\neq \emptyset\}.$$ Let $\restfa_{N}(\detfa_{M})$ be the set of all feature allocations of $[N]$ whose restriction to $[M]$ is $\detfa_{M}$.

Let $\mbp$ denote a probability measure on some probability space supporting $(\randfa_{n})$. We say that the sequence of random feature allocations $(\randfa_{n})$ is *consistent in distribution* if for all $M$ and $N$ such that $M < N$, we have $$%\label{eq:consistency_in_dist}
    \mbp(\randfa_{M} = \detfa_{M}) = \sum_{\detfa_{N} \in \restfa_{N}(\detfa_{M})} \mbp(\randfa_{N} = \detfa_{N}).$$ We say that the sequence $(\randfa_{n})$ is *strongly consistent* if for all $M$ and $N$ such that $M < N$, we have $$\randfa_{N} \stackrel{a.s.}{\in} \restfa_{N}(\randfa_{M}).$$ Given any $(\randfa_{n})$ that is consistent in distribution, the Kolmogorov extension theorem implies that we can construct a sequence of random feature allocations that is strongly consistent and has the same finite dimensional distributions. So henceforth we simply use the term “consistency” to refer to strong consistency.

With this consistency condition, we can define a random feature allocation $\randfa_{\infty}$ of $\mathbb{N}$ as a consistent sequence of finite feature allocations. Thus $\randfa_{\infty}$ may be thought of as a random element of the space of such sequences: $\randfa_{\infty} = (\randfa_{n})_{n=1}^{\infty}$. We say that $\randfa_{N}$ is a restriction of $\randfa_{\infty}$ to $[N]$ when it is the $N$th element in 