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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.

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Hint: Borel-Cantelli lemma shows that $\Bbb{P}(|X_{n}| > n \text{ i.o.}) = 0$. – Sangchul Lee Jul 7 '13 at 13:14
Crossposted: – cardinal Jul 7 '13 at 15:24
@cardinal so what? i can not post my question in two different sections? – johan paul Jul 7 '13 at 15:33
@johan: See this meta.SO answer. This is the quasi official policy on this topic. Cheers. – cardinal Jul 7 '13 at 15:36
@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.) – 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 $$\mathbb P\left(\limsup_{n\to\infty} \frac{|X_n|}n\leqslant 1 \right)=1. $$

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How do you know the first equation? From the contrapositive of the direct result of BCL1? Anyway answer looks good \m/ – 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.

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Thank you very much. May I ask you something more? How can we pass from the probability to the limit suprermum of the inequality? – johan paul Jul 7 '13 at 13:29
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. – Bunder Jul 7 '13 at 13:42
but this equality is true only if the events are independent, which in my case are not. Am I right? – johan paul Jul 7 '13 at 14:50
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 – Bunder Jul 7 '13 at 15:01

Important inequalities (Probability w/ Martingales):

1, 2

enter image description here

  1. $$\liminf x_n > z \to \liminf(x_n > z)$$

  2. $$\liminf x_n < z \to \limsup(x_n < z)$$

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)

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