# law of large number

• Jan 20th 2011, 08:16 PM
amb03
law of large number
I have problem with the following question

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

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

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

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

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

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

so finally
$\displaystyle E\left(\xi_1|S_n,S_{n+1},...\right)=\frac{S_n}{n} \ \ \ (a.s)$
• Jan 20th 2011, 09:57 PM
matheagle

$\displaystyle E(S_n|S_n)=S_n$

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

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

But each of these are the same, producing $\displaystyle nE(X_1|S_n)=S_n$
• Jan 20th 2011, 10:10 PM
amb03

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

Thanks
• Jan 20th 2011, 10:11 PM
matheagle
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)
• Jan 21st 2011, 12:17 AM
Moo
Same problem : http://www.mathhelpforum.com/math-he...-v-136142.html

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

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

equality of the expectations doesn't mean equality of the conditional expectations.
• Jan 21st 2011, 12:28 AM
Moo
As for proving that $\displaystyle E[S_n|S_n,S_{n+1},\dots]=E[S_n|S_n]$, remember that if we have two sigma-algebras (or sigma-algebras generated by random variables) $\displaystyle \mathcal G, \mathcal H$ such that $\displaystyle \mathcal H\subset \mathcal G$, then for a rv X, $\displaystyle E[E[X|\mathcal G]|\mathcal H]=E[X|\mathcal H]$

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

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

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