2. The model

2.1. Preliminaries

From now on, $\mathbb{X}$ is a complete separable metric space endowed with the Borel $\sigma$ -algebra $\mathcal{X}$ , which from standard results is countably generated. All random quantities are defined on a common probability space $(\Omega,\mathcal{H},\mathbb{P})$ unless otherwise specified. For all unexplained measure-theoretic details, we refer to [Reference Kallenberg20].

Given two arbitrary measures $\mu,\pi$ on $\mathbb{X}$ , we say that $\mu$ is absolutely continuous with respect to $\pi$ , denoted $\mu\ll\pi$ , if $\mu(A)=0$ whenever $\pi(A)=0$ for any $A\in\mathcal{X}$ , and we say that $\mu$ and $\pi$ are mutually singular, denoted $\mu\perp\pi$ , if there exists $S\in\mathcal{X}$ such that $\mu(S)=0=\pi(S^\textrm{c})$ .

A transition kernel on $\mathbb{X}$ is a function $R\colon\mathbb{X}\times\mathcal{X}\rightarrow\mathbb{R}_+$ such that (i) the map $x\mapsto R(x,B)\equiv R_x(B)$ is $\mathcal{X}$ -measurable, for all $B\in\mathcal{X}$ ; and (ii) the map $B\mapsto R_x(B)$ is a measure on $\mathbb{X}$ , for all $x\in\mathbb{X}$ . In addition, R is said to be finite if $R_x(\mathbb{X})<\infty$ , and is called a probability kernel if $R_x(\mathbb{X})=1$ for all $x\in\mathbb{X}$ .

Let $\nu$ be a probability measure on $\mathbb{X}$ , and $\mathcal{G}\subseteq\mathcal{X}$ a sub- $\sigma$ -algebra on $\mathbb{X}$ . A probability kernel R on $\mathbb{X}$ is said to be a regular version of the conditional distribution (r.c.d.) for $\nu$ given $\mathcal{G}$ , which for emphasis we will denote by

\[ R_x(\!\cdot\!)=\nu(\,\cdot\mid\mathcal{G})(x) \quad\text{for $\nu$-almost every (a.e.)} \ \textit{x},\]

if (i) the map $x\mapsto R_x(B)$ is $\mathcal{G}$ -measurable, for all $B\in\mathcal{X}$ ; and (ii) $\int_AR_x(B)\,\nu({\rm d}x)=\nu(A\cap B)$ for all $A\in\mathcal{G}$ and $B\in\mathcal{X}$ . The assumptions on $(\mathbb{X},\mathcal{X})$ guarantee that an r.c.d. for $\nu$ given $\mathcal{G}$ exists and is unique up to a $\nu$ -null set, and all conditional distributions must be understood as regular versions. Moreover, a regular version R of $\nu(\,\cdot\mid\mathcal{G})$ is called almost everywhere proper if there exists $F\in\mathcal{G}$ such that $\nu(F)=1$ and

\[R_x(A)=\delta_x(A) \quad\text{for all $A\in\mathcal{G}$ and $x\in F$}.\]

Following [Reference Berti and Rigo6, p. 649], an almost everywhere proper r.c.d. for $\nu$ given $\mathcal{G}$ exists if and only if $\mathcal{G}$ is countably generated (c.g.) under $\nu$ , in the sense that there exists $C\in\mathcal{G}$ such that $\nu(C)=1$ and $\mathcal{G}\cap C$ is c.g.

An $\mathbb{X}$ -valued sequence $(X_n)_{n\geq1}$ of random variables is said to be exchangeable if, for each $n=1,2,\ldots$ and all permutations $\sigma$ of $\{1,\ldots,n\}$ ,

\[(X_1,\ldots,X_n)\overset{\textrm{d}}{=}(X_{\sigma(1)},\ldots,X_{\sigma(n)}).\]

In that case, by the de Finetti representation theorem for exchangeable sequences (see, e.g., [Reference Aldous, Aldous, Ibragimov and Jacod1, Section 3]) there exists a random probability measure $\tilde{P}$ on $\mathbb{X}$ , called the directing random measure of the sequence, such that, given $\tilde{P}$ , the random variables $X_1,X_2,\ldots$ are conditionally independently and identically distributed (i.i.d.) according to $\tilde{P}$ ,

\begin{eqnarray*} X_n\mid\tilde{P}\ \ & \overset{\textrm{i.i.d.}}{\sim} &\ \ \tilde{P}, \\\tilde{P}\ \ & \sim &\ \ Q,\end{eqnarray*}

where Q is the (nonparameteric) prior distribution of $\tilde{P}$ . Moreover, $\tilde{P}$ is the almost sure weak limit of the empirical measure,

\[{1}/{n}\sum_{i=1}^n\delta_{X_i}\overset{\textrm{w}}{\longrightarrow}\tilde{P} \quad\text{almost surely (a.s.)},\]

and the predictive distributions of $(X_n)_{n\geq1}$ ,

\[ \mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n)\overset{\textrm{w}}{\longrightarrow}\tilde{P} \quad\text{(a.s.)}.\]

The most popular prior distribution in Bayesian nonparametric analysis is arguably the Dirichlet process, $\mbox{DP}(\theta,\nu)$ , which is parameterized by a probability measure $\nu$ on $\mathbb{X}$ and a constant $\theta>0$ . We recall that $\tilde{P}\sim\mbox{DP}(\theta,\nu)$ if, for every finite partition $\{B_1,\ldots,B_n\}\subseteq\mathcal{X}$ of $\mathbb{X}$ , the random vector $(\tilde{P}(B_1),\ldots,\tilde{P}(B_n))$ has a Dirichlet distribution with parameters $(\theta\nu(B_1),\ldots,\theta\nu(B_n))$ . Equivalently (see, e.g., [Reference Ghosal and van der Vaart15, Theorem 4.12]), the Dirichlet process is the distribution of an a.s. discrete random probability measure with so-called stick-breaking weights,

(2.1) \begin{equation} \sum_{j=1}^\infty V_j\delta_{Z_j}\sim\mbox{DP}(\theta\nu),\end{equation}

where $V_j=W_j\prod_{i=1}^{j-1}W_i$ , and $W_1,W_2,\ldots\overset{\textrm{i.i.d.}}{\sim}\text{Beta}(1,\theta)$ and $Z_1,Z_2,\ldots\overset{\textrm{i.i.d.}}{\sim}\nu$ are independent random variables.

Exchangeable processes having directing random measures $\tilde{P}$ as in (2.1), for arbitrary sequences of nonnegative random weights $(V_j)_{j\geq1}$ and independent atoms $(Z_j)_{j\geq1}$ coming from a diffuse distribution $\nu$ , are called (proper) species sampling sequences and have predictive distributions of the form (1.5) for more general functions $p_n$ and $r_n$ ; see [Reference Hansen and Pitman16, Reference Pitman, Ferguson, Shapley and MacQueen25] for a complete characterization. Because of the discrete nature of their $\tilde{P}$ and the fact that the atoms $Z_1,Z_2,\ldots$ are a.s. all distinct, species sampling models are, by Kingman’s theory, in one-to-one correspondence with exchangeable random partitions, up to a specification of $\nu$ (for more information, we refer to [Reference Pitman, Ferguson, Shapley and MacQueen25] and [Reference Ghosal and van der Vaart15, Chapter 14]). In fact, the exchangeable partition generated by a random sample from a Dirichlet process (2.1) with diffuse $\nu$ is the well-known Chinese restaurant process.

2.2. Exchangeable MVPS

Following [Reference Sariev and Savov27], we call any sequence $(X_n)_{n\geq1}$ of $\mathbb{X}$ -valued random variables a measure-valued Pólya (urn) sequence with parameters $\theta$ , $\nu$ , and R, denoted MVPS $(\theta,\nu,R)$ , if $\nu(\!\cdot\!)=\mathbb{P}(X_1\in\cdot\,)$ and, for each $n=1,2,\ldots$ ,

(2.2) \begin{equation} \mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n)=\frac{\theta\nu(\!\cdot\!)+\sum_{i=1}^n R_{X_i}(\!\cdot\!)}{\theta+\sum_{i=1}^n R_{X_i}(\mathbb{X})},\end{equation}

where R is a finite transition kernel on $\mathbb{X}$ , called the reinforcement kernel of the process, $\nu$ is a probability measure on $\mathbb{X}$ , known as the base measure, and $\theta>0$ is a positive constant. When $R_x(\mathbb{X})=m$ for $\nu$ -a.e. x and some $m>0$ , both R and the corresponding MVPS are said to be balanced.

