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For a Markov process $X$ on the Polish space $\mathscr X$ its transition probability is given by $$ P(x,A) :=\mathsf P_x (X_1\in A) $$ and $X$ is time-reversible if there is a probability measure $\pi$ such that for any $A,B$ in a Borel $\sigma$-algebra of $\mathscr X$ denoted by $\mathscr B(\mathscr X)$ it holds that $$ \int\limits_A P(x,B)\pi(dx) = \int\limits_B P(x,A)\pi(dx). $$ The discrete-time Laplacian $\Delta$ and the gradient $\nabla_{xy}$ are defined as $$ \nabla_{xy}f = f(y)-f(x) $$ and $$ \Delta f(x) = \int\limits_\mathscr Xf(y)P(x,dy)-f(x) = \int\limits_\mathscr X(\nabla_{xy}f)P(x,dy). $$

In the case when $\mathscr X$ is countable, for any measurable $A$ the following Green's formula holds: $$ \int\limits_A\Delta f(x)g(x)\pi(dx) = -\frac12\int\limits_{A}\nabla_{xy}f\cdot\nabla_{xy}g\;\pi(dx)+\int\limits_{A}\int\limits_{A^c}(\nabla_{xy}f)g(x)P(x,dy)\pi(dx). $$ This formula can be generalized for an uncountable state space as well. I wonder though if there is an analogue for the case when $X$ is not time-reverisble.

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  • $\begingroup$ I don't know if this is what you are looking for, but the left-hand side in your last identity is called the Dirichlet form $\mathcal{E}(f,g)$. For a Markov chain on a countable space and with invariant measure $\pi$, (not necessarily reversible), it is always true that $$\mathcal{E}(f,g) = \sum_{x,y} \pi(x) P(x,y) g(x) \nabla_{x,y} f.$$ But this is an easy calculation, so presumably you are aware of it. $\endgroup$
    – Nathanael Berestycki
    Commented Dec 20, 2011 at 22:05
  • $\begingroup$ @Nathanael: but this is only for the case $A=\mathscr X$, isn't it? In that case you obtain an answer just by writing $\Delta f(x)$ through the integral (or the sum in you example) $\endgroup$
    – SBF
    Commented Dec 21, 2011 at 7:06
  • $\begingroup$ @Nathanael: could you refer me to any book (or maybe, lecture notes) on Dirichlet forms for discrete time Markov processes with general state spaces (I guess you call them simply Markov chains)? $\endgroup$
    – SBF
    Commented Jan 12, 2012 at 13:03

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