# Independence of infinite dimensional random vectors

Hi everyone I have the following problem. I'd appreciate any help.

Let $\{Y_n:n\in \mathbb{N}\}$ be a sequence of real valued random variables on a probabilility space $(\Omega,\mathscr{F}, P)$. Let $n\ge 1$ a fixed integer. Define $Z_n=(Y_1,\ldots,Y_n)$ and $W_n=(Y_{n+1},\ldots)$.

i) Show that $W_n$ and $Z_n$ are independent.

ii) Assume furthermore that the random variables $Y_n$ are identically distributed, that is the distribution function satisfies $F_{Y_n}=F_{Y_m}$ for every $0\le n<m<\infty$. Then prove that for every $n\ge 1$, $Z_n$ and $W_n$ have the same distribution.

For i) what I proved using Dynkin classes is that $\mathcal{T}_n =\sigma \left( \bigcup_{k=n+1}^\infty \sigma(Y_k) \right)$ and $\mathcal{B}=\sigma\left(\bigcup_{k=1}^n \sigma(Y_k) \right)$ are independent under $P$, where $\sigma(Y_k)$ is the minimum sigma algebra which makes $Y_k$ measurable. But from here I'm not sure how to conclude.

For $ii)$ I don't understand what is the distribution of an infinite dimensional random vector.

I have no experience working with infinite dimensional random vectors. I'd appreciate any help please. (If someone considers it necessary I can add the proof of the independence using Dynkin classes, but this is standard.)

For i), what you need to conclude is that $\mathcal{T}_n$ is actually $\sigma(W_n)$.
Formally, the definition of a infinite dimensional random vector $(Y_n)_{n\in\mathbf N}$ in $\mathbf R^{\mathbf N}$ is given by $\mathbb P\{(Y_n)_{n\in\mathbf N}\in B\}$, where $B$ belongs to the Borel $\sigma$-algebra on $\mathbf R^{\mathbf N}$. Actually, to show that the distributions of two such infinite dimensional vectors are equal, you only have to check that they have the same finite dimensional distributions. In this context, this means that for each integer $d$ and each $B$ in the Borel $\sigma$-algebra of $\mathbf R^d$, we have $$\mathbb P\left\{\left(Y_1,\dots,Y_d\right)\in B\right\}=\mathbb P\left\{\left(Y_n,\dots,Y_{n+d-1} \right)\in B\right\}.$$
• Hi, thank you for take you time to answer it. The last point is because the sigma algebra in $\mathbb{R}^{\mathbb{N}}$ is generated by the Borel cylinders, and for that reason it is sufficient to check the equality for generators of the sigma algebra? – Jose Antonio Oct 26 '15 at 14:34