The predictive structure (2.2) admits a physical interpretation in terms of a generalized urn sampling scheme as follows. Suppose initially that we have an urn whose contents are described by the measure $\theta\nu$ in the sense that, for each $B\in\mathcal{X}$ , the quantity $\theta\nu(B)$ records the total mass of balls whose colors lie in B. At time $n=1$ , a ball is drawn at random from the urn with probability distribution $\nu$ . Depending on its color, $X_1$ , we reinforce the urn with additional balls according to the measure $R_{X_1}$ , so that the updated composition becomes $\theta\nu+R_{X_1}$ . At each subsequent time $n>1$ , a ball with color $X_n$ is drawn with probability proportional to $\theta\nu+\sum_{i=1}^{n-1}R_{X_i}$ , and the contents of the urn get ‘reinforced’ by $R_{X_n}$ . The model with $|\mathbb{X}|=k$ and $R_x=\delta_x$ , i.e. we only add one additional ball of the observed color, corresponds to the classical k-color Pólya urn scheme, and measure-valued Pólya urn processes have been proposed by [Reference Bandyopadhyay and Thacker3, Reference Fortini, Petrone and Sariev13, Reference Janson18, Reference Mailler and Marckert24] as extensions to infinite colors and more general reinforcement mechanisms. The case where the observation process $(X_n)_{n\geq1}$ is exchangeable has been investigated by [Reference Sariev and Savov27], and the main results from that paper are summarized in the following theorem.

Theorem 2.1. ([Reference Sariev and Savov27, Theorems 3.2, 3.7, and 3.9].) Let $(X_n)_{n\geq1}$ be an exchangeable, but not i.i.d., MVPS $(\theta,\nu,R)$ with directing random measure $\tilde{P}$ . Then:

1. (i) R is balanced and there exists a (c.g. under $\nu$ ) sub- $\sigma$ -algebra $\mathcal{G}$ of $\mathcal{X}$ such that, normalized, R is an r.c.d. for $\nu$ given $\mathcal{G}$ ,

   $$\frac{R_x(\!\cdot\!)}{R_x(\mathbb{X})}=\nu(\,\cdot\mid\mathcal{G})(x) \quad for\ \nu\text{-}a.e.\ x;$$

2. (ii) $\mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n)$ converges a.s. in total variation, as $n\rightarrow\infty$ , to $\tilde{P}$ ;

3. (iii) $\tilde{P}$ is equal in law to

   (2.3) \begin{equation} \sum_{j=1}^\infty V_j\frac{R_{Z_j}(\!\cdot\!)}{R_{Z_j}(\mathbb{X})}, \end{equation}
   where $(V_j)_{j\geq1}$ and $(Z_j)_{j\geq1}$ are given in (2.1).

Independently from the theory on measure-valued urn processes, [Reference Berti, Dreassi, Leisen, Pratelli and Rigo4] introduced the kernel-based Dirichlet sequence, which is a balanced MVPS with reinforcement kernel $R(\!\cdot\!)=\nu(\,\cdot\mid\mathcal{G})$ , for some sub- $\sigma$ -algebra $\mathcal{G}$ of $\mathcal{X}$ . In [Reference Berti, Dreassi, Leisen, Pratelli and Rigo4], the authors prove that kernel-based Dirichlet sequences are exchangeable, having directing random measures (2.3). Theorem 2.1 provides a converse statement and shows that, under exchangeability, R need not be assumed to be balanced a priori. In fact, it follows from Theorem 2.1 that any exchangeable MVPS $(\theta,\nu,R)$ can be reparameterized as a balanced exchangeable MVPS $(\tilde{\theta},\nu,\tilde{R})$ for some probability kernel $\tilde{R}$ and constant $\tilde{\theta}$ , so that

(2.4) \begin{equation} \mathbb{P}(X_{n+1}\in\cdot\mid X_1,\ldots,X_n) = \frac{\tilde{\theta}}{\tilde{\theta}+n}\nu(\!\cdot\!) + \frac{n}{\tilde{\theta}+n}\cdot\frac{1}{n}\sum_{i=1}^n\tilde{R}_{X_i}(\!\cdot\!).\end{equation}

Therefore, exchangeable MVPSs satisfy the sufficientness postulate (1.6), and in Section 3 we show that the opposite statement is also true. On the other hand, by Theorem 2.1(iii), the directing random measure $\tilde{P}$ of any MVPS inherits its properties directly from the reinforcement kernel $\tilde{R}$ , so that $\tilde{P}$ is a.s. diffuse and/or absolutely continuous with respect to some reference probability measure whenever $\tilde{R}$ is. This is in stark contrast to the discrete nature of the Dirichlet process (2.1) and species sampling models, in general, all while keeping the same simple form of the predictive distributions (2.4).

Theorem 2.1 can be refined when $\mathbb{X}$ is countable or R is dominated by $\nu$ , in which case exchangeable MVPSs turn out to have Dirichlet process mixture priors.

Theorem 2.2. ([Reference Sariev and Savov27, Theorems 3.10 and 3.13].) Let $(X_n)_{n\geq1}$ be as in Theorem 2.1. If $R_x\ll\nu$ for $\nu$ -a.e. x, then:

1. (i) there exists a countable partition $D_1,D_2,\ldots$ of $\mathbb{X}$ in $\mathcal{X}$ such that, for $\nu$ -a.e. x,

   \[\frac{R_x(\!\cdot\!)}{R_x(\mathbb{X})}=\sum_{k\geq1}\nu(\,\cdot\mid D_k)\cdot\boldsymbol{1}_{D_k}(x);\]

2. (ii) letting $m=R_x(\mathbb{X})$ ,

   \begin{eqnarray*} X_n\mid\tilde{P}\ \ & \overset{\textrm{i.i.d.}}{\sim} &\ \ \sum_{k\geq1}\tilde{P}(D_k)\,\nu(\,\cdot\mid D_k), \\ (\tilde{P}(D_k))_{k\geq1}\ \ & \sim & \ \ \textrm{DP}({\theta}/{m},\nu). \end{eqnarray*}

4. Proofs

Proof of Theorem 3.1. Theorem 3.1 will be proved in several steps. We start by deriving some relations between $\nu$ , R, and $(a_n)_{n\geq1}$ as a consequence of exchangeability.

Step 1. Let $A,B\in\mathcal{X}$ . It follows from the form of the predictive distributions (3.1) and $\mathbb{P}(X_1\in \textrm{d}x)=\nu(\textrm{d}x)$ that

\begin{align*} \mathbb{P}(X_1\in A,X_2\in B) & = \int_A\mathbb{P}(X_2\in B\mid X_1=x)\mathbb{P}(X_1\in \textrm{d}x) \\ & = \int_A\{(1-a_1)\nu(B)+a_1R_x(B)\}\,\nu(\textrm{d}x) \\ & = (1-a_1)\nu(A)\nu(B) + a_1\int_AR_x(B)\,\nu(\textrm{d}x). \end{align*}

Since $(X_1,X_2)\overset{\textrm{d}}{=}(X_2,X_1)$ by exchangeability, we get

\begin{align*} (1-a_1)\nu(A)\nu(B) + a_1\int_AR_x(B)\,\nu(\textrm{d}x) & = \mathbb{P}(X_1\in A,X_2\in B) \\ & = \mathbb{P}(X_1\in B,X_2\in A) \\ & = (1-a_1)\nu(A)\nu(B) + a_1\int_BR_x(A)\,\nu(\textrm{d}x); \end{align*}

therefore, $\int_{x\in A}\int_{y\in B}R_x(\textrm{d}y)\,\nu(\textrm{d}x) = \int_{x\in B}\int_{y\in A}R_x(\textrm{d}y)\,\nu(\textrm{d}x)$ or, as measures on $\mathbb{X}^2$ ,

(4.1) \begin{equation} R_x(\textrm{d}y)\nu(\textrm{d}x) = R_y(\textrm{d}x)\nu(\textrm{d}y). \end{equation}

