Therefore by (<ref>), (<ref>) and (<ref>), we conclude

\begin{equation}
\P \( \( S_n , V_n^2 \) \in \Omega_3 \) \le A R_x \[ 1 - \Phi( x + c) \] \Psi_x^* e^{ A_1 R_x }.  \label{Omega3}
\end{equation}

Consequently, the desired bound (<ref>) follows from (<ref>), (<ref>) and (<ref>). The proof is completed.

§.§ Proof of Proposition <ref>

The main idea is to apply the general Theorem <ref>. Recall the notation in (<ref>), based on the one-dependent structure, $ \{ (X_j, Y_j) \}_{j=1}^k $ is a sequence of independent random vectors. Further, the moment condition in Theorem <ref> indicates

\begin{align*}
\f 12 \Big( \f{\e  S_{n 1} ^2 }{ \e  V_{n 1}^2 }- 1 \Big)   =\rho_n+ O \big(\f{ a^2}{ n^{  \alpha  }} \big).
\end{align*}

\begin{equation}
\begin{split}
& \P \big( S_{n 1 }  \geq  x V_{n 1 } + d_1 n^{ - \alpha / 2  } x B_n  \big) \\
& =  \P \left (  \frac { S_{n 1 } } { ( \e  S_{ n 1 }^2 )^{ 1/2 } } > \frac {x  \big( 1 + O ( a^2 n^{ - \alpha  } )  \big) } { \sqrt{ 1 + 2 \rho_n} }   \frac {  V_{ n 1 } } {  ( \e    V_{ n 1}^2  )^{ 1/2 }  }  +  O \big( d_1 n^{ -\alpha /2   } x \big) \right) .
\end{split}
\label{normalized}
\end{equation}

We now focus on bounding the error terms involved in Theorem <ref>. Note that  $\{ \xi_i \}_{1\leq i \leq n}$ are one-dependent sequence of random variables, and Rosenthal's inequality yields for $2<p\le 4$,

\begin{align*}
\e |X_i|^p &= \e \Big[ \Big| \sum_{j\in H_i, odd} \xi_j + \sum_{j\in H_i, even} \xi_j   \Big|^p \Big] \\
& \leq 2^p \Big\{  \e \Big[ \Big| \sum_{j\in H_i, odd} \xi_j   \Big|^p  \Big]+  \e \Big[ \Big| \sum_{j\in H_i, even} \xi_j   \Big|^p  \Big]  \Big\}  \\
& \leq A \Bigg[ \Big( \sum_{j\in H_i, odd}\e  \xi_j^2  \Big)^{p/2} +\Big( \sum_{j\in H_i, even}\e  \xi_j^2  \Big)^{p/2}+  \sum_{j\in H_i}\e | \xi_j | ^p     \Bigg] \\
& \leq A  a_1^p n^{p \alpha /2 }.
\end{align*}

In  addition,

\begin{gather*}
\e |Y_i|^p \le \big( \e |Y_i|^4 \big)^{\f{p}{4}}    \le \Big[ k\sum_{j\in H_i} \e \xi_j^4  \Big]^{\f{p}{4}} \le  A  a_1^p n^{\f {p \alpha}{2} } , \quad
B_{n1}^2 := \sum_{j=1}^k \e Y_i^2 \ge  A  a_2^2 n , \\
\Big| \sum_{j = 1}^k \e  X_j Y_j^2   \Big| + \Big| \sum_{i = 1}^k \e  X_i^3  \Big|\le 7\sum_{i = 1}^k \sum_{j\in H_i} \big| \e \xi_j^3  \big|\leq Aa_1^3 n.
\end{gather*}

\begin{gather*}
L_{3, n} \le  A  a^3  n^{- \f {1-\alpha}{2}}, \quad
\de_x \le  A  a^4  \f {( 1 + x )^4}{ n^{ (1 - \alpha)} } , \quad
\Psi_{x}^{*} \leq \exp\big(A  a^3  x^3 n^{ - 1/2  } \big).
\end{gather*}

Thefore, applying Theorem <ref> to (<ref>) yields

\begin{align}
& \P \big( S_{n 1 }  \geq  x V_{n 1 } + d_1 n^{ - \alpha / 2  } x B_n  \big)   \label{st1} \\
&  =  \Big[ 1 - \Phi( \f{x}{ \sqrt{ 1 + 2 \rho_n } } ) \Big]   e^{ O_1  \sum r_{x, j}  } \Big( 1 + O_2 \big(  \frac { a^4 x^4  } { n^{ 1- \alpha } } +   \frac { a^2 x^2  } { n^{ \f{\alpha}{2} } } +\f{a^3x}{n^{\f{1-\alpha}{2}}} \big) \Big)  \nn
\end{align}

for $ x \in ( 2, a^{ - 1 }  \min \{ n^{ \alpha / 4 }, n^{ ( 1 - \alpha  ) / 4  } \} ) $, where

\begin{equation*}
r_{x,j} \leq \e \bigg[ \exp \bigg\{ \f {  (\sum_{i \in H_j} \xi_i)^2 } { \sum_{i \in H_j} (\xi_i^2 + c_0 \e \xi_i^2) } \bigg\} \IF \Big ( x \Big|\sum_{i \in H_j } \xi_i \Big| >  c(\rho)   a_2 n^{1/2} \Big) \bigg],
\end{equation*}

for some pre-assigned constant $c_0>0$ and constant $c(\rho)$. The derivation of the upper bound of $ r_{x, j} $ is rather complicated.  We exploit a technical lemma <ref> whose proof will be provided in Section <ref> to bound $ r_{x, j} $.

Let $ \{ X_i\}_{1\leq i\leq n}$ be a sequence of independent random variables with $ \e X_i = 0 $ and $ \e |X_i|^p < \infty $ for $ p > 2 $. Denote $ S_n = \sum_{i=1}^{n} X_i $, $ V_n^2 = \sum_{i=1}^{n} X_i^2 $ and $ B_n^2 = \e V_n^2 $. Assume $ b > 0 , A > 0  $. Then there exist positive constants $c_0$ and $ K $ depending on $ p $ and $ A $, such that

\begin{equation}
\e  \Big[  e^{ \f {A S_n^2} {V_n^2 + c_0 B_n^2}} \IF ( b |S_n| > 1)  \Big] \le K b^{p} \Big( \sum_{i=1}^{n} \e |X_i|^p + (\e S_n^2)^{\f p2} \Big) .  \label{rxi}
\end{equation}

Since $ \{\xi_i \}_{1 \leq i \leq n }$ are one-dependent and Lemma <ref> is built for independent random variables, we separate the odd-indexed terms and the even-indexed terms to apply Lemma <ref>. Specifically,

\begin{align}
r_{x,j} &\le \e \bigg[ \exp \Big\{ \f { 4 |\sum_{i \in H_j, odd} \xi_i |^2} {\sum_{i \in H_j, odd}(\xi_i^2 + c_0\e \xi_i^2) } \Big \} \IF \Big ( \Big|\sum_{i \in H_j,odd } \xi_i \Big| > \f {c(\rho ) a_2 n^{1/2}}{ 2 x } \Big  ) \bigg]  \label{separation} \\
& \quad + \e \bigg[ \exp \Big\{ \f { 4 |\sum_{i \in H_j, even} \xi_i |^2} {\sum_{i \in H_j, even}(\xi_i^2 + c_0\e \xi_i^2) }  \Big\} \IF \Big ( \Big|\sum_{i \in H_j, even} \xi_i \Big| > \f {c(\rho ) a_2 n^{1/2}}{ 2 x } \Big ) \bigg] \nn\\
& \le C(\rho) \f { x^4 } { a_2^4  n^2 } \Big[ \sum_{i \in H_j} \e |\xi_i|^4 + ( \sum_{i \in H_j, odd} \e \xi_i^2  )^2 + ( \sum_{i \in H_j, even} \e \xi_i^2  )^2 \Big] \nn\\
& \le C(\rho)  \f{a^4 x^4 }{ n^{2-2 \alpha} }  , \nn
\end{align}

where the second inequality is derived by applying Lemma <ref>. Therefore, $ r_x = \sum_{j=1}^k r_{x,j} \leq  C(\rho) a^4 x^4 / n^{1 - \alpha} $, which together with (<ref>) yields the desired result (<ref>).

In the same manner as the proof of (<ref>), using Chebyshev's inequality and Taylor expansion, it is easy to obtain (<ref>). To avoid redundancy, here we omit the proof.

§.§ Proof of Proposition <ref>

We start with the error bound (<ref>). Denote $\omega_j=\hat{\xi}_{j(l+1)}-\e \hat{\xi}_{j(l+1)}$ for $1\leq j\leq k$. Therefore, $\{ \omega_j\}_{1\leq j\leq k}$ are independent random variables with $\omega_j \leq 2\tau$ and $\e\omega_j=0$ for each $j$. Recalling the definitions in (<ref>) and $ \tau = B_{n 2 } / x $.
For $ d_1 = \kappa_{\rho} a^2 $ with $\kappa_{\rho} \geq 10$, routine calculation shows

\begin{align*}
\sum_{j=1}^{k} \e \omega_j^2 &\leq B_{n2}^2 \leq a_1^2 n^{1-\alpha} , \quad d_1 n^{ - \alpha /2  } x B_n -\sum_{j=1}^{k}\e \hat{\xi}_{j(l+1)}  \geq (\kappa_{\rho} - 1) a_1 x n^{\frac{1-\alpha}{2}} .
\end{align*}

Hence, it follows from Bernstein inequality (see for example, Theorem A in [12]) that

\begin{align*}
\P \big(\hatS_{n2} > d_1 n^{ - \alpha /2  } x B_n \big)
%		  \\
%		 =&   \P \Big( \sum_{j=1}^{k}\omega_j \geq d_1 n^{ - \alpha /2  }  x B_n - \sum_{j=1}^{k}\e \hat{\xi}_{j(l+1)} \Big) \\
&\leq   \P\Big( \sum_{j=1}^{k}\omega_j \geq (\kappa_{\rho}-1)a_1xn^{(1-\alpha)/2} \Big) \\
\leq \exp\Big( &-\frac{\kappa_{\rho}^2a_1^2 x^2 n^{1-\alpha}} { 3 B_{n2}^2+3\tau\kappa_{\rho} a_1 x n^{\f{(1-\alpha)}{2}} }  \Big)  \leq  \exp\Big( -\frac{\kappa_{\rho} x^2 } { 6  }   \Big).
\end{align*}

As for inequality (<ref>), similarly,

\begin{align*}
&\P \Big(\hatS_{n2} < - d_1 n^{ - \alpha /2  } x B_n + x (\hatV_n - V_{n1}), ~V_{n1}^2 > \f14 B_n^2 \Big) \\
= & \P \Big(\sum_{j=1}^{k} \hat{\xi}_{j(l+1)} < - d_1 n^{ - \alpha /2  } x B_n + x\frac{\sum_{j=1}^{k} \hat{\xi}_{j(l+1)}^2}{\hat{V}_n+V_{n1}}, V_{n1}^2 > \frac{1}{4} B_n^2 \Big) \\
\leq & \P \Big( \sum_{j=1}^{k} \big( \gamma_j - \e \gamma_j \big) > d_1 n^{ - \alpha /2  }  x B_n- \sum_{j=1}^{k} \e \gamma_j \Big) ,
\end{align*}

where $\gamma_j=\frac{x}{B_n}\hat{\xi}_{j(l+1)}^2 - \hat{\xi}_{j(l+1)}$, $1\leq j\leq k$, are independent random variables with $\gamma_j-\e \gamma_j \leq 4\tau$ for each $j$. Note that

\begin{align*}
\sum_{j=1}^{k} \e (\gamma_j-\e \gamma_j)^2 &\leq \sum_{j=1}^{k} \e \gamma_j^2 \leq  2\sum_{j=1}^{k} \Big(\frac{x^2\tau^2}{B_n^2} +1\Big)\e \hat{\xi}_{j(l+1)}^2\leq 4a_1^2 n^{1-\alpha} , \\
\Big |\sum_{j=1}^{k} \e \gamma_j \Big|
&= \Big| \frac{x}{B_n} \sum_{j=1}^{k} \e \hat{\xi}_{j(l+1)}^2 +\sum_{j=1}^{k} \e \xi_{j(l+1)} \IF (|\xi_{j(l+1)}|>\tau )\Big| \\
%&\leq \sum_{j=1}^{k} \big|\frac{x}{B_n}\e \hat{\xi}_{j(l+1)}^2 + \frac{1}{\tau} \e \xi_{j(l+1)}^2 I {\{ |\xi_{j(l+1)}|>\tau \}} \big| \\
&\leq \frac{1}{\tau} \sum_{j=1}^{k} \e \xi_{j(l+1)}^2 = xB_{n2}.
\end{align*}

Again, Bernstein inequality leads to

\begin{align*}
\P \Big(\hatS_{n2} < - d_1 n^{ - \alpha /2  } x B_n + x (\hatV_n - V_{n1}), ~V_{n1}^2 > \f14 B_n^2\Big)   \leq \exp \Big(   -  \frac  { \kappa_{\rho} x^2  }{  14 }     \Big).
\end{align*}

The proof is completed.

§.§ Proof of Proposition <ref>

Observe that for any $  a,b, \delta_2\in\mathbbm{R}$ satisfying $3b\geq a^2$ and $\delta_2\geq 0$, we have for arbitrary $x^2\geq 3$,

\begin{align*}
x\sqrt{b+\delta_2} &\geq \sqrt{x^2\Big(  \f{a^2}{3}+\delta_2    \Big)}=\sqrt{(x^2-3) \f{a^2}{3}+(x^2-3)\delta_2+ a^2+3\delta_2}\\
&\geq \sqrt{(x^2-3)\delta_2+ a^2+ 2a\sqrt{\delta_2(x^2-3)}}\geq a+\sqrt{(x^2-3)\delta_2}.
\end{align*}

Denote $H= \{ j ( \ell + 1  ): 1\leq j \leq k  \}$, the index set of small block terms. We set

\begin{align*}
&a:=\hat{\xi}_{j(\ell+1)}+\xi_{j(\ell+1)-1}+\xi_{j(\ell+1)+1},\quad \hat{S}_n^{(j)}:=\hat{S}_n-a,\\
&b:=\hat{\xi}_{j(\ell+1)}^2+\xi_{j(\ell+1)-1}^2+\xi_{j(\ell+1)+1}^2,\quad \delta_2:= \big(\hat{V}_n^{(j)}\big)^2 = \hat{V}_n^2-b,
\end{align*}

then $\{ \hat{S}_n>x\hat{V}_n  \}\subset \{  \hat{S}_n^{(j)} > (\sqrt{x^2-3}) \hat{V}_n^{(j)}        \}$ by the above analysis. Moreover, $ \hat{S}_n^{(j)} $ and $ \hat{V}_n^{(j)} $ are independent of $\xi_{j (l +1 )}$ since $ \{\xi_i, 1 \leq i \leq n\}  $ are one-dependent random variables. Thus it follows from (<ref>) that for $  x \in (c_{\rho} \sqrt{\log n}, d_0 a^{-1} n^{1/8} )$,

\begin{align}
&\P \Big( \hatS_n \geq x \hatV_n, \max_{1 \le j \le k} |\xi_{j(l+1)}| > \tau \Big)  \label{trun-error1} \\
& \le \sum_{i \in H} \P \Big( \hatS_n^{ (i ) } \geq \sqrt{x^2-3} \hatV_n^{( i )}  \Big) \P \big(|\xi_{ i }| > \tau\big) \nn \\
& \le A   \f{\sum_{ i \in H }  \e|\xi_{  i  } |^4}{\tau^4}  \Bigg[ 1 - \Phi\Big(\f{\sqrt{x^2-3}}{ \sqrt{ 1 + 2 \rho_n^{ (i) } } } \Big) \Bigg]   \Big( 1 + O ( \frac {a ^4  x^2  } { n^{ 1/4 } }  ) \Big) \nn \\
& \le  A  a^4 \f{(1+x)^4} {n^{1/2}}\Bigg[ 1 - \Phi\Big( \f{\sqrt{x^2-3}}{ \sqrt{ 1 + 2 \rho_n } } \Big) \Bigg]  \Big( 1 + O ( \frac {a ^4  x^2  } { n^{ 1/4 } }  ) \Big) \nn \\
& \le A e^{3/2(1-2\rho)} a^4 \f{(1+x)^4} {n^{1/2}}\Bigg[ 1 - \Phi\Big( \f{x}{ \sqrt{ 1 + 2 \rho_n } } \Big) \Bigg]  \Big( 1 + O ( \frac {a ^4  x^2  } { n^{ 1/4 } }  ) \Big). \nn
\end{align}

where $\rho_{ n }^{(i)}$ is defined parallel to $\rho_n$ and

\begin{align*}
\rho_{ n }^{(i)}&:=  \frac 1 2 \Big(  \frac { \e [ ( \hat{S}_n^{(i)})^2 ]   } { \e [ ( \hat{V}_n^{(i)} )^2 ] }  - 1 \Big) = \rho_n \Big(1+O\big( \f{ a^2}{n} \big)  \Big)
\end{align*}

because $ \e \xi_i^4 \leq a_1^4  $ and $ \e \xi_i^2 \geq a_2^2$ for $ i \geq 1 $. Then (<ref>) is derived.

