Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 215 / 1 / c / Solution

At $T_{{\tau }_{k-1}}^{(k)}$, part b and the strong-stationarity assumption make $X^{(k)}/2$ uniform on ${\mathbb{Z}}_{2^{k-1}}$ and independent of the stopping index. One further lazy step either stays or moves by one, each parity occurring with probability one half; conditional on the even residue already selected, this chooses uniformly between its two lifts to ${\mathbb{Z}}_{2^k}$. Thushas a uniform terminal state independent of ${\tau }_k$, and is a strong stationary time.