Solution
= Solution
For $0<x,y<1$, set $a=x^{-1/n}-1$ and $b=y^{-1/n}-1$. Conditional on $S$,
$$
\mathbb P(X_1\le x,X_2\le y\mid S)=e^{-(a+b)S}.
$$
Using part (a),
$$
F_{X_1,X_2}(x,y)=(x^{-1/n}+y^{-1/n}-1)^{-n}.
$$
<Differentiation> gives
$$
f_{X_1,X_2}(x,y)=\frac{n+1}{n}\frac{x^{-1-1/n}y^{-1-1/n}}{(x^{-1/n}+y^{-1/n}-1)^{n+2}}.
$$
Putting $y=1$ yields $F_{X_1}(x)=x$, so $X_1$ is uniform on $(0,1)$.