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Let $X_n$ be a sequence of non-negative iid random variables.

Is it true that the condition,

$$\limsup_{n\rightarrow\infty} \frac{X_n}{n} = \infty \text{ almost surely}$$

is equivalent to the condition,

$$\mathbb{P} \Bigg( \limsup_{n\rightarrow\infty} \Big\{ {\frac{X_n}{n} \geq x} \Big\} \Bigg) = 1 \ \text{ for all } \ x > 0$$

I am wondering because I would like to rephrase the initial condition so as to make use of the Borel Cantelli lemma.

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up vote 2 down vote accepted

EDIT (thanks, Learner):

For Borel-Cantelli, note that $\omega\in \{\text{limsup}_n \frac{X_n}{n}=\infty\}$ iff for every $k> 0$ there is $n\geq 1$ such that $\frac{X_n(\omega)}{n}>k$. So, if $A_k:= \{\frac{X_n}{n}>k\}$, then $\omega \in \cap_{k=1}^\infty \cup_{n=1}^\infty A_k = \text{limsup}_n A_n$ . Thus your initial condition is equivalent to $P(\text{limsup}_n A_n)=1$.

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