A.1. Proof of Proposition 1 and uniform integrability

For the proof of several lemmas and other auxiliary results we make use of a corollary to Stirling’s formula, which can be found as Corollary 6.1.46 in [Reference Abramowitz and Stegun1] and reads as follows.

Lemma 5.

\begin{align*}\lim_{n\to\infty}n^{b-a}\frac{\Gamma(a+n)}{\Gamma(b+n)}=1\quad\forall a,b\in\mathbb{R}.\end{align*}

This allows us to estimate binomial coefficients by their respective powers and vice versa.

Lemma 6. Under the assumptions of Proposition 1, we have

\begin{align*}\prod_{i=1}^r\frac{D_i^{k_i}(n)}{\Gamma(k_i+1)}\sim \prod_{i=1}^r\binom{D_i(n)+k_i-1}{k_i}\quad \text{ almost surely}\end{align*}

and there exist constants $C_1,C_2>0$ independent of n such that almost surely

(A.1) \begin{align}C_1\cdot\prod_{i=1}^r\binom{D_i(n)+k_i-1}{k_i}\leq \prod_{i=1}^rD_i^{k_i}(n) \leq C_2\cdot\prod_{i=1}^r\binom{D_i(n)+k_i-1}{k_i}.\end{align}

Proof. Remember from Lemma 2 that almost surely $D_i(n)\to\infty$ as $n\to\infty$ . Thus, by Lemma 5, we can find, for each $i=1, \ldots, r$ , constants $0<C^i_1<C_2^i$ (depending on $k_i$ ) such that

\begin{align*}C_1^i \binom{D_i(n)+k_i-1}{k_i}= C_1^i \frac{\Gamma(D_i(n)+k_i)}{\Gamma(k_i+1)\Gamma(D_i(n))}\leq \frac{D_i^{k_i}(n)}{\Gamma(k_i+1)} \leq C_2^i \binom{D_i(n)+k_i-1}{k_i}\end{align*}

holds for all $n \geq (r-s) \vee 0$ . Combining those constants leads to (A.1).

Now to Proposition 1. The following proof adapts a martingale approach from [Reference Móri24] (Theorem 2.1).

Proof of Proposition 1.

1. (1) We call the total ball count at time m in the corresponding urn model

   \begin{align*}S_m:\!=z+m+\beta\cdot\Bigl\lfloor\frac{m}{l}\Bigr\rfloor\quad \text{with }z:\!=\sum_{i=1}^{N(0)}C_i(0)+j\cdot \beta.\end{align*}
   Only one colour obtains a new ball at a time, so
   \begin{align*}&\mathbb{E}\Bigl(\prod_{i=1}^r\binom{C_i(m+1)+k_i-1}{k_i}\Big|\mathcal{F}_m\Bigr) \\&=\sum_{j=1}^r\mathbb{E}\Bigl(\underset{i\neq j}{\prod_{i=1}^r}\binom{C_i(m)+k_i-1}{k_i}\binom{C_j(m)+k_j}{k_j}\cdot{\unicode{x1D7D9}}_{\{C_j(m+1)=C_j(m)+1\}}\Big|\mathcal{F}_m\Bigr)\\ &\;\;\;+\mathbb{E}\Big(\prod_{i=1}^r\binom{C_i(m)+k_i-1}{k_i}\cdot{\unicode{x1D7D9}}_{\bigcap_{j=1}^r\{C_j(m+1)=C_j(m)\}}\Big|\mathcal{F}_m\Big),\end{align*}
   and using the transition probabilities of the urn model this equals
   \begin{align*}&\sum_{j=1}^r\underset{i\neq j}{\prod_{i=1}^r}\binom{C_i(m)+k_i-1}{k_i}\cdot \binom{C_j(m)+k_j-1}{k_j}\cdot\frac{C_j(m)+k_j}{C_j(m)}\cdot\frac{C_j(m)}{S_m}\\ \;\;\; &+\prod_{i=1}^r\binom{C_i(m)+k_i-1}{k_i}\cdot \Bigl(1-\sum_{j=1}^r\frac{C_j(m)}{S_m}\Bigr)\\&=\prod_{i=1}^r\binom{C_i(m)+k_i-1}{k_i}\Bigl(\sum_{j=1}^r\frac{C_j(m)+k_j}{S_m}+1-\sum_{j=1}^r\frac{ C_j(m)}{S_m}\Bigr)\\&=\prod_{i=1}^r\binom{C_i(m)+k_i-1}{k_i}\frac{S_m+k}{S_m}.\end{align*}
   Now consider times $m=n\cdot l$ . These are the critical times when a new colour is migrated into the urn (a new vertex is added to the graph). In between nothing of interest happens and we can iterate the above calculation l times to obtain
   \begin{align*}\mathbb{E}\Bigl(\prod_{i=1}^r\binom{C_i((n+1)\cdot l)+k_i-1}{k_i}\Big|\mathcal{F}_{n\cdot l}\Bigr)=\prod_{i=1}^r\binom{C(n\cdot l)+k_i-1}{k_i}\prod_{i=0}^{l-1}\frac{S_{n\cdot l+i}+k}{S_{n\cdot l+i}}.\end{align*}
   In the next step we see that
   \begin{align*}&\prod_{i=0}^{l-1}\frac{S_{n\cdot l+i}+k}{S_{n\cdot l+i}}=\prod_{i=0}^{l-1}\frac{z+n\cdot l+i+\beta\cdot n+k}{z+n\cdot l+i+\beta\cdot n}=\prod_{i=0}^{l-1}\frac{\frac{z+k+i}{l+\beta}+n}{\frac{z+i}{l+\beta}+n}\\&=\biggl(\prod_{i=0}^{l-1}\frac{\Gamma(\frac{z+k+i}{l+\beta}+n+1)}{\Gamma(\frac{z+i}{l+\beta}+n+1)}\biggr)\Big/\biggl(\prod_{i=0}^{l-1}\frac{\Gamma(\frac{z+k+i}{l+\beta}+n)}{\Gamma(\frac{z+i}{l+\beta}+n)}\biggr)=:\frac{c(n+1,k)}{c(n,k)},\end{align*}
   and the martingale property follows. Lastly, we apply Lemma 5 to each factor of c(n,k) to get
   (A.2) \begin{equation}c(n,k)=\prod_{i=0}^{l-1}\frac{\Gamma(\frac{z+k+i}{l+\beta}+n)}{\Gamma(\frac{z+i}{l+\beta}+n)}\sim \prod_{i=0}^{l-1}n^{k/(l+\beta)}=n^{k\cdot l/(l+\beta)}.\end{equation}

2. (2) Doob’s martingale convergence theorem (see Chapter XI.14 in [Reference Doob14]) guarantees the existence of an almost sure limit of the martingale given in (2.1) and that this limit is in $L^1(\mathbb{P})$ . Lemma 6 in combination with (A.2) then ensures the almost sure convergence of the sequence $n^{-k_i l/(l+\beta)}D_i(n)^{k_i}$ to a limit $\zeta_i^{k_i} \in L^1(\mathbb{P})$ , and a further application of Lemma 6 implies that the expression in (2.2) is indeed the limit of the process (2.1).

3. (3) Doob’s martingale convergence theorem (Chapter XI.14 in [Reference Doob14]) yields that (2.2) is the right closure to the martingale (2.1) if the latter is uniformly integrable. To show this uniform integrability we show that the process is $L^{1+\epsilon}$ -bounded for any $\epsilon>0$ . To this end, set $p_i:\!=k_i(1+\epsilon)$ and $p:\!=k(1+\epsilon)=p_1+\ldots+p_r$ . By Lemma 6 there exist constants $C_1,C_2>0$ independent of n such that

   \begin{align*}\Bigl[\frac{1}{c(n,k)}\prod_{i=1}^r\binom{D_i(n)+k_i-1}{k_i}\Bigr]^{1+\epsilon}&\leq C_1\cdot\frac{\prod_{i=1}^rD^{p_i}_i(n)}{n^{p\cdot l/(l+\beta)}}\\&\leq C_2\cdot \frac{1}{c(n,p)}\prod_{i=1}^r\binom{D_i(n)+p_i-1}{p_i}.\end{align*}
   Since the right-hand side is a multiple of a martingale, it has bounded expectation and the uniform integrability of (2.1), and thus the first part of (3), follows.

   This now allows us to iteratively trace back the expectation of $\prod_{i=1}^r\zeta_i^{k_i}$ to the times at which colour $i, 1 \leq i \leq r,$ was first introduced to the urn, that is, $n_0^i:\!=l(i-s) \vee 0$ , and the number of balls was still deterministic, namely either $\beta$ if $i>N(0)$ or $D_i(0)$ if $i\leq N(0)$ :

   \begin{align*}&\mathbb{E}\Bigl(\prod_{i=1}^r\frac{\zeta_i^{k_i}}{\Gamma(k_i+1)}\Bigr)=\mathbb{E}\Bigl(\mathbb{E}\Bigl(\prod_{i=1}^r\frac{\zeta_i^{k_i}}{\Gamma(k_i+1)}\Big|\mathcal{F}_{n_0^r}\Bigr)\Bigr)\\&=\mathbb{E}\Bigl(\frac{1}{c(n_0^r/l,k_1+\ldots+k_r)}\prod_{i=1}^r\binom{D_i(n_0^r/l)+k_i-1}{k_i}\Bigr)\end{align*}
   \begin{align*}&=\frac{c(n_0^r/l,k_1+\ldots+k_{r-1})}{c(n_0^r/l,k_1+\ldots+k_r)}\cdot\begin{cases}\binom{D_r(0)+k_r-1}{k_r}\quad &\text{if }r\leq N(0)\\\binom{\beta+k_r-1}{k_r}\quad &\text{if }r> N(0)\end{cases}\\&\phantom{=}\cdot\mathbb{E}\Bigl(\mathbb{E}\Bigl(\frac{1}{c(n_0^r/l,k_1+\ldots+k_{r-1})}\prod_{i=1}^{r-1}\binom{D_i(n_0^r/l)+k_i-1}{k_i}\Big|\mathcal{F}_{n_0^{r-1}}\Bigr)\Bigr)\\&=\prod_{i=2}^{r}\frac{c(n_0^i/l,k_1+\ldots+k_{i-1})}{c(n_0^i/l,k_1+\ldots+k_i)}\cdot\frac{1}{c(n_0^1/l,k_1)}\prod_{i=1}^r\begin{cases}\binom{D_i(0)+k_i-1}{k_i}\quad &\text{if }i\leq N(0),\\\binom{\beta+k_i-1}{k_i}\quad &\text{if }i> N(0).\end{cases}\end{align*}

We end this section with a uniform integrability property that will turn out to be useful on multiple occasions.

A.2. Proof of Theorem 3

We adapt the proof from Theorem 3 in [Reference Wang and Resnick37] to the case of a general separable Banach space. By our additional assumption about the continuity of $(X_t)$ we are able to straighten a previously not properly addressed subtlety in the proof about the application of Egorov’s theorem.

Proof of Theorem 3. By a Portmanteau theorem for $\mathbb{M}$ -convergence, see Theorem 2.1 in [Reference Lindskog, Resnick and Roy23], it is sufficient to show

(A.3) \begin{equation}\lim_{t\to\infty}t\mathbb{E} f\Bigl(\frac{X_T T}{b(t)}\Bigr)=\int_{0}^\infty\mathbb{E} f(X_\infty y)\,\nu_\alpha(\text{d} y)<\infty,\end{equation}

for all non-negative, bounded and uniformly continuous functions f on B whose support is bounded away from 0, i.e. there exists an $\epsilon_0>0$ such that $f(x)=0$ for all $x \in B_{\epsilon_0}(0)$ . Without loss of generality, we will assume that f is bounded by 1. In order to show (A.3), we write, for $\eta, M >0$ ,

\begin{align*}&\Bigl|t\mathbb{E} f\Bigl(\frac{X_TT}{b(t)}\Bigr)-\int_0^\infty\mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y)\Bigr|\\&\leq\Bigl|t\mathbb{E}\Bigl(f\Bigl(\frac{X_TT}{b(t)}\Bigr){\unicode{x1D7D9}}_{ \{M \geq T/b(t)\geq\eta\}}\Bigr)-\int_\eta^M\mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y)\Bigr| \\&+t\mathbb{E}\Bigl(f\Bigl(\frac{X_TT}{b(t)}\Bigr){\unicode{x1D7D9}}_{\{T/b(t)<\eta\}}\Bigr) +\int_0^\eta\mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y) \\&+t\mathbb{E}\Bigl(f\Bigl(\frac{X_TT}{b(t)}\Bigr){\unicode{x1D7D9}}_{\{T/b(t)>M\}}\Bigr) +\int_M^\infty\mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y) \\&=: I^1(t,\eta,M)+I^2(t,\eta)+I^3(t,M).\end{align*}