Next, for $\nu$ -a.e. x,

\begin{align*} \mathbb{P}(X_3 & \in B\mid X_1=x) \\ & = \int_\mathbb{X}\mathbb{P}(X_3\in B\mid X_1=x,X_2=y)\mathbb{P}(X_2\in \textrm{d}y\mid X_1=x) \\ & = \int_\mathbb{X}\bigg\{(1-a_2)\nu(B) + \frac{a_2}{2}R_x(B) + \frac{a_2}{2}R_y(B)\bigg\} ((1-a_1)\,\nu(\textrm{d}y) + a_1\,R_x(\textrm{d}y)) \\ & = (1-a_2)\nu(B) + \frac{a_2}{2}R_x(B) + \frac{(1-a_1)a_2}{2}\int_\mathbb{X}R_y(B)\,\nu(\textrm{d}y) + \frac{a_1a_2}{2}\int_\mathbb{X}R_y(B)\,R_x(\textrm{d}y) \\ & \overset{\textrm{(a)}}{=} (1-a_2)\nu(B) + \frac{a_2}{2}R_x(B) + \frac{(1-a_1)a_2}{2}\nu(B) + \frac{a_1a_2}{2}\int_\mathbb{X}R_y(B)\,R_x(\textrm{d}y) \end{align*}

for every $B\in\mathcal{X}$ , where (a) follows from (4.1) as $\int_\mathbb{X}R_y(B)\,\nu(\textrm{d}y)=\nu(B)$ . Since $(X_1,X_2)\overset{\textrm{d}}{=}(X_1,X_3)$ by exchangeability we get, for $\nu$ -a.e. x,

\begin{align*} (1-a_1)\nu(B) & + a_1 R_x(B) \\ & = \mathbb{P}(X_2\in B\mid X_1=x) \\ & = \mathbb{P}(X_3\in B\mid X_1=x) \\ & = (1-a_2)\nu(B) + \frac{a_2}{2}R_x(B) + \frac{(1-a_1)a_2}{2}\nu(B) + \frac{a_1a_2}{2}\int_\mathbb{X}R_y(B)\,R_x(\textrm{d}y); \end{align*}

thus, rearranging the terms,

\[ \int_\mathbb{X}R_y(B)\,R_x(\textrm{d}y) = \bigg(1-\frac{a_2-2a_1+a_1a_2}{a_1a_2}\bigg)R_x(B) + \frac{a_2-2a_1+a_1a_2}{a_1a_2}\nu(B). \]

Using that $\mathcal{X}$ is countably generated, we obtain, for $\nu$ -a.e. x, that, as measures on $\mathbb{X}$ ,

(4.2) \begin{equation} \int_{y\in\mathbb{X}}R_y(\textrm{d}z)\,R_x(\textrm{d}y) = (1-c^*)\,R_x(\textrm{d}z) + c^*\,\nu(\textrm{d}z), \end{equation}

where $c^*\,:\!=(a_2-2a_1+a_1a_2)/(a_1a_2)$ .

Let $n\geq2$ . It follows from (3.1) that, for a.e. $(x_1,\ldots,x_{n-1},x)$ in $\mathbb{X}^n$ with respect to the marginal distribution $\mathbb{P}_{(X_1,\ldots,X_n)}$ of $(X_1,\ldots,X_n)$ on $\mathcal{X}^n$ ,

(4.3) \begin{multline} \mathbb{P}(X_{n+1}\in B\mid X_1=x_1,\ldots,X_{n-1}=x_{n-1},X_n=x) \\ = (1-a_n)\,\nu(\textrm{d}y) + \frac{a_n}{n}\sum_{i=1}^{n-1}R_{x_i}(B) + \frac{a_n}{n}R_x(B) \end{multline}

for every $B\in\mathcal{X}$ . On the other hand, disintegrating $\mathbb{P}_{(X_1,\ldots,X_n)}$ repeatedly via (3.1), we obtain (see also (4.12)) $\mathbb{P}(X_1\in \textrm{d}x_1,\ldots,X_n\in\textrm{d}x_n) \geq \prod_{i=1}^{n-1}(1-a_i)\,\nu(\textrm{d}x_1)\cdots\nu(\textrm{d}x_n)$ . Thus, (4.3) remains true for $(\nu\times\cdots\times\nu)$ -a.e. $(x_1,\ldots,x_{n-1},x)$ in $\mathbb{X}^n$ , so integrating (4.3) with respect to $\nu(\textrm{d}x_1),\ldots,\nu(\textrm{d}x_{n-1})$ we get, for $\nu$ -a.e. x,

(4.4) \begin{align} & \int_\mathbb{X}\cdots\int_\mathbb{X} \mathbb{P}(X_{n+1}\in B\mid X_1=x_1,\ldots,X_{n-1}=x_{n-1},X_n=x)\,\nu(\textrm{d}x_{n-1})\cdots\nu(\textrm{d}x_1) \nonumber \\ & = (1-a_n)\nu(B) + \frac{a_n}{n}\sum_{i=1}^{n-1}\nu(B) + \frac{a_n}{n}R_x(B) = (1-a_n)\nu(B) + \frac{(n-1)a_n}{n}\nu(B) + \frac{a_n}{n}R_x(B), \end{align}

where we have used that $\int_\mathbb{X}R_{x_i}(B)\,\nu(\textrm{d}x_i)=\nu(B)$ from (4.1), for $i=1,\ldots,n-1$ .

On the other hand, for $(\nu\times\cdots\times\nu)$ -a.e. $(x_1,\ldots,x_{n-1},x)$ in $\mathbb{X}^n$ ,

(4.5) \begin{align} & \mathbb{P}(X_{n+2}\in B\mid X_1=x_1,\ldots,X_{n-1}=x_{n-1},X_n=x) \nonumber \\ & = \int_\mathbb{X}\Bigg\{(1-a_{n+1})\nu(B) + \frac{a_{n+1}}{n+1}\sum_{i=1}^{n-1}R_{x_i}(B) + \frac{a_{n+1}}{n+1}R_x(B) + \frac{a_{n+1}}{n+1}R_y(B)\Bigg\} \nonumber \\ & \qquad\quad\Bigg((1-a_n)\,\nu(\textrm{d}y) + \frac{a_n}{n}\sum_{i=1}^{n-1}R_{x_i}(\textrm{d}y) + \frac{a_n}{n}\,R_x(\textrm{d}y)\Bigg) \nonumber \\ & = (1-a_{n+1})\nu(B) + \frac{a_{n+1}}{n+1}\sum_{i=1}^{n-1}R_{x_i}(B) + \frac{a_{n+1}}{n+1}R_x(B) + \frac{(1-a_n)a_{n+1}}{n+1}\int_\mathbb{X}R_y(B)\,\nu(\textrm{d}y) \nonumber \\ & \quad + \frac{a_na_{n+1}}{n(n+1)}\sum_{i=1}^{n-1}\int_\mathbb{X}R_y(B)\,R_{x_i}(\textrm{d}y) + \frac{a_na_{n+1}}{n(n+1)}\int_\mathbb{X}R_y(B)\,R_x(\textrm{d}y) \nonumber \\ & \overset{\textrm{(a)}}{=} (1-a_{n+1})\nu(B) + \frac{a_{n+1}}{n+1}\sum_{i=1}^{n-1}R_{x_i}(B) + \frac{a_{n+1}}{n+1}R_x(B) + \frac{(1-a_n)a_{n+1}}{n+1}\nu(B) \nonumber \\ & \quad + \frac{a_na_{n+1}}{n(n+1)}\sum_{i=1}^{n-1}\int_\mathbb{X}R_y(B)\,R_{x_i}(\textrm{d}y) + (1-c^*)\frac{a_na_{n+1}}{n(n+1)}R_x(B) + c^*\frac{a_na_{n+1}}{n(n+1)}\nu(B) \end{align}

for every $B\in\mathcal{X}$ , where (a) follows from (4.1) and (4.2). Note that, for each $i=1,\ldots,n-1$ , the following identity is true from (4.1):

\begin{align*} \int_\mathbb{X}\int_\mathbb{X}R_y(B)\,R_{x_i}(\textrm{d}y)\,\nu(\textrm{d}x_i) & = \int_\mathbb{X}\int_\mathbb{X}R_y(B)\,R_y(\textrm{d}x_i)\,\nu(\textrm{d}y) \\ & = \int_\mathbb{X}R_y(B)R_y(\mathbb{X})\,\nu(\textrm{d}y) = \int_\mathbb{X}R_y(B)\,\nu(\textrm{d}y) = \nu(B). \end{align*}

Thus, integrating (4.5) with respect to $\nu(\textrm{d}x_1),\ldots,\nu(\textrm{d}x_{n-1})$ , we get, for $\nu$ -a.e. x,

(4.6) \begin{align} \int_\mathbb{X}\cdots & \int_\mathbb{X}\mathbb{P}(X_{n+2}\in B\mid X_1=x_1,\ldots,X_{n-1}=x_{n-1},X_n=x)\, \nu(\textrm{d}x_{n-1})\cdots\nu(\textrm{d}x_1) \nonumber \\ & = (1-a_{n+1})\nu(B) + \frac{a_{n+1}}{n+1}\sum_{i=1}^{n-1}\nu(B) + \frac{a_{n+1}}{n+1}R_x(B) + \frac{(1-a_n)a_{n+1}}{n+1}\nu(B) \nonumber \\ & \quad + \frac{a_na_{n+1}}{n(n+1)}\sum_{i=1}^{n-1}\nu(B) + (1-c^*)\frac{a_na_{n+1}}{n(n+1)}R_x(B) + c^*\frac{a_na_{n+1}}{n(n+1)}\nu(B) \nonumber \\ & = (1-a_{n+1})\nu(B) + \frac{(n-1)a_{n+1}}{n+1}\nu(B) + \frac{a_{n+1}}{n+1}R_x(B) + \frac{(1-a_n)a_{n+1}}{n+1}\nu(B) \nonumber \\ & \quad + \frac{(n-1)a_na_{n+1}}{n(n+1)}\nu(B) + (1-c^*)\frac{a_na_{n+1}}{n(n+1)}R_x(B) + c^*\frac{a_na_{n+1}}{n(n+1)}\nu(B). \end{align}

Since $(X_1,\ldots,X_n,X_{n+1})\overset{\textrm{d}}{=}(X_1,\ldots,X_n,X_{n+2})$ by exchangeability, we can equate (4.4) and (4.6) to obtain, for $\nu$ -a.e. x,

\begin{align*} (1-a_n) & \nu(B) + \frac{(n-1)a_n}{n}\nu(B) + \frac{a_n}{n}R_x(B) \\ & = \int_\mathbb{X}\cdots\int_\mathbb{X}\mathbb{P}(X_{n+1}\in B\mid X_1=x_1,\ldots,X_{n-1}=x_{n-1},X_n=x)\, \nu(\textrm{d}x_{n-1})\cdots\nu(\textrm{d}x_1) \\ & = \int_\mathbb{X}\cdots\int_\mathbb{X}\mathbb{P}(X_{n+2}\in B\mid X_1=x_1,\ldots,X_{n-1}=x_{n-1},X_n=x)\, \nu(\textrm{d}x_{n-1})\cdots\nu(\textrm{d}x_1) \\ & = (1-a_{n+1})\nu(B) + \frac{(n-1)a_{n+1}}{n+1}\nu(B) + \frac{a_{n+1}}{n+1}R_x(B) + \frac{(1-a_n)a_{n+1}}{n+1}\nu(B) \\ & \quad + \frac{(n-1)a_na_{n+1}}{n(n+1)}\nu(B) + (1-c^*)\frac{a_na_{n+1}}{n(n+1)}R_x(B) + c^*\frac{a_na_{n+1}}{n(n+1)}\nu(B) \end{align*}

for every $B\in\mathcal{X}$ ; therefore, after some simple algebra,

\begin{equation*} (na_{n+1}-(n+1)a_n+(1-c^*)a_na_{n+1})\nu(B) = (na_{n+1}-(n+1)a_n+(1-c^*)a_na_{n+1})R_x(B). \end{equation*}

Using that $\mathcal{X}$ is countably generated, we get, for $\nu$ -a.e. x, that, as measures on $\mathbb{X}$ ,

(4.7) \begin{multline} (na_{n+1}-(n+1)a_n + (1-c^*)a_na_{n+1})\,\nu(\textrm{d}y) \\ = (na_{n+1}-(n+1)a_n + (1-c^*)a_na_{n+1})\,R_x(\textrm{d}y). \end{multline}

Step 2. We proceed by showing that, for different specifications of R, (4.2) and (4.7) together imply the prescribed form of the coefficients $a_n$ and, in particular, that $c^*=0$ . To that end, let us consider, for each $x\in\mathbb{X}$ , the Lebesgue decomposition of the reinforcement kernel,

$$R_x=R_x^\perp+R_x^{a},$$

where $R_x^\perp$ and $R_x^{a}$ are substochastic measures on $\mathbb{X}$ such that $R_x^\perp\perp\nu$ for some $S_x\in\mathcal{X}$ with $R_x^\perp(S_x^\textrm{c})=0=\nu(S_x)$ , and $R_x^{a}\ll\nu$ . It follows from [Reference Lange21, Theorem 1] that the mappings $x\mapsto R_x^\perp$ and $x\mapsto R_x^{a}$ are measurable, hence the following sets are well-defined in $\mathcal{X}$ , since $\mathcal{X}$ is countably generated:

\[ D\,:\!=\{x\in\mathbb{X}\colon R_x\neq\nu\}, \qquad F\,:\!=\{x\in\mathbb{X}\colon R_x^\perp(S_x)>0\}. \]

Case 1: $\nu(D)>0$ . By definition, for every $x\in D$ , there exists $B_x\in\mathcal{X}$ such that $\nu(B_x)\neq R_x(B_x)$ . It follows from (4.7) that, for $\nu$ -a.e. x in D,

\[(na_{n+1}-(n+1)a_n+(1-c^*)a_na_{n+1})\nu(B_x)=(na_{n+1}-(n+1)a_n+(1-c^*)a_na_{n+1})R_x(B_x),\]

which is true if and only if $na_{n+1}-(n+1)a_n+(1-c^*)a_na_{n+1}=0$ , or, equivalently,

\[\frac{a_{n+1}}{n+1}=\frac{a_n}{n+(1-c^*)a_n}.\]

Using the substitution $b_n={n}/{a_n}$ , we obtain $b_{n+1}=b_n+(1-c^*)$ ; thus, $b_n=b_1+(n-1)(1-c^*)$ for each $n=1,2,\ldots$ , and so

(4.8) \begin{equation} a_n=\frac{n}{(n-1)(1-c^*)+({1}/{a_1})}\quad\mbox{for }n=1,2,\ldots \end{equation}

If $c^*=1$ then $a_n=na_1$ , which leads to a contradiction since, by assumption, $0<a_n<1$ for all $n=1,2,\ldots$ ; otherwise,

\[ \lim_{n\rightarrow\infty}a_n = \lim_{n\rightarrow\infty}\frac{1}{(1-c^*)-(({1-c^*})/{n})+({1}/{a_1n})}=\frac{1}{1-c^*}, \]

implying that $a_n\notin(0,1)$ for n large enough when $c^*\in(0,1)\cup(1,\infty)$ – absurd. Therefore, $c^*\leq0$ .

Sub-case 1a: $\nu(D)>0$ and $\nu(F)>0$ . Let $G\in\mathcal{X}$ be the $\nu$ -almost sure set in (4.2). Define $G_0\,:\!=\{x\in G\colon R_x(G)=1\}$ and $G_n\,:\!=\{x\in G_{n-1}\colon R_x(G_{n-1})=1\}$ for $n\geq1$ . It follows from (4.1) that

\[ 1 = \int_\mathbb{X}R_x(\mathbb{X})\,\nu(\textrm{d}x) = \int_GR_x(\mathbb{X})\,\nu(\textrm{d}x) = \int_\mathbb{X}R_x(G)\,\nu(\textrm{d}x), \]

which implies that $R_x(G)=1$ for $\nu$ -a.e. x, and so $\nu(G_0)=1$ . Proceeding by induction, we obtain $\nu(G_n)=1$ for all $n\geq1$ ; thus, letting $G^*\,:\!=\bigcap_{n=0}^\infty G_n$ , we have $\nu(G^*)=1$ and $R_x(G_n)=1$ for all $x\in G^*$ and $n\geq1$ . As a result, for every $x\in G^*$ we get $R_x(G^*)=1$ and, since $G^*\subseteq G$ ,

(4.9) \begin{equation} \int_{y\in\mathbb{X}}R_y(\textrm{d}z)\,R_x(\textrm{d}y) = (1-c^*)\,R_x(\textrm{d}z) + c^*\,\nu(\textrm{d}z). \end{equation}

