# A little confusion about conditional expectation

If $(X_i)$ are iid. Let $S_n=\sum^n_{i=1}X_i$, then how do we compute $E(X_i\mid S_n)$. Is it independent of $i$?

I know it is a random variable. I guess that is independent of $i$, but I don't know how to show.

• I am not sure it is worth writing more than it is obvious by symmetry. We do need to assume that $E(X_1)$ exists. Dec 16, 2015 at 22:30
• sure, but how do you calculate $E(X_i|S_n)$? Dec 16, 2015 at 22:34
• Hint: By the linearity of expectation, we have $E((X_1+\cdots+X_n)\mid S_n)=E(X_1\mid S_n)+\cdots +E(X_n\mid S_n)$. Dec 16, 2015 at 22:42
• I believe you're not asking the question quite correctly. A conditional expectation needs to be conditioned on an event, not on a random variable (which is what a sum of random variables is). I think what you want to ask is how to compute $E(X_i|S_n=s)$, where $s$ is a possible sum of the $X_i$'s. Dec 16, 2015 at 22:42
• Dec 16, 2015 at 22:45

The fact that $E(X_i\mid S_n)=E(X_j\mid S_n)$ is obvious by symmetry.
For the follow-up question about $E(X_i\mid S_n)$, note that by the linearity of (conditional) expectation we have $$E((X_1+\cdots+X_n)\mid S_n)=E(X_1\mid S_n)+\cdots +E(X_n\mid S_n).$$ But $E((X_1+\cdots +X_n)\mid S_n)=E(S_n\mid S_n)=S_n$. It follows that $E(X_i\mid S_n)=\frac{S_n}{n}$.