Regarding the proof of (<ref>), a more technical iterative argument will be exploited. We have

\begin{align}
&\P \Big( S_n \geq x V_n, \max_{1 \le j \le k} |\xi_{j(l+1)}| > \tau \Big)  \nn\\
& \le \sum_{i_1 \in H} \P \Big(S_n^{(i_1)} \geq \sqrt{x^2-3} V_n^{(i_1)}\Big) \P \big(|\xi_{i_1}| > \tau\big) \nn\\
& \le \sum_{i_1 \in H} A \f{a^4 (1+x)^4} {n} \P \Big(S_n^{(i_1)} \geq \sqrt{x^2-3} V_n^{(i_1)}\Big) \nn \\
& \leq \sum_{i_1\in H} A \f{a^4 (1+x)^4} {n} \Big[ \P \Big(\hat{S}_n^{(i_1)} \geq \sqrt{x^2-3} \hat{V}_n^{(i_1)}\Big) \nn\\
&\qquad \qquad \qquad +\P \Big(S_n^{(i_1)} \geq \sqrt{x^2-3} V_n^{(i_1)} , \max_{j\in H\setminus \{ i_1\}} |\xi_{j(l+1)}| > \tau \Big) \Big] \nn \\
& \leq \sum_{i_1\in H} A \f{a^4 (1+x)^4} {n} \Big[ \P \Big(\hat{S}_n^{(i_1)} \geq \sqrt{x^2-3} \hat{V}_n^{(i_1)}\Big) \nn\\
&\qquad \qquad \qquad +\sum_{ i_2\in H \setminus \{i_1 \} }\P \Big(S_n^{(i_1,i_2)} \geq \sqrt{x^2-6} V_n^{(i_1,i_2)} \Big) \P \big(|\xi_{ i_2} | > \tau \Big) \Big] \nn \\
& \leq  \sum_{i_1\in H} A \f{a^4 (1+x)^4} {n} \P \Big(\hat{S}_n^{(i_1)} \geq \sqrt{x^2-3} \hat{V}_n^{(i_1)}\Big) \nn\\
& \qquad +\sum_{i_1\in H} \sum_{ i_2\in H \setminus \{i_1 \} } \Big( A \f{a^4 (1+x)^4} {n} \Big)^2  \P \Big(S_n^{(i_1,i_2)} \geq \sqrt{x^2-6} V_n^{(i_1,i_2)} \Big). \label{trun-error2}
\end{align}

Put $ u = [\f{x^2}{6}] $ so that $\sqrt{x^2-3u}  \approx  x / \sqrt 2$ and we repeat the above procedure iteratively up to $ u $ times,

\begin{align*}
&\P \Big( S_n \geq x V_n, \max_{1 \le j \le k} |\xi_{j(l+1)}| > \tau \Big)  \nn\\
& \leq  \sum_{j=1}^{u-1} \sum_{i_1 \in H  } \cdots \sum_{i_j \in H  }  \Big( A \f{a^4 (1+x)^4} {n} \Big)^j \P \Big(\hatS_n^{(i_1,\ldots,i_j)} \geq \sqrt{x^2-3j} \hatV_n^{(i_1,\ldots,i_j)} \Big)  \nn\\
&~ \quad +\sum_{i_1 \in H  } \cdots \sum_{i_u  \in H  } \Big( A \f{a^4 (1+x)^4} {n}\Big)^u \P \Big(S_n^{ (  i_1,\ldots,i_u ) } \geq \sqrt{x^2-3u} V_n^{ ( i_1,\ldots,i_u ) }\Big).
\end{align*}

As we have chosen $ \alpha = 1/2 $ and $|H|=k\asymp n^{1-\alpha}=\sqrt{n}$, it follows by (<ref>) that for arbitrary $i_1,\cdots,i_j\in H$, and $1\leq j\leq x^2/6$, we have for $ x \in (c_{\rho } \sqrt{\log n } , d_0 a^{- 2} n^{1/8}) $,

\begin{align*}
& \sum_{i_1 \in H  } \cdots \sum_{i_j \in H  }   \Big( A \f{a^4 (1+x)^4} {n} \Big)^j \P \Big(\hatS_n^{(i_1,\ldots,i_j)} \geq \sqrt{x^2-3j} \hatV_n^{(i_1,\ldots,i_j)} \Big) \\
& \leq  \Big( A \f{a^4 (1+x)^4} {n^{1/2}} \Big)^j  \bigg[ 1-\Phi\bigg(  \f{\sqrt{x^2-3j}}{\sqrt{1+2\rho_{ n }^{(i_1,\cdots,i_j)}}}  \bigg) \bigg]     \Big( 1+O\Big(  \f{a^4 x^2}{n^{1/4}}   \Big)  \Big) \\
& \leq   A \Big( A \f{a^4 (1+x)^4} {n^{1/2}} \Big)^{j} e^{3j/(2+4\rho_{ n })} \bigg[ 1-\Phi\Big(  \f{x}{\sqrt{1+2\rho_{ n }}}  \Big) \bigg]     \Big( 1+O\Big(  \f{a^4 x^2}{n^{1/4}}   \Big)  \Big) \\
& \leq   A \f{a^4(1+x)^4} {n^{1/2}} \Big( A \f{a^4 (1+x)^4} {n^{1/2}}e^{3/(2+4\rho_{ n })} \Big)^{j-1}  \bigg[ 1-\Phi\Big(  \f{x}{\sqrt{1+2\rho_{ n }}}  \Big) \bigg]    \\
& \leq   A \f{a^4(1+x)^4} {n^{1/2}}\cdot \big( \f{1}{2} \big)^{j-1} \bigg[ 1-\Phi\Big(  \f{x}{\sqrt{1+2\rho_{ n }}}  \Big) \bigg]
\end{align*}

for sufficiently large $n$ and some constant $A''$ depending on $a,\rho$, where $\rho_{ n }^{(i_1,\cdots,i_j)}$ is defined as

\begin{align*}
\rho_{ n }^{(i_1,\cdots,i_j)} : =  \frac 1 2 \Big(  \frac { \e [ (\hat{S}_n^{( i_1, \ldots, i_j )})^2  ]  } { \e [ ( \hat{V}_n^{(i_1, \ldots, i_j )} )^2 ] }  - 1 \Big) = \rho_n \Big(1+O\big( \f{ a^2x^2}{n} \big)  \Big)
\end{align*}

Therefore, by combining the upper bounds above we conclude

\begin{align*}
&\P \Big( S_n \geq x V_n, \max_{1 \le j \le k} |\xi_{j(l+1)}| > \tau \Big)  \\
& \le  \sum_{j=1}^{u-1} A\f{a^4(1+x)^4} {n^{1/2}}\cdot \big( \f{1}{2} \big)^{j-1} \bigg[ 1-\Phi\Big(  \f{x}{\sqrt{1+2\rho_{ n }}}  \Big) \bigg] + \Big(A \f{ a^ 4 (1+x)^4}{n^{1/2}}\Big)^{u}  \nn \\
& \le  A_1 \f{a^4 x^4}{n^{1/2}}\bigg[ 1-\Phi\Big(  \f{x}{\sqrt{1+2\rho_{ n }}}  \Big) \bigg]\Big( 1 + \Big(A \f{ a^ 4 (1+x)^4}{n^{1/2}} e^{3/(1+2\rho_n)} \Big)^{x^2/3} \Big)\nn \\
& \le A_2  \f{a^4 x^4}{n^{1/2}}\bigg[ 1-\Phi\Big(  \f{x}{\sqrt{1+2\rho_{ n }}}  \Big) \bigg]  \nn
\end{align*}

for $ x \in (c_{\rho } \sqrt{\log n } , d_0 a^{- 2} n^{1/8}) $. The proof is completed.

§.§ Proof of Proposition <ref>

We apply Theorem <ref> to estimate $ \P ( S_{k,1} \ge x V_{k,1} - \vp x B_n ) $. As a preparation, we first calculate the relevant moments. By Lemma <ref>, there exist some positive constants $ A_1 $,$ A_2 $ depending on $ a_1, a_2 $ and $ \tau $ such that for $ 1 \leq u \leq k_1  $,

\begin{equation}
\e \xi_u^4 +\e \eta_u^4 \le A_1 n^{2 (\alpha +\alpha_1) } \mu_1^4,\quad  \e |\xi_u |^3 +\e |\eta_u |^3 \le A_2 n^{\f32 (\alpha +\alpha_1 ) } \mu_1^3. \label{4-m}
\end{equation}

In addition, by Lemma <ref> and some routine calculations we can obtain

\begin{align}
|\e \xi_u^3|  \le  A_3 n^{(\alpha + \alpha_1)} c_1^3 , \quad  |\e \xi_u \eta_u^2|  \le  A_3 n^{(\alpha + \alpha_1)} c_1^3,  \label{3-mm}
\end{align}

for a positive constant $ A_3 $ depending on $ a_1, a_2 $ and $ \tau $.
To apply the moderate deviation theorem for general self-normalized sums, it remains to estimate the error term

\begin{equation*}
r_{x, u } :=  \e \bigg[ \exp \Big( \f{  \xi_u ^2 } { \eta_u ^2 + c_0 \e \eta_u ^2 } \Big) \IF \Big( \f{1 + x} {B_{n,1}} |\xi_u| > 1 \Big) \bigg] .
\end{equation*}

We split the block $ I_u  $ into odd-indexed terms and even-indexed terms to construct a weakly dependent structure so that Lemma <ref> can be applied. To be specific, let  $ \mathcal{J}_1 = \{  j  : j~ \mbox{ is odd} \} $ and $ \mathcal{J}_2 = \{  j  : j ~ \mbox{is even} \} $.

Correspondingly, we define for $ t = 1, 2 $

\begin{equation*}
\mathcal{G}_t :=\sum_{j\in I_u \cap \mathcal{J}_t} Y_j, \quad \mathcal{K}_t^2 := \sum_{j\in I_u \cap \mathcal{J}_t} \big( Y_j^2+c_0 \e Y_j^2\big).
\end{equation*}

In the same manner as (<ref>), we have

\begin{align}
r_{x, u }
%	 &=   \int_{0}^{\infty} e^t \; \P \Big( \f { |\xi_u| } { \sqrt{ \eta_u^2  + c_0 \e \eta_u^2 } } > \sqrt{t}, x'|\xi_u| > B_{n,1} \Big) \; d t            \label{rxl}  \\
%	 &=  \sum_{s=0}^v \int_{0}^{\infty} e^t \; \P \Bigg( \f { |\sum_{s=0}^v p_s| } { \sqrt{ \sum_{s=0}^v q_s^2 } } > \sqrt{t}, \, \f{x'|\xi_u|}{B_{n,1}} > 1,\, | p_s |\geq \f{1}{v}|\xi_u| \Bigg) \; d t            \nn  \\
%	 &\leq  \sum_{s=0}^v \int_{0}^{\infty} e^t \; \P \Bigg( \f { |p_s| } { \sqrt{ \sum_{s=0}^v q_s^2 } } > \f{\sqrt{t}}{v}, \, \f{x'|p_s|}{B_{n,1}} > \f{1}{v} \Bigg) \; d t            \nn  \\
%	 &\leq  \sum_{s=0}^v \int_{0}^{\infty} e^t \; \P \Bigg( \f { |p_s| } { q_s  } > \f{\sqrt{t}}{v}, \, \f{x'|p_s|}{B_{n,1}} > \f{1}{v} \Bigg) \; d t            \nn  \\
&\le   \sum_{ t = 1}^{ 2 } \e \Big[ \exp \Big( \f{  4  \mathcal{G}_t^2 } { \mathcal{K}_t^2 } \Big) \IF \Big( \f{ (1 + x)| \mathcal{G}_t |} {B_{n,1}}  >  \f{1}{v} \Big) \Big] :  = \sum_{ t = 1}^{ 2 } r_{ x, u, t }. \label{rxl}
\end{align}

As $\{ r_{x, u, t}, t = 1, 2 \} $ share the same bound, we only present the analysis of $ r_{x, u ,1} $. Suppose the cardinality of $ I_{u}\cap \mathcal{J}_1 $ is $  [n^{\alpha_1}  / 2] $ for simplicity. By Lemma <ref>, $ \{Y_j, j \in I_u \cap \mathcal{J}_1 \} $ can be approximated by a sequence of independent random variables $ \{Y_j^*, 1 \le j \le  [n^{\alpha_1}  /2] \} $. Denote

\begin{equation*}
\mathcal{G}_1^*=\sum_{j=1}^{[n^{\alpha_1} / 2 ]} Y_j^*, \quad  (\mathcal{K}_1^{*} )^2 =  \sum_{j=1}^{[n^{\alpha_1} / 2]} \big( (Y_j^*)^2+c_0 \e (Y_j^*)^2\big).
\end{equation*}

It follows from Lemmas <ref> and <ref> that

\begin{align}
r_{x,u,1} &\le \e \Big[ \exp  \Big( \f{ 4 ( \mathcal{G}_1^* )^2 } {(\mathcal{K}_1^{*})^2 } \Big) \IF \Big(\f{ (1 + x )| \mathcal{G}_1^*  | } {B_{n,1}}  > \f {1} {2}  \Big) \Big] \label{rxl_1} \\
& \quad + \e \Big[ \exp \Big( \f{  4 \mathcal{G}_1^2 } {\mathcal{K}_1^2 } \Big) \IF \Big( \f{ (1 + x )| \mathcal{G}_1|} {B_{n,1}}  >  \f{1}{2} \Big) \IF \big( \, \exists j: Y_j \neq Y_j^*  \big) \Big] \nn \\
&\le \e \Big[ \exp  \Big( \f{ 4 ( \mathcal{G}_1^* )^2 } {(\mathcal{K}_1^{*})^2 } \Big) \IF \Big(\f{ (1 + x) | \mathcal{G}_1^*  | } {B_{n,1}}  > \f12  \Big) \Big] +  \f{a_1 n^{ \alpha_1 } }{2} e^{ 4 n^{ \alpha_1 }- a_2  n^{\alpha  \tau}  } \nn \\
& \le \f{ C(1 + x)^4 }{ n^{2(1-\alpha-\alpha_1)} } +  \f{a_1 n^{ \alpha_1 } }{ 2 } e^{ 4 n^{ \alpha_1 }- a_2 n^{ \alpha \tau} } , \nn
\end{align}

where  $ \alpha_1 $ should be chosen such that $  \alpha_1 \leq \alpha \tau $. Therefore, by Theorem <ref> and plugging in the bounds given in (<ref>), (<ref>), (<ref>) and (<ref>), we can obtain there exist a constant  $ c_0 $ depending on  $d_1,  \mu_1/\mu_2, a_1, a_2, \alpha $ and $\tau $ such that when $\epsilon=d_1 n^{-\alpha_1/2}\log n$,

\begin{align}
&\P \big( S_{k,1} \ge x V_{k,1} \pm \vp x B_n\big) \big/ \big[1- \Phi(x) \big]    \\
=& 1 + O\bigg(\f{(1 + x)^4}{ n^{1- \alpha- \alpha_1} }  + \f{1 + x }{n^{(1 - \alpha - \alpha_1 )/2 }} + \f{( 1 + x )^2 }{ n^{\alpha }  } + \f{ ( 1 + x )^2 \log n } {n^{ \alpha_1/2 }} \bigg)  \nn
%& \P( S_{k,1} \ge x V_{k,1} + \vp x B_n) \nn \\
%&=  [1- \Phi(x)] \big(1 + O(1) (\f{(1 + x)^4}{ n^{1- \alpha- \alpha_1} } + \f{(1 + x )^3}{\sqrt{n}} + \f{1 + x }{n^{(1 - \alpha - \alpha_1 )/2 }} + \f{( 1 + x )^2 }{ n^{\alpha }  } + \f{ ( 1 + x )^2 } {n^{ \alpha_1/2 }} ) \big)   \label{main2}
\end{align}

uniformly for $ x \in (0, c_0 \min \{ n^{( 1 - \alpha - \alpha_1 )/4 }, n^{\alpha \tau/2}, n^{\alpha/2}, (\log n)^{-1/2}n^{\alpha_1/4}\} )$. The proof of Proposition <ref> is completed.

§.§ Proof of Proposition <ref>

Note that $ \lim_{\ell \rightarrow - \infty} \e ( X_{i} \vert \mathscr{F}_{\ell} ) $ belongs to the tail $ \sigma $-field, hence $  \lim_{\ell \rightarrow - \infty} \e ( X_{i} \vert \mathscr{F}_{\ell} ) = \e (  X_{i} ) = 0 $ by the zero-one law. Therefore, we may write $X_i$ as

\begin{equation*}
X_i = \sum_{u=0}^{\infty} \mathcal{P}_{i-u} X_i , \qquad  {\rm where}~\ \mathcal{P}_{ i - u } ( \cdot ) =\e(\cdot|\mathscr{F}_{ i - u })-\e(\cdot|\mathscr{F}_{ i - u -1}).
\end{equation*}

Observe that,

\begin{equation*}
\parallel Y_j \parallel_4 = \Big \Vert  \sum_{u =0}^{\infty}\sum_{ i \in H_j}\mathcal{P}_{i -u }X_i  \Big \Vert_4 \leq \sum_{ u =0}^{\infty}  \Big\Vert \sum_{ i \in H_j}\mathcal{P}_{ i -u } X_i  \Big \Vert_4.
\end{equation*}

For fixed $ u \geq 0 $, denote $Z_i=\mathcal{P}_{ i- u }X_i $ and $ \mathscr{A}_i   =\mathscr{F}_{ i-u }$. It is obvious that $ Z_i \in \mathscr{A}_i $ and $ \mathscr{A}_{i - 1 } \subset  \mathscr{A}_i  $ for any $ i \ge 1   $.

\begin{align*}
\e ( Z_i |\mathscr{A}_{ i - 1 } ) &  =   \e ( X_i | \mathscr{A}_{i - 1 } )  -   \e ( X_i | \mathscr{A}_{i - 1 } )    = 0,
\end{align*}

thus for any fixed $ u \geq 0  $, $\left\{ ( Z_i ,\mathscr{A}_i ), i \geq 1  \right\} $ is a martingale difference sequence. It follows from Burkholder's martingale inequality that

\begin{align*}
\e \Big[  \Big|  \sum_{ i \in H_j  }  Z_i  \Big|^4  \Big]  \leq C  \e \Big[ \Big|  \sum_{ i \in H_j  } Z_i^2  \Big|^{  2  }  \Big]  \leq C m     \sum_{ i \in H_j } \e ( Z_i^4 ).
\end{align*}

In addition, by conditional Jensen's inequality and recalling the definition in (<ref>), we obtain

\begin{align*}
\e  ( Z_i^4  ) &  =  \e \Big[    \Big| \e\big( X_i  \big|\mathscr{F}_{ i - u } \big)-\e \big( X_ i \big|\mathscr{F}_{ i - u -1} \big)  \Big| ^4 \Big] \\
& =  \e\Big\{ \Big|  \e\big[ X_i - G_i \big( \ldots,\vp_{ i - u - 1 },\epsilon_{  i  - u }^{\ast},\vp_{ i - u +1},\ldots,\vp_{i }\big) \big| \mathscr{F}_{ i - u } \big] \Big|^4  \Big\} \\
& \leq    \e \big[  \big( X_i - G_i \big( \ldots,\vp_{ i - u - 1 },\epsilon_{  i  - u }^{\ast},\vp_{ i - u +1},\ldots,\vp_{i }\big)  \big)^4    \big ]   = \big [ \theta_4 (u) \big]^4 .
\end{align*}

Thus, we obtain by condition (<ref>) and Remark <ref> that

\begin{align*}
\parallel Y_j \parallel_4  &  \leq \sum_{ u =0}^{\infty}  \Big\Vert \sum_{ i \in H_j} Z_i \Big \Vert_4 \leq C  \sum_{ u = 0 }^{\infty} m^{ 1/ 2 }  a_1' e^{ - a_2' u^{ \tau  } }  \leq C_1 m^{ 1/2 }
\end{align*}

for $ C_1  $ being a constant depending on $ a_1, a_2 $ and $ \tau $. Further, it follows from Minkowski's inequality and (<ref>) that for sufficiently large $n$,

\begin{gather}
\parallel \tilde{Y}_j \parallel_4  \leq  \parallel Y_j \parallel_4 + \parallel \tilde{Y}_j-Y_j \parallel_4 \leq C_1 m^{1/2}+ m a_1 e^{-a_2m^{\tau}} \leq \tilde{C}_1 m^{1/2} ,  \label{dd1}\\
\parallel \tilde{Y}_j \parallel_2  \geq \parallel Y_j \parallel_2 - \parallel \tilde{Y}_j-Y_j \parallel_2 \geq c_1 m^{1/2} - ma_1e^{-a_2m^{\tau}} \geq \tilde{C}_2 m^{1/2} , \label{dd2}
\end{gather}

where $\tilde{C}_1$ and $\tilde{C}_2=\omega_1-n^{\f{\alpha}{2}}a_1e^{-a_2n^{\alpha\tau}}> \omega_1/2 $ are constants depending on $\tau, \alpha, \omega_1,  a_1$ and $a_2$.

