the both sides of ([Adr1]{}) and recalling Lemma \[min2\] we see that$$\lim \sup_{n\rightarrow \infty }\mathbf{P}\left( \alpha _{n}(\tau
(nt,n)+R)>1+\varepsilon ;\tau \left( n\right) <nt\right) =0.$$Hence the statement of the lemma follows. $\square $

\[Lprospect2\]Assume that $A1-$ $A2$ hold. Then for any $R\in \mathbb{Z}$, any $t\in (0,1),$ and $N\in \{n-1,$ $n\}$$$\left\{ f_{\tau (nt,n)+R,N}(0)|\tau (n)<nt\right\} \overset{d}{\rightarrow }\theta _{R}\text{ as }n\rightarrow \infty ,$$where $\theta _{R}$ is the same as in Theorem \[Tclose\].

**Proof**. Consider first $R<0.$ Clearly, $$\begin{aligned}
&&\mathbf{E}\left[ e^{-\lambda f_{\tau (nt,n)+R,N}(0)};\tau (n)<nt\right]  \\
&=&\mathbf{E}\left[ e^{-\lambda f_{\tau (nt,n)+R,N}(0)}\right] -\mathbf{E}\left[ e^{-\lambda f_{\tau (nt,n)+R,N}(0)};\tau (n)\geq nt\right] .\end{aligned}$$Introduce two independent environmental sequences $\left\{ f_{n}^{-},n\geq
0\right\} $ and $\left\{ f_{n}^{+},n\geq 0\right\} $ and the respective associated random walks $\left\{ S_{n}^{-},n\geq 0\right\} $ and $\left\{
S_{n}^{+},n\geq 0\right\} $. In the notation of Lemmas \[Lfran\] and [Cfran]{} we set $$T_{\tau (n(1-t)),n(1-t)}^{\ast }:=e^{-\lambda f_{\tau
(n(1-t))+R,N-n(1-t)}(0)},\,T_{l,n-l}:=e^{-\lambda
f_{R,0}^{-}(f_{0,N-n(1-t)-l}^{+}(0))}.$$Observing that** **$$e^{-\lambda f_{\tau (n(1-t))+R,N-n(1-t)}(0)}\overset{d}{=}e^{-\lambda
f_{\tau (nt,n)+R,N}(0)}$$and$$f_{R,0}^{-}(f_{0,N-n(1-t)-l}(0))\rightarrow f_{R,0}^{-}(q^{+})$$$\mathbf{\hat{P}}$ $-$a.s. as $\min (l,N-n(1-t)-l)\rightarrow \infty $ we conclude by Lemma \[Cfran\] and the generalized arcsine law that $$\begin{aligned}
\mathbf{E}\left[ e^{-\lambda f_{\tau (nt,n)+R,N}(0)}\right]  &=&\mathbf{E}\left[ e^{-\lambda f_{\tau (n(1-t))+R,N-n(1-t)}(0)}\right]  \\
&\rightarrow &\mathbf{\hat{E}}\left[ e^{-\lambda f_{R,0}^{-}(q^{+})}\right]
,\,n\rightarrow \infty .\end{aligned}$$Further we have $$\begin{aligned}
\lefteqn{\mathbf{E}\left[ e^{-\lambda f_{\tau (nt,n)+R,N}(0)};\tau (n)\geq nt\right] =\mathbf{E}\left[ e^{-\lambda f_{\tau (nt,n)+R,N}(0)};\tau (n)\geq nt\right] } \\
&=&\sum_{j=nt}^{n}\mathbf{E}\left[ e^{-\lambda f_{\tau (n)+R,N}(0)};\tau
(n)=j\right]  \\
&=&\sum_{j=nt}^{n}\mathbf{E}\left[ e^{-\lambda 