1. (i). We start by showing $I^1(t,\eta,M) \to 0, t \to \infty.$ To this end, first fix some $\epsilon>0$ . By Egorov’s theorem there exists a measurable set $A_\epsilon$ with $\mathbb{P}(A_\epsilon^c)<\epsilon$ such that $X_t\to X_\infty$ for $t\to\infty$ uniformly on $A_\epsilon$ . At this point we use the one-sided continuity of $(X_t)$ as a sufficient condition to ensure the measurability of $A_\epsilon$ . This allows us to bound $I^1(t,\eta,M)$ separately on $A_\epsilon$ and $A_\epsilon^c$ :

   \begin{align*}I^1(t,\eta,M)&\leq t\mathbb{E}\Bigl(f\Bigl(\frac{X_TT}{b(t)}\Bigr){\unicode{x1D7D9}}_{A_\epsilon^c \cap \{M \geq T/b(t)\geq\eta\}}\Bigr)+\int_\eta^M\mathbb{E} \Bigl(f(X_\infty y){\unicode{x1D7D9}}_{A_\epsilon^c}\Bigr) \nu_\alpha(\text{d} y) \\& + t\mathbb{E}\Bigl(\Bigl|f\Bigl(\frac{X_T T}{b(t)}\Bigr)-f\Bigl(\frac{X_\infty T}{b(t)}\Bigr)\Bigr|{\unicode{x1D7D9}}_{A_\epsilon\cap\{M \geq T/b(t)\geq\eta\}}\Bigr)\\& + \Bigl|t\mathbb{E}\Bigl(f\Bigl(\frac{X_\infty T}{b(t)}\Bigr){\unicode{x1D7D9}}_{A_\epsilon\cap\{M \geq T/b(t)\geq\eta\}}\Bigr)-\int_\eta ^M \mathbb{E}\Bigl( f(X_\infty y){\unicode{x1D7D9}}_{A_\epsilon}\Bigr) \nu_\alpha(\text{d} y)\Bigr|\\&=: I_a^1(t,\eta,M)+I_b^1(t,\eta,M)+I_c^1(t,\eta,M)+I_d^1(t,\eta,M).\end{align*}
   Starting with $I_a^1(t,\eta,M)$ , we use that $A_\epsilon^c \in\sigma(X_t,t\geq 0)$ which therefore is independent of T:
   \begin{align*}\limsup_{t \to \infty} I_a^1(t,\eta,M) \leq \limsup_{t \to \infty} t\mathbb{P}\Bigl(A_\epsilon^c \cap\Bigl\{\frac{T}{b(t)}\geq \eta \Bigr\}\Bigr)\leq \epsilon\cdot\eta^{-\alpha}.\end{align*}
   Second,
   \begin{align*}I_b^1(t,\eta,M) \leq \mathbb{P} (A_\epsilon^c)\int_\eta^M \nu_\alpha(\text{d} y) \leq \epsilon \cdot (\eta^{-\alpha}-M^{-\alpha}).\end{align*}
   For $I_c^1(t,\eta,M)$ , we note that $b(t)\to\infty$ for $t\to\infty$ , yielding an arbitrarily large lower bound for T on the set $\{T/b(t)\geq\eta\}$ . Then, intersecting with $A_\epsilon$ we obtain that $b(t)^{-1} T ||X_T - X_\infty ||\cdot{\unicode{x1D7D9}}_{A_\epsilon\cap\{M \geq T/b(t)\geq\eta\}}$ uniformly tends to 0 and combining this with the uniform continuity of f we find a c(t) with $c(t)\to 0$ for $t\to\infty$ such that, for $t \to \infty$ ,
   \begin{align*}I_c^1(t,\eta,M)&\leq c(t)\cdot t\mathbb{P}\Bigl(A_\epsilon\cap\Bigl\{M \geq \frac{T}{b(t)}\geq \eta \Bigr\}\Bigr)\leq c(t)\cdot t\mathbb{P}\Bigl(\frac{T}{b(t)}\geq\eta \Bigr)\to 0.\end{align*}
   Finally, for $I_d^1(t,\eta,M)$ we observe that
   \begin{align*} y \mapsto E(f(X_\infty y)\unicode{x1D7D9}_{A_\epsilon}) \unicode{x1D7D9}_{[\eta,M]}(y)\end{align*}
   is bounded and has support bounded away from 0. Even though it is not continuous in y, a standard approximation argument by continuous functions combined with dominated convergence gives
   \begin{eqnarray*} && t\mathbb{E}\Bigl(f\Bigl(\frac{X_\infty T}{b(t)}\Bigr){\unicode{x1D7D9}}_{A_\epsilon\cap\{M \geq T/b(t)\geq\eta\}}\Bigr) \\ & = & \int_\eta^M \mathbb{E}(f(X_\infty y ){\unicode{x1D7D9}}_{A_\epsilon})t\mathbb{P}^{T/b(t)}(\text{d} y) \to \int_\eta ^M \mathbb{E} (f(X_\infty y){\unicode{x1D7D9}}_{A_\epsilon}) \nu_\alpha(\text{d} y)\end{eqnarray*}
   for $t \to \infty$ , where we used regular variation of T in combination with Fubini’s theorem, as T is independent of $X_\infty$ . Thus, $I_d(t,\eta,M) \to 0$ for $t \to \infty$ . From the above and since we can choose $\epsilon>0$ arbitrarily small, we see that $I^1(t,\eta,M) \to 0, t \to \infty$ .

