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},
\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)

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Please log in using one of these methods to post your comment: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: