Exchangeability and independence of random variables

Question

I have a question on the relation between exchangeability and independence between random variables. Consider the random vectors $$u_1:= \begin{pmatrix} \epsilon_{1}\\ \epsilon_2\\ \epsilon_3 \end{pmatrix} $$ and $$u_2:= \begin{pmatrix} \epsilon_{4}\\ \epsilon_5\\ \epsilon_6 \end{pmatrix} $$ All random variables are defined on the same probability space and have the same support. Under which conditions (other than i.i.d.) we have

(1) $\epsilon_1,\epsilon_2,\epsilon_3,\epsilon_4,\epsilon_5,\epsilon_6$ exchangeable

and

(2) $u_1$ independent of $u_2$

?

I am confused on the following: of course, $\epsilon_1,\epsilon_2,\epsilon_3,\epsilon_4,\epsilon_5,\epsilon_6$ i.i.d. is sufficient to have (1) and (2). However, could there be other sufficient conditions? For example?

Answer 1

It seems to me that iid is the only way (1) and (2) can hold.

First, the vector comprising any three of the $\epsilon_k$s is independent of the triple comprising the other three. Thus, for example, $(\epsilon_1,\epsilon_3,\epsilon_4)$ is independent of $(\epsilon_2,\epsilon_5,\epsilon_6)$. Marginalizing, $\epsilon_1$ is independent of $\epsilon_2$; and then by (1), any two of the $\epsilon_k$s are independent. Next, $(\epsilon_1,\epsilon_2,\epsilon_4)$ is independent of $(\epsilon_3,\epsilon_5,\epsilon_6)$, and marginalizing again, $(\epsilon_1,\epsilon_2)$ is independent of $\epsilon_3$. Since we already know that $\epsilon_1$ is independnet of $\epsilon_2$, it follows that $\epsilon_1,\epsilon_2,\epsilon_3$ are mutually independent. By (1), the same follows for any three of the $\epsilon_k$s. It is now a short step to conclude that the six $\epsilon_k$s are mutually independent, and of course they all have the same distribution by exchangeability.