2
$\begingroup$

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.

$\endgroup$
8
  • 1
    $\begingroup$ 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. $\endgroup$ Dec 16, 2015 at 22:30
  • $\begingroup$ sure, but how do you calculate $E(X_i|S_n)$? $\endgroup$
    – user284873
    Dec 16, 2015 at 22:34
  • 1
    $\begingroup$ Hint: By the linearity of expectation, we have $E((X_1+\cdots+X_n)\mid S_n)=E(X_1\mid S_n)+\cdots +E(X_n\mid S_n)$. $\endgroup$ Dec 16, 2015 at 22:42
  • 2
    $\begingroup$ 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. $\endgroup$ Dec 16, 2015 at 22:42
  • 3
    $\begingroup$ @BarryCipra Conditioning on a r.v. is perfectly valid -- $\mathbb{E}[X\mid Y]$ is a random variable. $\endgroup$
    – Clement C.
    Dec 16, 2015 at 22:45

1 Answer 1

6
$\begingroup$

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 $E(X_i\mid S_n)=\frac{S_n}{n}$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .