Show $\sqrt{n}[\log(X_{(1)}) - \log(\alpha)] \overset{p}{\to} 0$ This is a follow-up question to $W_n = \frac{1}{n}\sum\log(X_i) - \log(X_{(1)})$ with Delta method. 
Note: $\log = \ln$. Note also that this is from a past qualifying exam, so I am looking for a solution which can be done under time constraints.
Please only use the following in your solutions: definitions of convergence in probability and distribution, Delta method, Central Limit Theorem, Slutsky's Theorem, Weak Law of Large Numbers, Continuous Mapping Theorem. I am not familiar with big-$O$ notation or being "bounded in probability." For those of you familiar with the literature, this is at the level of Casella and Berger, so any measure-theoretic approaches should not be used.
Suppose $X_1, \dots, X_n \sim \text{Pareto}(\alpha, \beta)$ with $n > \dfrac{2}{\beta}$ are independent. The Pareto$(\alpha, \beta)$ pdf is $$f(x) = \beta\alpha^{\beta}x^{-(\beta +1)}I(x > \alpha)\text{, } \alpha, \beta > 0\text{.}$$

If $X_{(1)}$ is the first order statistic, I wish to show
  $$\sqrt{n}[\log(X_{(1)}) - \log(\alpha)] \overset{p}{\to} 0\text{.}$$

What can I tell you about $X_{(1)}$ at this point of the exam?


*

*$X_{(1)} \overset{p}{\to} \alpha$.

*$n(X_{(1)} - \alpha)$ converges in distribution to an exponential distribution with mean $\alpha/\beta$.

*$F_{X_{(1)}}(x) = \begin{cases}
0, & x \leq \alpha \\
1-\left(\dfrac{\alpha}{x}\right)^{\beta n}, & x > \alpha\text{.}
\end{cases}$

*$\mathbb{E}[X_{(1)}] = \dfrac{\alpha\beta n}{\beta n - 1}$ 

*$\mathbb{E}[X_{(1)}^2] = \dfrac{\alpha^2\beta n}{\beta n - 2}$


One possible approach: let $G$ be the CDF of $\sqrt{n}[X_{(1)}-\alpha]$. Then the support is over $(0, \infty)$ and for $x$ in the support of $G$,
$$\begin{align}
G(x) &= F_{X_{(1)}}\left(\dfrac{x}{\sqrt{n}}+\alpha\right) \\
&= 1 - \left[ \dfrac{x/\sqrt{n}+\alpha}{\alpha}\right]^{-\beta n} \\
&= 1 - \left[ \dfrac{x/\alpha}{\sqrt{n}}+1\right]^{-\beta n} \\
\end{align}$$
From here, we could probably apply the Delta method if I could figure out how to compute the limit as $n \to \infty$. The $\sqrt{n}$ is problematic.
Given the above, I know that $\log(X_{(1)}) \overset{p}{\to} \log(\alpha)$, but I'm not sure if anything can be done with this. Alternative approaches are welcome as well.
 A: Not sure if this is mentioned in the previous post and if you accept this or not. Note that 
$$Y_n = \sqrt{n}(\ln X_{(1)} - \ln\alpha) \sim \text{Exp}(\sqrt{n}\beta)$$
and thus CDF of $Y_n$ is
$$ F_{Y_{n}}(y) = \Pr\{\sqrt{n}(\ln X_{(1)} - \ln\alpha)\leq y\} =
\begin{cases}  1 - e^{-\sqrt{n}\beta y} & \text{when} & y > 0 \\
0 & \text{when} & y \leq 0
\end{cases}$$
So obviously you can show that $Y_n$ converge to $0$ in distribution, and thus in turns you show that it converges in probability.
A: To elaborate on BGM's answer, $X_{(1)} \sim \text{Pareto}(\alpha, \beta n)$, as we can see from its CDF. It follows that $$\ln(X_{(1)})-\ln(\alpha)\sim\text{Exp}(\beta n)$$
as it can be seen from Finding $\mathbb{E}[X]$, $X \sim \text{Pareto}$ under exam conditions.
See Is the family of exponential distributions closed under scaling?. Thus,
$$Y_n = \sqrt{n}[\ln(X_{(1)})-\ln(\alpha)] \sim \text{Exp}(\beta n/\sqrt{n}) = \text{Exp}(\beta\sqrt{n})\text{.}$$
Hence, its CDF is given by 
$$1-e^{-y\beta\sqrt{n}}$$
for $y > 0$.
