n}R_i\ge H\theta n\right)\le e^{-c(H) \theta n}\text{ and
}\ \ {\widetilde}{\Pi}_n^{\varrho}\left(\sum_{1\le i\le \theta n}R_i\le h\theta
  n\right)\le e^{-c'(h) \theta n}.$$

Finally, we are ready to convert information about configurations under ${\widetilde}{\Pi}_n^{\varrho}$ to information under $Q_{0,y}$. For $1\le n\le k\le y$ let $$\begin{aligned}
  A_{n,k}:&=\left\{(r_1,r_2,\dots,r_k)\in{\mathbb{N}}^k\,:\,\sum_{i=1}^{n-1} r_i<y,\
\sum_{i=1}^n r_i\ge y\right\};\nonumber\\
  A_n:&=\left\{(r_1,r_2,\dots,r_n)\in{\mathbb{N}}^n\,:\,\sum_{i=1}^n
    r_i=y\right\}\subset A_{n,n}.\label{an}\end{aligned}$$

\[corbond\] For each $H>EY_1$ and $h< EZ_1$ there exist strictly positive $c_1(H)$ and $c_1'(h)$ such that for all $y$ large $$\begin{aligned}
    (1-q)^y\sum_{A_n}\prod_{i=1}^n {\varrho}F_M(r_{i-1}, r_{i}, u_{i-1}) &\leq e^{-c_1(H)
  y}\quad\text{for all $n\in [1,y/H]$};\\
(1-q)^y\sum_{A_n}\prod_{i=1}^n {\varrho}F_M(r_{i-1}, r_{i}, u_{i-1}) &\leq e^{-c_1'(h) y}\quad\text{for all
  $n\in[y/h,y]$}.
  \end{aligned}$$

We shall start with the second inequality. If $h< 1$ then there is nothing to prove as $[y/h,y]$ is empty. Assume that $1\le h<EZ_1$ and apply Corollary \[corexp\] to ${\widetilde}{\Pi}_y^{\varrho}$ with $n$ in place of $\theta n$. As $A_{n,y}\subset\{(r_1,\dots,r_y)\in{\mathbb{N}}^y\,:\, \sum_{i=1}^{n-1}r_i\le
  hn\}$, we get that for $n\in[y/h,y]$ $$\label{nn}
    e^{-c'(h)(y/h)}\ge e^{-c'(h)n}\ge \sum_{A_{n,y}} \prod_{i=1}^y
{\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i, u_{i-1}).$$ If $y\ge n+1$ then we write the right-hand side of the above inequality as $$\begin{gathered}
  \sum_{A_{n,n}} \prod_{i=1}^n
{\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i,
u_{i-1})\times\left(\sum_{r_{n+1},\dots,r_y\in{\mathbb{N}}} \prod_{i=n+1}^y
{\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i, u_{i-1})\right),\end{gathered}$$ and apply (\[co2\]) to the last summation. Thus, for all $n\in[y/h,y]$ we have $$\begin{aligned}
  e^{-c'(h)y/h}&\ge (1-e^{-M})\sum_{A_{n,n}} \prod_{i=1}^n
  {\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i,
  u_{i-1})\\&\overset{(\ref{an})}{\ge} (1-e^{-M})\sum_{A_n}
  \prod_{i=1}^n {\varrho}(1-q)^{r_i}F_M(r_{i-1}, r_i,
  u_{i-1})\\&=(1-e^{-M})(1-q)^y\sum_{A_n} \prod_{i=1}^n {\varrho}F_M(r_{i-1}, r_i, u_{i-1}).\end{aligned}$$

The proof of the 