$\mathbb{P}_1$, for some real index $\alpha$. We shall always assume here that $\alpha<0$. For all $t\geq 0$, let introduce the time-change $$\begin{aligned}
\label{Equation Lamperti tau_t}
        \tau_t:=\int_0^tX(s)^\alpha\mathrm{d}s.
    \end{aligned}$$ The well-known Lamperti’s transformation states that there exists a unique Lévy process $\xi=(\xi(t))_{t\geq 0}$ such that $$\begin{aligned}
        X(t)=\exp\left(\xi(\tau_t)\right),\quad \forall t\geq 0.
    \end{aligned}$$ Equivalently to , one can write $$\begin{aligned}
        \tau_t=\inf\left\{s\geq 0:\int_0^s\exp(-\alpha\xi(u))\mathrm{d}u\geq t\right\}.
    \end{aligned}$$ Let $\Lambda$ be the Lévy measure of $\xi$ and assume that $\int_{{\left(1\,,\infty\right)}}e^{y}\Lambda(\mathrm{d}y)<\infty$. We thus have, at least for $q\in{\left[0\,,1\right]}$, that $\mathbb{E}(\exp(q\xi(t)))=\exp(t\psi(q))$ for all $t\geq 0$, with $$\begin{aligned}
\label{Equation Laplace exponent psi of xi}
        \psi:q\mapsto bq+\frac{\sigma^2}{2}q^2+\int_{\mathbb{R}}\left(e^{qy}-1+q(1-e^y)\right)\Lambda(\mathrm{d}y),\quad q\geq 0,
    \end{aligned}$$ where $b,\sigma^2\geq 0$. We point out that this is not the traditional Lévy-Khintchin formula for the Laplace exponent $\psi$ of a Lévy process, in which $(1-e^y)$ above is usually replaced by $y\mathds{1}_{\left\{|y|<1\right\}}$, but is more convenient for our purposes. We assume throughout this paper that $\psi'(0)<0$. It is known that this entails that $\xi$ drifts to $-\infty$ almost surely, and in particular that the absorption time in $0$ of $X$, namely $\int_0^\infty\exp(-\alpha\xi(t))\mathrm{d}t$, is finite almost surely.

#### The cell-system.

We briefly recall Bertoin’s construction of the cell-system as in [@B17]. We use the classical Ulam-Harris-Neveu notation to label the particles of the 