How the perturbation of a Markov chain affect the stationary distribution

Question

I wonder whether there is such kind of relation between the scale of the perturbation and the stationary distribution of a Markov chain. Suppose $\hat{P}=P+F\in\mathbb{R}^{n\times n}$, where $F$ is the perturbation matrix, each of whose line is sum up to $0$. The corresponding stationary distribution is $\pi, \hat{\pi}$. Is there exist some $C(n)$, which only relate to $n$, so that $$\sum_{i,j=1}^n|F_{ij}|\ge C(n)||\hat{\pi}-\pi||_1$$

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 $|F_{ij}|\le\epsilon|P_{ij}|$, we have $||\hat{\pi}-\pi||_1\le2(n-1)\epsilon+\mathcal{O}(\epsilon^2)$. But still I can't derive what $C(n)$ is in my question. I also start doubting the existence of this C. Is there any counter example that can prove that this $C$ does not exist?

Answer 1

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?