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Start the random-to-top card shuffle from a fixed ordering and let $T$ be the first time every card has been selected. Once a card has been selected, its location relative to the other selected cards is determined by their most recent selection times. At $T$ these times have a uniformly random strict order, so the deck is uniform and independent of the event $\{T=t\}$. Thus $T$ is a strong stationary time, andThe coupon collector problem gives $T=n\log n+O_{\mathbb{P}}(n)$, so for $c\to \infty$,

Before all cards have been selected, the unselected cards form the bottom block of the deck in their original relative order. At time $n\log n-cn$, with $c\to \infty$ sufficiently slowly, the number $U$ of unselected cards tends to infinity in probability. Choose $k\to \infty$ with $\mathbb{P}(U\ge k)\to 1$. The bottom $k$ cards are then in their original relative order, whereas under the uniform distribution this has probability $1/k!$. HenceThe random-to-top shuffle therefore has cutoff for Markov chains at $n\log n$ with window $O(n)$.

The top-to-random card shuffle is the time reversal of the random-to-top card shuffle under the uniform stationary distribution. Equivalently, their step distributions on the symmetric group are carried into one another by permutation inversion, which preserves total variation from uniform. Their mixing profiles agree, so top-to-random has the same cutoff at $n\log n$ with window $O(n)$.

Track on the lifted integer line the coordinate refreshed at each step. It is a random walk $R_t$ with right-step probability $2/3$ and left-step probability $1/3$. Every visited residue modulo $n$ has had its bit replaced by an independent fair bit. Consequently the first time all residues have been visited is a strong stationary time for this shift-refresh chain on a hypercube.

For the upper bound, reaching level $n$ visits every residue, so the refresh time is at most ${\tau }_n$. The supplied moments giveFor any sequence $w_n$ with $w_n/\sqrt{n}\to \infty$, Chebyshev inequality and the strong-stationary-time bound imply

For the lower bound, choose integers $b_n$ such thatand consider ${\tau }_{n-2b_n}\wedge {\tau }_{-b_n}$. Its supplied mean is $3n-6b_n+o(1)$ and its variance is $O(n)$; after replacing $w_n$ by a sequence with $\sqrt{n}\ll w_n\ll n$, take for example $b_n=\lfloor \sqrt{w_n\sqrt{n}}\rfloor$. This satisfies the displayed scale separation, and the supplied moment bounds showBefore that exit, the visited interval has length less than $n-b_n$, so at least $b_n$ coordinates retain their initial values.

Start the chain from the all-zero vector. Conditional on the explored path, the number of one-bits is binomial with at most $n-b_n$ trials, whereas under stationarity it is $\mathrm{Bin}(n,1/2)$. Since $b_n/\sqrt{n}\to \infty$, standard binomial concentration separates these two laws; for instance the eventhas probability tending to one for the chain and to zero under $\pi$. ThereforeThe transition from one to zero occurs in an $o(n)$ window around $3n$, proving total-variation cutoff with cutoff time $3n$.

A stopping time $T$ is a strong stationary time iffor every $x,y,t$. Thus $X_T\sim \pi$ and is independent of $T$. The separation distance isFor any strong stationary time,Hence $1-P^t(x,y)/\pi (y)\le {\mathbb{P}}_x(T>t)$, and maximizing proves the claim.

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.

A randomized stopping time $\tau$ is a strong stationary time whenfor every initial state $x$, state $y$, and time $t$. Equivalently, $X_{\tau }$ has distribution $\pi$ and is independent of $\tau$.

The random-to-top shuffle chooses one uniformly random card and moves it to the top of the deck. The time at which every card has been chosen is a strong stationary time, yielding cutoff at $n\log n$ with a window of order $n$.

For a finite Markov chain with stationary distribution $\pi$, the separation distance from an initial state $x$ isIf $\tau$ is a strong stationary time, then $s_x(t)\le {\mathbb{P}}_x(\tau >t)$, and total variation distance is at most separation distance.

A shift-refresh chain cyclically moves a pointer through the coordinates of a bit vector and replaces the visited bit by an independent fair bit. Once the pointer walk has visited every residue, all coordinates have been refreshed, giving a strong stationary time.