that for all $N\geq N_{1}$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\big{(}\exists\
i\in\big{\\{}0,...,i^{*}_{N}-1\big{\\}}\text{ such that
}\xi^{(N)}_{2i+3}-\xi^{(N)}_{2i+1}\geq N^{\delta-2\varepsilon}\big{)}$
$\displaystyle\leq\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\Big{[}\mathbb{P}_{(x,y,0)}\big{(}i^{*}_{N}\geq\frac{3}{2}N^{1+2\varepsilon}T\big{)}+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\cdot
c^{\prime}_{5}e^{-c^{\prime}_{6}N^{\delta/5}}\Big{]}$ $\displaystyle\leq
e^{-\frac{3}{32}N^{1+2\varepsilon}T}+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\times
c^{\prime}_{5}e^{-c^{\prime}_{6}N^{\delta/5}}\leq
a_{1}e^{-b_{1}N^{\delta/5}},$ (E.9)

where $a_{1}$ and $b_{1}$ are appropriate constants. ∎

###### Proof.

Proof of Proposition 6.2. Define the event
$w^{(N)}\coloneqq\big{\\{}\sup_{0\leq t\leq
T(N,B)}Q^{(N)}_{2}(t)<N^{\frac{1}{2}+\varepsilon+\delta}\big{\\}}$. Then

$\begin{split}&\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
13N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}\\\
\leq&\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{0\leq
t\leq T(N,B)}Q^{(N)}_{2}(t)\geq N^{\frac{1}{2}+\varepsilon+\delta}\Big{)}\\\
&+\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+2}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
3N^{\frac{1}{2}-\varepsilon+\delta},w^{(N)}\Big{)}\\\
&+\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon}\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta},w^{(N)}\Big{)}.\end{split}$ (E.10)

By (E), for all large enough $N$,

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{0\leq
t\leq T(N,B)}Q^{(N)}_{2}(t)\geq N^{\frac{1}{2}+\varepsilon+\delta}\Big{)}\leq
c^{\prime}_{1}e^{-c^{\prime}_{2}N^{\delta/5}}.$ (E.11)

For the second term in the right hand side of (E.10),

$\begin{split}&\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+2}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
3N^{\frac{1}{2}-\varepsilon+\delta},w^{(N)}\Big{)}\\\
\leq&\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1}\geq
N^{-\frac{1}{2}-\varepsilon+\delta},w^{(N)}\Big{)}\\\
&+\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+2}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
3N^{\frac{1}{2}-\varepsilon+\delta},w^{(N)},\tilde{w}^{(N)}\Big{)},\end{split}$
(E.12)

where
$\tilde{w}^{(N)}\coloneqq\big{\\{}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1}<N^{-\frac{1}{2}-\varepsilon+\delta}\big{\\}}$.
Since the instantaneous rate of increase of $I^{(N)}$ when $I^{(N)}(t)=0$ is
$N-Q^{(N)}_{2}$, then for each $i\in\\{0,...,i^{*}_{N}-1\\}$,
$\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1}$ is upper bounded by an exponential random
variable, $\mathrm{Exp}(N-N^{\frac{1}{2}+\varepsilon+\delta})$. Hence,
proceeding along the same line as (E), (E.8) and (E), there exist constants
$c_{1}$, $c_{2}$ such that for large enough $N$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1}\geq
N^{-\frac{1}{2}-\varepsilon+\delta},w^{(N)}\Big{)}$
$\displaystyle\leq\mathbb{P}(i^{*}_{N}\geq\frac{3}{2}N^{1+2\varepsilon}T)+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\times
e^{-(N-N^{\frac{1}{2}+\varepsilon+\delta})N^{{}^{-\frac{1}{2}-\varepsilon+\delta}}}$
$\displaystyle\leq
e^{-\frac{3}{32}N^{1+2\varepsilon}T}+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\times
e^{-(N-N^{\frac{1}{2}+\varepsilon+\delta})N^{{}^{-\frac{1}{2}-\varepsilon+\delta}}}\leq
c_{1}e^{-c_{2}N^{\frac{1}{2}-\varepsilon+\delta}}.$ (E.13)

For each $i\in\\{0,...,i^{*}_{N}-1\\}$ and
$t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+2})$,

$\displaystyle\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|$
$\displaystyle\leq A\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})t\big{)}-A\big{(}(N-\beta
N^{\frac{1}{2}-\varepsilon})\xi^{(N)}_{2i+1}\big{)}+D\big{(}Nt\big{)}-D\big{(}N\xi^{(N)}_{2i+1}\big{)}$
$\displaystyle\leq_{st}\mathrm{Po}\big{(}2N(\xi^{(N)}_{2i+2}-\xi^{(N)}_{2i+1})\big{)},$

where $\mathrm{Po}(\lambda)$ is a Poisson random variable with mean $\lambda$.
Therefore, there exist constants $c_{3}$ and $c_{4}$ such that for all $N$
large enough,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+2}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
3N^{\frac{1}{2}-\varepsilon+\delta},w^{(N)},\tilde{w}^{(N)}\Big{)}$
$\displaystyle\leq\mathbb{P}(i^{*}_{N}\geq\frac{3}{2}N^{1+2\varepsilon}T)+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\mathbb{P}\Big{(}\mathrm{Po}(2N^{\frac{1}{2}-\varepsilon+\delta})\geq
3N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}$ $\displaystyle\leq
e^{-\frac{3}{32}N^{1+2\varepsilon}T}+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\times
e^{-\frac{3}{16}N^{\frac{1}{2}-\varepsilon+\delta}}\leq
c_{3}e^{-c_{4}N^{\frac{1}{2}-\varepsilon+\delta}}.$ (E.14)

