## 3 The Case of Finite Exchangeable Sequences
### Exact Finite Representation as a Mixture of Urn Sequences  
Another interesting result about finite exchangeable sequences is that they can be expressed as a mixture (in the usual convex sense) of so-called _urn sequences_, explained below. It seems, this provides the most direct analogy of [PERSON]'s Theorem for the finite case, yet this result did not receive the attention it deserves, as pointed out by [PERSON], [PERSON], and [PERSON] [18]. The idea goes back to [PERSON] [19]. Later it was used by several authors at various levels of generality as a proof technique, rather than a target result in itself, see, e.g., [PERSON] and [PERSON] [16], so it did not become a \"named\" theorem. Finally, the most general version, which applies to arbitrary random variables, appears in the book of [PERSON] (see [12], Proposition 1.8). _Urn sequences_ constitute a simple model of generating random sequences. As the most basic version, imagine an urn in which we place \(N\) balls, and each ball has a certain color. We randomly draw the balls from the urn one by one and observe the obtained random sequence of colors. We can distinguish two basic variants of the process: after drawing a ball, it is put back in the urn (urn process with replacement), or it is not put back (urn process without replacement). Consider the following simple example. Let us put \(N\) balls in the urn, \(K\) black, and \(N-K\) white balls. If we randomly draw them with replacement, then an i.i.d. sequence is obtained, in which each entry is black with probability \(K/N\), and white with probability \((N-K)/N\). The length of the sequence can be arbitrary (even infinite), as the drawing can continue indefinitely. On the other hand, if we do this experiment _without replacement_, then the maximum length of the obtained sequence is \(N\), since after that we run out of balls. The probability that among the first \(n\leq N\) draws (without replacement) there are precisely \(X\) black balls follows the _hypergeometric distribution_ (see, e.g., Rice [20]) given by  
\[\Pr(X=k)=\frac{\binom{K}{k}\binom{N-K}{n-k}}{\binom{N}{n}}. \tag{4}\]  
For our purposes the important variant is the case _without replacement_, and with \(n=N\), that is, all the balls are drawn out of the urn. Then the obtained sequence has length \(N=n\). Note that it cannot be i.i.d., as it contains precisely \(K\) black and \(N-K\) white balls. However, otherwise, it is completely random, so the distribution of \(X\) is the same as it were in an i.i.d. sequence, conditioned on including precisely \(K\) black balls. The number of colors can be more than two, even infinite. The obtained random sequence is still similar to an i.i.d. one, with the difference that each color occurs in it a fixed number of times. We can then formulate the general definition of the urn sequences of interest to us. For a short description, let us first introduce some notations. The set \(\{1,\ldots,n\}\) is abbreviated by \([n]\), and the family of all permutations of \([n]\) is denoted by \(\Sigma_{n}\). If a permutation \(\sigma\) is applied to a sequence \(X=(X_{1},\ldots,X_{n})\), then the resulting sequence is denoted by \(\sigma(X)\), which is an abbreviation of \((X_{\sigma(1)},\ldots,X_{\sigma(n)})\). We also use the following naming convention:  
**Convention 1** (Uniform random permutation). : _Let \(\sigma\in\Sigma_{n}\) be a permutation. We say that \(\sigma\) is a uniform random permutation, if it is chosen from the uniform distribution over \(\Sigma_{n}\)._  
Now the urn sequences of interest to us are defined as follows:  
**Definition 2** (Urn sequence). : _Let \(x=(x_{1},\ldots,x_{n})\) be a deterministic sequence, each \(x_{i}\) taking values from a set \(S\), and let \(\sigma\in\Sigma_{n}\) be a uniform random permutation. Then \(X=\sigma(x)\) is called an urn sequence._  
Here each \(x_{i}\) represents the color of a ball, allowing repeated occurrences. The meaning of \(\sigma(x)\) is simply that we list the balls in random order.