Multi-drawing P\'olya urns via labelled random DAGs

TL;DR

This paper extends the classical Pólya urn model to multi-draw scenarios using random DAGs and stochastic tensors, providing new convergence results for these complex processes.

Contribution

It introduces a novel approach linking multi-drawing Pólya urns to random DAGs and stochastic tensors, enabling convergence analysis.

Findings

Proves convergence of multi-drawing Pólya urns under practical assumptions.

Develops a new analytical framework using random DAGs and stochastic tensors.

Simplifies the verification of conditions for convergence.

Abstract

A P\'olya urn of replacement matrix $R=(R_{i,j})_{1\leq i,j\leq d}$ is a Markov process that encodes the following experiment: an urn contains balls of $d$ different colours and at every time-step, a ball is drawn uniformly at random in the urn, and if its colour is $i$, then it is replaced in the urn with an additional $R_{i,j}$ balls of colour $j$, for all $1\leq i, j\leq d$. We study a natural extension of this model in which, instead of drawing one ball at each time-step, we draw a set of $m\geq 2$ balls: in this case, the replacement matrix becomes a replacement tensor. Because of the multi-draws, this process can no longer be seen as a branching process, which makes its analysis much more intricate than in the classical P\'olya urn case. Partial results proved by stochastic approximation techniques exist in the literature. In this article, we introduce a new approach based on seeing the process as a stochastic process indexed by a random directed-acyclic graph (DAG) and use this approach, together with the theory of stochastic tensors, to prove a convergence theorem for these multi-drawing P\'olya urns, with assumptions that are straightforward to check in practice.

Equations

$$U_i(n+1)=U_i(n)+R(i,C_1(n),C_2(n)),$$

$$\mathbb{P}\mathopen{}\mathclose{{\left(C_{1}(n)=k}}\right)=\mathbb{P}\mathopen{}\mathclose{{\left(C_{2}(n)=k}}\right)=\frac{U_{k}(n)}{\sum_{i=1}^{d}U_{i}(n)},$$

$${\mathcal{R}}_k(x,y)=\sum _{i=1}^d\sum _{j=1}^d{R}_{kij}{x}_i{y}_j,$$

$$\mathcal{R}(e_i,e_j)=(R_{1ij},\dots ,R_{dij}),$$

$$\Delta_{d}=\mathopen{}\mathclose{{\left\{x\in[0,1]^{d}\nonscript\>\middle|\nonscript\>\mathopen{}\|x\|_{1}=1}}\right\}.$$

$$\frac{U(n)}{\Vert U(n){\Vert }_1}\to x^{\ast }.$$

$$\Vert U(n){\Vert }_1=\sum _{i=1}^d{U}_i(n)=\Vert U(0){\Vert }_1+\sigma n.$$

$$\bigg{(}m_{n}:=\sum_{i=1}^{d}U_{i}(n)\delta_{i}\bigg{)}_{n\geq 0},$$

$${\hat{m}}_n=\frac{m_n}{m_n(\{1,\dots ,d\})},$$

$$\rho (j,k)=\sum _{i=1}^d{R}_{ijk}{\delta }_i.$$

$$m_{n+1}=m_n+\rho (C_1(n),C_2(n)),$$

$$(A_n,B_n)\Rightarrow \nu \otimes \nu ,$$

$$X(n+1)=({\tilde{C}}_1(n),{\tilde{C}}_2(n)),$$

$$\big{(}\tilde{C}_{1}(n),\tilde{C}_{2}(n)\big{)}\sim\hat{m}_{n}\otimes\hat{m}_{n}.$$

$$X(n+1)=\big{(}\tilde{C}_{1}(n),\tilde{C}_{2}(n)\big{)}\Rightarrow\nu\otimes\nu,$$

$$X(n)\Rightarrow \nu \otimes \nu .$$

$${\mathcal{E}}_n$$

$${\mathcal{F}}_n$$

$$\displaystyle\mathbb{P}\mathopen{}\mathclose{{\left(\exists v\in L_{n}:v<n_{1}(n)}}\right)$$

$$\displaystyle\leq 2^{\ell(n)}\mathbb{P}\mathopen{}\mathclose{{\left(nX_{1}\cdots X_{\ell(n)}-\ell(n)<n_{1}(n)}}\right).$$

$$\displaystyle\mathbb{P}\mathopen{}\mathclose{{\left(nX_{1}\cdots X_{\ell(n)}-\ell(n)<n_{1}(n)}}\right)$$

$$\displaystyle=\mathbb{P}\bigg{(}\sum_{i=1}^{\ell(n)}\big{(}\log(1/X_{i})-1\big{)}>\log\bigg{(}\frac{n}{n_{1}(n)+\ell(n)}\bigg{)}-\ell(n)\bigg{)}.$$

$$\log\bigg{(}\frac{n}{n_{1}(n)+\ell(n)}\bigg{)}-\ell(n)=\log n-\log\bigg{(}\frac{n}{\log n}+o(\log\log n)\bigg{)}+o(\log\log n)=\log\log n+o(\log\log n).$$

$$\mathbb{P}\mathopen{}\mathclose{{\left(nX_{1}\cdots X_{\ell(n)}-\ell(n)<n_{1}(n)}}\right)\leq\frac{\ell(n)}{(1+o(1))(\log\log n)^{2}},$$

$$\mathbb{P}\mathopen{}\mathclose{{\left(\exists v\in L_{n}:v<n_{1}(n)}}\right)\leq\frac{\ell(n)2^{\ell(n)}}{((1+o(1))\log\log n)^{2}}\to 0,$$

$$\mathbb{P}(W_{i}^{\scriptscriptstyle(n)}=w\>|\>(W_{i}^{\scriptscriptstyle(n-1)})_{i\geq 1})=T\big{(}w,W^{\scriptscriptstyle(n-1)}_{2i},W^{\scriptscriptstyle(n-1)}_{2i+1}\big{)}.$$

$${\pi }_i^{(n)}(w)=\sum _{x,y\in \mathcal{S}}{T}_{wxy}{\pi }_{2i}^{(n-1)}(x){\pi }_{2i+1}^{(n-1)}(y).$$

$$\pi_{i}^{\scriptscriptstyle(n)}=\mathcal{T}\big{(}\pi_{2i}^{\scriptscriptstyle(n-1)},\pi_{2i+1}^{\scriptscriptstyle(n-1)}\big{)}.$$

$$\tau (\mathfrak{p})=\underset{x,y\in \Delta (\mathcal{S})}{\sup }\frac{\Vert \mathfrak{p}x-\mathfrak{p}y{\Vert }_1}{\Vert x-y{\Vert }_1},$$

$$\Delta(\mathcal{S})=\{x=(x_{i})_{i\in\mathcal{S}}\in[0,1]^{\mathopen{}\mathclose{{\left\lvert\mathcal{S}}}\right\rvert}\nonscript\>|\nonscript\>\mathopen{}\sum_{i\in\mathcal{S}}x_{i}=1\}.$$

$$(T_x^{(\ell )}{)}_{i,j}=\sum _{k=1}^d{T}_{ikj}{x}_k(T_x^{(r)}{)}_{i,j}=\sum _{k=1}^d{T}_{ijk}{x}_k.$$

$${\tau }_{\ell }(T):=\underset{x\in \Delta (\mathcal{S})}{\max }\tau (T_x^{(\ell )})\text{ and }{\tau }_r(T):=\underset{x\in \Delta (\mathcal{S})}{\max }\tau (T_x^{(r)}).$$

Full text

1 Introduction

A $d$-colour Pólya urn of replacement matrix $\mathfrak{r}=(\mathfrak{r}_{i,j})_{1\leq i,j\leq d}$ is a stochastic process that describes the contents of an urn containing balls of $d$ different colours. At every time step, we pick a ball uniformly at random, and if its colour is $i$, then we replace it in the urn together with an additional $\mathfrak{r}_{i,j}$ balls of colour $j$, for all $1\leq j\leq d$. The case when the replacement matrix is the identity (or an integer multiple of it) dates back to Markov [Mar06] as well as Pólya & Eggenberger [EP23]. In this case, if $U(n)=(U_{1}(n),\ldots,U_{d}(n))$, where $U_{i}(n)$ is the number of balls of colour $i$ in the urn at time $n$, then $U(n)/n$ converges almost surely to a random variable whose distribution depends on $U(0)$. A very different behaviour is observed when the replacement matrix $\mathfrak{r}$ is irreducible: in this case, by Perron-Frobenius theorem, the spectral radius $\lambda$ of $\mathfrak{r}$ is a simple eigenvalue of $\mathfrak{r}$, and it admits a unique eigenvector $v$ whose coordinates are all non-negative and such that $\|v\|_{1}=1$. Athreya & Karlin [AK68] proved that, in this case, almost surely as $n\uparrow\infty$, $U(n)/n\to\lambda v$. Contrary to the “identity case”, the limit here is deterministic and does not depend on $U(0)$. Since Athreya & Karlin’s work, the model has been largely studied and many generalisations have been considered: for example, Janson [Jan04] generalises the convergence result of Athreya and Karlin to the case when the replacement matrix is random and balls of different colours have different weights/activities, and provides functional limit theorems for the fluctuations of $U(n)/n$ around its limit.

The aim of this paper is to generalise the results of Athreya & Karlin [AK68] to the context of multi-drawing urns: the idea is that, instead of drawing one ball at random in the urn at every time step, we draw a couple of balls from the urn (with replacement) and, if the first ball is of colour $j$, and the second one of colour $k$, then we add $R_{ijk}$ balls of colour $i$, for all $1\leq i\leq d$. The collection of integers $(R_{ijk})_{1\leq i,j,k\leq d}$ is called the “replacement tensor” of the multi-drawing urn. Note that our aim is in fact to generalise to the case when an $m$-tuple of balls is drawn from the urn at every time step, and our results do apply to $m>2$. However, in order to keep notation as simple as possible we choose to state and prove our results in the case $m=2$; the extension to $m>2$ is outlined in Section 4. The main difficulty is that the methods of Athreya & Karlin [AK68] and Janson [Jan04] both rely on embedding the urn process into continuous time to get a multi-type branching process. Although this could be done in the multi-drawing case, the resulting continuous-time process is not a branching process.

Multi-drawing Pólya urns have been studied in the literature, mostly the two-colour case. For example, Chen & Wei [CW05], Chen & Kuba [CK13] and Kuba, Mahmoud & Panholzer [KMP13] and Aguech, Lasmar & Selmi [RNO19] all study particular examples. Kuba & Mahmoud [KM17] and Kuba & Sulzbach [KS17a] gave general convergence results (and up to third-order asymptotics) for the composition of a $2$-colour urn under the assumption that there exists two deterministic sequences $(\alpha_{n})_{n\geq 0}$ and $(\beta_{n})_{n\geq 0}$ such that, for all $n\geq 0$, $\mathbb{E}[U_{1}(n)|\mathcal{F}_{n}]=\alpha_{n}U_{1}(n)+\beta_{n}$; they call this assumption the “affinity” assumption. This assumption is arguably quite restrictive, but it has the advantage to make the problem amenable to standard martingale techniques. Later, Lasmar, Mailler & Selmi [LMS18] proved a convergence result for general $m$-drawing Pólya urns. Their results, proved using stochastic approximation methods, apply without the affinity assumption, but under assumptions that are difficult to verify in practice.

