Appendix 1: Proofs of One-Dimensional Results
Proof of Lemma 2
For any \(i\ge 1\), let \(S_i=-\sum _{j=1}^i\log (A_jA_{j-1})\) and define \(S_0=0\). As \(i\mapsto {\tilde{P}}_i^{\omega }({\tilde{\tau }}_{-1}> {\tilde{\tau }}_n)\) is the solution to the Dirichlet problem
$$\begin{aligned} {\left\{ \begin{array}{ll} \varphi (-1)=0,\ \varphi (n)=1\\ {\tilde{E}}^{\omega }_i(\varphi ({\tilde{\eta }}_1))=\varphi (i) &{} i\in {\llbracket } 0,n-1{\rrbracket }. \end{array}\right. } \end{aligned}$$
It follows that
$$\begin{aligned} {\tilde{P}}_{i}^{\omega }({\tilde{\tau }}_{-1}>{\tilde{\tau }}_{n})=\frac{\sum _{j=0}^{i}\exp (S_{j})}{\sum _{j=0}^{n}\exp (S_{j})}. \end{aligned}$$
(38)
As a consequence, for any \(0\le l\le n\),
$$\begin{aligned}&{\tilde{P}}_0^{\omega }({\tilde{\tau }}_l<{\tilde{\tau }}_{-1})=\frac{1}{\sum _{j=0}^l\exp (S_j)}\ge \frac{\exp \left( -\max _{0\le j\le l}S_j\right) }{l+1}\\&\quad {\tilde{P}}_{l+1}^{\omega }({\tilde{\tau }}_n<{\tilde{\tau }}_l)=\frac{\exp (S_{l+1})}{\sum _{j=l+1}^{n}\exp (S_{j})}\le \exp \left( -\max _{l+1\le j\le n}(S_j-S_{l+1})\right) \\&\quad {\tilde{P}}_{l-1}^{\omega }({\tilde{\tau }}_{-1}<{\tilde{\tau }}_l)=\frac{\exp (S_{l})}{\sum _{j=0}^l\exp (S_j)}\le \exp \left( -\max _{0\le j\le l}(S_j-S_l)\right) . \end{aligned}$$
We only need to consider \(n\) large, take \(l=\lfloor z_1n\rfloor \), note that
$$\begin{aligned} {\tilde{P}}_l^{\omega }({\tilde{\tau }}_l^*>{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_n)&=p(l,l+1){\tilde{P}}_{l+1}^{\omega }({\tilde{\tau }}_n<{\tilde{\tau }}_l)+p(l,l-1){\tilde{P}}_{l-1}^{\omega }({\tilde{\tau }}_{-1}<{\tilde{\tau }}_l)\\&\le \max ({\tilde{P}}_{l+1}^{\omega }({\tilde{\tau }}_n<{\tilde{\tau }}_l),{\tilde{P}}_{l-1}^{\omega }({\tilde{\tau }}_{-1}<{\tilde{\tau }}_l)). \end{aligned}$$
Therefore,
$$\begin{aligned}&{\tilde{P}}_0^{\omega }\left( {\tilde{\tau }}_n\wedge {\tilde{\tau }}_{-1}>m\right) \\&\quad \ge {\tilde{P}}_0^{\omega }\left( {\tilde{\tau }}_l<{\tilde{\tau }}_{-1}\right) {\tilde{P}}_l^{\omega }\left( {\tilde{\tau }}_l^*<{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_n\right) ^m\\&\quad \ge \frac{\exp \left( -\max _{0\le j\le l}S_j\right) }{l+1}\left( 1-{\tilde{P}}_l^{\omega }\left( {\tilde{\tau }}_l^*\ge {\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_n\right) \right) ^m\\&\quad \ge \frac{\exp \left( -\max _{0\le j\le l}S_j\right) }{l+1} \left( 1-\exp \left( -\max _{l+1\le k\le n}\left( S_k-S_{l+1}\right) \wedge \max _{0\le k\le l}\left( S_k-S_l\right) \right) \right) ^m\\&\quad \ge \frac{\mathbbm {1}_{\max _{0\le k\le l}S_k\le 0}}{l+1} \left( 1-e^{-zn}\right) ^m \mathbbm {1}_{\max _{l+1\le k\le n}\left( S_k-S_{l+1}\right) \ge zn}\mathbbm {1}_{\max _{0\le k\le l}\left( S_k-S_l\right) \ge zn}. \end{aligned}$$
As \(m\approx e^{zn}\), we have \((1-e^{-zn})^m=O(1)\); taking expectation under \({\mathbf {P}}(\cdot |A_0\in [a,\frac{1}{a}])\) yields
$$\begin{aligned}&{\tilde{{\mathbb {P}}}}_0\left( {\tilde{\tau }}_n\wedge {\tilde{\tau }}_{-1}>m|A_0\in \left[ a,\frac{1}{a}\right] \right) \\&\quad \ge \frac{c}{n}{\mathbf {P}}\left( \max _{0\le k\le l}S_k\le 0,\ \max _{0\le k\le l}\left( S_k-S_l\right) \ge zn |A_0\in \left[ a,\frac{1}{a}\right] \right) \\&\qquad \times {\mathbf {P}}\left( \max _{l+1\le k\le n}\left( S_k-S_{l+1}\right) \ge zn\right) \\&\quad \ge \frac{c}{n}{\mathbf {P}}\left( \max _{0\le k\le l}S_k\le 0,\ S_l\le -zn |A_0\in \left[ a,\frac{1}{a}\right] \right) {\mathbf {P}}\left( \left( S_n-S_{l+1}\right) \ge zn\right) . \end{aligned}$$
For \(k\ge 1\), write \({\mathscr {S}}_k=-\sum _{i=1}^k \log A_i\), then as \(S_k=-\log A_0 +{\mathscr {S}}_{k-1}+{\mathscr {S}}_k\),
$$\begin{aligned}&{\mathbf {P}}\left( \max _{0\le k\le l}S_k\le 0,\ S_l\le -zn |A_0\in \left[ a,\frac{1}{a}\right] \right) \\&\quad \ge {\mathbf {P}}\left( A_0\ge 1,A_l\ge 1, \max _{1\le k\le l-1}{\mathscr {S}}_k\le 0,\ {\mathscr {S}}_{l-1}\le -\frac{zn}{2}|A_0\in \left[ a,\frac{1}{a}\right] \right) \\&\quad ={\mathbf {P}}\left( A_0\ge 1|A_0\in \left[ a,\frac{1}{a}\right] \right) {\mathbf {P}}\left( A_l\ge 1\right) {\mathbf {P}}\left( \max _{1\le k\le l-1}{\mathscr {S}}_k\le 0,\ {\mathscr {S}}_{l-1}\le -\frac{zn}{2}\right) \end{aligned}$$
