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Assume $V$ is a countable state space and $L:V^2 \to \mathbb R$ the infinitesimal generator, and $\mu$ the initial distribution. Moreover, $(X_t)_{t \ge 0}$ is the associated continuous Markov chain on the probability space $(\Omega, \mathcal{G}, \mathbb{P})$. Given $\omega \in \Omega$, we define a sequence of random jump times $(\sigma_n)$ recursively as follows:

First, let $\sigma_0 := 0$. Second, let $i := X_{\sigma_n} (\omega) \in V$ and $L(i) := - L(i,i)$. Notice that $X_{\sigma_n} (\omega) := X_{\sigma_n (\omega)} (\omega)$. Then the time until transition out of state $i$ is $\sigma_{n+1} -\sigma_{n} \sim \operatorname{Exp}(L(i))$.

  • If $L(i) = 0$, then $\sigma_{n+1} -\sigma_{n} = +\infty$ a.s. Hence, $\sigma_{n+1} = +\infty$ a.s. and thus $i$ is an absorbing state. It follows that $X_t (\omega) = i$ for all $t \in[ \sigma_n (\omega), +\infty)$ and that $\sigma_{m} = +\infty$ a.s. for all $m \ge n+1$.

  • If $L(i) > 0$, then $i$ is not an absorbing state. It follows that $X_t (\omega) = i$ for all $t \in [\sigma_n (\omega),\sigma_{n+1} (\omega))$. In this case, the chain jumps to a new state, i.e., $X_{\sigma_{n+1}} (\omega) \neq i$.

Let $\sigma = \lim_{n \to \infty} \sigma_n$. If $\sigma (\omega) = +\infty$, then we know $X_t$ for all $t \ge 0$.

IMHO, $(V,L,\mu)$ completely determines $(X_t)_{t \ge 0}$.

In case $\sigma (\omega) < \infty$, how to recover $X_t(\omega)$ for $t \ge \sigma (\omega)$?

Akira
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1 Answers1

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If $\sigma<\infty$ then you have "finite time blowup" and the dynamics cannot be run any further. This is similar to the situation of finite time blowup in ordinary differential equations. For example, for the equation $y'=y^2,y(0)=1$ the solution is $y=\frac{1}{1-t}$ which has finite time blowup at $t=1$. After that there is no sense in continuing to run the dynamics. The same concept emerges in the context of exploding CTMCs.

Ian
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