In this paper, we use a different point of view on multi-drawing Pólya urns to provide general convergence results for multi-drawing Pólya urns under assumptions that are easy to verify in practice. Namely, inspired by the work of Mailler & Marckert [MM17] who see Pólya urns as branching Markov chains on the random recursive tree, we see the multi-drawing Pólya urns as “higher order Markov chains” or “$m$-dependent Markov chains” indexed by random DAGs. Our proof relies on both the theory of higher-order Markov chains and their transition tensors (equivalent of the transition matrix for a Markov chain), and asymptotic results on the shape of a typical large random DAG.

In the rest of the introduction, we first state our main result, before giving the main ideas behind the proof, and then discussing the result in light of the existing literature.

1.1 Main Result

Definition 1.1**.**

Let $(R_{ijk}=R(i,j,k))_{1\leq i,j,k\leq d}$ be a collection of non-negative real numbers. We define the 2-drawing d-colour Plya urn with replacement $R$ as a stochastic process $(U(n)=(U_{1}(n),\ldots,U_{d}(n)))_{n\geq 0}$ on $\mathbb{N}^{d}$ satisfying

[TABLE]

for all $n\geq 0$ and $1\leq i\leq d$, where $(C_{1}(n),C_{2}(n))$ is a couple of independent and identically-distributed random variables on $\{1,\ldots,d\}$ satisfying

[TABLE]

for all $k\in\{1,\ldots,d\}$.

In other words, if we interpret $(U(n))_{n\geq 0}$ as the composition of an urn as time increases, then this process encodes the following experiment: At every time-step, we pick two balls uniformly at random in the urn and let $C_{1}(n)$ denote the colour of the first ball, $C_{2}(n)$ the colour of the second ball (note that we put the first ball back into the urn before drawing the second ball, and then put the second ball back in the urn as well). We then add $R(i,C_{1}(n),C_{2}(n))$ balls of colour $i$ in the urn, for all $1\leq i\leq d$.

Remark 1.2**.**

In the interpretation as balls in an urn, we need that $R_{ijk}\in\mathbb{N}$ for all $1\leq i,j,k\leq d$. This assumption is not needed in Definition 1.1. Also note that we require that $R_{ijk}\geq 0$ for all $1\leq i,j,k\leq d$; in terms of balls in an urn, this means that we never remove balls from the urn. In practice, the model still makes sense if we remove balls from the urn, as long as all the coordinates of $U(n)$ stay non-negative. In the literature on classical Pólya urns, one can either work conditionally on the event that all coordinates of $(U(n))_{n\geq 0}$ stay non-negative, or choose a replacement rule such that this event holds almost surely.

Note that, just as a $d\times d$ matrix defines a linear map from $\mathbb{R}^{d}$ into itself, a collection $(R_{ijk}=R(i,j,k))_{1\leq i,j,k\leq d}$ of real numbers defines a bilinear map from $\mathbb{R}^{d}\times\mathbb{R}^{d}$ into $\mathbb{R}^{d}$. Indeed, for all $x=(x_{1},\ldots,x_{d})$ and $y=(y_{1},\ldots,y_{d})$ in $\mathbb{R}^{d}$, if we set, for all $1\leq k\leq d$,

[TABLE]

and $\mathcal{R}(x,y)=(\mathcal{R}_{1}(x,y),\ldots,\mathcal{R}_{d}(x,y))\in\mathbb{R}^{d}$, then $\mathcal{R}$ is a bilinear map from $\mathbb{R}^{d}\times R^{d}\to\mathbb{R}^{d}$. We refer to both $R=(R_{ijk})_{1\leq i,j,k\leq d}$ and $\mathcal{R}$ as ($3$-mode) tensors. Note that, if we let $(e_{1},\ldots,e_{d})$ be the canonical basis of $\mathbb{R}^{d}$, then, for all $1\leq i,j\leq d$,

[TABLE]

which correspond to what we add in the urn when we first draw a ball of colour $i$ and then a ball of colour $j$.

In this paper, we assume that the replacement tensor $(R_{ijk})_{1\leq i,j,k\leq d}$ satisfies the following assumptions:

• (T)

For all $1\leq i,j,k\leq d$, $R_{ijk}\geq 0$.

• (B)

There exists $\sigma\in(0,\infty)$ such that, for all $1\leq j,k\leq d$, $\sum_{i=1}^{d}R_{ijk}=\sigma$.

• (E)

\max_{1\leq j,j^{\prime},k,k^{\prime}\leq d}\sum_{i=1}^{d}\mathopen{}\mathclose{{\left\lvert R_{ijk}-R_{ij^{\prime}k^{\prime}}}}\right\rvert<\sigma.

We define

[TABLE]

Theorem 1.3**.**

Let $(U(n))_{n}$ be a $2$-drawing $d$-colour urn whose replacement tensor $\mathcal{R}$ satisfies assumptions (T), (B) and (E). Then,

• (i)

there exists a unique $x^{*}\in\Delta_{d}$ such that $\sigma x^{*}=\mathcal{R}(x^{*},x^{*})$ and,

• (ii)

as long as $\|U(0)\|_{1}>0$, in probability as $n\uparrow\infty$,

[TABLE]

Remark 1.4** (Discussion of the assumptions).**

The first assumption ensures that the urn is “tenable”, i.e. that all coordinates of $(U(n))_{n\geq 0}$ are non-negative at all times, and thus the urn process is well-defined. The second assumption is common (although not desirable) in the classical Pólya urn case and is called the “balance” assumption. It ensures that the total number of balls in the urn is a deterministic function of time; more precisely, for all $n\geq 0$,

[TABLE]

Assumption (E) ensures that the map $x\mapsto\mathcal{R}(x,x)/\sigma$ is a contraction: it implies that there exists $q\in(0,1)$ such that, for all $x,y\in\Delta_{d}$, $\|\mathcal{R}(x,x)-\mathcal{R}(y,y)\|_{1}\leq q\sigma\|x-y\|_{1}$. In the classical case of balanced uni-drawing urns (of replacement matrix $\mathfrak{r}$), this assumption would be $\max_{1\leq i,j\leq d}\mathfrak{r}_{i,j}<\sigma$, which, as explained in [FT20, Theorem 4.3], implies that the map $x\mapsto x\mathfrak{r}/\sigma$ is a contraction and thus that the Markov chain of transition matrix $\mathfrak{r}/\sigma$ is ergodic. This is why we call this assumption the ergodicity assumption.

A straightforward generalisation is for the case of (finitely-many-colour) $m$-drawing Pólya urns. The only reason we state our main theorem and its proof for $m=2$ is to keep the notation as simple as possible and thus make the result and its proof easier to read. No additional idea is needed for larger values of $m$. The equivalent of Theorem 1.3 and a sketch of its proof are provided in Section 4.

Interestingly, most of our proof holds even if the set of colours is infinite (possibly uncountable). However, Assumption (E) is much more abstract in this setting and difficult to check in practice: we thus only state our result for infinitely-many-colour, two-drawing Pólya urns in Section 6.

1.2 Ideas of the proof

As mentioned in the introduction, our proof relies on seeing the $m$-drawing Pólya urn as an “$m$-dependent” Markov chain indexed by a random DAG. For all $n\geq 0$, node $n$ in the DAG is labelled by the set of balls added at time $n$ in the urn. The $m$ parents of node $n$ are the nodes corresponding to the $m$ balls drawn at random at time $n$ in the urn. We will show that the DAG obtained is the “uniform recursive DAG”, and that the distribution of the label of each node only depends on the labels of its $m$ parents. When $m=1$, as seen in [MM17], we get a branching Markov chain on the random recursive tree; we call the analogous object when $m\geq 2$ an “$m$-dependent Markov chain on the uniform recursive DAG”.

We first use results of Janson [Jan24] to prove that the uniform recursive DAG “looks locally like a complete $m$-ary tree” and thus reduce our study to $m$-dependent Markov chains on the infinite $m$-ary tree. This removes the difficulty coming from the fact that a DAG has (non-directed) loops.

Ergodicity of $m$-dependent Markov chains on the infinite $m$-ary tree has been studied in the literature (see, e.g. Ching, Ng and Fung [CNF08]). As developed by Lim [Lim05] and Qi [Qi05], the transition probabilities of such a process can be represented by a transition tensor (instead of a transition matrix for classical $1$-dependent Markov chains). The question of existence and uniqueness of stationary distributions for $m$-dependent Markov chains leads to the study of fixed points, or Z-eigenvectors, of such tensors. As tensors are multilinear maps, solving fixed-point equations is in general much harder than for matrices (in fact, it is NP-hard for general tensors – see [SPY16, ĆP07]). Some sufficient criteria for uniqueness of a fixed point of a transition tensor are given in Li and Ng [LN14], Chang and Zhang [CZ13], Wang and Li [WL23], Fascino and Tudisco [FT20] (see also [CQZ13] for an early survey). A more general view is taken in Gautier, Tudisco and Hein [GTH23, GT19, GTH19a, GTH19b], where the authors generalise the Perron-Frobenius theory to multi-homogeneous mappings.

Further, the existence of a unique fixed point, or stationary distribution, of such a tensor is not sufficient to imply convergence (in distribution) of the $m$-dependent Markov chain. To get convergence of our $m$-dependent Markov chain on the infinite $m$-ary tree, we apply results of Fascino and Tudisco [FT20].

1.3 Discussion of our main result in view of the existing literature

As already mentioned, general results for the convergence of multi-drawing Pólya urns already exist in the literature. The aim of this section is to discuss our main result in view of this existing literature.

One strand of the literature focuses on “affine” Pólya urns: see, in particular, Kuba & Mahmoud [KM17] and Kuba & Sulzbach [KS17a]. They say that a Pólya urn $(U(n))_{n\geq 0}$ is affine if there exists two deterministic sequences $(\alpha_{n})_{n\geq 0}$ and $(\beta_{n})_{n\geq 0}$ such that, for all $n\geq 0$, $\mathbb{E}[U_{1}(n)|\mathcal{F}_{n}]=\alpha_{n}U_{1}(n)+\beta_{n}$. They are able to prove up to third-order asymptotic results for urns that satisfy this assumption. In Section 5.2, we show how to apply Theorem 1.3 to affine two-drawing urns and recover the first-order results of Kuba & Mahmoud [KM17] for a large class of affine urns (albeit not all of them, importantly). Because our result is not tailored for affine Pólya urns, but applies more generally, it should not be surprising that our result does not fully recover the first-order results of Kuba & Mahmoud [KM17], which use powerful martingale methods that apply only to affine urns. These martingale methods are also the reason why Kuba & Mahmoud [KM17] and Kuba & Sulzbach [KS17a] are able to prove second- and third-order asymptotic results for affine urns, while these remain completely open in our more general setting.

