Choose a shortest path $x=x_0,\ldots,x_m=y$. Glue the prescribed one-step edge couplings successively. The <triangle inequality> for the Wasserstein transportation metric gives
$$
\rho_K(P(x,\cdot),P(y,\cdot))
\leq\sum_{j=1}^me^{-\alpha}\rho(x_{j-1},x_j)
=e^{-\alpha}\rho(x,y).
$$
Now couple initial states $(U,V)$ optimally for $\mu,\nu$ and conditionally use these one-step couplings. Taking expectations and then the infimum gives
$$
\rho_K(\mu P,\nu P)\leq e^{-\alpha}\rho_K(\mu,\nu).
$$
Iteration with $\nu=\pi$ yields $\rho_K(P^t(x,\cdot),\pi)\leq e^{-\alpha t}\operatorname{diam}(V)$. Since this metric dominates total variation, it is at most $\varepsilon$ once
$$
t\geq\frac1\alpha\left(\log\operatorname{diam}(V)+\log(1/\varepsilon)\right).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 215 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-215/3/a/solution