note that, by construction, each word in $C(w_0)$ has the same family of blocks.

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Thus, roughly speaking, the second smallest block $Y_{j_0}$ has the same probability to be at every possible place among all the blocks, and this is why its position along the word $w$, divided by $n$, should follows approximately the uniform law on $[0,1]$. To formalize this intuition and to exploit this invariance property, we enlarge our probability space $\mathcal G_n $: consider a sequence $(U_{i})_{i \in {\mathbb N}}$ of independent identically distributed random variables on a probability space $(X, \mathcal X , {\mathbb Q})$, following the uniform distribution on $[0,1]$. We denote by ${\bar{\mathbb P}_n}$ the uniform probability on $\mathcal G_n $ and consider the product probability ${\mathbb Q}\otimes {\bar{\mathbb P}_n}$ on the product space $X \times \mathcal G_n$; this means that $(U_{i})_{i \in {\mathbb N}}$ are independent of the choice of the random Lyndon word in $\mathcal G_n$.

For $x \in X$, we define a uniform random permutation $\pi_x \in \mathfrak{S}_{K_n(w)-1}$ by the order statistics of $U_{1}(x),U_{2}(x),...,U_{K_n(w)-1}(x)$: $$U_{\pi_x(1)}(x) <
{U}_{\pi_x(2)}(x) < \dots < {U}_{\pi_x(K_n(w)-1)}(x).$$

Therefore from the previous lemma, the random variable $$W: \;
\left\{ \begin{array}{rll}
X \times \mathcal G_n &
\longrightarrow & \mathcal 