is it possible for an i.i.d. series of random variables $(X_n)_{n\in\mathbb N}$, that $$ \limsup_{n\to\infty} \frac{X_n}{n} = 0 \quad a.s.$$ does NOT hold?
Thanks in advance
i.i.d. random variables divided by $n, n \to \infty$
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random-variables
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0@Shalop I don't get it. Clearly $\sup_{n\in\mathbb{N}} \{ \frac{X_n}{n} \} = 1 >0$. But $\limsup_{n\to \infty} \frac{X_n}{n} = \limsup_{n\to \infty}\frac{1}{n} = 0$. Am I mistaken? – 2017-02-13
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0My mistake! If the $X_n$ have a finite first moment, then by SLLN it will indeed be true that $X_n/n = (S_n-S_{n-1})/n \to \mu-\mu=0$. – 2017-02-13
1 Answers
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Yes there is an example: let $X_n$ be i.i.d. with the common pdf, $f(x)=x^{-2}1_{\{x>1\}}$.
Note that $P\big(\frac{X_n}{n}>1\big) = 1/n$. Thus $\sum_n P(X_n/n>1)=\infty$, so by the second borel cantelli lemma, we see that $X_n/n>1$ infinitely often (a.s.). So clearly, $\limsup X_n/n \geq 1$ a.s.