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

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.

The two recursions use a synchronous coupling: they receive exactly the same fresh inputs. Subtraction cancels the additive term:The variable $U_n$ is independent of the preceding difference and has expected value $1/2$. Thus expectation of a product of independent random variables givesBy Markov's inequality, for every $\epsilon >0$,This proves the requested convergence in probability; in fact the difference has L1 convergence. It is synchronous contraction of a random affine recursion, and the estimate holds for every real $c$.

Every $X_n$ has the same exponential distribution, so $X_n\stackrel{d}{\to }X$ for $X\sim \mathrm{Exp}(1)$. Part (ii) gives $Z_n-X_n\stackrel{p}{\to }0$. Applying Slutsky's theorem to $Z_n=X_n+(Z_n-X_n)$ yieldsThe limiting probability distribution does not depend on $c$. This is precisely the attraction to the exponential invariant law of a random affine recursion.