Asymptotic distribution of the extreme, standardized order statistics of uniform distribution?

Question

Let $\{U_{k, n}\}_{k=1}^n$, denote the order statistics of a sample of $n$ iid uniform $[0, 1]$ variates.

Note that, marginally $U_{k, n}$ is distributed $\mathrm{Beta}(k, n+1 -k)$. Therefore, let us denote the corresponding mean and variance parameters, respectively, by $$ \mu_{k, n} = \frac{k}{n+1} \quad \mbox{and} \quad \sigma^2_{k,n} = \frac{k (n+1 - k)}{(n+1)^2(n+2)}. $$ Let us define the extreme values $$ Y_n = \max_{1 \leq k \leq n} \frac{U_{k, n} - \mu_{k, n}}{\sigma_{k, n}} \quad \mbox{and} \quad Z_n = \min_{1 \leq k \leq n} \frac{U_{k, n} - \mu_{k, n}}{\sigma_{k, n}} $$ Is it possible to derive the asymptotic distributions of $Y_n$ and $Z_n$,marginally, as $n \to \infty$?

Answer 1

A possible approach. Note that $U_{1,n}, \ldots, U_{n,n}$ is jointly uniformly distributed over the set $\left\{\mathbf{x} \in [0, 1]^n: 0 \le x_1 \le x_2 \le \cdots \le x_n \lt 1 \right\}|$ (proof)

Denote $F^n_U$ the joint CDF of $U_{1,n}, \ldots, U_{n,n}$ then we have

$$ F_Y(y) = P(Y_n < y) = P\left(\frac{U_{1, n} - \mu_{1, n}}{\sigma_{1, n}} < y \wedge \ldots \wedge \frac{U_{n, n} - \mu_{n, n}}{\sigma_{n, n}} < y\right) = F_U(y \sigma_{1,n} + \mu_{1,n}, \ldots, y \sigma_{n,n} + \mu_{n,n}) = \frac{|\left\{\mathbf{x} \in [0, 1]^n: 0 \le x_1 \le x_2 \le \cdots \le x_n \lt 1 \wedge x_1 \leq y \sigma_{1,n} + \mu_{1,n} \wedge \ldots \wedge x_n \leq y \sigma_{n,n} + \mu_{n,n} \right\}|}{|\left\{\mathbf{x} \in [0, 1]^n: 0 \le x_1 \le x_2 \le \cdots \le x_n \lt 1 \right\}|} $$

So we have transformed the problem into computing volumes of two bounded convex polytopes.

The denominator volume is easy - it is the probability of a draw of $n$ independent uniform distributions to be ordered which should be $\frac{1}{n!}$.

The numerator volume is a bit more annoying - lets denote $a_i = \min{1, y \sigma_{i,n} + \mu_{i,n}}$.

For arbitrary bounds $a_i$ where $0 \leq a_1 \leq \ldots \leq a_n \leq 1$ (which our choice satisfies) we have the volume as

$$ \int_0^{a_1}\int_{x_1}^{a_2}\int_{x_2}^{a_3}\ldots \int_{x_{n - 2}}^{a_{n - 1}} (a_n - x_{n-1}) \text{d}x_{n - 1} \ldots \text{d}x_1 $$

Now I didn't do this very rigorously, but it appears that this resolves to:

$$ \sum_{\mathbf{r} \in R, \sum{r_i} = n} (-1)^{\text{alternating as exponents change}} \frac{1}{\prod r_i!} \prod_{i=1}^n a_i^{r_i} $$

Where the indexing set $R$ is

$$ R = \{(r_1, \ldots, r_n) \in \mathbb{N_0}^n | \sum_{i=1}^n r_i = n \wedge r_i \leq n - i + 1 \wedge \text{some extra condition} \} $$

Sorry for the missing pieces, I'll try to edit this once I figure those out.

Hope that helps you move forward and that I didn't mess up completely.