Plugging (E) and (E) into (E.12), we have for all $N$ large enough,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}$
$\displaystyle\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+2}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
3N^{\frac{1}{2}-\varepsilon+\delta},w^{(N)}\Big{)}$ $\displaystyle\leq$
$\displaystyle c_{5}e^{-c_{6}N^{\frac{1}{2}-\varepsilon+\delta}},$ (E.15)

where $c_{5}$ and $c_{6}$ are constants. Now, consider the third term of the
right hand side of (E.10). For any $x\leq K_{1}N^{\frac{1}{2}-\varepsilon}$
and $BN^{\frac{1}{2}+\varepsilon}<y\leq K_{2}N^{\frac{1}{2}+\varepsilon}$,

$\displaystyle\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}$
$\displaystyle\leq\mathbb{P}_{(x,y,\underline{0})}(i^{*}_{N}>\frac{3}{2}N^{1+2\varepsilon}T)+\sum_{i=0}^{\lceil\frac{3}{2}N^{1+2\varepsilon}T\rceil-1}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta},i\leq i^{*}_{N}-1\Big{)}$
$\displaystyle\leq\mathbb{P}_{(x,y,\underline{0})}(i^{*}_{N}>\frac{3}{2}N^{1+2\varepsilon}T)+\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{t\in[0,N^{2\varepsilon}T]}Q^{(N)}_{3}(t)>0\Big{)}$
$\displaystyle+\sum_{i=0}^{\lceil\frac{3}{2}N^{1+2\varepsilon}T\rceil-1}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta},\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}Q^{(N)}_{3}(t)=0,i\leq i^{*}_{N}-1\Big{)}.$ (E.16)

Recall that, by (E.8),

$\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}(i^{*}_{N}>\frac{3}{2}N^{1+2\varepsilon}T)\leq
e^{-\frac{3}{32}N^{1+2\varepsilon}T}.$ (E.17)

By Proposition 3.9, there exist constants
$B_{0},c_{1},c_{2},N_{0}^{\prime}>0$, possibly depending on $T$, such that for
all $N\geq N_{0}^{\prime}$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{t\in[0,N^{2\varepsilon}T]}Q^{(N)}_{3}>0\Big{)}$
$\displaystyle\hskip 113.81102pt\leq\mathbb{P}_{(0,\lfloor
B_{0}N^{\frac{1}{2}+\varepsilon}\rfloor,\underline{0})}\Big{(}\sup_{t\in[0,N^{2\varepsilon}T]}Q^{(N)}_{2}(t)=N\Big{)}\leq
c_{7}e^{-c_{8}N^{(\frac{1}{2}-\varepsilon)/5}}.$ (E.18)

Next, for $i\leq i^{*}_{N}-1$ and $t\in[\xi^{(N)}_{2i+2},\xi^{(N)}_{2i+3})$,
observe that on the event $\sup_{t\in[0,N^{2\varepsilon}T]}Q_{3}^{(N)}(t)=0$,

$\displaystyle\sup_{t\in[\xi^{(N)}_{2i+2},\xi^{(N)}_{2i+3})}\left|S^{(N)}(t)-S^{(N)}(\xi_{2i+2})\right|$
$\displaystyle\leq\sup_{t\in[\xi^{(N)}_{2i+2},\xi^{(N)}_{2i+3})}\Big{(}I^{(N)}(t)+Q_{2}^{(N)}(\xi^{(N)}_{2i+2})-Q_{2}^{(N)}(\xi^{(N)}_{2i+3})\Big{)}$
$\displaystyle\leq\sup_{t\in[\xi^{(N)}_{2i+2},\xi^{(N)}_{2i+3})}I^{(N)}(t)+Q_{2}^{(N)}(\xi^{(N)}_{2i+2}).$

Indeed, the inequalities are due to the fact that when there is no queue with
length greater than $3$, the difference between $S^{(N)}(t)$ and
$S^{(N)}(\xi_{2i+2})$ is caused by the change of $I^{(N)}(t)$ and the change
of $Q_{2}^{(N)}(t)$, and for $t\in[\xi^{(N)}_{2i+2},\xi^{(N)}_{2i+3})$,
$I^{(N)}(t)$ is always positive so $Q_{2}^{(N)}(t)$ can only decrease during
$[\xi^{(N)}_{2i+2},\xi^{(N)}_{2i+3})$. Now, consider the last term of (E). For
any $x\leq K_{1}N^{\frac{1}{2}-\varepsilon}$ and
$BN^{\frac{1}{2}+\varepsilon}<y\leq K_{2}N^{\frac{1}{2}+\varepsilon}$, any
$i\geq 0$,

