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Consider a random variable $X$ with $\mathbb E(\vert X \vert)=\infty$. I am then wondering if this implies that for $X_i$ iid copies of $X$ $$\limsup_n \frac{\vert \sum_{i=1}^n X_i \vert}{n}=\infty?$$

This looks plausible. However, the proof seems only simple when looking at $ \sum_{i=1}^n \vert X_i \vert$ instead, since in this case we do not run into the problem of cancellations between the individual random variables.

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  • $\begingroup$ An argument I believe can be made formal: Your absolute value is going to be the difference of the magnitude of the ones that come out positive from the magnitude of the ones that come out non-positive, or vice-versa. but no matter the selection for which come out positive and which come out negative, the resulting difference has undefined or infinite expected value. from here you can get your limsup to infinity with a large numbers law $\endgroup$
    – Christian Chapman
    Commented Nov 9, 2020 at 5:38
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    $\begingroup$ The usual argument which shows that the limit does not exist can be adapted. Usually you consider that since $\Sigma P(|X_n| > n) = \infty$ then $|X_n| > n$ happens i.o., but replace $> n$ with $ > kn $ for some large k, and argue as before $\endgroup$
    – mike
    Commented Nov 9, 2020 at 8:01
  • $\begingroup$ @mike would this not just show that $limsup |X_n|/n=\infty? $ I do not see where this sum of iid copies enters in your case...Also note, the random variables are not assumed to be integer-valued by the OP $\endgroup$
    – Sascha
    Commented Nov 9, 2020 at 11:20
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    $\begingroup$ @Sascha yes, but if $|X_n|>100 n$, then either $|X_1+\ldots+X_n|$ or $|X_1+\ldots+X_{n-1}|$ is greater than $50n$. And the sum of these probabilities remains infinite even when $X_n$'s take real values. $\endgroup$
    – Fedor Petrov
    Commented Nov 9, 2020 at 16:02

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This very statement is a part of Theorem 46.3 in Sato's book Lévy processes and infinitely divisible distributions, in case you need a reference.

The proof only uses Borel–Cantelli, as in mike's and Fedor Petrov's comments: for every $a > 0$, we have $$\sum_{n=0}^\infty \mathbb{P}(|X_n| > n a) = \sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} (|X_1| / a) = \infty,$$ and hence $|X_n| > n a$ infinitely often. This means that $|S_n| > \tfrac{n a}{2}$ or $|S_{n-1}| > \tfrac{n a}{2}$. Thus, $n^{-1} |S_n| > \tfrac{a}{2}$ infinitely often.

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  • $\begingroup$ I would replace the inequality $\sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} |X_1|$ to $\sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} |X_1|/a$ (that is always true, whenever RHS is finite or infinite). $\endgroup$
    – Fedor Petrov
    Commented Nov 10, 2020 at 12:48
  • $\begingroup$ @FedorPetrov: Corrected, thanks! (And I also like the way you pointed out this typo.) $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 10, 2020 at 13:16

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