We leet $X, X_1, X_2,...$ be iid with $\mathbb{P}(X>t) = e^{-t}$ for $t >0$. Let $S_n = X_1+\cdots+ X_n$. Now Let $U, V$ be independent uniform on $[0, 1]$ Let $A = \min(U, V)$ and $B = \max(U, V)$. Show that \begin{equation} (A, B) \text{ and } \left(\frac{X_1}{S_3}, \frac{X_1+X_2}{S_3}\right) \end{equation} have the same joint distribution.
My approach is that we have to verify for any constants $a, b \in [0, 1]$,
\begin{equation} P(A\leq a, B \leq b) = P\left(\frac{X_1}{S_3} \leq a, \frac{X_1+X_2}{S_3}\leq b\right) \end{equation}
Is there a simpler way to do this?
The joint density of $(S_1,S_2,S_3)$ is $$f(x,y,z)=e^{-x}e^{-(y-x)}e^{-(z-y)}=e^{-z}$$ for $0<x<y<z$ and zero elsewhere. Integrating $dx dy$ we find the density of $S_3$ is $f_{S_3}(z)=z^2e^{-z}/2$ for $z>0$ (this is a $\Gamma(3,1)$ distribution, see [1]). Thus the conditional density of $(S_1, S_2)$ given $S_3$ is the density ratio $2/z^2$ for $0<x<y<z$. In other words, given $S_3=z$, the pair $(S_1,S_2)$ is uniformly distributed on the triangle $\{(x,y): 0<x<y<z \}$. It follows that, given $S_3=z$, the pair $(S_1/S_3,\, S_2/S_3)$ is uniformly distributed on the triangle $\{(x,y): 0<x<y<1\}$, with constant conditional density 2. Since this distribution does not depend on $z$, it is also the (unconditional) joint distribution of $(S_1/S_3,\, S_2/S_3)$. Clearly, this is also the joint distribution of $(A,B)$. See also Theorem 6.6(d) in [2].
[1] https://en.wikipedia.org/wiki/Gamma_distribution
[2] https://www.stat.purdue.edu/~dasgupta/orderstats.pdf