Question:
Let us consider the random walk $S_n=X_1+...+X_n$ for $n\ge1$ from I.I.D random variables $X_n$ with $E[\mid X_1\mid] < \infty$. Since $\sigma(X_{n+1},X_{n+2},...)$ and $\sigma(X_1,S_n)$ are independent, $E[X_1|G_n]=E[X_1|S_n]$ where $G_n$ is defined as $\sigma(S_n,X_{n+1},X_{n+2},...)=\sigma(S_n,S_{n+1},S_{n+2},...)$ ...
Confusion:
Could anyone please explain why $\sigma(S_n,X_{n+1},X_{n+2},...)=\sigma(S_n,S_{n+1},S_{n+2},...)$? I've tried to apply this definition of a sigma field $\sigma(X_i, i \in I)$= $\sigma(\cup_{i \in I} \sigma(X_i))$; However, I'm unable to make the connection. My intuition is that since both sigma fields have $S_n$, the only new information being added is $X_{n+1},X_{n+2},...$ which is captured on both sides. But I'm not sure if this reasoning is correct, and how to make this proof more rigorous.
Thanks in advance!
Let $\mathcal A:=\sigma(S_n,X_{n+1},X_{n+2},\dots)$ and $\mathcal A':=\sigma(S_n,S_{n+1},S_{n+2},\dots)$.
Then $S_n$ is measurable wrt $\mathcal A'$ and for $k=1,2,\dots$ also $X_{n+k}=S_{n+k}-S_{n+k-1}$ is measurable wrt $\mathcal A'$.
This justifies the conclusion that $\mathcal A\subseteq\mathcal A'$.
Conversely $S_n$ is measurable wrt $\mathcal A'$ and for $k=1,2,\dots$ also $S_{n+k}=S_n+\sum_{i=1}^{k}X_{n+i}$ is measurable wrt $\mathcal A$.
This justifies the conclusion that $\mathcal A'\subseteq\mathcal A$.
In short: every (generating) random variable used by the definition of $\mathcal A$ can be written by means of random variables that are measurable wrt $\mathcal A'$ (hence is measurable wrt $\mathcal A'$) and vice versa. This is enough.