the content of the next subsection.

Environment under the annealed measure
--------------------------------------

\[F\] Let $x_0:=0<x_1<\dots<x_n$, $r_i=x_i-x_{i-1}$, $i=1,2,\dots,n$, and consider an environment such that $\{x_1,x_2,\dots,x_n\}$, $n\in{\mathbb{N}}$, is the set of all occupied sites in $(0,x_n]$. Denote by $u_n$ the probability that a random walk starting at $x_{n-1}$ reaches $x_n$ before hitting $0$, i.e. $u_n={P}^{x_{n-1},\omega}(\tau_{x_n}<\tau_0)$, $n\in{\mathbb{N}}$. Then $$\label{u_n}
    u_1=\frac{e^{-V(0,\omega)}}{2r_1}=e^{M-V(0,\omega)}F_M(0,r_1,0); 
\  u_n=F_M(r_{n-1},r_n,u_{n-1}),\ n>1,$$ where $F_M:({\mathbb{N}}\cup\{0\})\times{\mathbb{N}}\times[0,1]\to [0,1]$, $$\label{defF}
    F_M(\ell,r,u)=
    \begin{cases}
      \dfrac{e^{-M}}{2r}\,\left(1- e^{- M}\left(1- \dfrac{1}{2 r}
        - \dfrac{1-u}{2 \ell} \right)\right)^{-1},&\text{if } \ell\ne 0;\\[3mm]
    \dfrac{e^{-M}}{2r},&\text{if } \ell=0.
    \end{cases}$$

The proof is given in the Appendix.

As $\ell,r\to\infty$ $$\label{Fb}
    F_M(\ell,r,u)\sim \frac{e^{- M}}{2 r (1- e^{- M})},
  \quad\text{uniformly in $u\in[0,1]$.}$$

From now on we shall identify every environment $\omega$ on $(0,y)$ with the vector $\overline{R}_N=(R_1,R_2,\dots,R_N)\in
\cup_{i=1}^\infty{\mathbb{N}}^i$ of $N=N(y,\omega)$ successive distances between occupied sites in $(0,y)$, where $R_1$ is the distance from the first positive occupied site in $(0,y)$ to the origin and $R_N$ is the distance from the last occupied site in $(0,y)$ to $y$. If the interval $(0,y)$ is empty then we set $N=1$ and $R_1=y$.

\[sp\] For any $n\in\{1,2,\dots,y\}$ and $(r_1,r_2,\dots,r_n)\in{\mathbb{N}}^n$ with $\sum_{i=1}^n r_i=y$, $r_0=0$, $u_0=0$, $$\begin{aligned}
\label{23}
Q_{0,y}&(N=n, \overline{R}_n=(r_1,r_2,\dots,r_n))\nonumber
\\=&Z_{0,y}^{-1}\left(\frac{e^M(1-p)}{p}+1
\right)\prod^n_{i=1} \left(p(1-p)^{r_i-1}\,F_M(r_{i-1}, r_{i},
  u_{i-1})\right)\\=&Z_{0,y}^{-1}(1-p)^y\left(\frac{e^M}{{\varrho}}+1
\right)\prod^n_{i=1} \left({\varrho}\,F_M(r_{i-1}, r_{i},
  u_{i-1})\right),\ \text{where ${\varrho}:=p/(1-p)$.}\nonumber\end{aligned}$$

**Heuristic derivation of (\[gap\]).** Before 