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The definition of exchangeable pairs is as follows:

If $(W,W')$ is an exchangeable pair, then $(W,W') = (W',W)$ in distribution.

Can we deduce from this definition that $W$ and $W'$ have the same marginal distributions?

My solution:

Let $F_{W,W'}(w,w')$ be the distribution function of $(W,W')$.

Then the marginal distribution function of $W$ is given by $F_W(w) =\int F_{W,W'}(w,w')\,dw' = \int F_{W',W}(w',w)\,dw = F_{W'}(w')$.

I'm not sure whether the second last equation is right or not. It does not seem to be rigorous.

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    $\begingroup$ What happened when you tried to find the marginal distributions from the exchangeable joint distribution? $\endgroup$
    – Henry
    Commented Nov 2, 2021 at 12:14

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We have $(W,W') = (W',W)$ in distribution. Consider $A = \{(x,y) \in \mathbf{R}^2 | x \in B\}$ where $B$ is an arbitrary Borel set. Hence $P(W \in B) = P( (W,W') \in A) = P( (W', W) \in A) = P(W' \in B)$. Hence $W = W'$ in distribution.

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  • $\begingroup$ @ Botnakov N. So can we treat an exchangeable pair as two copies of a r.v., but they may be correlated? $\endgroup$
    – XXX
    Commented Nov 3, 2021 at 10:04
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    $\begingroup$ @GGG, yes, if by "copies" we mean "copies in the sense of coincidence of distributions". $\endgroup$ Commented Nov 3, 2021 at 11:14

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