Denote $\rho_k =  \big( \sum_{j =1}^{k-1}\e( \tilde{Y}_j \tilde{Y}_{j +1} ) \big) \big/ \big( \sum_{j =1}^{k}\e (  \tilde{Y}_j ^2 )   \big) $. Note that

\begin{equation*}
\tilde{C}_2^2\cdot n =\tilde{C}_2^2\cdot  km \leq \sum_{i=1}^{k}\e(\tilde{Y}_i^2) \leq \tilde{C}_1^2 \cdot km = \tilde{C}_1^2 \cdot n,
\end{equation*}

\begin{align*}
& \bigg|\sum_{ j =1}^{k-1}\e \Big(\tilde{Y}_j \tilde{Y}_{ j +1}\Big) -\sum_{ j =1}^{k-1}\e\Big(Y_j Y_{j +1}\Big)  \bigg|\\
& \hspace{0.8cm} \leq  \sum_{ j =1}^{k-1}\bigg|\e\Big[Y_ j \Big(\tilde{Y}_{ j +1}-Y_{ j +1}\Big)\Big] +\e\Big[Y_{ j +1}\Big(\tilde{Y}_{ j }-Y_{ j }\Big)\Big]  \\
&\hspace{3cm}  +\e\Big[\Big(\tilde{Y}_j -Y_j \Big)\Big(\tilde{Y}_{j +1}-Y_{j +1}\Big) \Big] \bigg| \\
& \hspace{0.8cm} \leq \sum_{j =1}^{k-1}   \Big( \big\Vert Y_j \big\Vert_2 \cdot  \big\Vert \tilde{Y}_{j +1}-Y_{ j +1}\big\Vert_2 + \big\Vert Y_{ j +1}\big\Vert_2 \cdot  \big\Vert \tilde{Y}_{ j }-Y_{j }\big\Vert_2  \\
& \hspace{3cm} +  \big\Vert \tilde{Y}_j -Y_j \big\Vert_2 \cdot  \big\Vert \tilde{Y}_{j +1}-Y_{j +1}\big\Vert_2   \Big)  \\
& \hspace{0.8cm} \leq  k \left( 2 \tilde{C}_1 m^{1/2} \cdot ma_1e^{-a_2m^{\tau}} + m^2a_1^2e^{-2a_2m^{\tau}}   \right).
\end{align*}

Immediately we have

\begin{equation}
\bigg|\sum_{i=1}^{k-1}\e (\tilde{Y}_i\tilde{Y}_{i+1}  ) -\sum_{i=1}^{k-1}\e ( Y_iY_{i+1} )  \bigg| \bigg/ \bigg( \sum_{i=1}^{k}\e ( \tilde{Y}_i^2 )  \bigg)\leq C_3  e^{-C_4  a_2m^{\tau}} , \label{dif-1}
\end{equation}

with constants $C_3$ and $C_4$ depending on $\tau, \alpha, \omega_1,  a_1$ and $a_2$. Now, for further investigation into the ratio of $ \sum_{ j = 1 }^{ k - 1 } \e ( Y_j Y_{j + 1} ) $ against $ \sum_{j =1 }^{ k - 1 } \e   (\tilde{Y}_i^2 ) $, we make use of the martingale structure again by representing $Y_{j}$ and $Y_{j+1}$ as

\begin{equation*}
Y_j = \sum_{ i \in H_j }\sum_{u =0}^{\infty}\mathcal{P}_{i - u } X_i , \qquad  Y_{j+1}=\sum_{ i \in H_{j+1}}\sum_{ u =0}^{\infty}\mathcal{P}_{ i - u } X_i .
\end{equation*}

\begin{equation*}
\e(Y_jY_{j+1})= \sum_{ i_1 \in H_j } \sum_{ i_2 \in H_{j+1} } \sum_{u_1=0}^{ \infty}\sum_{ u_2=0}^{ \infty} \e \Big[ \Big( \mathcal{P}_{ i_1 - u_1 } X_{i_1} \Big) \cdot \Big( \mathcal{P}_{i_2- u_2 } X_{i_2} \Big)  \Big].
\end{equation*}

Note that $\mathcal{P}_{i_2-u_2}X_{i_2} \in \mathscr{F}_{i_2-u_2}$ and for any $\ell_1 < i_1 - u_1$, $\ell_2 < i_2-u_2$,

\begin{equation*}
\e\Big[ \mathcal{P}_{i_1-u_1}X_{i_1}\big|\mathscr{F}_{\ell_1}  \Big] = 0 , \quad \e\Big[ \mathcal{P}_{i_2-u_2}X_{i_2}\big|\mathscr{F}_{\ell_2}  \Big] = 0.
\end{equation*}

Therefore, for $ i_1- u_1\neq i_2- u_2$, we have $ \e \left[ \left( \mathcal{P}_{ i_1 - u_1 } X_{i_1} \right) \cdot \left( \mathcal{P}_{i_2- u_2 } X_{i_2} \right)  \right] = 0 $. Consequently,

\begin{align*}
\big|\e(Y_jY_{j+1}) \big| & = \bigg| \sum_{ i_1\in H_j } \sum_{ i_2\in H_{j+1} } \sum_{u_1=0}^{ \infty}  \e \Big[ \Big( \mathcal{P}_{i_1-u_1}X_{i_1} \Big)\cdot \Big( \mathcal{P}_{i_1-u_1} X_{i_2} \Big)  \Big] \bigg| \\
&\leq \sum_{i_1\in H_j}\sum_{i_2\in H_{j+1}} \sum_{u_1=0}^{+\infty} 4a_1'^2\cdot \exp\Big[ -a_2' \Big(u_1^{\tau}+(i_2-i_1+u_1)^{\tau} \Big)  \Big],
\end{align*}

where we have used the fact that $ \Vert   \mathcal{P}_{ i_1 - u_1 } X_{i_1} \Vert_2  \leq \theta_4 ( u_1 ) $ and $ \Vert   \mathcal{P}_{ i_1 - u_1 } X_{i_2} \Vert_2  \leq \theta_4 ( i_2 - i_1 + u_1 ) $.

It is trivial to see from the integrability of the function $ e^{ - a x^{\tau} } $ that

\begin{align}
&\sum_{ i_1\in H_j}\sum_{ i_2\in H_{j+1}} \sum_{u_1=0}^{ \infty} 4 a_1'^2\cdot \exp\Big[ -a_2' \Big(u_1^{\tau}+(i_2-i_1+u_1)^{\tau} \Big)  \Big] \label{dif-2} \\
\leq& \sum_{ u =0}^{ \infty}  \sum_{ \ell =1 }^{2 m}  \ell \cdot  \exp\Big[ - a_2' \Big( u ^{\tau}+( \ell + u )^{\tau} \Big) \Big]  \nn   \\
\leq&  \left( \sum_{ \ell =1}^{ 2 m }  \ell  e^{-a_2' \ell^{\tau}} \right)  \left(\sum_{ u =0}^{\infty} e^{-  a_2'  u ^{\tau}} \right) \leq C_5  ,    \nn
\end{align}

for some constant $ C_5 $ depending on $  \alpha, a_1, a_2 $ and $ \tau $. Finally, as a result of (<ref>), (<ref>) and (<ref>), we obtain

\begin{equation}
| \rho_k |:= \bigg|\sum_{i=1}^{k-1}\e (\tilde{Y}_i\tilde{Y}_{i+1}  )  \bigg| \bigg/ \bigg( \sum_{i=1}^{k}\e ( \tilde{Y}_i^2 )  \bigg) \leq  C_6 m^{ -1  }  \label{rho-k}
\end{equation}

for large enough $n$, where $C_6$ is a constant depending on $\tau, \alpha, \omega_1,   a_1 $ and $a_2$.

Now we are ready to apply Theorem <ref> to the self-normalized sum

\begin{equation*}
\tilde{T}_k=\frac{\sum_{j=1}^{k}\tilde{Y}_j}{ ( \sum_{ j = 1 }^k \tilde{Y}_j^2 )^{ 1/2 } }.
\end{equation*}

Notice that $ \{ \tilde{Y}_j\}_{ 1 \leq j \leq k } $ are one-dependent random variables and the bounds in (<ref>), (<ref>) and (<ref>), i.e., the conditions in Theorem <ref>, are satisfied. Recall that $ m = [ n^{ \alpha  } ] $ and $ k = [ n^{ 1 - \alpha }  ] $. We obtain by (<ref>) that there is a constant $ d_0 $ depending on $ \tau, \alpha, \omega_1, a_1 $ and $ a_2  $, such that

\begin{align*}
\P ( \tilde{T}_k \geq x ) & =  \left[  1 - \Phi \left( \frac { x  } { \sqrt{ 1 + O ( m^{ -1  } ) } } \right) \right] \Big( 1 + O \Big( \frac { 1 + x^2  } { n^{ ( 1 - \alpha  ) / 4 } }   \Big)  \Big)   \\
& = \Big[ 1 - \Phi (x) \Big] \Big( 1 + O \Big( \frac { 1 + x^2  }  { n^{ ( 1 - \alpha )/ 4  } } +  \frac { 1 +  x^2  } { n^{ \alpha } } \Big)  \Big),  \nn
\end{align*}

uniformly for $ x \in (0, d_0 \min \{ n^{ ( 1 - \alpha  )/ 8 }, n^{ \alpha / 2 }  \} ) $. This completes the proof of Proposition <ref>.

§.§ Proof of Proposition <ref>

For simplicity, we denote $ V_k^2 = \sum_{j = 1}^k Y_j^2  $ and $ \tilde{V}_k^2 = \sum_{j  = 1}^k \tilde{Y}_j^2  $. Recall the constant $\tilde{C}_2 $ in the lower bound shown in (<ref>). Set

\begin{align*}
\mathcal{B} = \left\{  \max_{ 1 \leq j  \leq k } \big| Y_i-\tilde{Y}_i  \big|\leq \f{1}{n^2}, ~\tilde{V}_k^2\geq \f{\tilde{C}_2^2}{2}\cdot n  \right\}.
\end{align*}

Within the set $ \mathcal{B} $, it holds that

\begin{align*}
| V_k  - \tilde{V}_k |  & \leq  \tilde{V}_k^{ -1 } | V_k^2 - \tilde{V}_k^2  | \leq  n^{ -2  } \tilde{V}_k^{ -1 }   \sum_{ j  = 1 }^k ( 2 | \tilde{Y}_j  | + n^{ -2  } ) \nn \\
& \leq    2 n^{ - 2  } k^{ 1/2 }  + \sqrt 2  \tilde{C}_2^{ -1 }   n^{ - 7 / 2 - \alpha  } \leq  ( 2 \tilde{C}_2^{ -1  } + 2  ) n^{ - \f 32 - \f {\alpha} { 2 } },
\end{align*}

hence when $ n  $ is sufficiently large,

\begin{align*}
\big| T_k - \tilde{T}_k   \big| & = \frac { \Big| \Big( \sum_{ j = 1 }^k Y_j  \Big) \tilde{V_k} -  \Big( \sum_{ j = 1 }^k \tilde{Y}_j \Big) V_k \Big|  } { V_k \tilde{V}_k } \nn \\
& \leq  \frac {   \sum_{ j = 1 }^k \big| Y_j  - \tilde{Y}_j  \big|  }  { V_k  } + \frac { \Big( \sum_{ j = 1 }^k \big| \tilde{Y}_j \big| \Big)   } {  \tilde{V}_k }\cdot \frac{\big| V_k - \tilde{V}_k \big| }{V_k} \nn \\
& \leq C \Big(  n^{ - 1/2 } k n^{  -2 } + k^{ 1/2 } n^{ - 1/2 } n^{ - \f 32 - \f { \alpha } { 2 } } \Big) \nn \\
& \leq  2 C n^{ - \f 32 - \alpha } \leq  2 C n^ { - 1 }.
\end{align*}

\begin{align}
\P \big( T_k \geq x  \big)  &  \leq \P \Big(  T_k \geq x , ~ \big| T_k - \tilde{T}_k  \big| \leq 2 C  n^{ - 1  }  \Big) + \P ( \mathcal{B}^c ) \label{upper}  \\
&  \leq \P \Big(  \tilde{T}_k \geq x - 2 C  n^{ -1  }  \Big)  + \P ( \mathcal{B}^c  ),\nn
\end{align}

and similarly,

\begin{align}
\P \big( T_k \geq x  \big) & \geq  \P \big( T_k \geq x,  ~ \mathcal{B} \big)\geq \P \Big( \tilde{T}_k \geq x + 2 C  n^{ -1 } , ~ \mathcal{B}  \Big) \label{lower} \\
& \geq \P \Big( \tilde{T}_k \geq x + 2 C n^{ -1 }   \Big) - \P ( \mathcal{B}^c  ).    \nn
\end{align}

As for the bound of  $ \P ( \mathcal{B}^c  ) $, notice that

\begin{align*}
\P ( \mathcal{B}^c  )  &\leq  \sum_{ j = 1 }^k  \P \big(  \big| Y_j - \tilde{Y}_j  \big| >   n^{  - 2   }  \big)  +  \P \big( \tilde{V }_k^2 \leq \tilde{C}_2^2 n / 2   \big).
%	&\leq  \sum_{ j = 1 }^k  \P \Big(  \big| Y_j - \tilde{Y}_j  \big| >   n^{  - 2   }  \Big)  \\
%	&\qquad \;\;+  \P\bigg( \sum_{ \ell =1}^{k/2} Y_{2 \ell-1}^2 \leq \f{\tilde{C}_2^2}{4} n \bigg)+\P\bigg( \sum_{ \ell =1}^{k/2} Y_{2 \ell}^2 \leq \f{\tilde{C}_2^2}{4} n \bigg) .
\end{align*}

By Chebyshev's inequality and (<ref>) with $r=2$, we obtain

\begin{align*}
\sum_{ j = 1 }^k  \P \big(  \big| Y_j - \tilde{Y}_j  \big| >   n^{  - 2   }  \big)  &  \leq  n^{ 1 - \alpha } n^{ 4 + 2 \alpha  }  a_1^2 e^{ - 2a_2 n^{ \alpha \tau  } } \leq C e^{ - a_2 n^{ \alpha \tau  }   }.
\end{align*}

In addition,

\begin{align*}
\P ( \tilde{ V }_k^2 \leq \tilde{C}_2^2 n / 2   ) & \leq  \P \Big( \sum_{ \ell = 1  }^{ k / 2  } \tilde{Y}_{ 2 \ell - 1  }^2  <  \tilde{C}_2^2 n / 4  \Big)  +   \P \Big( \sum_{ \ell = 1  }^{ k / 2  } \tilde{Y}_{ 2 \ell   }^2  <  \tilde{C}_2 ^2 n / 4  \Big) ,
\end{align*}

where, without loss of generality, we assumed that $ k /2  $ is an integer.
Denoting $ W_{\ell} = - \tilde{Y}_{2 \ell -1}^2+\e(\tilde{Y}_{2 \ell -1}^2)$, $1\leq \ell \leq k/2$,  (<ref>) suggests that

\begin{gather*}
W_{\ell} \leq  \e ( \tilde{Y}_{ 2 \ell - 1 }^2 ) \leq \tilde{C}_1^2 m ,\quad \quad \e ( W_{ \ell } )=0 ,     \\
{\rm and}\quad \sum_{ \ell = 1 }^{k/2} \e ( W_{\ell}^2 )=\sum_{i=1}^{k/2}\var(\tilde{Y}_{2i-1}^2)  \leq \sum_{\ell=1}^{k/2}\e(\tilde{Y}_{ 2 \ell - 1 }^4) \leq  \tilde{C}_1^4  k m^2 / 2  .
\end{gather*}

Observe that $ \{  Y_{ 2 \ell - 1  }, 1 \leq \ell \leq k / 2  \} $ is a sequence of independent random variables. It follows from Bernstein's inequality and (<ref>) that

\begin{align*}
\P\Big( \sum_{ \ell =1}^{k/2} &Y_{2 \ell-1}^2 \leq \f{\tilde{C}_2^2}{4} n \Big)\\
&= \P\Big( \sum_{\ell=1}^{k/2} W_{\ell} \geq -\f{\tilde{C}_2^2}{4}  n + \sum_{i=1}^{k/2} \e \big(\tilde{Y}_{2i-1}^2 \big) \Big) \leq \P\Big( \sum_{\ell=1}^{k/2}W_{\ell} \geq \f{\tilde{C}_2^2}{4} n \Big) \\
&\leq \exp \Big\{  -\f{ \big((\tilde{C}_2^2 n)/4\big)^2 } { 2 \big(  \tilde{C}_1^2 m \cdot \tilde{C}_2^2 n / 4  +  \tilde{C}_1^4 k m ^2 / 2   \big) }  \Big\}  \leq \exp \{ - C n^{1-\alpha} \},
\end{align*}

for some constant $C $ depends on $\tau, \alpha, \omega_1,   a_1$ and $a_2$. By the same token,

\begin{equation*}
\P\Big( \sum_{ \ell =1}^{k/2}Y_{2 \ell}^2\leq \f{\tilde{C}_2^2}{4} n \Big) \leq  \exp \{ - C n^{1-\alpha} \}.
\end{equation*}

Overall, we obtain

\begin{align}
\P ( \mathcal{B}^c ) \leq C e^{  - a_2 n^{ \alpha \tau  }   } +   2e^{ - C  n^{1-\alpha}}.
\end{align}

The proof of Proposition <ref> is completed.

§.§ Proof of Theorem <ref>

Define $ f (x) =  x \mathbbm{1} \{ | x | \leq \tau  \}  + \tau \mathbbm{1} \{ x > \tau \} - \tau \mathbbm{1} \{ x < - \tau \}$.
In view of the expression shown in (<ref>) and the notations presented above Theorem <ref>, the main idea is to apply Theorem <ref> and  estimate the corresponding bias-corrected term $ \Psi_x^* $ and error terms $ L_{3, n} $ and $ R_x $ under fourth moment, which are defined by

\begin{align*}
L_{3, n} & =    \frac { \e | f (Y) - \tilde{\mu} |^3 } {  \sqrt n  \sigma_1^3 }  + \frac { \e | f (Y) -  {\mu} |^3 } {  \sqrt n  \sigma_2^3 }    \\
\Psi_{x}^* & = \exp \Big\{  ( \frac {\sigma_2} {\sigma_1} )^3  x^3 n^{ - 1/2}  \Big ( \frac  43 \gamma^3     \frac { \e ( f (Y) - \tilde{\mu} )^3 } {     \sigma_1^3  } \\
&  \hspace{4cm} -  2  \gamma^2    \frac { \e [ (  f (Y) - \tilde{\mu} ) ( f (Y) -  {\mu} )^2 ] } {  \sigma_1   \sigma_2^2  } \Big )\Big \} \\
R_x & =  \frac { (1 + x)^4  } {  n} \Big(   \frac { \e | f (Y) - \tilde{\mu} |^4 } {    \sigma_1^4 }  + \frac { \e | f (Y) -  {\mu} |^4 } {   \sigma_2^4 }   \Big)  \\
& \quad + n  \e \bigg[ \exp \Big\{ \frac {  ( f ( Y ) - \tilde{\mu} )^2  } { ( f ( Y ) - \mu )^2   +  \sigma_2^2  }  \cdot \frac {\sigma_2^2} {\sigma_1^2 } \Big \}   \mathbbm{1} \big( (1 + x)  \frac { | f (Y) - \tilde{\mu} | } {\sqrt{ n }  \sigma_1 }   > 1  \big) \bigg],
\end{align*}

where $ \gamma = \f 12 (1 + \f c x )  $.

Throughout the rest of the proof, $ A  $ is an absolute positive constant which may vary at each appearance. When the higher moment $ \e [Y^4 ]  < \infty$ is assumed, we can figure out more accurate estimate for the bias-corrected term $ \Psi_x^* $. Let us collect the bounds for some crucial quantities in the proposition below. The proof of Proposition <ref> will be given in Section <ref>.

