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Jan 24, 2017 at 20:24 vote accept HolyMonk
Jan 24, 2017 at 19:43 answer added Ori Gurel-Gurevich timeline score: 3
Jan 24, 2017 at 16:04 comment added HolyMonk For every fixed $m$
Jan 24, 2017 at 16:03 comment added Fedor Petrov For every $m$? Is not $m$ fixed?
Jan 24, 2017 at 15:06 comment added HolyMonk The idea is to work with an ergodic process and view this far enough in time that we can assume that convergence to the stationary distribution has already occurred.
Jan 24, 2017 at 15:05 comment added HolyMonk I don't want to have that $X_{n+m}$ and $X_n$ are independent, but I want to have that for every $m$, the map $n \mapsto \mbox{Corr}(X_{n+m}, X_n)$ is constant
Jan 24, 2017 at 15:03 history edited HolyMonk CC BY-SA 3.0
added 18 characters in body
Jan 24, 2017 at 14:41 comment added Fedor Petrov It still seems to be not enough. Assume That $m=3$, $X,T,Z$ are independent i.i.d and your sequence is $ZTT\,TZX\,ZXT\,TZX\, ZXT\,...$. Then $X_n$ and $X_{n+3}$ are always independent, but $Y_1$ and $Y_3$ may have different distribution.
Jan 24, 2017 at 14:33 comment added HolyMonk I added a condition on $(X_n)_n$, namely that the correlation between $X_n$ and $X_{n+m}$ doesn't depend on $n$.
Jan 24, 2017 at 14:32 history edited HolyMonk CC BY-SA 3.0
added an assumption on $(X_n)_n$
Jan 24, 2017 at 14:30 comment added Fedor Petrov why are $Y$'s identically distributed?
Jan 24, 2017 at 12:53 history asked HolyMonk CC BY-SA 3.0