# Conditional expectation

• Nov 29th 2009, 11:16 PM
yavanna
Conditional expectation
How can I prove that if
$\displaystyle (X_{n})_{n \geq 1}$ are i.i.d. and $\displaystyle S_{n}=X_{1}+X_{2}+...+X_{n}$
then
$\displaystyle \mathbb{E}(X_{1}|S_{n},S_{n+1},...)=\frac{S_{n}}{n }, \forall n=1,2,...$?
• Nov 30th 2009, 12:24 AM
Moo
Hello,
Quote:

Originally Posted by yavanna
How can I prove that if
$\displaystyle (X_{n})_{n \geq 1}$ are i.i.d. and $\displaystyle S_{n}=X_{1}+X_{2}+...+X_{n}$
then
$\displaystyle \mathbb{E}(X_{1}|S_{n},S_{n+1},...)=\frac{S_{n}}{n }, \forall n=1,2,...$?

First prove that $\displaystyle \forall k=1\dots n ~,~ \text{ the } (X_k,S_n)$ follow the same distribution
(not very difficult since the rv's are iid)

Then we have, for any bounded (measurable) function f, and for $\displaystyle j\neq k$, $\displaystyle E[E[X_k|S_n]f(S_n)]=E[X_kf(S_n)]=E[X_jf(S_n)]=E[E[X_j|S_n]f(S_n)]$

So $\displaystyle E[X_k|S_n]=E[X_j|S_n]$ almost surely.

Thus $\displaystyle E[X_k|S_n]=\frac 1n \cdot \sum_{j=1}^n E[X_j|S_n]=\frac 1n \cdot E[S_n|S_n]=\frac{S_n}{n}$