2. (ii). We split up $I^2(t,\eta)$ into

   \begin{align*} t\mathbb{E}\Bigl(f\Bigl(\frac{X_TT}{b(t)}\Bigr){\unicode{x1D7D9}}_{\{T/b(t)<\eta\}}\Bigr) +\int_0^\eta\mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y)=:I^2_a(t,\eta)+I^2_b(\eta).\end{align*}
   Then, using Markov’s inequality as well as the independence of $(X_t)_{t\geq 0}$ and T, we arrive at the upper bound
   \begin{align*}&I^2_a(t,\eta) \leq t \mathbb{P}\Bigl(\Big\|\frac{X_TT}{b(t)}\Bigr\|>\epsilon_0,\frac{T}{b(t)}<\eta\Bigr)\\&=t\mathbb{P}\Bigl(\Big\|\frac{X_TT}{b(t)}{\unicode{x1D7D9}}_{\{T/b(t)<\eta\}}\Bigr\|>\epsilon_0 \Bigr)\leq \epsilon_0^{-\alpha'}t\mathbb{E}\Bigl(\Big\|\frac{X_TT}{b(t)}{\unicode{x1D7D9}}_{\{T/b(t)<\eta\}}\Bigr\|^{\alpha'}\Bigr)\\&= \epsilon_0^{-\alpha'}\int_{(0,\eta)}\mathbb{E}||X_{b(t)y}||^{\alpha'} y^{\alpha'}t\mathbb{P}(T/b(t)\in\text{d} y)\\&\leq \epsilon_0^{-\alpha'} \sup_{t\geq 0}\mathbb{E}||X_t||^{\alpha'} t\mathbb{E}\Bigl(\Bigl(\frac{T}{b(t)}\Bigr)^{\alpha'}{\unicode{x1D7D9}}_{\{T/b(t)<\eta\}}\Bigr).\end{align*}
   By Karamata’s theorem applied to truncated moments of regularly varying random variables, see Proposition 1.4.6 in [Reference Kulik and Soulier21], we get
   \begin{align*} & t\mathbb{E}\Bigl(\Bigl(\frac{T}{b(t)}\Bigr)^{\alpha'}{\unicode{x1D7D9}}_{\{T/b(t)<\eta\}}\Bigr) \\ &= t\mathbb{P}(T>b(t))\frac{\mathbb{E}\Bigl(T^{\alpha'}\unicode{x1D7D9}_{\{T<\eta b(t)\}}\Bigr)}{(b(t))^{\alpha'}\mathbb{P}(T>b(t))} \\ &\to \frac{\alpha}{\alpha'-\alpha}\eta^{\alpha'-\alpha}.\end{align*}
   Thus, for $t \to \infty, \eta \to 0$ , $I^2_a(t,\eta) \to 0$ .

In order to bound $I^2_b(\eta)$ , we first note that $\mathbb{E}||X_\infty||^{\beta}$ is finite for all $\alpha<\beta<\alpha'$ because $(||X_t||^\beta)_t$ is uniformly integrable (as $\sup_t\mathbb{E}\bigl[(||X_t||^\beta)^{\frac{\alpha'}{\beta}}\bigr]<\infty$ and $\frac{\alpha'}{\beta}>1$ ) and the Vitali convergence theorem implies that $||X_t||^\beta$ converges in $L^1$ to $||X_\infty||^\beta$ and so do the expected values. Then, again by Markov’s inequality, we get

\begin{align*}& \int_0^\eta\mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y) \leq \int_0^\eta \mathbb{P}(\|X_\infty\|> \epsilon_0 /y)\nu_\alpha(\text{d} y) \\&\leq \int_0^\eta \mathbb{E}\bigl(\|X_\infty\|^\beta\bigr) \epsilon_0^{-\beta} y^\beta \nu_\alpha(\text{d} y) \\&\leq \mathbb{E}\bigl(\|X_\infty\|^\beta\bigr) \epsilon_0^{-\beta} \frac{\alpha}{\beta-\alpha}\eta^{\beta-\alpha}.\end{align*}

Again, with $\eta \to 0$ we have shown that $I^2_b(\eta) \to 0$ and therefore $I^2(t,\eta) \to 0$ , and also that the right-hand side in (A.3) is finite, since $\int_\eta^\infty \mathbb{E} f(X_\infty y)\nu_\alpha(\text{d} y) \leq \eta^{-\alpha}$ .

(iii). Finally, regarding $I^3(t,M)$ we note that

\begin{align*} & \limsup_{t \to \infty} I^3(t,M) \leq \limsup_{t \to \infty} t\mathbb{P}(T>M b(t)) + \int_M^\infty \nu_\alpha (dy) \leq 2 M^{-\alpha},\end{align*}

and so $\lim_{M \to \infty} \limsup_{t \to \infty} I^3(t,M)=0$ . This finishes the proof of (A.3).

A.3. Moment conditions

The following statement extends Lemma 4.1 in [Reference Peköz, Röllin and Ross26] to arbitrary values of $\beta \geq 0$ .

Lemma 7. Assume the urn model of Section 2.1.2, fix $k \in \mathbb{N}$ and let

\begin{align*}U_k(n):\!=C_1(n)+\ldots+C_k(n)\end{align*}

denote the total number of balls of colours 1 to k at time n. Then for every $q\in\mathbb{N}$ there exist $c,C>0$ such that, for all $n\in\mathbb{N}_0$ ,

(A.4) \begin{equation} cn^{ql/(l+\beta)} < \mathbb{E}(U_k(n)^q) < Cn^{ql/(l+\beta)}.\end{equation}

Proof. Let $k \in \mathbb{N}$ be fixed throughout this proof. Note first that $U_k(n), n \in \mathbb{N}_0,$ is deterministic until there are at least $k+1$ different colours in the urn, i.e. for $n \leq n_0$ with

\begin{align*}n_0:\!=l(k+1-s) \vee 0,\end{align*}

and set

\begin{align*}n_p:\!=n_0+p,\,p\in\mathbb{N}.\end{align*}

It is then sufficient to show (A.4) for $n=n_p, p \in \mathbb{N},$ as for those finitely many $n \leq n_0$ there exist trivial bounds. Let $h_p=U_{(k+1)\vee s}(n_0)+p+\beta\lfloor{\frac{p}{l}}\rfloor$ be the (deterministic) total number of balls at time $n_p$ and define

