Suppose $X_1,X_2, \ldots$ be i.i.d. Show that $\mathbb{E}|X_1| < \infty \Leftrightarrow \frac{X_n}{n} \to 0$ a.s

I tried using Markov but I don't know anything about the $ \mathbb{E}X $. I was also thinking of borel Cantelli, to show $ \sum P\left[\frac{X_n}{n}>\varepsilon\right]< \infty $ for each $\varepsilon>0$, then invoke First Borel Cantelli but I am confused on how to even get to the sum is less than infinity part


2 Answers 2


This is easily proved using Borel-Cantelli Lemma and the following well known fact:

For a non-negative random variable $Y$ we have $EY <\infty$ iff $\sum P(Y>n) <\infty$.

Now let $\epsilon >0$. Taking $Y=\frac {|X_1|} \epsilon$ we see that $E|X_1|<\infty$ iff $\frac{E|X_1|}{\epsilon}<\infty$ iff $\sum P\left(\frac {|X_1|} {\epsilon} >n\right) <\infty$ iff $\sum P\left(\frac {|X_n|} n >\epsilon\right) <\infty$ iff $P\left( \frac {|X_n|} n >\epsilon\hspace{0.2cm} \text{ i.o.}\right)=0$. Can you finish the proof?

  • $\begingroup$ Perfect!! thank you so much $\endgroup$
    – foobar
    Jun 1, 2020 at 0:38
  • $\begingroup$ Can we somehow connect this a new question Q . For any sequence of r.v $X_n$, show that there exists constants $ c_n \to \infty $ s.t $ X_n/c_n \to 0 $ a.s $\endgroup$
    – foobar
    Jun 1, 2020 at 0:41
  • 1
    $\begingroup$ @foobar Choose $c_n$ such that $P(|X_n| >\frac {c_n} n) <\frac 1 {n^{2}}$ and apply Borel-Cantelli. $\endgroup$ Jun 1, 2020 at 5:23

Another solution is this:

Let $S_n=X_1+...+X_n$

Note that $$\frac{X_n}{n}=\frac{S_n}{n}-\frac{S_{n-1}}{n-1} \left(\frac{n-1}{n}\right)$$

Then use the strong law of large numbers.


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