$\begin{split}&\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta},\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}Q^{(N)}_{3}(t)=0,i\leq i^{*}_{N}-1\Big{)}\\\
&\leq\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}I^{(N)}(t)\geq 5N^{\frac{1}{2}-\varepsilon+\delta},i\leq
i^{*}_{N}-1\Big{)}\\\ &\hskip
56.9055pt+\mathbb{P}_{(x,y,\underline{0})}\Big{(}Q_{2}^{(N)}(\xi^{(N)}_{2i+2})\geq
5N^{\frac{1}{2}-\varepsilon+\delta},i\leq i^{*}_{N}-1\Big{)}.\end{split}$
(E.19)

Since $\delta<\frac{1}{2}-\varepsilon$, from Proposition 5.5, we have that for
$N\geq N_{0}$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}I^{(N)}(t)\geq 5N^{\frac{1}{2}-\varepsilon+\delta},i\leq
i^{*}_{N}-1\Big{)}$ $\displaystyle\leq\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\big{(}\sup_{0\leq
t\leq(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B))}I^{(N)}(t)\geq
5N^{\frac{1}{2}-\varepsilon+\delta}\big{)}\leq ae^{-bN^{\frac{\delta}{2}}},$
(E.20)

where $a$ and $b$ are positive constants depending on $B$, $\beta$ and $T$
only.

Proceeding as in (E), by Proposition 3.9, there exist
$c_{1}^{\prime},c_{2}^{\prime},N_{0}^{\prime}>0$, such that for all $N\geq
N_{0}^{\prime}$,

$\begin{split}&\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}Q^{(N)}_{2}(t)\geq
5N^{\frac{1}{2}+\varepsilon+\delta},i\leq i^{*}_{N}-1\Big{)}\\\
&\leq\sup_{\begin{subarray}{c}x\leq K_{1}N^{\frac{1}{2}-\varepsilon},\\\
BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{0\leq
t\leq(N^{2\varepsilon}T)\wedge\tau^{(N)}_{2}(B))}Q^{(N)}_{2}(t)\geq
5N^{\frac{1}{2}+\varepsilon+\delta}\Big{)}\leq
c^{\prime}_{1}e^{-c^{\prime}_{2}N^{\delta/5}}.\end{split}$ (E.21)

Hence, plugging (E) and (E.21) into (E.19), we have that for all sufficiently
large $N$, any $i\geq 0$,

$\mathbb{P}\Big{(}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta},\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}Q^{(N)}_{3}(t)=0,i\leq i^{*}_{N}-1\Big{)}\leq
c^{\prime\prime}_{1}e^{-c^{\prime\prime}_{2}N^{\frac{\delta}{5}}},$ (E.22)

where $c^{\prime\prime}_{1}$ and $c^{\prime\prime}_{2}$ are constants
dependent on $K_{2}$, $B$, $\beta$ and $T$. Plugging (E.17), (E), and (E.22)
into (E), we obtain for sufficiently large $N$,

$\displaystyle\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+2}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+2})\right|\geq
10N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}$ $\displaystyle\leq$
$\displaystyle
e^{-\frac{3}{32}N^{1+2\varepsilon}T}+c_{7}e^{-c_{8}N^{(\frac{1}{2}-\varepsilon)/5}}+\Big{\lceil}\frac{3}{2}N^{1+2\varepsilon}T\Big{\rceil}\times
c^{\prime\prime}_{1}e^{-c^{\prime\prime}_{2}N^{\frac{\delta}{5}}}\leq
c^{\prime\prime}_{3}e^{-c^{\prime\prime}_{4}N^{\frac{\delta}{5}}}.$ (E.23)

Finally, plugging (E.11), (E) and (E) into (E.10), we can choose appropriate
constants $a_{2}$ and $b_{2}$ such that for all sufficiently large $N$,

$\displaystyle\sup_{\begin{subarray}{c}x\leq
K_{1}N^{\frac{1}{2}-\varepsilon},\\\ BN^{\frac{1}{2}+\varepsilon}<y\leq
K_{2}N^{\frac{1}{2}+\varepsilon}\end{subarray}}\mathbb{P}_{(x,y,\underline{0})}\Big{(}\sup_{i\in\\{0,...,i^{*}_{N}-1\\}}\sup_{\xi^{(N)}_{2i+1}\leq
t\leq\xi^{(N)}_{2i+3}}\left|S^{(N)}(t)-S^{(N)}(\xi^{(N)}_{2i+1})\right|\geq
13N^{\frac{1}{2}-\varepsilon+\delta}\Big{)}\leq
a_{2}e^{-b_{2}N^{\frac{\delta}{5}}}.$

∎

######  Lemma E.2.

The following holds for all large enough $N$:

$\mathbb{P}\Big{(}\big{\\{}i^{*}_{N}\geq\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil\big{\\}}\cap\bar{E}^{(N)}\Big{)}\leq\frac{1}{9T}N^{-4\varepsilon-2\delta}.$

