Solution
= Solution
Put $W=X+Y$. The independent pair $(U,W)$ has the same <joint distribution> as $(V,W)$ from part (ii), because both first coordinates are uniform and independent of the same sum law. Hence
$$
U(X+Y)\ \stackrel d=\ VW=X.
$$
Consequently
$$
\boxed{U(X+Y)\sim\operatorname{Exp}(1).}
$$
The rate is one. This multiplication identity is the mechanism behind the <exponential invariant law of a random affine recursion>.