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A family exhibits pre-cutoff for Markov chains if there are constants $0<A<B<\infty$ and times $t_n$ such that its worst-case total-variation distance tends to one at $A{t}_n$ and to zero at $B{t}_n$. For irreducible reversible chains, a standard necessary condition for pre-cutoff is the product conditionThe hypothesis that $t_{\mathrm{mix}}^{(n)}/t_{\mathrm{rel}}^{(n)}$ is bounded contradicts this condition, while $t_{\mathrm{mix}}^{(n)}\to \infty$ excludes a bounded-time degeneracy. Hence the family cannot exhibit pre-cutoff.

A sequence of finite irreducible Markov chains exhibits pre-cutoff if there are constants $0<A<B<\infty$ and times $t_n$ such that its worst-case total variation distance satisfiesThis only bounds the transition between two fixed multiples of a scale; cutoff requires the transition window to be negligible relative to the mixing scale.

For reversible aperiodic chains with $t_n\to \infty$, pre-cutoff implies the product condition $t_{\mathrm{rel},n}=o(t_{\mathrm{mix},n}(1/4))$. In fact the analogous Absolute relaxation time must be negligible too. To see the mechanism, the two distance limits make $t_{\mathrm{mix},n}(1/4)$ comparable to $t_n$. If $r_n$ is the largest nontrivial eigenvalue modulus, the Eigenvalue lower bound for total variation mixing gives $d_n(t)\ge r_n^t/2$. The limit at $B{t}_n$ forces $r_n^{\lfloor B{t}_n\rfloor }\to 0$, which implies $(1-r_n)t_n\to \infty$ when $t_n\to \infty$. The ordinary spectral gap is at least $1-r_n$, giving the ordinary product condition.

The diverging-scale hypothesis matters. Nonlazy simple random walk on the complete graph has $d_n(0)\to 1$ and $d_n(1)=1/n\to 0$, so the displayed pre-cutoff convention holds on the bounded scale $t_n=1$, but its Relaxation time is not negligible relative to its mixing time.

Let $P$ be a transition matrix with Stationary distribution $\pi$, and let $\Pi$ have every row equal to $\pi$. For $0\le a\le 1$, the stationary-reset perturbationfollows the original transition with Probability $1-a$ and independently resamples from stationarity with Probability $a$. Since $P\Pi =\Pi P=\Pi$ and ${\Pi }^2=\Pi$,The total variation identity is exact: subtracting stationarity multiplies every signed row difference by the same nonnegative factor. Once the first reset occurs the distribution remains stationary, so the factor also has a direct no-reset Probability interpretation.

This can destroy pre-cutoff. Take the dimension-$m$ Shift-refresh chain on a hypercube, whose distance is $d_m(t)=1-2^{t-m}$ for $t<m$ and zero for $t\ge m$. Use reset Probability $a_m=c/m$ for fixed $c>0$ and $m>c$. For $0<s<1$, ${\tilde{d}}_m(\lfloor sm\rfloor )\to e^{-cs}$, while at $s\ge 1$ the distance is zero. A positive fraction of the mixing scale therefore already causes a nonvanishing decrease from one. Any scale capable of making the later distance tend to zero must be at least order $m$, and its earlier fixed positive multiple cannot have distance tending to one. Thus this perturbed family lacks pre-cutoff even though the original has a sharp cutoff.