note that
$$\begin{aligned} {\mathbf {P}}\left( \max _{1\le k\le l-1}{\mathscr {S}}_k\le 0,\ {\mathscr {S}}_{l-1}\le -\frac{zn}{2} \right) \ge \frac{1}{l}{\mathbf {P}}\left( {\mathscr {S}}_{l-1}\le -\frac{zn}{2}\right) \end{aligned}$$
and
$$\begin{aligned} S_n-S_{l+1}=-\log A_{l+1}-\log A_n-2\sum _{k=l+2}^{n-1}\log A_k. \end{aligned}$$
Therefore,
$$\begin{aligned}&{\tilde{{\mathbb {P}}}}_0\left( {\tilde{\tau }}_n\wedge {\tilde{\tau }}_{-1}>m|A_0\in \left[ a,\frac{1}{a}\right] \right) \\&\quad \ge \frac{c}{n^2}{\mathbf {P}}\left( {\mathscr {S}}_{l-1}\le -\frac{zn}{2}\right) {\mathbf {P}}\left( S_n-S_{l+1}\ge zn\right) \\&\quad \ge \frac{c}{n^2}{\mathbf {P}}\left( {\mathscr {S}}_{l-1}\le -\frac{zn}{2}\right) {\mathbf {P}}\left( A_{l+1}\le 1\right) {\mathbf {P}}\left( A_n\le 1\right) {\mathbf {P}}\left( -\sum _{k=l+2}^{n-1}\log A_k\ge \frac{zn}{2}\right) \\&\quad \ge \frac{c}{n^2}{\mathbf {P}}\left( {\mathscr {S}}_{l-1}\le -\frac{zn}{2}\right) {\mathbf {P}}\left( -\sum _{k=l+2}^{n-1}\log A_k\ge \frac{zn}{2}\right) \\&\quad \ge \frac{c}{n^{2}}{\mathbf {P}}\left( \sum _{k=1}^{l-1}\log A_{k}\ge \frac{zn}{2}\right) {\mathbf {P}}\left( \sum _{k=l+2}^{n-1}\log A_{k}\le -\frac{zn}{2}\right) \end{aligned}$$
Applying Cramér’s theorem to sums of i.i.d. random variables \(\log A_{k}\), we have
$$\begin{aligned}&{\tilde{{\mathbb {P}}}}_0\left( {\tilde{\tau }}_n\wedge {\tilde{\tau }}_{-1}>m|A_0\in \left[ a,\frac{1}{a}\right] \right) \\&\quad \gtrsim _{n} \exp \left( -n\left( z_1I\left( \frac{z}{2z_1}\right) +\left( 1-z_1\right) I\left( \frac{-z}{2\left( 1-z_1\right) }\right) \right) \right) \end{aligned}$$
where \(I(x)=\sup _{t\in {\mathbb {R}}}\{tx-\log {\mathbf {E}}(A^t)\}\) is the associated rate function. \(\square \)
Proof of Lemma 3
Replace \(I(\frac{-z}{2(1-z_{1})})\) using
$$\begin{aligned} I(-x)&=\sup _{t\in {\mathbb {R}}}\{-tx-\log {\mathbf {E}}(A^{t})\}=\sup _{t\in {\mathbb {R}}}\{-tx-\log {\mathbf {E}}(A^{1-t})\}\\&=\sup _{s\in {\mathbb {R}}}\{-(1-s)x-\log {\mathbf {E}}(A^{s})\}=I(x)-x. \end{aligned}$$
For fixed \(z\), by convexity of the rate function \(I\), the supremum of \(-z_{1}I(\frac{z}{2z_{1}})-(1-z_{1})I(\frac{z}{2(1-z_{1})})\) is obtained when \(z_1=\frac{1}{2}\); we are left to compute
$$\begin{aligned} \sup _{0<z}\left\{ \frac{\log q_1-I(z)}{z}+\frac{1}{2}\right\} , \end{aligned}$$
and clearly, \(\frac{\log q_1-I(z)}{z}\le -t^*\), when \(z\) is such that \((t\mapsto \log {\mathbf {E}}(A^t))'(t^*)=z>0\), the maximum is obtained. \(\square \)
Proof of Lemma 4
Observe that
$$\begin{aligned}&{\widetilde{P}}^{\omega }_{Y_1}({\tilde{\tau }}_y<{\tilde{\tau }}_{\overleftarrow{Y_1}}){\widetilde{G}}^{{\tilde{\tau }}_{Y_1}\wedge {\tilde{\tau }}_{Y_3}}(y,y)={\widetilde{P}}^{\omega }_{Y_1}({\tilde{\tau }}_y<{\tilde{\tau }}_{\overleftarrow{Y_1}}\wedge {\tilde{\tau }}_{Y_3}){\widetilde{E}}^{\omega }_y\bigg [\sum _{k=0}^{{\tilde{\tau }}_{Y_1}\wedge {\tilde{\tau }}_{Y_3}}1_{\{{\widetilde{\eta }}_k=y\}}\bigg ]\\&\quad \le {\widetilde{P}}^{\omega }_{Y_1}({\tilde{\tau }}_y<{\tilde{\tau }}_{\overleftarrow{Y_1}}\wedge {\tilde{\tau }}_{Y_3}){\widetilde{E}}^{\omega }_y\bigg [\sum _{k=0}^{{\tilde{\tau }}_{\overleftarrow{Y_1}}\wedge {\tilde{\tau }}_{Y_3}}1_{\{{\widetilde{\eta }}_k=y\}}\bigg ]={\widetilde{E}}^{\omega }_{Y_1}\bigg [\sum _{k=0}^{{\tilde{\tau }}_{\overleftarrow{Y_1}}\wedge {\tilde{\tau }}_{Y_3}}1_{\{{\widetilde{\eta }}_k=y\}}\bigg ]. \end{aligned}$$
Obviously,
$$\begin{aligned} {\widetilde{E}}^{\omega }_{Y_1}\bigg [\sum _{k=0}^{{\tilde{\tau }}_{\overleftarrow{Y_1}}\wedge {\tilde{\tau }}_{Y_3}}1_{\{{\widetilde{\eta }}_k=y\}}\bigg ]\le {\widetilde{E}}^{\omega }_{Y_1}\bigg [{\tilde{\tau }}_{\overleftarrow{Y_1}}\wedge {\tilde{\tau }}_{Y_3}\bigg ]. \end{aligned}$$
This gives us (11).
