almost surely

**symmetry7** · October 17th 2009, 11:36 PM

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

**symmetry7** · October 17th 2009, 11:39 PM

Here's the same file in pdf..

cheers

**Moo** · October 18th 2009, 12:54 AM

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

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

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

The idea is to apply Cesaro's mean to the divergent sequence, but I have quite some problems with finishing the proof

**symmetry7** · October 18th 2009, 05:06 AM

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.

**matheagle** · October 18th 2009, 07:47 AM

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.

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hmmm

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