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I have two claims that state some necessary conditions for exchangeability. I would like your help to understand whether they and relative proofs are correct. Consider three random variables $\epsilon_0,\epsilon_1, \epsilon_2$.


Claim 1: $\epsilon_1-\epsilon_0\sim \epsilon_2-\epsilon_0\sim \epsilon_1-\epsilon_2$ is necessary for having $\{\epsilon_1, \epsilon_2, \epsilon_0\}$ exchangeable.

Proof:

Assume that $\{\epsilon_1, \epsilon_2, \epsilon_0\}$ are exchangeable. Then, $(\epsilon_1,\epsilon_2,\epsilon_0)\sim (\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})$ for every permutation of labels $\pi$.

In particular, for any function $g:\mathbb{R}^3\rightarrow \mathbb{R}^k$, we have that $g(\epsilon_1, \epsilon_2, \epsilon_0)\sim g(\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})$.

Let $g:\mathbb{R}^3\rightarrow \mathbb{R}$ prescribed by $g(\epsilon_1, \epsilon_2, \epsilon_0)= \epsilon_1-\epsilon_0$.

Let $\pi$ be such that $\pi(1)=2, \pi(0)=0, \pi(2)=1$. Then, $g(\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})=\epsilon_2-\epsilon_0$.

It follows that $\epsilon_1-\epsilon_0\sim \epsilon_2-\epsilon_0$.

Let $\pi$ be such that $\pi(1)=1, \pi(0)=2, \pi(2)=0$. Then, $g(\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})=\epsilon_1-\epsilon_2$.

It follows that $\epsilon_1-\epsilon_0\sim \epsilon_1-\epsilon_2$.


Claim 2: Symmetry wrto zero of the distributions of $\epsilon_1-\epsilon_0$, $\epsilon_2-\epsilon_0$, $\epsilon_1-\epsilon_2$ is necessary for having $\{\epsilon_1, \epsilon_2, \epsilon_0\}$ exchangeable.

Proof:

Assume that $\{\epsilon_1, \epsilon_2, \epsilon_0\}$ are exchangeable. Then, $(\epsilon_1,\epsilon_2,\epsilon_0)\sim (\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})$ for every permutation of labels $\pi$.

In particular, for any function $g:\mathbb{R}^3\rightarrow \mathbb{R}^k$, we have that $g(\epsilon_1, \epsilon_2, \epsilon_0)\sim g(\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})$.

Let $g:\mathbb{R}^3\rightarrow \mathbb{R}$ prescribed by $g(\epsilon_1, \epsilon_2, \epsilon_0)= \epsilon_1-\epsilon_0$.

Let $\pi$ be such that $\pi(1)=0, \pi(0)=1, \pi(2)=2$. Then, $g(\epsilon_{\pi(1)},\epsilon_{\pi(2)},\epsilon_{\pi(0)})=\epsilon_0-\epsilon_1$.

It follows that $\epsilon_1-\epsilon_0\sim \epsilon_0-\epsilon_1$, i.e., the distribution of $\epsilon_1-\epsilon_0$ is symmetric wrto zero.

Similarly, $\epsilon_2-\epsilon_0\sim \epsilon_0-\epsilon_2$ and $\epsilon_1-\epsilon_2\sim \epsilon_2-\epsilon_1.$

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    $\begingroup$ Since the answer is "yes," this question really is not appropriate for this site. See our help center for more information about the kinds of questions we are looking for and can help with. $\endgroup$ – whuber Nov 2 '18 at 12:55

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