= Solution
Every $X_n$ has the same <exponential distribution>, so $X_n\xrightarrow d X$ for $X\sim\operatorname{Exp}(1)$. Part (ii) gives $Z_n-X_n\xrightarrow p0$. Applying <Slutsky's theorem> to $Z_n=X_n+(Z_n-X_n)$ yields
$$
\boxed{Z_n\xrightarrow d\operatorname{Exp}(1).}
$$
The limiting <probability distribution> does not depend on $c$. This is precisely the attraction to the <exponential invariant law of a random affine recursion>.
\Image[/past-exam-of-the-probability-course-of-the-university-of-oxford/2017-exponential-coupling.png]
{title=Common random inputs make different initial values approach the same stationary trajectory}
{description=Original simulations of the affine recursion with shared inputs. The right panel compares an exactly propagated path difference with its unconditional expected value; the geometric expectation estimate, rather than the simulation, proves convergence.}
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