invariant measure the particle spacings $(\chi_i)_{1\le i \le k}$ are distributed like independent geometric variables, with respective parameters ${\lambda}^{-i}$, conditioned to the event $\sum_{i=1}^k \chi_k\le N$.

\[lem:lbeq\] When holds we have for any ${\varepsilon}>0$ $$\begin{split}
  \lim_{N\to \infty}& \pi_{N,k}( \ell_N \le (N-k)-{\varepsilon}N)=0,\\
  \lim_{N\to \infty}& \pi_{N,k}( r_N \ge (N-k)+{\varepsilon}N)=0,
  \end{split}$$

By symmetry it is sufficient to prove the result for $\ell_N$ only, but for all $k\in{\llbracket}1,N-1{\rrbracket}$. Note that there is nothing to prove regarding $\ell_N$ if $\alpha:=\lim_{N\to \infty} k/N =1$, so we assume that $\alpha\in[0,1)$.\
Let $(X_i)_{1\le i \le k}$ be independent geometric variables, with respective parameters ${\lambda}^{-i}$. The sum of their means satisfies (recall that ${\lambda}-1$ is of order $b_N$) $$\begin{aligned}
\sum_{i=1}^k \frac{1}{1-{\lambda}^{-i}} = k+ \sum_{i=1}^k \frac{{\lambda}^{-i}}{1-{\lambda}^{-i}}\le k+ C b_N^{-1} \log \min( k, b_N^{-1})\;,\end{aligned}$$ for some constant $C>0$. The large bias assumption ensures that $b_N^{-1} \log \min( k_N, b_N^{-1})=o(N)$. Hence using the Markov inequality, we obtain that if $(X_i)_{1\le i \le k}$ is a sequence of such geometric variables, and if is satisfied, then for any ${\varepsilon}>0$ $${\mathbb{P}}\Big(\sum_{i=1}^k X_i\ge k+{\varepsilon}N\Big) \le \frac{{\mathbb{E}}\big[\big|\sum_{i=1}^k X_i - k\big|\big]}{{\varepsilon}N}=\frac{{\mathbb{E}}\big[\sum_{i=1}^k X_i - k\big]}{{\varepsilon}N}\;,$$ so that $$\lim_{N\to\infty}  {\mathbb{P}}\Big(\sum_{i=1}^k X_i\ge k+{\varepsilon}N\Big) = 0\;.$$ The above inequality for ${\varepsilon}<1-\alpha$ implies that ${\mathbb{P}}(\sum_{i=1}^k X_i\le N)\ge 1/2$ for all $N$ large 