Expectation of random variables coincides

Question

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.

Answer 1

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.