If $ \e |Y|^4  < \infty $, then we have there exists an absolute positive constant $A$ such that

\begin{gather}
|  \sigma_1^2 - \sigma^2 | \leq \frac { A \e | Y |^4   } { \tau^2 }, \quad  | \sigma_2^2 - \sigma^2 | \leq \frac { A \e | Y |^4   } { \tau ^2 } ,  \quad	| \mu - \tilde{\mu} | \leq \frac {  \e | Y |^4  } { \tau^3  }.  \label{eq_22}
\end{gather}

Moreover, when $\tau$ satisfies (<ref>), we have

\begin{equation}
\Big| \frac { \sigma_2^2 } {\sigma_1^2}  - 1 \Big|  \leq \frac { A (\e | Y |^4 )^2 } { \sigma^2 \tau^6  }, \label{eq_32}
\end{equation}

and there exist absolute positive constants $A$ and $ a_0 $ such that

\begin{equation}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_2 } { \sigma_1 }  x + c \big) }  { 1 - \Phi (x) }  - 1 \Big| \leq  \frac { A ( 1 + x   ) \sqrt n \e | Y |^4  } { \sigma  \tau^3 }  \label{eq_42}
\end{equation}

for $ 0 <  x \leq a_0 \frac {\tau^3 \sigma} { \sqrt n \e | Y |^4 } $, and

\begin{equation}
R_x \leq \frac { A  (1 + x)^4 \e | Y |^4  } { n  \sigma^4  }.  \label{eq_52}
\end{equation}

In addition, there exist positive constants $a_1 $ and $ C $ depending on $ \sigma $, $ \e  |Y|^3  $ and $ \e |Y|^4 $ such that

\begin{equation}
\Psi_x^*  = \exp \Big\{  -  \frac { x^3  \e ( Y - \mu )^3  } { 3 \sqrt n \sigma^3  }  \Big \} \Big[ 1 + O_1 \Big( \frac {  x^3  } { \sqrt n \tau  }  + \frac { x^2   } {\tau^3   } \Big) \Big] \label{eq_62}
\end{equation}

for $ 2 < x \leq  a_1 \min \{ \frac { \tau^3  } { \sqrt n },  (\sqrt n \tau )^{1/3}, \tau^{3/2} \}  $, where $ O_1 $ is a bounded quantity such that $ | O_1 | \leq C $.

We next consider two cases of $ 0 < x \leq 2  $ and $ x > 2  $, separately.
First for $ 0 < x \leq 2  $, it is immediate that $ 1 - \Phi (x)  \geq 1 - \Phi(2) $. Moreover, it follows from Proposition <ref>  and (<ref>) that

\begin{align*}
\P ( S_{\tau, n}^* > x  )  = [   1 - \Phi (x) ]   \Big[ 1 + O \Big( \frac { 1 } { \sqrt n } +  \frac {\sqrt n } {\tau^3 } \Big) \Big].
\end{align*}

Since the bias term $ \exp \{ - \frac { x^3 \e (Y - \mu)^3 } { 3 \sqrt n \sigma^3 } \} = 1 + O ( \frac {1 } {\sqrt n} ) $ for $ 0 < x \leq  2 $, we obtain

\begin{align}
\P ( S_{\tau, n}^* > x  )  = [   1 - \Phi (x) ]  \exp \Big\{ - \frac { x^3 \e (Y - \mu)^3 } { 3 \sqrt n \sigma^3 } \Big\}   \Big[ 1 + O \Big( \frac { 1 } { \sqrt n } +  \frac {\sqrt n } {\tau^3 } \Big) \Big]  \label{eq-14}
\end{align}

for $ 0 < x \leq 2  $, which completes the proof of (<ref>) for $ 0 < x \leq 2 $. Then we deal with the case of $ x > 2  $. Applying Theorem <ref> to (<ref>) yields

\begin{align}
\P ( S_{\tau, n}^* > x  )  = \Big[ 1 - \Phi ( \frac {\sigma_2} {\sigma_1} x + c   ) \Big] \Psi_x^* e^{ O_1 R_x } \Big( 1 + O_2 (1 + \frac {\sigma_2} {\sigma_1} x ) L_{3, n}  \Big) \label{xlarge-2}
\end{align}

uniformly for $ | c |  \leq x / 5$ and $ x  > 2 $ satisfying (<ref>) and (<ref>). Note that (<ref>) and (<ref>) are satisfied when $ 2 < x \leq  a_2 \frac {\sqrt n \sigma^4} {\e | Y |^4}  $ for some constant $a_2 > 0$. By plugging in the results in Proposition <ref>, we can obtain

\begin{align*}
\P ( S_{\tau, n}^* > x   ) & =  [   1 - \Phi (x) ]  \exp \Big\{ - \frac { x^3 \e (Y - \mu)^3 } { 3 \sqrt n \sigma^3 } \Big\} \\
& \quad \times \Big[  1 + O_1 \Big(  \frac { x \sqrt n } { \tau^3  } + \frac {  x ^3  } { \sqrt n \tau  } + \frac { x ^4 } { n } +  \frac { x  } { \sqrt n } + \frac {  x ^2  } {\tau^3 } \Big)    \Big]
\end{align*}

for $ 2 < x \leq a_3  \min \{ \tau^3 n^{-1/2} , ( \sqrt n \tau )^{1/3} , n^{1/4}, \tau^{3/2}\} $. Observe that when $ \tau  $ satisfies (<ref>), $ \tau^{3/2} \geq O (n^{1/4}) $ and $ \frac { 1 + x^2 } { \tau^3 }  \leq A \frac { (1 + x ) \sqrt n } { \tau^3  }$ for $ x = O (\sqrt n) $. Moreover, by the basic inequality that $   a + b >  (a^2 b)^{1/3} $ for $ a > 0$ and $ b> 0 $, we have $ \frac { x^3 } { \sqrt n \tau  } \leq \frac { x^4 } { n } + \frac { x \sqrt n  } { \tau^3 } $ and $ ( \sqrt n  \tau )^{1/3} \geq \min \{ n^{1/4},  \tau^{3} n^{ - 1/2} \} $.
Hence the desired result (<ref>) holds for $ x \in (2, c_2  \min \{ n^{1/4},   \tau^3 n^{- 1/2 }   \}  ) $, which together with the result (<ref>) for $  0 < x \leq 2 $ completes the proof of Theorem <ref>.

§.§ Proof of Theorem <ref>

Define $ g(Y) = (Y - \mu) \IF\{ | Y |  \leq \tau \}  $. Recall that

\begin{equation}
\P (U_{\tau, n}^* > x ) = \P \Big( \frac { S_n^{\circ}  - \delta } { V_n^{\circ} } > \frac { \sigma_4 } { \sigma_3 } x \Big), \label{self-trim}
\end{equation}

\begin{align*}
S_n^{\circ} &= \sum_{i = 1}^n \frac {  g(Y_i) - \mu_0 } { \sqrt n \sigma_3 }, \quad  (V_n^{\circ} )^2 = \sum_{i = 1}^n \frac { [ g(Y_i)]^2 } { n \sigma_4^2 } \quad
\mbox{and} \quad \delta = \frac { \sqrt n \mu_0 } { \sigma_3 }.
\end{align*}

The main idea is to apply Theorem <ref> and  estimate the corresponding bias-corrected term $ \Psi_x^* $ and error terms $ L_{3, n} $ and $ R_x $ under fourth moment, which are defined by

\begin{align*}
L_{3, n} & =    \frac { \e | g (Y) -  {\mu}_0 |^3 } {  \sqrt n  \sigma_3^3 }  + \frac { \e | g (Y)   |^3 } {  \sqrt n  \sigma_4^3 }    \\
\Psi_{x}^* & = \exp \Big\{  ( \frac {\sigma_4} {\sigma_3} )^3  x^3 n^{ - 1/2}  \Big ( \frac  43 \gamma^3     \frac { \e ( g(Y) - {\mu}_0 )^3 } {     \sigma_3^3  } \\
&  \hspace{4cm} -  2  \gamma^2    \frac { \e [ (  g (Y) - {\mu}_0 ) ( g(Y) )^2 ] } {  \sigma_3   \sigma_4^2  } \Big )\Big \} \\
R_x & =  \frac { (1 + x)^4  } {  n} \Big(   \frac { \e | g (Y) - {\mu}_0 |^4 } {    \sigma_3^4 }  + \frac { \e | g (Y)   |^4 } {   \sigma_4^4 }   \Big)  \\
& \quad + n  \e \bigg[ \exp \Big\{ \frac {  ( g ( Y ) - {\mu}_0 )^2  } { ( g(Y) )^2   +  \sigma_4^2  }  \cdot \frac {\sigma_4^2} {\sigma_3^2 } \Big \}   \mathbbm{1} \big( (1 + x)  \frac { | g (Y) - {\mu}_0 | } {\sqrt{ n }  \sigma_3 }   > 1  \big) \bigg],
\end{align*}

where $ \gamma = \f 12 (1 + \f \delta x )  $.

When the higher moment $ \e [Y^4 ]  < \infty$ is assumed, we can figure out more accurate estimate for the bias-corrected term $ \Psi_x^* $. Let us collect the bounds for some crucial quantities in the proposition below. The proof of Proposition <ref> will be given in Section <ref>.

If $ \e |Y|^4  < \infty $, then we have there exists an absolute positive constant $A$ such that

\begin{gather}
|  \sigma_3^2 - \sigma^2 | \leq \frac { A \e | Y |^4   } { \tau^2 }, \quad  | \sigma_4^2 - \sigma^2 | \leq \frac { A \e | Y |^4   } { \tau ^2 } ,  \quad	|  \mu_0 | \leq \frac {  \e | Y |^4  } { \tau^3  }.  \label{eq_22_t}
\end{gather}

Moreover, when $\tau$ satisfies (<ref>), we have

\begin{equation}
\Big| \frac { \sigma_4^2 } {\sigma_3^2}  - 1 \Big|  \leq \frac { A (\e | Y |^4 )^2 } { \sigma^2 \tau^6  }, \label{eq_32_t}
\end{equation}

and there exist absolute positive constants $A$ and $ a_0 $ such that

\begin{equation}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_4 } { \sigma_3 }  x + \delta \big) }  { 1 - \Phi (x) }  - 1 \Big| \leq  \frac { A ( 1 + x   ) \sqrt n \e | Y |^4  } { \sigma  \tau^3 }  \label{eq_42_t}
\end{equation}

for $ 0 <  x \leq a_0 \frac {\tau^3 \sigma} { \sqrt n \e | Y |^4 } $, and

\begin{equation}
R_x \leq \frac { A  (1 + x)^4 \e | Y |^4  } { n  \sigma^4  }.  \label{eq_52_t}
\end{equation}

In addition, there exist positive constants $a_1 $ and $ C $ depending on $ \sigma $, $ \e  |Y|^3  $ and $ \e |Y|^4 $ such that

\begin{equation}
\Psi_x^*  = \exp \Big\{  -  \frac { x^3  \e ( Y - \mu )^3  } { 3 \sqrt n \sigma^3  }  \Big \} \Big[ 1 + O_1 \Big( \frac {  x^3  } { \sqrt n \tau  }  + \frac { x^2   } {\tau^3   } \Big) \Big] \label{eq_62_t}
\end{equation}

for $ 2 < x \leq  a_1 \min \{ \frac { \tau^3  } { \sqrt n },  (\sqrt n \tau )^{1/3}, \tau^{3/2} \}  $, where $ O_1 $ is a bounded quantity such that $ | O_1 | \leq C $.

We next consider two cases of $ 0 < x \leq 2  $ and $ x > 2  $, separately.
First for $ 0 < x \leq 2  $, it is immediate that $ 1 - \Phi (x)  \geq 1 - \Phi(2) $. Moreover, it follows from Proposition <ref>  and (<ref>) that

\begin{align*}
\P ( U_{\tau, n}^* > x  )  = [   1 - \Phi (x) ]   \Big[ 1 + O \Big( \frac { 1 } { \sqrt n } +  \frac {\sqrt n } {\tau^3 } \Big) \Big].
\end{align*}

Since the bias term $ \exp \{ - \frac { x^3 \e (Y - \mu)^3 } { 3 \sqrt n \sigma^3 } \} = 1 + O ( \frac {1 } {\sqrt n} ) $ for $ 0 < x \leq  2 $, we obtain

\begin{align}
\P ( U_{\tau, n}^* > x  )  = [   1 - \Phi (x) ]  \exp \Big\{ - \frac { x^3 \e (Y - \mu)^3 } { 3 \sqrt n \sigma^3 } \Big\}   \Big[ 1 + O \Big( \frac { 1 } { \sqrt n } +  \frac {\sqrt n } {\tau^3 } \Big) \Big]   \label{eq-14-t}
\end{align}

for $ 0 < x \leq 2  $. Next we deal with the case of $ x > 2  $. Applying Theorem <ref> to (<ref>) yields

\begin{align}
\P ( U_{\tau, n}^* > x  )  = \Big[ 1 - \Phi ( \frac {\sigma_4} {\sigma_3} x + \delta   ) \Big] \Psi_x^* e^{ O_1 R_x } \Big( 1 + O_2 (1 + \frac {\sigma_4} {\sigma_3} x ) L_{3, n}  \Big) \label{xlarge-2-t}
\end{align}

uniformly for $ | \delta |  \leq x / 5$ and $ x  > 2 $ satisfying (<ref>) and (<ref>). Note that (<ref>) and (<ref>) are satisfied when $ 2 < x \leq  a_2 \frac {\sqrt n \sigma^4} {\e | Y |^4}  $ for some constant $a_2 > 0$. By plugging in the results in Proposition <ref>, we can obtain

\begin{align*}
\P ( U_{\tau, n}^* > x   ) & =  [   1 - \Phi (x) ]  \exp \Big\{ - \frac { x^3 \e (Y - \mu)^3 } { 3 \sqrt n \sigma^3 } \Big\} \\
& \quad \times \Big[  1 + O_1 \Big(  \frac { x \sqrt n } { \tau^3  } + \frac {  x ^3  } { \sqrt n \tau  } + \frac { x ^4 } { n } +  \frac { x  } { \sqrt n } + \frac {  x ^2  } {\tau^3 } \Big)    \Big]
\end{align*}

for $ 2 < x \leq a_3  \min \{ \tau^3 n^{-1/2} , ( \sqrt n \tau )^{1/3} , n^{1/4}, \tau^{3/2}\} $. Observe that when $ \tau  $ satisfies (<ref>), $ \tau^{3/2} \geq O (n^{1/4}) $ and $ \frac { 1 + x^2 } { \tau^3 }  \leq A \frac { (1 + x ) \sqrt n } { \tau^3  }$ for $ x = O (\sqrt n) $. Moreover, by the basic inequality that $   a + b >  (a^2 b)^{1/3} $ for $ a > 0$ and $ b> 0 $, we have $ \frac { x^3 } { \sqrt n \tau  } \leq \frac { x^4 } { n } + \frac { x \sqrt n  } { \tau^3 } $ and $ ( \sqrt n  \tau )^{1/3} \geq \min \{ n^{1/4},  \tau^{3} n^{ - 1/2} \} $.
Hence the desired result holds for $ x \in (2, c_2  \min \{ n^{1/4},   \tau^3 n^{- 1/2 }   \}  ) $, which together with the result (<ref>) for $  0 < x \leq 2 $ completes the proof of Theorem <ref>.

§.§ Proof of Theorem <ref>

Recall that $ f (x) =  x \mathbbm{1} \{ | x | \leq \tau  \}  + \tau \mathbbm{1} \{ x > \tau \} - \tau \mathbbm{1} \{ x < - \tau \}$. In view of the expression shown in (<ref>) and the notations presented above Theorem <ref>, the main idea is to apply Theorem <ref> and  estimate the corresponding error terms $ L_{3, n},  \Psi_x^* $ and $ R_x $, which are defined by

\begin{align*}
L_{3, n} & =    \frac { \e | f (Y) - \tilde{\mu} |^3 } {  \sqrt n  \sigma_1^3 }  + \frac { \e | f (Y) -  {\mu} |^3 } {  \sqrt n  \sigma_2^3 }    \\
\Psi_{x}^* & = \exp \Big\{  ( \frac {\sigma_2} {\sigma_1} )^3  x^3 n^{ - 1/2}  \Big ( \frac  43 \gamma^3     \frac { \e ( f (Y) - \tilde{\mu} )^3 } {     \sigma_1^3  } \\
&  \hspace{4cm} -  2  \gamma^2    \frac { \e [ (  f (Y) - \tilde{\mu} ) ( f (Y) -  {\mu} )^2 ] } {  \sigma_1   \sigma_2^2  } \Big )\Big \} \\
R_x & =  \frac { (1 + x)^3  } { \sqrt n} \Big(   \frac { \e | f (Y) - \tilde{\mu} |^3 } {    \sigma_1^3 }  + \frac { \e | f (Y) -  {\mu} |^3 } {   \sigma_2^3 }   \Big)  \\
& \quad + n  \e \bigg[ \exp \Big\{ \frac {  ( f ( Y ) - \tilde{\mu} )^2  } { ( f ( Y ) - \mu )^2   +  \sigma_2^2  }  \cdot \frac {\sigma_2^2} {\sigma_1^2 } \Big \}   \mathbbm{1} \big( (1 + x)  \frac { | f (Y) - \tilde{\mu} | } {\sqrt{ n }  \sigma_1 }   > 1  \big) \bigg],
\end{align*}

where $ \gamma = \f 12 (1 + \f c x )  $.

We collect the bounds for some crucial quantities in the proposition below. The proof of Proposition <ref> will be omited since it is similar to the proof of Proposition <ref>.

If $ \e [ | Y |^3 ] < \infty $, then we have there exists an absolute positive constant $A$ such that

\begin{gather}
\e | f (Y) - \mu |^3 \leq A \e  | Y |^3, \quad  \e | f (Y) - \tilde{\mu} |^3 \leq A \e  | Y |^3 \label{eq_1} \\
|  \sigma_1^2 - \sigma^2 | \leq \frac { A \e | Y |^3   } { \tau }, \quad  | \sigma_2^2 - \sigma^2 | \leq \frac { A \e | Y |^3   } { \tau } ,  \quad	| \mu - \tilde{\mu} | \leq \frac {  \e | Y |^3 } { \tau^2  } .  \label{eq_2}
\end{gather}

Moreover, when $\tau$ satisfies (<ref>), we have

\begin{equation}
\Big| \frac { \sigma_2^2 } {\sigma_1^2}  - 1 \Big|  \leq \frac { A (\e | Y |^3 )^2 } { \sigma^2 \tau^4  }, \label{eq_3}
\end{equation}

and there exist absolute constants $A$ and $ a_0 $ such that

\begin{equation}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_2 } { \sigma_1 }  x + c \big) }  { 1 - \Phi (x) }  - 1 \Big| \leq  \frac { A ( 1 + x   ) \sqrt n \e | Y |^3  } { \sigma  \tau^2 }  \label{eq_4}
\end{equation}

for $ 0 <  x \leq a_0 \frac {\tau^2 \sigma} { \sqrt n \e | Y |^3 } $,  and

\begin{equation}
R_x \leq \frac { A  (1 + x)^3 \e | Y |^3  } { \sigma^3 \sqrt n }. \label{eq_5}
\end{equation}

We next consider two cases of $ 0 < x \leq 2 $ and $ x > 2  $, separately.
First for $ 0 < x \leq 2 $, it follows from Proposition <ref> that

\begin{align}
\Big|  \P ( S_{\tau, n}^* > x  )  - \Big[ 1 - \Phi \big( \frac { \sigma_2 } {\sigma_1} x + c \big) \Big] \Big| \leq A L_{3, n}.
\end{align}

From Proposition <ref> that when $ \tau  $ satisfies (<ref>) with some large $a_1$, we can see that

\begin{align}
& L_{3, n} \leq \frac { A \e  |Y|^3 } { \sigma^3 \sqrt n }, \label{L_3n} \\
\mbox{and} \quad & \Big|  \frac {  1 - \Phi \big( \frac { \sigma_2 } {\sigma_1} x + c \big)  } { 1 - \Phi (x) } - 1 \Big| \leq \frac { A \sqrt n \e |Y|^3 } { \sigma \tau^2 }.
\end{align}

Therefore, it holds  for $ 0 < x \leq 2 $ that

\begin{equation}
\frac {\P ( S_{\tau, n} ^*  > x  ) } { 1 - \Phi(x) } = 1 + O_1  \Big( \f {  \e | Y|^3  } { \sigma^3 \sqrt{n} } + \frac { \sqrt{n} \e | Y|^3  } { \sigma \tau^2 }   \Big),
\end{equation}

with $ | O_1 | \leq A $. The desired result (<ref>) has been proved for $ 0 < x \leq 2 $.

