Distribution convergence of $n\Big(\log(n)\Big) X_{\min} /X_{\max}$ I'm figuring out the asymptotic distribution of the following.
$$
n\Big(\log(n)\Big) X_{\min} /X_{\max}
$$
where $X_1, \cdots, X_n \overset{iid}{\sim} \operatorname{Exp}(1)$
I know it converges in distribution to $\operatorname{Exp}(1)$. But it's hard for me to prove it.
I tried :
I tried using Slutsky's theorem, with $n X_{\min} \overset{d}{\to} \operatorname{Exp}(1)$ and $X_{\max} / \log(n) \overset{p}{\to} 1$.
Since $n X_{min}  \sim Exp(1)$ we have $nX_{\min} \overset{d}{\to} \operatorname{Exp}(1)$ and $P\Big(X_{\max} / \log(n) \le y \Big) = \Big(1 - (\frac{1}{n})^y \Big)^n$.
We have $X_{\max} / \log(n) \overset{p}{\to} 1$ only when $y > 1$. Otherwise it converges to 0 (when $0 <y < 1$) or $1/e$ (when $y = 1$).
 A: Let's clear this up.  You basically have the answer.  You already observed that $n X_{min} \sim Exp(1)$.  This follows from the fact that for any $t > 0$ we have
$$
\mathbb{P}(n X_{min} > t ) = \mathbb{P}(X_i > t/n,\; i=1,2,\ldots,n) = \mathbb{P}(X_i > t/n)^n = (e^{-t/n})^n = e^{-t}.
$$
Thus, trivially $n X_{min}\to Exp(1)$.
Similarly, as you mention, for any $y > 0$,
$$
\mathbb{P}(X_{max}/\ln(n) < y) = \left(1 - e^{-y \ln(n)}\right)^n
=\left(1 - n^{-y}\right)^n.
$$
At this point you need a little analysis for the right hand side, which you seem to have done.  It is more transparent to see after taking logs and using the fact that $\ln(1 + x) = x + o(x)$ as $x \to 0$ (e.g. Taylor's Theorem). In fact $-2x \le \ln(1 - x) \le -x$ for small enough $x$.
If $y<1$ then 
$$
\ln\mathbb{P}(X_{max}/\ln(n) < y) = n \ln (1 - n^{-y}) \le -n^{1-y} \to -\infty,
$$
from which it follows that $\mathbb{P}(X_{max}/\ln(n) < y) \to 0$.
If $y > 1$ then eventually (i.e. for $n$ large enough)
$$
\ln\mathbb{P}(X_{max}/\ln(n) < y) = n \ln (1 - n^{-y}) \ge -2 n^{1-y} \to 0,
$$
and so $\mathbb{P}(X_{max}/\ln(n) < y) \to 1$.
Thus, for any $\epsilon > 0$,
$$
\mathbb{P}(1 - \epsilon < X_{max}/\ln(n) < 1 + \epsilon)
 = \mathbb{P}(X_{max}/\ln(n) < 1 + \epsilon)
 - \mathbb{P}(X_{max}/\ln(n) \le 1 - \epsilon)
\to 1,
$$
which is the definition of $X_{max}/\ln(n) \to 1$ in probability.
Finally apply Slutsky's Theorem.
