-\lambda \frac{\zeta ^{-}(f_{0,R}^{+}(0))}{e^{S_{R}^{+}}}\right\} \right] .$$This and the generalized arcsine law give$$\begin{aligned}
\lefteqn{\lim_{n\rightarrow \infty }\mathbf{E}\left[ \exp \left\{ -\lambda 
\frac{1-f_{0,\tau (nt)+R}(0)}{e^{S_{\tau (nt)+R}}}\right\} ;\tau (n)<nt\right] } \\
&=&\mathbf{\hat{E}}\left[ \exp \left\{ -\lambda \frac{\zeta
^{-}(f_{0,R}^{+}(0))}{e^{S_{R}^{+}}}\right\} \right] \mathbf{P}\left( \tau
\leq t\right) .\end{aligned}$$Thus,$$\begin{aligned}
\lefteqn{\lim_{n\rightarrow \infty }\mathbf{E}\left[ \exp \left\{ -\lambda 
\frac{1-f_{0,\tau (nt)+R}(0)}{e^{S_{\tau (nt)+R}}}\right\} ;\tau (n)\geq nt\right] } \\
&=&\mathbf{\hat{E}}\left[ \exp \left\{ -\lambda \frac{\zeta
^{-}(f_{0,R}^{+}(0))}{e^{S_{R}^{+}}}\right\} \right] \mathbf{P}\left( \tau
\geq t\right)\end{aligned}$$and the statement of the lemma for $R\geq 0$ follows. The case $R<0$ can be treated in a similar way.$\square $

\[EstOprel\]Assume that $A1-$ $A2$ hold. For any $R\in \mathbb{Z}$, any $t\in (0,1)$ and $N\in \left\{ n-1,n\right\} $$$\left\{ O_{\tau (nt)+R,N}|\tau (n)\geq nt\right\} \overset{d}{\rightarrow }\frac{1}{\Theta _{R}}-1.$$

**Proof**. We have $$O_{\tau (nt)+R,N}=f_{\tau (nt)+R,N}(0)\beta _{N}(\tau (nt)+R)\left( \frac{e^{S_{\tau (nt)+R}}}{1-f_{0,\tau (nt)+R}(0)}-1\right) .$$Applying Lemmas \[Lpreelimit1\], \[Lprelim3\], and \[Lprelim4\] we get$$\left\{ O_{\tau (nt)+R,N}|\tau (n)\geq nt\right\} \overset{d}{\rightarrow }\frac{1}{\Theta _{R}}-1$$as desired.$\square$

\[DeltaVanish\]Assume that $A1-$ $A2$ hold. Then$$\Delta _{m,n}\overset{p}{\rightarrow }1\text{ as }\min \left( m,n-m\right)
\rightarrow \infty .$$

**Proof**. Clearly, it sufficies to show that$$\frac{1-f_{0,n}(0)}{1-f_{0,n-1}(0)}\overset{p}{\rightarrow }1\text{ as }n\rightarrow \infty .  \label{ratio1}$$To verify this write$$T_{\tau (n-1),n-1}^{\ast }:=\frac{1-f_{0,\tau (n-1)}(f_{\tau (n-1),n}(0))}{1-f_{0,\tau (n-1)}(f_{\tau (n-1),n-1}(0))}$$and$$T_{l,n-l-1}:=\frac{1-f_{l,0}^{-}(f_{0,n-l-1}^{+}(f_{n-1}(0)))}{1-f_{l,0}^{-}(f_{0,n-l-1}^{+}(0))}.$$Since, evidently, there exists $s\in \lbrack 0,1)$ such that $$\lim_{n-l\rightarrow \infty }f_{n-l-1}^{+}(s)\neq s,$$it follows from Lemma 1’ and Theorem 5 in [@AK71] and (\[asq\]) that $\mathbf{\hat{P}}^{+}$ a.s.$$\lim_{n-l\rightarrow \infty
}f_{0,n-l-1}^{+}(f_{n-1}(0))=\lim_{n-l\rightarrow \infty
}f_{0,n-l-1}^{+}(0)=q^{+}.$$Hence we 