A different general result for multi-drawing Pólya urns is given by Lasmar, Mailler and Selmi [LMS18]: they are able to prove up-to-second-order asymptotic results for balanced multi-drawing Pólya urns. The drawback of their first-order result is that the assumptions are difficult to check on practical examples; the advantage of it is its generality. In Section 5.3, we apply Theorem 1.3 to examples treated in [LMS18]. On one of these examples, our main result applies and gives convergence of the composition of the urn with much less effort than in [LMS18]. In many other of these examples, our main result simply does not apply. In conclusion, the main advantage of Theorem 1.3 is that its assumptions are easy to apply; however, these assumptions seem to be quite restrictive in practice.

Note that our theorem only applies to drawing balls “with replacement”: when drawing our set of $m$ balls at every time-step, we first draw the first ball, take note of its colour, put it back in the urn, then draw the second ball, etc. The results of [LMS18] hold both with or without replacement; those of [KM17] and [KS17a] hold without replacement (although their proofs would adapt straightforwardly to the with replacement case). Because the total number of balls in the urn tends deterministically to infinity (by Assumption (B)), drawing with or without replacement should not make any difference at large times111This heuristic is supported by results in the literature where distances between the two measures are estimated (see, e.g. [Fre77, Sta78]). and our result should hold also in the without-replacement case. However, our proof does not straightforwardly extend to the without-replacement case.

Also note that, in our framework, at every step, the order in which the $m$ balls are sampled can be taken into account. In other word, we do not assume that $R_{ijk}=R_{ikj}$ for al $1\leq i,j,k\leq d$. As far as we know, this is not a possibility in the existing literature mentioned so far ([KM17, KS17a, LMS18]). In Section 5.2, we include two examples in which the order of the $m$ balls matter in the replacement rule and call them “asymmetric examples”.

Finally, and importantly, a strength of our result and proof is that it extends to infinitely-many colour Pólya urns, as we show in Section 6. The only caveat is that the equivalent of Assumption (E) for infinitely-many colour Pólya urns is very hard to check, mainly because the paper of Fasino and Tudisco [FT20] only holds for $m$-dependent Markov chains on finite state-space: we leave it for future work to extend the results of Fasino and Tudisco [FT20] to infinite state-space and thus provide an easier-to-check sufficient assumption for (E).

Plan of the paper: In Section 2, we introduce our new point of view on multi-drawing Pólya urns: seeing them as a random labelling of a uniform random directed acyclic graph (DAG). We also give preliminary results on the typical shape of the uniform random DAG. In Section 3, we prove Theorem 1.3. In Section 4, we show how Theorem 1.3 can be adapted to larger values of $m$. In Section 5, we give a series of examples. Finally, in Section 6, we show how our model and Theorem 1.3 can be extended to multi-drawing Pólya urns with infinitely-many colours.

2 A Markov chain indexed by a uniform random DAG model

In this section, we show that the multi-drawing Pólya urn can be seen as a randomly-labelled uniform recursive DAG. This is the equivalent to the result of Mailler & Marckert [MM17], which says that Pólya urns can be seen as branching Markov chains on the random recursive tree. Because of the multi-drawing, the underlying graph is a DAG instead of a tree.

2.1 Seeing the urn process as a measure-valued process

In the following, it is convenient to consider the measure-valued process

[TABLE]

which contains all the information about $(U(n))_{n\geq 0}$. Under assumption (T), $m_{n}$ is a measure on $\{1,\ldots,d\}$ for all $n\geq 0$. For all $n\geq 0$, we let

[TABLE]

so that $\hat{m}_{n}$ is a probability measure for all $n\geq 0$.

With this new perspective, it is convenient to also have a measure version of the replacement tensor: for all $1\leq j,k\leq d$, we let

[TABLE]

With this notation, we have that, for all $n\geq 0$,

[TABLE]

where, given $m_{n}$, $(C_{1}(n),C_{2}(n))\sim\hat{m}_{n}\otimes\hat{m}_{n}$.

Lemma 2.1**.**

The following two statements are equivalent:

• (a)

$U(n)/\|U(n)\|_{1}\to x^{*}$* in probability as $n\uparrow\infty$.*

• (b)

$\hat{m}_{n}\to\nu$* in probability as $n\uparrow\infty$, for the topology of weak convergence on the set of probability measures on $\{1,\ldots,d\}$, where $\nu=\sum_{i=1}^{d}x_{i}^{*}\delta_{i}$.*

Remark 2.2**.**

Let $(\mu_{n})_{n\geq 0}$ and $\mu$ be probability measures on a measurable space $E$. We say that $\mu_{n}\to\mu$ for the topology of weak convergence if, for all continuous and measurable functions $f\colon E\to\mathbb{R}$, $\int_{E}f\mathrm{d}\mu_{n}\to\int_{E}f\mathrm{d}\mu_{n}$. If $E=\{1,\ldots,d\}$, it is equivalent to $\mu_{n}(\{i\})\to\mu(\{i\})$ for all $1\leq i\leq d$, and thus Lemma 2.1 is immediate.

Our aim is thus to show that $\hat{m}_{n}\to\nu=\sum_{i=1}^{d}x_{i}^{*}\delta_{i}$, in probability as $n\uparrow\infty$. To do so, we use the following lemma, which is folklore, and whose proof can be found in [MM17] (see Lemma 3.1 therein). We use the following notation: for any two probability measures $\mu$ and $\nu$, we write $(A,B)\sim\mu\otimes\nu$ to say that $A\sim\mu$ and $B\sim\nu$ are independent.

Lemma 2.3**.**

Let $(\nu_{n})_{n\geq 0}$ be a sequence of random probability measures on a measurable space $E$ and $\nu$ be a deterministic probability measure on $E$. For any integer $n\geq 0$, given $\nu_{n}$, we let $(A_{n},B_{n})\sim\nu_{n}\otimes\nu_{n}$. If, in distribution as $n\uparrow\infty$,

[TABLE]

then $\nu_{n}\to\nu$ in probability as $n\uparrow\infty$, for the topology of weak convergence on the set of probability measures on $E$.

Remark 2.4**.**

In Lemma 2.3, it is important to note that $A_{n}$ and $B_{n}$ are not independent random variables. Indeed they both depend on $\nu_{n}$, their random distribution.

2.2 Coupling of the urn process with a randomly labelled DAG

Definition 2.5**.**

For fixed $\pi$ a probability measure on $\{1,\ldots,d\}$ and $3$-mode tensor $\mathcal{R}$, which satisfies Assumptions (T) and (B), we jointly define a sequence of random DAGs $(G_{n}=(V_{n},E_{n}))_{n\geq 0}$, and a sequence of random labels $(X(n)=(X_{1}(n),X_{2}(n))\in\{1,\ldots,d\}^{2})_{n\geq 1}$ as follows: We first set $G_{0}=(\{0\},\varnothing)$ (this is the graph with one node and no edges). Node [math] has no label. Then, for all $n\geq 0$, given $G_{n}$ and $X(0),\ldots,X(n)$, we let $V_{n+1}=\{0,\ldots,n+1\}$ be the set of nodes of $G_{n+1}$ and:

• •

We sample $M_{n+1}$ and $D_{n+1}$ in $\{0,\ldots,n\}$ independently, uniformly at random in $\{0,\ldots,n\}$.

• •

We then sample $\tilde{C}_{1}(n)$ and $\tilde{C}_{2}(n)$ independently with the following distributions:

• –

If $M_{n+1}=0$, then $\tilde{C}_{1}(n)\sim\pi$, otherwise, $\tilde{C}_{1}(n)\sim\rho(X(M_{n+1}))/\sigma$.

• –

If $D_{n+1}=0$, then $\tilde{C}_{2}(n)\sim\pi$, otherwise, $\tilde{C}_{2}(n)\sim\rho(X(D_{n+1}))/\sigma$.

(See Equation (2) for the definition of $\rho$.)

We let $E_{n+1}=E_{n}\cup\{(n+1,M_{n+1})\}\cup\{(n+1,D_{n+1})\}$ be the set of edges of $G_{n+1}$. In other words, at every time-step, we add a node to the DAG, and create two directed edges from this new node to its two randomly-chosen parents (the notation $M(n)$ and $D(n)$ stand for “Mum” and “Dad”).

We let, for all $1\leq i\leq d$,

[TABLE]

where we interpret $X(n)$ as the “label of node $n$”. By Assumptions (T) and (B), $\rho(X(n))/\sigma$ is a probability distribution. Thus, the process $(G_{n},X(n))_{n\geq 0}$ is well-defined; we call this process the $\mathcal{R}$-labelled DAG process with initial distribution $\pi$.

Lemma 2.6**.**

Let $\mathcal{R}$ be a tensor satisfying Assumptions (T) and (B) and let $(G_{n},X(n))_{n\geq 0}$ be the $\mathcal{R}$-labelled DAG process with initial distribution $\pi$. For all $n\geq 0$, let $m_{n}=\pi+\sum_{i=1}^{n}\rho(X(n))$. Then, $(m_{n})_{n\geq 0}$ is the (measure-valued version of the) multi-drawing Pólya urn with initial composition $\pi$ and replacement tensor $\mathcal{R}$.

Proof.

By induction. ∎

Lemma 2.7**.**

Let $R$ be a tensor satisfying Assumptions (T) and (B) and let $(G_{n},X(n))_{n\geq 0}$ be the $\mathcal{R}$-labelled DAG process of initial distribution $\pi$. Then $(G_{n})_{n\geq 0}$ is the uniform recursive DAG.

Proof.

Immediate. ∎

The advantage of this coupling is the following fact: for all $n\geq 0$, given $(G_{n},X(0),\ldots,X(n))$,

[TABLE]

Thus, by Lemma 2.3, to prove Theorem 1.3(ii), it is enough to show that, in distribution as $n\uparrow\infty$,

[TABLE]

where $\nu=\sum_{i=1}^{d}x^{*}_{i}\delta_{i}$.

In other words, the following theorem implies Theorem 1.3$(ii)$:

Theorem 2.8**.**

Let $R$ be a tensor satisfying Assumptions (T), (B) and (I). Let $(G_{n},X(n))_{n\geq 0}$ be the $\mathcal{R}$-labelled DAG process. Then, in distribution as $n\uparrow\infty$,

[TABLE]

The proof of Theorem 2.8 is done in two independent parts: In the first part, we prove that nodes $M_{n}$ and $D_{n}$ are typically quite deep in $G_{n}$ (i.e. far from node [math]) and that, in a neighbourhood of node $n$, the DAG is locally a complete binary tree; this is done using results of Janson [Jan24] on the uniform recursive DAG. In the second part, we use the theory of $Z$-eigenvectors of tensors to prove convergence in distribution of $(X(M_{n}),X(D_{n}))$.

