Exp$(1),$ $EX=1$ and $EX^{2}=2,$ we obtain

$E_{\lambda^{\varepsilon}}\left(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)^{2}=E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{2}\rightarrow
2\text{ and }E_{\lambda^{\varepsilon}}\mathcal{Z}_{1}^{\varepsilon}\rightarrow
1.$

as $\varepsilon\rightarrow 0.$ This implies
Var${}_{\lambda^{\varepsilon}}(\mathcal{Z}_{1}^{\varepsilon})\rightarrow 1$ as
$\varepsilon\rightarrow 0.$ Obviously, there exists some
$\varepsilon_{0}\in(0,1)$ such that
$\inf\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\mathrm{Var}_{\lambda^{\varepsilon}}(\mathcal{Z}_{1}^{\varepsilon})\geq
1/2>0.$ This completes the proof. ∎

######  Remark 8.9.

Throughout the rest of this section, we will use $C$ to denote a constant in
$(0,\infty)$ which is independent of $\varepsilon$ but whose value may change
from use to use.

###  8.1 Chernoff bound

In this subsection we will provide upper bounds for

$\mathfrak{P}_{1}\doteq\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}+1}^{\infty}P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)\text{ and
}\mathfrak{R}_{1}\doteq\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}+1}^{\infty}\left(2n+1\right)P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)$

via a Chernoff bound. The following result is well known and its proof is
standard.

######  Lemma 8.10 (Chernoff bound).

Let $X_{1},\ldots,X_{n}$ be an iid sequence of random variables. For any
$a\in\mathbb{R}$ and for any $t\in(0,\infty)$

$P\left(X_{1}+\cdots+X_{n}\leq
a\right)\leq\left(Ee^{-tX_{1}}\right)^{n}e^{ta}.$

Recall that $\Gamma^{\varepsilon}\doteq
T^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}$ and
$\gamma^{\varepsilon}\doteq\left(\Gamma^{\varepsilon}\right)^{-\ell}$ with
some $\ell\in(0,1)$ which will be chosen later.

######  Lemma 8.11.

Given any $\delta>0$ and any $\ell>0,$ there exists $\varepsilon_{0}\in(0,1)$
such that for any $\varepsilon\in(0,\varepsilon_{0})$

$P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)\leq e^{-n\left(\Gamma^{\varepsilon}\right)^{-2\ell}}$

for any $n\geq\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}.$ In
addition,

$\mathfrak{P}_{1}\leq
C\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}\text{
and }\mathfrak{R}_{1}\leq
C\left(\Gamma^{\varepsilon}\right)^{1+2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}+C\left(\Gamma^{\varepsilon}\right)^{4\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}.$

###### Proof.

Given $\delta>0$, $\ell>0$ and $\varepsilon\in(0,1),$ we find that for
$n\geq\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}$

$\displaystyle P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)$
$\displaystyle=P_{\lambda^{\varepsilon}}\left(\frac{\tau_{1}^{\varepsilon}+\left(\tau_{2}^{\varepsilon}-\tau_{1}^{\varepsilon}\right)+\cdots+\left(\tau_{n}^{\varepsilon}-\tau_{n-1}^{\varepsilon}\right)}{E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\leq\Gamma^{\varepsilon}\right)$
$\displaystyle\leq
P_{\lambda^{\varepsilon}}\left(\frac{\tau_{1}^{\varepsilon}+\left(\tau_{2}^{\varepsilon}-\tau_{1}^{\varepsilon}\right)+\cdots+\left(\tau_{n}^{\varepsilon}-\tau_{n-1}^{\varepsilon}\right)}{E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\leq\frac{n}{1+2\gamma^{\varepsilon}}\right)$
$\displaystyle\leq\left(E_{\lambda^{\varepsilon}}e^{-\gamma^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}}\right)e^{\frac{n\gamma^{\varepsilon}}{1+2\gamma^{\varepsilon}}},$

where
$\mathcal{Z}_{1}^{\varepsilon}=\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}.$
We use the fact that
$\\{\tau_{n}^{\varepsilon}-\tau_{n-1}^{\varepsilon}\\}_{n\in\mathbb{N}}$ are
iid and apply Lemma 8.10 (Chernoff bound) with
$a=n/\left(1+2\gamma^{\varepsilon}\right)$ and $t=\gamma^{\varepsilon}$ for
the last inequality. Therefore, in order to verify the first claim, it
suffices to show that

$\left(E_{\lambda^{\varepsilon}}e^{-\gamma^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}}\right)e^{\frac{\gamma^{\varepsilon}}{1+2\gamma^{\varepsilon}}}\leq
e^{-\left(\gamma^{\varepsilon}\right)^{2}}=e^{-\left(\Gamma^{\varepsilon}\right)^{-2\ell}}.$

We observe that for any $x\geq 0,$ $e^{-x}\leq 1-x+x^{2}/2,$ and this gives

$\displaystyle
E_{\lambda^{\varepsilon}}e^{-\gamma^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}}\leq
1-E_{\lambda^{\varepsilon}}\left(\gamma^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}\right)+E_{\lambda^{\varepsilon}}\left(\gamma^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}\right)^{2}/2=1-\gamma^{\varepsilon}+\left(\gamma^{\varepsilon}\right)^{2}E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{2}/2.$

Moreover, since we can apply Lemma 8.8 to find
$E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{2}\rightarrow
2$ as $\varepsilon\rightarrow 0,$ there exists $\varepsilon_{0}\in(0,1)$ such
that for any $\varepsilon\in(0,\varepsilon_{0})$,
$E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{2}\leq
9/4.$ Thus, for any $\varepsilon\in(0,\varepsilon_{0})$

$\left(E_{\lambda^{\varepsilon}}e^{-\gamma^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}}\right)e^{\frac{\gamma^{\varepsilon}}{1+2\gamma^{\varepsilon}}}\leq\exp\left\\{\gamma^{\varepsilon}/(1+2\gamma^{\varepsilon})+\log(1-\gamma^{\varepsilon}+(9/8)\left(\gamma^{\varepsilon}\right)^{2})\right\\}.$

Using a Taylor series expansion we find that for all $\left|x\right|<1$

$1/(1+x)=1-x+O\left(x^{2}\right)\text{ and
}\log\left(1+x\right)=x-x^{2}/2+O\left(x^{3}\right),$

which gives

$\displaystyle\gamma^{\varepsilon}/(1+2\gamma^{\varepsilon})+\log(1-\gamma^{\varepsilon}+(9/8)\left(\gamma^{\varepsilon}\right)^{2})$
$\displaystyle\quad=\gamma^{\varepsilon}-2\left(\gamma^{\varepsilon}\right)^{2}+[(-\gamma^{\varepsilon}+(9/8)\left(\gamma^{\varepsilon}\right)^{2}]-[-\gamma^{\varepsilon}+(9/8)\left(\gamma^{\varepsilon}\right)^{2}]^{2}/2+O((\gamma^{\varepsilon})^{3})$
$\displaystyle\quad=-(11/8)\left(\gamma^{\varepsilon}\right)^{2}+O(\left(\gamma^{\varepsilon}\right)^{3})\leq-\left(\gamma^{\varepsilon}\right)^{2},$

for all $\varepsilon\in(0,\varepsilon_{0})$. We are done for part 1.

For part 2, we use the estimate from part 1 and find

$\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}+1}^{\infty}P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)\leq\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}+1}^{\infty}e^{-n\left(\gamma^{\varepsilon}\right)^{2}}\leq\frac{e^{-\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}\left(\gamma^{\varepsilon}\right)^{2}}}{1-e^{-\left(\gamma^{\varepsilon}\right)^{2}}}.$

Since $e^{-x}\leq 1-x+x^{2}/2$ for any $x\in\mathbb{R},$ we have $1-e^{-x}\geq
x-x^{2}/2\geq x-x/2=x/2$ for all $x\in(0,1),$ and thus $1/(1-e^{-x})\leq 2/x$
for all $x\in(0,1).$ As a result

$\displaystyle\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}+1}^{\infty}P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)\leq\frac{e^{-\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}\left(\gamma^{\varepsilon}\right)^{2}}}{1-e^{-\left(\gamma^{\varepsilon}\right)^{2}}}\leq\frac{2}{\left(\gamma^{\varepsilon}\right)^{2}}e^{-\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}\left(\gamma^{\varepsilon}\right)^{2}}\leq
2\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}.$

This completes the proof of part 2.

Finally, for part 3, we use the fact that for $x\in(0,1),$ and for any
$k\in\mathbb{N},$

$\sum\nolimits_{n=k}^{\infty}nx^{n}=kx^{k}(1-x)^{-1}+x^{k+1}(1-x)^{-2}\leq(k(1-x)^{-1}+(1-x)^{-2})x^{k}.$

Using the estimate from part 1 once again, we have

$\displaystyle\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}+1}^{\infty}nP_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)$
$\displaystyle\leq\sum_{n=\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}}^{\infty}ne^{-n\left(\gamma^{\varepsilon}\right)^{2}}$
$\displaystyle\leq\left(\frac{\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}}{1-e^{-\left(\gamma^{\varepsilon}\right)^{2}}}+\left(1-e^{-\left(\gamma^{\varepsilon}\right)^{2}}\right)^{-2}\right)e^{-\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}\left(\gamma^{\varepsilon}\right)^{2}}$
$\displaystyle\leq\left(4\left(\Gamma^{\varepsilon}\right)^{1+2\ell}+4\left(\Gamma^{\varepsilon}\right)^{4\ell}\right)e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}.$

We are done. ∎

######  Remark 8.12.

If $0<\ell<1/2,$ then $\mathfrak{P}_{1}$ and $\mathfrak{R}_{1}$ converge to
$0$ doubly exponentially fast as $\varepsilon\rightarrow 0$ in the sense that
for any $k\in(0,\infty)$

$\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[\left(\Gamma^{\varepsilon}\right)^{k}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}\right]=\infty.$

###  8.2 Berry-Esseen bound

In this subsection we will provide upper bounds for

$\mathfrak{P}_{2}\doteq\sum_{n=\left(1-2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}}^{\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}}P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)\text{ and
}\mathfrak{R}_{2}\doteq\sum_{n=\left(1-2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}}^{\left(1+2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}}\left(2n+1\right)P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)$

via the Berry-Esseen bound.

We first recall that
$\Gamma^{\varepsilon}=T^{\varepsilon}/E_{\lambda^{\varepsilon}}^{\varepsilon}\tau_{1}^{\varepsilon}$.
The following is Theorem 1 in [11, Chapter XVI.5].

######  Theorem 8.13 (Berry-Esseen).

Let $\left\\{X_{n}\right\\}_{n\in\mathbb{N}}$ be independent real-valued
random variables with a common distribution such that

$E\left(X_{1}\right)=0,\text{ }\sigma^{2}\doteq
E\left(X_{1}\right)^{2}>0,\text{ }\rho\doteq E\left|X_{1}\right|^{3}<\infty.$

Then for all $x\in\mathbb{R}$ and $n\in\mathbb{N},$

$\left|P\left(\frac{X_{1}+\cdots+X_{n}}{\sigma\sqrt{n}}\leq
x\right)-\Phi\left(x\right)\right|\leq\frac{3\rho}{\sigma^{3}\sqrt{n}},$

where $\Phi\left(\cdot\right)$ is the distribution function of
$N\left(0,1\right).$

######  Corollary 8.14.

For any $\varepsilon>0,$ let
$\left\\{X_{n}^{\varepsilon}\right\\}_{n\in\mathbb{N}}$ be independent real-
valued random variables with a common distribution such that

$E\left(X_{1}^{\varepsilon}\right)=0,\text{
}\left(\sigma^{\varepsilon}\right)^{2}\doteq
E\left(X_{1}^{\varepsilon}\right)^{2}>0,\text{ }\rho^{\varepsilon}\doteq
E\left|X_{1}^{\varepsilon}\right|^{3}<\infty.$

Assume that there exists $\varepsilon_{0}\in(0,1)$ such that

$\hat{\rho}\text{
}\doteq\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\rho^{\varepsilon}<\infty\text{
and
}\hat{\sigma}^{2}\doteq\inf\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\left(\sigma^{\varepsilon}\right)^{2}>0.$

Then for all $x\in\mathbb{R},n\in\mathbb{N}$ and
$\varepsilon\in(0,\varepsilon_{0}),$

$\left|P\left(\frac{X_{1}^{\varepsilon}+\cdots+X_{n}^{\varepsilon}}{\sigma^{\varepsilon}\sqrt{n}}\leq
x\right)-\Phi\left(x\right)\right|\leq\frac{3\rho^{\varepsilon}}{\left(\sigma^{\varepsilon}\right)^{3}\sqrt{n}}\leq\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{n}}.$

######  Lemma 8.15.

Given any $\delta>0$ and any $\ell>0,$ there exists $\varepsilon_{0}\in(0,1)$
such that for any $\varepsilon\in(0,\varepsilon_{0})$ and
$k\in\mathbb{N}_{0}$, $0\leq k\leq 2\gamma^{\varepsilon}\Gamma^{\varepsilon}$

$P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}+k}^{\varepsilon}\leq
T^{\varepsilon}\right)\leq
1-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}+k}}$

and

$P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}-k}^{\varepsilon}\leq
T^{\varepsilon}\right)\leq\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}-k}},$

where $(\sigma^{\varepsilon})^{2}\doteq
E_{\lambda^{\varepsilon}}\left(\mathfrak{X}_{1}^{\varepsilon}\right)^{2},$
$\hat{\rho}\doteq\sup_{\varepsilon\in(0,\varepsilon_{0})}E_{\lambda^{\varepsilon}}\left|\mathfrak{X}_{1}^{\varepsilon}\right|^{3}<\infty$
and
$\hat{\sigma}^{2}\doteq\inf_{\varepsilon\in(0,\varepsilon_{0})}(\sigma^{\varepsilon})^{2}>0$
with
$\mathfrak{X}_{1}^{\varepsilon}\doteq\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}^{\varepsilon}\tau_{1}^{\varepsilon}-1.$

###### Proof.

For any $n\in\mathbb{N},$ we define
$\mathfrak{X}_{n}^{\varepsilon}\doteq\mathcal{Z}_{n}^{\varepsilon}-E_{\lambda^{\varepsilon}}^{\varepsilon}\mathcal{Z}_{1}^{\varepsilon}$
with
$\mathcal{Z}_{n}^{\varepsilon}\doteq(\tau_{n}^{\varepsilon}-\tau_{n-1}^{\varepsilon})/E_{\lambda^{\varepsilon}}^{\varepsilon}\tau_{1}^{\varepsilon}.$
Obviously, $E_{\lambda^{\varepsilon}}\mathcal{Z}_{n}^{\varepsilon}=1$ and
$E_{\lambda^{\varepsilon}}\mathfrak{X}_{n}^{\varepsilon}=0$ and if we apply
Lemma 8.8, then we find that there exists some $\varepsilon_{0}\in(0,1)$ such
that
$\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{3}<\infty\text{
and
}\inf\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\mathrm{Var}_{\lambda^{\varepsilon}}(\mathcal{Z}_{1}^{\varepsilon})>0.$
Since $\mathcal{Z}_{1}^{\varepsilon}\geq 0$, Jensen’s inequality implies
$\left(E_{\lambda^{\varepsilon}}\mathcal{Z}_{1}^{\varepsilon}\right)^{3}\leq
E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{3}$, and
therefore

$\displaystyle\hat{\rho}\leq
4\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\left(E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{3}+\left(E_{\lambda^{\varepsilon}}\mathcal{Z}_{1}^{\varepsilon}\right)^{3}\right)\leq
8\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}E_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)^{3}<\infty,$

and

$\hat{\sigma}^{2}=\inf\nolimits_{\varepsilon\in(0,\varepsilon_{0})}E_{\lambda^{\varepsilon}}\left(\mathfrak{X}_{1}^{\varepsilon}\right)^{2}=\inf\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\mathrm{Var}_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}\right)>0.$

Therefore we can use Corollary 8.14 with the iid sequence
$\left\\{\mathfrak{X}_{n}^{\varepsilon}\right\\}_{n\in\mathbb{N}}$ to find
that for any $k\in\mathbb{N}_{0}$ and $0\leq k\leq
2\gamma^{\varepsilon}\Gamma^{\varepsilon}$

$\displaystyle
P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}+k}^{\varepsilon}\leq
T^{\varepsilon}\right)$
$\displaystyle=P_{\lambda^{\varepsilon}}\left(\mathcal{Z}_{1}^{\varepsilon}+\cdots+\mathcal{Z}_{\Gamma^{\varepsilon}+k}^{\varepsilon}\leq\Gamma^{\varepsilon}\right)$
$\displaystyle=P_{\lambda^{\varepsilon}}\left(\frac{\mathfrak{X}_{1}^{\varepsilon}+\cdots+\mathfrak{X}_{\Gamma^{\varepsilon}+k}^{\varepsilon}}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\leq\frac{-k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)$
$\displaystyle\leq
1-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}+k}},$

and similarly

$\displaystyle
P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}-k}^{\varepsilon}\leq
T^{\varepsilon}\right)$
$\displaystyle\leq\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}-k}}.$

∎

######  Lemma 8.16.

Given any $\delta>0$ and any $\ell\in(0,1/2),$ there exists
$\varepsilon_{0}\in(0,1)$ such that for any
$\varepsilon\in(0,\varepsilon_{0})$, $\mathfrak{P}_{2}\leq
C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+2\left(\Gamma^{\varepsilon}\right)^{1-\ell}.$

###### Proof.

We rewrite $\mathfrak{P}_{2}$ as

$\displaystyle\mathfrak{P}_{2}=\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}-k}^{\varepsilon}\leq
T^{\varepsilon}\right)+P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}}^{\varepsilon}\leq
T^{\varepsilon}\right)+\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}+k}^{\varepsilon}\leq
T^{\varepsilon}\right).$

Then we use the upper bounds from Lemma 8.15 to get

$\displaystyle\mathfrak{P}_{2}$
$\displaystyle\leq\sum_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left[\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}-k}}\right]+1+\sum_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left[1-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}+k}}\right]$
$\displaystyle\leq\frac{24\hat{\rho}}{\hat{\sigma}^{3}}\gamma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}+1+2\gamma^{\varepsilon}\Gamma^{\varepsilon}+\sum_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left[\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)\right].$

The sum of the first three terms is easily bounded above by
$C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+2\left(\Gamma^{\varepsilon}\right)^{1-\ell}$.
We will show that the last term is bounded above by a constant to complete the
proof.

