= Solution
The chain is not an <irreducible Markov chain>, since zero is absorbing and cannot reach a positive state. Write $q_i=-q_{ii}=\sum_{j\ne i}q_{ij}$. These are finite positive holding rates for the usual <continuous-time Markov chain>. Its first holding time in state $i\geq1$ is exponential with rate $q_i$, and its jump destination has probabilities $q_{ij}/q_i$. Therefore
$$
\boxed{\delta=\min_{1\leq i\leq M}\frac{q_{i0}}{q_i}>0,\qquad
\mathbb P_i(X_{J_1}=0)\geq\delta.}
$$
The recursively defined $T_m$ are <stopping times>: after $T_m$, wait until the path first reaches a different state in the finite set. Set all later times to infinity if a previous time is infinite. On $A_m$, the <Strong Markov property> applies at $T_m$. Its next jump goes to zero with probability at least $\delta$; if it does, $T_{m+1}=\infty$. Other jumps may also prevent another qualifying visit. Thus
$$
\mathbb P_i(A_{m+1}\mid\mathcal F_{T_m})\leq1-\delta\quad\text{on }A_m,
\qquad\mathbb P_i(A_{m+1})\leq(1-\delta)\mathbb P_i(A_m).
$$
Starting with $A_0$ certain gives \b[$\boxed{\mathbb P_i(A_m)\leq(1-\delta)^m}$].
Back to article page