2.3 Shape of the uniform recursive DAG

The aim of this section is to understand the genealogy of node $n$ in the uniform random DAG. We show that up to distance $o(\log\log\log n)$, this genealogy is a complete binary tree.

We look at the genealogy (set of all ancestors) of node $n$ in the uniform random DAG $G_{n}$ (i.e. its “mum” and “dad”, and by induction, its grandparents, great-grandparents, etc.) and then look at the graph $H_{n}$ induced by the ancestors $v$ of $n$ such that $v\geq n_{1}\coloneqq\lfloor n/\log n\rfloor$ (i.e. we keep only these nodes and the edges between them, and erase the rest of $G_{n}$). Janson [Jan24] shows that, with high probability, $H_{n}$ is a (binary) tree. In this section, we complete this result by showing that, if $\ell(n)=o(\log\log\log n)$, then the tree $H^{\prime}_{n}$ obtained by erasing all nodes of $H_{n}$ that are at (graph) distance more than $\ell(n)$ of $n$ is a complete binary tree (i.e. $n$ has degree 2 and all the nodes at distance at most $\ell(n)-1$ of $n$ have degree 3). We sketch this result on Figure 1. We let

[TABLE]

Lemma 2.9**.**

If $\ell(n)=o(\log\log\log n)$, then $\lim_{n\uparrow\infty}\mathbb{P}(\mathcal{E}_{n}\cap\mathcal{F}_{n})=1$.

Proof.

By Theorem 3.1 in [Jan24], we have $\mathbb{P}(\mathcal{E}_{n})\to 1$ as $n\uparrow\infty$. In the rest of the proof, we assume that $\mathcal{E}_{n}$ occurs. Recall that, by definition, for all $n\geq 1$, $M(n)$ (and $D(n)$) are two independent, uniform elements of $0,\ldots,n-1$. We can model this by sampling a sequence of random variables $(U_{n},V_{n})_{n\geq 1}$, i.i.d. and uniformly distributed on $[0,1]$: for all $n\geq 1$, we set $M(n)=\lfloor U_{n}n\rfloor$ and $D(n)=\lfloor V_{n}n\rfloor$. We let $L(n)$ the set of ancestors of $n$ that are at distance $\ell(n)$ of $n$. For any ancestor $u\in L(n)$, in distribution, $u=\lfloor\ldots\lfloor\lfloor nX_{1}\rfloor X_{2}\rfloor\ldots X_{\ell(n)}\rfloor$ where $(X_{i})_{i\geq 1}$ is a sequence of i.i.d. random variables, uniformly-distributed on $[0,1]$. And, on the event $\mathcal{E}_{n}$, $|L(n)|=2^{\ell(n)}$. Thus,

[TABLE]

Now note that

[TABLE]

Recall that $n_{1}(n)=\lfloor n/\log n\rfloor$ and $\ell(n)=o(\log\log n)$. Thus, as $n\uparrow\infty$,

[TABLE]

Thus, for all $n$ large enough, this is positive, which implies, by Chebyshev’s inequality,

[TABLE]

because \mathrm{Var}\big{(}\sum_{i=1}^{\ell(n)}\log(1/{X_{i}})\big{)}=\ell(n). Thus, by (5), as $n\uparrow\infty$,

[TABLE]

as $n\uparrow\infty$ because $\ell(n)=o(\log\log\log n)$. ∎

3 Random labelling of the DAG

In this section, we aim to understand the limit of $X(n)$, the random couple labelling node $n$ in the $\mathcal{R}$-labelled DAG, as $n\uparrow\infty$. First note that, in the construction of the $\mathcal{R}$-labelled DAG, one can first sample the unlabelled graph $(G_{n})_{n\geq 0}$, and then sample the labels $(X(n))_{n\geq 1}$.

Recall that, as defined in Section 2.3, $H_{n}$ is the set of all the ancestors $v$ of node $n$ such that $v\geq n_{1}$. Also, $H^{\prime}_{n}$ is the sub-graph of $H_{n}$ in which we only keep the nodes that are at graph distance at most $\ell(n)$ of node $n$. On $\mathcal{E}_{n}\cap\mathcal{F}_{n}$, $H^{\prime}_{n}$ is a complete binary tree and, in this tree, the label of each node is sampled at random, according to a distribution that is uniquely determined by the label of its parents: a “2-dependent Markov chain”. We therefore start this section by defining 2-dependent Markov chains on the infinite binary tree, and study their ergodicity.

3.1 Two-dependent Markov chains on the infinite binary tree

Definition 3.1**.**

Let $(\pi_{i})_{i\geq 1}$ be a sequence of probability distributions on a finite or countable space $\mathcal{S}$. Let $T=(T_{ijk}=T(i,j,k))_{i,j,k\in\mathcal{S}}$ a collection of non-negative real numbers satisfying $\sum_{k\in\mathcal{S}}T_{ijk}=1$ for all $i,j\in\mathcal{S}$. We sample $W^{\scriptscriptstyle(0)}_{i}\sim\pi_{i}$ for all $i\geq 1$, independently of each other. We then sample the values of $(W_{i}^{\scriptscriptstyle(n)})_{i\geq 1}$ for all $n\geq 1$ recursively as follows: for all $n\geq 1$, for all $i\geq 1$, for all $x,y,w\in\mathcal{S}$

[TABLE]

We call this process $(W_{i}^{\scriptscriptstyle(n)})_{i\geq 1,n\geq 0}$ the 2-dependent Markov chain of transition tensor $T$ and initial distribution $(\pi_{i})_{i\geq 1}$.

Note that a 2-dependent Markov chain can be interpreted as a random labelling of the infinite binary tree. For this interpretation, we index the nodes of the infinite binary tree as follows (see Figure 2): for all $n\geq 0$, $i\geq 1$, node $(n,i)$ is the $i$-th node in the $n$-th level. For all $n\geq 1$, $i\geq 1$, the two parents of node $(n,i)$ are nodes $(n-1,2i)$ and $(n-1,2i+1)$. The value $W_{i}^{\scriptscriptstyle(n)}$ can be interpreted as the label of node $(n,i)$: its distribution given the labels of its two parents is given by $T(\cdot,W_{2i}^{\scriptscriptstyle(n-1)},W_{2i+1}^{\scriptscriptstyle(n-1)})$.

Lemma 3.2**.**

Let $(\pi_{i})_{i\geq 1}$ be a sequence of probability distributions on a finite or countable space $\mathcal{S}$. Let $T=(T_{ijk}=T(i,j,k))_{i,j,k\in\mathcal{S}}$ a collection of non-negative real numbers satisfying $\sum_{k\in\mathcal{S}}T_{ijk}=1$ for all $i,j\in\mathcal{S}$. Let $(W_{i}^{\scriptscriptstyle(n)})_{i\geq 1,n\geq 0}$ be the 2-dependent Markov chain of transition tensor $T$ and initial distribution $(\pi_{i})_{i\geq 1}$. For all $n\geq 0,i\geq 1$, we let $\pi_{i}^{\scriptscriptstyle(n)}$ be the distribution of $W_{i}^{\scriptscriptstyle(n)}$. Then, for all $i\geq 1$, $\pi_{i}^{\scriptscriptstyle(0)}=\pi_{i}$ and, for all $n\geq 1$, $i\geq 1$, $w\in\mathcal{S}$,

[TABLE]

Note that, if we let $\mathcal{T}:\mathcal{S}\times\mathcal{S}\to\mathcal{S}$ be the 3-mode tensor associated to $T=(T_{ijk})_{i,j,k\in\mathcal{S}}$ (as we did in (1) for $(R_{ijk})_{1\leq i,j,k\leq d}$), and if we interpret the distributions $\pi_{i}^{\scriptscriptstyle(n)}$ as vectors whose entries are indexed by $\mathcal{S}$, then (6) can be written as

[TABLE]

With this in mind, it is natural to define invariant distributions as follow:

Definition 3.3**.**

We say that $\pi=(\pi(w))_{w\in S}$ is an invariant distribution of the 3-mode tensor $\mathcal{T}$ if and only if $\pi=\mathcal{T}(\pi,\pi)$.

It is now natural to ask under which conditions a 2-dependent Markov chain of transition tensor $T$ admits an invariant distribution and converges to it. To do so, we introduce the notion of ergodicity coefficients [Sen06]. The ($L^{1}$-norm) ergodicity coefficient of a stochastic matrix whose rows and columns are indexed by $\mathcal{S}$ is defined as

[TABLE]

where

[TABLE]

If $\tau(\mathfrak{p})<1$, then $x\mapsto\mathfrak{p}x$ admits a unique fixed point $x^{*}$ in $\Delta(\mathcal{S})$ and, for all $x_{0}\in\Delta(\mathcal{S})$, if $x_{n+1}=\mathfrak{p}x_{n}$ for all $n\geq 0$, then $x_{n}\to x^{*}$ as $n\uparrow\infty$.

To define the ergodicity coefficient of a $3$-mode tensor, Fasino and Tudisco [FT20] start by defining the following matrices: Given a $3$-mode tensor $T=(T_{ijk})_{i,j,k\in\mathcal{S}}$ and a vector $x\in\Delta(\mathcal{S})$, we let, for all $i,j\in\mathcal{S}$,

[TABLE]

(The superscripts stand for “left” and “right”.) If $T$ is a stochastic tensor (i.e. if $\sum_{i\in\mathcal{S}}T_{ijk}=1$ for all $j,k\in\mathcal{S}$), then $T^{(\ell)}_{x}$ and $T^{(r)}_{x}$ are stochastic matrices for all $x\in\Delta{(\mathcal{S})}$. We let

[TABLE]

By [FT20, Thm 5.1, Cor 5.2],

[TABLE]

which implies the following result:

Lemma 3.4**.**

For any stochastic tensor $T=(T_{ijk})_{i,j,k\in\mathcal{S}}$, the two following statements are equivalent:

• (i)

\displaystyle\max_{j,k,k^{\prime}\in\mathcal{S}}\sum_{i\in\mathcal{S}}\mathopen{}\mathclose{{\left\lvert T_{ijk}-T_{ijk^{\prime}}}}\right\rvert+\max_{j,j^{\prime},k\in\mathcal{S}}\sum_{i\in\mathcal{S}}\mathopen{}\mathclose{{\left\lvert T_{ijk}-T_{ij^{\prime}k}}}\right\rvert<2;

• (ii)

$\tau_{\ell}(T)+\tau_{r}(T)<1$.

Finally, the following lemma gives conditions under which a 2-dependent Markov chain admits an invariant distribution, and converges to it.

