Given the infinitesimal generator, how a continuous Markov chain behaves after the exploding time

markov chainsmarkov-processstochastic-processes

I'm reading paper Explosion, implosion, and moments of passage times for continuous-time Markov chains: a semimartingale approach:


Let $\mathbb X$ be the state space and $\Gamma=(\Gamma_{x y})_{x, y \in X}$ the infinitesimal generator of the continuous Markov chain. The stochastic Markovian matrix $P=(P_{x y})_{x, y \in \mathbb X}$ is defined by
$$
P_{x y}=\left\{\begin{array}{ll}
\frac{\Gamma_{x y}}{\gamma_{x}} & \text { if } \gamma_{x} \neq 0 \\
0 & \text { if } \gamma_{x}=0
\end{array} \text { for } y \neq x, \text { and } P_{x x}=\left\{\begin{array}{ll}
0 & \text { if } \gamma_{x} \neq 0 \\
1 & \text { if } \gamma_{x}=0
\end{array}\right.\right.
$$

The kernel $P$ defines a discrete-time $(\mathbb X, P)$-Markov chain $\tilde{\xi}=(\tilde{\xi}_{n})_{n \in \mathbb{N}}$ termed the Markov chain embedded at the moments of jumps. Define a sequence $\sigma=(\sigma_{n})_{n \geq 1}$ of random holding times distributed, conditionally on $\tilde{\xi}$, according to an exponential law. More precisely, consider $$\mathbb{P}\left(\sigma_{n} \in \mathrm{d} s | \tilde{\xi}\right)=\gamma_{\tilde{\xi}_{n-1}} \exp \left(-s \gamma_{\tilde{\xi}_{n-1}}\right) \mathbf{1}_{\mathbb{R}_{+}}(s) \,\mathrm{d} s$$ so that $\mathbb{E}\left(\sigma_{n} | \tilde{\xi}\right)=1 / \gamma_{\tilde{\xi}_{n-1}}$.

The sequence $J=(J_{n})_{n \in \mathbb{N}}$ of random jump times is defined accordingly by $J_{0}=0$ and for $n \geq 1$ by $J_{n}=\sum_{k=1}^{n} \sigma_{k}$. The life time is denoted $\zeta=\lim _{n \rightarrow \infty} J_{n}$. To have a unified description of both explosive and non-explosive processes, we can extend the state space into $\hat{\mathbb X}=\mathbb X \cup\{\partial\}$ by adjoining a special absorbing state $\partial$. The continuous-time Markov chain is then the càdlàg process $\xi=(\xi_{t})_{t \in[0, \infty]}$ defined by $$
\xi_{0}=\tilde{\xi}_{0} \text { and } \xi_{t}=\left\{\begin{array}{ll}
\sum_{n \in \mathbb{N}} \tilde{\xi}_{n} \mathbf{1}_{[J_{n}, J_{n+1})}(t) & \text { for } 0<t<\zeta \\
\partial & \text { for } t \geq \zeta
\end{array}\right.
$$


Let $(X_t)_{t \in[0, \infty]}$ be the Markov chain defined by $(\mathbb X, \Gamma)$.

In case $\zeta < \infty$, it seems to me that we don't know how the $X_t$ behaves when $t \ge \zeta$, so the authors introduce $\partial$. This goes against my understanding because we are given $(\mathbb X, \Gamma)$ and thus we know $(X_t)_{t \in[0, \infty]}$.

Could you pleas elaborate on my confusion?

Best Answer

I convert @lan's comment as answer to close this question.


If $\zeta<\infty$ then $(\mathbb{X},\Gamma)$ and the given entropy $\omega$ cannot specify what $X_\zeta(\omega)$ is, so you do not know $(X_t)_{t \in [0,\infty)}$. Intuitively, the system changes state infinitely often in an arbitrarily small interval to the left of $t=\zeta$ so there is no way to settle on what $X_\zeta$ should actually be. The convention is to say that at time $\zeta$ the process "exploded" and use this auxiliary state (with trivial dynamics associated to it) to track paths that have already exploded at various times.

Yes, if a chain can explode then $(\mathbb{X},\Gamma)$ does not determine it for all time. Basically it determines it for any finite number of jumps, but if infinitely many jumps can occur in finite time then the generator ceases to specify the process.