Do the terms of an iid sequence whose law has infinite expected value necessarily exceed the partial sums of the sequence infinitely often?

Question

Let $\mu$ be a probability measure on $(0,\infty)$, and let $(\mathbf X_n)_1^\infty$ be a sequence of independent $\mu$-distributed random variables. Fix $\kappa > 0$, and consider

A) $\int x \; d\mu(x) = \infty$

B) There almost surely exist infinitely many $n$ such that $$ \mathbf X_{n + 1} > \kappa \sum_{i = 1}^n \mathbf X_i. $$

Then (B) implies (A) (this follows from the law of large numbers together with the Borel-Cantelli lemma.) My question is: does (A) imply (B)?

According to the generalized Borel-Cantelli lemma, condition (B) is equivalent to

B$'$) The series $$ \sum_{n = 1}^\infty \mu\left(\left(\kappa\sum_{i = 1}^n \mathbf X_i,\infty\right)\right) $$ diverges almost surely. This fact was used to show that (B) holds when $d\mu(x) \sim x^{-(1 + \varepsilon)} dx$ ($\varepsilon\in (0,1)$.) I can also use it to show that (B) holds when $d\mu(x) \sim 1/(x^2\log(x))$, $d\mu(x) \sim 1/(x^2\log(x)\log\log(x))$, etc., using complicated calculations. This is highly suggestive that the result holds in general...

Answer 1

In 1970, Harry Kesten proposed essentially this question in the Advanced Problems section of The American Math Monthly.

Let $X_1, X_2, \ldots, X_n$ be iid random variables and $S_n = \sum_{i=1}^n X_i$. Show that $$ \limsup_{n\to\infty} \frac{|X_n|}{|S_{n-1}|} = \infty \quad \text{with probability 1,} $$ whenever the $\mathbb E |X_i|$, the expectation of $|X_i|$ is infinite.

The solution, also by Kesten, is in the March 1971 issue (The American Math Monthly, vol. 78, no. 3, 305–308). The proof relies on first reducing to the case where the $X_i$ take values in $\{2^k: k \in \mathbb N\}$.

He further uses it to prove a result of Chow and Robbins,

If $\mathbb E|X_i| = \infty$ and $(b_n)$ is any sequence of positive numbers, then either $\liminf_n |S_n|/b_n = 0$ almost surely or $\limsup_n |S_n|/b_n = \infty$ almost surely.