(j,nt-j)\rightarrow \infty $. Hence we see by Lemma \[Lfran\] that $$\lim_{\min (j,nt-j)\rightarrow \infty }\mathbf{E}\left[ \exp \left\{
-\lambda \frac{1-f_{j,0}^{-}(f_{0,nt-j}^{+}(0))}{e^{S_{j}^{-}}}\right\}
\left\vert A_{j,n-j}\right. \right] =\mathbf{\hat{E}}\left[ \exp \left\{
-\lambda \zeta \right\} \right] .$$This and the generalized arcsine law give$$\begin{aligned}
\lefteqn{\lim_{n\rightarrow \infty }\mathbf{E}\left[ \exp \left\{ -\lambda 
\frac{1-f_{j,0}^{-}(f_{0,nt-j}^{+}(0))}{e^{S_{j}^{-}}}\right\} ;\tau (n)<nt\right] } \\
&=&\mathbf{\hat{E}}\left[ \exp \left\{ -\lambda \zeta \right\} \right] 
\mathbf{P}\left( \tau \leq t\right) .\end{aligned}$$Thus,$$\begin{aligned}
\lefteqn{\lim_{n\rightarrow \infty }\mathbf{E}\left[ \exp \left\{ -\lambda 
\frac{1-f_{0,nt}(0)}{e^{S_{\tau (nt)}}}\right\} ;\tau (n)\geq nt\right] } \\
&=&\mathbf{\hat{E}}\left[ \exp \left\{ -\lambda \zeta \right\} \right] 
\mathbf{P}\left( \tau \geq t\right) \end{aligned}$$and the first statement of the lemma follows. The second statement can be checked in a similar way. $\square $

\[LOinf\] Under Assumptions $A1-$ $A2$, for any $t\in (0,1)$$$O_{nt,n}\overset{d}{\rightarrow }\infty \text{ \ as }n\rightarrow \infty .$$

**Proof**. This statement is a direct corollary of Lemmas [LOleftRight]{} and \[Comm1\].$\square $

**Proof of Theorem \[Tfar\].** Using Lemma \[BasicINeq\] we have $$\begin{aligned}
\frac{1}{1+O_{nt,n-1}(1-e^{-\lambda /O_{nt,n-1}})}\Delta _{nt,n}^{-1} &\leq &\mathbf{E}_{\mathbf{k}}\left[ \exp \left\{ -\lambda \frac{Z_{nt}}{O_{nt,n}}\right\} \left\vert T=n\right. \right] \\
&\leq &\frac{1}{1+O_{nt,n}(1-e^{-\lambda /O_{nt,n}})}\Delta _{nt,n}.\end{aligned}$$Now to complete the proof of the theorem it remains to observe that $$\lim_{n\rightarrow \infty }\Delta _{nt,n}\overset{p}{=}1$$by Lemmas \[Lratio\] and \[LOinf\] and (\[c3\]), and that by Lemma [LOinf]{} $$\lim_{n\rightarrow \infty }O_{nt,n}(1-e^{-\lambda /O_{nt,n}})\overset{d}{=}\lim_{n\rightarrow \infty }O_{nt,n-1}(1-e^{-\lambda /O_{nt,n}})\overset{d}{=}\lambda .$$

\[roughlyspeaking\] It follows from Lemmas \[LOleftRight\] and 