$\sigma(X_1,\dots, X_n) = \sigma(S_1,\dots, S_n)$ generated by partial sums is true that $\sigma(X_1,\dots, X_n) = \sigma(S_1,\dots,S_n)$ where $S_n=\sum_{i=1}^n X_i$ in general or I have to impose additional restrictions to the random variables (for instance, independence)?
If it is true, can somebody help me writing the proof?
 A: Since $S_k\in \Sigma^k_X:=\sigma(X_1,\dots,X_k)$ you obtain $\Sigma^n_S:=\sigma(S_1,\dots,S_n)\subseteq \Sigma_X^n$. Furthermore,
$$
  X_1 = S_1,X_2 = S_2 - S_1,\dots,X_n = S_n - S_{n-1}
$$
so $X_k \in \Sigma^k_S$ so that $\Sigma^n_X\subseteq \Sigma_S^n$. 
It's also worth giving some explanation, I believe. Let $\xi$ and $\eta$ be two random elements: that is, random variables with a pretty general range, say $\xi = (X_1,\dots,X_n)$ and $S = (S_1,\dots,S_n)$. Then, if $\eta = f(\xi)$ for some measurable map $f$, then $\Sigma_\eta\subseteq \Sigma_\xi$ just by definition:
$$
  \eta^{-1}(A) = \xi^{-1}(f^{-1}(A)) \in \Sigma_\xi
$$
for any measurable $A$, since $f^{-1}(A)$ is measurable by measurability of $f$. In simple terms, we want $\eta$ to be expressed as some function of $\xi$.
Now, what I've done in your setting is that I've constructed functions $f$ and $g$ such that $\eta = f(\xi)$ and $\xi = g(\eta)$. The function $f$ is easy to construct: that's how you express partial sums via the summands. The function $g$ is converse: how do you get summands if you are given their partial sums. 
This technique you can apply to a more general setting, so if you'll have some similar example you need help with, feel free to ping it here in the comment.