Moreover, to get (12), we only need to show that for any \(0\le p<m\), we have
$$\begin{aligned} {\widetilde{E}}^{\omega }_p[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]\le 1+A_{p}A_{p+1}+A_pA_{p+1}{\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_{p}\wedge {\tilde{\tau }}_m]. \end{aligned}$$
(39)
In fact, since \(0\le \lambda \le 1\), (39) implies that
$$\begin{aligned} {\widetilde{E}}^{\omega }_p[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]^{\lambda }\le 1+(A_{p}A_{p+1})^{\lambda }+(A_pA_{p+1})^{\lambda }{\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_{p}\wedge {\tilde{\tau }}_m]^{\lambda }. \end{aligned}$$
Applying this inequality a few times along the interval \({\llbracket }Y_1, \, Y_3{\rrbracket }\), we obtain (12). It remains to show (39). Observe that
$$\begin{aligned}&{\widetilde{E}}^{\omega }_p[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]={\widetilde{\omega }}(p,p-1)+{\widetilde{\omega }}(p,p+1)(1+{\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m])\\&\quad =1+{\widetilde{\omega }}(p,p+1){\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]\\&\quad =1+{\widetilde{\omega }}(p,p+1)\Big ({\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_m; {\tilde{\tau }}_m<{\tilde{\tau }}_p]+{\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_p; {\tilde{\tau }}_p< {\tilde{\tau }}_m]\\&\qquad +{\widetilde{P}}^{\omega }_{p+1}({\tilde{\tau }}_p<{\tilde{\tau }}_m){\widetilde{E}}^{\omega }_p[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]\Big ). \end{aligned}$$
It follows that
$$\begin{aligned} {\widetilde{E}}^{\omega }_p[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]&=\frac{1+{\widetilde{\omega }}(p,p+1){\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_p\wedge {\tilde{\tau }}_m]}{1-{\widetilde{\omega }}(p,p+1){\widetilde{P}}^{\omega }_{p+1}({\tilde{\tau }}_p<{\tilde{\tau }}_m)}\\&=\frac{1+{\widetilde{\omega }}(p,p+1){\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_p\wedge {\tilde{\tau }}_m]}{{\widetilde{\omega }}(p,p-1)+{\widetilde{\omega }}(p,p+1){\widetilde{P}}^{\omega }_{p+1}({\tilde{\tau }}_m<{\tilde{\tau }}_p)}\\&\le \frac{1+{\widetilde{\omega }}(p,p+1){\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_p\wedge {\tilde{\tau }}_m]}{{\widetilde{\omega }}(p,p-1)}. \end{aligned}$$
Therefore,
$$\begin{aligned} {\widetilde{E}}^{\omega }_p[{\tilde{\tau }}_{p-1}\wedge {\tilde{\tau }}_m]\le (1+A_pA_{p+1})+A_pA_{p+1}{\widetilde{E}}^{\omega }_{p+1}[{\tilde{\tau }}_p\wedge {\tilde{\tau }}_m]. \end{aligned}$$
\(\square \)
Proof of Lemma 5
Recall that \({\mathbf {E}}[A^t]<\infty \) for any \(t\in {\mathbb {R}}\). By Hölder’s inequality, it suffices to show that there exists some \(\delta '>0\) such that for all \(n\) large enough,
$$\begin{aligned} {\mathbf {E}}\Big [\Big ({\tilde{E}}^\omega _0[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]\Big )^{\lambda (1+\delta ')}\Big ]\le (q_1+\delta )^{-n}. \end{aligned}$$
(40)
It remains to prove (40). In fact, we only need to show that for \(1>\lambda '=\lambda (1+\delta )>0\),
$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\log {\mathbf {E}}\Big [\Big ({\tilde{E}}^\omega _0[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]\Big )^{\lambda '}\Big ]}{n}\le \psi (\lambda '+1/2) \end{aligned}$$
(41)
where \(\psi (t)=\log {\mathbf {E}}(A^{t})\). One therefore sees that if \(t^*-1/2>\lambda '\), then \(\psi (\lambda '+1/2)<\psi (t^*)=-\log q_1\). To show (41), recall that for any \(0\le i\le n-1\),
