# Independence of random variables

Let $\{X_n\}$ be a sequence of independent random variables on some probability space.

Then, by definition(according to the book that I am reading), I know that $\{\sigma(X_1),\sigma(X_2),\dots, \}$ is independent.

Then, I am wondering whether $\sigma(X_{n+1})$ is independent of $\sigma(\sigma(X_1),\sigma(X_2),\dots,\sigma(X_n))$.

Or in general, is $\sigma(\sigma(X_{n_1}),\sigma(X_{n_2}),\dots,\sigma(X_{n_k}))$ is independent of $\sigma(\sigma(X_{j_1}),\sigma(X_{j_2}),\dots,\sigma(X_{j_l}))$, where $\{n_1,n_2,\dots,n_k\} \cap \{j_1,j_2,\dots,j_l\} = \emptyset$?

• The answer is yes. Btw, it is better to write $\sigma(X_{n_1},\dots,X_{n_k})$ in stead of $\sigma(\sigma(X_{n_1}),\dots,\sigma(X_{n_k}))$; i.e. the smallest $\sigma$-algebra that makes, $X_{n_1},\dots,X_{n_k}$ measurable. – drhab Aug 19 '15 at 13:51
• @drhab Isn't $\sigma(X_{n_1},X_{n_2},\dots,X_{n_k})$ the same as $\sigma(\sigma(X_{n_1}),\dots,\sigma(X_{n_k}))$? Can you also give me an idea as to how I can prove it? – mononono Aug 19 '15 at 13:55
• The first notation is common (the second is not) and shorter. Most probably you mean the same. If the second is meant to be a $\sigma$-algebra generated by $\sigma$-algebras then there is no essential difference between both. – drhab Aug 19 '15 at 13:58

If $\{X_n:n\in\mathbb N\}$ is a sequence of independent random variables, then what we really mean is that the $\sigma$-algebras generated by those random variables, i.e. $\sigma(X_n) = \{X_n^{-1}(B) : B\in\mathcal B\}$ are independent. That is, for any finite subset $n_1, \ldots, n_k\subset\mathbb N$, $A_{n_i}\in\sigma(X_{n_i})$, $$\mathbb P\left(\bigcap_{i=1}^k A_{n_i}\right)=\prod_{i=1}^k\mathbb P(A_{n_i}),$$ and your question is just a special case of the above As for notation, $\sigma(\sigma(X_1), \ldots, \sigma(X_n))$ really denotes $$\sigma\left(\bigcup_{i=1}^n \sigma(X_i)\right).$$ But we write $\sigma(X_1, \ldots, X_n)$ for brevity - just as we say e.g. "$X$ and $Y$ are independent" instead of "$\sigma(X)$ and $\sigma(Y)$ are independent."
• So, to prove that $\sigma(X_{n+1})$ is independent of $\sigma(X_1,X_2,\dots,X_n)$, I need to show that for any $A \in \sigma(X_{n+1})$ and $B \in \sigma\left(\bigcup_{i=1}^n \sigma(X_i)\right)$, $A$ and $B$ are independent. But how do you prove it though? Your answer just explains the notation – mononono Aug 19 '15 at 15:24
• Take $n_1, \ldots, n_k$ as $1,2,\ldots,n+1$. – Math1000 Aug 19 '15 at 18:00
• Taking $n_1,\dots,n_k$, as $1,2,\dots,n+1$ does not prove anything, though. – mononono Aug 19 '15 at 19:33