Now we proceed to prove for $ x > 2 $. Applying Theorem <ref> to (<ref>) yields

\begin{align}
\P ( S_{\tau, n}^* > x  )  = \Big[ 1 - \Phi ( \frac {\sigma_2} {\sigma_1} x + c   ) \Big] \Psi_x^* e^{ O_1 R_x } \Big( 1 + O_2 (1 + \frac {\sigma_2} {\sigma_1} x ) L_{3, n}  \Big) \label{xlarge}
\end{align}

uniformly for $ | c |  \leq x / 5$ and $ x  > 2 $ satisfying (<ref>) and (<ref>). When $ \tau $ satisfies (<ref>), we can obtain from Proposition <ref> that $ | \sigma_1 / \sigma_2  - 1 | \leq A $ and

\begin{equation}
\Big|  \frac {  1 - \Phi \big( \frac { \sigma_2 } {\sigma_1} x + c \big)  } { 1 - \Phi (x) } - 1 \Big| \leq \frac { A ( 1 + x   ) \sqrt n \e | Y |^3  } { \sigma  \tau^2 }  \label{phi}
\end{equation}

for $  x \leq a_2 \frac {\tau^2 \sigma} { \sqrt n \e | Y |^3 } $, and

\begin{equation}
R_x \leq  \frac { A  (1 + x)^3 \e | Y |^3  } { \sigma^3 \sqrt n }.  \label{Rx}
\end{equation}

Moreover, since $ | \frac c x | \leq  1/ 5  $ as $ \tau  $ satisfies (<ref>) and $ x > 2 $, we have

\begin{equation}
\Psi_x^* \leq \exp \Big\{   \frac { A x^3 \e | Y |^3 } {\sigma^3 \sqrt n } \Big\}.  \label{Psix}
\end{equation}

In addition, note that the conditions of $x$ shown in (<ref>) and $ \eqref{G2}$ are satisfied when $ 0 <  x \leq a_2 \min \{  \frac { \sqrt n \sigma^3 } {\e |Y|^3 } , \frac {\tau^2 \sigma} { \sqrt n \e | Y |^3 } \} $. Consequently, the desired result (<ref>) for $ x > 2 $ are derived by substituting (<ref>) and (<ref>)–(<ref>) into (<ref>). This completes the proof of Theorem <ref>.

§.§ Proof of Theorem <ref>

Define $ g(Y) = (Y - \mu) \IF\{ | Y |  \leq \tau \}  $. Recall that

\begin{equation}
\P (U_{\tau, n}^* > x ) = \P \Big( \frac { S_n^{\circ}  - \delta } { V_n^{\circ} } > \frac { \sigma_4 } { \sigma_3 } x \Big), \label{self-trim-1}
\end{equation}

\begin{align*}
S_n^{\circ} &= \sum_{i = 1}^n \frac {  g(Y_i) - \mu_0 } { \sqrt n \sigma_3 }, \quad  (V_n^{\circ} )^2 = \sum_{i = 1}^n \frac { [ g(Y_i)]^2 } { n \sigma_4^2 } \quad
\mbox{and} \quad \delta = \frac { \sqrt n \mu_0 } { \sigma_3 }.
\end{align*}

The main idea is to apply Theorem <ref> and  estimate the corresponding bias-corrected term $ \Psi_x^* $ and error terms $ L_{3, n} $ and $ R_x $ under fourth moment, which are defined by

\begin{align*}
L_{3, n} & =    \frac { \e | g (Y) -  {\mu}_0 |^3 } {  \sqrt n  \sigma_3^3 }  + \frac { \e | g (Y)   |^3 } {  \sqrt n  \sigma_4^3 }    \\
\Psi_{x}^* & = \exp \Big\{  ( \frac {\sigma_4} {\sigma_3} )^3  x^3 n^{ - 1/2}  \Big ( \frac  43 \gamma^3     \frac { \e ( g(Y) - {\mu}_0 )^3 } {     \sigma_3^3  } \\
&  \hspace{4cm} -  2  \gamma^2    \frac { \e [ (  g (Y) - {\mu}_0 ) ( g(Y) )^2 ] } {  \sigma_3   \sigma_4^2  } \Big )\Big \} \\
R_x & =  \frac { (1 + x)^4  } {  n} \Big(   \frac { \e | g (Y) - {\mu}_0 |^4 } {    \sigma_3^4 }  + \frac { \e | g (Y)   |^4 } {   \sigma_4^4 }   \Big)  \\
& \quad + n  \e \bigg[ \exp \Big\{ \frac {  ( g ( Y ) - {\mu}_0 )^2  } { ( g(Y) )^2   +  \sigma_4^2  }  \cdot \frac {\sigma_4^2} {\sigma_3^2 } \Big \}   \mathbbm{1} \big( (1 + x)  \frac { | g (Y) - {\mu}_0 | } {\sqrt{ n }  \sigma_3 }   > 1  \big) \bigg],
\end{align*}

where $ \gamma = \f 12 (1 + \f \delta x )  $.

We collect the bounds for some crucial quantities in the proposition below. The proof of Proposition <ref> will be omited since it is similar to the proof of Proposition <ref>.

If $ \e [ | Y |^3 ] < \infty $, then we have there exists an absolute positive constant $A$ such that

\begin{gather}
\e | g (Y) - \mu_0 |^3 \leq A \e  | Y |^3, \quad  \e | g (Y)  |^3 \leq A \e  | Y |^3, \label{eq_1-t} \\
|  \sigma_3^2 - \sigma^2 | \leq \frac { A \e | Y |^3   } { \tau }, \quad  | \sigma_4^2 - \sigma^2 | \leq \frac { A \e | Y |^3   } { \tau } ,  \quad	| {\mu}_0  | \leq \frac {  \e | Y |^3 } { \tau^2  }.  \label{eq_2-t}
\end{gather}

Moreover, when $\tau$ satisfies (<ref>), we have

\begin{equation}
\Big| \frac { \sigma_4^2 } {\sigma_3^2}  - 1 \Big|  \leq \frac { A (\e | Y |^3 )^2 } { \sigma^2 \tau^4  }, \label{eq_3-t}
\end{equation}

and there exist absolute constants $A$ and $ a_0 $ such that

\begin{equation}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_4 } { \sigma_3 }  x + \delta \big) }  { 1 - \Phi (x) }  - 1 \Big| \leq  \frac { A ( 1 + x   ) \sqrt n \e | Y |^3  } { \sigma  \tau^2 }  \label{eq_4-t}
\end{equation}

for $ 0 <  x \leq a_0 \frac {\tau^2 \sigma} { \sqrt n \e | Y |^3 } $,  and

\begin{equation}
R_x \leq \frac { A  (1 + x)^3 \e | Y |^3  } { \sigma^3 \sqrt n }. \label{eq_5-t}
\end{equation}

We next consider two cases of $ 0 < x \leq 2 $ and $ x > 2  $, separately.
First for $ 0 < x \leq 2 $, it follows from Proposition <ref> that

\begin{align}
\Big|  \P ( U_{\tau, n}^* > x  )  - \Big[ 1 - \Phi \big( \frac { \sigma_4 } {\sigma_3} x + \delta \big) \Big] \Big| \leq A L_{3, n}
\end{align}

From Proposition <ref> that when $ \tau  $ satisfies (<ref>) with some large $a_1$, we can see that

\begin{align}
& L_{3, n} \leq \frac { A \e  |Y|^3 } { \sigma^3 \sqrt n }, \label{L_3n_t} \\
\mbox{and} \quad & \Big|  \frac {  1 - \Phi \big( \frac { \sigma_4 } {\sigma_3} x + \delta \big)  } { 1 - \Phi (x) } - 1 \Big| \leq \frac { A \sqrt n \e |Y|^3 } { \sigma \tau^2 }.
\end{align}

Therefore, it holds  for $ 0 < x \leq 2 $ that

\begin{equation}
\frac {\P ( U_{\tau, n} ^*  > x  ) } { 1 - \Phi(x) } = 1 + O_1  \Big( \f {  \e | Y|^3  } { \sigma^3 \sqrt{n} } + \frac { \sqrt{n} \e | Y|^3  } { \sigma \tau^2 }   \Big),
\end{equation}

with $ | O_1 | \leq A $. The desired result has been proved for $ 0 < x \leq 2 $.

Now we proceed to prove for $ x > 2 $. Applying Theorem <ref> to (<ref>) yields

\begin{align}
\P ( U_{\tau, n}^* > x  )  = \Big[ 1 - \Phi ( \frac {\sigma_4} {\sigma_3} x + \delta   ) \Big] \Psi_x^* e^{ O_1 R_x } \Big( 1 + O_2 (1 + \frac {\sigma_4} {\sigma_3} x ) L_{3, n}  \Big) \label{xlarge-t}
\end{align}

uniformly for $ | c |  \leq x / 5$ and $ x  > 2 $ satisfying (<ref>) and (<ref>). When $ \tau $ satisfies (<ref>), we can obtain from Proposition <ref> that $ | \sigma_3 / \sigma_4  - 1 | \leq A $ and

\begin{equation}
\Big|  \frac {  1 - \Phi \big( \frac { \sigma_4 } {\sigma_3} x + \delta \big)  } { 1 - \Phi (x) } - 1 \Big| \leq \frac { A ( 1 + x   ) \sqrt n \e | Y |^3  } { \sigma  \tau^2 }  \label{phi-t}
\end{equation}

for $  x \leq a_2 \frac {\tau^2 \sigma} { \sqrt n \e | Y |^3 } $, and

\begin{equation}
R_x \leq  \frac { A  (1 + x)^3 \e | Y |^3  } { \sigma^3 \sqrt n }.  \label{Rx-t}
\end{equation}

Moreover, since $ | \frac \delta x | \leq  1/ 5  $ as $ \tau  $ satisfies (<ref>) and $ x > 2 $, we have

\begin{equation}
\Psi_x^* \leq \exp \Big\{   \frac { A x^3 \e | Y |^3 } {\sigma^3 \sqrt n } \Big\}.  \label{Psix-t}
\end{equation}

In addition, note that the conditions of $x$ shown in (<ref>) and $ \eqref{G2}$ are satisfied when $ 0 <  x \leq a_2 \min \{  \frac { \sqrt n \sigma^3 } {\e |Y|^3 } , \frac {\tau^2 \sigma} { \sqrt n \e | Y |^3 } \} $. Consequently, the desired result (<ref>) for $ x > 2 $ are derived by substituting (<ref>) and (<ref>)–(<ref>) into (<ref>). This completes the proof of Theorem <ref>.

§.§ Proof of Theorem <ref>

Notice that $\Phi^{-1}(1-\alpha/2p)= 2 (1+o(1)) \sqrt{\log 2 p /\alpha } = o(n^{1/6} ) $ lays in the range of Theorem <ref>, hence by Theorem <ref>,

\begin{equation*}
\sum_{ j = 1 }^p \Big[ \P (\mu_j<L_j)+\P(\mu_j >U_j) \Big] \leq 2\sum_{ j = 1 }^p \P \big(S_{\tau, n} ^*  > \Phi^{-1} (1-\alpha/2p) \big)=\alpha+o(1),
\end{equation*}

which completes the proof of Theorem <ref>.

§.§ Proof of Proposition <ref>

From the definition of $ f (\cdot) $, it is obvious that $ | f (Y ) | \leq | Y |$ and $ | f (Y)  - Y| \leq |Y|  \mathbbm{1} ( | Y | >  \tau  ) $. First for the bound for $ | \sigma_1^2 - \sigma^2 | $, it holds that

\begin{align*}
| \sigma_1^2 - \sigma^2  | & = | \e ( f (Y) - \tilde{\mu} )^2 - \e ( Y - \mu )^2 | \\
& \leq \e \big\{ ( 2 | Y | + 2 \e | Y | ) ( | Y | \mathbbm{1} ( | Y | > \tau )  +  \e [ | Y | \mathbbm{1} ( | Y | > \tau ) ] ) \big\} \\
& \leq  \frac { A \e | Y |^4  } { \tau^2 }.
\end{align*}

Similarly, we can obtain the same bound for $ | \sigma_2^2 - \sigma^2 | $. Since $ | f (Y)  - Y| \leq |Y|  \mathbbm{1} ( | Y | >  \tau  ) $, we have

\begin{align*}
| \mu - \tilde{\mu} |  & = | \e f (Y) - \e Y | \\
& \leq \e | f (Y) - Y | \leq \e [  |Y|  \mathbbm{1} ( | Y | >  \tau  )  ] \leq \frac { \e | Y |^4 } { \tau^3 }.
\end{align*}

As for (<ref>), it follows by (<ref>) that when $ \tau  $ satisfies (<ref>),

\begin{align}
\Big| \frac {\sigma_2^2} {\sigma_1^2 } - 1  \Big| & = \frac { | \e[ (f (Y) - \mu)^2  ] - \e [ (f (Y) - \tilde{\mu})^2  ] | } {  \sigma_1^2  }  \nn \\
& =\frac { ( \mu - \tilde{\mu} )^2 } { \sigma_1^2  }  \leq \frac { ( \mu - \tilde{\mu} )^2   } { \sigma^2 - \frac {A \e | Y |^4  } {\tau^2 } } \nn  \\
& \leq \frac { A  ( \mu - \tilde{\mu} )^2  } { \sigma^2 } \leq  \frac { A ( \e | Y |^4 )^2 } { \sigma^2 \tau^6  }.  \label{eq_6}
\end{align}

Now we proceed to show (<ref>). First for $ 0 < x \leq 2 $, it holds that $ 1 - \Phi (2) \leq 1 - \Phi (x)  \leq 1 $. Then we can deduce that

\begin{align}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_2 } { \sigma_1 }  x + c \big) }  { 1 - \Phi (x) }  - 1 \Big|
&  \leq A \Big| \Phi \big( \frac { \sigma_2 } { \sigma_1 }  x + c \big) - \Phi (x)  \Big| \nn \\
& \leq A \Big[  \big( \frac {\sigma_2} {\sigma_1} - 1 \big) x + c  \Big]. \label{eq_7}
\end{align}

Recalling that $ | c  | = \frac { \sqrt n | \mu - \tilde{\mu} | } {\sigma_1} \leq \frac {A \sqrt n \e | Y |^4 } { \sigma \tau^3 } $, and $ | \frac {\sigma_1} {\sigma_2} - 1 | \leq |  \frac {\sigma_1^2} {\sigma_2^2} - 1  | \leq   \frac { A ( \e | Y |^4 )^2 } { \sigma^2 \tau^6  }$. Therefore, it follows that for $ 0 < x \leq 2 $,

\begin{align}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_2 } { \sigma_1 }  x + c \big) }  { 1 - \Phi (x) }  - 1 \Big|   & \leq A  \Big(  \frac {   (\e | Y |^4 )^2  } { \sigma^2  \tau^6 }    + \frac { \sqrt n \e | Y |^4   } { \sigma \tau^3  }  \Big) \nn \\
&  \leq \frac {  A \sqrt n \e | Y |^4   } { \sigma \tau^3  } ,
\end{align}

where the last inequality is derived by using the fact that $  \frac { \sqrt n \e | Y |^4   } { \sigma \tau^3  }  \geq  \frac {  A (\e | Y |^4 )^2  } { \sigma^2  \tau^6 }   $ for $\tau $ satisfying (<ref>). For $ x > 2  $, it holds that when $ x \leq a_0  \frac { \tau^3 \sigma } { \e | Y |^4  } $,

\begin{align*}
\Big|  \frac { 1 - \Phi \big( \frac { \sigma_2 } { \sigma_1 }  x + c \big) }  { 1 - \Phi (x) }  - 1 \Big|   & \leq A  \Big[  \big( \frac { \sigma_2 } {\sigma_1} - 1 \big) x^2 + c x  \Big] \exp\Big\{ A  \Big[  \big( \frac { \sigma_2^2 } {\sigma_1^2} - 1 \big)  x^2 + c x    \Big] \Big\} \nn  \\
& \leq A  \Big[  \big( \frac { \sigma_2 } {\sigma_1} - 1 \big) x^2 + c x  \Big]   \nn \\
& \leq \frac { A x \sqrt n \e | Y |^4 } {   \tau^3  \sigma } .
\end{align*}

The above results for the two cases of $ 0 < x \leq 2 $ and $ 2 < x \leq a_0   \frac { \tau^3 \sigma } { \e | Y |^4  }$ yield the desired result (<ref>).

Next we proceed to prove (<ref>).  Recall that

\begin{align}
R_x \leq A \frac { ( 1 + x  )^4 \e | Y |^4 } {  n \sigma^4 } + r_{x},  \label{Rx_bound}
\end{align}

\begin{align*}
r_x = n   \e \bigg[ \exp \Big\{ \frac {  ( f ( Y ) - \tilde{\mu} )^2  } { ( f ( Y ) - \mu )^2   +  \sigma_2^2  }  \cdot \frac {\sigma_2^2} {\sigma_1^2 } \Big \}   \mathbbm{1} \big( (1 + x)  \frac { | f (Y) - \tilde{\mu} | } {\sqrt{ n }  \sigma_1 }   > 1  \big) \bigg].
\end{align*}

The goal is to upper  bound the exponential part. By (<ref>) and the range (<ref>) for $\tau$, we can obtain

\begin{align}
\frac {  ( f ( Y ) - \tilde{\mu} )^2  } { ( f ( Y ) - \mu )^2   +  \sigma_2^2  }  \cdot \frac {\sigma_2^2} {\sigma_1^2 }  & \leq A \frac { (  f ( Y ) - \tilde{\mu} )^2 } { ( f ( Y ) - \mu )^2   +  \sigma_2^2  } \nn  \\
& = A \frac { (  f ( Y ) - \tilde{\mu} )^2 } { (  f ( Y ) - \tilde{\mu} )^2 + \kappa_1 ( f (Y) - \tilde{\mu} ) + \kappa_2  } \nn \\
& = \frac { A } { 1 + \kappa_1 ( f (Y) - \tilde{\mu} )^{-1} + \kappa_2 (  f (Y) - \tilde{\mu}  )^{ - 2} }, \label{eq-13}
\end{align}

\begin{align*}
\kappa_1 & =  2 ( \tilde{\mu}  - \mu  ) \leq \frac { A \e | Y |^4 } { \tau^3  }, \\
\kappa_2 & = \sigma_2^2 + ( \tilde{\mu} - \mu )^2 \geq A \sigma^2  - \frac { A ( \e | Y |^4 )^2  } { \tau^6 } \geq A \sigma^2
\end{align*}

for $ \tau $ satisfying (<ref>). Observing that for any $ x \in \mathbbm{R} $,

\begin{equation*}
1 + \kappa_1 x + \kappa_2 x^2 \geq 1 - \frac { \kappa_1^2 } { 4 \kappa_2 },
\end{equation*}

hence when (<ref>) hold with large number $ c_1 $ such that $ \kappa_1^2 \leq \kappa_2 $, we obtain

\begin{align}
\frac {  ( f ( Y ) - \tilde{\mu} )^2  } { ( f ( Y ) - \mu )^2   +  \sigma_2^2  }  \cdot \frac {\sigma_2^2} {\sigma_1^2 } \leq  A.
\end{align}

Consequently, it follows that

\begin{align*}
r_x \leq A \frac { (1 + x)^4 \e ( | f (Y) - \tilde{\mu} |^4 ) } {  n \sigma_1^4   } \leq \frac { A ( 1 + x )^4 \e | Y |^4  } { n \sigma^4    },
\end{align*}

which combining with (<ref>) yields the desired result (<ref>).

