Markov chain semigroup: action on densities

Question

I have some trouble working with continuous time Markov chains considering densities rather than distributions.

SETTING:

we have a continuous time Markov chain with finite state space. The generator is $Q= Q(x,y)$ and correspondingly the transition rate matrices are given by $$ P_t = e^{tQ}. $$ Call $\pi$ the invariant measure. Assume reversibility, i.e. $\pi(x)P_t(x,y) = \pi(y)P_t(y,x)$ for all $x,y$ in the state space and $t\geq 0$. Then I know that for any starting distribution $\mu_0$ we have that (with the convention that distributions are row vectors) $$ \mu_t := \mu_0 P_t $$ is the distribution of the Markov chain at time $t$.

WHAT I DON'T UNDERSTAND

I read that, if we work with densities with respect of $\pi$ instead of probability distributions, and so $\rho_0$ is the density of the starting distribution (as a column vector now), than $$ \rho_t = P_t \rho_0 $$ is the density of the Markov chain distribution at time $t$. However it is not clear to me:

1. Why we can work with densities instead of distributions?
2. Whether this works only with densities with respect to the invariant measure or also with respect to other measures.
3. Why we can change the order in matrix multiplication: for measures we had $\mu_0 P_t$ and for densities we have $P_t \rho_0$ and I would have expected a transpose in the second case intuitively. (I think this might have with reversibility but cannot show it).

Thanks!

Answer 1

1. If the invariant measure $\pi$ has full support, then there is a one-to-one correspondence between probability measures and probability densities given by $\mu\mapsto \frac{\mu}{\pi}$.
2. This inequality holds for the densities with respect to a probability measure $\sigma$ only if $\sigma$ is invariant. Indeed, if $\rho_0=1$, then $\rho_t=1$ and therefore $\sigma_t=\sigma$. Differentiating this at $t=0$ gives $\sigma=\sigma Q$.
3. This is indeed a consequence of the reversibility: $$ \rho_t(x)=\frac{\mu_t(x)}{\pi(x)}=\frac 1{\pi(x)}\sum_y\mu_0(y)P_t(y,x)=\frac 1{\pi(y)}\sum_y P_t(x,y)\mu_0(y)=\sum_y P_t(x,y)\rho_0(y). $$