Convergence in Distribution - find limiting distribution Question:
$X_1 , X_2 , ... X_n$ are unif(0, 1) random variables and 
$Y_n = {\sqrt n} \min \{{\sqrt X_1}, {\sqrt X_2}, ... {\sqrt X_n}\} $
Consider the sequence $Y_1 , Y_2 , ... Y_n$ 
and give the pdf or pmf of the limiting distribution, if it exists
My attempt:
The Support of $Y_n$ is $(0, {\sqrt n})$  , so for $ y \le 0$ $F_{Y_n}$ $(y) = 0$ and for $y \ge {\sqrt n}$ we have $ F_{Y_n}$ $(y) = 1$.
Then for $y \in (0, {\sqrt n})$ we have: 
$F_{Y_n}$ $(y) = P(Y_n \le y) \\ =P({\sqrt n} \min \{{\sqrt Y_1}, {\sqrt Y_2}, ... {\sqrt Y_n}\} \le y) \\ =P(\min \{{\sqrt Y_1}, {\sqrt Y_2}, ... {\sqrt Y_n}\} \le y/{\sqrt n}) \\ =[1 - (1 - (y/{\sqrt n}))]^n$
which gives you the cdf:
  $F_{Y_n}$ $(y) =
\begin{cases}
1,  & y \ge {\sqrt n} \\
[1 - (1 - (y/{\sqrt n}))]^n, & 0 \lt y\lt {\sqrt n} \\
0, & \ y\le 0
\end{cases}$  
Then: $\lim_{n\to\infty} F_{Y_n} (y) =
\begin{cases}
1 - e^{-y},  & y \gt 0 \\
0, & \ y\le 0
\end{cases}$  
And the pdf is then:
$f_{Y_n} (y) = e^{-y}$ where $y \ge 0$
Is this correct?  Not very confident in what I have done.
 A: It's best to proceed step by step.  Consider $$W_i = g(X_i) = \sqrt{X_i}, \quad i = 1, 2, \ldots.$$  Then $$\Pr[W_i > w] = \Pr[\sqrt{X_i} > w] = \Pr[X_i > w^2] = 1 - w^2, \quad 0 \le w \le 1.$$  Consequently, $$\Pr[Y_n > y] = \Pr[\sqrt{n} W_{(1)} > y] = \Pr[W_{(1)} \ge y/\sqrt{n}] = \prod_{i=1}^n \Pr[W_i > y/\sqrt{n}],$$ where $W_{(1)} = \min\{W_1, \ldots, W_n\}$ is the minimum order statistic of the root-transformed uniform $X_i$s.  Then the rest is straightforward computation:  $$S_{Y_n}(y) = \Pr[Y_n > y] = (1 - y^2/n)^n,$$ hence in the limit as $n \to \infty$, the survival function becomes $$S_{Y_\infty}(y) = e^{-y^2}, \quad y > 0,$$ and the asymptotic density is $$f_{Y_\infty}(y) = 2ye^{-y^2},$$ which implies $$Y_\infty \sim \operatorname{Weibull}(k = 2, \lambda = 1) \sim \operatorname{Rayleigh}(\sigma^2 = 1/2);$$ i.e., Weibull with shape $2$ and scale $1$, or equivalently, Rayleigh with scale $1/\sqrt{2}$.  This result is supported by simulation, using $10^4$ realizations of $Y_{100}$ as shown below.

