$X_k$, $k \geq 1$ are iid random variables such that $$\limsup_{n\rightarrow\infty} \frac{X_n}{n} < \infty$$ with probability 1. We want to show that $$\limsup_{n\rightarrow\infty} \frac{\sum_{i=1}^n X_i}{n} < \infty$$ with probability 1.

The hint says to apply the law of large numbers to the sequence $\max(X_k,0), k \geq 1$. SLLN gives that $$\frac{\sum_{i=1}^n \max(X_i,0)}{n} \rightarrow \mathbb{E}\max(X,0) = \mathbb{E}(X; X>0)$$ almost surely. I feel that the idea here is that $\limsup X_n/n < \infty$ a.s. implies that $\mathbb{E}(X;X>0)$, but I am not really sure how to approach this...

  • $\begingroup$ @Crostul Is that true? If you look at the sequence $(x_n)$ with $x_n=n$, then $\limsup_n x_n/n = 1$ but $\limsup_n \sum x_k/n = \limsup_n (n+1)/2 = \infty$... $\endgroup$ Sep 21, 2014 at 20:36
  • $\begingroup$ Ok, I will delete my comment. SOrry $\endgroup$
    – Crostul
    Sep 21, 2014 at 20:40

2 Answers 2


Starting where you stopped, assume that $E(X^+)$ is infinite, then $\sum\limits_{n=1}^\infty P(X_n\geqslant xn)=\sum\limits_{n=1}^\infty P(X\geqslant xn)$ diverges for every $x\gt0$ hence Borel-Canteli lemma implies that $X_n\geqslant xn$ for infinitely many $n$, almost surely, thus $\limsup\limits_{n\to\infty}\frac{X_n}n\geqslant x$, almost surely. This holds for every $x\gt0$ hence you are done.

To show that the series $\sum\limits_{n=1}^\infty P(X\geqslant xn)$ diverges, note that for every $x\gt0$ and every random variable $X$, one has $x\sum\limits_{n=1}^\infty \mathbf 1_{X\geqslant xn}\geqslant X^+$ almost surely hence $x\sum\limits_{n=1}^\infty P(X\geqslant xn)\geqslant E(X^+)$.


Consider $X_k^+ := \max(X_k,0)$. Then, \begin{align*} P\left(\limsup \frac{X_n}{n} < \infty\right)=1 &\Rightarrow P\left(\limsup \frac{X_n^+}{n} < \infty\right)=1 \\ &\Rightarrow \exists A: P\left(\frac{X_n^+}{n} > A \text{ i.o.}\right)=0 \text{ a.s.}\\ &\Rightarrow \sum_{i=1}^n P\left(\frac{X_i^+}{i} > A\right) < \infty \text{ a.s.} \\ &\Rightarrow \sum_{i=1}^n P\left( X^+ > iA \right) < \infty \text{ a.s.} \\ &\Rightarrow \mathbb{E}X^+ < \infty \text{ a.s.} \end{align*} By the Strong Law of Large Numbers, $$\frac{\sum_{i=1}^n X_i^+}{n} \rightarrow \mathbb{E}X^+ < \infty \text{ a.s.},$$ and so, $$\limsup \frac{\sum_{i=1}^n X_i}{n} \leq \frac{\sum_{i=1}^n X_i^+}{n} < \infty \text{ a.s.}$$

Lemma. $\limsup_n X_n < \infty$ a.s. if and only if $\sum P(X_n > A) < \infty$ for some $A$.

Proof of Lemma. We write $Y = \limsup_n X_n$ for notational simplicity. Since $X_n$ are independent, Borel-Cantelli Lemmas show that \begin{align*} \sum_{n=1}^\infty P(X_n > A) < \infty &\iff P(X_n > A \text{ i.o.}) = 0 \\ \sum_{n=1}^\infty P(X_n > A) = \infty &\iff P(X_n > A \text{ i.o.}) = 1. \end{align*} To relate this with the finiteness of $Y$, note that

$\bullet$ If $a_n > A$ i.o., then $\limsup_n a_n \geq A$.

$\bullet$ If $\limsup_n a_n \geq A$, then for any $\epsilon > 0$, we have $a_n \geq A-\epsilon$ i.o.

This yields the inequality $$P(Y \geq A+\epsilon) \leq P(X_n > A \text{ i.o.}) \leq P(Y \geq A)$$

This shows that

$\bullet$ If $P(Y < \infty) = 0$, then $P(Y \geq A) < 1$ for some constant $A$. Then, $P(X_n > A \text{ i.o.}) < 1$ and hence, $P(X_n > A \text{ i.o.}) = 0$.

$\bullet$ If $P(X_n > A \text{ i.o.}) = 0$, then $P(Y \geq A + \epsilon) = 0$, and thus, $P(Y= \infty) = 0$ as well.

These combine to show that $limsup_n X_n < \infty $ a.s. $\iff \sum_n P(X_n > A) < \infty$ for some $A$.

  • $\begingroup$ The second implication is wrong (it says that if a random variable is almost surely finite then it is almost surely bounded, which is untrue). $\endgroup$
    – Did
    Sep 22, 2014 at 8:31
  • $\begingroup$ @Did, is it not true that if $X_i$ are independent, then $\sup X_n < \infty$ a.s. iff $\limsup X_n < \infty$ a.s. iff $\sum P(X_n>A) < \infty$ for some constant $A$? $\endgroup$
    – tongtong
    Sep 22, 2014 at 8:57
  • $\begingroup$ ?? How is the question in your comment related to the assertion in your answer? If the independence of the sequence $(X_n)$ is crucial to show the second implication in your answer, you should explain in detail how. As the answer is written, it can only mean that the second implication always holds--and it does not, does it? $\endgroup$
    – Did
    Sep 22, 2014 at 9:13
  • $\begingroup$ @Did I scribbled down my claim as a lemma... I hope it's correct. $\endgroup$
    – tongtong
    Sep 23, 2014 at 6:10

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