# Exercise given a sequence of i.i.d. real RV and a distribution function $F$ need to verify three statements for $Z_n=n \min (X_1, \dots , X_n)$

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?

• Why not take the hint and relate $nX_n$ to $Z_n$, then apply the first results?
– whuber
Apr 29, 2016 at 18:43
• 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? Apr 29, 2016 at 18:46
• 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)?
– whuber
Apr 29, 2016 at 18:51
• 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. Apr 29, 2016 at 19:05
• (I meant "nonzero" rather than just "finite"!)
– whuber
Apr 29, 2016 at 19:56

$$\sum_{n} P(nX_n\leq \epsilon)=\sum_n P(X_n\leq \epsilon/n).$$
$$P(X_n\leq \epsilon/n)=F(\epsilon/n)=\frac{F(\epsilon/n)}{\epsilon/n}\frac{\epsilon}{n}\geq (\lambda-c(n))\frac{\epsilon}{n},$$
where you just need to pick $n$ large enough so that $c(n)$ is permanently smaller than $\lambda$.