$\limsup\limits_{n\to\infty}X_n / n = \infty$, if $X_n\geq 0$ is i.i.d. and has infinite mean.

Let $$(X_n)_{n\in\mathbb{N}}$$ be a sequence of i.i.d. real random variables with $$X_1\geq 0$$ a.s. and $$\mathbb{E}(X_1)=\infty$$ then it holds $$\limsup\limits_{n\to\infty}\frac{X_n}{n}=\infty \text{ a.s.}$$

I tried to prove this statement, but I'm not quite sure, whether my proof is correct, since the usage of $$\limsup$$ with respect to random variables and with respect to sets confuses me a bit.

My attempt:

We know (e.g. by Kolmogorov's zero-one law) that $$\limsup\limits_{n\to\infty}\frac{X_n}{n}$$ is constant a.s., i.e. $$\exists c \in\mathbb{R}\cup\{-\infty, \infty\}$$ such that $$P\bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}\leq c\bigr)=1$$.

If we assume $$P\bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}=\infty\bigr)<1$$, then we see that $$c<\infty$$ and since $$X_n\geq 0$$ a.s., we can conclude, that $$c\geq0$$, yielding

$$P\Bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}> c\Bigr)=0,$$ for a $$c\geq0$$.

This however means that $$P\bigl(\limsup\limits_{n\to\infty}\{\frac{X_n}{n}>c\}\bigr)=P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)=0$$. Because if $$P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)>0$$, then $$P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)=1$$ (by the Lemma of Borel-Cantelli) and hence we find a subsequence $$(n_k)_{k\in\mathbb{N}}\subset\mathbb{N}$$ with $$X_{n_k}>c \text{ a.s. } \forall k\in\mathbb{N}$$. But then $$\limsup\limits_{n\to\infty}\frac{X_n}{n}> c$$ a.s. (To be more specific: Per definition $$\limsup\limits_{n\to\infty}\frac{X_n}{n}=\inf\limits_{n\in\mathbb{N}}\sup\limits_{j\geq n}\{\frac{X_j}{j}\}$$ and the existence of the subsequence $$X_{n_k}$$, ensures that $$\sup\limits_{j\geq n}\{\frac{X_j}{j}\}>c$$ a.s.

Having now established, that $$P\bigl(\frac{X_n}{n}>c \text{ i.o.}\bigr)=0$$ has to hold, we can use the Lemma of Borel-Cantelli to infer $$\sum_{n=1}^{\infty}P\Bigl(\frac{X_n}{n}>c\Bigr)<\infty,$$ or equivalently $$\sum_{n=1}^{\infty}P\Bigl(\frac{X_n}{c}>n\Bigr)<\infty.$$

This however is equivalent to $$\mathbb{E}\bigl(\frac{X_1}{c}\bigr)<\infty$$ (since $$X_1\geq 0$$ a.s.), which is a contradiction to $$\mathbb{E}(X_1)=\infty$$ and hence the result follows.

My question:

-Is the proof correct? I'm not 100% sure, whether my conclusion above $$P\Bigl(\limsup\limits_{n\to\infty}\frac{X_n}{n}> c\Bigr)=0 \Longrightarrow P\bigl(\frac{X_n}{n}>c \text{ i.o.} \bigr)=0$$ is valid. It seems valid to me, but dealing with $$\limsup$$ of functions and of sets confuses me. Is there a relation between both concepts?

It looks valid to me. Maybe write with more care that your sequence $$(n_k)_{k\in\mathbb N}$$ depends on $$\omega$$, but this has no real impact on your proof. Also, just after the introduction of this sequence, you deduce that $$\limsup\limits_{n\to\infty}\frac{X_n}{n}\ge c$$, where the inequality is not necessarily strict because it can converge to $$c$$. But again, it does not invalidate the proof.
To recap, the key equality here is $$\left\{\limsup_{n\to+\infty}\frac{X_n}{n}=+\infty\right\}=\bigcap_{c\in\mathbb N^*}\limsup_{n\to+\infty}\left\{\frac{X_n}{n}>c\right\},$$ which connects the limit superior of a random variable with that of sets and paves the way to Borel-Cantelli's lemma. As you mentioned, for all $$c\in\mathbb N^*$$, because $$X_1$$ is nonnegative we have $$\mathbb E\left[\frac{X_1}{c}\right]=\sum_{n\in\mathbb N}\mathbb P\left(\frac{X_1}{c}>n\right)=\sum_{n\in\mathbb N}\mathbb P\left(\frac{X_n}{n}>c\right)=+\infty,$$ so by Borel-Cantelli's lemma, $$\mathbb P\left(\limsup_{n\to+\infty}\left\{\frac{X_n}{n}>c\right\}\right)=1$$. As a countable intersection of almost sure sets is itself almost sure, we deduce from the first equality above that $$\mathbb P\left(\left\{\limsup_{n\to+\infty}\frac{X_n}{n}=+\infty\right\}\right)=1$$. This is basically the proof you wrote, but presented in a more direct way.
• Thanks alot for pointing out where I was a bit careless. I like your proof way more than mine! So thanks for answering the question. Just a quick question: what do you mean by $\mathbb{N}^*$? Is it just a typo? Commented Dec 27, 2022 at 15:28
• Glad I could help. $\mathbb N^*$ means the set of positive natural numbers, see for instance en.wikipedia.org/wiki/Natural_number#Notation. This notation is considered unambiguous. And by $\mathbb N$ I mean the set of nonnegative natural numbers (so I include zero). This notation however is ambiguous, as many people understand it as what I denoted by $\mathbb N^*$, that is the set of natural numbers excluding zero.