$\limsup_n \dfrac{X_n}{n} = 0$ if $\mathbb{E}(X_1) < \infty$?

Question

Here is an exercise in the book of author Achim Klenke. Let $(X_n)$ be iid non-negative random variables. By using Borel-Cantelli lemma, show that: $$ \limsup_n \dfrac{X_n}{n} = 0 \text{ a.s} $$ if $\mathbb{E}(X_1) < \infty$. Otherwise, show that $\limsup_n \dfrac{X_n}{n} = \infty$ a.s

As suggested by the problem, I tried to express the event $\{\limsup_n X_n/n = 0\}$ as the limsup of events as follows: $$ \begin{align*} \left(\limsup_n \dfrac{X_n}{n} = 0 \right) &= \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \bigcap_{k \ge m} \left(\dfrac{X_k}{k} \le \dfrac{1}{n}\right) \end{align*} $$ Thus, $$ \mathbb{P}\left(\limsup_n \dfrac{X_n}{n} = 0\right) = 1 - \lim_{n \rightarrow \infty} \mathbb{P}\left(\limsup_m \left[\dfrac{X_m}{m} > \dfrac{1}{n}\right]\right) $$

From here, I want to show that $$ \sum_{m = 1}^\infty \mathbb{P}\left(\dfrac{X_m}{m} > \dfrac{1}{n}\right) < \infty \ \forall n $$ However, I can only prove this if we add the condition $\mathbb{E}(X_1^2) < \infty$, then $$ \sum_{m = 1}^\infty \mathbb{P}\left(\dfrac{X_m}{m} > \dfrac{1}{n}\right) = \sum_{m = 1}^\infty \mathbb{P}\left(\dfrac{X_m^2}{m^2} > \dfrac{1}{n^2}\right) \le n^2 \mathbb{E}(X_1^2)\sum_{m = 1}^\infty \dfrac{1}{m^2} < \infty $$ Without the assumption $\mathbb{E}(X_1^2) < \infty$, I'm pretty much stuck, so any hints for other ways are appreciated. Thanks

Answer 1

You have shown that the goal is to prove for any $\epsilon > 0$ (which is equivalent to $n^{-1}$ in your post), it holds that \begin{align} \sum_{m = 1}^\infty P(X_m > m\epsilon) < \infty. \end{align}

By the i.i.d. assumption (in fact, just "identically distributed" condition would suffice), this is equivalent to prove \begin{align} \sum_{m = 1}^\infty P(X_1 > m\epsilon) < \infty. \tag{1} \end{align}

Now the inequality (the first equality below uses $X_1$ is nonnegative) \begin{align} E[X_1] &= \int_0^\infty P[X_1 > t]dt = \sum_{m = 1}^\infty \int_{(m - 1)\epsilon}^{m\epsilon}P[X_1 > t]dt \geq \sum_{m = 1}^\infty \epsilon P[X_1 > m\epsilon] \end{align} and $E[X_1] < \infty$ imply that $(1)$ holds. This completes the proof.

The same trick can be used to prove $\limsup_n n^{-1}X_n = \infty$ almost surely given $E[X_1] = \infty$. But for this to hold, now we need the independent assumption as well (as opposed to the $E[X_1] < \infty$ case which only needs $X_1, X_2, \ldots$ are identically distributed) to deploy the second Borel-Cantelli lemma.