Inequality (\[aaa\]) hold because of (\[EQruns\]) and inequality (\[bbb\]) because $N_n \le n$. Taking $y= A\log_2n$, we obtain easily the announced upper bound.

To come back to Lyndon words, note that for a primitive word $w$, we have $M_n(\rho(w)) \leq \max\{M_n( w), X_1(w)+X_{N_n}(w)\}.$ Thus we obtain: $$\begin{aligned}
&& {\mathbb P_n}\left(  M_n(\rho(w)) \geq  A \log_2{n} \right) \\
& \leq & {\mathbb P_n}\left( M_n(w)  \geq  A \log_2{n}\right) + {\mathbb P_n}\left(
X_1+X_{N_n} \ge A \log_2{n}  \right)\\
& \leq & {\mathbb P_n}\left( A \log_2{n}  \geq  M_n(w) \right) +{\mathbb P_n}\left(
 X_1 \ge  \frac{A}2   \log_2{n}
   \right)
  +{\mathbb P_n}\left( X_{N_n} \ge  \frac{A}2  \log_2{n}
   \right).\end{aligned}$$ Note that $X_1$ and $X_{N_n}$ have the same law. Thus using (\[EQruns\]), we get: $${\mathbb P_n}\left(
 X_1 \ge   \frac{A}2   \log_2{n}\right)
\le {\mathbb P_n}\left( Z_1 \ge   \frac{A}2 
  \log_2{n}\right)=O(n^{-\frac{A}2})=o\left(n^{1-A}\right),$$ by the choice we made for $A$. Finally, by using Lemma \[usarg\], we get the desired result.

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Building of long blocks and short blocks of a word {#building}
==================================================

Let $0<\varepsilon<1$ and $B>2$ be fixed in this section. Our aim here is to find, in a word $w \in \mathcal A^n$, some *long blocks* beginning by a long run of “a” in a word $w \in \mathcal A^n$, and we moreover want to choose them 