$$\begin{aligned} {\widetilde{G}}^{{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_n}(i,i)= & {} {\widetilde{E}}^\omega _i\Big [\sum _{k=0}^{{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_n}1_{\eta =i}\Big ]\\= & {} \frac{1}{1-{\widetilde{\omega }}(i,i-1){\widetilde{P}}_{i-1}({\tilde{\tau }}_i<{\tilde{\tau }}_{-1})-{\widetilde{\omega }}(i,i+1){\widetilde{P}}_{i+1}({\tilde{\tau }}_i<{\tilde{\tau }}_n)}. \end{aligned}$$
Then, \({\widetilde{E}}^\omega _{0}[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]=1+\sum _{i=0}^{n-1}{\widetilde{P}}^\omega _0({\tilde{\tau }}_i<{\tilde{\tau }}_{-1}){\widetilde{G}}^{{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_n}(i,i)\) implies that
$$\begin{aligned} {\widetilde{E}}^\omega _{0}[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]=1+\sum _{i=0}^{n-1}\frac{{\widetilde{P}}^\omega _0({\tilde{\tau }}_i<{\tilde{\tau }}_{-1})}{{\widetilde{\omega }}(i,i-1){\widetilde{P}}^\omega _{i-1}({\tilde{\tau }}_{-1}<{\tilde{\tau }}_i)+{\widetilde{\omega }}(i,i+1){\widetilde{P}}^\omega _{i+1}({\tilde{\tau }}_n<{\tilde{\tau }}_i)}. \end{aligned}$$
Recall that by (38), if \(S_i:=\sum _{j=1}^i -\log (A_{j-1}A_j)\) for \(i\ge 1\) and \(S_0=0\), then
$$\begin{aligned} {\widetilde{P}}^\omega _0({\tilde{\tau }}_i<{\tilde{\tau }}_{-1})&=\frac{1}{\sum _{k=0}^i e^{S_k}}\\ {\widetilde{P}}^\omega _{i-1}({\tilde{\tau }}_{-1}<{\tilde{\tau }}_i)&=\frac{e^{S_i}}{\sum _{k=0}^i e^{S_k}}\\ {\widetilde{P}}^\omega _{i+1}({\tilde{\tau }}_n<{\tilde{\tau }}_i)&=\frac{1}{\sum _{k=i+1}^n e^{S_k-S_{i+1}}}. \end{aligned}$$
It is immediate that
$$\begin{aligned}&\frac{{\widetilde{P}}^\omega _0({\tilde{\tau }}_i<{\tilde{\tau }}_{-1})}{{\widetilde{\omega }}(i,i-1){\widetilde{P}}^\omega _{i-1}({\tilde{\tau }}_{-1}<{\tilde{\tau }}_i)+{\widetilde{\omega }}(i,i+1){\widetilde{P}}^\omega _{i+1}({\tilde{\tau }}_n<{\tilde{\tau }}_i)}\\&\quad =\frac{\frac{1}{\sum _{k=0}^i e^{S_k}}}{\frac{1}{1+A_iA_{i+1}}\frac{e^{S_i}}{\sum _{k=0}^i e^{S_k}}+\frac{A_iA_{i+1}}{1+A_iA_{i+1}} \frac{1}{\sum _{k=i+1}^n e^{S_k-S_{i+1}}}}\\&\quad \le \frac{1}{\frac{1}{1+A_iA_{i+1}}\frac{e^{S_i}}{\sum _{k=0}^i e^{S_k}}+\frac{A_iA_{i+1}}{1+A_iA_{i+1}} \frac{1}{\sum _{k=i+1}^n e^{S_k-S_{i+1}}}}. \end{aligned}$$
Let \(X_k=-\log A_k\). For any \(0\le i\le n\), define
$$\begin{aligned} H_i(-X)&:=\max _{0\le j\le i}\left( -X_j-X_{j+1}-\cdots -X_{i-1}\right) ,\\ H_{n-i-1}(X)&:=\max _{i+2\le j\le n}\left( X_{i+2}+\cdots +X_{j}\right) . \end{aligned}$$
Note that
$$\begin{aligned} S_k-S_i\le 2H_{i}(-X)+(-X_i)_+, \quad \forall 0\le k\le i, \end{aligned}$$
and that
$$\begin{aligned} S_k-S_{i+1}\le 2H_{n-i-1}(X)+(X_{i+1})_+, \quad \forall i+1\le k\le n. \end{aligned}$$
Then,
$$\begin{aligned} \frac{1}{1+A_iA_{i+1}}\frac{e^{S_i}}{\sum _{k=0}^i e^{S_k}}\ge & {} \frac{1}{1+A_iA_{i+1}}\frac{1}{(1+i)e^{2H_i(-X)+(-X_i)_+}}\\\ge & {} \frac{1}{n(A_i+1)(1+A_iA_{i+1})}e^{-2H_i(-X)}. \end{aligned}$$
Similarly,
$$\begin{aligned} \frac{A_iA_{i+1}}{1+A_iA_{i+1}} \frac{1}{\sum _{k=i+1}^n e^{S_k-S_{i+1}}}\ge \frac{(A_{i+1}\wedge 1)A_iA_{i+1}}{n(1+A_iA_{i+1})}e^{-2H_{n-i-1}(X)}. \end{aligned}$$
So,
$$\begin{aligned}&\frac{1}{1+A_iA_{i+1}}\frac{e^{S_i}}{\sum _{k=0}^i e^{S_k}}+\frac{A_iA_{i+1}}{1+A_iA_{i+1}} \frac{1}{\sum _{k=i+1}^n e^{S_k-S_{i+1}}} \\&\quad \ge \frac{1}{n(A_i+1)(1+A_iA_{i+1})}e^{-2H_i(-X)}+ \frac{(A_{i+1}\wedge 1)A_iA_{i+1}}{n(1+A_iA_{i+1})}e^{-2H_{n-i-1}(X)}\\&\quad \ge \frac{1}{n}\Big (\frac{1}{(A_i\vee 1)(1+A_iA_{i+1})}\wedge \frac{(A_{i+1}\wedge 1)A_iA_{i+1}}{1+A_iA_{i+1}}\Big ) e^{-2H_i(-X)}\vee e^{-2H_{n-i-1}(X)}. \end{aligned}$$
This implies that
$$\begin{aligned}&\frac{{\widetilde{P}}^\omega _0({\tilde{\tau }}_i<{\tilde{\tau }}_{-1})}{{\widetilde{\omega }}(i,i-1){\widetilde{P}}^\omega _{i-1}({\tilde{\tau }}_{-1}<{\tilde{\tau }}_i)+{\widetilde{\omega }}(i,i+1){\widetilde{P}}^\omega _{i+1}({\tilde{\tau }}_n<{\tilde{\tau }}_i)}\\&\quad \le n\Big ((A_i\vee 1)(1+A_iA_{i+1})+\frac{1+A_iA_{i+1}}{(A_{i+1}\wedge 1)A_iA_{i+1}}\Big )e^{2H_{i}(-X)\wedge H_{n-i-1}(X)}. \end{aligned}$$
Thus, for any \(\lambda \le 1\), \(n\ge 2\),
$$\begin{aligned}&{\widetilde{E}}^\omega _{0}[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]^\lambda \\&\quad \lesssim _{n} n+n^2\sum _{i=0}^{n-1} \Big ((A_i\vee 1)(1+A_iA_{i+1})+\frac{1+A_iA_{i+1}}{(A_{i+1}\wedge 1)A_iA_{i+1}}\Big )^\lambda e^{2\lambda H_{i}(-X)\wedge H_{n-i-1}(X)} \end{aligned}$$
By independence,
$$\begin{aligned} {\mathbf {E}}{\widetilde{E}}^\omega _{0}[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]^\lambda \lesssim _{n} n+n^3\max _{0\le i\le n-1}{\mathbf {E}}\left[ e^{2\lambda H_{i}(-X)\wedge H_{n-i-1}(X)}\right] \end{aligned}$$
(42)
Recall that \(\psi (\lambda )=\log {\mathbf {E}}[A^\lambda ]\) and \({\mathscr {S}}_{k}=-\sum _{i=1}^{k}\log A_{i}\). Let \(t>0\), for \(i\ge 1\), \(x>0\),
$$\begin{aligned} {\mathbf {P}}\left( H_i\left( -X\right) \ge xi\right)&\le {\mathbf {P}}\left( \max _{0\le k\le i}[-t {\mathscr {S}}_{k}-\psi \left( t\right) k] \ge xt i-\psi \left( t\right) i\right) \nonumber \\&\le {\mathbf {P}}\left( \max _{0\le k\le i}e^{-t {\mathscr {S}}_{k}-\psi \left( t\right) k}\ge e^{\left( xt-\psi \left( t\right) \right) i}\right) \nonumber \\&\le e^{-\left( xt-\psi \left( t\right) \right) i}, \end{aligned}$$
(43)
where the last inequality stems from Doob’s maximal inequality and the fact that \((e^{-t {\mathscr {S}}_{j}-\psi (t)j})_{j}\) is a martingale. Since \(x\ge {\mathbf {E}}(\log A)\), \(I(x)=\sup _{t>0}\{tx-\psi (t)\}\), we have
$$\begin{aligned} {\mathbf {P}}(H_i(-X)\ge xi)\le e^{-I(x)i}. \end{aligned}$$
(44)
Similarly, for any \(j\ge 1\) and \(x>{\mathbf {E}}[-\log A]\) .
$$\begin{aligned} {\mathbf {P}}\left( H_{j}\left( X\right) \ge xj\right)&\le {\mathbf {P}}\left( \max _{0\le k\le j}[t {\mathscr {S}}_{k}-\psi \left( -t\right) k]\ge xt j-\psi \left( -t\right) j\right) \nonumber \\&\le {\mathbf {P}}\left( \max _{0\le k\le j}e^{t {\mathscr {S}}_{k}-\psi \left( -t\right) k}\ge e^{\left( xt -\psi \left( -t\right) \right) j}\right) \nonumber \\&\le e^{-\left( xt-\psi \left( -t\right) \right) j}, \end{aligned}$$
(45)
which implies that
$$\begin{aligned} {\mathbf {P}}(H_{j}(X)\ge xj)\le e^{-I(-x)j}. \end{aligned}$$
(46)
Further, for \(0<x<{\mathbf {E}}[-\log A]\), one sees that by Cramér’s theorem,
$$\begin{aligned} {\mathbf {P}}(H_j(X)\le xj)&\le {\mathbf {P}}(X_1+\cdots +X_j\le xj)\nonumber \\&={\mathbf {P}}(-X_1-\cdots -X_j\ge -xj)\le e^{-I(-x)j}. \end{aligned}$$
(47)
Take \(\eta >0\). In (42), we can replace \(H_{i}(-X)\wedge H_{n-i-1}(X)\) by \(H_{i}(-X)\wedge H_{n-i-1}(X)\wedge K\eta n\) with some \(K\ge 1\) large enough. In fact,
$$\begin{aligned} {\mathbf {E}}[e^{2\lambda H_{i}(-X)\wedge H_{n-i-1}(X)}]\le & {} \underbrace{{\mathbf {E}}[e^{2\lambda H_{i}(-X)\wedge H_{n-i-1}(X)}; H_{i}(-X)\vee H_{n-i-1}(X)\le K\eta n]}_{\Xi ^-_K(i)}\\&+\underbrace{{\mathbf {E}}[e^{2\lambda H_{i}(-X)\wedge H_{n-i-1}(X)};H_{i}(-X)\vee H_{n-i-1}(X)\ge K\eta n ]}_{=:\Xi ^+_K(i)}. \end{aligned}$$
Observe that
$$\begin{aligned} \Xi ^+_K(i)&\le {\mathbf {E}}[e^{2\lambda H_{i}(-X)};H_{i}(-X)\ge K\eta n ]+{\mathbf {E}}[e^{2\lambda H_{n-i-1}(X)};H_{n-i-1}(X)\ge K\eta n ]\\&=:\Xi _1+\Xi _2 \end{aligned}$$
Let us bound \(\Xi _{1}\),
$$\begin{aligned} \Xi _1&={\mathbf {E}}\int _{-\infty }^{H_i(-X)}2\lambda e^{2\lambda x}\mathbf{1}_{H_i(-X)\ge K\eta n}dx=\int _{{\mathbb {R}}} 2\lambda e^{2\lambda x}{\mathbf {P}}(H_i(-X)\ge K\eta n\vee x)dx\\&=\int _{-\infty }^{K\eta n}2\lambda e^{2\lambda x}dx{\mathbf {P}}(H_i(-X)\ge K\eta n)+\int _{K\eta n}^\infty 2\lambda e^{2\lambda x}{\mathbf {P}}(H_i(-X)\ge x)dx\\&= e^{2\lambda K\eta n}{\mathbf {P}}(H_i(-X)\ge K\eta n)+\int _K^\infty 2\lambda \eta n e^{2\lambda t\eta n}{\mathbf {P}}(H_i(-X)\ge t\eta n)dt \end{aligned}$$
