5.2 Exchangeability
Exchangeability is essentially an expression of symmetry in the probabilistic behaviour of events. (Barnett 1999, 90)
Definition 5.1 Exchangeability by (Finetti 1930, 107). The operations \(R\) and \(S\) are interchangeable: \(RS=SR\), and therefore \(R^r S^s = S^s R^r = S^{s_1} R^{r_1} S^{s_2} R^{r_2} \ldots \;\;\) (\(s_1+s_2+\ldots = s\), \(r_1+r_2+\ldots = r\)).
Definition 5.2 Exchangeability by (Finetti 1974, 211). Given \(n\) events, the probabilities (…) that \(h\) of them will occur (\(h = 0,1,\ldots,n\)) are now arbitrary. (…) However, the combinations of \(h\) 1s and \(n-h\) 0s all have the same probability.
Example 5.2 (Cordani and Wechsler 2006) Consider a random sample without replacement of marbles from an urn. The composition is known, with 10 red (V) and 5 white (B) balls. The balls are selected one by one and the event \(V_i\) is defined to occur when the \(i\)-th ball sampled is red. Similarly, \(B\) can be used for the event related to the white ball.
The probabilities can be calculated from the tree diagram above.
\[\begin{align*} Pr(V_1) &= \frac{10}{15} = \frac{2}{3} \\ Pr(V_2) &= \frac{10}{15}\cdot\frac{9}{14} + \frac{5}{15}\cdot\frac{10}{14} = \frac{2}{3} \\ Pr(V_3) &= \frac{10}{15}\cdot\frac{9}{14}\cdot\frac{8}{13} + \frac{10}{15}\cdot\frac{5}{14}\cdot\frac{9}{13} + \frac{5}{15}\cdot\frac{10}{14}\cdot\frac{9}{13} + \frac{5}{15}\cdot\frac{4}{14}\cdot\frac{10}{13} = \frac{2}{3} \\ Pr(V_1 V_2) &= \frac{10}{15}\cdot\frac{9}{14} = \frac{3}{7} \\ Pr(V_2 V_3) &= \frac{10}{15}\cdot\frac{9}{14}\cdot\frac{8}{13} + \frac{5}{15}\cdot\frac{10}{14}\cdot\frac{9}{13} = \frac{3}{7} \\ Pr(V_1 V_3) &= \frac{10}{15}\cdot\frac{9}{14}\cdot\frac{8}{13} + \frac{10}{15}\cdot\frac{5}{14}\cdot\frac{9}{13} = \frac{3}{7} \\ Pr(V_2|V_1) &= \frac{9}{14} \end{align*}\]
As \(Pr(V_2|V_1) = \frac{9}{14} \ne \frac{2}{3} = Pr(V_2)\), \(V_1\) and \(V_2\) are not independent, i.e., knowledge about the occurrence of \(V_1\) changes the probability of \(V_2\).
Note that the probabilities \(P(V_i)\) are equal to 10/15, for \(i \in \{1,2,3\}\). Furthermore, the intersection probabilities are such that \(Pr(V_i V_j) = 3/7\) for \(i,j \in \{1,2,3\}\), \(i \ne j\). This happens because the aforementioned events are exchangeable. This is a well-known example of dependency, with exchangeability: the selections are indistinguishable, but still dependent.
Example 5.3 Again with the data from Example 5.2, withdrawals can be considered with replacement. Thus,
\[\begin{align*} Pr(V_1) &= Pr(V_2) = Pr(V_3) = \frac{10}{15} = \frac{2}{3} \\ Pr(V_1 V_2) &= Pr(V_2 V_3) = Pr(V_1 V_3) = \frac{4}{9} \\ Pr(V_2|V_1) &= \frac{10}{15} = \frac{2}{3} \end{align*}\]
In this example, we obtain the same probabilities \(Pr(V_1)\), \(Pr(V_2)\) and \(Pr(V_3)\) as in Example 5.2. The conditional probabilities, however, are now equal to the unconditional probabilities, for example, \(Pr(V_2|V_1) = P(V_2)\). This is therefore a situation in which events are not only exchangeable, but also independent.
It is worth mentioning that in both examples above the urn has a known composition. If the composition is unknown, the events are only exchangeable but not independent, even when draws are made with replacement.