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Assume $X_{1}, X_{2}, X_{3}, ...$ be pairwise independent random variables with $N(0,1)$ distribution. Prove:

$$ P \left( \limsup_{n \to \infty} \frac{X_{n}}{\sqrt{\log n}} = \sqrt{2} \right) = P \left( \liminf_{n \to \infty} \frac{X_{n}}{\sqrt{\log n}} = -\sqrt{2} \right) = 1 $$ and all points in $[-\sqrt{2}, \sqrt{2}]$ are accumulation points.

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  • $\begingroup$ To be honest I have no clue and would really appreciate any hint...By the way this problem is marked as very challenging on my assignment sheet... $\endgroup$
    – mathsstudentTUD
    Nov 14, 2017 at 22:09
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    $\begingroup$ Relevant: math.stackexchange.com/questions/987604/… $\endgroup$
    – Clement C.
    Nov 14, 2017 at 22:13
  • $\begingroup$ I know what $\lim$ is and what $\liminf$ is but what is $\limsup$? $\endgroup$
    – Mr Pie
    Nov 14, 2017 at 22:24
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    $\begingroup$ @user477343 if you know $\lim\inf$, you should know $\lim\sup$. It's the similar definition, but for the upper bound (supremum) instead of lower bound (infimum). $\endgroup$
    – Clement C.
    Nov 15, 2017 at 4:29
  • $\begingroup$ Now I am even more confused. What is meant by my Professor: $$P \left( \limsup_{n \to \infty} \frac{X_{n}}{\sqrt{\log n}} = \sqrt{2} \right) = P \left( \left\{ \omega \in \Omega : \limsup_{n \to \infty} \frac{X_{n}(\omega)}{\sqrt{\log n}} = \sqrt{2} \right\} \right) =1$$ or is it: $$P \left( \limsup_{n \to \infty} \frac{X_{n}}{\sqrt{\log n}} = \sqrt{2} \right) = P \left( \bigcap_{n=1}^{\infty} \bigcup_{ m \geq n }\left\{ \omega \in \Omega : \frac{X_{n}(\omega)}{\sqrt{\log n}} = \sqrt{2} \right\} \right) =1$$ $\endgroup$
    – mathsstudentTUD
    Nov 15, 2017 at 11:09

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