By applying (43), one sees that for any \(0\le i\le n-1\) and \(\mu =3>2\lambda \),
$$\begin{aligned} \Xi _1&\le e^{2\lambda K\eta n} e^{-\mu K\eta n+\psi (\mu )i}+\int _K^\infty 2\lambda \eta n e^{2\lambda t\eta n} e^{-\mu t\eta n+\psi (\mu )i}dt\\&\quad \le e^{-K\eta n+\psi (3)n}+2\lambda e^{\psi (3)n}\int _K^\infty \eta n e^{-t\eta n}dt\\&\quad \le 3e^{-K\eta n+\psi (3)n}, \end{aligned}$$
which is less than 1 when we choose \(K\) large enough. Similarly, we can show that for any \(i\le n-1\),
$$\begin{aligned} \Xi _2\le 1, \end{aligned}$$
for \(K\) large enough. Consequently, (42) becomes that
$$\begin{aligned} {\mathbf {E}}{\widetilde{E}}^\omega _{0}[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]^\lambda \lesssim _{n} 3n^3+n^3\max _{0\le i\le n-1}\Xi ^-_K(i). \end{aligned}$$
(48)
It remains to bound \(\Xi ^-_K(i)\). Take sufficiently small \(\varepsilon >0\) and let \(L=\lfloor \frac{1}{\varepsilon }\rfloor \). For any \(i\) such that \(l_1\lfloor \varepsilon n\rfloor \le i<(l_1+1)\lfloor \varepsilon n\rfloor \) and \(l_2\lfloor \varepsilon n\rfloor \le n-i-1<(l_2+1)\lfloor \varepsilon n\rfloor \) with \(0\le l_1, l_2\le L\), we have
$$\begin{aligned} \Xi ^-_K(i)&\le \sum _{0\le k_1,k_2\le K}e^{2\lambda k_1\wedge k_2 \eta n+2\lambda \eta n}{\mathbf {P}}(k_1\eta n\\&\le H_{i}(-X)< (k_1+1)\eta n){\mathbf {P}}(k_2\eta n\le H_{n-i-1}(X)<(k_2+1)\eta n)\\&\le \sum _{0\le k_1,k_2\le K}e^{2\lambda k_1\wedge k_2 \eta n+2\lambda \eta n}{\mathbf {P}}( H_{i}(-X)\ge k_1\eta n){\mathbf {P}}(k_2\eta n\\&\le H_{n-i-1}(X)<(k_2+1)\eta n). \end{aligned}$$
By (44), we have
$$\begin{aligned} {\mathbf {P}}( H_{i}(-X)\ge k_1\eta n)\le e^{-I(x_{1})i} \end{aligned}$$
where \(x_1\) is the point in \([\frac{k_1\eta n}{(l_1+1)\lfloor \varepsilon n\rfloor }, \frac{k_1\eta n}{l_1\lfloor \varepsilon n\rfloor }]\) where \(I\) reaches the minimum in this interval. By large deviation estimates (46) (47), we have
$$\begin{aligned} {\mathbf {P}}(k_2\eta n\le H_{n-i-1}(X)<(k_2+1)\eta n)\le e^{-I(x_{2})(n-i)} \end{aligned}$$
where \(x_2\) is the point in \([\frac{k_1\eta n}{(l_2+1)\lfloor \varepsilon n\rfloor }, \frac{(k_2+1)\eta n}{l_2 \lfloor \varepsilon n\rfloor }]\) where \(I\) reaches the minimum in this interval. Therefore,
$$\begin{aligned} \Xi ^-_K(i)\le&\sum _{0\le k_1,k_2\le K} e^{2\lambda k_1\wedge k_2 \eta n+2\lambda \eta n} e^{-I(x_1)l_1\lfloor \varepsilon n\rfloor }e^{-I(-x_2)l_2\lfloor \varepsilon n\rfloor } \end{aligned}$$
Taking maximum over all \(l_1,l_2,k_1,k_2\) yields that
$$\begin{aligned}&{\mathbf {E}}{\widetilde{E}}^\omega _{0}[{\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}]^\lambda \lesssim _{n} 3n^2+n^2 K^2 \max _{l_1,l_2,k_1,k_2} \exp \{2\lambda k_1\wedge k_2 \eta n+2\lambda \eta n\nonumber \\&\quad -I(x_1)l_1\lfloor \varepsilon n\rfloor -I(-x_2)l_2\lfloor \varepsilon n\rfloor \}. \end{aligned}$$
(49)
Observe that
$$\begin{aligned}&2\lambda k_1\wedge k_2 \eta n+2\lambda \eta n-I(x_1)l_1\lfloor \varepsilon n\rfloor -I(-x_2)l_2\lfloor \varepsilon n\rfloor \\&\quad \le 2\lambda (x_1l_1\wedge x_2 l_2)\lfloor \varepsilon n\rfloor -I(x_1)l_1\lfloor \varepsilon n\rfloor -I(-x_2)l_2\lfloor \varepsilon n\rfloor +3\lambda \eta n. \end{aligned}$$
Define
$$\begin{aligned} L(\lambda ):=\sup _{{\mathcal {D}}}\Big \{\Big (x_1z_1\wedge x_2z_2\Big )\lambda -I(x_1)z_1-I(-x_2)z_2\Big \}, \end{aligned}$$
where \({\mathcal {D}}:=\{x_1,x_2,z_1,z_2\ge 0, z_1+z_2\le 1\}\).
