Functional Large Deviations for Kac–Stroock Approximation to a Class of Gaussian Processes with Application to Small Noise Diffusions

Abstract

In this paper, we establish the functional large deviation principle (LDP) for the Kac–Stroock approximations of a wild class of Gaussian processes constructed by telegraph types of integrals with $L^2$-integrands under mild conditions and find the explicit form for their rate functions. Our investigation includes a broad range of kernels, such as those related to Brownian motions, fractional Brownian motions with whole Hurst parameters, and Ornstein–Uhlenbeck processes. Furthermore, we consider a class of non-Markovian stochastic differential equations driven by the Kac–Stroock approximation and establish their Freidlin–Wentzell type LDP. The rate function clearly indicates an interesting phase transition phenomenon as the approximating rate moves from one region to the other.

Appendix A

In this section, we first give an alternative proof of Proposition 3 explicitly. Then, some estimations of tail probability for small noise diffusions and deviation inequalities for Poisson random integral will be given.

Proof of Proposition 3

By Theorem 4.2 in [29], which addresses inhomogeneous Markov chains, our case being homogeneous simplifies the scenario. Consequently, the so-called H-functional reduces to a single variable without the time t. Thus, it is enough to prove that for any $\beta \in {\mathbb {R}}$,

$$\begin{aligned} H(\beta )=\sqrt{1+\beta ^{2}}-1. \end{aligned}$$

We note that $\xi (t)=(-1)^{N(t)}$ is a  $\{-1,1\}$-valued reversible Markov process with stationary distribution $\mu$ satisfying

$$\begin{aligned} \mu (1)=\mu (-1)=\frac{1}{2}. \end{aligned}$$

and Q-matrix

$$\begin{aligned} Q=\left( \begin{array}{ll}-1\\ ~~ 1\\ \end{array} \begin{array}{ll}~~ 1\\ -1\\ \end{array}\right) . \end{aligned}$$

By Theorem 4.2 of Friedlin and Wentzel [23] in p. 218 with $b(x,i)=i$, the desired result is obtained after simple calculation. $\square$

1.1 Estimations of Tail Probability for Small Noise Diffusions

Recall that $N^{\alpha }$ is a Poisson random measure with the intensity measure $\alpha dt$ on  $({\mathbb {R}}_{+}, {\mathcal {B}}({\mathbb {R}}_{+}))$ and $\tilde{N}^{\alpha }(ds)={N}^{\alpha }(ds)-\alpha ds$ is the compensated Poisson random measure. Now, as the main result of this subsection, the following proposition gives estimation of tail probability for small noise diffusions, which plays a crucial role in the proof of Theorem 2.

Proposition 6

Assume that $U^{\epsilon }=\{U^{\epsilon }(t), t\in [0,1]\}$ is a continuous stochastic process valued in $\mathbb {R}$. Define  $\hat{Y}^{\epsilon }=\{\hat{Y}^{\epsilon }(t), t\in [0,1]\}$ and  $\check{Y}^{\epsilon }=\{\check{Y}^{\epsilon }(t), t\in [0,1]\}$ as follows

$$\begin{aligned} \begin{aligned}&\hat{Y}^{\epsilon }(t)=\hat{y}^{\epsilon }_{0}+\int ^{t}_{0}\hat{h}^{\epsilon }(s)ds +\lambda (\epsilon )\epsilon \int ^{t}_{0}\hat{\gamma }^{\epsilon }(s)\tilde{N}^{\epsilon ^{-2}}(ds), \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\check{Y}^{\epsilon }(t)=\check{y}^{\epsilon }_{0}+\int ^{t}_{0}\check{h}^{\epsilon }(s)ds +\lambda (\epsilon )\epsilon \int ^{t}_{0}\check{\gamma }^{\epsilon }(s)\tilde{N}^{\epsilon ^{-2}}(ds). \end{aligned} \end{aligned}$$

For the stop** time $\tau \in [0,1]$ and some positive constants $\rho$, B and M, suppose the following assumptions hold.

