
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
$\displaystyle E\left(\xi_1S_n,S_{n+1},...\right)=\frac{S_n}{n} \ \ \ (a.s)$
where $\displaystyle S_n= \xi_1+...+\xi_n$
I proceed this way:
I want to know wether this is true or not.
$\displaystyle S_n = E(S_nS_n)=E\left(S_nS_{n},S_{n+1},S_{n+2},...\ri ght)$ ???
Now since $\displaystyle \xi_i$ are independent and identically distributed, then
$\displaystyle E(\xi_1)=E(\xi_2)=...=E(\xi_n) $
$\displaystyle E \left(S_nS_{n},S_{n+1},S_{n+2},...\right)=n.E\lef t(\xi_1S_{n},S_{n+1},S_{n+2},...\right)$
so finally
$\displaystyle E\left(\xi_1S_n,S_{n+1},...\right)=\frac{S_n}{n} \ \ \ (a.s)$

I'll use X's instead
$\displaystyle E(S_nS_n)=S_n$
So $\displaystyle E(X_1+\cdots +X_nS_n)=S_n$
Thus $\displaystyle E(X_1S_n)+\cdots +E(X_nS_n)=S_n$
But each of these are the same, producing $\displaystyle nE(X_1S_n)=S_n$

Thanks for your reply,
I want to know what about this, is this true ?
$\displaystyle E(S_nS_n)=E\left(S_nS_{n},S_{n+1},S_{n+2},...\ri ght)$
Thanks

Both are S_n, as long as you condition on S_n, you know S_n, the rest of that sequence is unnecessary.
E(XX)=X=E(XX,Y)

Same problem : http://www.mathhelpforum.com/mathhe...v136142.html
And I don't agree with you going to this step :
$\displaystyle E \left(S_nS_{n},S_{n+1},S_{n+2},...\right)=n.E\lef t(\xi_1S_{n},S_{n+1},S_{n+2},...\right)$
equality of the expectations doesn't mean equality of the conditional expectations.

As for proving that $\displaystyle E[S_nS_n,S_{n+1},\dots]=E[S_nS_n]$, remember that if we have two sigmaalgebras (or sigmaalgebras 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 sigmaalgebra is, then just forget my post :D