Let $\Pi_n(x,y)=\pi_n(y)$. Since $P_n\Pi_n=\Pi_nP_n=\Pi_n$ and $\Pi_n^2=\Pi_n$, the <stationary-reset perturbation of a Markov chain> satisfies
$$
\widetilde P_n^t
=\Pi_n+(1-a_n)^t(P_n^t-\Pi_n).
$$
Every row difference from stationarity is multiplied by the nonnegative scalar $(1-a_n)^t$, so
$$
\boxed{
\lVert\widetilde P_n^t(x,\mathord\cdot)-\pi_n\rVert_{\mathrm{TV}}
=(1-a_n)^t
\lVert P_n^t(x,\mathord\cdot)-\pi_n\rVert_{\mathrm{TV}}}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-215/2/b/i/solution