I am confused on the meaning of exchangeable random variables. The question is: consider the random variables $X_1,X_2,X_3$ defined one the same probability space $(\omega, \mathcal{F}, P)$; is sharing the same support a necessary condition for exchangeability?
I think the answer is yes, but I would like to receive a more formal confirmation.
In fact, let the support of $X_1$ be $\{1,4,7\}$, the support of $X_2$ be $\{2,5,8\}$ and the support of $X_3$ be $\{3,6,9\}$ and assume that $X_1,X_2,X_3$ are exchangeable. Let $\varphi$ be a permutation over $1,2,3$ such that $\varphi(1)=2$, $\varphi(2)=3$ and $\varphi(3)=1$. Then, one implication of exchangeability is that $$P(X_1=4, X_2=5, X_3=6)=P(X_{\varphi(1)}=4, X_{\varphi(2)}=5, X_{\varphi(3)}=6)= P(X_2=4, X_3=5, X_1=6)$$ but $P(X_2=4, X_3=5, X_1=6)=0$ and $P(X_1=4, X_2=5, X_3=6)>0$.
If I am not mistaken, exchangeability means (for two variables) that the law of $(X_1,X_2)$ is the same as the law of $(X_2,X_1)$.
Thus, the space of possible values for the couple $(X_1,X_2)$ must be the same as the space of possible values for the couple $(X_2,X_1)$.
In a nutshell, exchangeability is stronger than sharing the same support which is a necessary but not sufficient condition.