###### Proof.

Proof of Lemma E.2. On $\bar{E}^{(N)}$, we have that for $t\in[0,T(N,B)]$,

$\displaystyle 2N-S^{(N)}(t)-2I^{(N)}(t)$
$\displaystyle=N-I^{(N)}(t)-Q^{(N)}_{2}(t)\geq
N-5N^{\frac{1}{2}-\varepsilon+\delta}-N^{\frac{1}{2}+\varepsilon+\delta}\geq
N-6N^{\frac{1}{2}+\varepsilon+\delta}.$

On each $[\xi^{(N)}_{2i},\xi^{(N)}_{2i+1})$, we construct an M/M/1 queue
$M^{(N)}_{i}$ with rate of increase $N-6N^{\frac{1}{2}+\varepsilon+\delta}$
and rate of decrease $N$. We can couple $M^{(N)}_{i}$ with $I^{(N)}$ on
$[\xi^{(N)}_{2i},\xi^{(N)}_{2i+1})$ by setting
$M^{(N)}_{i}(\xi^{(N)}_{2i})=I^{(N)}(\xi^{(N)}_{2i})=1$ such that
$M^{(N)}_{i}(t)\leq I^{(N)}(t)$. Define
$\xi^{(N)}_{m,i}\coloneqq\inf\\{t\geq\xi^{(N)}_{2i}:M^{(N)}_{i}(t)=0\\}$ . Due
to the coupling, we have
$\xi^{(N)}_{m,i}-\xi^{(N)}_{2i}\leq\xi^{(N)}_{2i+1}-\xi^{(N)}_{2i}$. Now, we
have

$\displaystyle\mathbb{P}\Big{(}\big{\\{}i^{*}_{N}\geq\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil\big{\\}}\cap\bar{E}^{N}\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\Big{\\{}\sum_{i=1}^{\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil}\big{(}\xi^{(N)}_{2i+1}-\xi^{(N)}_{2i}\big{)}\leq
N^{2\varepsilon}T\Big{\\}}\cap\bar{E}^{N}\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil}\big{(}\xi^{(N)}_{m,i}-\xi^{(N)}_{2i}\big{)}\leq
N^{2\varepsilon}T\Big{)}.$

Let $H^{(N)}_{i}\coloneqq\xi^{(N)}_{m,i}-\xi^{(N)}_{2i}$. Since $H^{(N)}_{i}$
is the busy time of the M/M/1 queue $M^{(N)}_{i}$, we have
$\mathbb{E}(H^{(N)}_{i})=\frac{1}{6}N^{-\frac{1}{2}-\varepsilon-\delta}$ and
$\mathrm{Var}(H^{(N)}_{i})=\frac{2}{6^{3}}N^{-\frac{1}{2}-3\varepsilon-3\delta}-\frac{1}{6^{2}}N^{-1-2\varepsilon-2\delta}\leq\frac{2}{6^{3}}N^{-\frac{1}{2}-3\varepsilon-3\delta}$
[1, Section 7.9, Page 75]. Hence, by Doob’s $L^{2}$-maximal inequality, we
have for all large enough $N$,

$\displaystyle\mathbb{P}\Big{(}\big{\\{}i^{*}_{N}\geq\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil\big{\\}}\cap\bar{E}^{N}\Big{)}$
$\displaystyle\leq\mathbb{P}\Big{(}\sum_{i=1}^{\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil}\big{(}H^{(N)}_{i}-\mathbb{E}(H^{(N)}_{i})\big{)}\leq-N^{2\varepsilon}T\Big{)}$
$\displaystyle\leq\frac{\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil\times
2N^{-\frac{1}{2}-3\varepsilon-3\delta}}{6^{3}N^{4\varepsilon}T^{2}}\leq\frac{1}{9T}N^{-4\varepsilon-2\delta}.$

∎

The next lemma will be used in the proof of Lemma 6.5.

######  Lemma E.3.

Suppose that $Q$ is an M/M/1 queue with arrival rate $\alpha$ and service rate
$\mu$,with $\mu>\alpha$. Let $\tilde{T}$ be the length of the excursion of the
M/M/1 queue started from zero (that is, the time taken for the queue length to
become non-zero and then zero again). Then

$\mathbb{E}\Big{[}\int_{0}^{\tilde{T}}\big{(}Q(s)-\frac{\alpha}{\mu-\alpha}\big{)}ds\Big{]}=0.$
(E.24)

Moreover, there exists a universal positive constant $C$ not depending on
$\alpha,\mu$ such that for any integer $a\geq 2$,

$\begin{split}\mathbb{E}\Big{[}\Big{(}\int_{0}^{\tilde{T}}Q(s)ds\Big{)}^{2}\Big{]}\leq
C\bigg{[}a^{2}\left(\frac{1}{\alpha^{2}}+\frac{\mu}{(\mu-\alpha)^{3}}\right)&+\frac{(\mu/\alpha)}{(\log(\mu/\alpha))^{2}}\Big{(}\frac{\alpha}{\mu}\Big{)}^{a/8}\bigg{(}\frac{1}{\alpha^{2}}+\frac{\big{(}\mu/\alpha\big{)}^{1/4}}{(\sqrt{\mu}-\sqrt{\alpha})^{4}}\bigg{)}\bigg{]}.\end{split}$
(E.25)

###### Proof.

Proof. Throughout this proof, we will denote by $C^{\prime}$ a universal
positive constant, not depending on $\alpha,\mu$, whose value might change
between lines.