• (1) For any $t\in [0,\tau ]$,

  $$\begin{aligned}&|\hat{h}^{\epsilon }(t)|\vee |\check{h}^{\epsilon }(t)|\le B\left( \rho ^{2}+|U^{\epsilon }(t-)|^{2}\right) ^{\frac{1}{2}},\\&|\hat{\gamma }^{\epsilon }(t)|\vee |\check{\gamma }^{\epsilon }(t)|\le M\left( \rho ^{2}+|U^{\epsilon }(t-)|^{2}\right) ^{\frac{1}{2}}. \end{aligned}$$

• (2) For any $t\in [0,\tau )$, $\check{Y}^{\epsilon }(t)\le U^{\epsilon }(t)\le \hat{Y}^{\epsilon }(t)$. Take $\epsilon$ sufficient small such that $1-\epsilon \lambda (\epsilon )M>\frac{1}{\sqrt{2}}$. Then, for any $\delta >0$,

  $$\begin{aligned} \begin{aligned}&\lambda ^2(\epsilon )\log {\mathbb {P}}\Bigg (\sup _{t\in [0,\tau ]}|U^{\epsilon }(t)|>\delta \Bigg )\\&\quad \le 2\sqrt{2}B+4M^2\epsilon ^{2}(2+3\lambda ^{2}(\epsilon ))e^{4M\epsilon /{\lambda (\epsilon )}} +\log \Big (\frac{\rho ^{2}+|\hat{y}^{\epsilon }_{0}|^{2}+|\check{y}^{\epsilon }_{0}|^{2}}{\rho ^{2}+\delta ^{2}}\Big ). \end{aligned} \end{aligned}$$

To prove Proposition 6, we need the following lemma and its proof will be postponed to the end of this subsection.

Lemma 6

Define the $\mathbb {R}^{2}$-valued processes  $Y^{\epsilon }=\{Y^{\epsilon }(t), t\in [0,1]\}$ as follows:

$$\begin{aligned} \begin{aligned} Y^{\epsilon }(t)&=y^{\epsilon }_{0}+\int ^{t}_{0}h^{\epsilon }(s)ds +\lambda (\epsilon )\epsilon \int ^{t}_{0}\gamma ^{\epsilon }(s)\tilde{N}^{\epsilon ^{-2}}(ds). \end{aligned} \end{aligned}$$

(A1)

For the stop** time  $\tau \in [0,1]$, suppose that there exist some positive constants B, $\rho$ and M such that for any $t\in [0,\tau ]$,

$$\begin{aligned} \begin{aligned}&|h^{\epsilon }(t)|\le B\left( \rho ^{2}+|Y^{\epsilon }(t-)|^{2}\right) ^{\frac{1}{2}},\\&|\gamma ^{\epsilon }(t)|\le M\left( \rho ^{2}+|Y^{\epsilon }(t-)|^{2}\right) ^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

(A2)

Take $\epsilon$ sufficient small such that $1-\epsilon \lambda (\epsilon )M>\frac{1}{\sqrt{2}}$. Then, for any $\delta >0$

$$\begin{aligned} \begin{aligned}&\lambda ^{2}(\epsilon )\log {\mathbb {P}}\Bigg (\sup _{t\in [0,\tau ]}|Y^{\epsilon }(t)|>\delta \Bigg )\\&\quad \le 2B+2M^2\epsilon ^{2}(2+3\lambda ^{2}(\epsilon ))\cdot e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}}+\log \Big (\frac{\rho ^{2}+\Vert y^{\epsilon }_{0}\Vert ^{2}}{\rho ^{2}+\delta ^{2}}\Big ). \end{aligned} \end{aligned}$$

Proof of Proposition 6

For any $t\in [0,\tau ]$,

$$\begin{aligned} \begin{aligned}&|\hat{h}^{\epsilon }(t)|\vee |\check{h}^{\epsilon }(t)|\le B\Big (\rho ^{2}+|\hat{Y}^{\epsilon }(t-)|^{2}+|\check{Y}^{\epsilon }(t-)|^{2}\Big )^{\frac{1}{2}},\\&|\hat{\gamma }^{\epsilon }(t)|\vee |\check{\gamma }^{\epsilon }(t)| \le M\Big (\rho ^{2}+|\hat{Y}^{\epsilon }(t-)|^{2}+|\check{Y}^{\epsilon }(t-)|^{2}\Big )^{\frac{1}{2}}. \end{aligned} \end{aligned}$$

