Proof of Theorem 1
We start with the decomposition:
$$\begin{aligned} \mathbb {E}\big (b_{1}(\varepsilon )\big )&=\mathbb {E}\Big (\big (\sigma W_{h}+J_{h}\big )^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )\nonumber \\&=\sigma ^{2}\,\mathbb {E}\Big (W_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )\!+\!\mathbb {E}\Big (J_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )\!+\!2\sigma \,\mathbb {E}\Big (W_{h}J_{h}{} \mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big ).\quad \end{aligned}$$
(33)
The asymptotic behavior of each of the terms above is obtained in three steps.
Step 1
We first analyze the behavior of the first term in Eq. (33) as \(h\rightarrow 0\). By Eqs. (7) and (8), we first decompose it as
$$\begin{aligned} \mathbb {E}\Big (W_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )&=h-\mathbb {E}\Big (W_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|>\varepsilon \}}\Big )=h-\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}-\eta h}\,W_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|>\varepsilon \}}\Big ) \nonumber \\&=h-he^{-\eta h}\,\widetilde{\mathbb {E}}\Big (W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big ) \nonumber \\&\quad -he^{-\eta h}\,\widetilde{\mathbb {E}}\Big (\Big (e^{-\widetilde{U}_{h}}-1\Big )W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big ) \nonumber \\&=:h-he^{-\eta h}I_{1}(h)-he^{-\eta h}I_{2}(h). \end{aligned}$$
(34)
In what follows, we will analyze the asymptotic behavior of \(I_{1}(h)\) and \(I_{2}(h)\), as \(h\rightarrow 0\), respectively, in two sub-steps.
Step 1.1
Clearly, by Eq. (9) and the symmetry of \(W_{1}\),
$$\begin{aligned} I_{1}(h)=\widetilde{\mathbb {E}}\Big (W_{1}^{2}\,\mathbf{1}_{\{\sigma \sqrt{h}W_{1}+J_{h}>\varepsilon \}}\Big )+\widetilde{\mathbb {E}}\Big (W_{1}^{2}\,\mathbf{1}_{\{\sigma \sqrt{h}W_{1}-J_{h}>\varepsilon \}}\Big )=:I_{1}^{+}(h)+I_{1}^{-}(h). \end{aligned}$$
(35)
Denote by
$$\begin{aligned} \phi (x):=\frac{1}{\sqrt{2\pi }}e^{-x^{2}/2},\quad \overline{\Phi }(x):=\int _{x}^{\infty }\phi (x)\,dx,\quad x\in \mathbb {R}. \end{aligned}$$
By conditioning on \(J_{h}\) and using the fact that \(\widetilde{\mathbb {E}}(W_{1}^{2}{} \mathbf{1}_{\{W_{1}>x\}})=x\phi (x)+\overline{\Phi }(x)\), for all \(x\in \mathbb {R}\), we have that
$$\begin{aligned} I_{1}^{\pm }(h)=\widetilde{\mathbb {E}}\left( \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\mp \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\mp \frac{J_{h}}{\sigma \sqrt{h}}\bigg )+\overline{\Phi }\bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\mp \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) . \end{aligned}$$
(36)
Let \(p_{J,h}^{\pm }\) be the density of \(\pm J_{h}\) under \(\widetilde{\mathbb {P}}\), and recall the Fourier transform and its inverse transform defined by
$$\begin{aligned} (\mathcal {F}g)(z):=\frac{1}{\sqrt{2\pi }}\int _{\mathbb {R}}g(x)e^{-izx}\,dx,\quad \big (\mathcal {F}^{-1}g\big )(x):=\frac{1}{\sqrt{2\pi }}\int _{\mathbb {R}}g(z)e^{izx}\,dz. \end{aligned}$$
In what follows, we set
$$\begin{aligned} \psi (x)&:=\bigg (\mathcal {F}^{-1}\phi \bigg (\frac{\cdot }{\sigma \sqrt{h}}-\frac{\varepsilon }{\sigma \sqrt{h}}\bigg )\bigg )(x)=\frac{1}{\sqrt{2\pi }}\int _{\mathbb {R}}\phi \bigg (\frac{z}{\sigma \sqrt{h}}-\frac{\varepsilon }{\sigma \sqrt{h}}\bigg )e^{izx}\,dz \nonumber \\&\,=\frac{\sigma \sqrt{h}\,e^{i\varepsilon x}}{\sqrt{2\pi }}\int _{\mathbb {R}}\phi (\omega )e^{i\sigma \sqrt{h}\omega x}\,d\omega =\frac{\sigma \sqrt{h}}{\sqrt{2\pi }}\exp \bigg (i\varepsilon x-\frac{1}{2}\sigma ^{2}x^{2}h\bigg ). \end{aligned}$$
(37)
Then, we deduce that
$$\begin{aligned} \widetilde{\mathbb {E}}\left( \phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) =\int _{\mathbb {R}}\big (\mathcal {F}\psi \big )(z)p_{J,h}^{\mp }(z)\,dz=\int _{\mathbb {R}}\psi (u)\big (\mathcal {F}p_{J,h}^{\mp }\big )(u)\,du, \end{aligned}$$
where
$$\begin{aligned} \big (\mathcal {F}p_{J,h}^{\pm }\big )(u)&=\frac{e^{\mp iu\widetilde{\gamma }h}}{\sqrt{2\pi }}\exp \!\left( \!-(C_{+}\!+C_{-})\,\bigg |\!\cos \bigg (\!\frac{\pi Y}{2}\!\bigg )\!\bigg |\Gamma (-Y)|u|^{Y}h\bigg (\!1-i\frac{C_{+}\!-C_{-}}{C_{+}\!+C_{-}}\tan \!\bigg (\!\frac{\pi Y}{2}\!\bigg )\text {sgn}(u)\!\bigg )\!\right) \nonumber \\&=:\frac{1}{\sqrt{2\pi }}\exp \Big (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)\mp iu\widetilde{\gamma }h\Big ), \end{aligned}$$
(38)
with \(c_{1}:=(C_{+}\!+C_{-})\,\cos (\pi Y/2)\Gamma (-Y)\) and \(c_{2}:=(C_{-}-C_{+})\sin (\pi Y/2)\Gamma (-Y)\). Hence, we have
$$\begin{aligned} \widetilde{\mathbb {E}}\left( \phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right)&=\frac{\sigma \sqrt{h}}{2\pi }\!\!\int _{\mathbb {R}}\exp \!\bigg (\!c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{1}{2}\sigma ^{2}u^{2}h+iu\big (\varepsilon \!\pm \!\widetilde{\gamma }h\big )\!\bigg )du \nonumber \\&=\frac{\sigma \sqrt{h}}{\pi }\int _{0}^{\infty }\exp \bigg (c_{1}u^{Y}h-\frac{1}{2}\sigma ^{2}u^{2}h\bigg )\cos \Big (c_{2}u^{Y}h+u\big (\varepsilon \pm \widetilde{\gamma }h\big )\Big )du\\&=\frac{1}{\pi }\int _{0}^{\infty }\exp \!\bigg (c_{1}\cdot \frac{\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\cos \!\bigg (c_{2}\cdot \frac{\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}+\omega \cdot \frac{\varepsilon \!\pm \!\widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&=\frac{1}{\pi }\int _{0}^{\infty }\!\exp \!\bigg (c_{1}\!\cdot \!\frac{\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\!-\!\frac{\omega ^{2}}{2}\bigg )\cos \!\bigg (c_{2}\!\cdot \!\frac{\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\bigg )\cos \!\bigg (\omega \!\cdot \!\frac{\varepsilon \!\pm \!\widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\quad -\!\frac{1}{\pi }\!\int _{0}^{\infty }\!\exp \!\bigg (\!c_{1}\!\cdot \!\frac{\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\!-\!\frac{\omega ^{2}}{2}\bigg )\sin \!\bigg (\!c_{2}\!\cdot \!\frac{\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\bigg )\sin \!\bigg (\!\omega \!\cdot \!\frac{\varepsilon \!\pm \!\widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega . \end{aligned}$$
(39)
By expanding the Taylor series for \(\exp (c_{1}\sigma ^{-Y}|\omega |^{Y}h^{1-Y/2})\), as well as for \(\cos (c_{2}\sigma ^{-Y}\omega ^{Y}h^{1-Y/2})\) and \(\sin (c_{2}\sigma ^{-Y}\omega ^{Y}h^{1-Y/2})\), we deduce that
$$\begin{aligned}&\widetilde{\mathbb {E}}\left( \phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) \nonumber \\&\,\,\,\,=\frac{1}{\pi }\!\int _{0}^{\infty }\!\cos \!\bigg (\omega \!\cdot \!\frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )e^{-\omega ^{2}/2}d\omega \!+\!\frac{1}{\pi }\!\!\sum _{(k,j)\in \mathbb {Z}^{2}_{+}:\,(k,j)\ne (0,0)}\!\!(-1)^{j}a_{k,2j,0}^{\pm }(h)\!-\!\frac{1}{\pi }\sum _{k,j=0}^{\infty }(-1)^{j}d_{k,2j+1,0}^{\pm }(h) \nonumber \\&\,\,\,\,=\frac{1}{\sqrt{2\pi }}\exp \bigg (\!\!-\!\frac{(\varepsilon \pm \widetilde{\gamma }h)^{2}}{2\sigma ^{2}h}\bigg )+\frac{1}{\pi }\!\sum _{(k,j)\in \mathbb {Z}^{2}_{+}:\,(k,j)\ne (0,0)}\!(-1)^{j}a_{k,2j,0}^{\pm }(h)-\frac{1}{\pi }\sum _{k,j=0}^{\infty }(-1)^{j}d_{k,2j+1,0}^{\pm }(h), \end{aligned}$$
(40)
where, for \(m,n\in \mathbb {Z}_{+}\) and \(r\in \mathbb {R}_{+}\),
$$\begin{aligned} a_{m,n,r}^{\pm }(h)&:=\frac{c_{1}^{m}\,c_{2}^{n}\,h^{(m+n)(1-Y/2)}}{m!\,n!\,\sigma ^{(m+n)Y}}\int _{0}^{\infty }\omega ^{(m+n)Y+r}\cos \bigg (\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )e^{-\omega ^{2}/2}\,d\omega ,\end{aligned}$$
(41)
$$\begin{aligned} d_{m,n,r}^{\pm }(h)&:=\frac{c_{1}^{m}\,c_{2}^{n}\,h^{(m+n)(1-Y/2)}}{m!\,n!\,\sigma ^{(m+n)Y}}\int _{0}^{\infty }\omega ^{(m+n)Y+r}\sin \bigg (\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )e^{-\omega ^{2}/2}\,d\omega . \end{aligned}$$
(42)
By applying the formula for the integrals of \(\omega ^{kY}\cos (\beta \omega )\) and \(\omega ^{kY}\sin (\beta \omega )\) with respect to \(e^{-\omega ^{2}/2}\) on \(\mathbb {R}_{+}\), as well as the asymptotics for the Kummer’s function M(a, b, z), as \(h\rightarrow 0\), we deduce that
$$\begin{aligned} a_{m,n,r}^{\pm }(h)&=\frac{c_{1}^{m}\,c_{2}^{n}\,h^{(m+n)(1-Y/2)}}{m!\,n!\,\sigma ^{(m+n)Y}}\cdot 2^{((m+n)Y+r-1)/2}\,\Gamma \bigg (\frac{(m+n)Y+r+1}{2}\bigg )\\&\quad \,\cdot M\bigg (\frac{(m+n)Y+r+1}{2},\frac{1}{2},-\frac{\big (\varepsilon \pm \widetilde{\gamma }h\big )^{2}}{2\sigma ^{2}h}\bigg )\\&\sim \frac{c_{1}^{m}\,c_{2}^{n}\,h^{(m+n)(1-Y/2)}}{m!\,n!\,\sigma ^{(m+n)Y}}\cdot 2^{((m+n)Y+r-1)/2}\,\Gamma \bigg (\frac{(m+n)Y+r+1}{2}\bigg )\\&\quad \,\cdot \left( \frac{\Gamma (1/2)}{\Gamma \big (\!-((m+n)Y+r)/2\big )}\bigg (\frac{\varepsilon ^{2}}{2\sigma ^{2}h}\bigg )^{-((m+n)Y+r+1)/2}\right. \\&\qquad \,\,\left. +\frac{\Gamma (1/2)\,e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}}{\Gamma \big (((m+n)Y+r+1)/2\big )}\bigg (\frac{\varepsilon ^{2}}{2\sigma ^{2}h}\bigg )^{((m+n)Y+r)/2}\right) , \end{aligned}$$
and that
$$\begin{aligned} d_{m,n,r}^{\pm }(h)&=\frac{c_{1}^{m}\,c_{2}^{n}\,h^{(m+n)(1-Y/2)}}{m!\,n!\,\sigma ^{(m+n)Y}}\cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\cdot 2^{((m+n)Y+r)/2}\,\Gamma \bigg (\frac{(m+n)Y+r}{2}+1\bigg )\\&\quad \,\cdot M\bigg (\frac{(m+n)Y+r}{2}+1,\frac{3}{2},-\frac{\big (\varepsilon \pm \widetilde{\gamma }h\big )^{2}}{2\sigma ^{2}h}\bigg )\\&\sim \frac{c_{1}^{m}\,c_{2}^{n}\,h^{(m+n)(1-Y/2)}}{m!\,n!\,\sigma ^{(m+n)Y}}\cdot \frac{\varepsilon }{\sigma \sqrt{h}}\cdot 2^{((m+n)Y+r)/2}\,\Gamma \bigg (\frac{(m+n)Y+r}{2}+1\bigg )\\&\quad \,\cdot \left( \frac{\Gamma (3/2)}{\Gamma \big ((1-(m+n)Y-r)/2\big )}\cdot \bigg (\frac{\varepsilon ^{2}}{2\sigma ^{2}h}\bigg )^{-((m+n)Y+r)/2-1}\right. \\&\qquad \quad \left. +\frac{\Gamma (3/2)\,e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}}{\Gamma \big ((m+n)Y+r)/2+1\big )}\cdot \bigg (\frac{\varepsilon ^{2}}{2\sigma ^{2}h}\bigg )^{((m+n)Y+r-1)/2}\right) . \end{aligned}$$
In the asymptotic formulas for the Kummers function in the expression of \(a_{n,m,r}^{\pm }(h)\) above, the first term vanishes if \(\Gamma (-((m+n)Y+r)/2)\) are infinity. This happens when \(-((m+n)Y+r)/2\) is a nonpositive integer. Similarly, in the asymptotic formulas for the Kummers function in \(d_{m,n,r}^{\pm }(h)\), the first term vanishes if \((1-(m+n)Y-r)/2\) is a nonpositive integer. Hence, for \(m,n\in \mathbb {Z}_{+}\) and \(r\in \mathbb {R}_{+}\), as \(h\rightarrow 0\),
$$\begin{aligned} a_{m,n,r}^{\pm }(h)&=O\bigg (\frac{h^{m+n+(r+1)/2}}{\varepsilon ^{(m+n)Y+r+1}}\bigg )\!+O\bigg (\frac{\varepsilon ^{(m+n)Y+r}e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}}{h^{(m+n)(Y-1)+r/2}}\bigg )\!=O\bigg (\frac{h^{m+n+(r+1)/2}}{\varepsilon ^{(m+n)Y+r+1}}\bigg ), \end{aligned}$$
(43)
