Write $r_n=t_{\mathrm{mix}}^{(n)}/t_{\mathrm{rel}}^{(n)}\to\infty$. Then
$$
a_nt_{\mathrm{mix}}^{(n)}=\sqrt{r_n}\to\infty,
\qquad
a_n^{-1}=\sqrt{t_{\mathrm{rel}}^{(n)}t_{\mathrm{mix}}^{(n)}}=o(t_{\mathrm{mix}}^{(n)}).
$$
By cutoff, at every fixed multiple $c/a_n$ the original chain is still asymptotically unmixed, while part i gives
$$
\widetilde d_n(c/a_n)\longrightarrow e^{-c}.
$$
Thus the new chain crosses between fixed distance levels gradually on the full scale $a_n^{-1}$: for example its $\varepsilon$-mixing times are asymptotic to $a_n^{-1}\log(1/\varepsilon)$. No two fixed multiples of one scale can make the limiting distance respectively one and zero, so the family has no pre-cutoff.


Solution

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

 Solution

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