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

For $\widetilde P=(1-a)P+a\Pi$, where every row of $\Pi$ equals the stationary distribution $\pi$,
$$
\widetilde P^t-\Pi=(1-a)^t(P^t-\Pi).
$$
The perturbation superposes gradual exponential mixing on the original profile and can destroy pre-cutoff.


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