Conditional expectation of iid random variables If $Y_1, Y_2, \ldots$ are iid, with $S_n:= Y_1+\cdots+Y_n$ why do we have that $E(Y_1\mid\sigma(S_n, S_{n+1}, \ldots))=S_n/n$ for all $n \geq 1$?
Durrett's textbook asserts this is true when trying to give a simpler proof of the Strong Law of Large Numbers from the backwards Martingale theorem.  However, he merely says it is "by symmetry" that we have that $E(Y_1\mid\sigma(S_n, S_{n+1}, \ldots))=E(Y_k\mid \sigma(S_n, S_{n+1}, \ldots))$ whenever $k\leq n$.  But I'd like to see an actual proof of this.
 A: It really is "by symmetry". Because the Y_i are iid, the distribution of
$$
(Y_1,\ldots,Y_n, S_n, S_{n+1},\ldots)
$$
is unchanged when $Y_i$ and $Y_j$ are exchanged. It follows that
$$
\tag{1}E(Y_i|\sigma(S_n,S_{n+1},\ldots)) = E(Y_j|\sigma(S_n,S_{n+1},\ldots))
$$
for $i,j\in\{1,\ldots,n\}$.
Also,
$$
S_n = E(S_n |\sigma(S_n,S_{n+1},\ldots)) = \sum_{i=1}^n E(Y_i | \sigma(S_n,S_{n+1},\ldots)),
$$
so
$$
   S_n = n E(Y_1| \sigma(S_n,S_{n+1},\ldots)).
$$
 EDIT 
How does one deduce (1) above? Observe that for any constants $s_n,s_{n+1},\ldots$,
$$
E(Y_i 1_{S_n\le s_n,S_{n+1}\le s_{n+1},\ldots}) = E(Y_j 1_{S_n\le s_n,S_{n+1}\le s_{n+1},\ldots})
$$
for all $i,j\in\{1,\ldots,n\}$. This follows from the symmetry of the law of $(Y_1,\ldots,Y_n,S_n,S_{n+1},\ldots)$ with respect to exchanging the $Y_i$. We conclude that
$$
E(Y_i 1_B) = E(Y_j 1_B)
$$
for all $B\in\sigma(S_n,S_{n+1},\ldots)$. Staring at the definition of conditional expectation, we see
$$
E(Y_i | \sigma(S_n,S_{n+1},\ldots)) = E(Y_j | \sigma(S_n,S_{n+1},\ldots)).
$$
A: Note that by the independence of $(Y_j)_{j \geq 1}$ we have $$\mathbb{E}(Y_j \mid \sigma(S_n,S_{n+1},S_{n+2},\ldots)) = \mathbb{E}(Y_j \mid \sigma(S_n,Y_{n+1},Y_{n+2},\ldots))= \mathbb{E}(Y_j \mid \sigma(S_n))$$ for all $j \leq n$.
Moreover,
$$\mathbb{E}(Y_1 \cdot 1_B(S_n)) = \int \ldots \int_{y_1+\ldots+y_n \in B}y_1 \, \underbrace{d\mathbb{P}_{Y_1}(y_1)}_{d\mathbb{P}_{Y_j}(y_1)} \, d\mathbb{P}_{Y_2}(y_2) \ldots \underbrace{d\mathbb{P}_{Y_j}(y_j)}_{d\mathbb{P}_{Y_1}(y_j)} \ldots d\mathbb{P}_{Y_n}(y_n) \\ = \int \ldots \int_{y_1+\ldots+y_n \in B} y_1 \, d\mathbb{P}_{Y_j}(y_1) \, d\mathbb{P}_{Y_2}(y_2) \ldots d\mathbb{P}_{Y_1}(y_j) \ldots d\mathbb{P}_{Y_n}(y_n) = \mathbb{E}(Y_j \cdot 1_B(S_n))$$
for all Borel sets $B \in \mathcal{B}(\mathbb{R})$, $j \leq n$. Thus $$\mathbb{E}(Y_1 \mid \sigma(S_n)) = \mathbb{E}(Y_j \mid \sigma(S_n))$$ for all $j \leq n$. Therefore
$$S_n = \mathbb{E}(S_n \mid \sigma(S_n)) = n \cdot \mathbb{E}(Y_1 \mid \sigma(S_n)) = n \cdot \mathbb{E}(Y_1 \mid \sigma(S_n,S_{n+1},\ldots))$$
