Consider a finite sequence of random variables $X_1,...,X_n$
(1) SUFF COND: Suppose $X_1,...,X_n$ are exchangeable, meaning that the joint probability distribution of $X_1,...,X_n$ is equivalent to the joint probability distribution of $X_{\varphi(1)},...,X_{\varphi(n)}$ for any finite permutation $\varphi$ over $1,...,n$. Does this imply exchangeability of the elements of every finite subsequence?
(2) NEC COND: Does exchangeability of the elements of every finite subsequence implies exchangeability of the elements of the whole sequence?
I know that for infinite sequences the answer to both questions is yes. I don't know how to solve the finite case
The answer is yes. As a concrete example of (1), suppose $X_1, X_2, X_3$ are exchangeable. We show $X_1, X_3$ are exchangeable by filling in the omitted slots with the 'everything' event: $$\begin{align} P(X_1\in A, X_3\in B)&=P(X_1\in A, X_2\in{\mathbb R}, X_3\in B)\\ &\stackrel{(*)}=P(X_3\in A, X_2\in {\mathbb R}, X_1\in B)\\ &=P(X_3\in A, X_1\in B) \end{align}$$ In step (*) we use exchangeability of the entire sequence.
As for (2), the whole sequence is a subsequence of the whole sequence, so exchangeability follows trivially. (If we only assume every strict subsequence is exchangeable, then the whole sequence need not be exchangeable: $n=2$ provides an easy counterexample.)