(A.5) \begin{equation} M_{p,q}:\!=\prod_{i=0}^{q-1}(i+U_k(n_p)), p \in \mathbb{N}_0, q \in \mathbb{N}.\end{equation}

We start the proof by showing that

(A.6) \begin{equation}\mathbb{E} M_{p,q}=M_{0,q}\cdot\prod_{i=0}^{p-1}\Bigl(1+\frac{q}{h_i}\Bigr).\end{equation}

Given the value $U_k(n_{p-1})$ , there are only two possible values for $U_k(n_p)$ , which are $U_k(n_{p-1})+1$ and $U_k(n_{p-1})$ : either we draw a ball of colours 1 to k or we do not. Using this, we get

\begin{align*}\mathbb{E}(M_{p,q}|U_k(n_{p-1}))&=\frac{U_k(n_{p-1})}{h_{p-1}}\prod_{i=0}^{q-1}(i+U_k(n_{p-1})+1)+\frac{h_{p-1}-U_k(n_{p-1})}{h_{p-1}}M_{p-1,q}\\&= \frac{U_k(n_{p-1})}{h_{p-1}}\prod_{i=1}^{q}(i+U_k(n_{p-1}))+\frac{h_{p-1}-U_k(n_{p-1})}{h_{p-1}}M_{p-1,q}\\&=\frac{U_k(n_{p-1})}{h_{p-1}}\frac{M_{p-1,q}(U_k(n_{p-1})+q)}{U_k(n_{p-1})}+\frac{h_{p-1}-U_k(n_{p-1})}{h_{p-1}}M_{p-1,q}\\&=M_{p-1,q}\Bigl(1+\frac{q}{h_{p-1}}\Bigr).\end{align*}

By iterating this calculation p times, one obtains (A.6).

Now, by definition of $h_p, p \in \mathbb{N}$ , we have

\begin{align*}U_{(k+1)\vee s}(n_0)+p+\beta\Bigl(\frac{p}{l}-1\Bigr)\leq h_p\leq U_{(k+1)\vee s}(n_0)+p+\beta\frac{p}{l}.\end{align*}

Setting $x:\!=\frac{l}{l+\beta}$ and $y:\!=(U_{(k+1)\vee s}(n_0)-\beta)\frac{l}{l+\beta}$ (which is non-negative as for each colour there are at least $\beta$ balls) for brevity, this is equivalent to

(A.7) \begin{align}\frac{y+p}{x} &\leq h_p\leq \frac{y+p+\beta x}{x}.\end{align}

For the upper bound in (A.4) use (A.6) and (A.7) to get

\begin{align*}\mathbb{E} M_{p,q}\leq M_{0,q} \cdot\prod_{i=0}^{p-1}\Bigl(1+\frac{qx}{y+i}\Bigr)&= M_{0,q} \cdot \frac{\Gamma(y)}{\Gamma(qx+y)} \cdot\frac{\Gamma(qx+y+p)}{\Gamma(y+p)}.\end{align*}

Thus, using Lemma 5, there exist $\tilde{C}, C>0$ such that

\begin{align*}\mathbb{E} U_k(n_p)^q&\leq \mathbb{E} M_{p,q} \leq M_{0,q} \cdot \frac{\Gamma(y)}{\Gamma(qx+y)} \cdot\frac{\Gamma(qx+y+p)}{\Gamma(y+p)} \\&\leq \tilde{C} (y+p)^{qx} < C n_p^{qx}= C n_p^{ql/(l+\beta)}\end{align*}

for all $p \in \mathbb{N}$ , proving the upper bound.

For the lower bound, use Jensen’s inequality, (A.6), (A.7), Lemma 5 and similar reasoning as above to see that there exist $\tilde{c}, c>0$ such that

\begin{align*} \mathbb{E} U_k(n_p)^q&=\mathbb{E} M_{p,1}^q\geq(\mathbb{E} M_{p,1})^q \\ &\geq \left(M_{0,1}\cdot\prod_{i=0}^{p-1}\Bigl(1+\frac{x}{y+i+\beta x}\Bigr) \right)^q \\ &= M_{0,1}^q \left(\frac{\Gamma(y+\beta x)}{\Gamma(x+y+\beta x)}\right)^q \left(\frac{\Gamma(x+y+\beta x+p)}{\Gamma(y+\beta x+p)}\right)^q \\ &\geq \tilde{c} (y+\beta x +p)^{qx} > c n_p^{qx}=c n_p^{ql/(l+\beta)},\end{align*}

for all $p \in \mathbb{N}$ , which finishes the proof.

A.4. Proof of Lemma 4

1. (1) We adapt the proof of Lemma 3.3 from [Reference Móri24]. By definition, $B_k^\downarrow=\zeta_{k}/(\zeta_1+\ldots+\zeta_{k})$ and so the statement follows immediately for $k=1$ . Thus, keep $k>1$ fixed in the following. We disregard all colours in the urn model but 1 through k and further simplify the model. To find the distribution of $B_k^\downarrow$ it is sufficient to view all balls of colours 1 to $k-1$ as being ‘black’ and those of colour k as being ‘white’. At the time $n_0^k$ , when the last relevant colour is added to the urn, the number $w_k$ of white and $b_k$ of black balls is deterministic and given by

   \begin{align*} (w_k,b_k)=\begin{cases}(C_k(0),\sum_{i=1}^{k-1}C_i(0)) \quad&\text{if } 1 < k\leq s,\\ (\beta,\sum_{i=1}^{s}C_i(0)+(l+\beta)(k-1-s)+l) \quad &\text{if } k>s.\end{cases} \end{align*}
   Now at each time either the number of black and white balls does not change or a new ball is added to one of them. Conditionally on the latter, the transition probabilities are the same as in a traditional Pólya urn and therefore Lemma 3 yields that the vector $(B_k^\downarrow,1-B_k^\downarrow)$ has a Dirichlet distribution, whose marginals are known to be beta distributions with the given parameters.

2. (2) We provide an alternative construction for $C^r(n)$ to help us derive the desired independence properties. We start with the usual setup of an urn with s different colours and starting amounts $C_0(0),\ldots,C_s(0)$ . From there on, instead of picking a ball directly, in each time step we flip a sequence of independent Bernoulli coins, determined below, until we observe heads for the first time. Then we add a new ball of one colour, based on which flip resulted in heads, and proceed to the next time step. The individual coins may have different success probabilities, i.e. probabilities of showing heads. More precisely, for the ball to be selected at time n, let m be the currently largest numbered colour in the urn. If $m>r$ the first flip in a time step always represents the colours $>r$ . If that coin flip shows heads we add a new ball to one of those (which one specifically is irrelevant to us). If it shows tails, however, we start traversing the remaining r colours in backwards order. This means we start by flipping for colour r, then either add a new ball to it if the flip showed heads or otherwise continue to $r-1$ and so on. If $m\leq r$ we skip the flips for colours $m+1,m+2,\ldots,$ ‘ $>r$ ’. To also incorporate the migration of new colours in this model, after every lth time step, we add additional $\beta$ balls to colour $m+1$ or if $m\geq r$ to the ‘ $>r$ ’ category. Next, for this process to have the same distribution as $(C^r(n))_{n \in \mathbb{N}_0}$ , we need to find suitable success probabilities $p_i(n)$ when flipping for colour i in time step $n\mapsto n+1$ . To this end, we write

   \begin{align*} U_k(n):\!=\sum_{i=1}^k C_i(n), \;\;\; k \geq 1,\end{align*}
   for brevity and set
   \begin{align*}\mbox{if }i\leq r:&&p_i(n)&=\frac{C_i(n)}{C_1(n)+\ldots+C_i(n)}=1-\frac{U_{i-1}(n)}{U_i(n)}, \\\mbox{ and } && p_{>r}(n)&=\frac{C_{r+1}(n)+\ldots+C_m(n)}{C_1(n)+\ldots+C_m(n)}=1-\frac{U_{r}(n)}{U_m(n)}.\end{align*}
   To verify that this yields exactly the same transition probabilities as in the urn model, write (with $\mathcal{F}_n=\sigma(C_i(t):\,t\leq n,\,i,t\in\mathbb{N})$ ) for colours $i\leq r$ ,
   \begin{align*}&\mathbb{P}(C_i(n+1)-C_i(n)=1|\mathcal{F}_{n})=\frac{C_i(n)}{U_m(n)}=\frac{U_{r}(n)}{U_m(n)} \frac{U_{r-1}(n)}{U_r(n)}\ldots \frac{U_i(n)}{U_{i+1}(n)}\frac{C_i(n)}{U_i(n)}\\&\qquad=(1-p_{>r}(n))\cdot (1-p_r(n))\cdot\ldots\cdot (1-p_{i+1}(n))\cdot p_{i}(n)\end{align*}
   and for colour ‘ $>r$ ’,
   \begin{align*} \mathbb{P}([U_{m}(n+1)-U_r(n+1)]-[U_m(n)-U_r(n)]=1|\mathcal{F}_{n})=\frac{U_m(n)-U_r(n)}{U_m(n)}=p_{>r}(n). \end{align*}
   So we conclude that the two processes indeed have the same transition probabilities and proceed to show the independence of the $B_i^\downarrow $ .

For that let $(Y_j^i)_{j=1}^\infty$ be the random variables which contain the outcome of the jth flip for colour i, $i=1,\ldots,r$ . Define $T_j^i$ to be the step in the above urn model when coin $Y_j^i$ is flipped, noting that $T_j^i,\,j>0$ , is random but almost surely finite by Lemma 2. The success probability $p_i(n)$ at time $n=T_j^i$ is dependent on prior flips and given by

\begin{align*} \frac{C_i(n)}{C_1(n)+\ldots+C_i(n)}=\frac{C_i(T^i_1)+\sum_{k=1}^{j-1}Y_k^i}{U_i(T^i_1)+j-1} \end{align*}

for colours $i \leq r$ , as $U_i(T^i_1)$ balls of colours 1 to i and $C_i(T^i_1)$ balls of colour i are present when we first flip the coin for this colour, and at the jth flip a further $j-1$ balls of colours 1 to i have already been added to the urn, with $\sum_{k=1}^{j-1}Y_k^i$ of them to colour i. Therefore, the $p_i(n)$ at time $n=T_j^i$ are completely determined by deterministic starting values $U_i(T^i_1),C_i(T^i_1)$ and the previous flips for colour i and the sequences $(Y_j^1)_j, \ldots, (Y_j^r)_j$ are jointly independent.

With this, the independence of the $B_i^\downarrow$ follows since, for $j \to \infty$ ,

\begin{align*}\frac{C_i(T^i_1)+\sum_{k=1}^{j-1}Y_k^i}{U_i(T^i_1)+j-1}=\frac{C_i(T_j^i)}{C_1(T_j^i)+\ldots+C_i(T_j^i)} \to B_i^\downarrow \qquad\quad\text{ for }i=2,\ldots,r.\end{align*}

(3) In addition to the sequences introduced in (2), let $(Y_j^{>r})_{j=1}^\infty$ be the random variables which contain the outcome of the jth flip for colour $`>r$ ’ and let $T_j^{>r}, j>0$ , be the step in the urn model when coin $Y_j^{>r}$ is flipped. The first flip $T_1^{>r}$ takes place at time $n_0^{r+1}$ with $U_{(r+1) \vee s}(n_0^{r+1})$ total balls in the urn, and with $U_{(r+1) \vee s}(n_0^{r+1})-U_{r}(n_0^{r+1})$ of them of colour $`>r$ ’, both numbers being deterministic. After $n_0^{r+1}$ , the coin for colour $`>r$ ’ is flipped in every consecutive step of the urn model. Thus, at the jth flip for this colour, the total number of balls has grown to $U_{(r+1) \vee s}(n_0^{r+1})+j-1+\beta\lfloor{\frac{j-1}{l}}\rfloor$ (where the last summand is due to the immigration), and the number of colour $`>r$ ’ has grown to $U_{(r+1) \vee s}(n_0^{r+1})-U_{r}(n_0^{r+1})+\sum_{k=1}^{j-1}Y_k^{>r}+\beta\lfloor{\frac{j-1}{l}}\rfloor$ . Therefore, the success probability $p_{>r}(n)$ at time $n=T_j^{>r}$ is given by

