Pardon the title, I couldn't think of a better one. The three statements below are to prepare for a tougher exercise (which I could solve) however I am lost on 1 of the statements.

Problem: Let $(X_n)$ be a sequence of i.i.d. real RV with distribution function $F$ such that $F(t)/t \to \lambda>0$ as $t \to 0^+$. Define $Z_n:= n \min (X_1, \dots , X_n)$ and verify

1) For every $n \in \mathbb{N}, \ Z_n > 0$ almost surely
2) For every $t>0$, $P(Z_n >t) \to e^{- \lambda t}$ for $n \to \infty$
3) For every $\epsilon >0$ there is $n X_n \leq \epsilon$ inifinitely often, almost surely

I managed to verify the first two. If you want to see my arguments to show my effort I will gladly write it down here. However they are really easy statements.

3) I am stuck here, the exercises cries Borel-Cantelli lemma. But I don't see how to apply it since I cannot find convergent/divergent bounds for $$P(nX_n \leq \epsilon) $$

Maybe someone can provide me with a hint to get me unstuck?

  • $\begingroup$ Why not take the hint and relate $nX_n$ to $Z_n$, then apply the first results? $\endgroup$
    – whuber
    Apr 29, 2016 at 18:43
  • $\begingroup$ I was indeed looking for a correlation between $nX_n$ and $Z_n$ but the only one I could have found is that $nX_n$ (as an event) is obviously contained in $Z_n$ henceforth $P(nX_n \leq \epsilon) \leq P(Z_n \leq \epsilon ) $ do you mean something along these lines? $\endgroup$
    – Spaced
    Apr 29, 2016 at 18:46
  • 1
    $\begingroup$ Yes, but it looks a little trickier than that. One way to get a handle on it might be to contemplate what properties $Z_n$ would have if you assume there is a finite chance that $nX_n \le \epsilon$ only finitely often: could that contradict result (2)? $\endgroup$
    – whuber
    Apr 29, 2016 at 18:51
  • $\begingroup$ Thanks a lot, I will think about that. Mind if I ask you what does "finite chance" mean? I am sure I misunderstand this word, but if you mean the probability of an event, I can't see how this ever can be something else than between $0$ and $1$, i.e. finite. $\endgroup$
    – Spaced
    Apr 29, 2016 at 19:05
  • $\begingroup$ (I meant "nonzero" rather than just "finite"!) $\endgroup$
    – whuber
    Apr 29, 2016 at 19:56

1 Answer 1


Borel Cantelli shows that the probability of events happening infinitely often is 0. You want to use the converse which says that if your events are independent, and the sum diverges then the occur infinitely often with probability 1.

$$\sum_{n} P(nX_n\leq \epsilon)=\sum_n P(X_n\leq \epsilon/n).$$

Now use the fact that

$$P(X_n\leq \epsilon/n)=F(\epsilon/n)=\frac{F(\epsilon/n)}{\epsilon/n}\frac{\epsilon}{n}\geq (\lambda-c(n))\frac{\epsilon}{n},$$

where you just need to pick $n$ large enough so that $c(n)$ is permanently smaller than $\lambda$.


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