To prove this, we observe that for any $k\leq
2\gamma^{\varepsilon}\Gamma^{\varepsilon},$ we may assume
$k\leq\Gamma^{\varepsilon}/2$ by taking $\varepsilon$ sufficiently small. Then
we apply the Mean Value Theorem and find

$\displaystyle\left|\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)\right|\leq\sup\limits_{x\in\left[\frac{\sqrt{2/3}k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}},\frac{\sqrt{2}k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}}\right]}\phi\left(x\right)\cdot\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}-\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right),$

where $\phi\left(x\right)\doteq e^{-\frac{x^{2}}{2}}/\sqrt{2\pi}$ and since
$0\leq k\leq\Gamma^{\varepsilon}/2,$ we have

$\left[\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}},\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right]\subset\left[\frac{\sqrt{2/3}k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}},\frac{\sqrt{2}k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}}\right].$

Additionally, because $\phi\left(x\right)=e^{-\frac{x^{2}}{2}}/\sqrt{2\pi}$ is
a monotone decreasing function on $[0,\infty)$, we find that

$x\in[(\sqrt{2/3}k)(\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}),(\sqrt{2}k)(\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}})]\quad\mbox{implies}\quad\phi\left(x\right)\leq
e^{-\frac{k^{2}}{3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}}}/{\sqrt{2\pi}}.$

Also, $\sqrt{1+x}-\sqrt{1-x}\leq 2x$ for all $x\in[0,1]$ and
$k\leq\Gamma^{\varepsilon}/2$ and a little algebra give
${k}/{\sqrt{\Gamma^{\varepsilon}-k}}-{k}/{\sqrt{\Gamma^{\varepsilon}+k}}\leq{4k^{2}}/{\Gamma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}}.$
Therefore we find

$\displaystyle\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left[\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)\right]$
$\displaystyle\qquad\leq\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\frac{1}{\sqrt{2\pi}}e^{-\frac{k^{2}}{3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}}}\frac{4k^{2}}{\sigma^{\varepsilon}\Gamma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}}\leq\frac{4}{\sigma^{\varepsilon}\Gamma^{\varepsilon}}\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\int_{k-1}^{k}\frac{\left(1+x\right)^{2}}{\sqrt{2\pi\Gamma^{\varepsilon}}}e^{-\frac{x^{2}}{3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}}}dx$
$\displaystyle\qquad\leq\frac{4}{\Gamma^{\varepsilon}}\sqrt{\frac{3}{2}}\int_{0}^{\infty}\frac{\left(1+x\right)^{2}}{\sqrt{3\pi\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}}}e^{-\frac{x^{2}}{3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}}}dx\leq\frac{6}{\Gamma^{\varepsilon}}E\left(1+X^{+}\right)^{2},$

where $X\sim
N(0,3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}/2).$ Finally,
since $E\left(1+X^{+}\right)^{2}\leq
2+2E\left(X^{2}\right)=2+3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon},$
this implies that

$\displaystyle\sum_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left[\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)\right]$
$\displaystyle\leq\frac{6}{\Gamma^{\varepsilon}}\left(2+3\left(\sigma^{\varepsilon}\right)^{2}\Gamma^{\varepsilon}\right)\leq
12+18\hat{\rho}^{2/3},$ (8.4)

where the last inequality is from

$\displaystyle\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\sigma^{\varepsilon}$
$\displaystyle=\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}\left(E_{\lambda^{\varepsilon}}(\mathfrak{X}_{1}^{\varepsilon}\right)^{2})^{1/2}\leq\sup\nolimits_{\varepsilon\in(0,\varepsilon_{0})}(E_{\lambda^{\varepsilon}}\left|\mathfrak{X}_{1}^{\varepsilon}\right|^{3})^{1/3}=\hat{\rho}^{1/3}.$

Since according to Lemma 8.15 $\hat{\rho}^{1/3}$ is finite, we are done. ∎

######  Lemma 8.17.

Given any $\delta>0$ and any $\ell\in(0,1/2),$ there exists
$\varepsilon_{0}\in(0,1)$ and a constant $C<\infty$ such that for any
$\varepsilon\in(0,\varepsilon_{0})$, $\mathfrak{R}_{2}\leq
4\left(\Gamma^{\varepsilon}\right)^{2-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)}.$

###### Proof.

The proof of this lemma is similar to the proof of Lemma 8.16. We rewrite
$\mathfrak{R}_{2}$ as

$\displaystyle\mathfrak{R}_{2}$
$\displaystyle=\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}-2k+1\right)P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}-k}^{\varepsilon}\leq
T^{\varepsilon}\right)+\left(2\Gamma^{\varepsilon}+1\right)P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}}^{\varepsilon}\leq
T^{\varepsilon}\right)$
$\displaystyle\quad+\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}+2k+1\right)P_{\lambda^{\varepsilon}}\left(\tau_{\Gamma^{\varepsilon}+k}^{\varepsilon}\leq
T^{\varepsilon}\right).$

Then we use the upper bounds from Lemma 8.15 to get

$\displaystyle\mathfrak{R}_{2}\leq\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}-2k+1\right)\left[\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}-k}}\right]+\left(2\Gamma^{\varepsilon}+1\right)$
$\displaystyle\quad\qquad+\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}+2k+1\right)\left[1-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}+k}}\right].$

The next thing is to pair all the terms carefully and bound these pairs
separately. We start with

$\displaystyle\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}-2k+1\right)\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)-\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}+2k+1\right)\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)$
$\displaystyle\quad\leq\left(2\Gamma^{\varepsilon}+1\right)\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left[\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}-k}}\right)-\Phi\left(\frac{k}{\sigma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}+k}}\right)\right]\leq
C\Gamma^{\varepsilon}.$

We use (8.4) for the last inequality. The second pair is

$\displaystyle\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}-2k+1\right)\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}-k}}+\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}+2k+1\right)\frac{3\hat{\rho}}{\hat{\sigma}^{3}\sqrt{\Gamma^{\varepsilon}+k}}$
$\displaystyle\quad=\frac{6\hat{\rho}}{\hat{\sigma}^{3}}\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(\sqrt{\Gamma^{\varepsilon}-k}+\sqrt{\Gamma^{\varepsilon}+k}\right)+\frac{3\hat{\rho}}{\hat{\sigma}^{3}}\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(\frac{1}{\sqrt{\Gamma^{\varepsilon}-k}}+\frac{1}{\sqrt{\Gamma^{\varepsilon}+k}}\right)$
$\displaystyle\quad\leq\frac{6\hat{\rho}}{\hat{\sigma}^{3}}\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}2\sqrt{2\Gamma^{\varepsilon}}+\frac{3\hat{\rho}}{\hat{\sigma}^{3}}\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}2\leq
C\gamma^{\varepsilon}\Gamma^{\varepsilon}\sqrt{\Gamma^{\varepsilon}}+C\gamma^{\varepsilon}\Gamma^{\varepsilon}\leq
C\left(\Gamma^{\varepsilon}\right)^{\frac{3}{2}-\ell},$

where the first inequality holds due to $k\leq\Gamma^{\varepsilon}/2$. The
third term is

$\displaystyle\sum\nolimits_{k=1}^{2\gamma^{\varepsilon}\Gamma^{\varepsilon}}\left(2\Gamma^{\varepsilon}+2k+1\right)+\left(2\Gamma^{\varepsilon}+1\right)$
$\displaystyle=4\gamma^{\varepsilon}\left(\Gamma^{\varepsilon}\right)^{2}+2\gamma^{\varepsilon}\Gamma^{\varepsilon}+4\left(\gamma^{\varepsilon}\Gamma^{\varepsilon}\right)^{2}+2\gamma^{\varepsilon}\Gamma^{\varepsilon}+\left(2\Gamma^{\varepsilon}+1\right)$
$\displaystyle\leq
4\gamma^{\varepsilon}\left(\Gamma^{\varepsilon}\right)^{2}+C\left(\gamma^{\varepsilon}\Gamma^{\varepsilon}\right)^{2}=4\left(\Gamma^{\varepsilon}\right)^{2-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)},$

where the inequality holds since for $\ell\in(0,1/2),$ $2-2\ell\geq 1$ and
this implies that $\left(2\Gamma^{\varepsilon}+1\right)\leq
C\left(\gamma^{\varepsilon}\Gamma^{\varepsilon}\right)^{2}.$ Lastly, combining
all the pairs and the corresponding upper bounds, we find that for any
$\ell\in(0,1/2),$

$\displaystyle\mathfrak{R}_{2}$
$\displaystyle\leq[4\left(\Gamma^{\varepsilon}\right)^{2-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)}]+C\Gamma^{\varepsilon}+C\left(\Gamma^{\varepsilon}\right)^{\frac{3}{2}-\ell}\leq
4\left(\Gamma^{\varepsilon}\right)^{2-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)},$

where $C$ is a constant which depends on $\ell$ only (and in particular is
independent of $\varepsilon$). ∎

###  8.3 Asymptotics of moments of $N^{\varepsilon}(T^{\varepsilon})$

In this subsection, we prove Lemma 8.2 and Lemma 8.4.

###### Proof of Lemma 8.2.

First, recall that

$E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)=\sum\nolimits_{n=0}^{\infty}P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
T^{\varepsilon}\right)=\mathfrak{P}_{1}+\mathfrak{P}_{2}+\mathfrak{P}_{3},$

where the $\mathfrak{P}_{i}$ are defined in (8.2). We can simply bound
$\mathfrak{P}_{3}$ from above by
$\left(1-2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}$. Applying Lemma
8.11 and Lemma 8.16 for the other terms, we have for any $\ell\in(0,1/2)$ that

$\displaystyle
E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)$
$\displaystyle\leq
C\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}+(C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+2\left(\Gamma^{\varepsilon}\right)^{1-\ell})+\left(1-2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}$
$\displaystyle=T^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}+C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}.$

On the other hand, from the definition of
$N^{\varepsilon}\left(T^{\varepsilon}\right),$
$E_{\lambda^{\varepsilon}}\tau_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}\geq
T^{\varepsilon}.$ Using Wald’s first identity, we find

$E_{\lambda^{\varepsilon}}\tau_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}=E_{\lambda^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}\left(\tau_{n}^{\varepsilon}-\tau_{n-1}^{\varepsilon}\right)=E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)\cdot
E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}.$

Hence

$0\leq\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}-\frac{1}{E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\leq\frac{1}{T^{\varepsilon}}[C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}].$

Therefore,

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}-\frac{1}{E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right|\geq\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[\frac{1}{T^{\varepsilon}}\left(C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}\right)\right].$

It remains to find an appropriate lower bound for the liminf.

We use (7.2), Lemma 8.7 and Remark 8.12 to find that for any $\eta>0,$ there
exists $\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$ and
any $\ell\in(0,1/2)$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[\frac{1}{T^{\varepsilon}}\left(C\left(\Gamma^{\varepsilon}\right)^{\frac{1}{2}-\ell}+\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}\right)\right]$
$\displaystyle\quad\geq\liminf_{\varepsilon\rightarrow 0}\varepsilon\log
T^{\varepsilon}+\min\left\\{\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\Gamma^{\varepsilon}\right)^{1/2-\ell},\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\left(\Gamma^{\varepsilon}\right)^{2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}\right)\right\\}$
$\displaystyle\quad\geq
c+\min\left\\{\left(1/2-\ell\right)\left(h_{1}-c-\eta\right),\infty\right\\}=c+\left(1/2-\ell\right)\left(h_{1}-c-\eta\right).$

We complete the proof by sending $\ell$ to $1/2$. ∎

###### Proof of Lemma 8.4.

Recall that

$E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)^{2}=\sum\nolimits_{n=0}^{\infty}\left(2n+1\right)P_{\lambda^{\varepsilon}}\left(\tau_{n}^{\varepsilon}\leq
t\right)=\mathfrak{R}_{1}+\mathfrak{R}_{2}+\mathfrak{R}_{3}$

where the $\mathfrak{R}_{i}$ are defined in (8.3). We can bound
$\mathfrak{R}_{3}$ from above by

$\displaystyle\sum\nolimits_{n=0}^{\left(1-2\gamma^{\varepsilon}\right)\Gamma^{\varepsilon}-1}\left(2n+1\right)=(1-4\gamma^{\varepsilon}+4\left(\gamma^{\varepsilon}\right)^{2})\left(\Gamma^{\varepsilon}\right)^{2}.$

Applying Lemma 8.11 and Lemma 8.17, we have for any $\ell\in(0,1/2)$ that

$\displaystyle
E_{\lambda^{\varepsilon}}(N^{\varepsilon}\left(T^{\varepsilon}\right))^{2}$
$\displaystyle\leq
C\left(\Gamma^{\varepsilon}\right)^{1+2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}+4\left(\Gamma^{\varepsilon}\right)^{2-\ell}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)}+(1-4\gamma^{\varepsilon}+4\left(\gamma^{\varepsilon}\right)^{2})\left(\Gamma^{\varepsilon}\right)^{2}$
$\displaystyle\leq\left(\Gamma^{\varepsilon}\right)^{2}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)}+C\left(\Gamma^{\varepsilon}\right)^{1+2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}.$

As in the proof of Lemma 8.2
$E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)\geq\Gamma^{\varepsilon}.$
Thus for any $\ell\in(0,1/2)$

$\displaystyle\mathrm{Var}_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)$
$\displaystyle\leq
E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)^{2}-\left(\Gamma^{\varepsilon}\right)^{2}$
$\displaystyle\leq[\left(\Gamma^{\varepsilon}\right)^{2}+C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)}+C\left(\Gamma^{\varepsilon}\right)^{1+2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}]-\left(\Gamma^{\varepsilon}\right)^{2}$
$\displaystyle=C\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)}+C\left(\Gamma^{\varepsilon}\right)^{1+2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}.$

Again we use (7.2), Lemma 8.7 and Remark 8.12 to find that for any $\eta>0,$
there exists $\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$
and for any $\ell\in(0,1/2)$,

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{\mathrm{Var}_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}$
$\displaystyle\quad\geq\liminf_{\varepsilon\rightarrow 0}\varepsilon\log
T^{\varepsilon}+\min\left\\{\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\Gamma^{\varepsilon}\right)^{2\left(1-\ell\right)},\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\left(\Gamma^{\varepsilon}\right)^{1+2\ell}e^{-\left(\Gamma^{\varepsilon}\right)^{1-2\ell}}\right)\right\\}$
$\displaystyle\quad\geq
c+\min\left\\{2\left(1-\ell\right)\left(h_{1}-c-\eta\right),\infty\right\\}=2\left(1-\ell\right)(h_{1}-\eta)+\left(2\ell-1\right)c.$

We complete the proof by sending $\ell$ to $1/2$. ∎

###  8.4 Asymptotics of moments of $\hat{N}^{\varepsilon}(T^{\varepsilon})$

The proof of the following result is given in Section 10.

######  Theorem 8.18.

If $w\geq h_{1}$, then given any $m>0$ such that $m+h_{1}>w$ and for any
$\delta>0$ sufficiently small,

$\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}=m+\varkappa_{\delta}\text{
and
}\hat{\tau}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}\overset{d}{\rightarrow}\mathrm{Exp}(1).$

Moreover, there exists $\varepsilon_{0}\in(0,1)$ and a constant $\tilde{c}>0$
such that

$P_{\lambda^{\varepsilon}}\left(\hat{\tau}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}>t\right)\leq
e^{-\tilde{c}t}$

for any $t>0$ and any $\varepsilon\in(0,\varepsilon_{0}).$

Notice that Theorem 8.18 is a multicycle version of Theorem 8.5, which is the
key to the proofs of the asymptotics of moments of
$N^{\varepsilon}(T^{\varepsilon})$, namely, Lemma 8.2 and Lemma 8.4. Given
Theorem 8.18, the proofs of the following analogous results follow from
essentially the same arguments as those of Lemma 8.2 and Lemma 8.4.

######  Lemma 8.19.

Suppose that $w\geq h_{1}$, $m+h_{1}>w$, and that
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>w$. Then there exists
$\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$

$\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|\frac{E_{\lambda^{\varepsilon}}(\hat{N}^{\varepsilon}\left(T^{\varepsilon}\right))}{T^{\varepsilon}}-\frac{1}{E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right|\geq
c.$

######  Corollary 8.20.

Suppose that $w\geq h_{1}$, $m+h_{1}>w$ and that
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>w$. Then there exists
$\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$

$\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{E_{\lambda^{\varepsilon}}(\hat{N}^{\varepsilon}\left(T^{\varepsilon}\right))}{T^{\varepsilon}}\geq
m+\varkappa_{\delta}.$

######  Lemma 8.21.

Suppose that $w\geq h_{1}$, $m+h_{1}>w$ and that
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>w$. Then for any
$\eta>0,$ there exists $\delta_{0}\in(0,1)$ such that for any
$\delta\in(0,\delta_{0})$

$\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{\mathrm{Var}_{\lambda^{\varepsilon}}(\hat{N}^{\varepsilon}\left(T^{\varepsilon}\right))}{T^{\varepsilon}}\geq
m+h_{1}-\eta.$

##  9 Large Deviation Type Upper Bounds

In this section we collect results from the previous sections to prove the
main results of the paper, Theorems 4.3 and 4.5, which give large deviation
upper bounds on the bias under the empirical measure and the variance per unit
time. We also give the proof of Theorem 4.9, which shows how to simplify some
expressions appearing in the large deviation bounds. Before giving the proof
of the first result we establish Lemma 9.1 and Lemma 9.2 for the single cycle
case, and Lemma 9.3 and Lemma 9.4 for the multicycle case, which are needed in
the proof of Theorems 4.3. Recall that for any $n\in\mathbb{N}$

$S_{n}^{\varepsilon}\doteq\int_{\tau_{n-1}^{\varepsilon}}^{\tau_{n}^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt.$
(9.1)

######  Lemma 9.1.

If $h_{1}>w$, $A\subset M$ is compact and
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>h_{1}$, then for any
$\eta>0$, there exists $\delta_{0}\in(0,1)$ such that for any
$\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|\frac{E_{\lambda^{\varepsilon}}N^{\varepsilon}\left(T^{\varepsilon}\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|$
$\displaystyle\quad\geq\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+c-h_{1}-\eta.$

###### Proof.

To begin, by Lemma 4.1 with
$g\left(x\right)=e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)$,
we know that for any $\delta$ sufficiently small and $\varepsilon>0,$

$\displaystyle E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}$
$\displaystyle=E_{\lambda^{\varepsilon}}\left(\int_{0}^{\tau_{1}^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{s}^{\varepsilon}\right)}1_{A}\left(X_{s}^{\varepsilon}\right)ds\right)=E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\cdot\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right).$

This implies that

$\displaystyle\left|\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|=E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}\cdot\left|\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}-\frac{1}{E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right|.$

Hence, by (7.1), Lemma 7.21 and Lemma 8.2, we find that given $\eta>0,$ there
exists $\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|$
$\displaystyle\quad\geq\liminf_{\varepsilon\rightarrow 0}-\varepsilon\log
E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}+\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}-\frac{1}{E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right|$
$\displaystyle\quad\geq\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+c-h_{1}-\eta.$

∎

In the application of Wald’s identity a difficulty arises in that, owing to
the randomness of $N^{\varepsilon}\left(T^{\varepsilon}\right)$,
$S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}$ need not have
the same distribution as $S_{1}^{\varepsilon}$. Nevertheless, such minor term
can be dealt with by using technique in, for example, [18, Theorem 3.16]. The
proof of the following lemma can be found in the Appendix.

######  Lemma 9.2.

If $h_{1}>w$, $A\subset M$ is compact and
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>h_{1}$, then for any
$\eta>0$, there exists $\delta_{0}\in(0,1)$ such that for any
$\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}}\geq\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+c-h_{1}-\eta.$

We have similar results for multicycles. To be specific, we have the following
two lemmas.

######  Lemma 9.3.

Suppose that $w\geq h_{1}$, $m+h_{1}>w$, $A\subset M$ is compact and that
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>w$. Then for any
$\eta>0$, there exists $\delta_{0}\in(0,1)$ such that for any
$\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|\frac{E_{\lambda^{\varepsilon}}\hat{N}^{\varepsilon}\left(T^{\varepsilon}\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}\hat{S}_{1}^{\varepsilon}-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|$
$\displaystyle\quad\geq\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+c-(m+h_{1})-\eta.$

######  Lemma 9.4.

Suppose that $w\geq h_{1}$, $m+h_{1}>w$, $A\subset M$ is compact and that
$T^{\varepsilon}=e^{\frac{1}{\varepsilon}c}$ for some $c>w$. Then for any
$\eta>0$, there exists $\delta_{0}\in(0,1)$ such that for any
$\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{E_{\lambda^{\varepsilon}}\hat{S}_{\hat{N}^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}}\geq\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+c-(m+h_{1})-\eta.$

###### Proof of Theorem 4.3.

If $h_{1}>w$, then recall that

$\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)-1}S_{n}^{\varepsilon}\leq\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\leq\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon},$

where $S_{n}^{\varepsilon}$ is defined in (9.1). Then we apply Wald’s first
identity to obtain

$\displaystyle
E_{\lambda^{\varepsilon}}\left(\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)-1}S_{n}^{\varepsilon}\right)$
$\displaystyle=E_{\lambda^{\varepsilon}}\left(\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)-E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}$
$\displaystyle=E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}-E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}.$

Thus

$\displaystyle\left|E_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\right)-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|$
$\displaystyle\quad\leq\left|\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|+\frac{E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}}.$

Therefore, by Lemma 9.1 and Lemma 9.2 we have that for any $\eta>0$, there
exists $\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$,

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left|E_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\right)-\int_{M}e^{-\frac{1}{\varepsilon}f\left(x\right)}1_{A}\left(x\right)\mu^{\varepsilon}\left(dx\right)\right|$
$\displaystyle\quad\geq\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+c-h_{1}-\eta.$

The argument for $h_{1}\leq w$ is entirely analogous but uses by Lemma 9.3 and
Lemma 9.4. ∎

The following lemma bounds quantities that will arise in the proof of Theorem
4.5. Its proof is given in the Appendix.

######  Lemma 9.5.

Recall the definitions $R_{1}^{(2)}\doteq 2\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{1},x\right)\right]-h_{1},$ and for $j\in
L\setminus\\{1\\}$, $R_{j}^{(2)}\doteq 2\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right)-2W\left(O_{1}\right)+W\left(O_{1}\cup
O_{j}\right)$ with $h_{1}\doteq\min_{\ell\in
L\setminus\\{1\\}}V(O_{1},O_{\ell}).$ Then $2\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-2W\left(O_{1}\right)-h_{1}\geq\min_{j\in
L}R_{j}^{(2)}.$

###### Proof of Theorem 4.5.

We begin with the observation that is for any random variables $X,Y$ and $Z$
satisfying $0\leq Y-Z\leq X\leq Y,$

$\displaystyle\mathrm{Var}\left(X\right)$
$\displaystyle=EX^{2}-\left(EX\right)^{2}\leq
EY^{2}-\left(E\left(Y-Z\right)\right)^{2}$
$\displaystyle=\mathrm{Var}\left(Y\right)+2EY\cdot
EZ-\left(EZ\right)^{2}\leq\mathrm{Var}\left(Y\right)+2EY\cdot EZ.$

When $h_{1}>w$, since

$0\leq\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}-\frac{1}{T^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}\leq\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\leq\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon},$

we have

$\displaystyle\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\right)$
$\displaystyle\quad\leq\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)+2E_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)\frac{E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}},$

and with the help of (7.2)

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\right)T^{\varepsilon}\right)$
$\displaystyle\quad\geq\min\left\\{\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)T^{\varepsilon}\right],\right.$
$\displaystyle\quad\qquad\left.\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[E_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)\frac{E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}}T^{\varepsilon}\right]\right\\}.$

We complete the proof in the case of single cycle by showing both terms are
bounded below by $\min\nolimits_{j\in L}(R_{j}^{(1)}\wedge R_{j}^{(2)})-\eta$,
where we recall

$R_{j}^{(1)}\doteq\inf\nolimits_{x\in
A}\left[2f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right)-W\left(O_{1}\right),$
$R_{1}^{(2)}\doteq 2\inf\nolimits_{x\in
A}\left[f\left(x\right)+V\left(O_{1},x\right)\right]-h_{1},$

and for $j\in L\setminus\\{1\\}$

$\displaystyle R_{j}^{(2)}$ $\displaystyle\doteq 2\inf\nolimits_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right)-2W\left(O_{1}\right)+W\left(O_{1}\cup
O_{j}\right).$

For the second term, we apply Wald’s first identity, Lemma 7.21, Corollary 8.3
and Lemma 9.2 to find that given $\eta>0,$ there exists $\delta_{0}\in(0,1),$
such that for any $\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[E_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)\frac{E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}}T^{\varepsilon}\right]$
$\displaystyle\quad\geq\liminf_{\varepsilon\rightarrow 0}-\varepsilon\log
T^{\varepsilon}+\liminf_{\varepsilon\rightarrow 0}-\varepsilon\log
E_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)+\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{E_{\lambda^{\varepsilon}}S_{N^{\varepsilon}\left(T^{\varepsilon}\right)}^{\varepsilon}}{T^{\varepsilon}}$
$\displaystyle\quad\geq-c+(\inf\nolimits_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)-h_{1}-\eta/3)+\varkappa_{\delta}$
$\displaystyle\qquad+(\inf\nolimits_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-W\left(O_{1}\right)+\left(c-h_{1}\right)-\eta/3)$
$\displaystyle\quad\geq 2\inf\nolimits_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-2W\left(O_{1}\right)-h_{1}-\eta$
$\displaystyle\quad\geq\min\nolimits_{j\in
L}R_{j}^{(2)}-\eta\geq\min\nolimits_{j\in L}(R_{j}^{(1)}\wedge
R_{j}^{(2)})-\eta.$

The third inequality holds by choosing $\delta$ sufficiently small
$h_{\delta}\geq h_{1}-\eta/3$. The fourth inequality is from Lemma 9.5.

Turning to the first term, we can bound the variance by (6.3):

$\displaystyle\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)T^{\varepsilon}$
$\displaystyle\leq
2\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}\mathrm{Var}_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}+2\frac{\mathrm{Var}_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}\left(E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}\right)^{2}$
$\displaystyle\leq
2\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}\left(S_{1}^{\varepsilon}\right)^{2}+2\frac{\mathrm{Var}_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}\left(E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}\right)^{2}.$

If we use Corollary 8.3 and Lemma 7.23, then we know that given $\eta>0,$
there exists $\delta_{0}\in(0,1),$ such that for any $\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}E_{\lambda^{\varepsilon}}\left(S_{1}^{\varepsilon}\right)^{2}\right]$
$\displaystyle\geq\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{E_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}+\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log
E_{\lambda^{\varepsilon}}\left(S_{1}^{\varepsilon}\right)^{2}$
$\displaystyle\geq\min_{j\in L}(R_{j}^{(1)}\wedge R_{j}^{(2)})-\eta.$

In addition, we can apply Lemma 7.21 and Lemma 8.4 to show that given
$\eta>0,$ there exists $\delta_{0}\in(0,1),$ such that for any
$\delta\in(0,\delta_{0})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left[\frac{\mathrm{Var}_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}\left(E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}\right)^{2}\right]$
$\displaystyle\quad\geq\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\frac{\mathrm{Var}_{\lambda^{\varepsilon}}\left(N^{\varepsilon}\left(T^{\varepsilon}\right)\right)}{T^{\varepsilon}}+2\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log E_{\lambda^{\varepsilon}}S_{1}^{\varepsilon}$
$\displaystyle\quad\geq 2\inf_{x\in
A}\left[f\left(x\right)+W\left(x\right)\right]-2W\left(O_{1}\right)-h_{1}-\eta$
$\displaystyle\quad\geq\min_{j\in L}R_{j}^{(2)}-\eta\geq\min_{j\in
L}(R_{j}^{(1)}\wedge R_{j}^{(2)})-\eta.$

The second last inequality comes from Lemma 9.5.

Hence, we find that given $\eta>0,$ there exists $\delta_{0}\in(0,1),$ such
that for any $\delta\in(0,\delta_{0})$

$\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\sum\nolimits_{n=1}^{N^{\varepsilon}\left(T^{\varepsilon}\right)}S_{n}^{\varepsilon}\right)T^{\varepsilon}\right)\geq\min_{j\in
L}(R_{j}^{(1)}\wedge R_{j}^{(2)})-\eta,$

and we are done for the single cycle case.

For multicycle case, by using a similar argument and applying Lemmas 7.24,
7.25, 8.21, 9.4 and Corollary 8.20, we find that

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\mathrm{Var}_{\lambda^{\varepsilon}}\left(\frac{1}{T^{\varepsilon}}\int_{0}^{T^{\varepsilon}}e^{-\frac{1}{\varepsilon}f\left(X_{t}^{\varepsilon}\right)}1_{A}\left(X_{t}^{\varepsilon}\right)dt\right)T^{\varepsilon}\right)\geq\min_{j\in
L}(R_{j}^{(1)}\wedge R_{j}^{(2)}\wedge R_{j}^{(3,m)})-\eta,$

with

$R_{j}^{(3,m)}\doteq 2\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+2W\left(O_{j}\right)-2W\left(O_{1}\right)-(m+h_{1}).$

We complete the proof by sending $m\downarrow w-h_{1}$. ∎

###### Proof of Theorem 4.9.

Parts 1, 2 and 3 are from Theorem 4.3, Lemma 4.3 (b) and Theorem 6.1 in [12,
Chapter 6], respectively.

We now turn to part 4. Before giving the proof, we state a result from [12].
The result is Lemma 4.3 (c) in [12, Chapter 6], which says that for any
unstable equilibrium point $O_{j},$ there exists a stable equilibrium point
$O_{i}$ such that $W(O_{j})=W(O_{i})+V(O_{i},O_{j}).$

Now suppose that $\min\nolimits_{j\in L}(\inf\nolimits_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right))$ is
attained at some $\ell\in L$ such that $O_{\ell}$ is unstable (i.e., $\ell\in
L\setminus L_{s}$). Then since there exists a stable equilibrium point $O_{i}$
(i.e., $i\in L_{s}$) such that $W(O_{\ell})=W(O_{i})+V(O_{i},O_{\ell})$ we
find

$\displaystyle\min_{j\in L}\left(\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right)\right)$
$\displaystyle\quad=\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{\ell},x\right)\right]+W\left(O_{\ell}\right)=\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{\ell},x\right)\right]+V(O_{i},O_{\ell})+W(O_{i})$
$\displaystyle\quad\geq\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{i},x\right)\right]+W(O_{i})\geq\min_{j\in
L_{\rm{s}}}\left(\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right)\right)$
$\displaystyle\quad\geq\min_{j\in L}\left(\inf_{x\in
A}\left[f\left(x\right)+V\left(O_{j},x\right)\right]+W\left(O_{j}\right)\right).$

The first inequality is from a dynamic programming inequality. Therefore, the
minimum is also attained at $i\in L_{\rm{s}}$ and $\min_{j\in
L}R_{j}^{(1)}=\min_{j\in L_{\rm{s}}}R_{j}^{(1)}.$ ∎

##  10 Exponential Return Law and Tail Behavior

In this section we give the proof of Theorem 8.5, which was the key fact
needed to obtain bounds on the distribution of
$N^{\varepsilon}(T^{\varepsilon})$, and the related multicycle analogy. A
result of this type first appears in [6], which asserts that the time needed
to escape from an open subset of the domain of attraction of a stable
equilibrium point that contains the equilibrium point has an asymptotically
exponential distribution. [6] also proves a nonasymptotic bound on the tail of
the probability of escape before a certain time that is also of exponential
form. Theorem 8.5 is a more complicated statement, in that it asserts the
asymptotically exponential form for the return time to the neighborhood of
$O_{1}$. To prove this we build on the results of [6], and decompose the
return time into times of transitions between equilibrium points. This in turn
will require the proof of a number of related results, such as establishing
the independence of certain estimates with respect to initial distributions.

The existence of an exponentially distributed first hitting time is a central
topic in the theory of quasistationary distributions. For a recent book length
treatment of the topic we refer to [5]. However, so far as we can tell the
types of situations we encounter are not covered by existing results, and so
as noted we develop what is needed using [6] as the starting point. See Remark
3.15.

For any $j\in L,$ define $\upsilon_{j}^{\varepsilon}$ as the hitting time of
$\partial B_{\delta}(O_{k})$ for some $k\in L\setminus\\{j\\}$, i.e.,

$\upsilon_{j}^{\varepsilon}\doteq\inf\left\\{t>0:X_{t}^{\varepsilon}\in\cup_{k\in
L\setminus\\{j\\}}\partial B_{\delta}(O_{k})\right\\}.$ (10.1)

We will prove the following result for first hitting times of another
equilibrium point, and later extend to return times.

######  Lemma 10.1.

For any $j\in L_{\rm{s}}$, there exists $\delta_{0}\in(0,1)$ such that for any
$\delta\in(0,\delta_{0})$ and any distribution $\lambda^{\varepsilon}$ on
$\partial B_{\delta}(O_{j}),$

$\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}=\min_{y\in\cup_{k\in
L\setminus\\{j\\}}\partial B_{\delta}(O_{k})}V(O_{j},y)\text{ and
}\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}\overset{d}{\rightarrow}\rm{Exp}(1).$

Moreover, there exists $\varepsilon_{0}\in(0,1)$ and a constant $\tilde{c}>0$
such that

$P_{\lambda^{\varepsilon}}\left(\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}>t\right)\leq
e^{-\tilde{c}t}$

for any $t>0$ and any $\varepsilon\in(0,\varepsilon_{0}).$

The organization of this section is as follows. The first part of Lemma 10.1
that is concerned with mean first hitting times is proved in Subsection 10.1,
while the second part that is concerned with an asymptotically exponential
distribution but when starting with a special distribution is proved in
Subsection 10.2. The last part of the lemma, which focuses on bounds on the
tail of the hitting time of another equilibrium point but when starting with a
special distribution is proved in Subsection 10.3. We then extend the second
and third parts of Lemma 10.1 to general initial distributions in Subsection
10.4 and Subsection 10.5. The last two subsections then extend all of Lemma
10.1 to return times for single cycles and multicycles, respectively.

###  10.1 Mean first hitting time

######  Lemma 10.2.

For any $\delta>0$ sufficiently small and $x\in\partial B_{\delta}(O_{j})$
with $j\in L_{\rm{s}}$

$\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{x}\upsilon_{j}^{\varepsilon}=\min_{y\in\cup_{k\in
L\setminus\\{j\\}}\partial B_{\delta}(O_{k})}V(O_{j},y).$ (10.2)

###### Proof.

For the given $j\in L_{\rm{s}}$ let $D_{j}$ denote the corresponding domain of
attraction. We claim there is $k\in L\setminus L_{\rm{s}}$ such that

$q_{j}\doteq\inf_{y\in\partial D_{j}}V(O_{j},y)=V(O_{j},O_{k}).$

Since $V(O_{j},\cdot)$ is continuous and $\partial D_{j}$ is compact, there is
a point $y^{\ast}\in\partial D_{j}$ such that $q_{j}=V(O_{j},y^{\ast})$. If
$y^{\ast}\in\cup_{k\in L\setminus L_{\rm{s}}}O_{k}$ then we are done. If this
is not true, then since $y^{\ast}\notin(\cup_{k\in
L_{\rm{s}}}D_{k})\cup(\cup_{k\in L\setminus L_{\rm{s}}}O_{k})$, and since the
solution to $\dot{\phi}=b(\phi),\phi(0)=y^{\ast}$ must converge to $\cup_{k\in
L}O_{k}$ as $t\rightarrow\infty$, it must in fact converge to a point in
$\cup_{k\in L\setminus L_{\rm{s}}}O_{k}$, say $O_{k}$. Since such trajectories
have zero cost, by a standard argument for any $\varepsilon>0$ we can
construct by concatenation a trajectory that connects $O_{j}$ to $O_{k}$ in
finite time and with cost less than $q_{j}+\varepsilon$. Since $\varepsilon>0$
is arbitrary we have $q_{j}=V(O_{j},O_{k})$.

There may be more than one $l\in L\setminus L_{\rm{s}}$ such that
$O_{l}\in\partial D_{j}$ and $q_{j}=V(O_{j},O_{l})$, but we can assume that
for some $k\in L\setminus L_{\rm{s}}$ and $\bar{y}\in\partial
B_{\delta}(O_{k})$ we attain the min in (10.2). Then $\bar{q}_{j}\doteq
V(O_{j},\bar{y})\leq q_{j}$, and we need to show $\lim_{\varepsilon\rightarrow
0}\varepsilon\log E_{x}\upsilon_{j}^{\varepsilon}=\bar{q}_{j}$.

Given $s<\bar{q}_{j}$, let $D_{j}(s)=\\{x:V(O_{j},x)\leq s\\}$ and assume $s$
is large enough that $B_{\delta}(O_{j})\subset D_{j}(s)^{\circ}$. Then
$D_{j}(s)\subset D_{j}^{\circ}$ is closed and contained in the open set
$D_{j}\setminus\cup_{l\in L\setminus\\{j\\}}B_{\delta}(O_{l})$, and thus the
time to reach $\partial D_{j}(s)$ is never greater than
$\upsilon_{j}^{\varepsilon}$. Given $\eta>0$ we can find a set
$D_{j}^{\eta}(s)$ that is contained in $D_{j}(s)$ and satisfies the conditions
of [12, Theorem 4.1, Chapter 4], and also $\inf_{z\in\partial
D_{j}^{\eta}(s)}V(O_{j},z)\geq s-\eta$. This theorem gives the equality in the
following display:

$\displaystyle\liminf_{\varepsilon\rightarrow 0}\varepsilon\log
E_{x}\upsilon_{j}^{\varepsilon}\geq\liminf_{\varepsilon\rightarrow
0}\varepsilon\log E_{x}\inf\\{t\geq 0:X_{t}^{\varepsilon}\in\partial
D_{j}^{\eta}(s)\\}=\inf_{z\in\partial D_{j}^{\eta}(s)}V(O_{j},z)\geq s-\eta.$

Letting $\eta\downarrow 0$ and then $s\uparrow\bar{q}_{j}$ gives
$\liminf_{\varepsilon\rightarrow 0}\varepsilon\log
E_{x}\upsilon_{j}^{\varepsilon}\geq\bar{q}_{j}$.

For the reverse inequality we also adapt an argument from the proof of [12,
Theorem 4.1, Chapter 4]. One can find $T_{1}<\infty$ such that the probability
for $X_{t}^{\varepsilon}$ to reach $\cup_{l\in L}B_{\delta}(O_{l})$ by time
$T_{1}$ from any $x\in M\setminus\cup_{l\in L}B_{\delta}(O_{l})$ is bounded
below by $1/2$. (This follows easily from the law of large numbers and that
all trajectories of the noiseless system reach $\cup_{l\in
L}B_{\delta/2}(O_{l})$ in some finite time that is bounded uniformly in $x\in
M\setminus\cup_{l\in L}B_{\delta}(O_{l})$.) Also, given $\eta>0$ there is
$T_{2}<\infty$ and $\varepsilon_{0}>0$ such that $P_{x}\\{X_{t}^{\varepsilon}$
reaches $\cup_{k\in L\setminus\\{j\\}}\partial B_{\delta}(O_{k})$ before
$T_{2}\\}\geq\exp-(\bar{q}_{j}+\eta)/\varepsilon$ for all $x\in\partial
B_{\delta}(O_{j})$. It then follows from the strong Markov property that for
any $x\in M\setminus\cup_{l\in L}B_{\delta}(O_{l})$

$P_{x}\\{\upsilon_{j}^{\varepsilon}\leq T_{1}+T_{2}\\}\geq
e^{-\frac{1}{\varepsilon}(\bar{q}_{j}+\eta)}/2.$

Using the ordinary Markov property we have

$\displaystyle E_{x}\upsilon_{j}^{\varepsilon}$
$\displaystyle\leq\sum\nolimits_{n=0}^{\infty}(n+1)(T_{1}+T_{2})P_{x}\\{n(T_{1}+T_{2})<\upsilon_{j}^{\varepsilon}\leq(n+1)(T_{1}+T_{2})\\}$
$\displaystyle=(T_{1}+T_{2})\sum\nolimits_{n=0}^{\infty}P_{x}\\{\upsilon_{j}^{\varepsilon}>n(T_{1}+T_{2})\\}$
$\displaystyle\leq(T_{1}+T_{2})\sum\nolimits_{n=0}^{\infty}\left[1-\inf_{x\in
M\setminus\cup_{l\in
L}B_{\delta}(O_{l})}P_{x}\\{\upsilon_{j}^{\varepsilon}\leq
T_{1}+T_{2}\\}\right]^{n}$ $\displaystyle=(T_{1}+T_{2})\left(\inf_{x\in
M\setminus\cup_{l\in
L}B_{\delta}(O_{l})}P_{x}\\{\upsilon_{j}^{\varepsilon}\leq
T_{1}+T_{2}\\}\right)^{-1}$ $\displaystyle\leq
2(T_{1}+T_{2})e^{\frac{1}{\varepsilon}(\bar{q}_{j}+\eta)}.$

Thus $\limsup_{\varepsilon\rightarrow 0}\varepsilon\log
E_{x}\upsilon_{j}^{\varepsilon}\leq\bar{q}_{j}+\eta$, and letting
$\eta\downarrow 0$ completes the proof. ∎

######  Remark 10.3.

By the standard Freidlin-Wentzell theory, the convergence asserted in Lemma
10.2 is uniform on $\partial B_{\delta}(O_{j})$. Therefore, we have the first
part of Lemma 10.1.

###  10.2 Asymptotically exponential distribution

######  Lemma 10.4.

For each $j\in L_{\rm{s}}$ there is a distribution $u^{\varepsilon}$ on
$\partial B_{2\delta}(O_{j})$ such that
$\upsilon_{j}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{j}^{\varepsilon}\overset{d}{\rightarrow}\rm{Exp}(1).$

###### Proof.

To simplify notation and since it plays no role, we write $j=1$ throughout the
proof. We call $\partial B_{\delta}\left(O_{1}\right)$ and $\partial
B_{2\delta}\left(O_{1}\right)$ the inner and outer rings of $O_{1}$. We can
then decompose the hitting time as

$\upsilon_{1}^{\varepsilon}=\sum\nolimits_{k=1}^{\mathcal{N}^{\varepsilon}-1}\theta_{k}^{\varepsilon}+\zeta^{\varepsilon},$
(10.3)

where $\theta_{k}^{\varepsilon}$ is the $k$-th amount of time that the process
travels from the outer ring to the inner ring and back without visiting
$\cup_{j\in L\setminus\\{1\\}}\partial B_{\delta}(O_{j})$,
$\zeta^{\varepsilon}$ is the amount of time that the process travels from the
outer ring directly to $\cup_{j\in L\setminus\\{1\\}}\partial
B_{\delta}(O_{j})$ without visiting the inner ring, and
$\mathcal{N}^{\varepsilon}-1$ is the number of times that the process goes
back and forth between the inner ring and outer ring. (It is assumed that
$\delta>0$ is small enough that $B_{2\delta}\left(O_{1}\right)\subset
M\setminus\cup_{j\in L\setminus\\{1\\}}B_{2\delta}(O_{j})$.) Note that
$\theta_{k}^{\varepsilon}$ grows exponentially of the order $\delta$, due to
the time taken to travel from the inner ring to the outer ring, and
$\zeta^{\varepsilon}$ is uniformly bounded in expected value.

For any set $A,$ define the first hitting time by
$\tau\left(A\right)\doteq\inf\left\\{t>0:X_{t}^{\varepsilon}\in A\right\\}.$
Consider the conditional transition probability from $x\in\partial
B_{2\delta}\left(O_{1}\right)$ to $y\in\partial B_{\delta}\left(O_{1}\right)$
given by

$\psi_{1}^{\varepsilon}\left(dy|x\right)\doteq P\left(X_{\tau\left(\partial
B_{\delta}\left(O_{1}\right)\right)}^{\varepsilon}\in
dy|X_{0}^{\varepsilon}=x,\text{ }X_{t}^{\varepsilon}\notin\cup_{j\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{j}),t\in[0,\tau(\partial
B_{\delta}\left(O_{1}\right)))]\right),$

and the transition probability from $y\in\partial
B_{\delta}\left(O_{1}\right)$ to $x\in\partial B_{2\delta}\left(O_{1}\right)$
given by

$\psi_{2}^{\varepsilon}\left(dx|y\right)\doteq P\left(X_{\tau\left(\partial
B_{2\delta}\left(O_{1}\right)\right)}^{\varepsilon}\in
dx|X_{0}^{\varepsilon}=y\right).$ (10.4)

Then we can create a transition probability from $x\in\partial
B_{2\delta}\left(O_{1}\right)$ to $y\in\partial B_{2\delta}\left(O_{1}\right)$
by

$\psi^{\varepsilon}\left(dy|x\right)\doteq\int_{\partial
B_{\delta}\left(O_{1}\right)}\psi_{2}^{\varepsilon}\left(dy|z\right)\psi_{1}^{\varepsilon}\left(dz|x\right).$
(10.5)

Since $\partial B_{2\delta}\left(O_{1}\right)$ is compact and
$\\{X_{t}^{\varepsilon}\\}_{t}$ is non-degenerate and Feller, there exists an
invariant measure $u^{\varepsilon}\in\mathcal{P}\left(\partial
B_{2\delta}\left(O_{1}\right)\right)$ with respect to the transition
probability $\psi^{\varepsilon}\left(dy|x\right).$ If we start with the
distribution $u^{\varepsilon}$ on $\partial B_{2\delta}\left(O_{1}\right)$,
then it follows from the definition of $u^{\varepsilon}$ and the strong Markov
property that the
$\\{\theta_{k}^{\varepsilon}\\}_{k<\mathcal{N}^{\varepsilon}}$ are iid.
Moreover, the indicators of escape (i.e., $1_{\\{\tau(\cup_{j\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{j}))=\tau(\cup_{j\in L}\partial
B_{\delta}(O_{j}))\\}}$) are iid Bernoulli, and we write them as
$Y_{k}^{\varepsilon}$ with
$P_{u^{\varepsilon}}(Y_{k}^{\varepsilon}=1)=e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon},$
where $\delta>0$ is from the construction,
$h_{1}^{\varepsilon}(\delta)\rightarrow h_{1}(\delta)$ as
$\varepsilon\rightarrow 0$ and $h_{1}(\delta)\uparrow h_{1}$ as
$\delta\downarrow 0$ with $h_{1}=\min_{j\in L\setminus\\{1\\}}V(O_{1},O_{j})$.
Note that
$\mathcal{N}^{\varepsilon}=\inf\left\\{k\in\mathbb{N}:Y_{k}^{\varepsilon}=1\right\\}.$
We therefore have

$P_{u^{\varepsilon}}(\mathcal{N}^{\varepsilon}=k)=(1-e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon})^{k-1}e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon},$

and thus

$E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}=E_{u^{\varepsilon}}\left[\sum\nolimits_{j=1}^{\mathcal{N}^{\varepsilon}-1}\theta_{j}^{\varepsilon}\right]+E_{u^{\varepsilon}}\zeta^{\varepsilon}=E_{u^{\varepsilon}}(\mathcal{N}^{\varepsilon}-1)E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}+E_{u^{\varepsilon}}\zeta^{\varepsilon},$

where the second equality comes from Wald’s identity. Using
$\sum_{k=1}^{\infty}ka^{k-1}={1/(1-a)^{2}}$ for $a\in[0,1)$, we also have

$\displaystyle
E_{u^{\varepsilon}}\mathcal{N}^{\varepsilon}=\sum\nolimits_{k=1}^{\infty}k(1-e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon})^{k-1}e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}=e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}e^{2h_{1}^{\varepsilon}(\delta)/\varepsilon}=e^{h_{1}^{\varepsilon}(\delta)/\varepsilon},$

and therefore

$E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}=e^{h_{1}^{\varepsilon}(\delta)/\varepsilon}E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}+(E_{u^{\varepsilon}}\zeta^{\varepsilon}-E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}).$
(10.6)

Next consider the characteristic function of
$\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}$

$\phi^{\varepsilon}(t)=E_{u^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}=\phi_{\upsilon}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}),$

where $\phi_{\upsilon}^{\varepsilon}$ is the characteristic function of
$\upsilon_{1}^{\varepsilon}.$ By (10.3), we have

$\displaystyle\phi_{\upsilon}^{\varepsilon}(s)$
$\displaystyle=E_{u^{\varepsilon}}e^{is\left(\sum_{k=1}^{\mathcal{N}^{\varepsilon}-1}\theta_{k}^{\varepsilon}+\zeta^{\varepsilon}\right)}=E_{u^{\varepsilon}}e^{is\zeta^{\varepsilon}}E_{u^{\varepsilon}}e^{is\left(\sum\nolimits_{k=1}^{\mathcal{N}^{\varepsilon}-1}\theta_{k}^{\varepsilon}\right)}$
$\displaystyle=\phi_{\zeta}^{\varepsilon}(s)\sum\nolimits_{k=1}^{\infty}(1-e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon})^{k-1}e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\phi_{\theta}^{\varepsilon}(s)^{k-1}$
$\displaystyle=\phi_{\zeta}^{\varepsilon}(s)e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}(1-[(1-e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon})\phi_{\theta}^{\varepsilon}(s)])^{-1},$

where $\phi_{\theta}^{\varepsilon}$ and $\phi_{\zeta}^{\varepsilon}$ are the
characteristic functions of $\theta_{1}^{\varepsilon}$ and
$\zeta^{\varepsilon}$, respectively. We want to show that for any
$t\in\mathbb{R}$

$\phi^{\varepsilon}(t)=\phi_{\upsilon}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\rightarrow
1/(1-it)\text{ as }\varepsilon\rightarrow 0.$

We first show that
$\phi_{\zeta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\rightarrow
1.$ By definition,
$\phi_{\zeta}^{\varepsilon}\left(t/E_{u^{\varepsilon}\upsilon_{1}^{\varepsilon}}\right)=E_{u^{\varepsilon}}\cos\left(t\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)+iE_{u^{\varepsilon}}\sin\left(t\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right).$
According to [12, Lemma 1.9, Chapter 6], we know that there exist
$T_{0}\in(0,\infty)$ and $\beta>0$ such that for any $T\in(T_{0},\infty)$ and
for all $\varepsilon$ sufficiently small

$P_{u^{\varepsilon}}\left(\zeta^{\varepsilon}>T\right)\leq
e^{-\frac{1}{\varepsilon}\beta\left(T-T_{0}\right)},$ (10.7)

and therefore for any bounded and continuous function
$f:\mathbb{R}\rightarrow\mathbb{R}$

$\displaystyle\left|E_{u^{\varepsilon}}f\left(t\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)-f\left(0\right)\right|\leq
2\left\|f\right\|_{\infty}P_{u^{\varepsilon}}\left(\zeta^{\varepsilon}>T\right)+E_{u^{\varepsilon}}\left[\left|f\left(t\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)-f\left(0\right)\right|1_{\left\\{\zeta^{\varepsilon}\leq
T\right\\}}\right].$

The first term in the last display goes to $0$ as $\varepsilon\rightarrow 0$.
For any fixed $t,$ $t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\rightarrow
0$ as $\varepsilon\rightarrow 0$. Since $f$ is continuous, the second term in
the last display also converges to $0$ as $\varepsilon\rightarrow 0.$
$\phi_{\zeta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\rightarrow
1$ follows by taking $f$ to be $\sin x$ and $\cos x.$

It remains to show that for any $t\in\mathbb{R}$

$e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\left(1-\left[(1-e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon})\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\right]\right)^{-1}\rightarrow
1/(1-it)$

as $\varepsilon\rightarrow 0.$ Observe that

$e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\left(1-\left[(1-e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon})\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\right]\right)^{-1}=\left({\frac{1-\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})}{e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}}+\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}})\right)^{-1},$

so it suffices to show that
$\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\rightarrow
1$ and
$[1-\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})]/e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\rightarrow-
it$ as $\varepsilon\rightarrow 0.$

For the former, note that by (10.6)

$0\leq
E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)\leq\frac{tE_{u^{\varepsilon}}\theta_{1}^{\varepsilon}}{\left(e^{h^{\varepsilon}_{1}(\delta)/\varepsilon}-1\right)E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}}\rightarrow
0$

as $\varepsilon\rightarrow 0,$ and so
$t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}$
converges to $0$ in distribution. Moreover, since $e^{ix}$ is bounded and
continuous, we find
$\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\rightarrow
1$. For the second part, using

$x-{x^{3}}/{3!}\leq\sin x\leq x\text{ and }1-{x^{2}}/{2}\leq\cos x\leq 1$

for $x\in\mathbb{R}$ we find that

$0\leq\frac{1-E_{u^{\varepsilon}}\cos\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}\leq\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)^{2}}{2e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}$

and

$\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}-\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)^{3}}{3!e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}\leq\frac{E_{u^{\varepsilon}}\sin\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}\leq\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}.$

From our previous observation regarding the distribution of
$\zeta^{\varepsilon}$ and (10.6)

$\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}\rightarrow
t\text{ as }\varepsilon\rightarrow 0.$

In addition, since $\theta_{1}^{\varepsilon}$ can be viewed as the time from
the outer ring to the inner ring without visiting $\cup_{j\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{j})$ plus the time from the inner
ring to the outer ring, by applying (10.7) to the former and using [6, Theorem
4 and Corollary 1] under Condition 3.13 to the later, we find that

$P_{u^{\varepsilon}}\left(\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}>t\right)\leq
2e^{-t}$ (10.8)

for all $t\in[0,\infty)$ and $\varepsilon$ sufficiently small. This implies
that

$\displaystyle
E_{u^{\varepsilon}}\left(\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}\right)^{2}=2\int_{0}^{\infty}t^{2}P_{u^{\varepsilon}}\left(\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}>t\right)dt\leq
4\int_{0}^{\infty}t^{2}e^{-t}dt=8$

and similarly
$E_{u^{\varepsilon}}\left(\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}\right)^{3}=3\int_{0}^{\infty}t^{3}\leq
36$. Then combined with (10.6), we have

$0\leq\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)^{2}}{2e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}\leq\frac{t^{2}E_{u^{\varepsilon}}\left(\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}\right)^{2}}{2e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}(e^{h_{1}^{\varepsilon}(\delta)/\varepsilon}-1)^{2}}\rightarrow
0$

and

$0\leq\frac{E_{u^{\varepsilon}}\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)^{3}}{3!e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}}\leq\frac{t^{3}E_{u^{\varepsilon}}\left(\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}\right)^{3}}{3!e^{-h_{1}^{\varepsilon}(\delta)/\varepsilon}(e^{h_{1}^{\varepsilon}(\delta)/\varepsilon}-1)^{3}}\rightarrow
0.$

Therefore, we have shown that for any $t\in\mathbb{R}$

$\frac{1-\phi_{\theta}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})}{e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}}=\frac{1-E_{u^{\varepsilon}}\cos\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}}-i\frac{E_{u^{\varepsilon}}\sin\left(t\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}{e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}}\rightarrow-
it.$

∎

######  Remark 10.5.

From the proof of Lemma 10.4, we actually know that

$\phi_{\upsilon}^{\varepsilon}(t/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon})\rightarrow
1/(1-it)$

uniformly on any compact set in $\mathbb{R}$ as $\varepsilon\rightarrow 0$.

###  10.3 Tail probability

The goal of this subsection is to prove the following.

######  Lemma 10.6.

For each $j\in L_{\rm{s}}$ there is a distribution $u^{\varepsilon}$ on
$\partial B_{2\delta}(O_{j})$ and $\tilde{c}>0$ such that for any
$t\in[0,\infty)$,
$P_{u^{\varepsilon}}(\upsilon_{j}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{j}^{\varepsilon}>t)\leq
e^{-\tilde{c}t}$ (here $\upsilon_{j}^{\varepsilon}$ and $u^{\varepsilon}$ are
defined as in the last subsection).

###### Proof.

As in the last subsection we give the proof for the case $j=1$. To begin we
note that for any $\alpha>0$ Chebyshev’s inequality implies

$P_{u^{\varepsilon}}\left(\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t\right)=P_{u^{\varepsilon}}(e^{\alpha\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}>e^{\alpha
t})\leq e^{-\alpha t}\cdot
E_{u^{\varepsilon}}e^{\alpha\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}.$

By picking $\alpha=\alpha^{\ast}\doteq 1/8$, it suffices to show that
$E_{u^{\varepsilon}}e^{{\alpha^{\ast}\upsilon_{1}^{\varepsilon}}/{E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}}$
is bounded by a constant. We will do this by showing how the finiteness of
$E_{u^{\varepsilon}}e^{{\alpha^{\ast}\upsilon_{1}^{\varepsilon}}/{E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}}$
is implied by the finiteness of
$E_{u^{\varepsilon}}e^{{\alpha^{\ast}\theta_{1}^{\varepsilon}}/{E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}}$
and
$E_{u^{\varepsilon}}e^{{\alpha^{\ast}\zeta^{\varepsilon}}/{E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}}.$

Using (10.8) we find that for any $\alpha>0$

$P_{u^{\varepsilon}}(e^{\alpha\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}}>t)\leq
2e^{-\frac{1}{\alpha}\log t}=2t^{-\frac{1}{\alpha}}$

for all $t\in[1,\infty)$ and $\varepsilon$ sufficiently small. Then (10.6)
implies
$E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\geq\left(e^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-1\right)E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}$
and therefore

$\displaystyle
E_{u^{\varepsilon}}e^{\alpha^{\ast}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\theta_{1}^{\varepsilon}$
$\displaystyle\leq\int_{0}^{1}P_{u^{\varepsilon}}\left(\exp\left(\alpha^{\ast}\theta_{1}^{\varepsilon}/[{(e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-1{)}E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}]\right)>t\right)dt$
$\displaystyle\quad+\int_{1}^{\infty}P_{u^{\varepsilon}}\left(\exp\left(\alpha^{\ast}\theta_{1}^{\varepsilon}/[{(e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-1{)}E_{u^{\varepsilon}}\theta_{1}^{\varepsilon}]\right)>t\right)dt$
$\displaystyle\leq
1+2\int_{1}^{\infty}t^{-{(e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-1{)/\alpha}^{\ast}}dt$
$\displaystyle=1+2[{(e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-1{)/\alpha}^{\ast}-1]^{-1}=1+2\alpha^{\ast}[{e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-\alpha^{\ast}-1]^{-1}.$

To estimate $\zeta^{\varepsilon},$ we use that by (10.7) there are
$T_{0}\in(0,\infty)$ and $\beta>0$ such that for any $t\in(T_{0},\infty)$ and
for all $\varepsilon$ sufficiently small
$P_{u^{\varepsilon}}\left(\zeta^{\varepsilon}>t\right)\leq
e^{-\frac{1}{\varepsilon}\beta\left(t-T_{0}\right)},$ so that for any
$\alpha>0$
$P_{u^{\varepsilon}}\left(e^{\alpha\zeta^{\varepsilon}}>t\right)\leq
e^{-\frac{1}{\varepsilon}\beta\left(\frac{1}{\alpha}\log t-T_{0}\right)}$ for
any $t\geq e^{\alpha T_{0}}.$ Given $n\in\mathbb{N},$ for all sufficiently
small $\varepsilon$ we have
$\alpha^{\ast}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\leq 1/n$, and
thus

$P_{u^{\varepsilon}}\left(e^{\alpha^{\ast}\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}>t\right)\leq
P_{u^{\varepsilon}}\left(e^{\zeta^{\varepsilon}/n}>t\right)\leq
e^{-\frac{1}{\varepsilon}\beta\left(n\log t-T_{0}\right)}.$

Hence for any $n$ such that $e^{T_{0}/n}\leq 3/2$ and $\left(-\beta
n+1\right)\log\left(3/2\right)+\beta T_{0}<0,$ and for $\varepsilon$ small
enough that $\alpha^{\ast}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}\leq
1/n,$ we have

$\displaystyle
E_{u^{\varepsilon}}e^{\alpha^{\ast}\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}$
$\displaystyle\leq\int_{0}^{\infty}P_{u^{\varepsilon}}\left(e^{\alpha^{\ast}\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}>t\right)dt\leq
3/2+\int_{\frac{3}{2}}^{\infty}P_{u^{\varepsilon}}\left(e^{\alpha^{\ast}\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}>t\right)dt$
$\displaystyle\leq
3/2+\int_{\frac{3}{2}}^{\infty}e^{-\frac{1}{\varepsilon}\beta\left(n\log
t-T_{0}\right)}dt=3/2+e^{\frac{1}{\varepsilon}\beta T_{0}}(\beta
n/{\varepsilon}-1)^{-1}\left(3/2\right)^{\frac{1}{\varepsilon}\left(-\beta
n+\varepsilon\right)}$ $\displaystyle=3/2+(\beta
n/{\varepsilon}-1)^{-1}e^{\frac{1}{\varepsilon}\left[\left(-\beta
n+\varepsilon\right)\log\left(3/2\right)+\beta T_{0}\right]}\leq 3/2+(\beta
n/{\varepsilon}-1)^{-1}\leq 2.$

We have shown that for such $\alpha^{\ast},$
$E_{u^{\varepsilon}}e^{\alpha^{\ast}\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}$
and
$E_{u^{\varepsilon}}e^{\alpha^{\ast}\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}$
are uniformly bounded for all $\varepsilon$ sufficiently small. Lastly, using
the same calculation as used for the characteristic function

$\displaystyle
E_{u^{\varepsilon}}e^{\alpha^{\ast}\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}$
$\displaystyle=E_{u^{\varepsilon}}e^{\alpha^{\ast}\zeta^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\cdot
e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\left(1-\left[(1-e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon})E_{u^{\varepsilon}}e^{\alpha^{\ast}\theta_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\right]\right)^{-1}$
$\displaystyle\leq
2e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\left(1-\left[(1-e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon})\left(1+\frac{2\alpha^{\ast}}{{e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-\alpha^{\ast}-1}\right)\right]\right)^{-1}$
$\displaystyle=2e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\left(e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}-\frac{2\alpha^{\ast}}{{e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-\alpha^{\ast}-1}+\frac{2\alpha^{\ast}}{{e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-\alpha^{\ast}-1}e^{-h^{\varepsilon}_{1}(\delta)/\varepsilon}\right)^{-1}$
$\displaystyle=2\left(1-2\alpha^{\ast}\frac{e^{h^{\varepsilon}_{1}(\delta)/\varepsilon}-1}{{e}^{h^{\varepsilon}_{1}\left(\delta\right)/\varepsilon}-\alpha^{\ast}-1}\right)^{-1}\leq
2/(1-4\alpha^{\ast})=4.$

∎

###  10.4 General initial condition

This subsection presents results that will allow us to extend the results in
the previous two subsections to arbitrary initial distribution
$\lambda^{\varepsilon}\in\mathcal{P}(\partial B_{\delta}\left(O_{1}\right))$.
Under our assumptions, for any $j\in L_{\rm{s}}$ we observe that the process
model

$dX_{t}^{\varepsilon}=b\left(X_{t}^{\varepsilon}\right)dt+\sqrt{\varepsilon}\sigma\left(X_{t}^{\varepsilon}\right)dW_{t}$
(10.9)

has the property that $b(x)=A(x-O_{j})[1+o(1)]$ and
$\sigma\left(x\right)=\bar{\sigma}[1+o\left(1\right)]$, where $o(1)\rightarrow
0$ as $\left\|x-O_{j}\right\|\rightarrow 0$, $A$ is stable and $\bar{\sigma}$
is invertible. By an invertible change of variable we can arrange so that
$O_{j}=0$ and $\bar{\sigma}=I$, and to simplify we assume this in the rest of
the section.

Since $A$ is stable there exists a positive definite and symmetric solution
$M$ to the matrix equation $AM+MA^{T}=-I$ (we can in fact exhibit the solution
in the form $M=\int_{0}^{\infty}e^{At}e^{A^{T}t}dt)$. To prove the ergodicity
we introduce some additional notation: $U(x)\doteq\langle x,Mx\rangle$,
$B_{i}\doteq\\{x:U(x)<b_{i}^{2}\\}$ and
$\mathcal{S}_{i}(\varepsilon)\doteq\\{x:U(x)<a_{i}^{2}\varepsilon\\},$ for
$i=1,2$, where $0<a_{1}<a_{2}$, $0<b_{0}<b_{1}<b_{2}$. If
$\varepsilon_{0}=(b_{0}^{2}/a_{2}^{2})/2$ then with cl denoting closure, ${\rm
cl}(\mathcal{S}_{2}(\varepsilon_{0}))\subset B_{0},$ and we will assume
$\varepsilon\in(0,\varepsilon_{0})$ henceforth. For a use later on, we will
also assume that $a_{1}^{2}=2\sup\nolimits_{x\in
B_{2}}\mbox{tr}[\sigma(x)\sigma(x)^{T}M].$

######  Remark 10.7.

The sets $B_{1}$ and $B_{2}$ will play the roles that $B_{\delta}(O_{1})$ and
$B_{2\delta}(O_{1})$ played previously in this section. Although elsewhere in
this paper as well as in the reference [12] these sets are taken to be balls
with respect to the Euclidean norm, in this subsection we take them to be
level sets of $U(x)$. The shape of these sets and the choice of the factor of
$2$ relating the radii play no role in the analysis of [12] or in our prior
use in this paper. However in this subsection it is notationally convenient
for the sets to be level sets of $U$, since $U$ is a Lyapunov function for the
noiseless dynamics near $0$. After this subsection we will revert to the
$B_{\delta}(O_{1})$ and $B_{2\delta}(O_{1})$ notation.

In addition to the restrictions $a_{1}<a_{2}$ and
$a_{2}^{2}\varepsilon_{0}\leq b_{0}^{2}$, we also assume that $a_{1},a_{2}$
and $\varepsilon_{0}>0$ are such that if $\phi^{x}$ is the solution to the
noiseless dynamics $\dot{\phi}=b(\phi)$ with initial condition $x$, then: (i)
for all $x\in\partial\mathcal{S}_{2}(\varepsilon)$, $\phi^{x}$ never crosses
$\partial B_{1}$; (i) for all $x\in\partial\mathcal{S}_{1}(\varepsilon)$,
$\phi^{x}$ never exits $\mathcal{S}_{2}(\varepsilon)$.

The idea that will be used to establish asymptotic independence from the
starting distribution is the following. We start the process on $\partial
B_{1}$. With some small probability it will hit $\partial B_{2}$ before
hitting $\partial\mathcal{S}_{2}(\varepsilon)$. This gives a contribution to
$\psi_{2}^{\varepsilon}(dz|x)$ defined in (10.4) that will be relatively
unimportant. If instead it hits $\partial\mathcal{S}_{2}(\varepsilon)$ first,
then we do a Freidlin-Wentzell type analysis, and decompose the trajectory
into excursions between $\partial\mathcal{S}_{2}(\varepsilon)$ and
$\partial\mathcal{S}_{1}(\varepsilon)$, before a final excursion from
$\partial\mathcal{S}_{2}(\varepsilon)$ to $\partial B_{2}$.

To exhibit the asymptotic independence from $\varepsilon$, we introduce the
scaled process $Y^{\varepsilon}_{t}=X^{\varepsilon}_{t}/\sqrt{\varepsilon}$,
which solves the SDE

$dY^{\varepsilon}_{t}=\frac{1}{\sqrt{\varepsilon}}b(\sqrt{\varepsilon}Y^{\varepsilon}_{t})dt+\sigma(\sqrt{\varepsilon}Y^{\varepsilon}_{t})dW_{t}.$

Let $\mathcal{\bar{S}}_{1}=\partial\mathcal{S}_{1}(1)\text{ and
}\mathcal{\bar{S}}_{2}=\partial\mathcal{S}_{2}(1).$ Let
$\omega^{\varepsilon}(w|x)$ denote the density of the hitting location on
$\mathcal{\bar{S}}_{2}$ by the process $Y^{\varepsilon}$, given
$Y^{\varepsilon}_{0}=x\in\mathcal{\bar{S}}_{1}$. The following estimate is
essential. The density function can be identified with the normal derivative
of a related Green’s function, which is bounded from above by the boundary
gradient estimate and bounded below by using the Hopf lemma [13].

######  Lemma 10.8.

Given $\varepsilon_{0}>0$, there are $0<c_{1}<c_{2}<\infty$ such that
$c_{1}\leq\omega^{\varepsilon}(w|x)\leq c_{2}$ for all
$x\in\mathcal{\bar{S}}_{1}$, $w\in\mathcal{\bar{S}}_{2}$ and
$\varepsilon\in(0,\varepsilon_{0})$.

Next let $p^{\varepsilon}(u|w)$ denote the density of the return location for
$Y^{\varepsilon}$ on $\mathcal{\bar{S}}_{2}$, conditioned on visiting
$\mathcal{\bar{S}}_{1}$ before $\partial B_{2}/\sqrt{\varepsilon}$, and
starting at $w\in\mathcal{\bar{S}}_{2}$. The last lemma then directly gives
the following.

######  Lemma 10.9.

For $\varepsilon_{0}>0$ and $c_{1},c_{2}$ as in the last lemma $c_{1}\leq
p^{\varepsilon}(u|w)\leq c_{2}$ for all $u,w\in\mathcal{\bar{S}}_{2}$ and
$\varepsilon\in(0,\varepsilon_{0})$.

Let $r^{\varepsilon}(w)$ denote the unique stationary distribution of
$p^{\varepsilon}(u|w)$, and let $p^{\varepsilon,n}(u|w)$ denote the $n$-step
transition density. The preceding lemma, [14, Theorem 10.1 Chapter 3], and the
existence of a uniform strictly positive lower bound on $r^{\varepsilon}(u)$
for all sufficiently small $\varepsilon>0$ imply the following.

######  Lemma 10.10.

There is $K<\infty$ and $\alpha\in(0,1)$ such that for all
$\varepsilon\in(0,\varepsilon_{0})$

$\sup_{w\in\mathcal{\bar{S}}_{2}}\left|p^{\varepsilon,n}(u|w)-r^{\varepsilon}(u)\right|/r^{\varepsilon}(u)\leq
K\alpha^{n}.$

Let $\eta^{\varepsilon}(dx|w)$ denote the distribution of $X^{\varepsilon}$
upon first hitting $\partial B_{2}$ given that $X^{\varepsilon}$ reaches
$\partial\mathcal{S}_{1}(\varepsilon)$ before $\partial B_{2}$ and starts at
$w\in\partial\mathcal{S}_{2}(\varepsilon)$.

######  Lemma 10.11.

There is $\kappa>0$ and $\varepsilon_{0}>0$ such that for all
$\varepsilon\in(0,\varepsilon_{0})$

$\sup_{x\in\partial B_{1}}P_{x}\left\\{X^{\varepsilon}\mbox{ reaches }\partial
B_{2}\mbox{ before }\mathcal{S}_{2}(\varepsilon)\right\\}\leq
e^{-\kappa/\varepsilon}.$

######  Lemma 10.12.

There are $\bar{\eta}^{\varepsilon}(dz)\in$ $\mathcal{P}(\partial B_{2})$,
$s^{\varepsilon}$ that tends to $0$ as $\varepsilon\rightarrow 0$ and
$\varepsilon_{0}>0$, such that for all $A\in\mathcal{B}(\partial
B_{2}),w\in\partial\mathcal{S}_{2}(\varepsilon)$ and
$\varepsilon\in(0,\varepsilon_{0})$

$\bar{\eta}^{\varepsilon}(A)[1-s^{\varepsilon}K/(1-\alpha)]\leq\eta^{\varepsilon}(A|w)\leq\bar{\eta}^{\varepsilon}(A)[1+s^{\varepsilon}K/(1-\alpha)],$

where $K$ and $\alpha$ are from Lemma 10.10.

###### Proof of Lemma 10.11.

Recall that $a_{1}^{2}=2\sup_{x\in B_{2}}$tr$[\sigma(x)\sigma(x)^{T}M]$. We
then use that $AM+MA^{T}=-I$ to get that with $U(x)\doteq\left\langle
x,Mx\right\rangle$,

$\left\langle DU(x),b(x)\right\rangle\leq-\varepsilon a_{1}^{2}$ (10.10)

for $x\in B_{2}\setminus\mathcal{S}_{2}(\varepsilon)$, and

$\left\langle DU(x),b(x)\right\rangle\leq-\frac{1}{8}b_{0}^{2}$ (10.11)

for $B_{2}\setminus(B_{0}/2)$. By Itô’s formula

$\displaystyle dU(X^{\varepsilon}_{t})=\left\langle
DU(X^{\varepsilon}_{t}),b(X^{\varepsilon}_{t})\right\rangle
dt+\frac{\varepsilon}{2}\text{tr}[\sigma(X^{\varepsilon}_{t})\sigma(X^{\varepsilon}_{t})^{T}M]dt+\sqrt{\varepsilon}\left\langle
DU(X^{\varepsilon}_{t}),\sigma(X^{\varepsilon}_{t})dW_{t}\right\rangle.$
(10.12)

Starting at $x\in\partial B_{1}$, we are concerned with the probability

$P_{x}\left\\{U(X^{\varepsilon}_{t})\text{ reaches }b_{2}^{2}\text{ before
}a_{2}^{2}\varepsilon\right\\},$

where $U(x)=b_{1}^{2}$. However, according to (10.12) and (10.11), reaching
$b_{2}^{2}$ before $b_{0}^{2}/4$ is a rare event, and its probability decays
exponentially in the form $e^{-\kappa/\varepsilon}$ for some $\kappa>0$ and
uniformly in $x\in\partial B_{1}$. Once the process reaches $B_{0}/2$, (10.12)
and (10.10) imply $U(X^{\varepsilon}_{t})$ is a supermartingale as long as it
is in the interval $[0,b_{0}^{2}]$, and therefore after $X^{\varepsilon}_{t}$
reaches $B_{0}/2$, the probability that $U(X^{\varepsilon}_{t})$ reaches
$a_{2}^{2}\varepsilon$ before $b_{0}^{2}$ is greater than $1/2$. ∎

###### Proof of Lemma 10.12.

Consider a starting position $w\in\partial\mathcal{S}_{2}(\varepsilon)$, and
recall that $\eta^{\varepsilon}(dz|w)$ denotes the hitting distribution on
$\partial B_{2}$ after starting at $w$. Let $\theta_{k}^{\varepsilon}$ denote
the return times to $\partial\mathcal{S}_{2}(\varepsilon)$ after visiting
$\partial\mathcal{S}_{1}(\varepsilon)$, and let $q_{n}^{\varepsilon}(w)$
denote the probability that the first $k$ for which $X^{\varepsilon}$ visits
$\partial B_{2}$ before visiting $\partial\mathcal{S}_{1}(\varepsilon)$ during
$[\theta_{k}^{\varepsilon},\theta_{k+1}^{\varepsilon}]$ is $n$. Then by the
strong Markov property and using the rescaled process

$\int_{\partial
B_{2}}g(z)\eta^{\varepsilon}(dz|w)=\sum\nolimits_{n=0}^{\infty}\int_{\partial
B_{2}}g(z)q_{n}^{\varepsilon}(w)\int_{\partial\mathcal{S}_{2}(\varepsilon)}\eta^{\varepsilon}(dz|u)J^{\varepsilon}(u)p^{\varepsilon,n}(\sqrt{\varepsilon}u|\sqrt{\varepsilon}w)du,$

where $J^{\varepsilon}(u)$ is the Jacobian that accounts for the mapping
between $\partial\mathcal{S}_{2}(\varepsilon)$ and
$\partial\mathcal{S}_{2}(1)$ and is given by $u/\sqrt{\varepsilon}$. We next
use that uniformly in $w\in\partial\mathcal{S}_{2}(\varepsilon)$

$p^{\varepsilon,n}(\sqrt{\varepsilon}u|\sqrt{\varepsilon}w)\leq
r^{\varepsilon}(\sqrt{\varepsilon}u)[1+K\alpha^{n}]$

to get

$\displaystyle\sum\nolimits_{n=0}^{\infty}\int_{\partial
B_{2}}g(z)q_{n}^{\varepsilon}(w)\int_{\partial\mathcal{S}_{2}(\varepsilon)}\eta^{\varepsilon}(dz|u)J^{\varepsilon}(u)p^{\varepsilon,n}(\sqrt{\varepsilon}u|\sqrt{\varepsilon}w)du$
$\displaystyle\quad\leq\left(\sum\nolimits_{n=0}^{\infty}\int_{\partial
B_{2}}g(z)q_{n}^{\varepsilon}(w)\int_{\partial\mathcal{S}_{2}(\varepsilon)}\eta^{\varepsilon}(dz|u)J^{\varepsilon}(u)r^{\varepsilon}(\sqrt{\varepsilon}u)du\right)[1+K\alpha^{n}]$
$\displaystyle\quad=\int_{\partial
B_{2}}g(z)\int_{\partial\mathcal{S}_{2}(\varepsilon)}\eta^{\varepsilon}(dz|u)J^{\varepsilon}(u)r^{\varepsilon}(\sqrt{\varepsilon}u)du\left[1+K\sum\nolimits_{n=0}^{\infty}q_{n}^{\varepsilon}(w)\alpha^{n}\right].$

Now use that $K\sum_{n=0}^{\infty}\alpha^{n}=K/(1-\alpha)<\infty$ and
$\sup_{w\in\partial\mathcal{S}_{2}(\varepsilon)}\sup_{n\in\mathbb{N}_{0}}q_{n}^{\varepsilon}(w)\rightarrow
0$ as $\varepsilon\rightarrow 0$ to get the upper bound with

$\bar{\eta}^{\varepsilon}(dz)\doteq\int_{\partial\mathcal{S}_{2}(\varepsilon)}\eta^{\varepsilon}(dz|u)J^{\varepsilon}(u)r^{\varepsilon}(\sqrt{\varepsilon}u)du.$

When combined with the lower bound which has an analogous proof, Lemma 10.12
follows. ∎

###  10.5 Times to reach another equilibrium point after starting with general
initial distribution

######  Lemma 10.13.

For each $j\in L_{\rm{s}}$, there exist $\tilde{c}>0$ and
$\varepsilon_{0}\in(0,1)$ such that for any distribution
$\lambda^{\varepsilon}$ on $\partial B_{\delta}(O_{j})$,

$P_{\lambda^{\varepsilon}}(\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}>t)\leq
e^{-\tilde{c}t}$

for all $t>0$ and $\varepsilon\in(0,\varepsilon_{0}).$

###### Proof.

We give the proof for the case $j=1$. We first show for any $r\in(0,1)$ there
is $\varepsilon_{0}>0$ such that for any $\varepsilon\in(0,\varepsilon_{0})$
and $\lambda^{\varepsilon},\theta^{\varepsilon}\in\mathcal{P}(\partial
B_{\delta}(O_{1}))$

$E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}/E_{\theta^{\varepsilon}}\upsilon_{1}^{\varepsilon}\geq
r.$ (10.13)

We use that $\upsilon_{1}^{\varepsilon}$ can be decomposed into
$\bar{\upsilon}_{1}^{\varepsilon}+\hat{\upsilon}_{1}^{\varepsilon}$, where
$\bar{\upsilon}_{1}^{\varepsilon}$ is the first hitting time to $\partial
B_{2\delta}(O_{1})$. Since by standard large deviation theory the exponential
growth rate of the expected value of $\upsilon_{1}^{\varepsilon}$ is strictly
greater than that of $\bar{\upsilon}_{1}^{\varepsilon}$ (uniformly in the
initial distribution)
$E_{\lambda^{\varepsilon}}\bar{\upsilon}_{1}^{\varepsilon}$ (respectively
$E_{\theta^{\varepsilon}}\bar{\upsilon}_{1}^{\varepsilon}$) is negligible
compared to $E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}$
(respectively $E_{\theta^{\varepsilon}}\upsilon_{1}^{\varepsilon}$), and so it
is enough to show
$E_{\lambda^{\varepsilon}}\hat{\upsilon}_{1}^{\varepsilon}/E_{\theta^{\varepsilon}}\hat{\upsilon}_{1}^{\varepsilon}\geq
r.$ Owing to Lemma 10.11 (and in particular because $\kappa>0$) the
contribution to either
$E_{\lambda^{\varepsilon}}\hat{\upsilon}_{1}^{\varepsilon}$ or
$E_{\theta^{\varepsilon}}\hat{\upsilon}_{1}^{\varepsilon}$ from trajectories
that reach $\partial B_{2\delta}(O_{1})$ before
$\partial\mathcal{S}_{2}(\varepsilon)$ can be neglected. Using Lemma 10.12 and
the strong Markov property gives

$\inf_{w_{1},w_{2}\in\partial\mathcal{S}_{2}(\varepsilon)}\frac{E_{w_{1}}\hat{\upsilon}_{1}^{\varepsilon}}{E_{w_{2}}\hat{\upsilon}_{1}^{\varepsilon}}\geq\frac{[1-s^{\varepsilon}K/(1-\alpha)]}{[1+s^{\varepsilon}K/(1-\alpha)]},$

and the lower bound follows since $s^{\varepsilon}\rightarrow 0$.

We next claim that a suitable bound can be found for
$P_{\lambda^{\varepsilon}}(\hat{\upsilon}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t)$.
Recall that $u^{\varepsilon}\in\mathcal{P}(\partial B_{2\delta}(O_{1}))$ is
the stationary probability for $\psi^{\varepsilon}$ defined in (10.5). Let
$\beta^{\varepsilon}$ be the probability measure on $\partial
B_{\delta}(O_{1})$ obtained by integrating the transition kernel
$\psi_{1}^{\varepsilon}$ with respect to $u^{\varepsilon}$, and note that
integrating $\psi_{2}^{\varepsilon}$ with respect to $\beta^{\varepsilon}$
returns $u^{\varepsilon}$. Since the diffusion matrix is uniformly
nondegenerate, by using well known “Gaussian type” bounds on the transition
density for the process [2] there are $K\in(0,\infty)$ and $p\in(0,\infty)$
such that

$P_{x}\left\\{X_{\theta}^{\varepsilon}\in
A|X_{\theta}^{\varepsilon}\in\partial B_{2\delta}(O_{1})\right\\}\leq
Km(A)/\varepsilon^{p}$

for all $x\in\partial B_{\delta}(O_{1})$, where $m$ is the uniform measure on
$\partial B_{2\delta}(O_{1})$ and
$\theta=\inf\\{t>0:X_{t}^{\varepsilon}\in\partial
B_{2\delta}(O_{1})\cup\mathcal{S}_{2}(\varepsilon)\\}$. Together with Lemmas
10.11 and 10.12, this implies that for all sufficiently small $\varepsilon>0$
and any bounded measurable function $h:\partial
B_{2\delta}(O_{1})\rightarrow\mathbb{R}$,

$\displaystyle\int_{\partial B_{2\delta}(O_{1})}\int_{\partial
B_{\delta}(O_{1})}h(y)\psi_{2}^{\varepsilon}(dy|x)\lambda^{\varepsilon}(dx)$
$\displaystyle\leq 2\int_{\partial B_{2\delta}(O_{1})}\int_{\partial
B_{\delta}(O_{1})}h(y)\psi_{2}^{\varepsilon}(dy|x)\beta^{\varepsilon}(dx)$
$\displaystyle\leq 2\int_{\partial
B_{2\delta}(O_{1})}h(y)u^{\varepsilon}(dy).$

Using the last display for the first inequality, (10.13) for the second, that
$\bar{\upsilon}_{1}^{\varepsilon}$ is small compared with
$\hat{\upsilon}_{1}^{\varepsilon}$ for the third and Lemma 10.6 for the last,
there is $\varepsilon_{1}>0$ such that

$\displaystyle
P_{\lambda^{\varepsilon}}(\hat{\upsilon}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t)$
$\displaystyle=E_{\lambda^{\varepsilon}}(P_{X_{\bar{\upsilon}_{1}^{\varepsilon}}^{\varepsilon}}(\hat{\upsilon}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t))\leq
2P_{u^{\varepsilon}}(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t)$
$\displaystyle\leq
2P_{u^{\varepsilon}}(\upsilon_{1}^{\varepsilon}/E_{\beta^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t/2)\leq
2P_{u^{\varepsilon}}(\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t/4)\leq
2e^{-\tilde{c}t/4}$

for all $\varepsilon\in(0,\varepsilon_{1})$ and $t\geq 0$.

Since as noted previously
$E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}\geq
E_{\lambda^{\varepsilon}}\bar{\upsilon}_{1}^{\varepsilon}$ and since by [6,
Theorem 4 and Corollary 1] there exists $\varepsilon_{2}\in(0,1)$ such that
$P_{\lambda^{\varepsilon}}(\bar{\upsilon}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\bar{\upsilon}_{1}^{\varepsilon}>t)\leq
2e^{-t/2}$ for any $t>0$ and $\varepsilon\in(0,\varepsilon_{2})$, we conclude
that for any
$t>0$$P_{\lambda^{\varepsilon}}(\bar{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t/2)\leq
P_{\lambda^{\varepsilon}}(\bar{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\bar{\upsilon}^{\varepsilon}_{1}>t/2)\leq
2e^{-t/2}.$ The conclusion of the lemma follows from these two bounds and

$\displaystyle
P_{\lambda^{\varepsilon}}(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}$
$\displaystyle>t)\leq
P_{\lambda^{\varepsilon}}(\bar{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t/2)+P_{\lambda^{\varepsilon}}(\hat{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}>t/2).$

∎

######  Lemma 10.14.

For any $j\in L_{\rm{s}}$ and any distribution $\lambda^{\varepsilon}$ on
$\partial B_{\delta}(O_{j})$,
$\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}$
converges in distribution to an Exp(1) random variable under
$P_{\lambda^{\varepsilon}}.$ Moreover,
$E_{\lambda^{\varepsilon}}e^{it\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}}\rightarrow
1/(1-it)$ uniformly on any compact set in $\mathbb{R}$.

###### Proof.

We give the proof for the case $j=1$. Recall that
$E_{u^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{u^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\rightarrow
1/(1-it)$ uniformly on any compact set in $\mathbb{R}$ as
$\varepsilon\rightarrow 0$ from Remark 10.5. We would like to show that
$E_{\lambda^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\rightarrow
1/(1-it)$ uniformly on any compact set in $\mathbb{R}$. Since
$\upsilon_{1}^{\varepsilon}=\bar{\upsilon}^{\varepsilon}_{1}+\hat{\upsilon}^{\varepsilon}_{1}$
with $\bar{\upsilon}^{\varepsilon}_{1}$ the first hitting time to $\partial
B_{2\delta}(O_{1}),$ we know that
$E_{\lambda^{\varepsilon}}\bar{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}\rightarrow
0$ and thus
$E_{\lambda^{\varepsilon}}\hat{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}\rightarrow
1.$ Observe that

$\displaystyle
E_{\lambda^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}=E_{\lambda^{\varepsilon}}\left[e^{it\bar{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\cdot
E_{X^{\varepsilon}\left(\bar{\upsilon}^{\varepsilon}_{1}\right)}\left(e^{it\hat{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\right)\right],$
$E_{\lambda^{\varepsilon}}\left[E_{X^{\varepsilon}\left(\bar{\upsilon}^{\varepsilon}_{1}\right)}\left(e^{it\hat{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\right)\right]\leq\frac{[1+s^{\varepsilon}K/(1-\alpha)]}{[1-s^{\varepsilon}K/(1-\alpha)]}E_{u^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\rightarrow
1/(1-it)$

and

$E_{\lambda^{\varepsilon}}\left[E_{X^{\varepsilon}\left(\bar{\upsilon}^{\varepsilon}_{1}\right)}\left(e^{it\hat{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\right)\right]\geq\frac{[1-s^{\varepsilon}K/(1-\alpha)]}{[1+s^{\varepsilon}K/(1-\alpha)]}E_{u^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\rightarrow
1/(1-it).$

Since
$E_{\lambda^{\varepsilon}}\bar{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}\rightarrow
0$ and $e^{ix}$ is a bounded and continuous function, a conditioning argument
gives

$\left|E_{\lambda^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}-E_{\lambda^{\varepsilon}}\left[E_{X^{\varepsilon}\left(\bar{\upsilon}^{\varepsilon}_{1}\right)}\left(e^{it\hat{\upsilon}^{\varepsilon}_{1}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\right)\right]\right|\leq
E_{\lambda^{\varepsilon}}\left|e^{it\bar{\upsilon}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}-1\right|\rightarrow
0.$

We conclude that
$E_{\lambda^{\varepsilon}}e^{it\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}}\rightarrow
1/(1-it)$ uniformly on any compact set in $\mathbb{R}$. ∎

###  10.6 Return times (single cycles)

In this subsection, we will extend all the three results to return times for
the single cycle case (i.e. when $h_{1}>w$).

######  Lemma 10.15.

There exists $\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$
and any distribution $\lambda^{\varepsilon}$ on $\partial B_{\delta}(O_{1}),$

$\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}=\min_{y\in\cup_{k\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{k})}V(O_{1},y).$

###### Proof.

We have
$E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}=E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}+E_{\lambda^{\varepsilon}}(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})$,
and by Lemma 10.2 we know that

$\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}=\min_{y\in\cup_{k\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{k})}V(O_{1},y).$

Moreover, observe that $W(O_{j})>W(O_{1})$ for any $j\in L\setminus\\{1\\}$
due to Remark 3.14. Note that $\upsilon_{1}^{\varepsilon}$ as defined in
(10.1) coincides with $\sigma_{0}^{\varepsilon}$ defined in (3.4). We can
therefore apply Remark 7.22 with $f=0$, $A=M$ and $\eta=[\min_{j\in
L\setminus\\{1\\}}W(O_{j})-W(O_{1})]/3,$ we find that there exists
$\delta_{1}\in(0,1)$ such that for any $\delta\in(0,\delta_{1})$

$\displaystyle\liminf_{\varepsilon\rightarrow
0}-\varepsilon\log\left(\sup_{z\in\partial
B_{\delta}(O_{1})}E_{z}\left(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon}\right)\right)$
$\displaystyle\geq\min_{j\in L\setminus\\{1\\}}W(O_{j})-W(O_{1})-\min_{j\in
L\setminus\\{1\\}}V(O_{1},O_{j})-\eta$ $\displaystyle=-\min_{j\in
L\setminus\\{1\\}}V(O_{1},O_{j})+2\eta.$

On the other hand, by continuity of $V(O_{1},\cdot),$ for this given $\eta,$
there exists $\delta_{2}\in(0,1)$ such that for any $\delta\in(0,\delta_{2})$

$\min_{y\in\cup_{k\in L\setminus\\{1\\}}\partial
B_{\delta}(O_{k})}V(O_{1},y)\geq\min_{j\in
L\setminus\\{1\\}}V(O_{1},O_{j})-\eta.$

Thus, for any $\delta\in(0,\delta_{0})$ with
$\delta_{0}\doteq\delta_{1}\wedge\delta_{2}$

$\displaystyle\limsup_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})$
$\displaystyle\leq\limsup_{\varepsilon\rightarrow
0}\varepsilon\log\left(\sup_{z\in\partial
B_{\delta}(O_{1})}E_{z}(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})\right)$
$\displaystyle\leq\min_{j\in
L\setminus\\{1\\}}V(O_{1},O_{j})-2\eta\leq\min_{y\in\cup_{k\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{k})}V(O_{1},y)-\eta$
$\displaystyle=\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}-\eta$

and

$\displaystyle\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}$
$\displaystyle=\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}=\min_{y\in\cup_{k\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{k})}V(O_{1},y).$

∎

######  Lemma 10.16.

Given $\delta>0$ sufficiently small, and for any distribution
$\lambda^{\varepsilon}$ on $\partial B_{\delta}(O_{1}),$ there exist
$\tilde{c}>0$ and $\varepsilon_{0}\in(0,1)$ such that

$P_{\lambda^{\varepsilon}}(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}>t)\leq
e^{-\tilde{c}t}$

for all $t\geq 1$ and $\varepsilon\in(0,\varepsilon_{0}).$

###### Proof.

For any $t>0,$
$P_{\lambda^{\varepsilon}}(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}>t)\leq
P_{\lambda^{\varepsilon}}(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}>t/2)+P_{\lambda^{\varepsilon}}((\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}>t/2)$.
It is easy to see that the first term has this sort of bound due to Lemma
10.13 and $E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\geq
E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}.$

It suffices to show that this sort of bound holds for the second term, namely,
there exists a constant $\tilde{c}>0$ such that

$P_{\lambda^{\varepsilon}}\left((\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}>t\right)\leq
e^{-\tilde{c}t}$

for all $t\in[0,\infty)$ and $\varepsilon$ sufficiently small. By Chebyshev’s
inequality,

$P_{\lambda^{\varepsilon}}\left((\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}>t\right)=P_{\lambda^{\varepsilon}}(e^{(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}>e^{t})\leq
e^{-t}E_{\lambda^{\varepsilon}}e^{(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}},$

and it therefore suffices to prove that
$E_{\lambda^{\varepsilon}}e^{(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}$
is less than a constant for all $\varepsilon$ sufficiently small. Observe that

$\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon}=\sum\nolimits_{j\in
L\setminus\\{1\\}}\sum\nolimits_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k),$

where $N_{j}$ is the number of visits of $\partial B_{\delta}(O_{j})$, and
$\upsilon_{j}^{\varepsilon}(k)$ is the $k$-th copy of the first hitting time
to $\cup_{k\in L\setminus\\{j\\}}\partial B_{\delta}(O_{k})$ after starting
from $\partial B_{\delta}(O_{j}).$

If we consider $\partial B_{\delta}(O_{j})$ as the starting location of a
regenerative cycle, as was done previously in the paper for $\partial
B_{\delta}(O_{1})$, then there will be a unique stationary distribution, and
if the process starts with that as the initial distribution then the times
$\upsilon_{j}^{\varepsilon}(k)$ are independent from each other and from the
number of returns to $\partial B_{\delta}(O_{j})$ before first visiting
$\partial B_{\delta}(O_{1})$. While these random times as used here do not
arise from starting with such a distribution, we can use Lemma 10.12 to bound
the error in terms of a multiplicative factor that is independent of
$\varepsilon$ for small $\varepsilon>0$, and thereby justify treating $N_{j}$
as though it is independent of the $\upsilon_{j}^{\varepsilon}(k)$.

Recalling that $l\doteq|L|$,

$\displaystyle
E_{\lambda^{\varepsilon}}e^{(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}$
$\displaystyle=E_{\lambda^{\varepsilon}}\prod\nolimits_{j\in
L\setminus\\{1\\}}e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}$
$\displaystyle\leq\prod\nolimits_{j\in
L\setminus\\{1\\}}\left(E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right]\right)^{1/(l-1)},$

where we use the generalized Hölder’s inequality for the last line. Thus, if
we can show for each $j\in L\setminus\\{1\\}$ that
$E_{\lambda^{\varepsilon}}\exp[{(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k))(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}]$
is less than a constant for all $\varepsilon$ sufficiently small, then we are
done.

Such an estimate is straightforward for the case of an unstable equilibrium,
i.e., for $j\in L\backslash L_{\rm{s}}$, and so we focus on the case $j\in
L_{\rm{s}}\backslash\\{1\\}$. For this case, we apply Lemma 10.13 to find that
there exists $\tilde{{c}}>0$ and $\varepsilon_{0}\in(0,1)$ such that for any
$j\in L$ and any distribution $\tilde{\lambda}^{\varepsilon}$ on $\partial
B(O_{j}),$

$P_{\tilde{\lambda}^{\varepsilon}}(\upsilon_{j}^{\varepsilon}/E_{\tilde{\lambda}^{\varepsilon}}\upsilon_{j}^{\varepsilon}>t)\leq
e^{-\tilde{{c}}t}$ (10.14)

for any $t>0$ and $\varepsilon\in(0,\varepsilon_{0}).$ Hence, given any
$\eta>0$, there is $\bar{\varepsilon}_{0}\in(0,\varepsilon_{0})$ such that for
all $\varepsilon\in(0,\bar{\varepsilon}_{0})$ and any $j\in L\setminus\\{1\\}$

$\displaystyle
E_{\lambda^{\varepsilon}}\left[e^{\upsilon_{j}^{\varepsilon}(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right]$
$\displaystyle\leq
1+\int_{1}^{\infty}P_{\lambda^{\varepsilon}}(e^{(l-1)\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}>t)dt$
$\displaystyle\leq
1+\int_{1}^{\infty}P_{\lambda^{\varepsilon}}\left(\upsilon_{j}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon}>\log
t\cdot
E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}/((l-1)E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon})\right)dt$
$\displaystyle\leq
1+\int_{1}^{\infty}t^{-\tilde{{c}}E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}/((l-1)E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon})}dt$
$\displaystyle=1+\left(\tilde{{c}}E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}/((l-1)E_{\lambda^{\varepsilon}}\upsilon_{j}^{\varepsilon})-1\right)^{-1}\leq
1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)},$

where the last inequality comes from Lemma 10.2 and Lemma 10.15, and by
picking the range of $\varepsilon$ small if it needs to be.

By using induction and a conditioning argument, it follows that for any
$\eta>0$, for any $j\in L\setminus\\{1\\}$ and for any $n\in\mathbb{N},$

$E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{n}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right]\leq\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)^{n}.$

This implies that

$E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right]\leq
E_{\lambda^{\varepsilon}}\left[\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)^{N_{j}}\right].$

The next thing we need to know is the distribution of $N_{j}$, i.e.,
$P_{\lambda^{\varepsilon}}(N_{j}=n)$ for $n\in\mathbb{N}$. Following a similar
argument as in the proof of Lemma 7.3 and the proof of Lemma 7.6, for
sufficiently small $\varepsilon>0$ we find

$\displaystyle P_{\lambda^{\varepsilon}}(N_{j}=n)$
$\displaystyle\leq\left(1\wedge e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)(1-q_{j})^{n-1}q_{j},$

where

$\displaystyle q_{j}\doteq\frac{\inf_{x\in\partial
B_{\delta}(O_{j})}P_{x}(\tilde{{T_{1}}}<{\tilde{T}}_{j}^{+})}{1-\sup_{y\in\partial
B_{\delta}(O_{j})}p(y,\partial B_{\delta}(O_{j}))}\geq
e^{-\frac{1}{\varepsilon}(W(O_{1})-W(O_{1}\cup O_{j})-h_{j}+\eta)}.$ (10.15)

Therefore,

$\displaystyle
E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right]$
$\displaystyle\quad\leq
E_{\lambda^{\varepsilon}}\left[\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)^{N_{j}}\right]=\sum\nolimits_{n=1}^{\infty}\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)^{n}P_{\lambda^{\varepsilon}}(N_{j}=n)$
$\displaystyle\quad\leq\sum_{n=1}^{\infty}\left(1\wedge
e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)^{n}(1-q_{j})^{n-1}q_{j}$
$\displaystyle\quad=\frac{\left(1\wedge
e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)q_{j}\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)}{1-\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)(1-q_{j})}$
$\displaystyle\quad\leq\frac{\left(1\wedge
e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)}{-e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}/q_{j}+1}$
$\displaystyle\quad\leq\frac{\left(1\wedge
e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)}{-e^{-\frac{1}{\varepsilon}(h_{1}+W(O_{1}\cup
O_{j})-W(O_{1})-2\eta)}+1}.$

The second equality holds since $h_{1}>w\geq h_{j}$ and (10.15) imply
$(1-q_{j})(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)})<1$ for all
$\varepsilon$ sufficiently small; the last inequality is from (10.15).

Then we use the fact that for $x\in(0,1/2)$, $1/(1-x)\leq 1+2x$ to find that

$\displaystyle
E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}}\right]$
$\displaystyle\leq\left(1\wedge e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)\left(1+e^{-\frac{1}{\varepsilon}(h_{1}-h_{j}-\eta)}\right)\left(1+2e^{-\frac{1}{\varepsilon}(h_{1}+W(O_{1}\cup
O_{j})-W(O_{1})-2\eta)}\right)$ $\displaystyle\leq\left(1\wedge
e^{-\frac{1}{\varepsilon}(W(O_{j})-W(O_{1}\cup
O_{j})-h_{1}-\eta)}\right)\left(1+5e^{-\frac{1}{\varepsilon}(h_{1}+W(O_{1}\cup
O_{j})-W(O_{1})-2\eta)}\right)$ (10.16) $\displaystyle\leq 1\cdot 6=6.$

The third inequality holds due to the fact that $W(O_{1})\geq W(O_{1}\cup
O_{j})+h_{j}$ and the last inequality comes from the assumption that $h_{1}>w$
and by picking $\eta$ to be smaller than $(h_{1}-w)/2$. This completes the
proof. ∎

######  Lemma 10.17.

Given $\delta>0$ sufficiently small, and for any distribution
$\lambda^{\varepsilon}$ on $\partial B_{\delta}(O_{1})$,
$\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}$
converges in distribution to an Exp(1) random variable under
$P_{\lambda^{\varepsilon}}.$ Moreover,
$E_{\lambda^{\varepsilon}}(e^{it\left(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)})\rightarrow
1/(1-it)$ uniformly on any compact set in $\mathbb{R}$.

###### Proof.

Note that

$E_{\lambda^{\varepsilon}}\left(e^{it\left(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}\right)=E_{\lambda^{\varepsilon}}\left(e^{it\left(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}E_{X^{\varepsilon}(\upsilon_{1}^{\varepsilon})}\left(e^{it\left((\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}\right)\right).$

Since

$E_{\lambda^{\varepsilon}}\left(e^{it\left(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}\right)=E_{\lambda^{\varepsilon}}\left(e^{it\left(E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)\left(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}\right)}\right)$

and we know that
$E_{\lambda^{\varepsilon}}\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\rightarrow
1$ from the proof of Lemma 10.15, by applying Lemma 10.14 we have
$E_{\lambda^{\varepsilon}}(e^{it\left(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)})\rightarrow
1/(1-it)$ uniformly on any compact set in $\mathbb{R}$. Also

$\displaystyle\left|E_{\lambda^{\varepsilon}}\left(e^{it\left(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}\right)-E_{\lambda^{\varepsilon}}\left(e^{it\left(\upsilon_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}\right)\right|\leq
E_{\lambda^{\varepsilon}}\left|E_{X^{\varepsilon}(\upsilon_{1}^{\varepsilon})}\left(e^{it\left((\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)}\right)-1\right|,$

where the right hand side converges to $0$ using
$E_{\lambda^{\varepsilon}}(\tau_{1}^{\varepsilon}-\upsilon_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\rightarrow
0$ and the dominated convergence theorem. The convergence of
$\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}$ to an
Exp(1) random variable under $P_{\lambda^{\varepsilon}}$ and uniform
convergence of
$E_{\lambda^{\varepsilon}}(e^{it\left(\tau_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\right)})$
to $1/(1-it)$ on compact set in $\mathbb{R}$ follows. ∎

###  10.7 Return times (multicycles)

In this subsection, we will extend all the three results to multi regenerative
cycles (when $w\geq h_{1}$). Recall that the multicycle times
$\hat{\tau}^{\varepsilon}_{i}$ are defined according to (6.4) where
$\\{\mathbf{M}^{\varepsilon}_{i}\\}_{i\in\mathbb{N}}$ is a sequence of
independent and geometrically distributed random variables with parameter
$e^{-m/\varepsilon}$ for some $m>0$ such that $m+h_{1}>w$. In addition,
$\\{\mathbf{M}^{\varepsilon}_{i}\\}$ is independent of
$\\{\tau^{\varepsilon}_{n}\\}$.

######  Lemma 10.18.

There exists $\delta_{0}\in(0,1)$ such that for any $\delta\in(0,\delta_{0})$
and any distribution $\lambda^{\varepsilon}$ on $\partial B_{\delta}(O_{1}),$

$\lim_{\varepsilon\rightarrow 0}\varepsilon\log
E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}=m+\min_{y\in\cup_{k\in
L\setminus\\{1\\}}\partial B_{\delta}(O_{k})}V(O_{1},y).$

###### Proof.

Since $\\{\mathbf{M}^{\varepsilon}_{i}\\}$ is independent of
$\\{\tau^{\varepsilon}_{n}\\}$ and
$E_{\lambda^{\varepsilon}}\mathbf{M}^{\varepsilon}_{i}=e^{m/\varepsilon}$, we
apply Lemma 10.15 to complete the proof. ∎

######  Lemma 10.19.

Given $\delta>0,$ for any distribution $\lambda^{\varepsilon}$ on $\partial
B_{\delta}(O_{1}),$ there exist $\tilde{c}>0$ and $\varepsilon_{0}\in(0,1)$
such that

$P_{\lambda^{\varepsilon}}(\hat{\tau}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}>t)\leq
e^{-\tilde{c}t}$

for all $t\geq 1$ and $\varepsilon\in(0,\varepsilon_{0}).$

###### Proof.

We divide the multicycle into a sum of two terms. The first term is the sum of
all the hitting times to $\cup_{j\in L\setminus\\{1\\}}\partial
B_{\delta}(O_{j})$, and the second term is the sum of all residual times. That
is,
$\hat{\tau}_{1}^{\varepsilon}=\hat{\upsilon}_{1}^{\varepsilon}+(\hat{\tau}_{1}^{\varepsilon}-\hat{\upsilon}_{1}^{\varepsilon})$,
where

$\hat{\upsilon}_{1}^{\varepsilon}=\sum\nolimits_{i=1}^{\mathbf{M}^{\varepsilon}_{1}}\upsilon_{1}^{\varepsilon}(i)\text{
and
}\hat{\tau}_{1}^{\varepsilon}-\hat{\upsilon}_{1}^{\varepsilon}=\sum\nolimits_{i=1}^{\mathbf{M}^{\varepsilon}_{1}}\left(\sum\nolimits_{j\in
L\setminus\\{1\\}}\sum\nolimits_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(i,k)\right).$

As discussed many times, it suffices to show that there exist $\tilde{c}>0$
and $\varepsilon_{0}\in(0,1)$ such that

$P_{\lambda^{\varepsilon}}\left(\hat{\upsilon}_{1}^{\varepsilon}/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}>t\right)\leq
e^{-\tilde{c}t}\text{ and
}P_{\lambda^{\varepsilon}}\left((\hat{\tau}_{1}^{\varepsilon}-\hat{\upsilon}_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}>t\right)\leq
e^{-\tilde{c}t}$

for all $t\geq 1$ and $\varepsilon\in(0,\varepsilon_{0}).$

The first bound is relatively easy since $\upsilon_{1}^{\varepsilon}(i)$ is a
sum of approximate exponentials with a tail bound of the given sort, and since
the sum of geometrically many independent and identically distributed
exponentials is again an exponential distribution.

For the second bound, we use Chebyshev’s inequality again as in the proof of
Lemma 10.16 to find that it suffices to prove that
$E_{\lambda^{\varepsilon}}e^{(\hat{\tau}_{1}^{\varepsilon}-\hat{\upsilon}_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}$
is less than a constant for all $\varepsilon$ sufficiently small. Now due to
the independence of $\mathbf{M}^{\varepsilon}_{1}$ and
$\\{\upsilon_{j}^{\varepsilon}(i,k)\\}$, we have

$\displaystyle
E_{\lambda^{\varepsilon}}e^{(\hat{\tau}_{1}^{\varepsilon}-\hat{\upsilon}_{1}^{\varepsilon})/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}$
$\displaystyle\qquad=\sum\nolimits_{i=1}^{\infty}\left(E_{\lambda^{\varepsilon}}\left[\prod\nolimits_{j\in
L\setminus\\{1\\}}e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right]\right)^{i}\cdot
P_{\lambda^{\varepsilon}}(\mathbf{M}^{\varepsilon}_{1}=i)$
$\displaystyle\qquad=e^{-m/\varepsilon}\cdot
E_{\lambda^{\varepsilon}}\left[\prod\nolimits_{j\in
L\setminus\\{1\\}}e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right]$
$\displaystyle\qquad\qquad\cdot\sum\nolimits_{i=1}^{\infty}\left(E_{\lambda^{\varepsilon}}\left[\prod\nolimits_{j\in
L\setminus\\{1\\}}e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right](1-e^{-m/\varepsilon})\right)^{i-1}.$
(10.17)

To do a further computation, we have to at least make sure that

$E_{\lambda^{\varepsilon}}\left[\prod\nolimits_{j\in
L\setminus\\{1\\}}e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right](1-e^{-m/\varepsilon})<1.$
(10.18)

To see this, we first use the generalized Hölder’s inequality to find

$\displaystyle E_{\lambda^{\varepsilon}}\left[\prod\nolimits_{j\in
L\setminus\\{1\\}}e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right]\leq\prod\nolimits_{j\in
L\setminus\\{1\\}}\left(E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right]\right)^{1/(l-1)}.$

Moreover, since $m+h_{1}>w$ and
$E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}=E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}\cdot
E_{\lambda^{\varepsilon}}\mathbf{M}^{\varepsilon}_{1}=e^{m/\varepsilon}E_{\lambda^{\varepsilon}}\tau_{1}^{\varepsilon}$,
by the same argument that gives (10.6), for any $\eta>0$ and $j\in
L\setminus\\{1\\}$

$\displaystyle
E_{\lambda^{\varepsilon}}\left[e^{\left(\sum_{k=1}^{N_{j}}\upsilon_{j}^{\varepsilon}(k)\right)(l-1)/E_{\lambda^{\varepsilon}}\hat{\tau}_{1}^{\varepsilon}}\right]$
$\displaystyle\quad\leq\left(1\wedge