Proposition 3.5**.**

Let $T=(T_{ijk})_{i,j,k\in\mathcal{S}}$ be a stochastic tensor satisfying condition (i) of Lemma 3.4. Then there exists a unique $\pi\in\Delta(\mathcal{S})$ such that $\mathcal{T}(\pi,\pi)=\pi$. Furthermore, any sequence $(\pi_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ of vectors of $\Delta(\mathcal{S})$ that satisfies (6) is such that, for all $n\geq 0$,

[TABLE]

for $q=\tau_{\ell}(T)+\tau_{r}(T)<1$. Therefore, $\lim_{n\to\infty}\pi_{1}^{\scriptscriptstyle(n)}=\pi$.

Note that, by Lemma 3.2, $\lim_{n\to\infty}\pi_{1}^{\scriptscriptstyle(n)}=\pi$ implies that $W_{1}^{\scriptscriptstyle(n)}\Rightarrow\pi$ in distribution as $n\uparrow\infty$, where $(W_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ is the 2-dependent Markov chain of transition tensor $T=(T_{ijk})_{i,j,k\in\mathcal{S}}$.

Proof.

Under condition (i) of Lemma 3.4, $\tau_{\ell}(T)+\tau_{r}(T)<1$ implies the existence of a unique fixed point $\pi$ of $\mathcal{T}$ via [FT20, Thms 6.1 and 6.3]. Further, for all $x,y,u,v\in\Delta(\mathcal{S})$, we have

[TABLE]

(See (8) for the definition of the matrices $T^{(\ell)}_{x}$ and $T^{(r)}_{v}$; also recall that, by definition, for all $x,y\in\mathbb{R}^{\mathcal{S}}$, for all $i\in\mathcal{S}$, $\mathcal{T}(x,y)_{i}=\sum_{j,k\in\mathcal{S}}T_{ijk}x_{j}y_{k}$.) This implies that, for all $x,y\in\Delta(\mathcal{S})$,

[TABLE]

by definition of $\tau(\cdot)$ (see (7)). Now using the definition of $\tau_{\ell}(T)$ and $\tau_{r}(T)$ (see (9)), we get

[TABLE]

where we have set $q=\tau_{\ell}(T)+\tau_{r}(T)$. We now prove inductively on $n\geq 0$, that for all $n\geq 0$, for all $i\geq 1$,

[TABLE]

The base case holds straighforwardly; for the induction, note that, for all $n\geq 1$, for all $i\geq 1$, because $\pi=\mathcal{T}(\pi,\pi)$ and (6), we have

[TABLE]

by (10). Now, by the induction hypothesis, we get

[TABLE]

which concludes the proof of (11). The result follows by taking $i=1$ in (11). ∎

3.2 The randomly labelled DAG is “locally” a 2-dependent Markov chain

In this section, we consider the $\mathcal{R}$-labelled DAG as defined in Section 2.2. First recall that the shape of the $\mathcal{R}$-labelled DAG is that of a uniform recursive DAG, As in Section 2.3, $H_{n}$ is the set of all the ancestors $v$ of node $n$ such that $v\geq n_{1}$. Also, $H^{\prime}_{n}$ is the sub-graph of $H_{n}$ in which we only keep the nodes that are at graph distance at most $\ell(n)$ of node $n$. On $\mathcal{E}_{n}\cap\mathcal{F}_{n}$ (see (4)), which occurs with high probability when $n$ is large (see Lemma 2.9), $H^{\prime}_{n}$ is a complete binary tree.

For all $k\geq 0$, we let $L(k)$ be the set of all nodes in $H_{n}$ at distance $k$ of node $n$. (Note that $L(k)$ is empty for all $k$ large enough.)

Lemma 3.6**.**

Conditionally on $G_{n_{1}(n)}$ and $(X(i))_{0\leq i\leq n_{1}(n)}$, and on the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$, for all $1\leq k\leq\ell(n)-1$, $(X(i))_{i\in L(k)}$ is a sequence of independent random variables.

Proof.

Let $u,v\in L(k)$. On $\mathcal{E}_{n}\cap\mathcal{F}_{n}$, $H_{u}$ and $H_{v}$ are two disjoint sub-trees of $H_{n}$. The label of $u$ is sampled by applying the following procedure:

• •

First, complete the tree by adding leaves so that every node in $H_{u}$ has degree 3.

• •

By definition of the $\mathcal{R}$-labelled DAG, these leaves are merged with nodes chosen uniformly and independently at random in $G_{n_{1}(n)}$.

• •

Now, given the labels of the leaves of $H_{u}$, we sample the labels of all nodes in $H_{u}$ recursively using the fact that, if a node’s parents are labelled by the couples $x=(x_{1},x_{2})$ and $y=(y_{1},y_{2})$, then its label is distributed as $\rho(x)\otimes\rho(y)$, and independent of the rest.

The label of $v$ is sampled similarly, and because $H_{u}$ and $H_{v}$ do not intersect, this can be done independently for each of the two trees, implying that $X(u)$ and $X(v)$ are independent, as claimed. ∎

Lemma 3.7**.**

Fix $n\geq 0$; we work conditionally on $G_{n_{1}(n)}$ and $(X(i))_{1\leq i\leq n_{1}(n)}$, and on the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$. We order the nodes in $L(\ell(n)-1)$ from left to right so that $H^{\prime}_{n}$ is a plane binary tree. For all $1\leq i\leq 2^{\ell(n)-1}$, we let $\pi_{i}^{\scriptscriptstyle(0)}$ be the distribution of the label of the $i$-th node in level $L(\ell(n)-1)$ (from left to right); if $i>2^{\ell(n)-1}$, we let $\pi_{i}^{\scriptscriptstyle(0)}$ be a fixed, and arbitrary distribution on $\{1,\ldots,d\}^{2}$.

Then, in distribution, $X(n)=W_{1}^{\scriptscriptstyle(n)}$, where $(W_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ is the 2-dependent Markov chain on $\{1,\ldots,d\}^{2}$, of transition tensor $T$ given by

[TABLE]

for all $i_{1},i_{2},j_{1},j_{2},k_{1},k_{2}\in\{1,\ldots,d\}$.

Proof.

This is straightforward from Lemma 3.6 (which gives independence of the labels of the leaves - needed in the definition of a 2-dependent Markov chain), and from the definition of the $\mathcal{R}$-labelled DAG (which gives the replacement tensor, see Section 2.2). ∎

Lemma 3.8**.**

If $R=(R_{ijk})_{1\leq i,j,k\leq d}$ satisfies Assumption (E), then $T$ satisfies Assumption (i) of Lemma 3.4.

Proof.

Let $i,j,j^{\prime},k,k^{\prime}\in\{1,\ldots,d\}^{2}=\mathcal{S}$. In what follows, we write $i=(i_{1},i_{2})$ (and similarly for $j,j^{\prime},k,k^{\prime}$), where $i_{1},i_{2}\in\{1,\ldots,d\}$. We have

[TABLE]

where we have used Assumption (B). A similar calculation gives

[TABLE]

In total, we thus get that

[TABLE]

by Assumption (E). This concludes the proof. ∎

3.3 Proof of Theorem 2.8

Together with Proposition 3.5, Lemma 3.8 implies that there exists a unique $\pi\in\Delta(\{1,\ldots d\}^{2})$ such that $\mathcal{T}(\pi,\pi)=\pi$, and $W_{1}^{\scriptscriptstyle(n)}$ converges in distribution to $\pi$. We thus get, by Lemma 3.7, that $X(n)\Rightarrow\pi$ in distribution as $n\uparrow\infty$, conditionally on $G_{n_{1}(n)}$ and $(X(i))_{1\leq i\leq n_{1}(n)}$, and on the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$. This means that, for all $1\leq i,j\leq d$,

[TABLE]

as $n\uparrow\infty$. (This is because the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$ is independent of $G_{n_{1}(n)}$ and $(X(i))_{1\leq i\leq n_{1}(n)}$, by definition.) Now, by Lemma 2.9, $\mathbb{P}(\mathcal{E}_{n}\cap\mathcal{F}_{n})\to 1$ as $n\uparrow\infty$. Hence,

[TABLE]

which, by the tower rule, implies

[TABLE]

Using again the fact that $\mathbb{P}(\mathcal{E}_{n}\cap\mathcal{F}_{n})\to 1$ as $n\uparrow\infty$, we get

[TABLE]

Hence, to prove Theorem 2.8, it thus only remains to prove that $\pi=\nu\otimes\nu$, where $\nu$ is the unique solution of $\sigma\nu=\mathcal{R}(\nu,\nu)$ in $\Delta_{d}$, where we interpret $\nu$ as a $d$-dimensional vector $\nu=(\nu(\{1\}),\ldots,\nu(\{d\}))$. We know that this solution exists and is unique by Assumption (E): Indeed, Assumption (E) implies that the 3-mode tensor $R/\sigma$ satisfies condition (i) of Lemma 3.4, which, by Proposition 3.5, implies that $\nu=\mathcal{R}(\nu,\nu)/\sigma$ admits a unique solution in $\Sigma_{d}$.

To prove that $\pi=\nu\otimes\nu$, it is enough to prove that $\pi^{\prime}:=\nu\otimes\nu$ satisfies $\pi^{\prime}=\mathcal{T}(\pi^{\prime},\pi^{\prime})$ (because the solution of this equation is unique, by Proposition 3.5). First recall that we interpret the measure $\nu$ as a vector $(\nu_{i})_{1\leq i\leq d}$. Similarly, we interpret $\pi^{\prime}$ as a vector whose coordinates are $\pi^{\prime}_{(i,j)}=\nu_{i}\nu_{j}$, for all $1\leq i,j,\leq d$. By definition of $\mathcal{T}$, for all $x=(x_{1},x_{2})\in\{1,\ldots,d\}^{2}$,

[TABLE]

by (12). We thus get, by definition of $\pi^{\prime}=\nu\otimes\nu$ (i.e. $\pi^{\prime}_{(i,j)}=\nu_{i}\nu_{j}$, for all $1\leq i,j,\leq d$),

[TABLE]

because $\mathcal{R}(\nu,\nu)=\sigma\nu$, and by definition of $\pi^{\prime}=\nu\otimes\nu$. We thus get that $\pi^{\prime}=\mathcal{T}(\pi^{\prime},\pi^{\prime})$, which, by uniqueness, implies that $\pi=\pi^{\prime}=\nu\otimes\nu$, as desired. This concludes the proof of Theorem 2.8.

3.4 Proof of Theorem 1.3

We have proved in Section 2.2 that Theorem 2.8 implies Theorem 1.3$(ii)$. Part $(i)$ is proved in Section 3.3.

4 Drawing more than two balls at a time

So far, we have restricted ourselves to drawing two balls at every time step in the urn process. This was mainly to keep the proofs as simple as possible, but extending the results and the proofs to drawing $m$ balls at each time step is straightforward. In this section, we explain how to do so.