Recall that $F=\{x\in\mathbb{X}\colon R_x^\perp(S_x)>0\}$ . Let us define $\beta\,:\!=\sup_{x\in F\cap G^*}R_x^\perp(S_x)$ . Assuming that $\nu(F)>0$ , we have $\beta>0$ . Fix $\varepsilon\in(0,\beta)$ . From the properties of suprema, there exists $x_0\in F\cap G^*$ such that $R_{x_0}^\perp(S_{x_0})\geq\beta-\varepsilon$ . Moreover, $\nu(S_{x_0})=0$ , which implies that $R_y^{a}(S_{x_0})=0$ for all $y\in\mathbb{X}$ . It follows that, first,

\[(1-c^*)R_{x_0}(S_{x_0})+c^*\nu(S_{x_0})=(1-c^*)R_{x_0}^\perp(S_{x_0})\geq(1-c^*)(\beta-\varepsilon),\]

since $c^*\leq0$ from the previous part, and second,

\begin{align*} \int_\mathbb{X}R_y(S_{x_0})\,R_{x_0}(\textrm{d}y) & = \int_\mathbb{X}R_y^\perp(S_{x_0})\,R_{x_0}(\textrm{d}y) \\ & \overset{\textrm{(a)}}{\leq} \int_\mathbb{X}R_y^\perp(S_y)\,R_{x_0}(\textrm{d}y) \overset{\textrm{(b)}}{=} \int_{F\cap G^*}R_y^\perp(S_y)\,R_{x_0}(\textrm{d}y) \leq \beta, \end{align*}

where we have used in (a) that $R_y^\perp(S_y^\textrm{c})=0$ , so $R_y^\perp(S_{x_0})=R_y^\perp(S_{x_0}\cap S_y)\leq R_y^\perp(S_y)$ for $y\in\mathbb{X}$ ; and in (b) that (i) $R_y^\perp(S_y)=0$ for $y\in F^\textrm{c}$ by construction, and (ii) $R_{x_0}(G^*)=1$ , since $x_0\in G^*$ . Therefore, by (4.9),

\[ \beta \geq \int_\mathbb{X}R_y(S_{x_0})\,R_{x_0}(\textrm{d}y) = (1-c^*)R_{x_0}(S_{x_0}) + c^*\nu(S_{x_0}) \geq (1-c^*)(\beta-\varepsilon). \]

Letting $\varepsilon\downarrow0$ , we get $1-c^*\leq{\beta}/({\beta-\varepsilon})\rightarrow 1$ , and so $c^*\geq 0$ . But $c^*\leq0$ from the previous part, hence $c^*=0$ , which implies from (4.8) that

\[a_n=\frac{n}{n+({1}/{a_1})-1}.\]

Writing $\theta\,:\!=a_1^{-1}-1$ , we obtain $\theta>0$ and $a_n=n/(n+\theta)$ for each $n=1,2,\ldots$ , so

\begin{equation*} \mathbb{P}(X_{n+1}\in \textrm{d}x\mid X_1,\ldots,X_n) = \frac{\theta}{\theta+n}\,\nu(\textrm{d}x) + \frac{n}{\theta+n}\cdot\frac{1}{n}\sum_{i=1}^n R_{X_i}(\textrm{d}x) = \frac{\theta\,\nu(\textrm{d}x)+\sum_{i=1}^n R_{X_i}(\textrm{d}x)}{\theta+n}, \end{equation*}

that is, $(X_n)_{n\geq1}$ is an MVPS with parameters $(\theta,\nu,R)$ . Sub-case 1b: $\nu(D)>0$ and $\nu(F)=0$ . If $\nu(F)=0$ instead, then $R_x\ll\nu$ for $\nu$ -a.e. x, so $R_x(\textrm{d}y)\,\nu(\textrm{d}x)\ll\nu(\textrm{d}x)\,\nu(\textrm{d}y)$ and there exists a jointly measurable function $r\colon\mathbb{X}^2\rightarrow\mathbb{R}_+$ such that

(4.10) \begin{equation} R_x(\textrm{d}y)\,\nu(\textrm{d}x) = r(x,y)\,\nu(\textrm{d}x)\,\nu(\textrm{d}y). \end{equation}

In that case, (4.1) implies that, as measures on $\mathbb{X}^2$ ,

(4.11) \begin{equation} r(x,y)\,\nu(\textrm{d}x)\,\nu(\textrm{d}y) = r(y,x)\,\nu(\textrm{d}x)\,\nu(\textrm{d}y). \end{equation}

Let $A,B,C\in\mathcal{X}$ . It follows from the form of the predictive distributions (3.1) that

(4.12) \begin{align} & \mathbb{P}(X_1\in A,X_2\in B,X_3\in C) \nonumber \\ & = \int_A\int_B\mathbb{P}(X_3\in C\mid X_1=x,X_2=y)\mathbb{P}(X_2\in \textrm{d}y\mid X_1=x)\, \nu(\textrm{d}x) \nonumber \\ & = \int_A\int_B\bigg\{(1-a_2)\nu(C) + \frac{a_2}{2}R_x(C) + \frac{a_2}{2}R_y(C)\bigg\} ((1-a_1)\,\nu(\textrm{d}y) + a_1\,R_x(\textrm{d}y))\,\nu(\textrm{d}x) \nonumber \\ & = \int_A\bigg\{(1-a_1)(1-a_2)\nu(B)\nu(C) + a_1(1-a_2)\nu(C)R_x(B) + \frac{(1-a_1)a_2}{2}\nu(B)R_x(C) \nonumber \\ & \qquad\quad + \frac{a_1a_2}{2}R_x(B)R_x(C) + \frac{(1-a_1)a_2}{2}\int_BR_y(C)\,\nu(\textrm{d}y) + \frac{a_1a_2}{2}\int_BR_y(C)\,R_x(\textrm{d}y)\bigg\}\,\nu(\textrm{d}x) \nonumber \\ & = (1-a_1)(1-a_2)\nu(A)\nu(B)\nu(C) + a_1(1-a_2)\nu(C)\int_AR_x(B)\,\nu(\textrm{d}x) \nonumber \\ & \quad + \frac{(1-a_1)a_2}{2}\nu(B)\int_AR_x(C)\,\nu(\textrm{d}x) + \frac{a_1a_2}{2}\int_AR_x(B)R_x(C)\,\nu(\textrm{d}x) \nonumber \\ & \quad + \frac{(1-a_1)a_2}{2}\nu(A)\int_BR_y(C)\,\nu(\textrm{d}y) + \frac{a_1a_2}{2}\int_A\int_BR_y(C)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x). \end{align}

By exchangeability, $\mathbb{P}(X_1\in A,X_2\in B,X_3\in C)=\mathbb{P}(X_1\in A,X_2\in C,X_3\in B)$ , so from (4.12) we obtain, after some simplification,

\begin{align*} & a_1(1-a_2)\nu(C)\int_AR_x(B)\,\nu(\textrm{d}x) + \frac{(1-a_1)a_2}{2}\nu(B)\int_AR_x(C)\,\nu(\textrm{d}x) \\ & \quad + \frac{(1-a_1)a_2}{2}\nu(A)\int_BR_y(C)\,\nu(\textrm{d}y) + \frac{a_1a_2}{2}\int_A\int_BR_y(C)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x) \\ & = a_1(1-a_2)\nu(B)\int_AR_x(C)\,\nu(\textrm{d}x) + \frac{(1-a_1)a_2}{2}\nu(C)\int_AR_x(B)\,\nu(\textrm{d}x) \\ & \quad + \frac{(1-a_1)a_2}{2}\nu(A)\int_CR_y(B)\,\nu(\textrm{d}y) + \frac{a_1a_2}{2}\int_A\int_CR_y(B)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x). \end{align*}

Using that $\int_BR_y(C)\,\nu(\textrm{d}y)=\int_CR_y(B)\,\nu(\textrm{d}y)$ from (4.1), we further get

\begin{align*} & a_1(1-a_2)\nu(C)\int_AR_x(B)\,\nu(\textrm{d}x) + \frac{(1-a_1)a_2}{2}\nu(B)\int_AR_x(C)\,\nu(\textrm{d}x) \\ & \quad + \frac{a_1a_2}{2}\int_A\int_BR_y(C)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x) \\ & = a_1(1-a_2)\nu(B)\int_AR_x(C)\,\nu(\textrm{d}x) + \frac{(1-a_1)a_2}{2}\nu(C)\int_AR_x(B)\,\nu(\textrm{d}x) \\ & \quad + \frac{a_1a_2}{2}\int_A\int_CR_y(B)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x), \end{align*}

which after some simple algebra becomes

\begin{multline*} \int_A\int_BR_y(C)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x) - c^*\cdot\nu(C)\int_AR_x(B)\,\nu(\textrm{d}x) \\ = \int_A\int_CR_y(B)\,R_x(\textrm{d}y)\,\nu(\textrm{d}x) - c^*\cdot\nu(B)\int_AR_x(C)\,\nu(\textrm{d}x), \end{multline*}

where $c^*=({a_2-2a_1+a_1a_2})/{a_1a_2}\leq 0$ from before. In terms of the representation in (4.10), the above becomes

\begin{align*} & \int_A\int_B\int_C(r(x,y)r(y,z)-c^*r(x,y))\,\nu(\textrm{d}z)\,\nu(\textrm{d}y)\,\nu(\textrm{d}x) \\ & \qquad = \int_A\int_B\int_C(r(x,z)r(z,y)-c^*r(x,z))\,\nu(\textrm{d}z)\,\nu(\textrm{d}y)\,\nu(\textrm{d}x) \\ & \qquad = \int_A\int_B\int_C(r(x,z)r(y,z)-c^*r(x,z))\,\nu(\textrm{d}z)\,\nu(\textrm{d}y)\,\nu(\textrm{d}x), \end{align*}

where the last equation follows from (4.11). As a result, for $(\nu\times\nu\times\nu)$ -a.e. (x,y,z),

\[r(x,y)(r(y,z)-c^*)=r(x,z)(r(y,z)-c^*).\]

