## 3 The Case of Finite Exchangeable Sequences
### Exact Expression of a Finite Exchangeable Sequence by a Signed Mixture  
Another interesting result on the finite case is due to [PERSON] and [PERSON] [16]. They proved that any finite exchangeable sequence, taking values from an arbitrary measurable space, can always be expressed _exactly_ as an i.i.d. mix. This would not hold in the original setting. However, the twist that [PERSON] and [PERSON] have introduced is that the mixing measure is a so-called _signed measure_. The latter means that it may also take negative values. In the notation recall that \(\mathcal{P}(S)\) denotes the set of all probability distributions on \(S\).  
**Theorem 8**.: _Let \(X_{1},\ldots,X_{n}\) be a sequence of exchangeable random variables, taking values from an arbitrary measurable space \(S\). Then there exists a signed measure \(\
u\) on \(\mathcal{P}(S)\), such that for any measurable \(A_{1},\ldots,A_{n}\subseteq S\) the following holds:_  
\[\Pr(X_{1}\in A_{1},\ldots,X_{n}\in A_{n})=\int_{\mathcal{P}(S)}\pi(A_{1}) \cdot\ldots\cdot\pi(A_{n})\,dv(\pi) \tag{3}\]  
_where \(\pi\) runs over \(\mathcal{P}(S)\), integrated according to the signed measure \(\
u\)._  
Here the mixing measure \(\
u\) does not have to be unique, in contrast to the traditional versions of the theorem.  
A harder question, however, is this: comparing with the traditional versions, the right-hand side of (3) means that \(\pi\) is drawn according to a signed measure from \(\mathcal{P}(S)\). What does this mean from the probability interpretation point of view?  
Formally, the integral on the right-hand side of (3) is just a mixture (linear combination, with weights summing to 1) of the values \(\pi(A_{1})\cdot\ldots\cdot\pi(A_{n})\), where \(\pi\) runs over \(\mathcal{P}(S)\). The only deviation from the classical case is that some \(\pi\in\mathcal{P}(S)\) can be weighted with negative weights. Thus, _formally_, everything is in order, we simply deal with a mixture of _probability distributions_, allowing negative weights, but insisting that at the end, a non-negative function must result. However, if we want to interpret it as a mixture of _random sequences_, rather than just probability distributions, then the signed measure amounts to a selection via a probability distribution incorporating _negative probabilities_.  
What does it mean? How can we pick a value of a random variable with negative probability? To answer this meaningfully is not easy. There are some attempts in the literature to interpret negative probabilities, for a short introduction see [PERSON] [17]. Nevertheless, it appears that negative probabilities are neither widely accepted in probability theory, nor usually adopted in applications, apart from isolated attempts. Therefore, we rather stay with the formal interpretation: \"drawing\" \(\pi\in\mathcal{P}(S)\) according to a signed measure for the integral just means taking a mixture (linear combination) of probability distributions with weights summing to 1, also allowing negative weights, while insisting that the result is still a non-negative probability distribution. This makes Theorem 8 formally correct, avoiding troubles with interpretation. Nevertheless, the interpretation still remains a challenging philosophical problem, given that Theorem 8 has been the only version to date that provides an exact expression of the _distribution_ any finite exchangeable sequence as a mix of i.i.d distributions, but it does not correspond to the mixture of random _sequences_ in the usual (convex) sense.