the length of the vacant interval between these two sites.

Quenched speed
==============

The results of Theorem \[vque\] are rather straightforward. The fact that the speed is at most of order $p$ comes immediately from the fact that no matter what the value of $M$, the time for a conditioned random walk to traverse an empty interval of size $R$ will be of order $R^2$, corresponding to speed of order $1/R$. The upper bound on speed follows since most sites lie in vacant intervals of size of order $1/p$.

We shall need two facts. The first is a very basic fact about the standard random walk but we do not have a reference at hand and, thus, give a proof in the Appendix.

\[basic1\] Let $(S_j)_{j\ge 0}$ be the simple symmetric random walk, $n\in{\mathbb{N}}$, and $S_0=k\in\{1,2,\dots,n-1\}$. Then $$E^k(\tau_n;\tau_n<\tau_0)=\frac{k(n-k)(n+k)}{3n}.$$

The second is mostly a consequence of the ergodic theorem (see [@Sz94 (1.30) and Theorem 2.6] for a treatment of Brownian motion among Poissonian obstacles). The proof is given in the Appendix.

\[basic2\] Let $V(x,\cdot)$, $x\in{\mathbb{Z}}$ be i.i.d. random variables, which satisfy (\[pot\]). Then there exists limit $$\begin{aligned}
    \frac{1}{v^{\mathrm{que}}}&:=\lim_{y\to\infty}\frac{E_{Q^\omega_y}(\tau_y)}{y}
    ={\mathbb{E}}(E^\omega(\tau_1\,|\,\tau_1<\infty))<\infty
    \quad({\mathbb{P}}\text{-a.s.}),\quad \text{and, moreover,}\\
    \frac{1}{v^{\mathrm{que}}}&=\frac{d}{d\lambda}\alpha_{\lambda+V}(1)\Big|_{\lambda=0+}.
  \end{aligned}$$

The key idea for calculation of the 