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The problem is to show that $P(|X_n| \ge c \sqrt{\ln n}\text{ i.o.}) = 0$ for standard normal $X_n$ that are not necessarily independent. Also, identify the smallest such $c$.

I am thinking that the Borel-Cantelli lemmas are the way to go here but I can't figure out how to get a bound on this probability. Chebychev's inequality does not work. I suppose I will need to use Markov's inequality and the 4th moment. But even in that case how would I find the smallest $c$? Any suggestions?

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    Borel-Cantelli indeed. Hint: when $x\to+\infty$, $1-\Phi(x)\approx\exp(-x^2/2)$, in a loose sense which can be made precise. // But, if every $X_n$ is **standard** normal (please make clear whether this hypothesis holds), there is no *smallest* $c$ such that this holds.2012-12-12
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    Does "i.o." mean "infinitely often"?2012-12-12
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    Actually when my question was edited the "ln" under the sqrt was removed but it should be there.2012-12-12
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    Then you need to find a simple equivalent of $P(|X|\geqslant c\sqrt{\log n})$ when $n\to\infty$, for $X$ standard normal. Did you reach that?2012-12-13
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    Any luck with the hints above?2012-12-16

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