# Expectation of random variables coincides

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.

## 1 Answer

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.