\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{|\xi_{n+1}| \leq (\lambda - \delta)n\}}\Big]\\
&\leq \tilde{E} \Big[ \mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{\bar{\xi}_n \geq \delta n\}} \Big]\\
&\leq \Big( \tilde{E} \Big[ \mathrm{e}^{2\hat{\beta}_0 \tilde{L}_{n+1} + 2\hat{\beta}(n+1)} \Big]
\Big)^{\frac{1}{2}} \Big( \tilde{P} \big( \bar{\xi}_n \geq \delta n \big) \Big)^{\frac{1}{2}}\end{aligned}$$ using Cauchy-Schwarz inequality in the last line.Then as we know from $$\frac{1}{n} \log
\Big( \tilde{E} \Big[ \mathrm{e}^{2\hat{\beta}_0 \tilde{L}_{n+1} + 2\hat{\beta}(n+1)} \Big]
\Big)^{\frac{1}{2}} \to \hat{\beta}_0^2 + \hat{\beta} \qquad \text{ as }n \to \infty$$ while, since $\bar{\xi}_n \stackrel{d}{=} |\mathcal{N}(0,1)|$, $$\frac{1}{n^2} \log \Big( \tilde{P} \big( \bar{\xi}_n \geq \delta n \big) \Big)^{\frac{1}{2}} \to 
\frac{-\delta^2}{4} \qquad \text{ as } n \to \infty.$$ Thus $$\label{i1}
\limsup_{n \to \infty} \frac{1}{n^2} \log I_1 \leq - \frac{\delta^2}{4}.$$ On the other hand, $$\begin{aligned}
I_2 &= \tilde{E} \Big[ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{|\xi_{n+1}| > (\lambda - \delta)n\}}
\Big]\\
&\leq \tilde{E} \Big[ \mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{|\xi_{n+1}| > (\lambda - \delta)n\}}\Big]\\
&= 2\tilde{E} \Big[ \mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{\xi_{n+1} > (\lambda - \delta)n\}}\Big]\\
&= 2E|N_{n+1}^{(\lambda - \delta)n}|\end{aligned}$$ using symmetry in the third line and identity in the fourth line. Thus from we can see (just as we did in - ) that $$\label{i2}
\limsup_{n \to \infty} \frac{1}{n} \log I_2 \leq \Delta_{\lambda - \delta}.$$ From and we 