\Pi^{\varrho}_n(\{r_1,\dots,r_n\})=\prod_{i=1}^n {\varrho}(1-q)^{r_i}
    F_M(r_{i-1},r_i,u_{i-1})=\\\prod_{i\in A_R}{\varrho}(1-q)^{r_i}
    F_M(r_{i-1},r_i,u_{i-1}) \times
    \prod_{i\in\bar{A}_R}{\varrho}(1-q)^{r_i} F_M(r_{i-1},r_i,u_{i-1})\ge
    \\ (1-e^{-M})^{|A_R|}(1-{\varepsilon}/4)^{n-|A_R|}\prod_{i=1}^n
    \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}.
  \end{gathered}$$ Choose $\delta>0$ small enough to have $(1-e^{-M})^\delta\ge
  1-{\varepsilon}/4$. Then $$\begin{aligned}
    1&\ge \Pi^{\varrho}_n(|A_R|\le \delta n)=\sum_{|A_R|\le \delta n}
    \prod_{i=1}^n {\varrho}(1-q)^{r_i} F_M(r_{i-1},r_i,u_{i-1})\\&\ge
    (1-{\varepsilon}/4)^{2n}\sum_{|A_R|\le \delta n} \prod_{i=1}^n
    \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}
    \\&=(1-{\varepsilon}/4)^{2n}\left(\sum_{r_1,\dots,r_n\in{\mathbb{N}}}
      \prod_{i=1}^n \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}-\sum_{|A_R|>
        \delta n} \prod_{i=1}^n \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}\right)\\
    &\ge
    (1-{\varepsilon}/4)^{2n}(1+{\varepsilon})^n-(1-{\varepsilon}/4)^{2n}\sum_{|A_R|>
      \delta n} \prod_{i=1}^n \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}
    \\&\ge (1+{\varepsilon}/4)^n-(1-{\varepsilon}/4)^{2n}\sum_{|A_R|>
      \delta n} \prod_{i=1}^n \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}.
  \end{aligned}$$ To get a contradiction, it is enough to show that the last sum is bounded uniformly in $n$. Such a bound is easily obtained from basic large deviations for i.i.d. Bernoulli random variables. Notice that by (\[rho\]) there is a constant $C$, $1-e^{-M}\le C\le 1$, such that $C{\varrho}(1-q)^r/(2r(e^M-1))$, $r\in{\mathbb{N}}$, is a probability distribution on ${\mathbb{N}}$. Consider a sequence of i.i.d. random variables $(Y_i)_{i\in{\mathbb{N}}}$ with this distribution. Then $P(Y_i\le R)\le C{\varrho}R/(2(e^M-1))\to 0$ as ${\varrho}\to 0$. Thus, for an arbitrary $c>0$ we can choose ${\varrho}$ small enough so that for all sufficiently large $n$ $$P\left(\sum_{i=1}^n{\mathbbm{1}}_{\{Y_i\le R\}}>\frac{\delta n}{2}\right) \le
  e^{-cn}.$$ Since $\{|A_R|>\delta
  n\}\subset\{\sum_{i=1}^n{\mathbbm{1}}_{\{r_i\le R\}}>\delta n/2\}$, $$\sum_{|A_R|>
   \delta n} \prod_{i=1}^n \frac{{\varrho}(1-q)^{r_i}}{2(e^M-1)r_i}\le
C^{-n}P\left(\sum_{i=1}^n{\mathbbm{1}}_{\{Y_i\le R\}}>\frac{\delta n}{2}
     \right)\le (Ce^c)^{-n}\le
 1$$ for $c>-\log(1-e^{-M})$ and all large $n$, and we are done.

We shall need the following comparison lemma. The notation $\le_{\mathrm{st}}$ (resp. $\ge_{\mathrm{st}}$) means “stochastically smaller” (resp. “stochastically larger”), where we use 