# 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.

• "$\leq \sqrt n/y$" in the third line of $F_{Y_n}(y) =$ should be $\leq y/\sqrt n$, but you got it right in the next line, so it looks like a typo. Oct 4, 2016 at 18:27
• What is the relationship between $X_i$ and $Y_i$? Furthermore, your formula $$Y_n = \sqrt{n} \min\{ \sqrt{Y_1}, \sqrt{Y_2}, \ldots, \sqrt{Y_n} \}$$ is self-referential. Oct 4, 2016 at 18:41
• Your evaluation of the limit does not look correct. Note that $$\left(1-\frac{y}{\sqrt{n}}\right)^n=\left(\left(1-\frac{y}{\sqrt{n}}\right)^{\sqrt{n}/y}\right)^{y\sqrt{n}}.$$ Letting $m=\sqrt{n}/y$, the inner part is in the form $(1-1/m)^m$, whose limit as $n\to\infty$ indeed is $\exp(-1)$, but it remains to raise that to the $y\sqrt{n}$ power.
– whuber
Oct 4, 2016 at 19:36

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.