Proof that if $|X_n|>n $ i.o. then $|S_n|>n $ i.o.

probabilityprobability theory

I’m having trouble with this problem: given $(X_n)$ a sequence of independent and identically distributed random variables such that $E|X_n|=\infty$. First it asks to prove that $|X_n|>n$ i.o., which I did using

$$E|X_1|=\infty \iff \sum_{n=1}^{\infty}P(|X_1|>n)=\infty \iff \sum_{n=1}^{\infty}P(|X_n|>n)=\infty \iff P(\limsup(|X_n|>n))=1$$

So $|X_n|>n$ i.o. happens almost surely. Then it asks to show that $|S_n|>n$ i.o. where $S_n = \sum_{i=1}^n X_i$. I think the argument could be using Borell-Cantelli, but since the $|S_n|$ might not be independent I don’t know how to continue.

The last step of the exercise is to prove that $\limsup |S_n|/n = \infty$ almost surely. Again I think I can play with the series and use Borell-Cantelli but I’m not sure how.

Best Answer

By the same argument of yours, we can see that $|X_n|>2n$ i.o
So if for a $n$ we have $X_n>2n$, then either $S_{n-1}+X_n>n$ or $S_{n-1} < -n$. Hence either $|S_n| >n$ or $|S_{n-1}|>n$
Likewise for the case $X_n<-2n$.
Both of them imply that $$|S_n| >n$$ infinitely often.

I owe this answer to this link: https://mathoverflow.net/questions/376007/random-variables-with-no-first-moment

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