Define  $Y^{\epsilon }(t)=\big (\hat{Y}^{\epsilon }(t), \check{Y}^{\epsilon }(t)\big )$. Then,  $Y^{\epsilon }$ satisfies the conditions of Lemma 6 with $y^{\epsilon }_{0}=\big (\hat{y}^{\epsilon }_{0}, \check{y}^{\epsilon }_{0}\big )$. Consequently, for the stop** time $\tau \in [0,1]$ and any $\delta >0$,

$$\begin{aligned}{} & {} \lambda ^{2}(\epsilon )\log {\mathbb {P}}\Bigg (\sup _{t\in [0,\tau ]}\Vert Y^{\epsilon }(t)\Vert >\delta \Bigg )\\{} & {} \quad \le 2\sqrt{2}B+4M^2\epsilon ^{2}(2+3\lambda ^{2}(\epsilon ))e^{4M\epsilon /{\lambda (\epsilon )}} +\log \Big (\frac{\rho ^{2}+\Vert y^{\epsilon }_{0}\Vert ^{2}}{\rho ^{2}+\delta ^{2}}\Big ). \end{aligned}$$

Together with the fact that  $\sup _{t\in [0,\tau ]}|U^{\epsilon }(t)|^{2}\le \sup _{t\in [0,\tau ]}(|\hat{Y}^{\epsilon }(t)|^{2}+|\check{Y}^{\epsilon }(t)|^{2})=\sup _{t\in [0,\tau ]}|Y^{\epsilon }(t)|^{2}$, we can complete the proof of this proposition. $\square$

To end this subsection, we will give the proof to Lemma 6. Firstly, we give the following auxiliary result.

Lemma 7

Assume that $\rho \ge 0$, $0<\beta <1$, and $a,b\in \mathbb {R}^d$ satisfy

$$\begin{aligned} |b|^{2}\le \beta (\rho ^{2}+|a|^{2}). \end{aligned}$$

Then, we have

$$\begin{aligned} \rho ^{2}+|a|^{2}\le \frac{\rho ^{2}+|a+b|^{2}}{(1-\sqrt{\beta })^{2}}. \end{aligned}$$

Proof

For any $\epsilon >0$, it holds

$$\begin{aligned} |a|^{2}&\le |a+b|^{2}+|b|^{2}+2|a+b||b|\\&\le |a+b|^{2}+|b|^{2}+\epsilon |a+b|^{2}+\frac{|b|^{2}}{\epsilon }\\&\le (1+\epsilon )|a+b|^{2}+\Big (1+\frac{1}{\epsilon }\Big )|b|^{2}\\&\le (1+\epsilon )|a+b|^{2}+\beta \Big (1+\frac{1}{\epsilon }\Big )(\rho ^{2}+|a|^{2}), \end{aligned}$$

which implies immediately

$$\begin{aligned} \rho ^{2}+|a|^{2}\le (1+\epsilon )(\rho ^{2}+|a+b|^{2})+\beta \Big (1+\frac{1}{\epsilon }\Big )(\rho ^{2}+|a|^{2}). \end{aligned}$$

Therefore, provided that $\epsilon (1-\beta )-\beta >0$, we obtain

$$\begin{aligned} \rho ^{2}+|a|^{2}\le \frac{\epsilon (1+\epsilon )}{\epsilon (1-\beta )-\beta }\Big (\rho ^{2}+|a+b|^{2}\Big ). \end{aligned}$$

Taking $\epsilon =\frac{\beta +\sqrt{\beta }}{1-\beta }$, we have

$$\begin{aligned} \rho ^{2}+|a|^{2} \le \frac{(1+\sqrt{\beta })^{2}}{(1-\beta )^{2}}\left( \rho ^{2}+|a+b|^{2}\right) =\frac{\rho ^{2}+|a+b|^{2}}{(1-\sqrt{\beta })^{2}}. \end{aligned}$$

$\square$

Proof of Lemma 6

Define for $u\in \mathbb {R}^d$

$$\begin{aligned} \Phi (u)= & {} \log (\rho ^{2}+|u|^{2}),\\ \psi (u)= & {} \exp \Big \{\frac{\Phi (u)}{\lambda ^{2}(\epsilon )}\Big \}=(\rho ^{2}+|u|^{2})^{\frac{1}{\lambda ^2(\epsilon )}}. \end{aligned}$$

By Itô’s formula, we can write

$$\begin{aligned} \psi (Y^{\epsilon }(t))= & {} \psi (y^{\epsilon }_{0})+\int _0^t\mathcal {B}^{\epsilon }(s)\tilde{N}^{\epsilon ^{-2}}(ds)\nonumber \\{} & {} +\,\int _0^t\langle \nabla \psi (Y^{\epsilon }(s-)),h^{\epsilon }(s)\rangle ds +\int _0^t\mathcal {A}^{\epsilon }(s)ds, \end{aligned}$$

(A3)

where

$$\begin{aligned} \mathcal {A}^{\epsilon }(s)= & {} \exp \Big \{\frac{\Phi (Y^{\epsilon }(s-)+\lambda (\epsilon )\epsilon \gamma ^{\epsilon }(s))}{\lambda ^2(\epsilon )}\Big \}-\exp \Big \{\frac{\Phi (Y^{\epsilon }(s-)}{\lambda ^2(\epsilon )}\Big \}\nonumber \\{} & {} -\frac{\epsilon \langle \nabla \Phi (Y^{\epsilon }(s-),\gamma ^{\epsilon }(s)\rangle }{\lambda (\epsilon )} \exp \Big \{\frac{\Phi (Y^{\epsilon }(s-)}{\lambda ^2(\epsilon )}\Big \} \end{aligned}$$

(A4)

and

$$\begin{aligned} \mathcal {B}^{\epsilon }(s)=\exp \Big \{\frac{\Phi (Y^{\epsilon }(s-)+\lambda (\epsilon )\epsilon \gamma ^{\epsilon }(s))}{\lambda ^2(\epsilon )}\Big \} -\exp \Big \{\frac{\Phi (Y^{\epsilon }(s-)}{\lambda ^2(\epsilon )}\Big \}. \end{aligned}$$

Firstly, Taylor formula gives

$$\begin{aligned} \mathcal {A}^{\epsilon }(s)= & {} \frac{\epsilon ^{2}}{2\lambda ^2(\epsilon )}\psi (\tilde{Y}^{\epsilon }(s))\langle \nabla \Phi (\tilde{Y}^{\epsilon }(s)\otimes \nabla \Phi (\tilde{Y}^{\epsilon }(s), \gamma ^{\epsilon }(s)\otimes \gamma ^{\epsilon }(s)\rangle \nonumber \\{} & {} -\frac{\epsilon ^{2}}{2}\psi (\tilde{Y}^{\epsilon }(s))\langle \nabla \otimes \nabla \Phi \big (\tilde{Y}^{\epsilon }(s)\big ), \gamma ^{\epsilon }(s)\otimes \gamma ^{\epsilon }(s)\rangle , \end{aligned}$$

(A5)

where $\tilde{Y}^{\epsilon }(s)=Y^{\epsilon }(s-)+\theta \lambda (\epsilon )\epsilon \gamma (s)$, $\theta \in [0,1]$. Now, we will give the estimates of above terms, respectively.

Let $a=Y^{\epsilon }(s-)$,  $b=\theta \lambda (\epsilon )\epsilon \gamma ^{\epsilon }(s)$. For $s\in [0,\tau ]$, we have by (A2)

$$\begin{aligned} |b|^{2}\le \lambda ^2(\epsilon )\epsilon ^{2}\Vert \gamma ^{\epsilon }(s)\Vert ^{2}\le M^2\epsilon ^{2}\lambda ^2(\epsilon )(\rho ^{2}+|a|^{2}). \end{aligned}$$

Then, from Lemma 7, it follows

$$\begin{aligned} |\gamma ^{\epsilon }(s)|^{2}\le & {} M^2\left( \rho ^{2}+|Y^{\epsilon }(s-)|^{2}\right) \nonumber \\\le & {} \frac{M^2}{\big (1- \epsilon \lambda (\epsilon )M\big )^{2}} \cdot \left( \rho ^{2}+|\tilde{Y}^{\epsilon }(s)|^{2}\right) \nonumber \\\le & {} 2M^{2}\left( \rho ^{2}+|\tilde{Y}^{\epsilon }(s)|^{2}\right) , \end{aligned}$$

(A6)

where the last inequality is obtained by $1-\epsilon \lambda (\epsilon )M>\frac{1}{\sqrt{2}}$ and thus for any $s\le \tau$,

$$\begin{aligned}{} & {} \Big |\langle \nabla \Phi (\tilde{Y}^{\epsilon }(s)\otimes \nabla \Phi (\tilde{Y}^{\epsilon }(s), \gamma ^{\epsilon }(s)\otimes \gamma ^{\epsilon }(s)\rangle \Big |\nonumber \\{} & {} \quad \le \frac{4|\gamma ^{\epsilon }(s)|^{2}\big |\tilde{Y}^{\epsilon }(s)\big |^{2}}{\Big (\rho ^{2}+\big |\tilde{Y}^{\epsilon }(s)\big |^{2}\Big )^{2}}\nonumber \\{} & {} \quad \le \frac{8M^2\big |\tilde{Y}^{\epsilon }(s)\big |^{2}}{\rho ^{2}+\big |\tilde{Y}^{\epsilon }(s)\big |^{2}}\le 8M^2. \end{aligned}$$

(A7)

Moreover, there exists $\theta '\in [0,1]$ such that

$$\begin{aligned} \psi (\tilde{Y}^{\epsilon }(s)) =\psi (Y^{\epsilon }(s)) \exp \left( \frac{2\epsilon \theta }{\lambda (\epsilon )} \cdot \frac{<Y^{\epsilon }(s-)+\theta '\theta \lambda (\epsilon )\epsilon \gamma ^{\epsilon }(s), \gamma ^{\epsilon }(s)>}{\rho ^{2}+|Y^{\epsilon }(s-)+\theta '\theta \lambda (\epsilon )\epsilon \gamma ^{\epsilon }(s)|^{2}}\right) . \end{aligned}$$

Following the same line as in the proof of (A6), we have for $s\in [0,\tau ]$

$$\begin{aligned} |\gamma ^{\epsilon }(s)|^{2}\le 2M^2\left( \rho ^{2}+|Y^{\epsilon }(s-)+\theta '\theta \lambda (\epsilon )\epsilon \gamma ^{\epsilon }(s)|^{2}\right) . \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \psi (\tilde{Y}^{\epsilon }(s))\le \psi (Y^{\epsilon }(s))e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}},\quad s\le \tau . \end{aligned}$$

(A8)

Now, the fact $\nabla \otimes \nabla \Phi _{i,j}(u)=\frac{2\delta _{i,j}}{\rho ^{2}+\Vert u\Vert ^{2}}-\frac{4u_{i}u_{j}}{(\rho ^{2}+\Vert u\Vert ^{2})^{2}}$ gives the following operator norm

$$\begin{aligned} \Big \Vert \nabla \otimes \nabla \Phi \big (\tilde{Y}^{\epsilon }(s)\big )\Big \Vert \le \frac{6}{\rho ^{2}+\big |\tilde{Y}^{\epsilon }(s)\big |^{2}}, \end{aligned}$$

which together with (A6) implies

$$\begin{aligned} \Big |\langle \nabla \otimes \nabla \Phi \big (\tilde{Y}^{\epsilon }(s)\big ), \gamma ^{\epsilon }(s)\otimes \gamma ^{\epsilon }(s)\rangle \Big | \le 12M^2. \end{aligned}$$

(A9)

Combing  (A7)–(A9), we have

$$\begin{aligned} \big |\mathcal {A}^{\epsilon }(s)\big |\le \left( 4M^2\epsilon ^{2}\lambda ^{-2}(\epsilon )+6M^2\epsilon ^{2}\right) e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}}\psi (Y^{\epsilon }(s)). \end{aligned}$$

(A10)

Secondly, it holds for any $s\le \tau$,

$$\begin{aligned} \big |\langle \nabla \psi (Y^{\epsilon }(s-)),h^{\epsilon }(s)\rangle \big |\le \frac{2B}{\lambda ^2(\epsilon )}\psi (Y^{\epsilon }(s-)). \end{aligned}$$

Together with (A3) and (A10), we can get for any $t\le \tau$,

$$\begin{aligned} \psi (Y^{\epsilon }(t))&\le \psi (y^{\epsilon }_{0})+\int _0^t\mathcal {B}^{\epsilon }(s)\tilde{N}^{\epsilon ^{-2}}(ds)\\&\quad +\left( \Big (4M^2\epsilon ^{2}\lambda ^{-2}(\epsilon )+6M^2\epsilon ^{2}\Big )e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}} +2B\lambda ^{-2}(\epsilon )\right) \int ^{t}_{0}\psi (Y^{\epsilon }(s))ds. \end{aligned}$$

Finally, letting $\tau _{1}:=\inf \left\{ t\ge 0; |Y^{\epsilon }(t)|\ge \delta \right\}$, we have

$$\begin{aligned}&\mathbb {E}\psi (Y^{\epsilon }(\tau \wedge \tau _{1}\wedge t))\\&\quad \le \psi (y^{\epsilon }_0)+\left( \Big (4M^2\epsilon ^{2}\lambda ^{-2}(\epsilon )+6M^2\epsilon ^{2}\Big )e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}} +2B\lambda ^{-2}(\epsilon )\right) \\&\qquad \int ^{t}_{0}\psi (Y^{\epsilon }(\tau \wedge \tau _{1}\wedge s))ds, \end{aligned}$$

which implies by Gronwall inequality

$$\begin{aligned} \mathbb {E}\psi (Y^{\epsilon }(\tau \wedge \tau _{1})) \le \psi (y^{\epsilon }_0)\exp \left\{ \Big (4M^2\epsilon ^{2}\lambda ^{-2}(\epsilon )+6M^2\epsilon ^{2}\Big )e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}} +2B\lambda ^{-2}(\epsilon )\right\} . \end{aligned}$$

Therefore,

$$\begin{aligned}&{\mathbb {P}}\Bigg (\sup _{t\in [0,\tau ]}|Y^{\epsilon }(t)|>\delta \Bigg )\\&\quad \le {\mathbb {P}}\Big (\psi (Y^{\epsilon }(\tau \wedge \tau _{1}))\ge \psi (\delta )\Big )\\&\quad \le \frac{\mathbb {E}\psi (Y^{\epsilon }(\tau \wedge \tau _{1}))}{\psi (\delta )}\\&\quad =\Big (\frac{\rho ^{2}+|y^{\epsilon }_0|^{2}}{\rho ^{2}+\delta ^{2}}\Big )^{\frac{1}{\lambda ^2(\epsilon )}} \exp \left\{ \Big (4M^2\epsilon ^{2}\lambda ^{-2}(\epsilon )+6M^2\epsilon ^{2}\Big )e^{2\sqrt{2}M\epsilon /{\lambda (\epsilon )}} +2B\lambda ^{-2}(\epsilon )\right\} , \end{aligned}$$

which completes the proof of this lemma. $\square$

1.2 Deviation Inequalities for Poisson Random Integrals

Proposition 7

For the stop** time $\tau$, assume that there exist some constants $\rho$, $\iota _1$ and $\iota _2$ such that

$$\begin{aligned} \iota _1\le \tau \le \iota _2,\quad \sup _{s\in [\iota _1,\tau ]}|\gamma (s)|\le \rho . \end{aligned}$$