(E.24) is proved in [16, Lemma EC.19]. To prove (E.25), write
$\overline{Q}:=\sup_{0\leq s\leq\tilde{T}}Q(s)$. Note that, for any integer
$a\geq 2$,

$\displaystyle\mathbb{E}\Big{[}\left(\int_{0}^{\tilde{T}}Q(s)ds\right)^{2}\Big{]}$
$\displaystyle\leq\mathbb{E}\left[\left(a\tilde{T}+\tilde{T}\overline{Q}\mathds{1}\left(\overline{Q}\geq
a\right)\right)^{2}\right]\leq
2a^{2}\mathbb{E}\left[\tilde{T}^{2}\right]+2\mathbb{E}\left[\left(\tilde{T}\overline{Q}\mathds{1}\left(\overline{Q}\geq
a\right)\right)^{2}\right]$ $\displaystyle\leq
2a^{2}\mathbb{E}\left[\tilde{T}^{2}\right]+2\sqrt{\mathbb{E}\left[\tilde{T}^{4}\right]}\sqrt{\mathbb{E}\left[\overline{Q}^{4}\mathds{1}\left(\overline{Q}\geq
a\right)\right]},$ (E.26)

where the last inequality follows from the Cauchy-Schwarz inequality. We can
write $\tilde{T}=\tilde{T}_{1}+\tilde{T}_{2}$ where $\tilde{T}_{1}$ is an
$\mathrm{Exp}(\alpha)$ random variable denoting the arrival time of the first
task and $\tilde{T}_{2}$ denotes the time taken after $\tilde{T}_{1}$ for the
queue to become empty again (busy time). Using
$\mathbb{E}[\tilde{T}_{1}^{2}]=2\alpha^{-2}$ and [1, Section 7.9, Page 75],

$\mathbb{E}\left[\tilde{T}^{2}\right]\leq
2\mathbb{E}\left[\tilde{T}_{1}^{2}\right]+2\mathbb{E}\left[\tilde{T}_{2}^{2}\right]\leq
4\left[\frac{1}{\alpha^{2}}+\frac{\mu}{(\mu-\alpha)^{3}}\right].$ (E.27)

Recall from Lemma E.1 that $\mathbb{P}\left(\tilde{T}_{2}\geq
t\right)\leq\sqrt{\frac{\mu}{\alpha}}e^{-(\sqrt{\mu}-\sqrt{\alpha})^{2}t},\,t\geq
0.$ From this, we obtain

$\mathbb{E}\left[\tilde{T}^{4}\right]\leq
2^{4}\mathbb{E}\left[\tilde{T}_{1}^{4}\right]+2^{4}\mathbb{E}\left[\tilde{T}_{2}^{4}\right]\leq
C^{\prime}\left[\frac{1}{\alpha^{4}}+\sqrt{\frac{\mu}{\alpha}}\frac{1}{(\sqrt{\mu}-\sqrt{\alpha})^{8}}\right].$
(E.28)

Moreover, using [16, Lemma EC.17], for any $x\in\mathbb{N}$,
$\mathbb{P}\left(\overline{Q}\geq
x\right)=\frac{1}{\sum_{n=1}^{x}\left(\frac{\mu}{\alpha}\right)^{n-1}}\leq\left(\frac{\alpha}{\mu}\right)^{x-1}.$
From the above bound, we obtain

$\displaystyle\mathbb{E}\left[\overline{Q}^{4}\mathds{1}\left(\overline{Q}\geq
a\right)\right]\leq\sum_{x=a^{4}}^{\infty}\left(\frac{\alpha}{\mu}\right)^{\lfloor
x^{1/4}\rfloor-1}\leq\frac{\mu^{2}}{\alpha^{2}}\int_{a^{4}-1}^{\infty}e^{-\log(\mu/\alpha)z^{1/4}}dz\leq\frac{C^{\prime}}{(\log(\mu/\alpha))^{4}}\frac{\mu^{2}}{\alpha^{2}}\left(\frac{\alpha}{\mu}\right)^{a/4}.$
(E.29)

Using (E.27), (E.28) and (E.29) in (E), we obtain

$\mathbb{E}\Big{[}\left(\int_{0}^{\tilde{T}}Q(s)ds\right)^{2}\Big{]}\leq
8a^{2}\left[\frac{1}{\alpha^{2}}+\frac{\mu}{(\mu-\alpha)^{3}}\right]+2C^{\prime}\left[\frac{1}{\alpha^{2}}+\frac{\left(\mu/\alpha\right)^{1/4}}{(\sqrt{\mu}-\sqrt{\alpha})^{4}}\right]\frac{\mu/\alpha}{(\log(\mu/\alpha))^{2}}\left(\frac{\alpha}{\mu}\right)^{a/8},$

which proves the lemma. ∎

###### Proof.

Proof of Lemma 6.5. Throughout this proof, we will denote by $C,C^{\prime}$
universal positive constants, possibly depending on $B,T$ but not $N$, whose
values might change between lines.

