Let $D=\max_{x,y}\rho(x,y)$ and let $\pi$ be the invariant distribution. Iterating the assumed <Wasserstein contraction> gives
$$
\rho_K(P^t(x,\cdot),\pi)
=\rho_K(P^t(x,\cdot),\pi P^t)
\leq e^{-\alpha t}\rho_K(\delta_x,\pi)
\leq e^{-\alpha t}D.
$$
Since $\rho(x,y)\geq\mathbf1_{\{x\ne y\}}$, the <coupling characterization of total variation distance> implies
$$
\lVert P^t(x,\cdot)-\pi\rVert_{\mathrm{TV}}
\leq\rho_K(P^t(x,\cdot),\pi)
\leq e^{-\alpha t}D.
$$
The right side is at most $\varepsilon$ when
$$
t\geq\frac1\alpha(\log D-\log\varepsilon),
$$
which proves the claimed mixing bound.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 215 / 4 / a / ii / Solution

The total variation distance is the least possible mismatch probability over all couplings:
$$
\lVert\mu-\nu\rVert_{\mathrm{TV}}=\inf\mathbb P(X\ne Y).
$$


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 Incoming links: Coupling characterization of total variation distance