## Continuous time Markov process

Let $\Sigma=\{1,\cdots, d\}^{\mathbb{N}}$, $B:\Sigma\to\mathbb{R}$ be a Lipschitz potential, and $L_B(f):x\mapsto \sum_{\sigma y=x} e^{B(y)}f(y)$. The potential $B$ is said to be normalized, if $\sum_{\sigma y=x} e^{B(y)}=1$ for all $x\in\Sigma$.

If $B$ is normalized, then its topological pressure $P(\sigma, B)=0$, and its equilibrium state $\mu_B$ is an $L_B^\ast$-invariant Gibbe measure. This induces a Markov process $(X_t)$ with values on the state space $\Sigma$. That is, suppose $X_t=x\in\Sigma$. Then it stays at this state for a while, waits for $T\sim \text{Exp}(1)$ and jumps to a point $X_{t+T}=y\in\sigma^{-1}x$ with probability $e^{B(y)}$. Then $\mu_B$ is a stationary measure for this Markov process.

More generally, we can assign different jump rates (exponential clocks) at different states. That is, let $r:\Sigma\to[c,C]$. Let $(X_t^r)$ be the modified Markov process with clock $r$. That is, suppose $X_t^r=x\in\Sigma$. Then it waits a time $T\sim \text{Exp}(r(x))$ and jumps to a point $X_{t+T}^r=y\in\sigma^{-1}x$ with probability $e^{B(y)}$. Then the naturally related measure is $\mu^r_B:E\mapsto \frac{1}{\mu_B(1/r)}\cdot\int_E \frac{1}{d}d\mu_B$.

Another setting is consider a system of $N$ sites, each carrying an energy $x_i$. Assume the neighboring sites $s_i$ and $s_{i+1}$ exchange energy when an exponential clock $\text{Exp}(\lambda_i(x_i+x_{i+1}))$ rings: $(\hat x_i,\hat x_{i+1})=(a,1-a)(x_i+x_{i+1})$, where $\alpha\sim U([0,1])$.
Now consider a function $f:\Sigma\to\mathbb{R}$, and its evolution $f(X_t)$ with $X_0=x$. Then the generator $L$ is defined by
$Lf:x\mapsto \mathbb{E}_x\lim_{t\to 0+}\frac{f(X_t)}{t} =\sum_{i}\lim_{t\to 0+}\frac{1}{t}\mathbb{P}(T_t(i))\cdot \mathbb{E}_x(f(X_t)-f(x)|T_t(i))$,
where $T_t(i)$ describes the event that only the i-th clock rings during the time $(0,t)$. For independent exponential clocks, we have
$\mathbb{P}(T_t(i))=e^{-\lambda_i t}\lambda_i t\cdot\prod_{j\neq i}e^{-\lambda_j t}$,
and
$\mathbb{E}_x(f(X_t)-f(x)|T_t(i))=\int_I [f(T_{ia}x)-f(x)]\cdot U(da)$. So
$Lf: x\mapsto\sum_i \lambda_i(x_i+x_{i+1})\cdot\int_I [f(T_{ia}x)-f(x)]\cdot U(da)$