We will only prove the first case since the second case is similar. As before,
due to Lemma 6.3, we will prove the convergence to zero on the event
$\bar{E}^{(N)}$. Note that the process $I^{(N)}_{u,i}(t)$ is defined only on
the time interval $[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$. However, let us
assume that $I^{(N)}_{u,i}(t)$ is allowed to continue past time
$\xi^{(N)}_{2i+3}$, $i\in\\{0,...,i^{*}_{N}-1\\},$ and let
$\xi^{(N)}_{u,i}\coloneqq\inf\\{t\geq\xi^{(N)}_{2i+3}:I^{(N)}_{u,i}(t)=0\\}$.
Further, let
$Z^{(N)}_{i}=\int_{\xi^{(N)}_{2i-1}}^{\xi^{(N)}_{u,i-1}}\Big{(}I^{(N)}_{u,i-1}(s)-\frac{\lambda^{(N)}_{u,i-1}}{\mu^{(N)}_{u,i-1}-\lambda^{(N)}_{u,i-1}}\Big{)}ds,\
i\geq 1.$ We expand the first integral in the lemma for $j\geq 0$ as follows:

$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{2i+3}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds$
$\displaystyle=$
$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}Z^{(N)}_{i+1}-\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+3}}^{\xi^{(N)}_{u,i}}\Big{(}I^{(N)}_{u,i}(s)-\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\Big{)}ds$
$\displaystyle\leq$
$\displaystyle\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=1}^{j+1}Z^{(N)}_{i}+\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=0}^{j}\int_{\xi^{(N)}_{2i+3}}^{\xi^{(N)}_{u,i}}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}ds.$
(E.30)

Define $M^{(N)}_{0}=0$ and
$M^{(N)}_{j}:=\frac{1}{N^{\frac{1}{2}+\varepsilon}}\sum_{i=1}^{j}Z^{(N)}_{i},\
j\geq 1.$ Write $\tilde{\mathcal{G}}_{i}$ for the stopped natural filtration
of the original queueing process up till time $\xi^{(N)}_{2i+1}$, $i\geq 0$.
Define $\mathcal{G}_{0}:=\tilde{\mathcal{G}}_{0}$ and
$\mathcal{G}_{i}:=\tilde{\mathcal{G}}_{i}\cup\left(\cup_{l=0}^{i-1}\sigma\\{I^{(N)}_{u,l}(t):t\geq\xi^{(N)}_{2l+1}\\}\right),\
i\geq 1.$ By (E.24), $\mathbb{E}\left(Z^{(N)}_{i+1}|\mathcal{G}_{i}\right)=0$
for any $i\geq 0$. Hence, $M^{(N)}$ is a martingale. We will show that, for
any $\eta>0$,

$\mathbb{P}\left(\left\\{\sup_{j\leq
i^{*}_{N}}\left|M^{(N)}_{j}\right|>\eta\right\\}\cap\bar{E}^{(N)}\right)\rightarrow
0\ \text{ as }\ N\rightarrow\infty.$ (E.31)

Define the stopping time $\tilde{\tau}$ with respect to
$\\{\mathcal{G}_{i}\\}_{i\geq 0}$ by $\tilde{\tau}:=\inf\\{i\geq
0:\xi^{(N)}_{2i+1}\geq T(N,B)\ \text{ or }\
Q^{(N)}_{2}(\xi^{(N)}_{2i+1})>N^{\frac{1}{2}+\varepsilon+\delta}\\}.$ Now, we
estimate $\mathbb{E}\left[(Z^{(N)}_{i+1})^{2}|\mathcal{G}_{i}\right]$ on the
event $\\{\tilde{\tau}\geq i+1\\}$. Note that for $i\geq 0$,

$\mathbb{E}\left[(Z^{(N)}_{i+1})^{2}|\mathcal{G}_{i}\right]\leq
2\mathbb{E}\left[\left(\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{u,i}}I^{(N)}_{u,i}(s)ds\right)^{2}\Big{|}\mathcal{G}_{i}\right]+2\mathbb{E}\left[\left(\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\right)^{2}\left(\xi^{(N)}_{u,i}-\xi^{(N)}_{2i+1}\right)^{2}\Big{|}\mathcal{G}_{i}\right].$
(E.32)

Recall
$\lambda^{(N)}_{u,i}=2N-S^{(N)}(\xi^{(N)}_{2i+1})+13N^{\frac{1}{2}-\varepsilon+\delta}$
and $\mu^{(N)}_{u,i}=N-\beta N^{\frac{1}{2}-\varepsilon}$. By (E.25), there
exists a deterministic universal constant $C>0$ such that for any integer
$a\geq 2$,