$$\begin{aligned} d_{m,n,r}^{\pm }(h)&=O\bigg (\frac{h^{m+n+(r+1)/2}}{\varepsilon ^{(m+n)Y+r+1}}\bigg )\!+O\bigg (\frac{\varepsilon ^{(m+n)Y+r}e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}}{h^{(m+n)(Y-1)+r/2}}\bigg )\!=O\bigg (\frac{h^{m+n+(r+1)/2}}{\varepsilon ^{(m+n)Y+r+1}}\bigg ). \end{aligned}$$
(44)
Therefore, by combining Eqs. (40), (43), and (44), we obtain that
$$\begin{aligned} \widetilde{\mathbb {E}}\left( \phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) =\frac{e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}}{\sqrt{2\pi }}+O\big (h^{3/2}\varepsilon ^{-1-Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(45)
Next, we note that
$$\begin{aligned} \widetilde{\mathbb {E}}\left( \mp J_{h}\,\phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) =\int _{\mathbb {R}}(\mathcal {F}\psi )(z)zp_{J,h}^{\mp }(z)\,dz=\int _{\mathbb {R}}\psi (u)\mathcal {F}\big (zp_{J,h}^{\mp }(z)\big )(u)\,du, \end{aligned}$$
where by Eq. (38),
$$\begin{aligned} \mathcal {F}\big (zp_{J,h}^{\pm }(z)\big )(u)=i\frac{d}{du}\big (\mathcal {F}p_{J,h}^{\pm }\big )(u)&=\frac{i}{\sqrt{2\pi }}\exp \Big (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)\mp iu\widetilde{\gamma }h\Big )\\&\quad \,\cdot \Big (c_{1}Y|u|^{Y-1}\,\text {sgn}(u)h+ic_{2}Y|u|^{Y-1}h\mp i\widetilde{\gamma }h\Big ). \end{aligned}$$
Together with Eq. (37), we obtain that
$$\begin{aligned}&\widetilde{\mathbb {E}}\left( \mp J_{h}\,\phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) \nonumber \\&\quad =\frac{ic_{1}\sigma Yh^{3/2}}{2\pi }\int _{\mathbb {R}}\,{\mathrm{sgn}}(u)|u|^{Y-1}\exp \bigg (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{\sigma ^{2}u^{2}h}{2}+iu\big (\varepsilon \pm \widetilde{\gamma }h\big )\bigg )du \nonumber \\&\qquad -\frac{c_{2}\sigma Yh^{3/2}}{2\pi }\int _{\mathbb {R}}\,|u|^{Y-1}\exp \bigg (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{\sigma ^{2}u^{2}h}{2}+iu\big (\varepsilon \pm \widetilde{\gamma }h\big )\bigg )du \nonumber \\&\qquad \mp \frac{\widetilde{\gamma }\sigma h^{3/2}}{2\pi }\int _{\mathbb {R}}\,\exp \bigg (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{\sigma ^{2}u^{2}h}{2}+iu\big (\varepsilon \pm \widetilde{\gamma }h\big )\bigg )du \end{aligned}$$
(46)
The asymptotics of the last term above follows from Eqs. (39) and (45), namely
$$\begin{aligned}&\frac{\widetilde{\gamma }\sigma h^{3/2}}{2\pi }\int _{\mathbb {R}}\,\exp \bigg (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{\sigma ^{2}u^{2}h}{2}+iu\big (\varepsilon \pm \widetilde{\gamma }h\big )\bigg )du \nonumber \\&\quad =O\Big (h\,e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )\!+\!O\big (h^{5/2}\varepsilon ^{-1-Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(47)
For the first term in Eq. (46), by expanding the Taylor series for \(\exp (c_{1}\sigma ^{-Y}|\omega |^{Y}h^{1-Y/2})\), as well as for \(\cos (c_{2}\sigma ^{-Y}\omega ^{Y}h^{1-Y/2})\) and \(\sin (c_{2}\sigma ^{-Y}\omega ^{Y}h^{1-Y/2})\) below, and using Eqs. (41) and (42), we have
$$\begin{aligned}&\frac{ic_{1}\sigma Yh^{3/2}}{2\pi }\int _{\mathbb {R}}\,{\mathrm{sgn}}(u)|u|^{Y-1}\exp \bigg (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{\sigma ^{2}u^{2}h}{2}+iu\big (\varepsilon \pm \widetilde{\gamma }h\big )\bigg )du \nonumber \\&\quad = -\frac{c_{1}\sigma Yh^{3/2}}{\pi }\int _{0}^{\infty }u^{Y-1}\exp \bigg (c_{1}u^{Y}h-\frac{\sigma ^{2}u^{2}h}{2}\bigg )\sin \Big (c_{2}u^{Y}h+u\big (\varepsilon \pm \widetilde{\gamma }h\big )\Big )du \nonumber \\&\quad = -\frac{c_{1}Yh^{(3-Y)/2}}{\sigma ^{Y-1}\pi }\int _{0}^{\infty }\omega ^{Y-1}\exp \bigg (\frac{c_{1}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\sin \bigg (\frac{c_{2}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}+\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\quad = -\frac{c_{1}Yh^{(3-Y)/2}}{\sigma ^{Y-1}\pi }\int _{0}^{\infty }\omega ^{Y-1}\exp \bigg (\frac{c_{1}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\sin \bigg (\frac{c_{2}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\bigg )\cos \bigg (\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\qquad -\frac{c_{1}Yh^{(3-Y)/2}}{\sigma ^{Y-1}\pi }\int _{0}^{\infty }\omega ^{Y-1}\exp \bigg (\frac{c_{1}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\cos \bigg (\frac{c_{2}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\bigg )\sin \bigg (\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\quad = -\frac{c_{1}Yh^{(3-Y)/2}}{\sigma ^{Y-1}\pi }\left( \sum _{k,j=0}^{\infty }(-1)^{j}a_{k,2j+1,Y-1}^{\pm }(h)+\sum _{k,j=0}^{\infty }(-1)^{j}d_{k,2j,Y-1}^{\pm }(h)\right) =O\big (h^{3/2}\varepsilon ^{-Y}\big ), \end{aligned}$$
(48)
as \(h\rightarrow 0\), where we have used the asymptotic formulas (43) and (44) in the last equality. Finally, for the second term in Eq. (46), again by expanding the Taylor series for \(\exp (c_{1}\sigma ^{-Y}|\omega |^{Y}h^{1-Y/2})\), \(\cos (c_{2}\sigma ^{-Y}\omega ^{Y}h^{1-Y/2})\), and \(\sin (c_{2}\sigma ^{-Y}\omega ^{Y}h^{1-Y/2})\) below, and using Eqs. (41), (42), (43), and (44), we deduce that
