= Exponential invariant law of a random affine recursion
{title2=$X_n=U_n(X_{n-1}+E_n),\quad X_\infty\sim\operatorname{Exp}(\lambda)$}
Let $U_n$ have the <uniform distribution> on $[0,1]$ and let $E_n$ have an <exponential distribution> of rate $\lambda>0$, with all these inputs independent. The <random affine recursion> $X_n=U_n(X_{n-1}+E_n)$ has an invariant exponential law of the same rate. Indeed, for independent exponential $X,E$, <beta-gamma independence> says that $V=X/(X+E)$ is uniform and independent of $W=X+E$. Thus an independent uniform $U$ gives $UW$ the same law as $VW=X$.
For any deterministic real initial value $c$, couple the recursion to a stationary copy. <Synchronous contraction of a random affine recursion> gives an expected difference $2^{-n}\mathbb E|c-X_0|\leq2^{-n}(|c|+\lambda^{-1})$. Hence the one-time laws converge to $\operatorname{Exp}(\lambda)$, independently of $c$. An invariant initialization keeps every one-time law unchanged while the chain continues to fluctuate.
\Image[/past-exam-of-the-probability-course-of-the-university-of-oxford/2017-exponential-coupling.png]
{title=Shared random inputs contract the affine recursion toward its invariant exponential law}
{description=Original calculations and simulations. The accompanying proof establishes the limiting result; finitely many simulated paths illustrate it.}
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