Finally, we deal with the proof of (<ref>) when $ \tau  $ satisfies (<ref>). It follows from (<ref>) that

\begin{align}
\Big| \Big( \frac { \sigma_2 } {\sigma_1} \Big)^3 - 1  \Big| \leq \frac { A ( \e | Y |^4  )^2  } { \sigma^2 \tau^6 }. \label{eq-9}
\end{align}

Since $ \gamma = \frac 12 ( 1 + \frac c x ) $ and $ | c | \leq  \frac {A \sqrt n | \mu - \tilde{\mu} | } { \sigma } \leq \frac {  A \sqrt n \e | Y |^4  } { \sigma \tau^3   }  \leq A_0 $ for some absolute constant $A_0$ when $ \tau $ satisfies (<ref>), we have $ | c |  / x \leq A_0 / 2 $  for $ x > 2 $ and $\tau $ satisfying (<ref>). In addition,

\begin{align}
\Big|   \gamma  - \frac 1 2 \Big|   \leq \frac {  A \sqrt n \e | Y |^4 } { x \sigma \tau^3  }, \label{eq-10}
\end{align}

and we can obtain from (<ref>) that

\begin{align}
&  \Big| \frac { \e ( f(Y)  - \tilde{\mu})^3 } { \sigma_1^3 }  - \frac { \e ( Y - \mu )^3 } { \sigma^3 } \Big| \nn \\
&  \leq   \frac { | \e (  f(Y) - \tilde{\mu} )^3  - \e (Y- \mu)^3   | } { \sigma_1^3  }  + | \e ( Y - \mu )^3 | \Big| \frac { 1 } { \sigma_1^3} - \frac {1} {\sigma^3} \Big| \nn \\
& \leq A \Big( \frac { \e | Y |^4  } {  \sigma^3 \tau  }  + \frac { | \e ( Y - \mu )^3 | ( \e | Y |^4 )^2 } { \sigma^5 \tau^6 }  \Big) = O \Big( \frac { 1 } { \tau  } \Big). \label{eq-11}
\end{align}

\begin{align}
\Big| \frac { \e [ ( f(Y)  - \tilde{\mu}) (f (Y) - \mu)^2 ] } { \sigma_1 \sigma_2^2 }  - \frac { \e ( Y - \mu )^3 } { \sigma^3 } \Big|
& = O \Big( \frac { 1 } { \tau  } \Big).  \label{eq-12}
\end{align}

Then the desired result (<ref>) is derived by plugging in the above bounds in (<ref>)–(<ref>). This completes the proof of Proposition <ref>.

§.§ Proof of Proposition <ref>

Recall the definition that $ g (Y) = (Y -\mu) \IF\{ | Y | \leq \tau \} $.  First for the bound for $ | \sigma_1^2 - \sigma^2 | $, it holds that

\begin{align*}
| \sigma_1^2 - \sigma^2  | & = | \e \{ [ g (Y) - {\mu}_0 - ( Y - \mu ) ] [ g (Y) - {\mu}_0 + ( Y - \mu )]  \} | \\
& \leq A  \tau^{-2}  \e \big\{  \big( | Y |^3 +   | \mu | Y^2   +  \e | Y |^3 + |\mu| \e Y^2  \big) ( |Y| + |\mu| + \e|Y| \big) \big\} \\
& \leq  \frac { A \e | Y |^4  } { \tau^2 }.
\end{align*}

Similarly, we can obtain the same bound for $ | \sigma_2^2 - \sigma^2 | $. Moreover, we have

\begin{equation*}
| \mu_0 |   = | \e g (Y)  |  \leq  \e [ | Y - \mu | \IF \{ |Y|  > \tau \} ]   \leq \frac { \e | Y |^4 } { \tau^3 }.
\end{equation*}

As for (<ref>), it follows by (<ref>) that when $ \tau  $ satisfies (<ref>),

\begin{align}
\Big| \frac {\sigma_4^2} {\sigma_3^2 } - 1  \Big| & = \frac { | \e[ (g (Y) - \mu_0)^2  ] - \e [ (g (Y) )^2  ] | } {  \sigma_3^2  }  \nn \\
& =\frac {  \mu_0^2 } { \sigma_3^2  }  \leq \frac {  \mu_0^2   } { \sigma^2 - \frac {A \e | Y |^4  } {\tau^2 } } \nn  \\
& \leq \frac { A   \mu_0^2 } { \sigma^2 } \leq  \frac { A ( \e | Y |^4 )^2 } { \sigma^2 \tau^6  }.  \label{eq_6_t}
\end{align}

Therefore, (<ref>) can be proved by the same procedure of proving (<ref>).

Next we proceed to prove (<ref>).  Recall that

\begin{align}
R_x \leq A \frac { ( 1 + x  )^4 \e | Y |^4 } {  n \sigma^4 } + r_{x},  \label{Rx_bound-t}
\end{align}

\begin{align*}
r_x = n   \e \bigg[ \exp \Big\{ \frac {  ( g ( Y ) - {\mu}_0 )^2  } { ( g ( Y )  )^2   +  \sigma_4^2  }  \cdot \frac {\sigma_4^2} {\sigma_3^2 } \Big \}   \mathbbm{1} \big( (1 + x)  \frac { | g (Y) - {\mu}_0 | } {\sqrt{ n }  \sigma_3 }   > 1  \big) \bigg].
\end{align*}

The key point is to upper  bound the exponential part. By (<ref>) and the range (<ref>) for $\tau$, we can obtain

\begin{align}
\frac {  ( g ( Y ) - {\mu}_0  )^2  } { ( g ( Y )   )^2   +  \sigma_4^2  }  \cdot \frac {\sigma_4^2} {\sigma_3^2 }  & \leq A \frac { (  g ( Y ) - {\mu}_0 )^2 } { ( g ( Y )  )^2   +  \sigma_4^2  } \nn  \\
& = A \frac { (  g ( Y ) - {\mu}_0 )^2 } { (  g ( Y ) -\mu_0  )^2 + \kappa_1 ( g (Y) - {\mu}_0 ) + \kappa_2  } \nn \\
& = \frac { A } { 1 + \kappa_1 ( g (Y) - {\mu}_0 )^{-1} + \kappa_2 (  g (Y) - {\mu}_0 )^{ - 2} }, \label{eq-13-t}
\end{align}

\begin{align*}
\kappa_1 & =  2  {\mu}_0 \quad \mbox{satisfying} ~| \kappa_1 | \leq \frac { A \e | Y |^4 } { \tau^3  }, \\
\kappa_2 & = \sigma_4^2 + ( {\mu}_0 )^2 \geq A \sigma^2  - \frac { A ( \e | Y |^4 )^2  } { \tau^6 } \geq A \sigma^2
\end{align*}

for $ \tau $ satisfying (<ref>). Observing that for any $ x \in \mathbbm{R} $,

\begin{equation*}
1 + \kappa_1 x + \kappa_2 x^2 \geq 1 - \frac { \kappa_1^2 } { 4 \kappa_2 },
\end{equation*}

hence when (<ref>) hold with large number $ c_1 $ such that $ \kappa_1^2 \leq \kappa_2 $, we obtain

\begin{align}
\frac {  ( g ( Y ) - {\mu}_0  )^2  } { ( g ( Y )   )^2   +  \sigma_4^2  }  \cdot \frac {\sigma_4^2} {\sigma_3^2 }  \leq  A.
\end{align}

Consequently, it follows that

\begin{align*}
r_x \leq A \frac { (1 + x)^4 \e ( | g (Y) - {\mu}_0 |^4 ) } {  n \sigma_3^4   } \leq \frac { A ( 1 + x )^4 \e | Y |^4  } { n \sigma^4    },
\end{align*}

which combining with (<ref>) yields the desired result (<ref>).

Finally, we deal with the proof of (<ref>) when $ \tau  $ satisfies (<ref>). It follows from (<ref>) that

\begin{align}
\Big| \Big( \frac { \sigma_4 } {\sigma_3} \Big)^3 - 1  \Big| \leq \frac { A ( \e | Y |^4  )^2  } { \sigma^2 \tau^6 }. \label{eq-9-t}
\end{align}

Since $ \gamma = \frac 12 ( 1 + \frac \delta x ) $ and $ | \delta | \leq  \frac {A \sqrt n | \mu_0 | } { \sigma } \leq \frac {  A \sqrt n \e | Y |^4  } { \sigma \tau^3   }  \leq A_0 $ for some absolute constant $A_0$ when $ \tau $ satisfies (<ref>), we have $ | \delta |  / x \leq A_0 / 2 $  for $ x > 2 $ and $\tau $ satisfying (<ref>). In addition,

\begin{align}
\Big|   \gamma  - \frac 1 2 \Big|   \leq \frac {  A \sqrt n \e | Y |^4 } { x \sigma \tau^3  }, \label{eq-10-t}
\end{align}

and we can obtain from (<ref>) that

\begin{align}
&  \Big| \frac { \e ( g(Y)  - {\mu}_0 )^3 } { \sigma_3^3 }  - \frac { \e ( Y - \mu )^3 } { \sigma^3 } \Big| \nn \\
&  \leq  A  \frac { | \e (  g(Y) - {\mu}_0 )^3  - \e (Y- \mu)^3   | } { \sigma_3^3  }  + A [  \e | Y - \mu |^3 ] \Big| \frac { 1 } { \sigma_3^3} - \frac {1} {\sigma^3} \Big| \nn \\
& \leq A \Big( \frac { \e | Y |^4  } {  \sigma^3 \tau  }  + \frac { | \e ( Y - \mu )^3 | ( \e | Y |^4 )^2 } { \sigma^5 \tau^6 }  \Big) = O \Big( \frac { 1 } { \tau  } \Big). \label{eq-11-t}
\end{align}

\begin{align}
\Big| \frac { \e [ ( g(Y)  - {\mu}_0 ) (g (Y))^2 ] } { \sigma_3 \sigma_4^2 }  - \frac { \e ( Y - \mu )^3 } { \sigma^3 } \Big|
& = O \Big( \frac { 1 } { \tau  } \Big).  \label{eq-12-t}
\end{align}

Then the desired result (<ref>) is derived by plugging in the above bounds in (<ref>)–(<ref>). This completes the proof of Proposition <ref>.

§.§ Proof of Corollary <ref>

We apply Theorem <ref> with $ Y_i^2 = \e X_i^2  $, $ c_0 = 0 $ and $ c = 0 $. In this case,

\begin{align*}
\Psi_x^* = \exp \Big \{   \frac { x^3   \e X_1^3 } { 6 \sigma^3 \sqrt{n}  }   \Big\} \quad , L_{3, n} \leq  \frac { 2   \e | X_1 |^3 } { \sigma^3 \sqrt n    }  , \quad  \delta_{x, 1} \leq \frac { 4 (1 + x)^4 \e X_1^4  }  { n^2 \sigma^4  }.
\end{align*}

For $ x \leq t_0 \sigma \sqrt n / 4 $, we have $ 2 x / (\sigma \sqrt n )  \leq t_0 / 2$ and thus

\begin{align*}
r_{x, 1}  & = \e \bigg[ \exp \Big\{ \min \Big( \frac { X_1^2  } { \sigma^2   } , \frac { 2 x X_1 } { \sigma \sqrt n  } \Big) \Big\}  \IF ( | (1 + x) X_1 | > \sigma \sqrt n  ) \bigg]  \\
& \leq  \e  \bigg[ \exp \Big\{  \frac { 2 x X_1 } { \sigma \sqrt n  }   \Big\}  \IF ( | (1 + x) X_1 | > \sigma \sqrt n  ) \bigg] \\
& \leq  \e  \big[  e^{ t_0 X_1 / 2  } \IF \{  (1 + x ) | X_1| > \sigma \sqrt n \} \big]  \\
& \leq (1 + x)^4 n^{-2 } \e \big[ | X_1 |^4 e^{ t_0 X_1 / 2  }   \big] \\
& \leq C t_0^{-4}  \e \big[ e^{ 3 t_0 X_1 / 4  }  \big]  ( 1 + x  )^4 n^{ - 2  } .
\end{align*}

Therefore it follows from Theorem <ref> that

\begin{align*}
\frac { \P ( S_n > x  \sigma \sqrt n  ) } { 1 - \Phi (x) }  = \exp \Big\{ \frac { x^3   \e X_1^3 } { 6 \sigma^3 \sqrt{n}  }  +  O \Big( \frac { ( 1 + x )^4  } { n  } \Big) \Big \} \Big[ 1 + O \Big( \frac { 1 + x } { \sqrt n } \Big) \Big]
\end{align*}

uniformly for $ 0 < x \leq O (n^{1/2}) $. Hence (<ref>) holds for $ 0 < x \leq O (n^{1/4}) $. This completes the proof of Corollary <ref>.

§ ADDITIONAL LEMMAS AND PROOFS

§.§ Proof of Lemma <ref>

Recall the notation $W_i=2x{X_i} - x^{2}{Y_i}^2$. Define

\nu_i = 	\min\left\{  \frac{{X_i}^2}{{Y_i}^2+c_0\e {Y_i}^2} +c_0x^2\e {Y_i}^2 , 2xX_i \right\}  .

Observe that $ W_i \leq \nu_i $.
For $ 1/4 \leq  \lambda \leq 3 /4 $ and $ k \in \{ 0, 1, 2, 3 \}$, by the elementary inequalities $ |s|^{k} e^{- s} \leq  c_1(k) $  and $ s^{k} e^{\lambda s }  \leq c_2(k) e^{s}$ for $ s \geq 0 $, we obtain

\begin{align}
&\e \{ |W_i|^k   e^{\la W_i} \mathbbm{1}(|(1+x)X_i| \geq 1 ) \} \label{Wtail}\\
& \hspace{1cm}   \leq   A \e \{e^{(0\lor W_i)} \mathbbm{1}(|(1+x)X_i| \geq 1 ) \} \nn \\
& \hspace{1cm}  \leq   A \big( \e \{  e^{\nu_i} \mathbbm{1}(|(1+x)X_i| \geq 1 ) \}  + \delta_{x,i} \big) \nn \\
& \hspace{1cm} \leq A  R_{x,i} ,  \nn
\end{align}

where the last inequality comes from the fact that $ c_0 x^2 \e {Y_i}^2 \le  1/4  $ for $x$ satisfying (<ref>). Additionally, there is a constant $A_1(k, s_0) $ such that $ |s|^k e^{s} \leq A_1 (k, s_0) $ if $ s \leq s_0 $. Hence,

\begin{align*}
&\e \{ |W_i|^k   e^{\la W_i} \mathbbm{1}(|(1+x)X_i| \leq 1, |(1+x)Y_i| \geq 1 ) \}   \\
& \leq   A \P(|(1+x)X_i| \leq 1, |(1+x)Y_i| \geq 1 ) \\
& \leq  A \de_{x,i}.
\end{align*}

For simplicity, denote $\mathcal{U}_i = \{ |(1+x) X_i| \leq 1, |(1+x) Y_i| \leq 1   \}$. We have

\begin{align}
\e \{ {W_i}^k e^{\la W_i}  \} =  \e \{ {W_i}^k e^{\la W_i} \mathbbm{1}(\mathcal{U}_i ) \} + O(1) R_{x,i}, \label{e1}
\end{align}

where and hereafter $ O(1) $ is a bounded quantity.

By Taylor expansion,

\begin{align}
\e \{ e^{\la W_i} \mathbbm{1}(\mathcal{U}_i ) \} &=  1+\la \e \{ W_i \mathbbm{1}(\mathcal{U}_i  )\}+\frac{{\la}^2}{2} \e \{  {W_i}^2 \mathbbm{1}(\mathcal{U}_i )\}  \label{0th} \\
& \quad   +\frac{{\la}^3}{6} \e \{  {W_i}^3 \mathbbm{1}(\mathcal{U}_i  )\}  + O(1) {\la}^4 \e \{  {W_i}^4   \mathbbm{1}(\mathcal{U}_i  )\}. \nn
\end{align}

The analysis for the terms on the right hand side of the above equality is similar. We calculate the third term for example. Notice that

\begin{align*}
\e \{{W_i}^2 \mathbbm{1}(\mathcal{U}_i )\} &= 4x^2\e {X_i}^2 - 4x^3\e X_i{Y_i}^2 +x^4\e \{ {Y_i}^4 \mathbbm{1}(\mathcal{U}_i ) \}\\
& \quad  -4x^2\e \{ {X_i}^2 \mathbbm{1}(|(1+x)X_i| \geq 1) \} \\
& \quad -4x^2\e \{{X_i}^2 \mathbbm{1}(|(1+x)X_i| \leq 1,|(1+x)Y_i| \geq 1) \} \\
& \quad +4x^3\e \{ X_i{Y_i}^2 \mathbbm{1}(|(1+x)X_i| \leq 1,|(1+x)Y_i| \geq 1) \} \\
& \quad +4x^3\e \{ X_i{Y_i}^2 \mathbbm{1}(|(1+x)X_i| \geq 1,|(1+x)Y_i| \leq 1) \} \\
& \quad +4x^3\e \{ X_i{Y_i}^2 \mathbbm{1}(|(1+x)X_i| \geq 1,|(1+x)Y_i| \geq 1) \}.
\end{align*}

By the basic inequality $ ab^2 \leq a^3 + b^3  $ for $ a > 0 , b> 0 $ and Chebyshev's inequality, we obtain

\begin{align*}
\e \{{W_i}^2 \mathbbm{1}(\mathcal{U}_i )\} = 4x^2\e {X_i}^2 - 4x^3\e X_i{Y_i}^2 +O(1)\de_{x,i}.
\end{align*}

In the same manner,

\begin{gather*}
\e \{W_i \mathbbm{1}(\mathcal{U}_i)\} = -x^2\e {Y_i}^2 +O(1)\de_{x,i}, \quad \e \{{W_i}^3 \mathbbm{1}(\mathcal{U}_i )\} = 8x^3\e {X_i}^3 +O(1)\de_{x,i}, \\
\quad {\rm and} \hspace{1cm} |\e \{{W_i}^4   \mathbbm{1}(\mathcal{U}_i )\}| \leq O(1)\de_{x,i}.
\end{gather*}

Therefore (<ref>)–(<ref>) yield

\begin{align}
\e e^{\la W_i} &= 1 - \la x^2\e {Y_i}^2 +2 {\la}^2x^2 \e {X_i}^2  \label{e2} \\
&\qquad -2{\la}^2x^3 \e {X_i}{Y_i}^2 +\frac{4}{3}{\la}^3x^3 \e {X_i}^3 +O(1)R_{x,i}. \nn
\end{align}