Definition 4.1**.**

Let $R=(R_{i_{1}\ldots i_{m+1}}=R(i_{1},\ldots,i_{m+1}))_{1\leq i_{1},\ldots,i_{m}\leq d}$ be a collection of non-negative integers. We define the m-drawing d-colour Plya urn with replacement tensor R as a Markov process $(U(n)=(U_{1}(n),\ldots,U_{d}(n)))_{n\geq 0}$ on $\mathbb{N}^{d}$ that satisfies, for all $n\geq 0$,

[TABLE]

where, given $U(n)$, $(C_{1}(n),\ldots,C_{m}(n))$ is a vector of i.i.d. random variables such that, for all $1\leq k\leq d$,

[TABLE]

Let $\mathcal{S}=\{1,\ldots,d\}^{m}$. We assume that the replacement tensor $R$ satisfies the following assumptions:

• (T)

For all $1\leq i_{1},\ldots,i_{m+1}\leq d$, $R_{i_{1},\ldots,i_{m+1}}\geq 0$.

• (B)

There exists $\sigma\in(0,\infty)$ such that, for all $1\leq i_{1},\ldots,i_{m}\leq d$, $\sum_{i=1}^{d}R_{ii_{1}\ldots i_{m}}=\sigma$.

• (E)

\max_{j,j^{\prime}\in S}\sum_{i=1}^{d}\mathopen{}\mathclose{{\left\lvert R_{ij_{1}\ldots j_{m}}-R_{ij^{\prime}_{1}\ldots j^{\prime}_{m}}}}\right\rvert<2\sigma/m.

We define $\mathcal{R}=\mathbb{R}^{d}\times\cdots\times\mathbb{R}^{d}\to\mathbb{R}^{d}$ as follows: for all $x=(x_{1},\ldots,x_{m})\in\mathbb{R}^{d}\times\cdots\times\mathbb{R}^{d}$,

[TABLE]

and $\mathcal{R}(x)=(\mathcal{R}_{1}(x),\ldots,\mathcal{R}_{d}(x))$. The multi-linear map $\mathcal{R}$ is called an $(m+1)$-mode tensor.

Theorem 4.2**.**

Let $(U(n))_{n}$ be an $m$-drawing $d$-colour balanced Plya urn with replacement tensor $R$. If $R$ satisfies Assumptions (T), (B), and (E), then there exists a unique $x^{*}\in\Delta_{d}$ satisfying $\sigma x^{*}=\mathcal{R}(x^{*},\ldots,x^{*})$ and, in probbaility as $n\uparrow\infty$,

[TABLE]

The proof of Theorem 4.2 is a straightforward extension of the proof of Theorem 1.3. We briefly explain how the proof can be extended to this case.

The construction of the labelled DAG extends naturally, with each node having $m$ parents instead of $2$. The events $\mathcal{E}_{n}$ and $\mathcal{F}_{n}$ become

[TABLE]

It is straightforward to prove that Lemma 2.9 also holds in this context (the fact that $\mathbb{P}(\mathcal{E}_{n})\to 1$ as $n\uparrow\infty$ is proved in [Jan24]). Thus, Lemma 3.6 extends straightforwardly, and we get, conditionally on $G_{n_{1}(n)}$ and $(X(i))_{1\leq i\leq n_{1}(n)}$, and on the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$, $X(n)=W_{1}^{\scriptscriptstyle(n)}$ in distribution, where $(W_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ is an $m$-dependent Markov chain taking values in $\{1,\ldots,d\}^{m}$ (defined on the infinite $m$-ary tree, as in Definition 3.1 for the binary case). The replacement tensor of this $m$-dependent Markov chain is given by

[TABLE]

for all $x,a^{\scriptscriptstyle(1)},\ldots,a^{\scriptscriptstyle(m)}\in\mathcal{S}=\{1,\ldots,d\}^{m}$. One can check that, if $R$ satisfies (E), then

[TABLE]

Similarly to Proposition 3.5, this implies that $W_{1}^{\scriptscriptstyle(n)}$ converges in distribution to $\pi$, the unique vector in $\Delta(\mathcal{S})$ satisfying $\pi=\mathcal{T}(\pi,\ldots,\pi)$. To get existence and uniqueness of this vector, one needs to extend the proof of [FT20, Thm 5.1, Cor 5.2]; this can be done using [FT20, Thm 4.4] and via similar arguments as the ones presented in the proof of [FT20, Thm 5.1].

Finally, by (E), there exists a unique vector $\nu$ in $\Delta_{d}$ satisfying $\nu=\mathcal{R}(\nu,\ldots,\nu)$. As in Section 3.3, one can check that $\pi=\nu\otimes\cdots\otimes\nu$ ($m$ times), and conclude the proof similarly as for the binary case.

5 Examples

To denote the replacement tensor of a $2$-colour, $2$-drawing Pólya urn, we write

[TABLE]

where we recall that $R_{ijk}$ is the number of balls of colour $i$ added in the urn when the colours of the couple of balls picked is $(j,k)$.

5.1 Simple examples

In this first section, we check that our main result gives the expected result (or non result) for three simple cases.

Example 1**.**

For our first example, we consider the following $2$-colour, $2$-drawing case:

[TABLE]

i.e. $R_{ijk}={\bf 1}_{i=j}+{\bf 1}_{i=k}$. In other words, if we pick a couple of balls of colour $(j,k)$, then we add one ball of colour $j$ and one ball of colour $k$ into the urn. Intuitively, this is a generalisation of the (classical) Pólya urn with identity replacement matrix (at every time step, we add a ball of the same colour as the one picked at that step). For this classical Pólya urn, it is well known that $U(n)$ converges almost surely to a random variable. So we do not expect Theorem 1.3 to apply. Indeed, Assumption (E) is not satisfied as

[TABLE]

Example 2**.**

We now consider the following $2$-colour, $2$-drawing case:

[TABLE]

i.e. $R_{ijk}=1$ for all $1\leq i,j,k\leq 2$. In words, this means that, at every time step, we add $1$ ball of colour $1$ and one ball of colour $2$, independently from the draw. This case is a thus a bit trivial: one expects that $U(n)/n\to(\nicefrac{{1}}{{2}},\nicefrac{{1}}{{2}})$ almost surely as $n\uparrow\infty$. Indeed, Theorem 1.3 applies: Assumptions (T) and (B) hold trivially, and

[TABLE]

so (E) holds as well. Now note that, by definition, $\mathcal{R}(x,y)_{i}=\sum_{j=1}^{2}\sum_{k=1}^{2}x_{j}y_{k}=(x_{1}+x_{2})(y_{1}+y_{2})$, and

[TABLE]

So Theorem 1.3 gives that $U(n)/n\to(\nicefrac{{1}}{{2}},\nicefrac{{1}}{{2}})$ in probability as $n\uparrow\infty$, as expected.

Example 3**.**

Consider a matrix for a single-drawing Plya urn

[TABLE]

This can be transformed into a multi-drawing Plya urn that only considers the first draw for its replacement as follows:

[TABLE]

The ergodicity coefficient for a single-drawing Plya urn [FT20] is given by

[TABLE]

We recover this coefficient when applying Assumption (E):

[TABLE]

However, the upper bound given by Assumption (E), $2\sigma/m$, is stronger than the upper bound for a single-drawing Plya urn, which is just $2\sigma$. A possible explanation for this gap is that, even though only the first draw matters for the distribution of the balls present in the urn, our result considers the distribution of all $m$ draws and their joint convergence.

5.2 Affine urns and non-symmetric urns

Example 4**.**

In [KM17], Kuba and Mahmoud prove convergence theorems for $2$-colour, $m$-drawing balanced “affine” Pólya urns. We focus here in the $2$-drawing case ($m=2$). They say that such an urn is affine if

[TABLE]

where there exists an integer $h$ such that $a_{i}=a_{0}+ih$, for all $i\in\{1,2\}$. To apply our results, we assume that $0\leq a_{0},a_{1},a_{2}\leq\sigma$ (which is not required in [KM17]), which implies $h<\sigma$. Then both Assumptions (T) and (B) hold. Now, Assumption (E) is equivalent to

[TABLE]

Under this extra assumption, not required in [KM17], we get that $U(n)/n\to x^{*}$ in probability as $n\uparrow\infty$, where $x^{*}=\mathcal{R}(x^{*},x^{*})/\sigma$. In this case, for all $x\in\Delta_{2}$,

[TABLE]

One can check that the unique solution of $x^{*}=\mathcal{R}(x^{*},x^{*})/\sigma$ is given by x^{*}=\big{(}\frac{a_{0}+2h}{\sigma+2h},1-\frac{a_{0}+2h}{\sigma+2h}\big{)}. This recovers the result of [KM17, Proposition 5], albeit under stronger assumptions, and with convergence in probability instead of almost-surely. Note that, in the case of balanced affine urns, martingale theory can be applied to get stronger results (and it is thus not surprising that our result gives a weaker result than the results specific to affine urns in the literature): Kuba and Mahmoud [KM17] get second order convergence results, Kuba and Sulzbach [KS17b] prove a law of the iterated logarithm, and Sparks, Kuba, Balaji and Mahmoud [SKBM25] prove second order results for the $d$-colour case, $d\geq 2$.

A strength of our approach is that the order in which the balls were drawn from the urn may matter for the replacement tensor $R$, which has not been considered in existing literature on multi-drawing Plya urns.

Example 5**.**

Consider a two-colour, two-drawing Plya urn with the following replacement tensor:

[TABLE]

This urn is non-symmetric, as when we draw first a ball of type $1$, then a ball of type $2$, we replace $2$ balls of the first and one of the second type, whereas this is reversed if the order of the draws is reversed. The urn fulfills Assumptions (T) and (B), with $\sigma=3$.

Assumption (E) is also satisfied since

[TABLE]

so a unique solution $x^{*}\in\Delta_{2}$ for $\mathcal{R}(x,x)/\sigma=x$ exists that the proportion of balls in the urn will converge to in distribution. Solving the fixed-point equation $x^{*}=\mathcal{R}(x^{*},x^{*})/\sigma$ yields $x^{*}=(\sqrt{2}-1,2-\sqrt{2})$ as the unique solution in $\Delta_{2}$.

Example 6**.**

Consider a two-colour, two-drawing Plya urn with the following replacement tensor:

[TABLE]

This urn is again non-symmetric and fulfills Assumptions (T) and (B), with $\sigma=5$.

Assumption (E) is also satisfied since

[TABLE]

so a unique solution $x^{*}\in\Delta_{2}$ for $\mathcal{R}(x,x)/\sigma=x$ exists that the proportion of balls in the urn will converge to in distribution. The unique solution to the fixed-point equation $x^{*}=\mathcal{R}(x^{*},x^{*})/\sigma$ is given by $x^{*}=(\sqrt{11}-3,4-\sqrt{11})$.

