= Solution
Reverse time under the uniform invariant law. The reverse shuffle selects $k$ cards uniformly without replacement and moves them to the top in random order. Track cards once selected. The forward time $\tau$ at which the initially bottom card first lies among the next top $k$ corresponds in reverse to the first time every card has been selected. At that <coupon collector problem>[coupon-collector] time, the order of all marked cards is uniform and independent of the marking time, by induction over the random insertions. Reversing again shows that $X_\tau$ is uniform and independent of $\tau$, so part (a) makes $\tau$ a strong stationary time.
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