# Problem in Conditional expectation of Shiryaev

Let $X_1, X_2, \ldots$ be independent identically distributed random variables with $E\left | X_i \right |<\infty$. Show that

$E(X_1\mid S_n,S_{n+1},\ldots)=\dfrac{S_n}{n}$ (a.s.),

where $S_n=X_1+\cdots+ X_n$

• possible duplicate of Help with conditional expectation question – Nate Eldredge May 6 '12 at 16:16
• This doesn't seem to be an exact duplicate, as Nate Eldredge suggests, since it's got the whole infinite sequence $S_n,S_{n+1},S_{n+2},\ldots$, and it seems to be part of the problem to show that the whole thing is the same as $\mathbb{E}(X_1\mid S_n)$. – Michael Hardy May 6 '12 at 17:01
• It's not an exact duplicate, but the idea is the same: use symmetry to show that $E[X_k \mid S_n, S_{n+1}, \dots]$ is the same for every $1 \le k \le n$. Then add them all together and you get $S_n$. – Nate Eldredge May 6 '12 at 17:08
• @NateEldredge : The fact that this conditional probability depends on $S_n,S_{n+1},S_{n+2},\ldots$ only through $S_n$ may seem to have a trivial proof, which you just gave, but does that mean it's a trivial fact? Perhaps it has non-trivial consequences. And that fact is not mentioned in that other question. – Michael Hardy May 6 '12 at 17:18

## 1 Answer

Observe that the information given by $S_n,S_{n+1},S_{n+2},S_{n+3},\ldots$ is the same as that given by $S_n,X_{n+1},X_{n+2},X_{n+3},\ldots$, i.e., if you know either, you can compute the other. Or, if you like, both sequences generate the same sigma-algebra. So you're looking for $$\mathbb{E}(X_1\mid S_n,S_{n+1},S_{n+2},S_{n+3},\ldots) = \mathbb{E}(X_1\mid S_n,X_{n+1},X_{n+2},X_{n+3},\ldots),$$ and the $X$s on which you're conditioning in the second expression are independent of $X_1$ and so can be dropped.

The rest is just the symmetry argument cited in comments above by Nate Eldredge, i.e. this expectation is the same as $\mathbb{E}(X_k\mid S_n)$ for $k=1,\ldots,k$, and it's easy to find the sum of those $k$ expectations.