I have a question regarding the exchangeable random variable

consider ($x_{m}$) be a (infinite) sequence of random variable, if ($x_{m}$) is stationary, does it implies that ($x_{m}$) is exchangeable?

There is no reference for this result, so I am wondering it should have a counterexample, but I can not make up one

any help will be extremely appreciated!!


Of course not. Assume that every $x_m$ is uniform on $\{1,2,3\}$ and that $x_{m+1}=x_m+1\pmod{3}$ for every $m$, then the distributions of $(x_m,x_{m+1})$ and $(x_{m+1},x_m)$ are mutually singular hence $(x_m)_m$ is stationary and not exchangeable.

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  • $\begingroup$ Thank you for your reply, but consider the following, since $x_{m}$ is uniform on {1,2,3}, if $x_{1}$ = 2, then $x_{2}$ = 0, which is not in {1,2,3}, which contradict your construction that every $x_{m}$ is uniform on {1,2,3}.. $\endgroup$ – MLEGMM Apr 13 '13 at 15:44
  • $\begingroup$ Modulo 3. If you prefer, replace $\{1,2,3\}$ by $\{0,1,2\}$. Seriously, what are you playing at? $\endgroup$ – Did Apr 13 '13 at 16:00
  • $\begingroup$ Thank you again! My previous question pops out because I am trying my best to understand why the distribution of ($x_{m}$,$x_{m+1}$) and ($x_{m+1}$,$x_{m}$) are mutually singular, since for me, it is not trivial. So could you please explain a bit further about that? $\endgroup$ – MLEGMM Apr 13 '13 at 16:28
  • $\begingroup$ The distributions of $(x_m,x_{m+1})$ and $(x_{m+1},x_m)$ are uniform on two subsets $I$ and $J$ of $\{1,2,3\}\times\{1,2,3\}$. Can you identify $I$ and $J$ and check that $I\cap J$ is empty? $\endgroup$ – Did Apr 13 '13 at 16:35
  • $\begingroup$ I = {{1,2}, {2,0},{3,1}}, J = {{2,1}, {0,2},{1,3}}, so I∩J is empty..I see! $\endgroup$ – MLEGMM Apr 13 '13 at 16:42

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