Solution
= Solution
The answer is
$$
\boxed{X_n\sim\operatorname{Exp}(1)\quad\text{for every }n\geq0.}
$$
It holds initially by construction. If it holds at step $n-1$, then $X_{n-1}$ is a function of $E_0,\ldots,E_{n-1},U_1,\ldots,U_{n-1}$, so it is independent of the fresh inputs $E_n,U_n$. Part (a)(iii) therefore gives $U_n(X_{n-1}+E_n)\sim\operatorname{Exp}(1)$. This completes the <mathematical induction> and identifies the <stationary distribution> of the recursion.