5.3 Examples from [LMS18]

In [LMS18], Lasmar, Mailler and Selmi consider the following multi-drawing model: At each timestep $n$, draw $m$ balls from the urn and denote by $\xi_{n,i}$ the number of balls of colour $i$ drawn. If the urn has $d$ colours, $\xi_{n}$ is a $d$-dimensional vector with $\sum_{i}\xi_{n,i}=m$. Then, $\hat{R}(\xi_{n})_{i}$ balls of type $i$ are replaced into the urn for each colour $i$. Again, the urn is called balanced, if for all such $\xi_{n}$, $\sum_{i}\hat{R}(\xi_{n,i})=\sigma$. In fact, this model is subsumed in our approach, as we can transfer this replacement function $\hat{R}$ into a tensor $(R_{ij_{1}\ldots j_{m}})_{ij_{1}\ldots j_{m}=1}^{d}$ as follows:

[TABLE]

The set of all such vectors $v$ is also the set of all possible draws in the model of Lasmar, Mailler and Selmi and defined as

[TABLE]

In the analysis of Lasmar, Mailler and Selmi [LMS18], the set of zeros of the following function

[TABLE]

plays a central role. We consider the $i$th component of $h(x)$ and rewrite for $x\in\{x\in[0,1]^{d}\nonscript\>|\nonscript\>\mathopen{}\sum_{i}x_{i}=1\}$,

[TABLE]

Our analysis requires the study of

[TABLE]

for $x\in\{x\in[0,1]^{d}\nonscript\>|\nonscript\>\mathopen{}\sum_{i=1}^{d}x_{i}=1\}$. Expanding the tensor multiplication for the $i$th component leads us to

[TABLE]

We see that Equations (13) and (5.3) reduce to the same expression for $x\in\{x\in\mathbb{R}^{d}\nonscript\>|\nonscript\>\mathopen{}\sum_{i=1}^{d}x_{i}=1\}$ and therefore have the same solution set in this space. We highlight this by looking at some of the examples presented in [LMS18].

Example 7**.**

Lasmar, Mailler and Selmi [LMS18] consider the 2-drawing Pólya urn with replacement tensor

[TABLE]

This is a non-affine urn with $\sigma=3$, fulfilling Assumptions (T) and (B). Assumption (E) is also satisfied since

[TABLE]

so a unique solution $x^{*}\in\Delta_{d}$ for $\mathcal{R}(x,x)/\sigma=x$ exists. Solving the resulting system of equations

[TABLE]

yields $x_{1}=x_{2}=\nicefrac{{1}}{{2}}$. Theorem 1.3 thus implies that the proportion of balls of each of the two colours converges in probability to $\nicefrac{{1}}{{2}}$, recovering partially the result from [LMS18] (only partially because the convergence is almost sure in [LMS18]).

While [LMS18, Examples 2-5] also all fulfill Assumptions (T) and (B), none of them fulfill Assumption (E) and as such we cannot treat them with our result. This is again not surprising, as the theory of stochastic approximations can be used to control the entire evolution of a process, while our result only looks at the final phase of the process. We still give an exemplary treatment of Examples 2 and 3 in the following.

Example 8**.**

Consider the 2-drawing urn of replacement tensor

[TABLE]

It satisfies assumptions (B) with $\sigma=4$ and (T). However,

[TABLE]

i.e. Assumption (E) is not satisfied. Note that the equation $\mathcal{R}(x,x)=4x$ admits two solutions in $\Delta_{2}$: $(\nicefrac{{1}}{{3}},\nicefrac{{2}}{{3}})$ and $(1,0)$. Lasmar, Mailler and Selmi [LMS18] are able to prove that, almost surely as $n\uparrow\infty$, $U(n)/n\to x^{*}=(\nicefrac{{1}}{{2}},\nicefrac{{2}}{{3}})$. Our result does not apply to this example.

Example 9**.**

Another example in Lasmar, Mailler and Selmi [LMS18] is the 2-drawing urn of replacement tensor

[TABLE]

It satisfies Assumptions (T) and (B) with $\sigma=8$. However, Assumption (E) is not satisfied since

[TABLE]

Note that $\mathcal{R}(x,x)=8x$ admits a unique solution in $\Delta_{2}$: $(1-\nicefrac{{1}}{{\sqrt{2}}},\nicefrac{{1}}{{\sqrt{2}}})$. Lasmar, Mailler and Selmi [LMS18] are able to prove that, almost surely as $n\uparrow\infty$, $U(n)/n\to x^{*}=(1-\nicefrac{{1}}{{\sqrt{2}}},\nicefrac{{1}}{{\sqrt{2}}})$. Our result does not apply to this example.

5.4 Further examples from the literature

As we have seen, the presence of a unique solution to $\mathcal{R}(x,x)=\sigma x$ is not sufficient for Assumption (E) to hold and thus for our result to apply. This also holds when considering higher-order Markov chains with a simpler dependency structure, leading to the analysis of power iteration schemes such as $x_{t+1}=\mathcal{R}(x_{t},x_{t})$. This is illustrated by the following example.

Example 10** ([LN14, Ex. 3]).**

The urn given by

[TABLE]

satisfies Assumptions (T) and (B) with $\sigma=1$ and admits a unique solution to $\mathcal{R}(x,x)=x$ with $x=\frac{1}{2}(1,1)$. However, Assumption (E) is not satisfied since

[TABLE]

Li and Ng [LN14] prove that a simple power iteration scheme already does not converge to this fixed point.

Further, for a stochastic matrix $P$, positivity is sufficient to imply contractivity of the map $x\mapsto Px$ and the existence of a unique fixed point $x^{*}$. This is not the case with stochastic tensors, as illustrated by the following example.

Example 11**.**

A 3-draw, 2-colour case (see [CZ13, Example 1.7]):

[TABLE]

This tensor has strictly positive entries and satisfies Assumption (B) with $\sigma=1$, however, $\mathcal{R}(x,x,x)=x$ is solved by both $v_{1}=(0.2,0.8)$ and $v_{2}=(0.6,0.4)$. When we check Assumption (E) by comparing all pairs $(R_{1j_{1}j_{2}j_{3}},R_{2j_{1}j_{2}j_{3}})$ for maximal absolute value difference, we find

[TABLE]

6 Infinitely-many colour case

As mentioned in the introduction, one can easily generalise our model to having infinitely-many colours. In this section, we define this more general model, and state and prove an equivalent of Theorem 1.3 for this case.

6.1 Definition of infinitely-many colour, multi-drawing urns

Let $\mathcal{C}$ be a measurable space. Let $(\rho_{x,y})_{x,y\in\mathcal{C}}$ be a kernel of non-negative measures on $\mathcal{C}$ such that, for all $x,y\in\mathcal{C}$, $\rho_{x,y}(\mathcal{C})=\sigma>0$.

We define a process of random non-negative measures on $\mathcal{C}$ recursively as follows: Let $m_{0}$ be a finite non-negative measure on $\mathcal{C}$ and, for all $n\geq 0$, given $m_{n}$, we sample $(C_{1}(n+1),C_{2}(n+1))$ a random variable of distribution $\hat{m}_{n}\otimes\hat{m}_{n}$, where $\hat{m}_{n}=m_{n}/m_{n}(\mathcal{C})$, and set

[TABLE]

The process $(m_{n})_{n\geq 0}$ is the two-drawing Pólya urn of initial composition $m_{0}$ and replacement kernel $(\rho_{x,y})_{x,y\in\mathcal{C}}$.

Remark 6.1**.**

This model is indeed an extension of the finitely-many colour case: in Section 2.1, we have already showed how to see the finitely-many colour case as a measure-valued process.

6.2 Two-dependent Markov chains on the infinite binary tree

The theory of two-dependent Markov chains introduced in Section 3.1 can be extended to Markov chains taking values on an infinite set $\mathcal{S}$. The definition is as follows:

Definition 6.2**.**

Let $(\pi_{i})_{i\geq 1}$ be a sequence of probability distributions on a finite or countable space $\mathcal{S}$. Let $(\tau_{x,y})_{x,y\in\mathcal{S}}$ be a kernel of non-negative measures on $\mathcal{S}$ satisfying $\tau_{x,y}(\mathcal{S})=1$ for all $x,y\in\mathcal{S}$. We sample $W^{\scriptscriptstyle(0)}_{i}\sim\pi_{i}$ for all $i\geq 1$, independently of each other. We then sample the values of $(W_{i}^{\scriptscriptstyle(n)})_{i\geq 1}$ for all $n\geq 1$ recursively as follows: for all $n\geq 1$, for all $i\geq 1$, for all $x,y,\in\mathcal{S}$, given $(W^{\scriptscriptstyle(n-1)}_{2i},W^{\scriptscriptstyle(n-1)}_{2i+1})=(x,y)$, sample $W_{i}^{\scriptscriptstyle(n)}$ a random variable of distribution $\tau_{x,y}$. We call this process $(W_{i}^{\scriptscriptstyle(n)})_{i\geq 1,n\geq 0}$ the 2-dependent Markov chain of transition tensor $(\tau_{x,y})_{x,y\in\mathcal{S}}$ and initial distribution $(\pi_{i})_{i\geq 1}$.

Similarly to the finitely-many colour case, a 2-dependent Markov chain can be interpreted as a random labelling of the infinite binary tree. The value $W_{i}^{\scriptscriptstyle(n)}$ can be interpreted as the label of node $(n,i)$: its distribution given the labels of its two parents is $\tau_{(W_{2i}^{\scriptscriptstyle(n-1)},W_{2i+1}^{\scriptscriptstyle(n-1)})}$.

In the following, we let $\mathcal{M}(\mathcal{S})$ be the set of all signed measures on $\mathcal{S}$, and $\mathcal{P}(\mathcal{S})$ be the set of all probability measures on $\mathcal{S}$.

We define the operator $\mathcal{T}=\mathcal{M}(\mathcal{S})\times\mathcal{M}(\mathcal{S})\to\mathcal{M}(\mathcal{S})$ as follows: for any two signed measures $\mu$ and $\nu$ on $\mathcal{S}$

[TABLE]

Lemma 6.3**.**

Let $(\pi_{i})_{i\geq 1}$ be a sequence of probability distributions on a finite or countable space $\mathcal{S}$. Let $(\tau_{x,y})_{x,y\in\mathcal{S}}$ be a kernel of non-negative measures on $\mathcal{S}$ satisfying $\tau_{x,y}(\mathcal{S})=1$ for all $x,y\in\mathcal{S}$. Let $(W_{i}^{\scriptscriptstyle(n)})_{i\geq 1,n\geq 0}$ be the 2-dependent Markov chain of transition tensor $(\tau_{x,y})_{x,y\in\mathcal{S}}$ and initial distribution $(\pi_{i})_{i\geq 1}$. For all $n\geq 0,i\geq 1$, we let $\pi_{i}^{\scriptscriptstyle(n)}$ be the distribution of $W_{i}^{\scriptscriptstyle(n)}$. Then, for all $i\geq 1$, $\pi_{i}^{\scriptscriptstyle(0)}=\pi_{i}$ and, for all $n\geq 1$, $i\geq 1$, $w\in\mathcal{S}$,

[TABLE]

Definition 6.4**.**

We say that $\pi\in\mathcal{P}(\mathcal{S})$ is an invariant distribution of the operator $\mathcal{T}$ if and only if $\pi=\mathcal{T}(\pi,\pi)$.

