What makes random variables exchangeable and what is implied by exchangeability?

Solution 1:

Exchangeability implies being identically distributed, but not the reverse. An exchangeable sequence has extra symmetry.

For instance, below I've written the joint mass function of $(X,Y)$, where $X$ and $Y$ take values in $\{0,1,2\}$. Checking the marginals shows that they are identically distributed, but $$\mathbb{P}(X=0,Y=2)=1/7\neq \mathbb{P}(Y=0,X=2)=0/7.$$ That is, the random variables $X$ and $Y$ have the same distribution, but the random vectors $(X,Y)$ and $(Y,X)$ don't.

$$\begin{array}{c|ccc} x\backslash y&0&1&2\cr \hline 0&0/7&1/7&1/7\\ 1&2/7&0/7&1/7\\ 2&0/7&2/7&0/7\\ \end{array} $$

See here and here for further applications of exchangeability.

Solution 2:

A simple and rather general way to build an exchangeable random vector $X=(X_k)_{1\leqslant k\leqslant n}$ is to start from an i.i.d. sequence $(Y_k)_{1\leqslant k\leqslant n}$ and to define, for every $k$, $X_k=\Phi(U,Y_k)$, where $U$ is a random variable independent on $(Y_k)_{1\leqslant k\leqslant n}$ and $\Phi$ a measurable function.

And, in some (loose) sense, this is the only way... as shown by Diaconis and Freedman for finite sequences, using mixtures of urn sequences rather than mixtures of i.i.d. sequences, and earlier, by de Finetti for infinite sequences.