If $\{X_n\}, n \in \mathbb{N}$ are an independent sequence of random variables, why is it that $\sigma(X_1, …, X_n)$ and $\sigma(X_{n+1},X_{n+2}, …)$ are independent? I'm trying to prove the Kolmogorov Zero One Law and this is the only part of the proof I don't understand. Obviously each of these sigma algebras is by definition a sigma algebra generated by the union of independent sigma algebras.
Thus, $\cup_{i=1}^n \sigma(X_i)$ and $\cup_{i=n+1}^{\infty} \sigma(X_i)$ are independent sets of events. But they are not necessarily $\pi$ systems by this post: Union of $\sigma$-algebras is a $\pi$-system…? (since I cannot find a way to show they are an increasing union of sigma algebras – I don't think they are). Therefore I can't necessarily say anything about $\sigma(\cup_{i=1}^n \sigma(X_i)) = \sigma(X_1, …, X_n)$ and $\sigma(X_{n+1},X_{n+2}, …) = \sigma(\cup_{i=n+1}^{\infty} \sigma(X_i))$
$\textbf{EDIT:}$
Would the following be a sufficient argument?
Let $S_1 = \{A : A = \cap_{j=1}^k\{X_{i_j} \in B_j\}, i_j \in \{1,…,n \} \}$ and $S_2 = \{A : A = \cap_{j=1}^k\{X_{i_j} \in B_j\}, i_j \in \{n+1, n+2,…\} \}$
Then $S_1$ and $S_2$ are independent $\pi$ systems whose sigma algebras are thus independent (this is a separate result that is easily provable) and contain $\sigma(\cup_{i=1}^n \sigma(X_i))$ and $\sigma(\cup_{i=n+1}^{\infty} \sigma(X_i))$ respectively. Hence the sub sigma algebras are trivially independent as they are subsets of independent collections.
Best Answer
Well, you "just" find another $\pi$-system. Note that $\sigma(X_1,...,X_n)$ is generated by sets of the form $(X_{j_1}\in B_1,...x_{j_k}\in B_k)$ for $(B_m)_{1\leq m\leq k}$ Borel sets and $1\leq j_1<...<j_k\leq n$, and the system of such sets is a $\pi$ system. Similarly, $\sigma(X_j)_{j>n}$ is generated by sets of the form $(X_{j_1}\in B_1,..., X_{j_k}\in B_k)$ again for a finite sequence of $(B_m)_{1\leq m\leq k}$ of Borel sets and $n<j_1<...<j_k$. Again this is a $\pi$ system.
Hence, fix Borel sets $(B_m)_{1\leq m\leq k}$ and $n<j_1<...j_k$ and let $(C_r)_{1\leq r\leq K}$ be any other finite sequence of Borel sets and $1\leq l_1<...<l_K\leq n$ be a finite increasing set of indices. Then, we see, by the independence assumption, that
$$ \mathbb{P}((\cap_{m=1}^k (X_{j_m}\in B_m))\cap (\cap_{r=1}^K X_{l_r}\in C_r))=\mathbb{P}(\cap_{m=1}^k (X_{j_m}\in B_m))\mathbb{P}(\cap_{r=1} (X_{l_r}\in C_r)), $$ implying by Dynkin's lemma that the two measures on $\sigma((X_j)_{1\leq j\leq n})$ given by $$ \mathbb{P}_1(A)=\mathbb{P}((\cap_{m=1}^k (X_{j_m}\in B_m))\cap A) $$ and $$ \mathbb{P}_2(A)=\mathbb{P}(\cap_{m=1}^k (X_{j_m}\in B_m))\mathbb{P}(A) $$ are equal.
Since our choice of Borel sets $B_m$ and indices $j_m$ was arbitrary, this now yields, by another application of Dynkin's Lemma, that for any $A\in \sigma(X_j)_{1\leq j\leq n},$ the two measures on $\sigma(X_j)_{j>n}$ given by $$ \mathbb{P}_3(D)=\mathbb{P}(D\cap A) $$ and $$ \mathbb{P}_4(D)=\mathbb{P}(D)\mathbb{P}(A) $$ are equal.
However, since $A$ was arbitrary, this implies that the two $\sigma$-algebras are independent.