Similarly to the finitely-many colour case, we define the following two maps: for all $\mu,\nu\in\mathcal{M}(\mathcal{S})$, we let

[TABLE]

If $\tau_{x,y}(\mathcal{S})=1$ for all $x,y\in\mathcal{S}$, then $\mathcal{T}^{(\ell)}_{\mu}$ and $\mathcal{T}^{(r)}_{\mu}$ are Markovian kernels on $\mathcal{S}$, for all $\mu\in\mathcal{P}(\mathcal{S})$. We can define the norm-based ergodicity coefficient (sometimes called Doeblin or Dobrushin coefficient) of a Markovian kernel $\mathcal{K}$ on $\mathcal{S}$ as follows:

[TABLE]

We let

[TABLE]

6.3 Main result for infinitely-many colours

As in the finitely-many colour case, we assume that the urn process is balanced:

• (B)

There exists $\sigma\in(0,\infty)$ such that, for all $x,y\in\mathcal{C}$, $\rho_{x,y}(\mathcal{C})=\sigma$.

Tenability is already ensured by the fact that, for all $x,y\in\mathcal{C}$, $\rho_{x,y}$ is a non-negative measure. The ergodicity assumption however, is much harder to state, mostly because the results of [FT20] are not available in this setting. To state it, we consider the operator $\mathcal{T}$ for the 2-dependent Markov chain on $\mathcal{S}=\mathcal{C}\times\mathcal{C}$: For all $\mu,\nu\in\mathcal{M}(\mathcal{C}\times\mathcal{C})\times\mathcal{M}(\mathcal{C}\times\mathcal{C})$,

[TABLE]

• (E)

$\tau_{\ell}(\mathcal{T})+\tau_{r}(\mathcal{T})<1$.

Analogously to the finitely-many colour case, we also define $\mathcal{R}=\mathcal{M}(\mathcal{C})\times\mathcal{M}(\mathcal{C})\to\mathcal{M}(\mathcal{C})$ as follows: for all $\mu,\nu\in\mathcal{M}(\mathcal{C})$,

[TABLE]

Theorem 6.5**.**

Let $(m_{n})_{n\geq 0}$ be a two-drawing Pólya process on a measurable space $\mathcal{C}$, with initial composition $m_{0}$ and replacement kernel $(\rho_{x,y})_{x,y\in\mathcal{C}}$. If $(\rho_{x,y})_{x,y\in\mathcal{C}}$ satisfies Assumptions (B) and (E), then there exists a unique $\nu\in\mathcal{P}(\mathcal{C})$ satisfying

[TABLE]

and, in probability as $n\uparrow\infty$, $\hat{m}_{n}\to\nu$.

Remark 6.6**.**

Because of Assumption (E) being hard to check in practice, we are not able to provide any (non-trivial) example of infinitely-many colour multi-drawing Pólya urn to which Theorem 6.5. We leave it for further work to prove an equivalent of Fasino and Tudisco [FT20]’s criterion for $2$-dependent Markov chains on infinite state spaces.

6.4 Proof of Theorem 6.5

Proposition 6.7**.**

Let $\mathcal{T}$ be a 2-dependent Markov operator on $\mathcal{S}$ satisfying condition (E). Then there exists a unique $\pi\in\mathcal{P}(S)$ such that $\mathcal{T}(\pi,\pi)=\pi$. Furthermore, any sequence $(\pi_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ of measures in $\mathcal{P}(\mathcal{S})$ that satisfies (15) is such that, for all $n\geq 0$,

[TABLE]

for $q=\tau_{\ell}(\mathcal{T})+\tau_{r}(\mathcal{T})<1$. Therefore, $\lim_{n\to\infty}\pi_{1}^{\scriptscriptstyle(n)}=\pi$.

Note that, by Lemma 6.3, $\lim_{n\to\infty}\pi_{1}^{\scriptscriptstyle(n)}=\pi$ implies that $W_{1}^{\scriptscriptstyle(n)}\Rightarrow\pi$ in distribution as $n\uparrow\infty$, where $(W_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ is the 2-dependent Markov chain with operator $\mathcal{T}$.

Proof.

Let $\tilde{\mathcal{T}}(\pi)=\mathcal{T}(\pi,\pi)$. Then $\tilde{\mathcal{T}}$ is a map from $\mathcal{P}(\mathcal{S})$ to $\mathcal{P}(\mathcal{S})$. By linearity of the integral, for all $\mu,\nu\in\mathcal{M}(\mathcal{S})$,

[TABLE]

(See (16) for the definition of the operators $T^{(\ell)}_{\mu}$ and $T^{(r)}_{\nu}$.) Therefore,

[TABLE]

Therefore, $\tilde{\mathcal{T}}$ is a contraction on $\mathcal{M}(S)$ and has a unique fixed point $\pi\in\mathcal{P}(\mathcal{S})$. In other words, there exists a unique probability measure $\pi$ such that $\mathcal{T}(\pi,\pi)=\pi$.

Similarly, for all $\mu_{1},\mu_{2},\nu_{1},\nu_{2}\in\mathcal{M}(\mathcal{S})$,

[TABLE]

Applying this to $\nu_{1}=\nu_{2}=\pi$ (the unique probability measure satisfying $\pi=\mathcal{T}(\pi,\pi)$), we get that, for all $\mu_{1},\mu_{2}\in\mathcal{M}(\mathcal{S})$,

[TABLE]

where we have used $q=\tau_{\ell}(\mathcal{T})+\tau_{r}(\mathcal{T})$. We now prove inductively on $n\geq 0$, that for all $n\geq 0$, for all $i\geq 1$,

[TABLE]

The base case holds straightforwardly; for the induction, note that, for all $n\geq 1$, for all $i\geq 1$, because $\pi=\mathcal{T}(\pi,\pi)$ and (15), we have

[TABLE]

by (21). Now, by the induction hypothesis, we get

[TABLE]

which concludes the proof of (22). The result follows by taking $i=1$ in (22). ∎

Note that Lemma 3.6 holds in the infinitely-many colour case because the proof was independent of the set of colours.

Lemma 6.8**.**

Fix $n\geq 0$; we work conditionally on $G_{n_{1}(n)}$ and $(X(i))_{1\leq i\leq n_{1}(n)}$, and on the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$. We order the nodes in $L(\ell(n)-1)$ from left to right so that $H^{\prime}_{n}$ is a plane binary tree. For all $1\leq i\leq 2^{\ell(n)-1}$, we let $\pi_{i}^{\scriptscriptstyle(0)}$ be the distribution of the label of the $i$-th node in level $L(\ell(n)-1)$ (from left to right); if $i>2^{\ell(n)-1}$, we let $\pi_{i}^{\scriptscriptstyle(0)}$ be a fixed, and arbitrary distribution on $\mathcal{C}^{2}$.

Then, in distribution, $X(n)=W_{1}^{\scriptscriptstyle(n)}$, where $(W_{i}^{\scriptscriptstyle(n)})_{n\geq 0,i\geq 1}$ is the 2-dependent Markov chain on $\mathcal{C}^{2}$ with operator $\mathcal{T}$ given by (18).

Proof.

This is straightforward from Lemma 3.6 (which gives independence of the labels of the leaves - needed in the definition of a 2-dependent Markov chain), and from the definition of the $\mathcal{R}$-labelled DAG, which gives the Markov operator $\mathcal{T}$. ∎

Proposition 6.7 states that under Assumptions (B) and (E), there exists a unique $\pi\in\mathcal{P}(\mathcal{C})$ such that $\mathcal{T}(\pi,\pi)=\pi$, and $W_{1}^{\scriptscriptstyle(n)}$ converges in distribution to $\pi$. We thus get, by Lemma 6.8, that $X(n)\Rightarrow\pi$ in distribution as $n\uparrow\infty$, conditionally on $G_{n_{1}(n)}$ and $(X(i))_{1\leq i\leq n_{1}(n)}$, and on the event $\mathcal{E}_{n}\cap\mathcal{F}_{n}$.

This means that, for all measurable sets $A$ of $\mathcal{C}\times\mathcal{C}$,

[TABLE]

as $n\uparrow\infty$.

As in Subsection 3.3, we get

[TABLE]

Hence, to prove Theorem 6.5, it only remains to prove that there is a unique $\nu\in\mathcal{P}(\mathcal{C})$ such that $\sigma\nu=\mathcal{R}(\nu,\nu)$, and that $\pi=\nu\otimes\nu$.

By definition of $\mathcal{T}$ (see (18)) and by Lemma 6.3, for any $A_{1},A_{2}$ two measurable subsets of $\mathcal{C}$, for any $n\geq 1$,

[TABLE]

where we have set $\nu(A)=\frac{1}{\sigma}\int_{\alpha_{1},\alpha_{2}}\rho_{\alpha_{1},\alpha_{2}}(A)\mathrm{d}\pi(\alpha_{1},\alpha_{2})$, for all $A\in\mathcal{C}$; in other words,

[TABLE]

Because the product topology on $\mathcal{C}\times\mathcal{C}$ is generated by “boxes” (subsets of the form $A_{1}\times A_{2}$ as above), and because the set of all boxes is stable by intersection (and thus a $\pi$-system), this implies that $\mathcal{T}(\pi,\pi)(A)=(\nu\otimes\nu)(A)$ for any measurable subset of $\mathcal{C}\times\mathcal{C}$. We thus get that $\pi=\mathcal{T}(\pi,\pi)=\nu\otimes\nu$.

Note that, by definition (see (24) for the definition of $\nu$, and (19) for the definition of $\mathcal{R}$), we have $\sigma\nu=\mathcal{R}(\nu,\nu)$. It only remains to show that this solution is unique. Assume that there exists $\nu^{\prime}\in\mathcal{P}(\mathcal{C})$ such that $\sigma\nu^{\prime}=\mathcal{R}(\nu^{\prime},\nu^{\prime})$. Then the same calculation as in (23) shows that $\mathcal{T}(\nu^{\prime}\otimes\nu^{\prime},\nu^{\prime}\otimes\nu^{\prime})=\nu^{\prime}\otimes\nu^{\prime}$. Because $\pi=\nu\otimes\nu$ is the unique solution of $\mathcal{T}(\pi,\pi)=\pi$ in $\mathcal{P}(\mathcal{C}\times\mathcal{C})$, we get that $\pi=\nu^{\prime}\otimes\nu^{\prime}$ and thus that $\nu^{\prime}=\nu$, proving uniqueness of the solution of (20).