Let $X_k, k \geq 1$, be i.i.d. random variables such that $\limsup_{n\rightarrow\infty} \frac{X_n}{n}<\infty$ a.s, then show that $\limsup_{n\rightarrow\infty} \frac{\sum_{k=1}^{n} X_k}{n}<\infty$ a.s.

I'm thinking to use Borel-Cantelli lemma but don't know where to start. Any hints would be helpful. Thanks.

  • $\begingroup$ Probably too simple but subadditivity of limsup? en.wikipedia.org/wiki/… $\endgroup$
    – BCLC
    Dec 12, 2015 at 20:45
  • $\begingroup$ @BCLC But it's finite almost surely so I'm not sure whether we can use the property of limsup directly here. $\endgroup$ Dec 12, 2015 at 22:56
  • $\begingroup$ What do you mean? $\endgroup$
    – BCLC
    Dec 13, 2015 at 2:24
  • $\begingroup$ @BCLC I mean this is almost surely, not pointwise. $\endgroup$ Dec 13, 2015 at 2:58
  • $\begingroup$ I don't follow. Why can't we just split up the sum on the right? $\endgroup$
    – BCLC
    Dec 13, 2015 at 3:02

2 Answers 2


Suppose $\limsup_{n\rightarrow\infty} X_n/n < \infty$ with prob 1. Define the nonnegative i.i.d. random variables $Y_n = \max[X_n,0]$ for all $n \in \{1, 2, 3, …\}$. Then with prob 1 we have $\limsup_{n\rightarrow\infty} Y_n/n < \infty$ and: $$ \limsup_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n X_i \leq \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{i=1}^n Y_i = E[Y] \quad \mbox{(by LLN)}$$ where $Y=Y_1$. We either have $E[Y]<\infty$ or $E[Y] = \infty$. If $E[Y]<\infty$ then we are done. Suppose $E[Y]=\infty$ (we reach a contradiction). Fix $m$ as a positive integer. Since $Y$ is nonnegative we have \begin{align} \infty &= E[Y] \\ &= \int_0^{\infty} P[Y>t]dt \\ &=\sum_{n=0}^{\infty} \int_{mn}^{m(n+1)} P[Y>t]dt \\ &\leq \sum_{n=0}^{\infty}mP[Y>mn] \\ &= mP[Y>0] + m\sum_{n=1}^{\infty} P[Y_n>mn] \\ &= mP[Y>0] + m \sum_{n=1}^{\infty} P[Y_n/n > m] \end{align} So $$\sum_{n=1}^{\infty} P[Y_n/n > m] = \infty $$ and by Borel-Cantelli it means that $\{Y_n/n>m\}$ for infinitely many indices $n$ (with prob 1), so that $\limsup_{n\rightarrow\infty} Y_n/n \geq m$ with probability 1. This holds for all $m>0$, so $\limsup_{n\rightarrow\infty} Y_n/n = \infty$ with probability 1, a contradiction. $\Box$

  • $\begingroup$ The main idea is to show that if $Z_n$ are i.i.d. random variables that satisfy $E[Z_i]=\infty$, then $\limsup_{n\rightarrow\infty} Z_n/n = \infty$ with prob 1. $\endgroup$
    – Michael
    Jun 30, 2018 at 19:22

I think the Stolz-Cesaro's lemma will solve the problem. https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem

  • $\begingroup$ Welcome to the site! This is better suited as a comment; an answer should first of all be certain in what it is stating; that you think it is useful is not enough, because it actually doesn't answer the question - it is simply a suggestion (which is fine in the form of a comment). And second, an answer should show why the lemma could be used, not just link to it. Cheers! $\endgroup$ Feb 28, 2016 at 11:56
  • $\begingroup$ Third, the suggestion to use Stolz-Cesaro does not work. $\endgroup$
    – Did
    Jun 30, 2018 at 20:54

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