Hi, I was wondering if anyone could help me with this question:

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- Oct 17th 2009, 11:36 PMsymmetry7almost surely
Hi, I was wondering if anyone could help me with this question:

I have posted a pdf version of the question in the next reply - Oct 17th 2009, 11:39 PMsymmetry7Reattached question in pdf format
Here's the same file in pdf..

cheers - Oct 18th 2009, 12:54 AMMoo
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 AMsymmetry7hmmm
Thanks for your reply Moo.

I'm not exactly sure I see where you are going with this... Does this Lemma have a name? And could you please explain why the expectation of the modulus of X_1 is infinite? (i.e. what belongs in the dots you posted - is it X_n?) I don't see it! Sorry my understanding is somewhat limited at this pont. - Oct 18th 2009, 07:47 AMmatheagle
It was really late last night, but I was going to say you can go with borel cantelli or another thought is the Weak Law.

It should be easy to show with Feller's Weak law that the sample mean goes to infinity in probability.

Thus a subsequence amost surely goes to infinity.