Past exam of the probability course of the University of Oxford / 2017 / 1 / b / i / Solution

The answer isIt holds initially by construction. If it holds at step $n-1$, then $X_{n-1}$ is a function of $E_0,\dots ,E_{n-1},U_1,\dots ,U_{n-1}$, so it is independent of the fresh inputs $E_n,U_n$. Part (a)(iii) therefore gives $U_n(X_{n-1}+E_n)\sim \mathrm{Exp}(1)$. This completes the mathematical induction and identifies the stationary distribution of the recursion.