\begin{align*} \frac{U_{(r+1) \vee s}(n_0^{r+1})-U_{r}(n_0^{r+1})+\sum_{k=1}^{j-1}Y_k^{>r}+\beta\lfloor{\frac{j-1}{l}}\rfloor}{U_{(r+1) \vee s}(n_0^{r+1})+j-1+\beta\lfloor{\frac{j-1}{l}}\rfloor}, \end{align*}

and the sequence $(Y_j^{r>})_{j=1}^\infty$ is independent of the sequences introduced in (2).

Observe that

\begin{align*}\frac{1}{j^{l/(l+\beta)}}U_r(n_0^{r+1}+j)=\frac{1}{j^{l/(l+\beta)}}(U_r(n_0^{r+1})+\sum_{k=1}^j(1-Y_k^{>r}))\overset{j\to\infty}{\to}\zeta_1+\ldots+\zeta_r,\end{align*}

and by independence of the coin flip sequences the limit $\zeta_1+\ldots+\zeta_r$ is thus independent of $(B_1^\downarrow,\ldots,B_r^\downarrow)$ and thus independent of $(B_1,\ldots,B_r)$ as well.

A.5. Proof of Proposition 2

We start by showing (2).

(2) Case $\alpha \leq p $ : Let $\alpha'>p\geq \alpha$ , then by Hölder’s inequality

\begin{align*}\sup_n\mathbb{E}\Bigl(\Bigl\lVert\frac{D(n)}{n^{l/(l+\beta)}}\Bigr\rVert_p^\alpha\Bigr)\leq \sup_n\Bigl(\mathbb{E}\Bigl(\Bigl\lVert\frac{D(n)}{n^{l/(l+\beta)}}\Bigr\rVert_p^{\alpha'}\Bigr)\Bigr)^{\frac{\alpha}{\alpha'}}=\Bigl(\sup_n\mathbb{E}\Bigl(\Bigl\lVert\frac{D(n)}{n^{l/(l+\beta)}}\Bigr\rVert_p^{\alpha'}\Bigr)\Bigr)^{\frac{\alpha}{\alpha'}}\end{align*}

and this case follows whenever there is a bound for the next one.

Case $\alpha > p$ : Because of the inverse Minkowski’s inequality (see, e.g., III 2.4 Theorem 9 in [Reference Bullen8]), p-quasinorms with $0<p<1$ are concave on $\mathbb{R}_{\geq 0}^d$ . We can thus apply Jensen’s inequality to

\begin{align*}\mathbb{E}\Bigl(\Bigl\lVert\frac{D(n)}{n^{l/(l+\beta)}}\Bigr\rVert_p^\alpha\Bigr)=\mathbb{E}\Bigl(\Bigl\lVert\Bigl(\frac{D(n)}{n^{l/(l+\beta)}}\Bigr)^\alpha\Bigr\rVert_{\frac{p}{\alpha}}\Bigr)\leq \Bigl(\sum_{i=1}^{N(n)}\Bigl(\mathbb{E}\frac{D_i^\alpha(n)}{n^{\alpha\cdot l/(l+\beta)}}\Bigr)^{\frac{p}{\alpha}}\Bigr)^{\frac{\alpha}{p}}.\end{align*}

An application of Proposition 1() and Lemma 6 yields that there exists a $C>0$ such that the right-hand side is bounded above by

\begin{align*}C\cdot\Bigg(\sum_{i=1}^{N(n)}\Bigg(c(n,\alpha)^{-1}\mathbb{E} \left(\begin{array}{c}{D_i(n)+\alpha-1}\\ {\alpha}\end{array}\right)\Bigg)^{\frac{p}{\alpha}}\Bigg)^{\frac{\alpha}{p}}.\end{align*}

Now, each base in the summands on the right is the expectation of a closed martingale and by Proposition 1(),

(A.8) \begin{align}\Biggl(\sum_{i=1}^{N(n)}\Biggl(c(n,\alpha)^{-1}\mathbb{E} \left(\begin{array}{c}{D_i(n)+\alpha-1}\\ {\alpha}\end{array}\right)\Biggr)^{\frac{p}{\alpha}}\Biggr)^{\frac{\alpha}{p}}=\Bigl(\sum_{i=1}^{N(n)}(\mathbb{E} \zeta_i^\alpha)^{\frac{p}{\alpha}}\Bigr)^{\frac{\alpha}{p}}.\end{align}

By (2.3) and Proposition 1() we have, for any fixed $k \in \mathbb{R}_{\geq 0}$ and all $i>N(0)$ ,

(A.9) \begin{align} \mathbb{E} (\zeta_i^k) = C(k) c((i-s),k)^{-1} \sim C(k) (i-s)^{-k\cdot l/(l + \beta)},\end{align}

where C(k) denotes a constant that does not depend on i. Thus, the limit as $n \to \infty$ of the right-hand side of (A.8) is finite if and only if

\begin{align*}\frac{\alpha \cdot l}{l+\beta} \cdot \frac{p}{\alpha} > 1 \;\;\; \Leftrightarrow \;\;\; p > \frac{l+ \beta}{l},\end{align*}

which is guaranteed by our assumption.

1. (1) (a) Assume first that $\frac{l+\beta}{l} < p < \infty$ . Use monotone convergence to get

   (A.10) \begin{align}\mathbb{E}(\|\zeta\|_p^p)=\mathbb{E}\sum_{i=1}^\infty\zeta_i^p =\sum_{i=1}^\infty\mathbb{E}\zeta_i^p.\end{align}
   Again from (A.9), this sum is finite under our assumption that $p > (1+\beta)/l$ and so $\|\zeta\|_p^p<\infty $ almost surely. Furthermore, since $\| \zeta\|_\infty \leq \| \zeta\|_p$ , the result also follows for $p=\infty$ .

