# almost surely

• Oct 17th 2009, 11:36 PM
symmetry7
almost surely
Hi, I was wondering if anyone could help me with this question:

• Oct 17th 2009, 11:39 PM
symmetry7
Reattached question in pdf format
Here's the same file in pdf..

cheers
• Oct 18th 2009, 12:54 AM
Moo
pagi pagi

*all credits to my lecture notes (and a bit of thinking too lol)*

There's a lemma that states the equivalence between the convergence of $\displaystyle \mathbb{E}(|X|)$ and the convergence of $\displaystyle \sum_{k\geq 0} \mathbb{P}(|X|\geq k)$
I can provide a proof, but that'll be for later...

Here, we have :
$\displaystyle \sum_{n\geq 0} \mathbb{P}(|X_n|\geq n)=\sum_{n\geq 0} \mathbb{P}(|X_1|\geq n)$ because all the Xi's follow the same distribution.
But since $\displaystyle \mathbb{E}(|X_1|)=\infty$ (because $\displaystyle \mathbb{E}(|...|)\geq |\mathbb{E}(X_1)|$), we get that :
$\displaystyle \sum_{n\geq 0} \mathbb{P}(|X_n|\geq n)=\infty$

By applying the second lemma of Borel-Cantelli, $\displaystyle \mathbb{P}(\limsup_n \{|X_n|\geq n\})=1$

The idea is to apply Cesaro's mean to the divergent sequence, but I have quite some problems with finishing the proof (Crying)
• Oct 18th 2009, 05:06 AM
symmetry7
hmmm