Dgdrxrt

I wonder whether there is such kind of relation between the scale of the perturbation and the stationary distribution of a Markov chain. Suppose $hatP=P+FinmathbbR^ntimes n$, where $F$ is the perturbation matrix, each of whose line is sum up to $0$. The corresponding stationary distribution is $pi, hatpi$. Is there exist some $C(n)$, which only relate to $n$, so that
$$sum_i,j=1^n|F_ij|ge C(n)||hatpi-pi||_1$$

I know there is some existing result like $C$ is some kind of condition number. I also find "A Note on Entrywise Perturbation Theory for Markov Chains", which shows that if $|F_ij|leepsilon|P_ij|$, we have $||hatpi-pi||_1le2(n-1)epsilon+mathcalO(epsilon^2)$. But still I can't derive what $C(n)$ is in my question.

I think I answered this here, but I'll translate into the setting of that question: Stationary distribution of convex combination of Markov chains

I believe there is no formula that only depends on $n$.

Here is my reasoning:

Instead of writing $P + F$, write $lambda P + (1 - lambda) Q$, with $F_lambda = ( lambda - 1) P + (1 - lambda)Q$.

Then we have $|F_lambda| leq (|P| + |Q|) ( 1- lambda) = 2n ( 1 - lambda)$, where $|M| = sum |M_ij|$, since $|P| = n$ for any transition Kernel. So
we can make $|F_lambda|$ as small as we want, by choosing the appropriate $lambda$.

So adding $F_lambda$ amounts to mixing with $Q$ by a factor of $lambda$. This is now in the setting of the link, which basically explained the following:

The original $P$ chain may be very lazy, and then adding $F_lambda$ will have a huge impact even for small $lambda$ (and make the stationary distribution immediately look a lot like $pi_Q$), or $Q$ might be extremely lazy, and adding it will have little impact, so the stationary distribution will continue to look like $pi_P$.

In light of this, the condition $|F_ij| leq epsilon |P_ij|$ seems to be ruling out the extremely lazy $P$ case : if $P$ is extremely lazy, then $F$ is basically the identity matrix, and adding it shouldn't change the stationary distribution too much. The inequality also rules out the case that $P + F$ can make transitions that $P$ couldn't: $F$ can't be positive where $P$ is zero.

In general the bound you give seems to be saying that if $Q$ is lazier than $P$, and doesn't introduce new jumps, you can get some control over the stationary distribution. That sounds reasonable.

What do you think?