$\sigma$-algebra generated by several random variables

Question

I find it difficult to work with $\sigma$-algebra generated by several variables... Here's my question :

Let $\lbrace X_{j}, j\in \mathbb{Z} \rbrace$ be a sequence of random variables, such that for all finite set $J \subset \mathbb{Z}$, and for all $m \in \mathbb{N}$, the random vectors $(X_{j}, j\in J)$ and $(X_{j+m}, j\in J)\, $ follow the same distribution.

For all $j \in \mathbb{Z}$, let $\, \mathcal{M}^{j}_{-\infty} \,$ be the $\sigma$-algebra generated by $\lbrace X_{i}, i \leq j\rbrace \,$, and $\mathcal{M}^{+\infty}_{j} \,$ the $\sigma$-algebra generated by $\lbrace X_{i}, i \geq j\rbrace \,$.

We define $\alpha_{n} = \alpha(\mathcal{M}^{0}_{-\infty}, \mathcal{M}^{+\infty}_{n})$, where for all sub-tribes $\mathcal{A}, \mathcal{B}$, $$\alpha(\mathcal{A}, \mathcal{B}) = \sup_{A \in \mathcal{A}, B \in \mathcal{B}} |P(A \cap B) - P(A)P(B) |$$

Now I want to show that for all $j \in \mathbb{Z}$, $$\alpha(\mathcal{M}^{0}_{-\infty}, \mathcal{M}^{+\infty}_{n}) = \alpha(\mathcal{M}^{j}_{-\infty}, \mathcal{M}^{+\infty}_{j+n})$$

I understand the idea : it follows from the fact that $(X_{i}, 1 \leq i \leq j)$ and $(X_{i}, n-1\leq i \leq j+n)$ follow the same distribution. But I don't know how to formalize it...

Edit : I don't know if it helps, but I forgot an assumption : $X_{0}$ has a zero mean.

Answer 1

For each positive integer $N$, the equality $$\alpha\left(\mathcal{M}^{0}_{-N}, \mathcal{M}^{N}_{n}\right) = \alpha\left(\mathcal{M}^{j}_{-N+j}, \mathcal{M}^{N+j}_{n+j}\right)$$ takes place by the assumption of strict stationarity.

To conclude, we have to show that if $\left(\mathcal A_N\right)_{N\geqslant 1}$ and $\left(\mathcal B_N\right)_{N\geqslant 1}$ are non-decreasing sequences of $\sigma$-algebra, then $\alpha\left(\mathcal A,\mathcal B\right)=\lim_{N\to\infty}\alpha\left(\mathcal A_N,\mathcal B_N\right)$, where $\mathcal A$ (respectively $\mathcal B$) is the $\sigma$-algebra generated by the $A_N$ (resp. $\mathcal B_N$), $N\geqslant 1$. To see this, notice that $\mathcal A$ and $\mathcal B$ are generated by the algebras $\bigcup_{N\geqslant 1}\mathcal A_N$ and $\bigcup_{N\geqslant 1}\mathcal B_N$ respectively. Then use an approximation argument.