if we denote these by $\pi _{1}^{\prime },\pi
_{2}^{\prime },\ldots $, $$P(\widetilde{\pi}=\pi _{k}^{\prime }|(|\pi _{j}^{\prime }|,j\geq 1))=|\pi
_{k}^{\prime }|/\sum_{j}|\pi _{j}^{\prime }|.$$

For any sequence of partitions $(\pi ^{(i)},i\geq 1),$ define $\pi=\bigcap_{i\geq 1}\pi ^{(i)}$ by $$k\overset{\pi}{\sim }j\iff k\overset{\pi ^{(i)}}{\sim }j\quad \forall i\geq
1.$$

\[Lemmafreqasym\] Let $(\pi^{(i)},i\geq 1)$ be a sequence of independent exchangeable partitions and set $\pi :=\bigcap_{i\geq 1}\pi ^{(i)}.$ Then, a.s. for every $j\in \mathbb{N},$$$\left| \pi _{j}\right| =\prod_{i\geq 1}\left| \pi _{k(i,j)}^{(i)}\right| ,$$ where $(k(i,j),j\geq 1)$ is defined so that $\pi_{j}=\bigcap_{i\geq 1}\pi
_{k(i,j)}^{(i)}.$

[**Proof.** ]{} First notice that $k(i,j)\leq j$ for all $i\geq
1$ a.s. This is clear when $\pi _{j}\neq \varnothing $, since $j\in \pi _{j}$ and then $j\in \pi _{k(i,j)}^{(i)}.$ When $\pi _{j}=\varnothing ,$ $j\in \pi
_{m}$ for some $m<j $ and then $m$ and $j$ belong to the same block of $\pi
^{(i)}$ for all $i\geq 1.$ Thus $k(i,j)\leq m<j.$ Using then the paintbox construction of exchangeable partitions explained above and the independence of the $\pi ^{(i)}$’s$,\ $we see that the r.v. $\prod_{i\geq 1}\mathbf{1}_{\{m\in \pi _{k(i,j)}^{(i)}\}},$ $m\geq j+1,$ are iid conditionally on $(|\pi _{k(i,j)}^{(i)}|,i\geq 1)$ with a mean equal to $\prod_{i\geq 1}|\pi
_{k(i,j)}^{(i)}|.$ The law of large numbers therefore gives $$\prod_{i\geq 1}\left| \pi _{k(i,j)}^{(i)}\right| =\lim_{n\rightarrow \infty }\frac{1}{n}\sum_{j+1\leq m\leq 