$$\begin{aligned}&\frac{c_{2}\sigma Yh^{3/2}}{2\pi }\int _{\mathbb {R}}\,|u|^{Y-1}\exp \bigg (c_{1}|u|^{Y}h+ic_{2}|u|^{Y}h\,\text {sgn}(u)-\frac{\sigma ^{2}u^{2}h}{2}+iu\big (\varepsilon \pm \widetilde{\gamma }h\big )\bigg )du\nonumber \\&\quad =\frac{c_{2}\sigma Yh^{3/2}}{\pi }\int _{0}^{\infty }u^{Y-1}\exp \bigg (c_{1}u^{Y}h-\frac{\sigma ^{2}u^{2}h}{2}\bigg )\cos \Big (c_{2}u^{Y}h+u\big (\varepsilon \pm \widetilde{\gamma }h\big )\Big )du \nonumber \\&\quad =\frac{c_{2}Yh^{(3-Y)/2}}{\pi \sigma ^{Y-1}}\int _{0}^{\infty }\omega ^{Y-1}\exp \bigg (\frac{c_{1}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\cos \bigg (\frac{c_{2}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}+\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\quad =\frac{c_{2}Yh^{(3-Y)/2}}{\pi \sigma ^{Y-1}}\int _{0}^{\infty }\omega ^{Y-1}\exp \bigg (\frac{c_{1}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\cos \bigg (\frac{c_{2}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\bigg )\cos \bigg (\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\qquad -\frac{c_{2}Yh^{(3-Y)/2}}{\pi \sigma ^{Y-1}}\int _{0}^{\infty }\omega ^{Y-1}\exp \bigg (\frac{c_{1}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}-\frac{\omega ^{2}}{2}\bigg )\sin \bigg (\frac{c_{2}\,\omega ^{Y}h^{1-Y/2}}{\sigma ^{Y}}\bigg )\sin \bigg (\omega \cdot \frac{\varepsilon \pm \widetilde{\gamma }h}{\sigma \sqrt{h}}\bigg )d\omega \nonumber \\&\quad =\frac{c_{2}Yh^{(3-Y)/2}}{\pi \sigma ^{Y-1}}\left( \sum _{k,j=0}^{\infty }(-1)^{j}a_{k,2j,Y-1}^{\pm }(h)-\sum _{k,j=0}^{\infty }(-1)^{j}d_{k,2j+1,Y-1}^{\pm }(h)\right) =O\big (h^{3/2}\varepsilon ^{-Y}\big ), \end{aligned}$$
(49)
as \(h\rightarrow 0\). Therefore, by combining Eqs. (46), (47), (48), and (49), we obtain that
$$\begin{aligned} \widetilde{\mathbb {E}}\left( \mp J_{h}\phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) =O\Big (he^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )+O\big (h^{3/2}\varepsilon ^{-Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(50)
It remains to analyze the asymptotic behavior of \(\mathbb {E}(\overline{\Phi }((\varepsilon \pm J_{h})/(\sigma \sqrt{h})))\). We first note that there exists a universal constant \(K>0\), such that
$$\begin{aligned}&\widetilde{\mathbb {E}}\left( \overline{\Phi }\bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\mathbf{1}_{\{\varepsilon \pm J_{h}\ge 0\}}\right) \le K\,\widetilde{\mathbb {E}}\left( \phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\mathbf{1}_{\{\varepsilon \pm J_{h}\ge 0\}}\right) \nonumber \\&\quad =O\left( \widetilde{\mathbb {E}}\left( \phi \bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) \right) =O\Big (e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )+O\big (h^{3/2}\varepsilon ^{-1-Y}\big ),\quad \text {as }\,h\rightarrow 0, \end{aligned}$$
(51)
where the last inequality above follows from Eq. (45). Moreover, by Eq. (11), as \(h\rightarrow 0\),
$$\begin{aligned}&\widetilde{\mathbb {E}} \ \left( \overline{\Phi }\bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\mathbf{1}_{\{\varepsilon \pm J_{h}\le 0\}}\!\right) =\int _{\mathbb {R}}\phi (u)\,\widetilde{\mathbb {P}}\Big (\varepsilon \pm J_{h}\le \sigma \sqrt{h}u,\,\varepsilon \pm J_{h}\le 0\Big )du \nonumber \\&\quad =\int _{0}^{\infty }\phi (u)\,\widetilde{\mathbb {P}}\Big (\pm Z_{h}\le -\varepsilon \mp \widetilde{\gamma }h\Big )du+\int _{-\infty }^{0}\phi (u)\,\widetilde{\mathbb {P}}\Big (\pm Z_{h}\le \sigma \sqrt{h}u-\varepsilon \mp \widetilde{\gamma }h\Big )du \nonumber \\&\quad =\frac{1}{2}\,\widetilde{\mathbb {P}}\bigg (\!\pm Z_{1}\le \frac{-\varepsilon \mp \widetilde{\gamma }h}{h^{1/Y}}\bigg )+\int _{-\infty }^{0}\phi (u)\,\widetilde{\mathbb {P}}\bigg (\pm Z_{1}\le \frac{\sigma \sqrt{h}u-\varepsilon \mp \widetilde{\gamma }h}{h^{1/Y}}\bigg )du \nonumber \\&\quad \le \frac{1}{2}\,\widetilde{\mathbb {P}}\bigg (\!\pm Z_{1}\le \frac{-\varepsilon \mp \widetilde{\gamma }h}{h^{1/Y}}\bigg )+\frac{\widetilde{K}h}{\varepsilon ^{Y}}\int _{-\infty }^{0}\phi (u)\bigg (1-\frac{\sigma \sqrt{h}\pm \widetilde{\gamma }h}{\varepsilon }u\bigg )^{-Y}du=O\big (h\varepsilon ^{-Y}\big ). \end{aligned}$$
(52)
Therefore, by combining Eqs. (51) and (52), we obtain that
$$\begin{aligned} \widetilde{\mathbb {E}}\left( \overline{\Phi }\bigg (\frac{\varepsilon }{\sigma \sqrt{h}}\pm \frac{J_{h}}{\sigma \sqrt{h}}\bigg )\right) =O\Big (e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )+O\big (h\varepsilon ^{-Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(53)
Finally, by combining Eqs. (35), (36), (45), (50), and (53), we conclude that
$$\begin{aligned} I_{1}(h)={\frac{\sqrt{2}}{\sigma \sqrt{\pi }}\cdot \frac{\varepsilon }{\sqrt{h}}\,e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}}+O\Big (e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )+O\big (h\varepsilon ^{-Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(54)
Step 1.2
We now study the asymptotic behavior of \(I_{2}(h)\), defined in Eq. (34), as \(h\rightarrow 0\). Let us first consider the following decomposition
$$\begin{aligned} I_{2}(h)&=\widetilde{\mathbb {E}}\Big (\Big (e^{-\widetilde{U}_{h}}-1+\widetilde{U}_{h}\Big )W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big )- \widetilde{\mathbb {E}}\Big (\widetilde{U}_{h}W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big ) \nonumber \\&=:I_{21}(h)-I_{22}(h). \end{aligned}$$
(55)
The first term \(I_{21}(h)\) can be bounded as follows: as \(h\rightarrow 0\),
$$\begin{aligned} 0\le I_{21}(h)\le \widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}}\!-1+\widetilde{U}_{h}\Big )=\exp \bigg (h\int _{\mathbb {R}_{0}}\!\Big (e^{-\phi (x)}\!-1+\phi (x)\Big )\widetilde{\nu }(dx)\bigg )-1=O(h). \end{aligned}$$
(56)
To deal with \(I_{22}\), for any \(t\in \mathbb {R}_{+}\), we further decompose \(\widetilde{U}_{t}\) as
$$\begin{aligned} \widetilde{U}_{t}=\int _{0}^{t}\int _{\mathbb {R}_{0}}\big (\varphi (x)+\alpha _{\text {sgn}(x)}x\big )\widetilde{N}(ds,dx)-\int _{0}^{t}\int _{\mathbb {R}_{0}}\alpha _{\text {sgn}(x)}x\widetilde{N}(ds,dx)=:\widetilde{U}^{\text {BV}}_{t}-\alpha _{+}{Z}_{t}^{+}-\alpha _{-}{Z}^{-}_{t}, \end{aligned}$$
where the first integral is well-defined in light of Assumption 1-(i) & (ii), so that
$$\begin{aligned} I_{22}(h)&=\widetilde{\mathbb {E}}\Big (\widetilde{U}^{\text {BV}}_{h}W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big )-\alpha _{+}\widetilde{\mathbb {E}}\Big (Z^{+}_{h}W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big ) \nonumber \\&\quad -\alpha _{-}\widetilde{\mathbb {E}}\Big (Z^{-}_{h}W_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|>\varepsilon \}}\Big )=:I_{22}^{\text {BV}}(h)-\alpha _{+}I_{22}^{+}(h)-\alpha _{-}I_{22}^{-}(h). \end{aligned}$$
For the first term \(I_{22}^{\text {BV}}(h)\), note that
$$\begin{aligned} \big |I_{22}^{\text {BV}}(h)\big |\le \widetilde{\mathbb {E}}\Big (\big |\widetilde{U}^{\text {BV}}_{h}\big |\Big )\le 2h\int _{\mathbb {R}_{0}}\big |\varphi (x)+\alpha _{\text {sgn}(x)}x\big |\,\widetilde{\nu }(dx), \end{aligned}$$
where the last integral is finite since in a neighborhood of the origin,
$$\begin{aligned} \big |\varphi (x)+\alpha _{\text {sgn}(x)}x\big |=\big |-\ln q(x)+\alpha _{\text {sgn}(x)}x\big |=O\Big (\big |1-q(x)+\alpha _{\text {sgn}(x)}x\big |\Big ), \end{aligned}$$
which is integrable with respect to \(\widetilde{\nu }(dx)\) in view of Assumption 1-(ii). As for the terms \(I_{22}^{\pm }\), due to the self-similarity of \(Z_{t}^{\pm }\) and the fact that \(\varepsilon h^{-1/Y}\rightarrow \infty\) (since \(Y\in (1,2)\)), the monotone convergence theorem implies that \(I_{22}^{\pm }(h)=o(h^{1/Y})\), as \(h\rightarrow 0\). Hence, we obtain that
$$\begin{aligned} I_{22}(h)=o\big (h^{1/Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(57)
By combining Eqs. (55), (56), and (57), we conclude that
$$\begin{aligned} I_{2}(h)=o\big (h^{1/Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(58)
Finally, from Eqs. (34), (54), and (58), we obtain that
$$\begin{aligned} \mathbb {E}\Big (\!W_{h}^{2}{} \mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\!\Big )\!=\!h\!-\!\frac{\sqrt{2h}\,\varepsilon }{\sigma \sqrt{\pi }}e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\!+\!O\Big (\!he^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\!\Big )\!+\!O\big (h^{2}\varepsilon ^{-Y}\big )\!+{\!o\big (h^{1+1/Y}\big )},\quad \end{aligned}$$
(59)
as \(h\rightarrow 0\), which completes the analysis in Step 1.
Step 2
In this step, we will study the asymptotic behavior of the second term in Eq. (33), as \(h\rightarrow 0\). By Eqs. (7), (8), and (9), we first have
$$\begin{aligned} \mathbb {E}\Big (J_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )&=\widetilde{\mathbb {E}} \ \Big (e^{-\widetilde{U}_{h}-\eta h}\,J_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big ) \nonumber \\&=e^{-\eta h}\,\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}}Z_{h}^{2}{} \mathbf{1}_{\{|W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )+2\widetilde{\gamma }he^{-\eta h}\,\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}}Z_{h}{} \mathbf{1}_{\{|W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big ) \nonumber \\&\quad +\widetilde{\gamma }^{2}h^{2}e^{-\eta h}\,\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}}\,\mathbf{1}_{\{|W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big ) \nonumber \\&=:e^{-\eta h}I_{3}(h)+2\widetilde{\gamma }he^{-\eta h}I_{4}(h)+\widetilde{\gamma }^{2}h^{2}e^{-\eta h}\,\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}}\,\mathbf{1}_{\{|W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big ). \end{aligned}$$
(60)
Clearly,
$$\begin{aligned} \widetilde{\gamma }^{2}h^{2}e^{-\eta h}\,\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}}\,\mathbf{1}_{\{|W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )=O\big (h^{2}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(61)
It remains to analyze the asymptotic behavior of the first two terms in Eq. (60).
Step 2.1
We begin with the analysis of \(I_{3}(h)\). Clearly,
$$\begin{aligned} I_{3}(h)&=\widetilde{\mathbb {E}}\Big (Z_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )+\widetilde{\mathbb {E}}\Big (\Big (e^{-\widetilde{U}_{h}}-1\Big )Z_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big ) \nonumber \\&=h^{2/Y}\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )+\widetilde{\mathbb {E}}\Big (\Big (e^{-\widetilde{U}_{h}}-1\Big )Z_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\Big ) \nonumber \\&=:h^{2/Y}I_{31}(h)+I_{32}(h). \end{aligned}$$
(62)
By the symmetry of \(W_{1}\), we note that
$$\begin{aligned} \widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{-\varepsilon \le \sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h\le 0\}}\Big )=\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}-h^{1/Y}\!Z_{1}-\widetilde{\gamma }h\le \varepsilon \}}\Big ). \end{aligned}$$
In what follows, we let \(h>0\) small enough so that \(\varepsilon -|\widetilde{\gamma }|h>0\).