By Lemma 8.1 in [1], one concludes that
$$\begin{aligned} \limsup _{n\rightarrow \infty }\frac{\log {\mathbf {E}}{\widetilde{E}}^\omega _{0}\left[ {\tilde{\tau }}_{-1}\wedge {\tilde{\tau }}_{n}\right] ^\lambda }{n}\le L(2\lambda )=\psi \left( \frac{1+2\lambda }{2}\right) . \end{aligned}$$
\(\square \)
Appendix 2: Some Observations on Random Walks on Random Trees
Proof of Lemma 10
As \(\beta (x)\) is identically distributed under \({\mathbb {P}}\),
$$\begin{aligned} {\mathbb {E}}_{\rho }\left( \sum _{|x|=n}\mathbbm {1}_{\tau _x<\infty }\right) {\mathbb {E}}(\beta )&={\mathbb {E}}\left[ \sum _{|x|=n}P_{\rho }^{\omega ,T}(\tau _x<\infty )\right] {\mathbb {E}}(\beta )\\&={\mathbb {E}}\left( \sum _{|x|=n}{\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )){\mathbb {E}}^T(\beta (x))\right) . \end{aligned}$$
Here we used the fact that \({\mathbb {E}}^{T}{P_{\rho }^{\omega ,T}(\tau _{x}<\infty )}\) and \({\mathbb {E}}^{T}{(\beta (x))}\) are independent. Now \(P_{\rho }^{\omega ,T}(\tau _x<\infty )\) is an increasing function of \(A_x\) since
$$\begin{aligned} P_{\rho }^{\omega ,T}(\tau _x<\infty )&=P_{\rho }^{\omega ,T}(\tau _{\overset{\longleftarrow }{x}}<\infty )\left( \sum _{k\ge 0}P_{\overset{\longleftarrow }{x}}^{\omega ,T}(\tau _{\overset{\longleftarrow }{x}}^*<\min (\tau _x,\infty ))^k\right) p(\overset{\longleftarrow }{x},x)\\&=\frac{P_{\rho }^{\omega ,T}(\tau _{\overset{\longleftarrow }{x}}<\infty )}{1-P_{\overset{\longleftarrow }{x}}^{\omega ,T}(\tau _{\overset{\longleftarrow }{x}}^*<\min (\tau _x,\infty ))}\frac{A_{\overset{\longleftarrow }{x}}A_x}{1+A_{\overset{\longleftarrow }{x}}B_{\overset{\longleftarrow }{x}}}, \end{aligned}$$
recall that \(\beta (x)\) is also an increasing function of \(A_x\); moreover, conditionally on \(A_x\), \(P_{\rho }^{\omega ,T}(\tau _x<\infty )\) and \(\beta (x)\) are independent, thus by FKG inequality,
$$\begin{aligned} {\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )\beta (x))&={\mathbb {E}}^T( {\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )\beta (x)|A_x ))\\&={\mathbb {E}}^T({\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )|A_x){\mathbb {E}}^T(\beta (x)|A_x) )\\&\ge {\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )){\mathbb {E}}^T(\beta (x)). \end{aligned}$$
Therefore,
$$\begin{aligned} {\mathbb {E}}\left( \sum _{|x|=n}{\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )){\mathbb {E}}^T(\beta (x))\right)&\le {\mathbb {E}}\left( \sum _{|x|=n}{\mathbb {E}}^T(P_{\rho }^{\omega ,T}(\tau _x<\infty )\beta (x))\right) \\&={\mathbb {E}}\left( \sum _{|x|=n}P_{\rho }^{\omega ,T}(\tau _x<\infty )\beta (x)\right) . \end{aligned}$$
For any GW tree and any trajectory on the tree, there is at most one regeneration time at the \(n\)th generation, therefore,
$$\begin{aligned} \sum _{|x|=n}\mathbbm {1}_{\tau _x<\infty ,\ \eta _k\ne \overset{\longleftarrow }{x},\forall k>\tau _x}\le 1. \end{aligned}$$
By taking expectation w.r.t. \(E_{\rho }^{\omega ,T}\) and using the Markov property at \(\tau _x\),
$$\begin{aligned} \sum _{|x|=n}P_{\rho }^{\omega ,T}(\tau _x<\infty )\beta (x)\le 1. \end{aligned}$$
Whence
$$\begin{aligned} {\mathbb {E}}\left( \sum _{|x|=n}\mathbbm {1}_{\tau _x<\infty }\right) {\mathbb {E}}(\beta )\le 1. \end{aligned}$$
By transient assumption, it suffices to take \(c_{11}=\frac{1}{{\mathbb {E}}(\beta )}<\infty \). \(\square \)
Proof of Lemma 11 and Corollary 3
Let \(T_i,\ i\ge 1\) be independent copies of GW tree with offspring distribution \((q)\), each endowed with independent environment \((\omega _x,x\in T_i)\). Let \(\rho ^{(i)}\) be the root of \(T_i\). In such setting, \(\beta (\rho ^{(i)}),\ i\ge 1\) are i.i.d. sequence with common distribution \(\beta \).
