Questions about the order statistics of uniform distributions I refer to the Simes (1986) paper found here.  In this setting, $P_{(1)}$ through $P_{(n)}$ are the order statistics of $n$ independent Uniform$[0,1]$ random variables and, for $0\le \alpha \le n$,
$$A_n(\alpha) = \Pr\{P_{(j)}\gt j\alpha/n; j = 1, 2, \ldots, n\}.$$
On page 2, I'm unsure of how the integral in the proof is obtained.  It asserts that because (a) the $P_{(j)}/P_{(n)}$ are the order statistics of $n-1$ independent uniform variables and (b) the distribution function of $P_{(n)}$ is $p^n$ ($0\le p \le 1$), then
$$A_n(\alpha) = \int_\alpha^1 A_{n-1}\left(\frac{\alpha(n-1)}{p n}\right) n p^{n-1} dp.$$
Simes mentioned that he thanks the referee for a shorter version of the proof and I'm assuming that it cut off a corner that would lead to my better understanding.
I suppose that my two specific questions are these…
1: Why consider the $P_{(1)}/P_{(n)} ..... P_{(n-1)}/P_{(n)}$ order statistics of $n-1$ independent uniform random variables on $(0,1)$ independent of $P_{(n)}$?
2: How is the $A_{n-1}$ {$\alpha$$(n-1)/np$} obtained in the integral?
 A: Preparing for final exams, I happen to run into the exactly same question.
Let's show $P_{(i)}/P_{(n)}$ is the $i^{th}$ order statistic of $n-1$ samples as well as independent of $P_{(n)}$.
Find the joint pdf of $(P_{(i)},P_{(n)})$, by definition
$$
f_{P_{(i)},P_{(n)}}(p_{(i)},p_{(n)})= \frac{n!}{(i-1)!(n-i-1)!}p_{(i)}^{i-1}(p_{(n)}-p_{(i)})^{n-i-1}
$$
Change variables, $U=P_{(i)}/P_{(n)},V=P_{(n)}$
$$
\begin{aligned}
f_{U,V}(u,v)&=\frac{n!}{(i-1)!(n-i-1)!}(uv)^{i-1}(v-uv)^{n-i-1}det(J)\\
&=\left\{\frac{(n-1)!}{(i-1)!(n-1-i)!}u^{i-1}(1-u)^{n-1-i}\right\}\left\{nv^{n-1}\right\}
\end{aligned}
$$
left part is the pdf of $i^{th}$ order statistics of $(n-1)$ samples, right part is the pdf of $n^{th}$ order statistics. Integral the above pdf, you can derive the marginal distribution. Obviously $U$ and $V$ are independent.
Then, let's show the equation in the paper.
$$
\begin{aligned}
A_n(\alpha) &= Pr\left(\bigcap^n_{j=1}\left\{P_{(j)}>\dfrac{j\alpha}{n}\right\}\right)\\
&=\int_\alpha^1 Pr\left(\dfrac{P_{(1)}}{P_{(n)}}>\dfrac{\alpha}{np},...,\dfrac{P_{(n-1)}}{P_{(n)}}>\dfrac{(n-1)\alpha}{np} \mid P_{(n)} = p\right)f_{P_{(n)}}(p)dp\\
&=\int_\alpha^1 Pr\left(\bigcap^{n-1}_{j=1}\left\{P_{(j)}'>\dfrac{j\alpha'}{(n-1)}\right\}\right)np^{n-1}dp~~~~\alpha' =\dfrac{(n-1)\alpha}{np}\\
&=\int_\alpha^1A_{n-1}\left\{\dfrac{(n-1)\alpha}{np}\right\}np^{n-1}dp\\
&=\int_\alpha^1\left(1-\dfrac{(n-1)\alpha}{np}\right)np^{n-1}dp = 1-\alpha\\
\end{aligned}
$$