$\displaystyle\mathbb{E}\bigg{[}\bigg{(}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{u,i}}I^{(N)}_{u,i}(s)ds\bigg{)}^{2}\Big{|}\mathcal{G}_{i}\bigg{]}\leq
C\bigg{[}a^{2}\bigg{(}\frac{1}{\big{(}\lambda^{(N)}_{u,i}\big{)}^{2}}+\frac{\mu^{(N)}_{u,i}}{(\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i})^{3}}\bigg{)}$
$\displaystyle\qquad+\bigg{(}\frac{1}{\big{(}\lambda^{(N)}_{u,i}\big{)}^{2}}+\bigg{(}\frac{\mu^{(N)}_{u,i}}{\lambda^{(N)}_{u,i}}\bigg{)}^{1/4}\frac{1}{\big{(}\sqrt{\mu^{(N)}_{u,i}}-\sqrt{\lambda^{(N)}_{u,i}}\big{)}^{4}}\bigg{)}\frac{1}{(\log(\mu^{(N)}_{u,i}/\lambda^{(N)}_{u,i}))^{2}}\frac{\mu^{(N)}_{u,i}}{\lambda^{(N)}_{u,i}}\bigg{(}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}}\bigg{)}^{a/8}\bigg{]}.$

Note that, on the event $\\{\tilde{\tau}\geq i+1\\}$, for sufficiently large
$N$, $N-N^{\frac{1}{2}+\varepsilon+\delta}\leq\lambda^{(N)}_{u,i}\leq
N-\frac{B}{2}N^{\frac{1}{2}+\varepsilon}.$ Using these bounds, on the event
$\\{\tilde{\tau}\geq i+1\\}$, for sufficiently large $N$,

$\displaystyle\frac{1}{\left(\lambda^{(N)}_{u,i}\right)^{2}}+\frac{\mu^{(N)}_{u,i}}{(\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i})^{3}}$
$\displaystyle\leq CN^{-\frac{1}{2}-3\varepsilon},$
$\displaystyle\bigg{(}\frac{1}{\big{(}\lambda^{(N)}_{u,i}\big{)}^{2}}+\bigg{(}\frac{\mu^{(N)}_{u,i}}{\lambda^{(N)}_{u,i}}\bigg{)}^{1/4}\frac{1}{\big{(}\sqrt{\mu^{(N)}_{u,i}}-\sqrt{\lambda^{(N)}_{u,i}}\big{)}^{4}}\bigg{)}\frac{1}{(\log(\mu^{(N)}_{u,i}/\lambda^{(N)}_{u,i}))^{2}}\frac{\mu^{(N)}_{u,i}}{\lambda^{(N)}_{u,i}}$
$\displaystyle\leq CN^{1-6\varepsilon},$
$\displaystyle\bigg{(}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}}\bigg{)}^{a/8}$
$\displaystyle\leq e^{-C^{\prime}aN^{-\frac{1}{2}+\varepsilon}}.$

Therefore, choosing $a=N^{\frac{1}{2}-\varepsilon+\delta}$, for sufficiently
large $N$, on the event $\\{\tilde{\tau}\geq i+1\\}$,

$\mathbb{E}\bigg{[}\bigg{(}\int_{\xi^{(N)}_{2i+1}}^{\xi^{(N)}_{u,i}}I^{(N)}_{u,i}(s)ds\bigg{)}^{2}\Big{|}\mathcal{G}_{i}\bigg{]}\leq
CN^{\frac{1}{2}-5\varepsilon+2\delta}.$ (E.33)

Now, we estimate the second term in (E.32). Note that, for sufficiently large
$N$,
$\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\leq\frac{4}{B}N^{\frac{1}{2}-\varepsilon}.$
Moreover, using [1, Section 7.9, Page 75], on the event $\\{\tilde{\tau}\geq
i+1\\}$, for sufficiently large $N$,

$\mathbb{E}\left[\left(\xi^{(N)}_{u,i}-\xi^{(N)}_{2i+1}\right)^{2}\right]\leq
4\bigg{[}\frac{1}{\left(\lambda^{(N)}_{u,i}\right)^{2}}+\frac{\mu^{(N)}_{u,i}}{(\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i})^{3}}\bigg{]}\leq
4CN^{-\frac{1}{2}-3\varepsilon}.$

Hence, on the event $\\{\tilde{\tau}\geq i+1\\}$, for sufficiently large $N$,

$\mathbb{E}\bigg{[}\bigg{(}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\bigg{)}^{2}\left(\xi^{(N)}_{u,i}-\xi^{(N)}_{2i+1}\right)^{2}\Big{|}\mathcal{G}_{i}\bigg{]}\leq
CN^{\frac{1}{2}-5\varepsilon}.$ (E.34)

Using the bounds (E.33) and (E.34) in (E.32), we conclude that, on the event
$\\{\tilde{\tau}\geq i+1\\}$, for sufficiently large $N$,

$\mathbb{E}\left[(Z^{(N)}_{i+1})^{2}|\mathcal{G}_{i}\right]\leq
CN^{\frac{1}{2}-5\varepsilon+2\delta}.$ (E.35)

The quadratic variation of the stopped martingale
$\\{M^{(N)}_{j\wedge\tilde{\tau}}\\}_{j\geq 0}$ is given by: $\langle
M^{(N)}\rangle_{0}=0$ and $\langle
M^{(N)}\rangle_{j\wedge\tilde{\tau}}=N^{-1-2\varepsilon}\sum_{i=1}^{j\wedge\tilde{\tau}}\mathbb{E}\left[(Z^{(N)}_{i})^{2}|\mathcal{G}_{i-1}\right],\
j\geq 1.$ Using (E.35), we obtain $\langle
M^{(N)}\rangle_{j\wedge\tilde{\tau}}\leq
CN^{-\frac{1}{2}-7\varepsilon+2\delta}j,\ j\geq 1.$ Hence, by Doob’s $L^{2}$
inequality, for any $\eta>0$, for large enough $N$,