Suppose by contradiction that $c^*<0$ . In such a case, after canceling the common term above, we get, for $(\nu\times\nu\times\nu)$ -a.e. (x, y, z) in $\mathbb{X}^3$ , $r(x,y)=r(x,z)$ . Together with (4.11), this implies that, for $(\nu\times\nu\times\nu)$ -a.e. (x, y, z) in $\mathbb{X}^3$ ,

(4.13) \begin{equation} r(x,y)r(y,z)=r(x,z)r(z,y). \end{equation}

Now, recall that for every $x\in D$ there exists $B_x\in\mathcal{X}$ such that $\nu(B_x)\neq R_x(B)$ . It follows from (4.13) that, for $\nu$ -a.e. x in D,

\begin{align*} \int_\mathbb{X}R_y(B_x)\,R_x(\textrm{d}y) & = \int_{y\in\mathbb{X}}\int_{z\in B_x}r(x,y)r(y,z)\,\nu(\textrm{d}z)\,\nu(\textrm{d}y) \\ & = \int_{y\in B_x}\int_{z\in\mathbb{X}}r(x,y)r(y,z)\,\nu(\textrm{d}z)\,\nu(\textrm{d}y) = R_x(B_x). \end{align*}

On the other hand, by (4.2), for $\nu$ -a.e. x in D,

$$\int_\mathbb{X}R_y(B_x)\,R_x(\textrm{d}y)=(1-c^*)R_x(B_x)+c^*\nu(B_x);$$

therefore, $R_x(B_x)=\nu(B_x)$ – absurd. As a result, $c^*=0$ , which implies as in the previous sub-case that $(X_n)_{n\geq1}$ is an MVPS with parameters $(a^{-1}-1,\nu,R)$ .

Case 2: $\nu(D)=0$ . If $\nu(D)=0$ , then $R_x\equiv\nu$ for $\nu$ -a.e. x. Since $\mathbb{P}(X_1\in\cdot\,)=\nu(\!\cdot\!)$ , it follows that, for every $B_1,B_2\in\mathcal{X}$ ,

\begin{align*} \mathbb{P}(X_1\in B_1,X_2\in B_2) & = \int_{B_1}\mathbb{P}(X_2\in B_2\mid X_1=x)\,\nu(\textrm{d}x) \\ & = \int_{B_1}\{(1-a_1)\nu(B_2)+a_1R_x(B_2)\}\,\nu(\textrm{d}x) = \nu(B_1)\nu(B_2); \end{align*}

thus, $\mathbb{P}_{(X_1,X_2)}(\!\cdot\!)=(\nu\times\nu)(\!\cdot\!)$ . By induction, for every $B_1,\ldots,B_n\in\mathcal{X}$ ,

\[\mathbb{P}(X_1\in B_1,\ldots,X_n\in B_n)=\prod_{i=1}^n\nu(B_i);\]

therefore, $(X_n)_{n\geq1}$ is i.i.d. with marginal distribution $\nu$ , and thus trivially an MVPS with parameters $(a^{-1}-1,\nu,R)$ . This concludes the proof of Theorem 3.1.

Proof of Theorem 3.2. Suppose that $(X_n)_{n\geq1}$ satisfies (3.3) and (3.4). By assumption, $\mathcal{G}$ is c.g. under $\nu$ , so there exists (see [Reference Berti and Rigo6, p. 649]) an almost everywhere proper regular version R of $\nu(\,\cdot\mid\mathcal{G})$ , in the sense that, for some $F\in\mathcal{G}$ such that $\nu(F)=1$ ,

\[ R_x(A)=\delta_x(A) \quad\text{for all $A\in\mathcal{G}$ and $x\in F$}. \]

Moreover, since $(X_n)_{n\geq1}$ is identically distributed with marginal distribution $\nu$ , we have $\mathbb{P}(X_1\in F,\ldots,X_n\in F)=1$ for each $n=1,2,\ldots$ , which implies that, on a set of probability 1,

(4.14) \begin{equation} \sum_{i=1}^n\delta_{X_i}(A)=\sum_{i=1}^n R_{X_i}(A)\quad\mbox{for all }A\in\mathcal{G}. \end{equation}

Let $B_1,\ldots,B_k\in\mathcal{G}$ form a partition of $\mathbb{X}$ with $k>2$ and $\#\{j\colon\nu(B_j)>0\}>2$ . It follows from (3.3) that the sequence $(X_n)_{n\geq1}$ on the level of $\{B_1,\ldots,B_k\}$ satisfies Johnson’s sufficientness postulate (1.2), so from [Reference Zabell29, Theorem 2.1 and Corollary 2.1], there exist $\alpha_{B_1},\ldots,\alpha_{B_k}\geq0$ such that $\sum_{i=1}^k\alpha_{B_i}>0$ and, for each $j=1,\ldots,k$ ,

\[ \mathbb{P}(X_{n+1}\in B_j\mid X_1,\ldots,X_n) = \frac{\alpha_{B_j}+N_{n,B_j}}{\sum_{i=1}^k\alpha_{B_i}+n} \quad\mbox{a.s.} \]

In particular, $\mathbb{P}(X_1\in B_j)=\alpha_{B_j}/\sum_{i=1}^k\alpha_{B_i}$ , which implies that $a_{B_j}=\nu(B_j)\sum_{i=1}^k\alpha_{B_i}$ . Moreover, $\theta\,:\!=\sum_{i=1}^k\alpha_{B_i}$ does not depend on the particular partition chosen. Indeed, let $\{C_1,\ldots,C_m\}\subseteq\mathcal{G}$ be another partition of $\mathbb{X}$ with $m>2$ and $\#\{l\colon\nu(C_l)>0\}>2$ . If $\{C_1,\ldots,C_m\}$ is finer than $\{B_1,\ldots,B_k\}$ and, say, $B_j=\bigcup_{l=1}^LC_{j_l}$ for some j with $\nu(B_j)>0$ , we obtain, using the aforementioned results in [Reference Zabell29],

\begin{align*} \frac{\theta\nu(B_j)+N_{n,B_j}}{\theta+n} & = \sum_{l=1}^L\mathbb{P}(X_{n+1}\in C_{j_l}\mid X_1,\ldots,X_n) \\ & = \sum_{l=1}^L\frac{\nu(C_{j_l})\sum_{i=1}^m\alpha_{C_i}+N_{n,C_{j_l}}}{\sum_{i=1}^m\alpha_{C_i}+n} = \frac{\nu(B_j)\sum_{i=1}^m\alpha_{C_i}+N_{n,B_j}}{\sum_{i=1}^m\alpha_{C_i}+n} \quad\mbox{a.s.} \end{align*}

After simplification, we get $n\big(\theta-\sum_{i=1}^m\alpha_{C_i}\big)\nu(B_j)=\big(\theta-\sum_{i=1}^m\alpha_{C_i}\big)N_{n,B_j}$ a.s. for all $n=1,2,\ldots$ , which implies that $\theta=\sum_{i=1}^m\alpha_{C_i}$ . In any other case, we can consider the partition $\{B_i\cap C_l,i=1,\ldots,k,l=1,\ldots,m\}$ and conclude that the three partitions produce the same constant.

