Let $U_{1}, \, ... \, ,U_{n}$ be a random sample of uniform random variables $U_i \sim \mathrm{Uniform}(0,1)$. Let $U_{(1)}, \, ... \, , U_{(n)}$ be the order statistics of the sample. The goal is to prove that:
$$ W = U_{(s)}-U_{(r)} \sim \textrm{Beta}(s-r, \, n - s + r +1) \qquad 1 \leq r < s \leq n $$
My "proof"
The joint pdf of $U_{(r)}, U_{(s)}$ is:
$$ f_{U_{(r)}, U_{(s)}} (u, v) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} u^{r-1} (v-u)^{s-1-r} (1 -v)^{n-s} \cdot \textbf{1}_{\{u < v\}} $$
Consider the transformation:
$$ \begin{Bmatrix} W = U_{(s)} - U{(r)} & U_{(r)} = Z \\ Z = U_{(r)} & U_{(s)} = W + Z \\ \end{Bmatrix} $$
The absolute value of the Jacobian determinant is 1. Therefore, the joint pdf of the transformation is:
$$ f_{W,Z}(w,z) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} z^{r-1} w^{s-1-r} (1 - w - z)^{n-s} $$
$$ f_W(w) = \int_{S} f_{W,Z}(w,z) \, \textrm{d}z $$
For fixed $w$, we have $S = \{z \, : \, 0 \leq z \leq 1 - w\}$. Thus:
$$ f_W(w) = \frac{n!}{(r-1)!(s-1-r)!(n-s)!} w^{s-1-r} \int_{0}^{1-w} z^{r-1}(1-w-z)^{n-s} \, \textrm{d} z $$
The problem is that I do not know how to evaluate the integral, but Wolfram Alpha does:
$$ \int_{0}^{1-w} z^{r-1}(1-w-z)^{n-s} \, \textrm{d} z= \frac{\Gamma(r)\Gamma(n-s+1)}{\Gamma(n+r-s+1)} (1- w)^{n+r-s} $$
Now it is really easy to figure out the distribution of $W$. Indeed, the only thing that has to be done is to rewrite the factorials using gamma functions.
Initially I was going to ask how to evaluate the integral without the help of Wolfram Alpha, but then I realized that maybe there is a "better" proof that avoids it. I have tried to find an alternative proof by using multiple transformations which involved $W$ and another transformation, but it was useless.
Edit
I have corrected some mistakes I think I made:
1) The joint pdf I wrote was incorrect.
2) I wrote that I the pdf of $W$ could be found as a convolution, but $U_{(r)}, U_{(s)}$ are clearly dependent. What I actually did was an implicit change of variables.