2. (1) (b) Next we prove the almost sure convergence in $c_{||\cdot||_p}$ . We need to show that

   \begin{align*}\Bigl\lVert\frac{D(n)}{n^{l/(l+\beta)}}-\zeta\Bigr\rVert_p^p=\sum_{i=1}^\infty\Bigl|\frac{D_i(n)}{n^{l/(l+\beta)}}-\zeta_i\Bigr|^p\to 0,\,n\to\infty\quad\text{almost surely}.\end{align*}
   By Scheffé’s theorem the above convergence follows from componentwise convergence, i.e.
   (A.11) \begin{align} \frac{D_i(n)}{n^{l/(l+\beta)}} \to \zeta_i, \, n \to \infty \quad\text{almost surely}, \end{align}
   for all $i \in \mathbb{N}$ in combination with convergence of the norms, which is
   (A.12) \begin{align}\sum_{i=1}^{N(n)}\frac{D_i(n)^p}{n^{pl/(l+\beta)}} \to\sum_{i=1}^\infty\zeta_i^p,\,n\to\infty. \end{align}
   Proposition 1() implies (A.11), so we are left to show (A.12). To this end, introduce the random variables $X_i(n):\!=c(n,p)^{-1}\left(\begin{array}{c}{D_i(n)+p-1}\\ {p}\end{array}\right)$ , which form, for each $i \in \mathbb{N}$ , a martingale, according to Proposition 1(). Their cumulative sums form submartingales, since
   \begin{align*}\mathbb{E}\Bigl(\sum_{i=1}^{N(n)}X_i(n)\Big|\mathcal{F}_{(n-1)l}\Bigr)=\sum_{i=1}^{N(n)}\mathbb{E}(X_i(n)|\mathcal{F}_{(n-1)l})\geq\sum_{i=1}^{{N(n-1)}}X_i(n-1), \; i \in \mathbb{N}.\end{align*}
   By Lemma 6 there exists, for each $k \in \mathbb{N}$ , a C such that
   \begin{align*}\mathbb{E}\Bigl(\Bigl(\sum_{i=1}^{N(n)}X_i(n)\Bigr)^k\Bigr)\leq C\mathbb{E}\Bigl(\sum_{i=1}^{N(n)}\frac{D_i(n)^p}{n^{pl/(l+\beta)}}\Bigr)^k.\end{align*}
   By (2) with $\alpha=pk$ , we observe that the right-hand side is bounded uniformly in n. This means the submartingale is uniformly integrable and therefore convergent almost surely and in $L_1$ (Chapter XI.14 in [Reference Doob14]). To derive the limit we first note that, by Proposition 1(2),
   \begin{align*}\lim_{n \to \infty} \sum_{i=1}^{N(n)}X_i(n) - \sum_{i=1}^l \frac{\zeta_i^p}{\Gamma(p+1)}=\lim_{n \to \infty} \sum_{i=l+1}^{N(n)}X_i(n) \geq 0\end{align*}
   for all $l \in \mathbb{N}$ . Let $l \to \infty$ to conclude that
   \begin{align*} \lim_{n \to \infty} \sum_{i=1}^{N(n)}X_i(n) - \sum_{i=1}^\infty \frac{\zeta_i^p}{\Gamma(p+1)} \geq 0.\end{align*}
   This implies that
   \begin{align*}&E\Bigl( \Bigl| \lim_{n \to \infty} \sum_{i=1}^{N(n)}X_i(n) - \sum_{i=1}^\infty \frac{\zeta_i^p}{\Gamma(p+1)} \Bigr| \Bigr) \\&= E\Bigl( \lim_{n \to \infty} \sum_{i=1}^{N(n)}X_i(n) - \sum_{i=1}^\infty \frac{\zeta_i^p}{\Gamma(p+1)} \Bigr) \\&= \lim_{n \to \infty} \sum_{i=1}^{N(n)}E(X_i(n))- \sum_{i=1}^\infty E\Bigl(\frac{\zeta_i^p}{\Gamma(p+1)} \Bigr) = 0,\end{align*}
   where we used the $L_1$ -convergence of our submartingale together with (A.10) in the penultimate step and Proposition 1() in the last step. Thus,
   (A.13) \begin{align} \lim_{n \to \infty} \sum_{i=1}^{N(n)}c(n,p)^{-1}\left(\begin{array}{c}{D_i(n)+p-1}\\ {p}\end{array}\right)= \sum_{i=1}^\infty \frac{\zeta_i^p}{\Gamma(p+1)} \;\; \mbox{almost surely}. \end{align}
   Returning to the original process, we can use Proposition 1() and Lemma 6 to find a $C>0$ such that
   \begin{align*} \frac{D_i(n)^p}{n^{pl/(l+\beta)}} \leq C \cdot c(n,k)^{-1}\left(\begin{array}{c}{D_i(n)+p-1}\\ {p}\end{array}\right)\end{align*}
   for all $i=1, \ldots, N(n)$ , and thus apply Pratt’s lemma [Reference Pratt29] to finally arrive at (A.12) and therefore conclude (1)(b).

3. (3) For $p\in (1,\infty)$ the p-norm on $\mathbb{R}^d$ among all vectors of equal $\ell_1$ -norm is minimised if each component takes the same value. Let c be the cumulated starting weight of the initial N(0) vertices. Then

   \begin{align*}\Bigl\lVert\frac{D(n)}{n^{l/(l+\beta)}}\Bigr\rVert_p^p=\frac{1}{n^{p\cdot l/(l+\beta)}}\sum_{i=1}^{{N(n)}}D_i^p(n)\geq \frac{(N(0)+n)\cdot (\frac{c+n(l+\beta)}{N(0)+n})^p}{n^{p\cdot l/(l+\beta)}}\sim C\cdot n^{1-\frac{p\cdot l}{l+\beta}}\end{align*}
   for some $C>0$ and, since by assumption $p<\frac{l+\beta}{l}$ , the right-hand side diverges to $\infty$ . Being almost surely unbounded, the sequence $D(n)/n^{l/(l+\beta)}$ thus fails to converge.