Suppose $\{X_n\}$ a sequence of random variables. If $\sum_{n=1}^\infty P(|X_n|>n)< \infty$

Prove that $$\limsup_{n\to\infty}\frac{ |X_n|}{n} \le1 $$ almost surely

What i have done so far:

I thought using the Borel-Cantelli lemma could lead me somewhere, but i didn't have any luck.

From Borel-Cantelli lemma we know that if $\sum_{n=1}^\infty P(|X_n|>n)< \infty$ then $P(|X_n|>n)=0$

How could I proceed? I would appreciate any help, advice. Thank you all very much in advance for your time and concern.

  • 3
    $\begingroup$ Hint: Borel-Cantelli lemma shows that $\Bbb{P}(|X_{n}| > n \text{ i.o.}) = 0$. $\endgroup$ – Sangchul Lee Jul 7 '13 at 13:14
  • 2
    $\begingroup$ Crossposted: stats.stackexchange.com/q/63561/2970 $\endgroup$ – cardinal Jul 7 '13 at 15:24
  • 1
    $\begingroup$ @cardinal so what? i can not post my question in two different sections? $\endgroup$ – johan paul Jul 7 '13 at 15:33
  • 1
    $\begingroup$ @johan: See this meta.SO answer. This is the quasi official policy on this topic. Cheers. $\endgroup$ – cardinal Jul 7 '13 at 15:36
  • 1
    $\begingroup$ @johan: It's ok. I just wanted you to be aware of the prevailing "policy". I generally believe it's good to have the content in one location since one objective is the site is to provide a long-term repository of questions and answers. (+1 to your question, in particular for supplying your initial thoughts on the problem. Cheers.) $\endgroup$ – cardinal Jul 7 '13 at 15:43

The first Borel-Cantelli lemma yields $$\mathbb P\left(\limsup_{n\to\infty}|X_n|>n\right)=0. $$ As for each $n$ $$\{|X_n|>n\}\subset \bigcup_{k=n}^\infty \{|X_k|>k\}, $$ it follows that \begin{align} 0&=\mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty \{|X_k|>k\}\right) \\&=\mathbb P\left(\lim_{n\to\infty}\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\ &=\lim_{n\to\infty}\mathbb P\left(\bigcup_{k=n}^\infty \{|X_k|>k\} \right)\\ &\geqslant\lim_{n\to\infty}\mathbb P(|X_n|>n) \end{align} and hence $\lim_{n\to\infty}\mathbb P(|X_n|>n)=0$. Further, $$\bigcap_{k=n}^\infty \{|X_k|>k\}\subset\{|X_n|>n\} $$ so that \begin{align} \mathbb P\left(\limsup_{n\to\infty} |X_n|\leqslant n\right) &= \mathbb P\left(\bigcap_{n=1}^\infty\bigcup_{k=n}^\infty\{|X_k|\leqslant k\}\right)\\ &=1 - \mathbb P\left(\bigcup_{n=1}^\infty\bigcap_{k=n}^\infty \{|X_k|>k\} \right)\\ &=1 - \mathbb P\left(\lim_{n\to\infty} \bigcap_{k=n}^\infty\{|X_k|>k\} \right)\\ &=1 - \lim_{n\to\infty}\mathbb P\left(\bigcap_{k=n}^\infty\{|X_k|>k\}\right)\\ &\geqslant1 - \lim_{n\to\infty}\mathbb P(|X_n|>n)\\ &= 1, \end{align} which implies that $$\mathbb P\left(\limsup_{n\to\infty} \frac{|X_n|}n\leqslant 1 \right)=1. $$

  • $\begingroup$ How do you know the first equation? From the contrapositive of the direct result of BCL1? Anyway answer looks good \m/ $\endgroup$ – BCLC Nov 26 '15 at 4:43

By Borel Cantelli lemma we have that $$ P( \liminf_{n \to \infty} \{ |X_n| \leq n \}) = P( \{|X_n| \leq n \text{ eventually } \} )= 1$$ In words this means than almost surely, the sequence $|X_n|$ is below $n$ for all $n$ sufficently large. I think you can take it from here.

  • $\begingroup$ Thank you very much. May I ask you something more? How can we pass from the probability to the limit suprermum of the inequality? $\endgroup$ – johan paul Jul 7 '13 at 13:29
  • $\begingroup$ If $|X_n|$ can't surpass $n$ for $n$, then what happens to $\frac{|X_n|}{n}$?. Also remember that the limsup of a sequence is its largest acummulation point. $\endgroup$ – Bunder Jul 7 '13 at 13:42
  • $\begingroup$ but this equality is true only if the events are independent, which in my case are not. Am I right? $\endgroup$ – johan paul Jul 7 '13 at 14:50
  • $\begingroup$ We used the "original" Borel Cantelli which does not need independence of the events. The converse (i.e $\sum P(\ldots) = \infty \Rightarrow P( \ldots i.o) = 1$) requires independence. Check en.wikipedia.org/wiki/Borel%E2%80%93Cantelli_lemma $\endgroup$ – Bunder Jul 7 '13 at 15:01

Important inequalities

Williams - Probability with Martingales

enter image description here

Deduced similarly:

(iii) If $\liminf x_n > z$, then

$ \ \ \ \ \ \ \ (x_n > z)$ eventually (that is, for infinitely many n)

(iv) If $\liminf x_n < z $, then

$ \ \ \ \ \ \ \ (x_n < z)$ infinitely often (that is, for infinitely many n)

By BCL1, we have $P(\limsup (|X_n| > n)) = 0$

$$\to P(\liminf (|X_n| \le n)) = 1$$

$$\to P(\liminf (|X_n|/n \le 1)) = 1$$

$$\to P([\limsup |X_n|/n] \le 1)) = 1$$

The last step follows by the contrapositive of 2 (see above)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.