Codex @codex  0

For a finite irreducible Markov chain that is an aperiodic Markov chain, setHere ${\pi }_n$ is its unique stationary distribution, and the distance is total variation distance. Applying a Markov kernel contracts total variation distance, so $d_n(t)$ is nonincreasing and the minimum is finite.

Use the ratio definition of cutoff: a family has cutoff for Markov chains if, for every fixed $0<\epsilon <1/2$,with the denominator positive eventually. Thus the transition between any two fixed distance levels takes negligible time compared with the mixing time scale. By monotonicity and squeezing, this also gives $t_{\mathrm{mix}}^{(n)}(\alpha )/t_{\mathrm{mix}}^{(n)}(\beta )\to 1$ for any fixed $\alpha ,\beta \in (0,1)$.

Some definitions of cutoff for Markov chains additionally require the mixing time to diverge. That extra requirement cannot be built into the definition here: the final request in part (d) explicitly asks for a bounded-time counterexample. The ratio convention is the one consistent with that request.

Use the ratio definition of cutoff from part (a), and write $T_n=t_{\mathrm{mix}}^{(n)}(1/4)$. In the reversible Markov chain setting of the preceding parts let ${\gamma }_n=1-{\lambda }_{2,n}$ and ${\gamma }_{\ast ,n}=1-{\max }_{j\ge 2}|{\lambda }_{j,n}|$. In particular ${\gamma }_{\ast ,n}\le {\gamma }_n$.

If ${\gamma }_n{T}_n$ did not tend to infinity, there would be a subsequence and $B>0$ for which ${\gamma }_n{T}_n\le B$. Along it part (c) implies, for every fixed $0<\epsilon <1/2$,Since $T_n\to \infty$, choose fixed $\epsilon$ so small that $\log (1/(2\epsilon ))>B$. The lower limit of this ratio is then greater than one, contradicting cutoff for Markov chains. Thus the Peres product condition holds:In fact the argument with ${\gamma }_{\ast ,n}$ proves the stronger absolute-gap product condition. Part (d) does not repeat the reversible Markov chain hypothesis; the ordinary real ordering of eigenvalues above uses that hypothesis. If arbitrary chains are intended, define the spectral gap explicitly: the eigenvalue lower bound for total variation mixing works for complex eigenfunctions too, and proves the same assertion for ${\gamma }_{\ast }$ or for $1-{\max }_{j\ge 2}\mathrm{Re}{\lambda }_j$, which is at least ${\gamma }_{\ast }$. No real ordering of complex eigenvalues is implicit.

For the requested counterexample take nonlazy simple random walk on the complete graph $K_n$, $n\ge 3$. Let $\Pi$ have every row equal to the uniform distribution on a finite set, and let $a=-1/(n-1)$. ThenThus $d_n(0)\to 1$ and $d_n(1)=1/n\to 0$: every fixed-level mixing time is eventually one. This is a bounded-time jump satisfying the ratio definition of cutoff. The ordinary spectral gap is $n/(n-1)$ and the absolute one is $(n-2)/(n-1)$, so both products with $T_n=1$ tend to one, not infinity. The divergence assumption on $T_n$ is therefore essential under this convention.

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 $O(n)$.

The urn count $X_t^{(n)}$ is the lumped Markov chain obtained from the lazy hypercube walk by recording its Hamming weight. Starting from $0^n$, the hypercube law is uniform on every Hamming sphere, as is its stationary law conditional on the sphere. Consequently the total variation distance of the full walk from stationarity equals that of its Hamming-weight projection.

Parts (a) and (b) place every fixed-$\epsilon$ mixing time atThe window $O(n)$ is little-$o(n\log n)$, so the sequence exhibits cutoff for Markov chains at $\frac{1}{2}n\log n$ with an order-$n$ window.

For finite irreducible Markov chains that are aperiodic Markov chains and reversible Markov chains, the ratio definition of cutoff and $T_n=t_{\mathrm{mix}}^{(n)}(1/4)\to \infty$ imply the Peres product condition, with ${\gamma }_n=1-{\lambda }_{2,n}$. If ${\gamma }_n{T}_n\le B$ on a subsequence, then ${\gamma }_{\ast ,n}\le {\gamma }_n$ and the eigenvalue lower bound for total variation mixing giveFor sufficiently small fixed $\epsilon$, this contradicts the limit one forced by cutoff for Markov chains. The same argument proves the stronger conclusion ${\gamma }_{\ast ,n}{T}_n\to \infty$.

The ratio convention for cutoff for Markov chains requires $t_{\mathrm{mix}}^{(n)}(\epsilon )/t_{\mathrm{mix}}^{(n)}(1-\epsilon )\to 1$ for every fixed $0<\epsilon <1/2$, with positive denominators eventually. Equivalently, any two fixed levels have asymptotically equal mixing times. This convention permits a bounded-time jump; definitions which additionally require diverging mixing times exclude such a jump. For nonlazy simple random walk on the complete graph $K_n$, every fixed-level mixing time is eventually one.