Let $P$ be an irreducible Markov matrix, and $\pi$ its stationary distribution. Let $D$ be a perturbation matrix which is zero except for two entries in row $r$: $$D_{rg}=+1 \qquad D_{r\ell}=-1.$$ Let $\widetilde{P}(\delta)=P+\delta D$ and let $\widetilde{\pi}(\delta)$ be its stationary distribution. The matrix $\widetilde{P}(\delta)$ is a perturbed version of $P$ where transitions from $r$ happen a little more to the gaining state $g$ and a little less to the losing state $\ell$.

In [1], Conlisk derived, based on identities in [2], the following remarkable formula for the effect on the stationary probability of the gaining state: $$ \frac{d}{d\delta} \widetilde{\pi}_g(\delta) = \pi_r\frac{m_{\ell g}}{m_{gg}}.$$ Here $M$ is the mean first-passage time matrix, with $m_{ij}$ defined to be the expected number of steps, when the chain starts at $i$, that it takes to first visit $j$ after that.

Conlisk's proof [1], and Schweitzer's calculations that underlie it, use the fundamental matrix of the chain. **It seems that such a simple formula should have a fairly simple, and probabilistic, proof.** Is there one out there?

[1] Conlisk, J., 1985. Comparative statics for Markov chains. Journal of Economic Dynamics and Control, 9(2), pp.139-151.

[2] Schweitzer, P.J., 1968. Perturbation theory and finite Markov chains. Journal of Applied Probability, 5(2), pp.401-413.