For each \(T_i\), take the leftmost infinite ray, denoted \(v_0^{(i)}=\rho ^{(i)},v_1^{(i)},\ldots ,v_n^{(i)},\ldots \) Let \(\Omega (x)=\{y\ne x;\ \overset{\longleftarrow }{x}=\overset{\longleftarrow }{y}\}\) be the set of all brothers of \(x\). Fix some constant \(C\), define
$$\begin{aligned} R_i=\inf \left\{ n\ge 1;\ \exists z\in \Omega (v_n^{(i)}),\ \frac{1}{A_z\beta (z)}\le C\right\} . \end{aligned}$$
By Eq. (15),
$$\begin{aligned} \frac{1}{\beta \left( v_{R_{i-1}}^{(i)}\right) }\le 1+\frac{1}{A_{v_{R_i-1}^{(i)}}A_z\beta (z)}\le 1+\frac{C}{A_{v_{R_i-1}^{(i)}}}. \end{aligned}$$
Also \(R_i\) and \(\{A_{v_n^{(i)}},\ n\ge 0\}\) are independent under \(Q\). By iteration,
$$\begin{aligned} \frac{1}{\beta \left( \rho ^{(i)}\right) }&\le 1+\frac{1}{A_{v_0^{(i)}}A_{v_1^{(i)}}\beta \left( v_1^{(i)}\right) }\le 1+\frac{1}{A_{v_0^{(i)}}A_{v_1^{(i)}}}\left( 1+\frac{1}{A_{v_1^{(i)}}A_{v_2^{(i)}}\beta \left( v_2^{(i)}\right) }\right) \\&\le \cdots \\&\le 1+\sum _{k=1}^{R_i-1}\frac{1}{A_{v_0^{(i)}}A_{v_k^{(i)}}}\prod _{j=1}^{k-1}A_{v_j^{(i)}}^{-2}+\frac{C}{A_{v_0^{(i)}}}\prod _{l=1}^{R_i-1}A_{v_l^{(i)}}^{-2}. \end{aligned}$$
For any \(n\ge 0\), denote
$$\begin{aligned} {\mathcal {C}}(n)=1+\sum _{k=1}^{n}\frac{1}{A_{v_0^{(i)}}A_{v_k^{(i)}}}\prod _{j=1}^{k-1}A_{v_j^{(i)}}^{-2}+\frac{C}{A_{v_0^{(i)}}}\prod _{l=1}^{n}A_{v_l^{(i)}}^{-2}. \end{aligned}$$
(50)
Thus \(\displaystyle \frac{1}{\beta (\rho ^{(i)})}\le {\mathcal {C}}(R_i-1)\), note also that, since \(\xi _{2}={\mathbf {E}}(A^{-2})=1+\frac{3}{c^2}+\frac{3}{c^4}\), \(E({\mathcal {C}}(n))\le c_{34}\xi _2^{n+1}\). Therefore, for any \(K\ge 1\),
$$\begin{aligned} \frac{1}{\sum _{i=1}^K\beta (\rho ^{(i)})}\le {\mathcal {C}}\left( \min _{1\le i\le K}R_i-1\right) . \end{aligned}$$
Taking expectation under \({\mathbb {P}}\) yields (as \(R_i\) i.i.d. let \(R\) be a r.v. with the common distribution)
$$\begin{aligned} {\mathbb {E}}\left( \frac{1}{\sum _{i=1}^K\beta (\rho ^{(i)})}\right)&\le {\mathbb {E}}( {\mathbb {E}}({\mathcal {C}}(\min _{1\le i\le K}R_i-1) | R_i;\ 1\le i\le K ))\\&\le c_{34}{\mathbb {E}}\left( \xi _2^{\min _{1\le i\le K}R_i}\right) \le c_{34}\sum _{n=0}^{\infty }\xi _2^{n+1} {\mathbb {P}}(R \ge n+1)^K\\&\le c_{34} \sum _{n\ge 0}\xi _2^{n+1}{\mathbb {E}}\left( \delta _C^{\sum _{k=0}^{n-1}(d(v_k)-2)}\right) ^K \end{aligned}$$
where \(\delta _C={\mathbb {P}}(\frac{1}{A_{\rho }\beta _{\rho }}>C)\). Let \(f(s)=\sum _{k\ge 1}q_ks^k\), as \(f(s)/s\downarrow q_1\) as \(s\downarrow 0\), for any \(\varepsilon >0\), we can take \(C\) large enough to ensure \(\frac{f(\delta _C)}{\delta _C}\le q_1(1+\varepsilon )\), thus
$$\begin{aligned} {\mathbb {E}}\left( \frac{1}{\sum _{i=1}^K\beta (\rho ^{(i)}) }\right)&\le c_{34}\sum _{n\ge 0}\xi _2^{n+1}\left( \frac{f(\delta _C)}{\delta _C}\right) ^{nK}\le c_{34}\sum _{n\ge 0}\xi _2^{n+1} \left( q_1(1+\varepsilon )\right) ^{nK}. \end{aligned}$$
Now take \(\varepsilon \) such that \(q_1(1+\varepsilon )<1\), then take \(K\) large enough such that \(\xi _2(q_1(1+\varepsilon ))^{K}<1\) leads to
$$\begin{aligned} {\mathbb {E}}\left( \frac{1}{\sum _{i=1}^K\beta (\rho ^{(i)}) }\right)<c_{12}<\infty \end{aligned}$$
Similarly, the following also holds
$$\begin{aligned} {\mathbb {E}}\left( \frac{1}{\sum _{i=1}^KA_{\rho ^{(i)}}\beta (\rho ^{(i)}) }\right)<c_{12}<\infty . \end{aligned}$$
In particular, if \(q_1\xi _2<1\), we can take \(K=1\) in \(\xi _2(q_1(1+\varepsilon ))^{K}<1\). Further, it follows from (50) and Chauchy–Schwartz inequality that
$$\begin{aligned} {\mathcal {C}}(n)^2\le (n+2)\bigg (1+\sum _{k=1}^{n}\frac{1}{A^2_{v_0^{(i)}}A^2_{v_k^{(i)}}}\prod _{j=1}^{k-1}A_{v_j^{(i)}}^{-4}+\frac{C}{A_{v_0^{(i)}}}\prod _{l=1}^{n}A_{v_l^{(i)}}^{-4}\bigg ). \end{aligned}$$
Thus,
$$\begin{aligned} {\mathbb {E}}[{\mathcal {C}}^2(n)]\le c_{35}(n+2)\xi _4^{n+1}. \end{aligned}$$
As soon as \(\xi _4<\infty \), the previous argument works again to conclude that for \(K\) large enough,
$$\begin{aligned} {\mathbb {E}}\left( \frac{1}{\sum _{i=1}^{K}\beta ^2(\rho ^{(i)}) }\right) +{\mathbb {E}}\left( \frac{1}{\sum _{i=1}^{K_0}A^2_{\rho ^{(i)}}\beta ^2(\rho ^{(i)}) }\right)<c_{13}<\infty . \end{aligned}$$
\(\square \)
