Let $Y\_1:=(X\_i)\_{i \in \mathbb Z}$ be a family of random variables that are identically distributed but not necessarily independent.


We can then also define the shifted sequence $Y\_2:=(X\_{i+1})\_{i \in \mathbb Z}.$


If the $X\_i$ were also independent then $f(Y\_1)$ would have the same law as $f(Y\_2).$


In particular, we can conclude from this that also $$\mathbb E(f(Y\_1))=\mathbb E(f(Y\_2)).$$


I am wondering whether we can still conclude 
$$\mathbb E(f(Y\_1))=\mathbb E(f(Y\_2))$$


if we assume the $X\_i$ only to be identically distributed but not necessarily independent.



Expectation of random variables coincides
Let $U$ and $V$ be two i.i.d. random variables having finite expectation. Let $X\_{3k}=X\_{3k+1}:=U$, $X\_{3k+2}:=V$ and $f\left(\left(x\_i\right)\_{i\in\mathbb Z}\right)=x\_0x\_1$. Then $\mathbb E\left[f\left(Y\_1\right)\right]=\mathbb E\left[U^2\right]$ and $\mathbb E\left[f\left(Y\_2\right)\right]=\mathbb E\left[UV\right]=\left(\mathbb E\left[U\right]\right)^2$ hence taking $U$ non-degenerated gives a counter-example.