Let $A\in\mathcal{G}$ . Arguing through the partition $\{A\cap B_j,A^\textrm{c}\cap B_j,j=1,\ldots,k\}$ , we get

\begin{equation*} \mathbb{P}(X_{n+1}\in A\mid X_1,\ldots,X_n) = \sum_{j=1}^k\frac{\theta\nu(A\cap B_j)+N_{n,A\cap B_j}}{\theta+n} = \frac{\theta\nu(A)+\sum_{i=1}^n\delta_{X_i}(A)}{\theta+n} \quad\mbox{a.s.} \end{equation*}

Therefore, using that $\mathcal{G}$ is c.g. under $\nu$ , we obtain, as measures on $(\mathbb{X},\mathcal{G})$ ,

(4.15) \begin{equation} \mathbb{P}(X_{n+1}\in \textrm{d}x\mid X_1,\ldots,X_n) = \frac{\theta\,\nu(\textrm{d}x)+\sum_{i=1}^n\delta_{X_i}(\textrm{d}x)}{\theta+n}\quad\mbox{a.s.} \end{equation}

Let $A\in\mathcal{G}$ and $B\in\mathcal{X}$ . Recall that $\mathcal{G}_n=X_n^{-1}(\mathcal{G})$ for $n\geq1$ . Then

\begin{equation*} \mathbb{E}[\boldsymbol{1}_A(X_{n+1})\cdot\mathbb{P}(X_{n+1}\in B\mid \mathcal{G}_{n+1})] = \nu(A\cap B) = \int_AR_x(B)\,\nu(\textrm{d}x) = \mathbb{E}[\boldsymbol{1}_A(X_{n+1})\cdot R_{X_{n+1}}(B)], \end{equation*}

and, since $\omega\mapsto R_{X_{n+1}(\omega)}(B)$ is $\mathcal{G}_{n+1}$ -measurable,

(4.16) \begin{equation} \mathbb{P}(X_{n+1}\in B\mid \mathcal{G}_{n+1})=R_{X_{n+1}}(B)\quad\mbox{a.s.} \end{equation}

As a result, using (3.4), (4.15), and (4.16), we obtain, on a set of probability 1,

\begin{align*} \mathbb{P}(X_{n+1}\in B\mid X_1,\ldots,X_n) & = \mathbb{E}[\mathbb{P}(X_{n+1}\in B\mid X_1,\ldots,X_n,\mathcal{G}_{n+1})\mid X_1,\ldots,X_n] \\ & = \mathbb{E}[\mathbb{P}(X_{n+1}\in B\mid\mathcal{G}_{n+1})\mid X_1,\ldots,X_n] \\ & = \mathbb{E}[R_{X_{n+1}}(B)\mid X_1,\ldots,X_n] \\ & = \int_\mathbb{X}R_x(B)\biggl\{ \frac{\theta\,\nu(\textrm{d}x)+\sum_{i=1}^n\delta_{X_i}\,(\textrm{d}x)}{\theta+n}\biggr\} \\ & = \frac{\theta\int_\mathbb{X}R_x(B)\,\nu(\textrm{d}x) + \sum_{i=1}^n\int_\mathbb{X}R_x(B)\,\delta_{X_i}(\textrm{d}x)}{\theta+n} \\ & = \frac{\theta\nu(B)+\sum_{i=1}^n R_{X_i}(B)}{\theta+n}. \end{align*}

Note that the above result does not contradict (4.15) when $B\in\mathcal{G}$ because of (4.14). Therefore, we can conclude that $(X_n)_{n\geq1}$ is an MVPS with parameters $(\theta,\nu,R)$ .

Conversely, suppose that $(X_n)_{n\geq1}$ is an exchangeable MVPS with reinforcement kernel $R(\!\cdot\!)=\nu(\,\cdot\mid\mathcal{G})$ . Recall from Theorem 2.1 and the discussion around (2.4) that any exchangeable MVPS admits such a representation. Moreover, we may assume that R is almost everywhere proper, without loss of generality. Arguing as in (4.14), we obtain, for all $A\in\mathcal{G}$ , on a set of probability 1,

\begin{equation*} \mathbb{P}(X_{n+1}\in A\mid X_1,\ldots,X_n) = \frac{\theta\nu(A)+\sum_{i=1}^n R_{X_i}(A)}{\theta+n} = \frac{\theta\nu(A)+\sum_{i=1}^n\delta_{X_i}(A)}{\theta+n} = \frac{\theta\nu(A)+N_{n,A}}{\theta+n}; \end{equation*}

thus, $(X_n)_{n\geq1}$ satisfies (3.3). Let $A_1,\ldots,A_n,B\in\mathcal{X}$ and $A\in\mathcal{G}$ . Then

\begin{align*} \mathbb{E}[\boldsymbol{1}_{A_1}(X_1) & \cdots\boldsymbol{1}_{A_n}(X_n)\boldsymbol{1}_{A}(X_{n+1})\boldsymbol{1}_{B}(X_{n+1})] \\ & = \mathbb{E}[\boldsymbol{1}_{A_1}(X_1)\cdots\boldsymbol{1}_{A_n}(X_n)\cdot \mathbb{P}(X_{n+1}\in A\cap B\mid X_1,\ldots,X_n)] \\ & = \mathbb{E}\biggl[\boldsymbol{1}_{A_1}(X_1)\cdots\boldsymbol{1}_{A_n}(X_n) \frac{\theta\nu(A\cap B)+\sum_{i=1}^n R_{X_i}(A\cap B)}{\theta+n}\biggr] \\ & \overset{\textrm{(a)}}{=} \mathbb{E}\biggl[\boldsymbol{1}_{A_1}(X_1)\cdots\boldsymbol{1}_{A_n}(X_n) \frac{\theta\nu(A\cap B)+\sum_{i=1}^n R_{X_i}(B)\delta_{X_i}(A)}{\theta+n}\biggr] \\ & = \mathbb{E}\biggl[\boldsymbol{1}_{A_1}(X_1)\cdots\boldsymbol{1}_{A_n}(X_n) \int_AR_x(B)\biggl\{\frac{\theta\,\nu(\textrm{d}x)+\sum_{i=1}^n\delta_{X_i}(\textrm{d}x)}{\theta+n}\biggr\} \biggr] \\ & = \mathbb{E}[\boldsymbol{1}_{A_1}(X_1)\cdots\boldsymbol{1}_{A_n}(X_n)\cdot \mathbb{E}[\boldsymbol{1}_A(X_{n+1})R_{X_{n+1}}(B)\mid X_1,\ldots,X_n]] \\ & = \mathbb{E}[\boldsymbol{1}_{A_1}(X_1)\cdots\boldsymbol{1}_{A_n}(X_n)\boldsymbol{1}_A(X_{n+1})\cdot R_{X_{n+1}}(B)] \\ & = \mathbb{E}[\mathbb{P}(X_1\in A_1,\ldots,X_n\in A_n\mid \mathcal{G}_{n+1})\cdot \boldsymbol{1}_A(X_{n+1})\cdot R_{X_{n+1}}(B)] \\ & \overset{\textrm{(b)}}{=} \mathbb{E}[\mathbb{P}(X_1\in A_1,\ldots,X_n\in A_n\mid\mathcal{G}_{n+1})\cdot \boldsymbol{1}_A(X_{n+1})\cdot\mathbb{P}(X_{n+1}\in B\mid\mathcal{G}_{n+1})], \end{align*}

where (a) follows from $R_x(A\cap B)=\nu(A\cap B\mid\mathcal{G})(x)=\nu(B\mid\mathcal{G})(x)\delta_x(A)=R_x(B)\delta_x(A)$ for $\nu$ -a.e. x, since $A\in\mathcal{G}$ ; and for (b) we have used that $\mathbb{P}(X_{n+1}\in B\mid\mathcal{G}_{n+1})=R_{X_{n+1}}(B)$ a.s., arguing as in (4.16). Therefore, $X_{n+1}$ and $(X_1,\ldots,X_n)$ are conditionally independent given $\mathcal{G}_{n+1}$ . This concludes the proof of Theorem 3.2.

Proof of Proposition 3.1. Let us define, for every $n\geq0$ , $l=1,\ldots,m$ , and $n_l=0,1,\ldots,n$ , the quantity

$$g_{n,l}(n_l)\,:\!=\sum_{j\in D_l}f_{n,j}(n_l).$$

It follows from (3.7) that $f_{n,j}(n_l)>0$ , and so $g_{n,l}(n_l)>0$ . Recall that m(j) denotes, for each $j=1,\ldots,k$ , the index $l\in\{1,\ldots,m\}$ such that $j\in D_l$ . Define $Y_n\,:\!=m(X_n)$ for $n\geq1$ . Then $(Y_n)_{n\geq1}$ is an exchangeable process with predictive distributions

(4.17) \begin{equation} \mathbb{P}(Y_{n+1}=l\mid Y_1,\ldots,Y_n)=P_n(D_l)=g_{n,l}(N_{n,l})\quad\mbox{a.s., for }l=1,\ldots,m. \end{equation}

Therefore, $(Y_n)_{n\geq1}$ satisfies Johnson’s sufficientness postulate (1.2), so from [Reference Zabell29, Theorem 2.1 and Corollary 2.1] there exist $\bar{w}_1,\ldots,\bar{w}_m>0$ such that

(4.18) \begin{equation} g_{n,l}(n_l)=\frac{\bar{w}_l+n_l}{\bar{w}+n}, \end{equation}

where $\bar{w}\,:\!=\sum_{l=1}^m\bar{w}_l$ ; thus, in particular,

(4.19) \begin{equation} \mathbb{P}_0(D_l)=\mathbb{P}(Y_1=l)=g_{0,l}(0)=\frac{\bar{w}_l}{\bar{w}}\quad\mbox{for }l=1,\ldots,m. \end{equation}

Let $j\in\{1,\ldots,k\}$ be such that $D_{m(j)}=\{j\}$ . Then $f_{n,j}(n_{m(j)})=g_{n,m(j)}(n_{m(j)})$ , so from (4.18) and (4.19),

(4.20) \begin{equation} \mathbb{P}_n(\{j\}) = g_{n,m(j)}(N_{n,m(j)}) = \frac{\bar{w}\cdot\mathbb{P}_0(\{j\})+\sum_{i=1}^n\boldsymbol{1}_{D_{m(j)}}(X_i)}{\bar{w}+n} \quad\mbox{a.s.} \end{equation}

Let $i,j\in\{1,\ldots,k\}$ be such that $m(i)=m(j)$ . Fix $n\geq0$ , where for $n=0$ the expressions involving $\mathbb{P}(\,\cdot\mid X_1,\ldots,X_n)$ are meant as unconditional statements. It follows that, for every $h=1,2,\ldots$ ,

\begin{align*} & \mathbb{P}(X_{n+1}=j,X_{n+2}\notin D_{m(j)},\ldots,X_{n+h+1}\notin D_{m(j)}\mid X_1,\ldots,X_n) \\ & \qquad\qquad\qquad = f_{n,j}(N_{n,m(j)})\prod_{s=1}^h(1-g_{n+s,m(j)}(N_{n,m(j)}+1)) \quad\mbox{a.s.}, \\ & \mathbb{P}(X_{n+1}\notin D_{m(j)},\ldots,X_{n+h}\notin D_{m(j)},X_{n+h+1}=j\mid X_1,\ldots,X_n) \\ & \qquad\qquad\qquad = f_{n+h,j}(N_{n,m(j)})\prod_{s=0}^{h-1}(1-g_{n+s,m(j)}(N_{n,m(j)})) \quad\mbox{a.s.} \end{align*}

By exchangeability, the two expressions are equal, so we obtain

\[ \frac{f_{n,j}(N_{n,m(j)})}{f_{n+h,j}(N_{n,m(j)})} = \frac{\prod_{s=0}^{h-1}(1-g_{n+s,m(j)}(N_{n,m(j)}))}{\prod_{s=1}^h(1-g_{n+s,m(j)}(N_{n,m(j)}+1))} \quad\mbox{a.s.} \]

Since the right-hand side depends on j only through m(j), the same ratio ${f_{n,i}}/{f_{n+h,i}}$ for i is equal to the same quantity; therefore, with $n_{m(j)}\equiv n_{m(i)}\in\{0,1,\ldots,n\}$ fixed,

(4.21) \begin{equation} \frac{f_{n,i}(n_{m(i)})}{f_{n,j}(n_{m(j)})} = \frac{f_{n+h,i}(n_{m(i)})}{f_{n+h,j}(n_{m(j)})} \quad\mbox{for every }h=1,2,\ldots \end{equation}

On the other hand, for every $h=1,\ldots,n$ ,

\begin{multline*} \mathbb{P}(X_1\notin D_{m(j)},\ldots,X_n\notin D_{m(j)},X_{n+1}=j,X_{n+2}\in D_{m(j)},\ldots,X_{2n+1}\in D_{m(j)}) \\ = f_{n,j}(0)\prod_{s=0}^{n-1}(1-g_{s,m(j)}(0))\prod_{s=1}^ng_{n+s,m(j)}(s) \quad\mbox{a.s.} \end{multline*}

and, applying the permutation $\sigma$ of $\{1,\ldots,2n+1\}$ that swaps the indices $(1,\ldots,h)$ with $(n+2,\ldots,n+h+1)$ ,

\begin{align*} & \mathbb{P}(X_{\sigma(1)}\notin D_{m(j)},\ldots,X_{\sigma(n)}\notin D_{m(j)}, X_{\sigma(n+1)}=j,\ldots,X_{\sigma(2n+1)}\in D_{m(j)}) \\ & \qquad\qquad\qquad = f_{n,j}(h)\prod_{s=0}^{h-1}g_{s,m(j)}(s)\prod_{s=0}^{n-h-1}(1-g_{h+s,m(j)}(h)) \\ & \qquad\qquad\qquad\quad \times \prod_{s=1}^h(1-g_{n+s,m(j)}(h))\prod_{s=1}^{n-h}g_{n+h+s,m(j)}(h+s-1) \quad\mbox{a.s.} \end{align*}

Since $(X_1,\ldots,X_{2n+1})\overset{\textrm{d}}{=}(X_{\sigma(1)},\ldots,X_{\sigma(2n+1)})$ by exchangeability, we obtain, arguing as in (4.21), with n fixed,

(4.22) \begin{equation} \frac{f_{n,i}(0)}{f_{n,j}(0)}=\frac{f_{n,i}(h)}{f_{n,j}(h)} \quad\mbox{for every }h=1,\ldots,n. \end{equation}

It follows from (4.21) and (4.22) that, for any such pair i, j, the ratio

\[\frac{f_{n,i}(n_{m(i)})}{f_{n,j}(n_{m(j)})}=:d_{i,j}\]

is independent of n and $n_{m(j)}\equiv n_{m(i)}$ . Summing over all i in $D_{m(j)}$ and letting $w_j\,:\!={\bar{w}_{m(j)}}/{\sum_{i\in D_{m(j)}}d_{i,j}}$ gives

\[f_{n,j}(n_{m(j)})=\frac{1}{\sum_{i\in D_{m(j)}}d_{i,j}}g_{n,m(j)}(n_{m(j)})=\frac{w_j}{\bar{w}_{m(j)}}g_{n,m(j)}(n_{m(j)});\]

thus, in particular, from (4.19),

\[\mathbb{P}_0(\{j\})=f_{0,j}(0)=\frac{w_j}{\bar{w}_{m(j)}}g_{0,m(j)}(0)=\frac{w_j}{\bar{w}}.\]

In that case, $\mathbb{P}_0(\{j\}\mid D_{m(j)})={w_j}/{\bar{w}_{m(j)}}$ ; therefore, using (4.18),

(4.23) \begin{align} \mathbb{P}_n(\{j\}) = f_{n,j}(N_{n,m(j)}) & = \frac{w_j}{\bar{w}_{m(j)}}\frac{\bar{w}_{m(j)}+N_{n,m(j)}}{\bar{w}+n} \nonumber \\ & = \frac{\bar{w}\cdot\mathbb{P}_0(\{j\})+\mathbb{P}_0(\{j\}\mid D_{m(j)})\sum_{i=1}^n\boldsymbol{1}_{D_{m(j)}}(X_i)} {\bar{w}+n} \quad\mbox{a.s.} \end{align}

We can now conclude from (4.20) and (4.23) that, as measures on $\mathbb{X}$ ,

\begin{align*} \mathbb{P}(X_{n+1}\in \textrm{d}y\mid X_1,\ldots,X_n) & = \frac{\bar{w}\cdot\mathbb{P}_0(\textrm{d}y) + \sum_{i=1}^n\sum_{l=1}^m\mathbb{P}_0(\textrm{d}y\mid D_l)\cdot\boldsymbol{1}_{D_l}(X_i)}{\bar{w}+n} \\ & = \frac{\bar{w}\cdot\mathbb{P}_0(\textrm{d}y) + \sum_{i=1}^n\mathbb{P}_0(\textrm{d}y\mid\sigma(D_1,\ldots,D_m))(X_i)}{\bar{w}+n} \quad\mbox{a.s.}, \end{align*}

which implies that $(X_n)_{n\geq1}$ is an MVPS with parameters $(\bar{w},\mathbb{P}_0,\mathbb{P}_0(\cdot\mid\sigma(D_1,\ldots,D_m)))$ .

Proof of Proposition 3.2. For every $l=1,\ldots,m$ and $p_l\in\{({\bar{w}_l+n_l})/({\bar{w}+n})\colon n_l=0,\ldots,n;\,n=0,1,\ldots\bigr\}$ , let us define the quantity $g_l(p_l)\,:\!=\sum_{j\in D_l}f_j(p_l)$ . Then a straightforward extension of the arguments in [Reference Hill, Lane and Sudderth17] to m colors implies that $g_l(p_l)=p_l$ . The rest of the proof follows the same steps as in the proof of Proposition 3.1, starting from (4.18).