Because $x$ satisfies (<ref>),

\begin{align}
|\e e^{\la W_i} - 1| & \leq 1/2 .  \nn
\end{align}

Furthermore, by Lemma <ref> we have

\begin{align*}
|\e e^{\la W_i} - 1|^2 & \leq A \big[ x^4\e \{{Y_i}^4 \mathbbm{1}(|(1+x)Y_i|\leq 1)\} + x^4 \e \{{X_i}^4 \mathbbm{1}(|(1+x)X_i|\leq 1)\} \nn\\
& \qquad + x^6 \e \{({X_i}^6+{Y_i}^6)\mathbbm{1}(\mathcal{U}_i)\} + R_{x,i} \big] \nn\\
& \leq A R_{x, i},
\end{align*}

Because $ | \log (1+a) - a | \leq  a^2$ whenever $|a|\leq 1/2$, it follows from (<ref>) that

\begin{align*}
& \log \e e^{\la W_i} = - \la x^2\e {Y_i}^2 +2 {\la}^2x^2 \e {X_i}^2 -2{\la}^2x^3 \e {X_i}{Y_i}^2 +\frac{4}{3}{\la}^3x^3 \e {X_i}^3 +O(1)R_{x,i},
\end{align*}

which completes the proof of (<ref>).
And by (<ref>), for $k=1,2$ we have

\begin{align}
\e W_i^k e^{\la W_i} & = \e \{W_i^k e^{\la W_i}\mathbbm{1}(\mathcal{U})\} +O(1)R_{x,i} \nn \\
& = \e \Big\{ W_i^k \Big(\sum_{j=0}^{3-k} \frac{\la^j}{j!} W_i^j \Big) \mathbbm{1}(\mathcal{U}) \Big\} \nn \\
&\quad \; +\e \Big\{W_i^k \Big(e^{\la W_i}-\sum_{j=0}^{3-k} \frac{\la^j}{j!} W_i^j \Big) \mathbbm{1}(\mathcal{U})  \Big\} +O(1)R_{x,i} \nn \\
& := P_{k1} +P_{k2}+O(1)R_{x,i}, \nn
%	\e {W_i}^2 e^{\la W_i} & = \e \{{W_i}^2 e^{\la W_i}I(|xX_i|\leq 1,|xY_i|\leq 1)\} +O(1)R_{x,i} \nn \\
%	& = \e \{{W_i}^2 (1+\la W_i )I(|xX_i|\leq 1,|xY_i|\leq 1)  \} \nn \\
%	&\quad \; +\e \{{W_i}^2 (e^{\la W_i}-1-\la W_i )I(|xX_i|\leq 1,|xY_i|\leq 1)  \} +O(1)R_{x,i} \nn \\
%	& = P_{21} +P_{22}, \nn \\
\end{align}

\begin{align}
\e |W_i|^3 e^{\la W_i} & = \e \{|W_i|^3 \mathbbm{1}(\mathcal{U})\}  +\e \{|W_i|^3 (e^{\la W_i}-1) \mathbbm{1}(\mathcal{U})\}+O(1)R_{x,i} \nn \\
& := P_{31} +P_{32}+O(1)R_{x,i}, \nn
\end{align}

where we have

\begin{align*}
|P_{12}|+|P_{22}|+|P_{32}| & \leq O(1)\e \{|W_i|^4 e^{\la |W_i|}\mathbbm{1}(\mathcal{U})\}  \leq O(1) \e \{ |W_i|^4 \mathbbm{1}(\mathcal{U})\} \} \\
& \leq O(1) [\e \{ x^4 {X_i}^4 \mathbbm{1}(\mathcal{U}) \} + \e \{ x^8 {Y_i}^8 \mathbbm{1}(\mathcal{U}) \}] \leq O(1)\de_{x,i}.
\end{align*}

By the same token,

\begin{align*}
\e W_ie^{\la W_i} & = \e \{W_i(1+\la W_i +\frac{{\la}^2}{2}{W_i}^2) \mathbbm{1}(\mathcal{U}_i)  \} +O(1)R_{x,i} \\
& = -x^2 \e {Y_i}^2 +4\la x^2 \e {X_i}^2 -4\la x^3 \e X_i{Y_i}^2 +4{\la}^2 x^3 \e {X_i}^3 +O(1)R_{x,i}, \\
\e {W_i}^2 e^{\la W_i} & = \e \{{W_i}^2 (1+\la W_i ) \mathbbm{1}(\mathcal{U}_i)  \}+O(1)R_{x,i} \\
& = 4x^2\e {X_i}^2 -4x^3\e X_i{Y_i}^2 +8\la x^3 \e {X_i}^3 +O(1)R_{x,i}, \\
\e |W_i|^3 e^{\la W_i} & = \e \{|W_i|^3  \mathbbm{1}(\mathcal{U}_i )\}+O(1)R_{x,i} \\
& = O(1) x^3 (\e |X_i|^3 +\e |Y_i|^3)+ O(1)R_{x,i}.
\end{align*}

By a similar procedure to the proof of (<ref>), we arrive at (<ref>)–(<ref>). The proof of Lemma <ref> is completed.

§.§ Proof of Lemma <ref>

Recall the definition $ m \(\la\) = \sum_{i=1}^n \log{ \e e^{\la W_i} } $. Note that $m(\la_{\de})$ is well-defined under condition (<ref>) and $m''(\la)=\sum_{i=1}^{n} Var \wt W_i	>0$ for $\la>0$, $x \neq 0$ and nondegenerate $X_i$ and $Y_i$. It follows from Lemmas <ref> and <ref> that for $1/4 \leq \la \leq 3/4$ and $x$ satisfying (<ref>)–(<ref>),

\begin{align}
m'(\la) & = \sum_{i=1}^{n} \e W_i e^{\la W_i} /\e e^{\la W_i} \label{term3} \\
& = (4\la -1)x^2 +4{\la}^2 x^3 \sum_{i=1}^{n} \e {X_i}^3 -4\la x^3 \sum_{i=1}^{n} \e X_i{Y_i}^2 +O(1)R_{x} ,
\nn
\end{align}

where $ | O(1) | \leq A $ for some absolute constant $A$.
Therefore, under (<ref>) with sufficiently small constant $ c_1 $, we have for $ | \de (x) | \leq x^2 /2  $ that

\begin{align*}
m'(1/4) <  x^2+\de (x) <  m'(3/4),
\end{align*}

which combined with the fact $ m'' (\la) > 0  $ implies that the equation $m'(\la)=x^2+\de (x)$ has a unique solution $\la_{\de}$ such that $1/4 < \la_{\de} < 3/4$. Furthermore, by virtue of (<ref>), it holds that

\begin{align*}
\la_{\de} = \frac{1}{2} + \frac{\de (x)}{4x^2} - {\la_{\de}}^2 x \sum_{i=1}^{n} \e {X_i}^3 + \la_{\de} x \sum_{ i =1 }^n \e X_i{Y_i}^2 + O (1) x^{ - 2 } R_{x},
\end{align*}

and hence (<ref>) follows. Again by (<ref>) and $1/4  < \la_{\de} <  3/4  $, $ 1/4  < \la_{\de_0} <  3/4   $, we obtain

\begin{align*}
| \la_{\de} - \la_{\de_0} | \leq A \big(  x^{-2 } |  \de (x ) - \de_0 (x)  |   +  x L_{3, n}  +  x^{- 2} R_x      \big) .
\end{align*}

\begin{align*}
\Big|  \la_{\de} - \la_{\de_0} - \frac { \de (x)  - \de_0 (x)  } { 4 x ^2  } \Big| &  \leq  A \big(  |\la_{\de} - \la_{\de_0} | x L_{3, n} +  x^{ - 2  } R_x \big) \\
&   \leq A \big(  x^{ -2 } R_x +  | \de (x)  - \de_{0} (x) | x^{ -1 } L_{2, n} \big).
\end{align*}

Thus we complete the proof of (<ref>).
The result (<ref>) directly follows from (<ref>), and (<ref>) follows from (<ref>) and (<ref>). This completes the proof of Lemma <ref>.

§.§ Proof of Lemma <ref>

In the sequel,  $ C, C_1, C_2, \ldots $ are positive constants that may depend on $ \omega $ and $ r_0 $ and may take different values at each appearance. Note that (<ref>) is a special case of (<ref>).   Denote $ Z_i = Y_i^2 - \e Y_i^2 $ and $ G_i = 2 r x X_i - w r  x^2 Y_i^2  $. Regarding (<ref>), we have

\begin{align}
&\e \left\{\left(V_n^2-1\right)^2 e^{\sum_{i=1}^{n} G_i  }\right\} \label{in-1} \\
& \hspace{0.8cm} =\e \bigg\{\Big(\sum_{i=1}^{n}Z_i\Big)^2 e^{\sum_{i=1}^{n} G_i } \bigg\} \nonumber \\
& \hspace{0.8cm} = \e e^{\sum_{i=1}^{n} G_i } \bigg[\sum_{i=1}^n\frac{ \e Z_i^2e^{ G_i }}{\e e^{G_i }}
+\sum_{i\ne j}\frac{\e Z_ie^{ G_i}}{\e e^{G_i}}\frac{\e Z_je^{G_j}}{\e e^{G_j}}\bigg] \nn \\
&\hspace{0.8cm}  \leq  \e e^{\sum_{i=1}^{n} G_i} \bigg[\sum_{i=1}^n\frac{ \e Z_i^2e^{G_i}}{\e e^{G_i}} + \bigg( \sum_{i=1}^n \frac{\e Z_ie^{G_i}}{\e e^{G_i}} \bigg)^2  \bigg] .\nn
\end{align}

We first treat $ \e [ Z_i e^{G_i} ] $. Recall that $ 0 < r <   r_0 $ for some number $r_0$. For $x$ satisfying (<ref>), $ x^2 \e Y_i^2  \leq 1/16 $ and $ c_0 x^2 \e Y_i^2 \leq 1/4 $. It follows from the basic inequality $ | e^{s} - 1 |  \leq |s| e^{s \lor 0} $ that

\begin{align}
& x^2 \big| \e [ Z_i e^{G_i} \IF ( x | X_i | \leq 1 , x | Y_i  | \leq 1  ) ] \big|  \label{bound-mo1} \\
& \hspace{0.5cm} = x^2 \big|  \e    \big[ Z_i \big(1  + O ( x | X_i |  + x^2 Y_i^2  )  \big) \IF  ( x | X_i | \leq 1 , x | Y_i  | \leq 1  ) \big]    \big| \nn \\
&  \hspace{0.5cm}  \leq  C \big\{ x^2 \e [ (Y_i^2 + \e Y_i^2 ) \IF \{ x | Y_i  | > 1  \} ]  \nn \\
& \hspace{1.5cm}  +   x^2  \e [ (Y_i^2 + \e Y_i^2 ) \IF \{ x | X_i  | > 1 , x | Y_i  | \leq  1  \} ] \nn \\
& \hspace{1.5cm}  + x^2 \e [ (Y_i^2 + \e Y_i^2 )  ( x| X_i | + x^2 Y_i^2  ) \IF ( x| X_i | \leq 1, x | Y_i | \leq 1  ) ] \big\}  \nn \\
&  \leq C x^3 ( \e | X_i|^3 + \e | Y_i |^3  + \e [ | X_i  Y_i^2 |  ]  ) + C x^4  (\e Y_i^2)^2     \nn  \\
& \leq C x^3 ( \e |X_i|^3 + \e |Y_i |^3  ) ,\nn
\end{align}

where the last inequality results from Lemma <ref> and the basic inequality $ a b^2 \leq a^3 + b^3  $ for $ a, b > 0 $. In addition, as $ G_i \leq 2 r < 2 $ when $ x | X_i  | \leq 1  $, we obtain for $x > 3  $

\begin{align}
& x^2 | \e [ Z_i e^{G_i} \IF (x | X_i  | \leq 1 ,   x | Y_i | > 1 ) ] | \label{bound-mo2} \\
&  \leq C x^2   \e [ Y_i^2 \IF ( x | Y_i  |  > 1  )  ] +  C x^2  \e [ Y_i^2  ] \P ( x | Y_i  | > 1 )  \nn \\
& \leq C x^3 \e [ |  Y_i |^3 \IF ( x | Y_i | > 1  ) ] \leq C \delta_{x, i} \nn.
\end{align}

Moreover, as $   x^2 \e Y_i^2 \leq  1/16 $,

\begin{align}
& x^2 | \e [ Z_i e^{G_i} \IF  ( x | X_i  | > 1 ) |   \label{bound-mo3} \\
& \leq x^2 \e [ Y_i^2 e^{ G_i  } \IF  ( x | X_i  | > 1  ) ] +  \e [ e^{ G_i  } \IF  ( x | X_i  | > 1  )  ] \nn.
\end{align}

As for the second error term, we have

\begin{align*}
& \e [e^{ G_i } \IF ( x | X_i  | > 1  ) ]  \\
&   = \e [ e^{G_i} \IF (  x | X_i  | > 1, X_i \leq  0 )  ] + \e [e^{ G_i} \IF (  x | X_i  | > 1, X_i > 0 ) ] \nn\\
&  \leq x^3 \e [ | X_i^3  | \IF( x | X_i | > 1 ) ] + \e [ e^{ G_i } \IF ( x  X_i   > 1 ) ] \nn.
\end{align*}

Observe that when $ x X_i > 1 $,

\begin{align*}
e^{ G_i  } = e^{ 2 r x  X_i - \omega r  x^2 Y_i^2  } \leq e^{ 2 r x X_i } \leq e^{ 2 x X_i }
\end{align*}

and by Cauchy inequality for $0  < r < r_0 < \omega $,

\begin{align}
2 r  x X_i  - \omega r x^2 Y_i^2           & \leq 2 (\omega r)^{1/2} x X_i - \omega r x^2 Y_i^2  \label{cauchy}  \\
& \leq \frac {  x^2 X_i^2 } { x^2 Y_i^2 + c_0  x^2 \e Y_i^2 } +  \omega r  c_0 x^2 \e  Y_i^2 . \nn
\end{align}

\begin{align}
&  \e [ e^{G_i}  \IF ( x |X_i| > 1 ) ] \label{mo-31} \\
& \leq x^3 \e [ | X_i |^3 \IF ( x | X_i | > 1  ) ] +  C_1  \e \Big[ e^{ \min \big\{ \frac { X_i^2  } { Y_i^2 + c_0 \e Y_i^2  } , 2 x X_i\big \} } \IF ( x | X_i | > 1  ) \Big] \nn \\
& \leq C_2 R_{x, i}. \nn
\end{align}

As for the first error term in (<ref>),  it follows from the basic inequality $ |x|^k e^{  - \delta  x }  \leq C(k, \delta)$ for some constant $ C(k, \delta) $ depending on $k$ and $ \delta $ that

\begin{align*}
& x^2 \e [ Y_i^2  e^{ G_i } \IF (x | X_i  |  > 1 ) ] \\
\leq&  C_1    \e [ e^{ 2 r x X_i  -  ( \omega/2  + r_0 /2 ) r x^2 Y_i^2 } \IF ( x | X_i  | > 1   ) ] \nn \\
\leq&  C_1    x^3 \e [ x | X_i |^3 \IF ( x | X_i | > 1  ) ] \\
& \hspace{0.5cm}  +   C_1     \e [ e^{ 2 r x X_i  -  ( \omega/2  + r_0 /2 ) r x^2 Y_i^2 } \IF ( x  X_i   > 1   ) ] \nn.
\end{align*}

As $ 0 < r < r_0 < \omega/2 + r_0 /2 $, by a similar procedure to (<ref>) we obtain

\begin{align}
x^2 \e [ Y_i^2  e^{ G_i } \IF (x | X_i  |  > 1 )  ] \leq  C_1   R_{x, i  }. \label{mo-32}
\end{align}

Consequently, it follows from the bounds (<ref>)–(<ref>) and (<ref>)–(<ref>) that

\begin{align}
| \e [ Z_i e^{ G_i } ] | \leq  C \big[ x ( \e | X_i |^3 + | Y_i |^3    ) + x^{ - 2} R_{x, i} \big].  \label{mo-1th}
\end{align}

By similar arguments, we obtain

\begin{align}
\e [ Z_i^2 e^{ G_i  } ] \leq  C x^{ -4 } R_{x, i}.  \label{mo-2th}
\end{align}

In addition,

\begin{align}
\e [ e^{ G_i   } ] & = \prod_{i = 1}^n \Big(  \e [ e^{ G_i  } \IF ( x | X_i  |  > 1 ) ] + \e [ e^{ G_i  } \IF ( x | X_i  | \leq 1 , x | Y_i  | \leq 1  ) ] \label{mo-0th} \\
& \hspace{2cm}  + \e [ e^{ G_i  }  \IF ( x | X_i  | \leq 1 , x  | Y_i |  > 1  ) ] \Big) \nn .
\end{align}

Note that

\begin{align*}
\e [ e^{ G_i  }  \IF ( x | X_i  | \leq 1 , x  | Y_i |  > 1  ) ] \leq e^{ 2 r_0 }  x^3 \e [ | Y_i |^3 \IF ( x | Y_i  | \geq 1  ) ]  \leq  C \delta_{x, i},
\end{align*}

and (<ref>) has shown

\begin{align*}
\e [ e^{ G_i  } \IF ( x | X_i  |  > 1 ) ] \leq C R_{x, i},
\end{align*}

moreover, by inequality $ | e^{s} - ( 1 + s + s^2 /2 + s^3 / 6  ) | \leq s^4 e^{s \lor 0}$ and $ G_i \leq 2 r_0  $ when $  x | X_i  | \leq 1 $,

\begin{align}
& \e [ e^{ G_i  } \IF ( x |X_i | \leq 1 , x | Y_i  | \leq 1 ) ]  \label{0th-expan} \\
&   =   \e \big[  \big( 1 + G_i + G_i ^2 /2 + G_i ^3 / 6 + O( G_i^4 ) \big)  \IF ( x | X_i  | \leq 1 , x | Y_i   | \leq 1  ) \big]\nn  \\
&  = 1 + ( 2 r^2 - \omega r^2  ) x^2 \e X_i^2 - 2 \omega r^2 x^3 \e [ X_i Y_i^2  ] + \frac { 4 } {3} r^3 x^3 \e X_i^3  + O (\delta_{x, i}).  \nn
\end{align}

Under condition (<ref>), we have

\delta_{x, i}  \leq R_{x, i} \leq x^3 ( \e | X_i |^3 + \e | Y_i |^3 ) + r_{x, i} \leq c_1 + 1/64.

As a result for small constant $c_1$,

\begin{align*}
1/2 \leq  \e [ e^{ G_i  }   ]  \leq 3/2.
\end{align*}

Thus it follows by substituting (<ref>) and (<ref>) into (<ref>) that

\begin{align}
& \e \left\{\left(V_n^2-1\right)^2 e^{\sum_{i=1}^{n} G_i  }\right\} \\
& \leq 2 \prod_{i = 1}^n \e [ e^{ G_i  } ] \Big(  x^{ - 4} R_{x} +  ( x L_{3, n} + x^{ - 2} R_{x} )^3  \Big)  \nn \\
& \leq  A x^{ - 2} R_x \exp \Big\{ ( 2 r^2 - \omega r^2  ) x^2 \e X_i^2 - 2 \omega r^2 x^3 \e [ X_i Y_i^2  ] + \frac { 4 } {3} r^3 x^3 \e X_i^3  +  A \delta_{x, i}   \Big\}  \nn,
\end{align}

hence the desired result (<ref>) is derived. (<ref>) is a special case of (<ref>), with $ r = \lambda_1 $ and $ \omega = 1 $.

