Q:

Let $S_n = X_1 +\cdots +X_n$. Is $\sigma\left(X_j, 1 \leq j \leq n \right) = \sigma\left(S_j, 1 \leq j \leq n \right)$? 

Let $X_1, \cdots, X_n$ be $\mathcal{L}_1$ random variables on a probability space $\left( \Omega, \mathscr{F}, \mathbb{P}\right)$. Define for $n \geq 1$, $S_n:= X_1 + \cdots + X_n$.  Is $\sigma\left(X_j, 1 \leq j \leq n \right) = \sigma\left(S_j,  1 \leq j \leq n \right)$?  Intuitively, I think this is true.  If not, what are the conditions needed for this to be true?  What if I remove the $\mathcal{L}_1$ condition on the $X_j's$? 

A:

Set $\mathcal{A}=\sigma(X_j)_{1\leq j\leq n}$ and $\mathcal{B}=\sigma(S_j)_{1\leq j \leq n}$. Let me first show that $\mathcal{A}\subset\mathcal{B}$. Since all the $S_j$ are $\mathcal{B}$-measurable, so is $X_j=S_j-S_{j-1}$. Thus all $X_j$ are $\mathcal{B}$-measurable and therefore $\mathcal{A}\subset\mathcal{B}$ ($\mathcal{A}$ is the smallest sigma-algebra that makes all the $X_j$-measurable).  In the same fashion, can you show that $\mathcal{B}\subset\mathcal{A}$? 
PS: Measurability only deals with the underlying sigma-algebras (on the domain and co-domain).  So omitting the $\mathcal{L}^1$ condition does not change anything :)