To study the asymptotic behavior of \(I_{31}(h)\), as \(h\rightarrow 0\), let us first consider
$$\begin{aligned} E_{1}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\ge 0,\,\pm Z_{1}\ge 0\}}\Big )\!=\!\!\int _{0}^{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}\!\!\bigg (\!\int _{0}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!\!\!u^{2}p_{Z}(\pm u)du\!\bigg )\phi (x)dx \nonumber \\&\,\,=\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{2}p_{Z}(\pm u)\,du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \nonumber \\&\,\,=\frac{C_{\pm }\big (\varepsilon \mp \widetilde{\gamma }h\big )}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{1-Y}du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \nonumber \\&\,\,\quad +\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{2}\Big (p_{Z}(\pm u)-C_{\pm }u^{-1-Y}\Big )du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega . \end{aligned}$$
(63)
For the first term in Eq. (63), we have
$$\begin{aligned} \int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}\!u^{1-Y}du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega&=\frac{\big (\varepsilon \mp \widetilde{\gamma }h\big )^{2-Y}}{(2-Y)h^{(2-Y)/Y}}\int _{0}^{1}(1-\omega )^{2-Y}\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \nonumber \\&\sim \frac{\big (\varepsilon \mp \widetilde{\gamma }h\big )^{2-Y}}{2(2-Y)h^{(2-Y)/Y}}\cdot \frac{\sigma \sqrt{h}}{\varepsilon \mp \widetilde{\gamma }h},\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(64)
For the second term in Eq. (63), since \(Y\in (1,2)\), we first observe that, for any \(z>0\),
$$\begin{aligned} \int _{0}^{z}u^{2}\big (u^{-Y-1}\!\wedge u^{-2Y-1}\big )du=\frac{z^{2-Y}}{2\!-\!Y}{} \mathbf{1}_{(0,1]}(z)+\frac{1\!-\!z^{2-2Y}}{2(Y\!-\!1)}{} \mathbf{1}_{(1,\infty )}(z)\le \frac{Y}{2(Y\!-\!1)(2\!-\!Y)}. \end{aligned}$$
(65)
Hence, we deduce from Eq. (13) that
$$\begin{aligned}&\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{2}\big |p_{Z}(\pm u)-C_{\pm }u^{-1-Y}\big |du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \\&\quad \le \frac{\widetilde{K}Y}{2(Y-1)(2-Y)}\int _{0}^{1}\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega =O\big (\sqrt{h}\,\varepsilon ^{-1}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
Therefore, we obtain that
$$\begin{aligned} E_{1}^{\pm }(h)=\frac{C_{\pm }}{2(2-Y)}\,h^{1-2/Y}\varepsilon ^{2-Y}+O(1),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(66)
Using the same argument as above and since \(\varepsilon \gg h\), we also obtain that, when \(\pm \widetilde{\gamma }>0\), as \(h\rightarrow 0\),
$$\begin{aligned} E_{2}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\le 0,\,\pm Z_{1}\le 0\}}\Big ) \nonumber \\&\,\,=\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\le \mp \widetilde{\gamma }h,\,W_{1}\le 0,\,\pm Z_{1}\le 0\}}\Big )=O\big (h^{3-Y-2/Y}\big )+O(1). \end{aligned}$$
(67)
Next, we consider
$$\begin{aligned} E_{3}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\ge 0,\,\pm Z_{1}\le 0\}}\Big ) \nonumber \\&\,\,=\!\int _{0}^{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}\!\!\bigg (\!\int _{-\frac{\sigma \sqrt{h}x}{h^{1/Y}}}^{0}\!\!u^{2}p_{Z}(\pm u)du\!\bigg )\phi (x)dx\!+\!\!\int _{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}^{\infty }\!\!\bigg (\!\int _{-\frac{\sigma \sqrt{h}x}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!\!\!\!u^{2}p_{Z}(\pm u)du\!\bigg )\phi (x)dx.\quad \end{aligned}$$
(68)
By Eq. (12), the first term in Eq. (68) is such that
$$\begin{aligned}&\int _{0}^{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}\bigg (\int _{-\frac{\sigma \sqrt{h}x}{h^{1/Y}}}^{0}u^{2}p_{Z}(\pm u)\,du\bigg )\phi (x)\,dx\le \widetilde{K}\int _{0}^{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}\bigg (\int _{0}^{\frac{\sigma \sqrt{h}x}{h^{1/Y}}}u^{1-Y}du\bigg )\phi (x)\,dx\\&\quad =\frac{\widetilde{K}\sigma ^{2-Y}h^{2-Y/2-2/Y}}{2-Y}\int _{0}^{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}x^{2-Y}\phi (x)\,dx=O\big (h^{2-Y/2-2/Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
Similarly, the second term in Eq. (68) can be estimated as follows:
$$\begin{aligned}&\int _{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}^{\infty }\bigg (\int _{-\frac{\sigma \sqrt{h}x}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}u^{2}p_{Z}(\pm u)\,du\bigg )\phi (x)\,dx\le \widetilde{K}\int _{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}^{\infty }\bigg (\int _{\frac{\sigma \sqrt{h}x-(\varepsilon \mp \widetilde{\gamma }h)}{h^{1/Y}}}^{\frac{\sigma \sqrt{h}x}{h^{1/Y}}}u^{1-Y}du\bigg )\phi (x)\,dx\\&\quad \le \frac{\widetilde{K}\sigma ^{2-Y}h^{2-2/Y-Y/2}}{2-Y}\int _{\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}}^{\infty }x^{2-Y}\phi (x)\,dx=o\big (h^{2-2/Y-Y/2}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
Therefore, we obtain that
$$\begin{aligned} E_{3}^{\pm }(h)=O\big (h^{2-Y/2-2/Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(69)
To complete the analysis for \(I_{3}(h)\), it remains to study
$$\begin{aligned} E_{4}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\le 0,\,\pm Z_{1}\ge 0\}}\Big )=\!\int _{-\infty }^{0}\!\bigg (\!\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!\!u^{2}p_{Z}(\pm u)du\!\bigg )\phi (x)dx \nonumber \\&\,\,=C_{\pm }\int _{-\infty }^{0}\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}u^{1-Y}du\bigg )\phi (x)\,dx \nonumber \\&\quad \,\,+\int _{-\infty }^{0}\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}u^{2}\Big (p_{Z}(\pm u)-C_{\pm }u^{-1-Y}\Big )du\bigg )\phi (x)\,dx. \end{aligned}$$
(70)
For the first term in Eq. (70), we have
$$\begin{aligned}&C_{\pm }\int _{-\infty }^{0}\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}u^{1-Y}du\bigg ) \ \phi (x)\,dx \nonumber \\&\quad =\frac{C_{\pm }\,\varepsilon ^{2-Y}}{(2-Y)h^{2/Y-1}}\int _{-\infty }^{0}\left( \bigg (1-\frac{\sigma \sqrt{h}x\pm \widetilde{\gamma }h}{\varepsilon }\bigg )^{2-Y}\!-\bigg (\!-\frac{\sigma \sqrt{h}x\pm \widetilde{\gamma }h}{\varepsilon }\bigg )^{2-Y}\right) \ \phi (x)\,dx \nonumber \\&\quad \sim \frac{C_{\pm }\,h^{1-2/Y}\varepsilon ^{2-Y}}{2(2-Y)},\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(71)