Next we prove (<ref>). We have by conditional Cauchy-Schwarz inequality that

\begin{align}
&\sum_{i=1}^n \e \Big\{\Big|W_i\Big(Z_i^2+2Z_i
\sum_{i\ne j} Z_j   \Big)\Big|e^{\lambda_1\sum_{i=1}^nW_i}\Big\}  \label{err-2} \\
\le& \sum_{i=1}^n \e \left|W_iZ_i^2e^{\lambda_1\sum_{i=1}^nW_i}\right|+2\sum_{i=1}^n \e \Big[\Big|Z_iW_i\sum_{j\ne i}Z_j\Big|e^{\lambda_1\sum_{i=1}^nW_j }\Big] \nonumber \\
\le& Q_1+Q_2,\nonumber
\end{align}

\begin{align*}
Q_1&= \prod_{i = 1}^n \e [ e^{ \lambda_1 W_i } ] \sum_{i=1}^n\frac{ \e \left|W_iZ_i^2e^{\lambda_1W_i}\right|}{ \e e^{\lambda_1W_i}}, \\
Q_2&=2\sum_{i=1}^n \e \left(|Z_iW_i|e^{\lambda_1W_i}\right)\Big\{ \e \Big[ \Big(\sum_{j\ne i}Z_j\Big)^2e^{\lambda_1\sum_{j\ne i}W_j} \Big] \Big\}^{\frac 1 2} \left[ \e e^{\lambda_1\sum_{j\ne i}W_j}\right]^{\frac 1 2}.
\end{align*}

Recalling (<ref>) and (<ref>), we have

\begin{align}
\prod_{i = 1}^n \e [ e^{ \lambda_1 W_i } ]  \leq  \exp \{ m ( \lambda_1 ) \} e^{ A R_{x} } \label{part1}
\end{align}

and when $ x  $ satisfies (<ref>) with small constant $ c_1 $,

\begin{align}
1/2 \leq   \e [ e^{ \lambda_1 W_i } ] \leq 3 /2 .  \label{part2}
\end{align}

Further, through an analogous proof to (<ref>), we obtain

\begin{align*}
&  x^2  \e [ | W_i Z_i | e^{\lambda_1 W_i  }  | ] \\
& \leq   C \e \big[ (  x^3  | X_i  Y_i^2 | + x^4Y_i^4 + x^3 | X_i | \e Y_i^2  +  x^4 Y_i^2 \e Y_i^2 )  e^{ \lambda_1 W_i  }\big] \\
& \leq C \e \big[ (  x^3  | X_i  Y_i^2 | + x^4 Y_i^4 + x | X_i |   +  x^2 Y_i^2  )  e^{ \lambda_1 W_i  } \IF ( x  X_i   > 1 ) \big]  \\
& \hspace{0.5cm}  + C \e \big[ (  x^3  | X_i  Y_i^2 | + x^3 |Y_i|^3 + x^3 | X_i | \e Y_i^2  +  x^4 Y_i^2 \e Y_i^2 )  e^{ \lambda_1 W_i  } \IF ( x  X_i   \leq  1 ) \big] \\
& \leq   C \e \big[ e^{ \frac {10 \lambda_1 } {9} x X_i  - \frac { 9 \lambda_1} { 10  }  x^2 Y_i^2 } \IF ( x  X_i  > 1  ) \big] + C x^3 ( \e | X_i |^3 + \e | Y_i |^3 ),
\end{align*}

where in the last inequality we used the fact that $ ab^2 \leq a^3 + b^3 $	for $  a> 0, b> 0$ and $ x^4 ( \e [Y_i^2] ) ^2 \leq 2 \delta_{x, i }  $ by Lemma <ref>.
By (<ref>). we obtain for $ \lambda_{1} \leq 3/4 $,

\begin{align*}
\e \big[ e^{ \frac {10 \lambda_1 } {9} x X_i  - \frac { 9 \lambda_1} { 10  }  x^2 Y_i^2 } \IF ( x | X_i  | > 1  ) \big]  \leq C \e \Big[ e^{ \min \big \{  \frac { X_i^2  }  { Y_i^2 + c_0 \e Y_i^2  } , 2 x X_i \big\} } \IF ( x  X_i  > 1  )  \Big]  .
\end{align*}

\begin{align}
\e [ | W_i Z_i | e^{ \lambda_1 W_i  }   ] \leq x^{- 2} R_{x, i} + x ( \e | X_i |^3 + \e | Y_i |^3 ). \label{part3}
\end{align}

Similarly, we can obtain

\begin{align}
\e \big( | W_i Z_i^2  | e^{ \lambda_1 W_i  } \big) \leq C  x^{ - 4 } R_{x, i },  \label{part4}
\end{align}

which together with (<ref>) and  (<ref>) gives

\begin{align*}
Q_1  \leq   C_1 \exp \{ m(\lambda_1) \} x^{ - 4 } R_{x} e^{ C_2 R_{x} }.
\end{align*}

As for $ Q_2 $, noting the bound (<ref>)–(<ref>) and (<ref>), we have

\begin{align*}
Q_2  & \leq  C  ( x^{ - 2} R_x  + x L_{3, n} ) \big( \exp \{ m ( \lambda_1 ) \} x^{ - 2} R_x e^{ A R_{x} }\big)^{ 1/2  } \big( \exp \{ m ( \lambda_1 ) \}  e^{ A R_x }  \big)^{ 1/2 } \\
&  \leq C ( x^{ - 2 } R_x  + x L_{3, n} ) \exp \{ m ( \lambda_1 )  \} x^{^{ - 1 } }R_x^{ 1 /2  } e^{ A R_{x} }\\
& \leq C \exp {m (\lambda_1)} x^{- 2} R_x e^{ A R_x }.
\end{align*}

Here we apply the fact $ x^4 L_{3, n}  \leq 2 \delta_x \leq 2 R_x $ in Lemma <ref> to derive the last inequality. Thus (<ref>) is proved. The proof of Lemma <ref> is completed.

§.§ Proof of Lemma <ref>

Assume $A=1$ without loss of generality, as the proof for general $A> 0$ is similar. By Cauchy-Schwarz inequality, it is obvious that $ S_n^2 \le n V_n^2 $ and hence $ \e [ e^{ \frac { S_n^2  } { V_n^2 + c_0 B_n^2  } }   ] < \infty $. We have

\begin{align}
&  \e \Big[ e^{ \f {S_n^2} {V_n^2 + c_0 B_n^2}}  \IF  ( b |S_n| > 1 )   \Big]   \label{e-1} \\
& \hspace{0.5cm} = \int_{0}^{n} e^t \, \P \Big( \f { |S_n| } { \sqrt{ V_n^2  + c_0 B_n^2 } } > \sqrt{t}, b |S_n| > 1 \Big)  d t \nn \\
& \hspace{0.5cm} = \int_{0}^{n} e^t \, \Big[ \P \Big( \f { S_n } { \sqrt{ V_n^2  + c_0 B_n^2 } } > \sqrt{t}, b |S_n| > 1 \Big)  \nn   \\
& \hspace{2.5cm}+ \P \Big( \f { -S_n } { \sqrt{ V_n^2  + c_0 B_n^2 } } > \sqrt{t}, b |S_n| > 1 \Big) \Big]  d t. \nn
\end{align}

The integral in $ 0< t <1 $ is bounded by $ e b^p \e | S_n |^p $  and hence is dominated by the right hand side of (<ref>) by Rothenthal's inequality.
Thus it remains to consider the integral for $ t>1 $. Let $ \nu =   \f{ c B_n}{\sqrt{t}} $, where $ c $ is a large positive number. Denote $ \hatX_i = X_i \IF ( |X_i| \le \nu ) $, then

\begin{align}
&  \P \Big( \f { S_n } { \sqrt{ V_n^2  + c_0 B_n^2 } } > \sqrt{t}, b |S_n| > 1 \Big)  \label{e-4} \\
& \le \P \Big( \f {\hatS_n } { \sqrt{ V_n^2 + c_0 B_n^2 } } > \f{\sqrt{t}}{2} , b |S_n| >1  \Big)  \nn \\
& \hspace{2cm} + \P \Big( \f {\sum_{i=1}^{n} X_i \IF ( |X_i| > \nu ) } { \sqrt{ V_n^2 + c_0 B_n^2 } } > \f{\sqrt{t}}{2} , b |S_n| >1  \Big) \nn \\
& \le \P  \Big( \f{ \hatS_n } {\sqrt {c_0} B_n } > \f{\sqrt{t}}{2}, b |S_n| > 1  \Big) + \P  \Big(  \sum_{i=1}^{n} \IF (  |X_i| > \nu )> \f t4 , b |S_n| > 1  \Big) \nn \\
& \le b^p e^{ - \f { c_0^{1/4} t }{ 2 } } \e  \Big[ |S_n|^p \,  \exp \big\{ c_0^{ - \f 14 } \sqrt{t} \f{\hatS_n} { B_n } \big\}  \Big] + b^p e^{-2t} \e  \Big [ |S_n|^p e^{ 8\sum_{i=1}^{n} \IF ( |X_i| > \nu )  }  \Big] \nn \\
& :=  b^p e^{ - \f { c_0^{1/4} t }{ 2 } } E_1 + b^p e^{-2t} E_2 , \nn
\end{align}

where the second inequality is obtained by Cauchy inequality. Now we turn to estimating $  E_1 $ and $ E_2 $. Let $ A_{n, p} = \f{1}{2 (4e^8)^{1/p}} \max \{ B_n, (\sum_{i=1}^{n} \e |X_i|^p)^{1/p} \} $. Denote $ \bar{X_i} = X_i \IF  ( |X_i| \le A_{n, p} ) $, $\bar{S}_n = \sum_{i = 1}^n \bar{X}_i $ and $ S_n^{ (i) } = S_n - X_i $. Observing that

\begin{align}
\IF ( |S_n| > x ) & \le \IF ( | \bar{S}_n | > x ) + \sum_{i=1}^{n} \IF  ( |S_n| > x, | X_i | > A_{n, p} )  \\
& \le  \IF  ( | \bar{S}_n | > x ) + \sum_{i=1}^{n} \IF  ( |X_i| > \f 12 x)  \nn \\
&  \hspace{1cm} + \sum_{i=1}^{n} \IF  ( |S_n^{(i)}| > \f 12 x, | X_i | >  A_{n, p} ) , \nn
\end{align}

we have

\begin{align}
E_2 & = \int_{0}^{\infty} p x^{p-1} \e [ e^{8 \sum_{i=1}^n \IF  ( |X_i| > \nu ) } \IF ( |S_n| > x )  ]  d x   \label{E2} \\
& \le \int_{0}^{\infty} p x^{p-1} \e [ e^{8 \sum_{i=1}^n \IF ( |X_i| > \nu ) } \IF ( |\bar{S}_n| > x ) ] d  x \nn \\
& \quad  + \sum_{i=1}^n  \int_{0}^{\infty} p x^{p-1} \e [ e^{8 \sum_{i=1}^n \IF ( |X_i| > \nu ) } \IF  ( |X_i| > \f 12 x )  ] d  x \nn \\
& \quad + \sum_{i=1}^n  \int_{0}^{\infty} p x^{p-1} \e [ e^{8 \sum_{i=1}^n \IF ( |X_i| > \nu ) } \IF ( S_n^{(i)} > \f 12 x, |X_i | > A_{n,p}  ) ] d x \nn \\
& \le Q_1 + \sum_{i=1}^n   e^8 \e (e^{8 \sum_{j \neq i} \IF ( |X_j| > \nu ) }) \int_{0}^{\infty} p x^{p-1} \P (|X_i| > \f 12 x )   d x \nn \\
& \quad + \sum_{i=1}^n e^8 ({A_{n,p}})^{-p} \e |X_i|^p \int_{0}^{\infty} p x^{p-1} \e [ e^{8 \sum_{j \neq i} \IF ( |X_j| > \nu ) } \IF ( S_n^{(i)} > \f 12 x ) ]  d  x  \nn \\
& \le Q_1 + e^8 2^p \exp \{ \f t4 \}   \sum_{i=1}^n \e |X_i|^p + e^8 2^p A_{n,p}^{-p} (\sum_{i=1}^{n} \e |X_i|^p ) E_2  \nn \\
& \le Q_1 + e^8 2^p \exp \{ \f t4 \}   \sum_{i=1}^n \e |X_i|^p + \f 14 E_2 , \nn
\end{align}

\begin{equation*}
Q_1 = \int_{0}^{\infty} p x^{p-1} \e [ e^{8 \sum_{i=1}^n \IF ( |X_i| > \nu ) }  \IF ( |\bar{S}_n| > x )  ] d  x .
\end{equation*}

The second to last inequality is based on the fact that $ |x + u|^p $ is a convex function with respect to $x$ and that there exists some large constant $c$ such that for $ \nu =   \f{  c B_n}{\sqrt{t}} $,

\begin{align*}
\e e^{8 \sum_{i=1}^{n} \IF ( |X_i| > \nu  ) } &\le \textstyle \prod_{i=1}^n [ 1 + e^8 \P ( |X_i| > \nu ) ] \nn \\
& \le \exp \{  \textstyle  \sum_{i=1}^n e^8 \P ( |X_i| > \nu  ) \} \nn \\
& \le \exp \{ \f{e^8}{\nu^2} B_n^2  \} \le \exp \{ \f t4 \}.
\end{align*}

As for $ Q_1 $,   it holds that

\begin{equation}
Q_1 \le \int_{0}^{\infty} p x^{p-1} e^{- \f x A_{n,p}} \prod_{i=1}^{n} \e e^{ 8 \IF ( |X_i | > \nu  ) + \f {1}{A_{n,p}} \bar{X}_i}  d x . \label{e-3}
\end{equation}

Furthermore, as $ |\bar{X}_i| / A_{n,p} \leq 1  $,  we obtain from Taylor expansion that

\begin{align}
\lefteqn{ \prod_{i=1}^{n} \e e^{ 8 \IF ( |X_i | > \nu  ) + \f {1} {A_{n,p}} \bar{X}_i} } \label{e-2} \\
& \le \prod_{i=1}^{n} \e \left( 1 + e^8 \IF ( |X_i| > \nu ) \right) \left( 1 + \f {\bar{X}_i }{A_{n,p}} +   \f{e^2 \bar{X}_i^2}{ 2 A_{n,p}^2 } \right) \nn \\
& \le C \exp \Big\{   \sum_{i=1}^{n} \P( |X_i| > \nu ) +  \f{1}{A_{n,p}}   \e |X_i| \IF  ( |X_i| > \nu ) \nn \\
& \hspace{4cm} + \f {  1}{ A_{n,p}^2 } \e X_i^2 \IF ( |X_i| > \nu ) \Big\} \nn \\
& \le C \exp \Big\{ \f{  B_n^2}{\nu^2} +  \f { B_n^2}{A_{n,p} \nu} \sum_{i=1}^n \e |X_i|^p  + \f{1}{2 A_{n,p}^2 } B_n^2 \Big\}  \le A \exp \big\{ \f t4 \big\}, \nn
\end{align}

which yields

\begin{align}
Q_1 &\le A\, (A_{n,p})^p \,\exp\{ \f t4 \} \int_{0}^{\infty} p x^{p-1} e^{-x}  d  x \label{Q1} \\
&\le A_1 \exp\{ \f t4 \} \Big( \sum_{i=1}^n \e |X_i|^p + (\e S_n^2)^{\f p2} \Big). \nn
\end{align}

Consequently, it follows from (<ref>) and (<ref>) that

\begin{equation}
E_2 \le A_2 \exp \{ \f t4 \}  \Big[ \sum_{i=1}^{n} \e |X_i|^p + (\e S_n^2)^{\f p2}   \Big]. \label{E-2}
\end{equation}

Next we deal with $ E_1 $. Recalling that $ | \hat{X}_i  |\leq \nu = C B_n / \nu $, we have for large constant $c_0$

\begin{align}
\e \exp \Big\{ c_0^{- \f 14 } \sqrt{t} \f {\hatS_n}{B_n} \Big\}
& \le \exp \bigg\{ \sum_{i=1}^n \Big[ \f{\sqrt{t}}{ c_0^{1/4} B_n } \e |X_i| \IF ( |X_i| > \nu )  \\
& \hspace{1.5cm}+ \exp \{\f{C}{c_0^{1/4}}\} \f{t}{2 c_0^{1/2} B_n^2} \e X_i^2   \Big]  \bigg\} \nn \\
& \le \exp \bigg\{  \f{\sqrt{t} B_n^2 }{ c_0^{\f14} B_n \nu } + \exp  \Big\{\f{C}{c_0^{1/4}} \Big\} \f{t}{2 c_0^{1/2}}  \bigg\}  \nn \\
& \le \exp \big\{\f t4 \big\} \nn  .
\end{align}

Denote $ \hat{S}_n^{(i)} = \hat{S}_n - \hat{X}_i $ and

\begin{equation*}
Q_2 =  \int_0^{\infty} p x^{p-1} e^ {- x/a} \e \Big[\exp \Big\{ c_0^{- \f 14 } \sqrt{t} \f {\hatS_n}{B_n} + \f {\bar{S}_n}{A_{n,p}}  \Big\}  \Big] d x .
\end{equation*}

By a similar procedure to (<ref>)–(<ref>),   we have for large $c_0$

\begin{align*}
E_1 & = \int_0^{\infty} p x^{p-1} \e \Big[\exp \Big\{ c_0^{- \f 14 } \sqrt{t} \f {\hatS_n}{B_n}  \Big\} \IF ( |S_n| > x )  ] dx \nn \\
& \leq Q_2 + C_1 \sum_{i=1}^n \e \Big[ \exp \Big(  c_0^{- \f 14 } \sqrt{t} \f {\hatS_n}{B_n} \Big) \Big] \cdot \e |X_i|^p \nn \\
& \quad + 2^p e^{c/c_0^{\f{1}{4}}} \sum_{i=1}^n \f{\e |X_i|^p}{A_{n,p}^{p}} \cdot  \e \Big[ |S_n^{(i)}|^p \exp \Big(  c_0^{- \f 14 } \sqrt{t} \f {\hatS_n^{(i)}}{B_n} \Big)  \Big] \nn  \\
& \le Q_2 + C_2 \exp\{\f t4\} \sum_{i=1}^n \e |X_i|^p + \f 14 E_1.
\end{align*}

Moreover, it follows from Taylor expansion that

\begin{align}
& \le C (A_{n,p})^p \prod_{i=1}^n  \e \Bigg[ \bigg( 1 + \f {\sqrt{t}}{ c_0^{1/4} B_n } \hatX_i +   \f{ t e^ { c / c_0^{1/4} } }{2 c_0^{1/2} B_n^2 }  \hatX_i^2 \bigg)\bigg( 1 + \f{ e \bar{X}_i}{A_{n,p}} \bigg) \Bigg] \nn \\
& \le Ce^{t/4}(A_{n,p})^p \le C e^{t/4}\Big[ \sum_{i=1}^{n} \e |X_i|^p + (\e S_n^2)^{p/2}\Big]. \nn
\end{align}

\begin{equation*}
E_1 \le  C\exp\{ \f t4 \} \Big[ \sum_{i=1}^{n} \e |X_i|^p + (\e S_n^2)^{p/2} \Big] ,
\end{equation*}

which together with (<ref>), (<ref>) and (<ref>) gives the result (<ref>). The proof is completed.

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