$t \in [n, n+1]$ it is always true that $|N_t^{\lambda t}| \leq |\hat{N}^{\lambda n}_n|$.

By the Many-to-One Lemma we have $$\begin{aligned}
E |\hat{N}^{\lambda n}_n| &= E \sum_{u \in N_{n+1}} 
\mathbf{1}_{\{\sup_{s \in [n, n+1]}X^u_s \geq \lambda n\}}\\
&= \tilde{E} \Big[ \mathbf{1}_{\{\sup_{s \in [n, n+1]}\xi_s \geq \lambda n\}} 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)}\Big]\\
&= \tilde{E} \Big[ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)}\Big],\end{aligned}$$ where $\bar{\xi}_n := \sup_{s \in [n, n+1]} (\xi_s - \xi_{n + 1})$ and $\bar{\xi}_n \stackrel{d}{=} \sup_{s \in [0,1]} \xi_s \stackrel{d}{=} |\mathcal{N}(0,1)|$ under $\tilde{P}$.

Then for any $\delta \in (0, \lambda)$ we can split the latter expectation as $$\begin{aligned}
 \tilde{E} \Big[ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)}\Big] &= 
 \tilde{E} \Big[ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{|\xi_{n+1}| \leq (\lambda - \delta)n\}}
\Big]\\
&+ \tilde{E} \Big[ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} 
\mathrm{e}^{\hat{\beta}_0 \tilde{L}_{n+1} + \hat{\beta}(n+1)} 
\mathbf{1}_{\{|\xi_{n+1}| > (\lambda - \delta)n\}}
\Big]\end{aligned}$$ We shall refer to the first term in the sum as $I_1$ and the second one as $I_2$. First we show that the contribution of $I_1$ is negligibly small as it has a faster than exponential decay rate. $$\begin{aligned}
I_1 &= \tilde{E} \Big[ \mathbf{1}_{\{\xi_{n+1} + \bar{\xi}_n \geq \lambda n\}} 
