# Thread: law of large number

1. ## law of large number

I have problem with the following question

Let $\xi_1, \xi_2,...$ be independent and identically distributed random variables with $E|\xi_i|< \infty$. Show that

$E\left(\xi_1|S_n,S_{n+1},...\right)=\frac{S_n}{n} \ \ \ (a.s)$
where $S_n= \xi_1+...+\xi_n$

I proceed this way:
I want to know wether this is true or not.

$S_n = E(S_n|S_n)=E\left(S_n|S_{n},S_{n+1},S_{n+2},...\ri ght)$ ???

Now since $\xi_i$ are independent and identically distributed, then
$E(\xi_1)=E(\xi_2)=...=E(\xi_n)$

$E \left(S_n|S_{n},S_{n+1},S_{n+2},...\right)=n.E\lef t(\xi_1|S_{n},S_{n+1},S_{n+2},...\right)$

so finally
$E\left(\xi_1|S_n,S_{n+1},...\right)=\frac{S_n}{n} \ \ \ (a.s)$

$E(S_n|S_n)=S_n$

So $E(X_1+\cdots +X_n|S_n)=S_n$

Thus $E(X_1|S_n)+\cdots +E(X_n|S_n)=S_n$

But each of these are the same, producing $nE(X_1|S_n)=S_n$

$E(S_n|S_n)=E\left(S_n|S_{n},S_{n+1},S_{n+2},...\ri ght)$

Thanks

4. Both are S_n, as long as you condition on S_n, you know S_n, the rest of that sequence is unnecessary.

E(X|X)=X=E(X|X,Y)

5. Same problem : http://www.mathhelpforum.com/math-he...-v-136142.html

And I don't agree with you going to this step :

$E \left(S_n|S_{n},S_{n+1},S_{n+2},...\right)=n.E\lef t(\xi_1|S_{n},S_{n+1},S_{n+2},...\right)$

equality of the expectations doesn't mean equality of the conditional expectations.

6. As for proving that $E[S_n|S_n,S_{n+1},\dots]=E[S_n|S_n]$, remember that if we have two sigma-algebras (or sigma-algebras generated by random variables) $\mathcal G, \mathcal H$ such that $\mathcal H\subset \mathcal G$, then for a rv X, $E[E[X|\mathcal G]|\mathcal H]=E[X|\mathcal H]$

now consider $\mathcal H=\sigma(S_n)$ and $\mathcal G=\sigma(S_n,S_{n+1},\dots)$ and you have your equality.

(why is there the inclusion ? because if $S_n$ is a measurable function of $\xi_1,\dots,\xi_n$ and $(S_n,S_{n+1},\dots)$ is a measurable function of $\xi_1,\dots,\xi_n,\dots$)

If you don't know what a sigma-algebra is, then just forget my post