For any $a>0$ and $\eta >0$, we have

$$\begin{aligned}{} & {} P\Bigg (\sup _{t\in [\iota _1,\tau ]}\lambda (\epsilon )\epsilon \int ^{t}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)>a\Bigg )\nonumber \\{} & {} \quad \le \exp \left\{ -\lambda ^{-2}(\epsilon )\Big (\eta a-\frac{1}{2}\eta ^2\rho ^2(\iota _2-\iota _1)e^{\epsilon \lambda ^{-1}(\epsilon )\eta \rho }\Big )\right\} . \end{aligned}$$

(A11)

Proof

Define the stop** time

$$\begin{aligned} \hat{\tau }=\inf \Big \{t\ge \iota _1; \lambda (\epsilon )\epsilon \int ^{t}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)\ge a\Big \}. \end{aligned}$$

It holds that

$$\begin{aligned}&\Big \{\sup _{t\in [\iota _1,\tau ]}\lambda (\epsilon )\epsilon \int ^{t}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)>a\Big \} \subset \Big \{\lambda (\epsilon )\epsilon \int ^{\tau \wedge \hat{\tau }}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)\ge a\Big \}. \end{aligned}$$

For $\eta >0$, Chebyshev inequality gives that

$$\begin{aligned}&{\mathbb {P}}\Big (\lambda (\epsilon )\epsilon \int ^{\tau \wedge \hat{\tau }}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)\ge a\Big )\\&\quad \le \exp \Big \{-\lambda ^{-2}(\epsilon )\eta a\Big \} {\mathbb {E}}\exp \Big \{\frac{\epsilon \eta }{\lambda (\epsilon )}\int ^{\tau \wedge \hat{\tau }}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)\Big \}. \end{aligned}$$

Notice that for any $p>1$

$$\begin{aligned} \bigg \{\exp \Big \{\frac{p\epsilon \eta }{\lambda (\epsilon )}\int ^{\tau \wedge \hat{\tau }\wedge t}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds) -\epsilon ^{-2}\int ^{\tau \wedge \hat{\tau }\wedge t}_{\iota _1} \big (e^{\frac{p\epsilon \eta }{\lambda (\epsilon )}\gamma (s)}-1-\frac{p\epsilon \eta }{\lambda (\epsilon )}\gamma (s)\big )ds\Big \}, t\ge 0\bigg \} \end{aligned}$$

is an exponential martingale. Therefore, by Hölder’s inequality, for $\frac{1}{p}+\frac{1}{q}=1$ with $p>0$, $q>0$,

$$\begin{aligned}&{\mathbb {E}}\exp \Big \{\frac{\epsilon \eta }{\lambda (\epsilon )}\int ^{\tau \wedge \hat{\tau }}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)\Big \}\\&\quad \le {\mathbb {E}}^{\frac{1}{q}}\exp \Big \{\frac{q\epsilon ^{-2}}{p}\int ^{\tau \wedge \hat{\tau }}_{\iota _1} \big (e^{\frac{p\epsilon \eta }{\lambda (\epsilon )}\gamma (s)}-1-\frac{p\epsilon \eta }{\lambda (\epsilon )}\gamma (s)\big )ds\Big \}\\&\quad \le \exp \left\{ \frac{1}{2}\lambda ^{-2}(\epsilon )p\eta ^2\rho ^2(\iota _2-\iota _1)e^{\epsilon \lambda ^{-1}(\epsilon )p\eta \rho }\right\} , \end{aligned}$$

where the last inequality is derived from the fact that

$$\begin{aligned} e^x-1-x\le \frac{x^2}{2}e^{|x|},\quad x\in \mathbb {R}. \end{aligned}$$

Consequently, it holds that

$$\begin{aligned}&P\Bigg (\sup _{t\in [\iota _1,\tau ]}\lambda (\epsilon )\epsilon \int ^{t}_{\iota _1}\gamma (s)\tilde{N}^{\epsilon ^{-2}}(ds)>a\Bigg )\\&\quad \le \exp \left\{ -\lambda ^{-2}(\epsilon )\Big (\eta a-\frac{1}{2}p\eta ^2\rho ^2(\iota _2-\iota _1)e^{\epsilon \lambda ^{-1}(\epsilon )p\eta \rho }\Big )\right\} . \end{aligned}$$

Letting $p\rightarrow 1$, we can get (A11) and complete the proof of this proposition. $\square$