Skip to main content
MathOverflow

Return to Answer

added 6 characters in body
Source Link
Mateusz Kwaśnicki
Mateusz Kwaśnicki
  • 17.2k
  • 1
  • 33
  • 55

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 $$\sum_{n=0}^\infty \mathbb{P}(|X_n| > n a) = \sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} |X_1| = \infty,$$$$\sum_{n=0}^\infty \mathbb{P}(|X_n| > n a) = \sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} (|X_1| / a) = \infty,$$ 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.

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 $$\sum_{n=0}^\infty \mathbb{P}(|X_n| > n a) = \sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} |X_1| = \infty,$$ 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.

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.

Source Link
Mateusz Kwaśnicki
Mateusz Kwaśnicki
  • 17.2k
  • 1
  • 33
  • 55

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 $$\sum_{n=0}^\infty \mathbb{P}(|X_n| > n a) = \sum_{n=0}^\infty \mathbb{P}(|X_1| > n a) \geqslant \mathbb{E} |X_1| = \infty,$$ 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.