For the second term in Eq. (70), we deduce from Eq. (65) that
$$\begin{aligned} \int _{-\infty }^{0}\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}u^{2}\Big |p_{Z}(\pm u)-C_{\pm }u^{-1-Y}\Big |du\bigg )\phi (x)\,dx=O(1),\quad h\rightarrow 0. \end{aligned}$$
Therefore, we obtain that
$$\begin{aligned} E_{4}^{\pm }(h)=\frac{C_{\pm }}{2(2-Y)}h^{1-2/Y}\varepsilon ^{2-Y}+O(1),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(72)
By combining Eqs. (66), (67), (69), and (72), we conclude that
$$\begin{aligned} I_{31}(h)=\sum _{i=1}^{4}\big (E_{i}^{+}(h)\!+\!E_{i}^{-}(h)\big )=\frac{C_{+}\!+\!C_{-}}{2-Y}h^{1-2/Y}\varepsilon ^{2-Y}+O\big (h^{2-Y/2-2/Y}\big ),\quad \text {as }\,h\rightarrow 0.\quad \end{aligned}$$
(73)
Next, we will study the asymptotic behavior of \(I_{32}(h)\), as \(h\rightarrow 0\). Clearly, by Cauchy-Schwarz inequality and self-similarity of \(Z_{h}\) and \(W_{h}\) under \(\widetilde{\mathbb {P}}\), we have that
$$\begin{aligned} I_{32}(h)\le h^{2/Y}\left( \widetilde{\mathbb {E}}\bigg (\!\Big (e^{-\widetilde{U}_{h}}-1\Big )^{2}\bigg )\right) ^{1/2}\bigg (\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )\bigg )^{1/2}. \end{aligned}$$
(74)
By Assumption 1-(v) and denoting \(\widetilde{C}_{\ell }=\int _{\mathbb {R}_{0}}\big (e^{-\ell \varphi (x)}-1+\ell \varphi (x)\big )\tilde{\nu }(dx)\), \(\ell =1,2\), we first have
$$\begin{aligned} \widetilde{\mathbb {E}}\bigg (\Big (e^{-\widetilde{U}_{h}}-1\Big )^{2}\bigg )=e^{\widetilde{C}_{2}h}-2e^{\widetilde{C}_{1}h}+1\sim \big (\widetilde{C}_{2}-2\widetilde{C}_{1}\big )h,\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(75)
The analysis of the asymptotic behavior, as \(h\rightarrow 0\), of the second factor in Eq. (74) is similar to that of \(I_{31}(h)\). More precisely, we first consider
$$\begin{aligned} F_{1}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\ge 0,\,\pm Z_{1}\ge 0\}}\Big )\\&\,\,=\frac{C_{\pm }\big (\varepsilon \mp \widetilde{\gamma }h\big )}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{3-Y}du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \\&\,\,\quad +\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{4}\Big (p_{Z}(\pm u)-C_{\pm }u^{-1-Y}\Big )du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega . \end{aligned}$$
A similar argument as in Eq. (64) shows that
$$\begin{aligned} \frac{C_{\pm }\big (\varepsilon \mp \widetilde{\gamma }h\big )}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{3-Y}du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \sim \frac{C_{\pm }}{2(4-Y)}h^{1-4/Y}\varepsilon ^{4-Y},\quad \text {as }\,h\rightarrow 0, \end{aligned}$$
and by Eq. (13),
$$\begin{aligned}&\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\int _{0}^{1}\bigg (\int _{0}^{\frac{(\varepsilon \mp \widetilde{\gamma }h)(1-\omega )}{h^{1/Y}}}u^{4}\big |p_{Z}(u)-Cu^{-1-Y}\big |du\bigg )\phi \bigg (\frac{\varepsilon \mp \widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega \\&\quad \le \frac{\varepsilon \!\mp \!\widetilde{\gamma }h}{\sigma \sqrt{h}}\cdot \frac{\widetilde{K}\big (\varepsilon \mp \widetilde{\gamma }h\big )^{4-2Y}}{2(2-Y)h^{(4-2Y)/Y}}\!\int _{0}^{1}(1-\omega )^{4-2Y}\phi \bigg (\frac{\varepsilon \!\mp \!\widetilde{\gamma }h}{\sigma \sqrt{h}}\omega \bigg )d\omega =O\big (h^{2-4/Y}\varepsilon ^{4-2Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
Hence, we obtain that
$$\begin{aligned} F_{1}^{\pm }(h)=\frac{C_{\pm }}{2(4-Y)}h^{1-4/Y}\varepsilon ^{4-Y}+O\big (h^{2-4/Y}\varepsilon ^{{4}-2Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(76)
Using the same argument as above and since \(\varepsilon \gg h\), we also obtain that, when \(\pm \widetilde{\gamma }>0\), as \(h\rightarrow 0\),
$$\begin{aligned} F_{2}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\le 0,\,\pm Z_{1}\le 0\}}\Big ) \nonumber \\&\,\,=\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\le \mp \widetilde{\gamma }h,\,W_{1}\ge 0,\,\pm Z_{1}\le 0\}}\Big )=O\big (h^{6-4/Y-2Y}\big ). \end{aligned}$$
(77)
Moreover, using arguments similar to those for \(E_{3}^{\pm }(h)\), we deduce that
$$\begin{aligned} F_{3}^{\pm }(h):=\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\ge 0,\,\pm Z_{1}\le 0\}}\Big )=O\big (h^{3-4/Y-Y/2}\big ),\quad \text {as }\,h\rightarrow 0.\quad \end{aligned}$$
(78)
Finally, we consider
$$\begin{aligned} F_{4}^{\pm }(h)&:=\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{0\le \sigma \sqrt{h}W_{1}\pm h^{1/Y}Z_{1}\pm \widetilde{\gamma }h\le \varepsilon ,\,W_{1}\le 0,\,\pm Z_{1}\ge 0\}}\Big )=\!\int _{-\infty }^{0}\!\bigg (\!\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!\!u^{4}p_{Z}(\pm u)du\!\bigg )\phi (x)dx\\&\,\,=C_{\pm }\!\int _{-\infty }^{0}\!\!\!\bigg (\!\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!\!\!u^{3-Y}\!du\!\bigg )\phi (x)dx\!+\!\!\int _{-\infty }^{0} \left( \!\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!\!\!u^{4}\bigg (\!p_{Z}(\pm u)\!-\!\frac{C_{\pm }}{u^{1+Y}}\!\bigg )du \right) \ \phi (x)dx. \end{aligned}$$
A similar argument as in Eq. (71) shows that
$$\begin{aligned} C_{\pm }\int _{-\infty }^{0}\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}u^{3-Y}du\bigg )\phi (x)\,dx\sim \frac{C_{\pm }}{2(4-Y)}h^{1-4/Y}\varepsilon ^{4-Y},\quad \text {as }\,h\rightarrow 0, \end{aligned}$$
and by Eq. (13),
$$\begin{aligned}&\int _{-\infty }^{0}\!\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!u^{4}\Big |p_{Z}(\pm u)-C_{\pm }u^{-1-Y}\Big |du\bigg )\phi (x)\,dx\le \widetilde{K}\int _{-\infty }^{0}\!\bigg (\int _{\frac{-\sigma \sqrt{h}x\mp \widetilde{\gamma }h}{h^{1/Y}}}^{\frac{\varepsilon \mp \widetilde{\gamma }h-\sigma \sqrt{h}x}{h^{1/Y}}}\!u^{3-2Y}du\bigg )\phi (x)\,dx\\&\quad =\frac{\widetilde{K}\varepsilon ^{4-2Y}}{2(2\!-\!Y)h^{4/Y-2}}\int _{-\infty }^{0}\!\left( \!\bigg (\!1\!-\!\frac{\sigma \sqrt{h}x\!\pm \!\widetilde{\gamma }h}{\varepsilon }\bigg )^{4-2Y}\!\!\!-\!\bigg (\!\!-\!\frac{\sigma \sqrt{h}x\!\pm \!\widetilde{\gamma }h}{\varepsilon }\bigg )^{4-2Y}\right) \ \phi (x)\,dx=O\bigg (\frac{\varepsilon ^{4-2Y}}{h^{4/Y-2}}\bigg ). \end{aligned}$$
