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 $At_n$ and to zero at $Bt_n$. For irreducible reversible chains, a standard necessary condition for pre-cutoff is the product condition
$$
t_{\mathrm{rel}}^{(n)}=o(t_{\mathrm{mix}}^{(n)}).
$$
The 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.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 215 / 2 / a / Solution

A sequence of chains exhibits pre-cutoff if there are constants $0<A<B<\infty$ and times $t_n$ such that the worst-case distance tends to one at $At_n$ and to zero at $Bt_n$. For reversible chains, the product condition $t_{\mathrm{rel}}=o(t_{\mathrm{mix}})$ is necessary for pre-cutoff.

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 Incoming links: Pre-cutoff for Markov chains