Sigma algebra associated with sum of random variables Let $Y_n$ be iid random varibles with $P(Y_n=-1)=p,$ $P(Y_n = 1) = q$, where $p,q>0$ and $p+q = 1$.
Set $X_0 = 0$ and $X_n = \sum_{i=1}^n Y_i$. 
If I let $F_n = \sigma(Y_1,Y_2,\dots,Y_n)$, then clearly $X_n$ is measurable with respect to $F_n$, but is it true that $\sigma(X_n) = F_n$?
 A: No, but because   $Y_k=X_k-X_{k-1}$ for $k\ge 1$, you have $\sigma(X_1,\ldots,X_n)=\sigma(Y_1,\ldots,Y_n)$.
A: No.
Let $\Omega:=\{\langle1,1\rangle,\langle1,-1\rangle,\langle-1,1\rangle,\langle-1,-1\rangle\}$ and let $\mathcal A:=\wp(\Omega)$. Let $Y_1$ and $Y_2$ be the projections and define a suitable probability $P:\wp(\Omega)\to[0,1]$. 
Then $\sigma(Y_1,Y_2)=\wp(\Omega)$.
However $\sigma(X_2)$ does not contain sets like $\{\langle-1,1\rangle\}$ and $\{\langle1,-1\rangle\}$.
This because no Borel-measurable set $A\subseteq\mathbb R$ can be found such that $\{X_2\in A\}=\{\langle-1,1\rangle\}$.
A: It is true that $\sigma(X_n) \subset \mathscr{F}_n$, as you proved. However, in general $\mathscr{F}_n$ is much bigger than $\sigma(X_n)$. For example, let $n = 2$ and consider the event $E = [\omega: Y_1(\omega) = 1, Y_2(\omega) = -1]$, which belongs to $\mathscr{F}_2$, but it does not belong to $\sigma(X_2)$, since if so, there exists an integer $k$ such that $$E = [\omega: Y_1(\omega) + Y_2(\omega) = k].$$
By the definition of $E$, $k = 0$. Therefore $A = [\omega: Y_1(\omega) = -1, Y_2(\omega) = 1] \subset E$. But on the other hand, $[\omega: Y_1(\omega) = -1, Y_2(\omega) = 1] \cap [\omega: Y_1(\omega) = 1, Y_2(\omega) = -1] = \varnothing.$
