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I'm having some trouble with an exercise. Suppose $(X_n)_n$ is a sequence of independent r.v with $\mathbb{P}(X_n > x) = e^{-x}$ for positive x and one otherwise. The exercise first asks for what values of $\alpha>0$ is it almost surely the case the $X_n > \alpha \log \ n$ i.o. which I have managed.

Then it asks if $L= \limsup_n \frac{X_n}{\log \ n}$ show that $\mathbb{P}(L=1)=1$. So far I can show $\mathbb{P}(L \geq 1)=1$ and now want to show $\mathbb{P}(L > 1)=0$. But this part is giving me some problems.

  • 0
    It is natural that you have some trouble: what you want to prove is false (with probability $1$, the r.v. $X_1$ is larger than $\ln (1) = 0$). If you put $L := \limsup_{n \to + \infty} X_n / \ln (n)$, on the other hand...2012-02-09
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    And now, the hint : try to prove that for all $n \geq 1$, one has $\mathbb{P} (L \geq 1 + 1/n) = 0$ (with my definition of $L$).2012-02-09
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    in the definition of the distribution, didn't you mean `and one otherwise`?2012-02-09
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    Sorry about that. It was meant to be limsup and one as you both rightly said.2012-02-09
  • 0
    What does "i.o." stand for?2012-02-09
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    i.o. means "infinitely often", as $n \to \infty$.2012-02-09

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