$\mathbb{P}\bigg{(}\sup_{j\leq\tilde{\tau}\wedge\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil}\left|M^{(N)}_{j}\right|>\eta\bigg{)}=\mathbb{P}\bigg{(}\sup_{j\leq\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil}\left|M^{(N)}_{j\wedge\tilde{\tau}}\right|>\eta\bigg{)}\leq
12CT\eta^{-2}N^{-4\varepsilon+3\delta}.$ (E.36)

Note that, on the event $\bar{E}^{(N)}$, $\tilde{\tau}=i^{*}_{N}+1$.
Therefore, for any $\eta>0$, for sufficiently large $N$,

$\displaystyle\mathbb{P}\bigg{(}\bigg{\\{}\sup_{j\leq
i^{*}_{N}}\left|M^{(N)}_{j}\right|>\eta\bigg{\\}}\cap\bar{E}^{(N)}\bigg{)}$
$\displaystyle\leq\mathbb{P}\bigg{(}\sup_{j\leq\tilde{\tau}\wedge\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil}\left|M^{(N)}_{j}\right|>\eta\bigg{)}+\mathbb{P}\Big{(}\big{\\{}i^{*}_{N}\geq\lceil
12TN^{\frac{1}{2}+3\varepsilon+\delta}\rceil\big{\\}}\cap\bar{E}^{(N)}\Big{)}$
$\displaystyle\leq
12CT\eta^{-2}N^{-4\varepsilon+3\delta}+\frac{1}{9T}N^{-4\varepsilon-2\delta},$

where the last bound follows from (E.36) and Lemma E.2. Recalling
$\delta<\varepsilon/2$, (E.31) follows from the above bound. Now consider the
second term of the right hand side of the inequality (E). For $i\leq
i^{*}_{N}-1$, for sufficient large $N$, on $\bar{E}^{(N)}$,
$0\leq\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}\leq\frac{4N^{\frac{1}{2}-\varepsilon}}{B}.$
Additionally, let
$\xi^{(N)}_{l,1,i}\coloneqq\inf\\{t\geq\xi^{(N)}_{2i+1}:I^{(N)}_{l,i}(t)>0\\}$
and
$\xi^{(N)}_{l,i}\coloneqq\inf\\{t\geq\xi^{(N)}_{l,1,i}:I^{(N)}_{l,i}(t)=0\\}$,
$i\in\\{0,...,i^{*}_{N}-1\\}$. We have that for sufficient large $N$,

$\displaystyle\sum_{i=1}^{j}\int_{\xi^{(N)}_{2i+3}}^{\xi^{(N)}_{u,i}}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}ds$
$\displaystyle\leq\sum_{i=1}^{j}\int_{\xi^{(N)}_{l,i}}^{\xi^{(N)}_{u,i}}\frac{\lambda^{(N)}_{u,i}}{\mu^{(N)}_{u,i}-\lambda^{(N)}_{u,i}}ds$
$\displaystyle\leq\frac{4N^{\frac{1}{2}-\varepsilon}}{B}\sum_{i=1}^{i^{*}_{N}-1}\Big{(}\xi^{(N)}_{u,i}-\xi^{(N)}_{l,i}\Big{)},$
(E.37)

where the first inequality is due to the fact that for
$t\in[\xi^{(N)}_{2i+1},\xi^{(N)}_{2i+3})$, $I^{(N)}_{l,i}(t)\leq
I^{(N)}(t)\leq I^{(N)}_{u,i}(t)$, which implies that
$\xi^{(N)}_{l,i}\leq\xi^{(N)}_{2i+3}\leq\xi^{(N)}_{u,i}$. For an M/M/1 with
arrival rate $\alpha$ and departure rate $\mu$,
$\mathbb{E}(\tilde{T})=\frac{1}{\alpha}+\frac{1}{\mu-\alpha}$ where
$\tilde{T}$ is same as in Lemma E.3. Then, for each
$i\in\\{0,...,i^{*}_{N}-1\\}$, we have for all sufficiently large $N$, on
$\bar{E}^{(N)}$,

$\displaystyle\mathbb{E}\Big{(}\xi^{(N)}_{u,i}-\xi^{(N)}_{l,i}\Big{)}=$
$\displaystyle\mathbb{E}\Big{(}\xi^{(N)}_{u,i}-\xi^{(N)}_{2i+1}\Big{)}-\mathbb{E}\Big{(}\xi^{(N)}_{l,i}-\xi^{(N)}_{2i+1}\Big{)}$
$\displaystyle=$
$\displaystyle\frac{1}{2N-S^{(N)}(\xi^{(N)}_{2i+1})+13N^{\frac{1}{2}-\varepsilon+\delta}}+\frac{1}{S^{(N)}(\xi^{(N)}_{2i+1})-N-\beta
N^{\frac{1}{2}-\varepsilon}-13N^{\frac{1}{2}-\varepsilon+\delta}}$