Hence, we obtain that
$$\begin{aligned} F_{4}^{\pm }(h)=\frac{C_{\pm }}{2(4-Y)}h^{1-4/Y}\varepsilon ^{4-Y}+O\big (h^{2-4/Y}\varepsilon ^{4-2Y}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(79)
Combining Eqs. (76), (77), (78), and (79), leads to
$$\begin{aligned}&\widetilde{\mathbb {E}}\Big (Z_{1}^{4}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )=\sum _{i=1}^{4}\big (F_{i}^{+}(h)+F_{i}^{-}(h)\big ) \nonumber \\&\quad =\frac{{\big (C_{+}+C_{-}\big )}h^{1-4/Y}\varepsilon ^{4-Y}}{4-Y}+O\big (h^{2-4/Y}\varepsilon ^{4-2Y}\big )+O\big (h^{3-4/Y-Y/2}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(80)
Therefore, by combining Eqs. (74), (75), and (80), we have
$$\begin{aligned} I_{32}(h)=O\big (h\,\varepsilon ^{2-Y/2}\big )+O\big (h^{3/2}\varepsilon ^{2-Y}\big )+O\big (h^{2-Y/4}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(81)
Finally, by combining Eqs. (62), (73), and (81), we obtain that
$$\begin{aligned} I_{3}(h)=\frac{C_{+}+C_{-}}{2-Y}\,h\varepsilon ^{2-Y}+O\big (h\varepsilon ^{2-Y/2}\big )+O\big (h^{2-Y/2}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(82)
Step 2.2
In this step, we will investigate the asymptotic behavior of \(I_{4}(h)\), as \(h\rightarrow 0\). Note that
$$\begin{aligned} I_{4}(h)&=h^{1/Y}\widetilde{\mathbb {E}}\Big (Z_{1}{} \mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )+\widetilde{\mathbb {E}}\bigg (\Big (e^{-\widetilde{U}_{h}}\!-1\Big )Z_{h}{} \mathbf{1}_{\{|\sigma W_{h}+Z_{h}+\widetilde{\gamma }h|\le \varepsilon \}}\bigg )\\&\,\,\le h^{1/Y}\bigg (\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )\bigg )^{1/2}\left( 1+\left( \widetilde{\mathbb {E}}\bigg (\Big (e^{-\widetilde{U}_{h}}-1\Big )^{2}\bigg )\right) ^{1/2}\right) \\&\,\,=O\left( h^{1/Y}\bigg (\widetilde{\mathbb {E}}\Big (Z_{1}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+h^{1/Y}\!Z_{1}+\widetilde{\gamma }h|\le \varepsilon \}}\Big )\bigg )^{1/2}\right) ,\quad h\rightarrow 0, \end{aligned}$$
where the second inequality above follows from Cauchy-Schwarz inequality. Therefore, by Eqs. (62) and (73), we obtain that
$$\begin{aligned} I_{4}(h)=O\big (\sqrt{h}\,\varepsilon ^{1-Y/2}\big )+O\big (h^{1-Y/4}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(83)
Finally, by combining Eqs. (60), (61), (82), and (83), we conclude that
$$\begin{aligned} \mathbb {E}\Big (J_{h}^{2}\,\mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )=\frac{C_{+}+C_{-}}{2-Y}\,h\varepsilon ^{2-Y}+O\big (h\varepsilon ^{2-Y/2}\big )+O\big (h^{2-Y/2}\big ),\quad \text {as }\,h\rightarrow 0, \end{aligned}$$
(84)
which completes the analysis of Step 2.
Step 3
In this last step, we will study the asymptotic behavior of the third term in Eq. (33), as \(h\rightarrow 0\). By Eqs. (7) and (8), we first decompose it as
$$\begin{aligned}&\mathbb {E}\Big (W_{h}J_{h}{} \mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )=\widetilde{\mathbb {E}}\Big (e^{-\widetilde{U}_{h}-\eta h}\,W_{h}J_{h}{} \mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big ) \nonumber \\&\quad =\sqrt{h}\,e^{-\eta h}\,\widetilde{\mathbb {E}}\Big (W_{1}J_{h}{} \mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|\le \varepsilon \}}\Big )+\sqrt{h}\,e^{-\eta h}\,\widetilde{\mathbb {E}}\Big (\Big (e^{-\widetilde{U}_{h}}-1\Big )W_{1}J_{h}{} \mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|\le \varepsilon \}}\Big )\nonumber \\&\quad =:e^{-\eta h}\sqrt{h}\,I_{5}(h)+e^{-\eta h}\sqrt{h}\,I_{6}(h). \end{aligned}$$
(85)
For \(I_{5}(h)\), by conditioning on \(J_{h}\), and using the fact that, for any \(x_{1},x_{2}\in \mathbb {R}\) with \(x_{1}<x_{2}\),
$$\begin{aligned} \widetilde{\mathbb {E}}\Big (W_{1}{} \mathbf{1}_{\{W_{1}\in [x_{1},x_{2}]\}}\Big )=\phi (x_{1})-\phi (x_{2}), \end{aligned}$$
we obtain from Eq. (50) that, as \(h\rightarrow 0\),
$$\begin{aligned} I_{5}(h)=\widetilde{\mathbb {E}}\left( J_{h}\bigg (\phi \bigg (\frac{\varepsilon +J_{h}}{\sigma \sqrt{h}}\bigg )-\phi \bigg (\frac{\varepsilon -J_{h}}{\sigma \sqrt{h}}\bigg )\bigg )\right) =O\Big (h\,e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )+O\big (h^{3/2}\varepsilon ^{-Y}\big ). \end{aligned}$$
(86)
As for \(I_{6}(h)\), by Cauchy-Schwarz inequality, Eq. (9), (62), (73), and (75), we obtain that
$$\begin{aligned} \big |I_{6}(h)\big |&\le \bigg (\widetilde{\mathbb {E}}\bigg (\Big (e^{-\widetilde{U}_{h}}-1\Big )^{2}\bigg )\bigg )^{1/2}\bigg (\widetilde{\mathbb {E}}\Big (J_{h}^{2}\,\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|\le \varepsilon \}}\Big )\bigg )^{1/2} \nonumber \\&\le \bigg (\widetilde{\mathbb {E}}\bigg (\Big (e^{-\widetilde{U}_{h}}-1\Big )^{2}\bigg )\bigg )^{1/2}\bigg (\widetilde{\mathbb {E}}\Big (2\big (Z_{h}^{2}+\widetilde{\gamma }^{2}h^{2}\big )\mathbf{1}_{\{|\sigma \sqrt{h}W_{1}+J_{h}|\le \varepsilon \}}\Big )\bigg )^{1/2} \nonumber \\&=O\big (h\varepsilon ^{1-Y/2}\big ),\quad \text {as }\,h\rightarrow 0. \end{aligned}$$
(87)
Therefore, by combining Eqs. (85), (86), and (87), we obtain that, as \(h\rightarrow 0\),
$$\begin{aligned} \mathbb {E}\Big (W_{h}J_{h}{} \mathbf{1}_{\{|\sigma W_{h}+J_{h}|\le \varepsilon \}}\Big )&=O\Big (h^{3/2}e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )+O\big (h^{2}\varepsilon ^{-Y}\big )+O\big (h^{3/2}\varepsilon ^{1-Y/2}\big ), \end{aligned}$$
(88)
which completes the analysis in Step 3.
Finally, by combining Eqs. (33), (59), (84), and (88), we conclude that, as \(h\rightarrow 0\),
$$\begin{aligned} \mathbb {E}\big (b_{1}(\varepsilon )\big )\!=\!\sigma ^{2}h\!-\!{\frac{\sigma \varepsilon \sqrt{2h}}{\sqrt{\pi }}e^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\!+\!\frac{C_{+}\!\!+\!C_{-}}{2-Y}h\varepsilon ^{2-Y}}\!\!+\!O\Big (he^{-\varepsilon ^{2}/(2\sigma ^{2}h)}\Big )\!+\!O\big (h\varepsilon ^{2-Y/2}\big )\!+\!O\big (h^{2-Y/2}\big ), \end{aligned}$$
(89)
which completes the proof of the theorem.
