we turn to rigorous analysis of (\[23\]) we would like to present a “back of the envelope derivation” of the gap asymptotics (\[gap\]). When $y\to\infty$ we might expect that measures $Q_{0,y}$ converge to a limiting measure, under which the consecutive gaps are essentially i.i.d.. It is reasonable to assume that if we let $M\to\infty$ or $p\to 0$ then the distances between consecutive occupied sites under this limiting measure will also go to infinity. Thus, we replace $F_M(r_{i-1}, r_{i},
    u_{i-1})$ in (\[23\]) with its limit as $r_{i-1}, r_{i}\to
    \infty$ given by (\[Fb\]). We get that for $y\to \infty$ $$\label{eu2}
    Q_{0,y}(N=n, \overline{R}_n=(r_1,r_2,\dots,r_n)) \ \asymp\ 
Z_{0,y}^{-1}\prod^n_{i=1} \left(p(1-p)^{r_i-1}\,\frac{e^{- M}}{2 r_i
    (1- e^{- M})}\right).$$ By [@KM12 Lemma 5.5], $\lim_{y\to\infty}y^{-1}\ln
  Z_{0,y}=\beta$, where $\beta:=\beta_V(1)$ is the annealed Lyapunov exponent (see (\[ann\])), and we replace $Z_{0,y}^{-1}$ in (\[eu2\]) with $e^{\beta y}=e^{\beta\sum_{i=1}^n r_i}$ to arrive at $$\begin{gathered}
    Q_{0,y}(N=n, \overline{R}_n=(r_1,r_2,\dots,r_n)) \ \asymp\  
    \prod^n_{i=1} \left(\frac{p}{2 (1-p)(e^M-
      1)}\, \frac{(e^{\beta}(1-p))^{r_i}}{r_i}\right) \\= \prod^n_{i=1}
  \left(\frac{1}{K} \frac{(e^{\beta}(1-p))^{r_i}}{r_i}\right),
  \end{gathered}$$ where $K=K(p,M)$ is the same as in (\[gap\]). For $Q_{0,y}$ to be a probability measure it should hold that $$\label{eu3}
    \frac1{K}\sum_{r=1}^\infty\frac{(e^{\beta}(1-p))^r}{r}=1.$$ In other words, the limiting gap size appears to have the so-called log-series distribution. Summing up the series in (\[eu3\]) we see that $\ln(1-e^{\beta}(1-p))=-K$, i.e. $e^\beta(1-p)=1-e^{-K}$, and 