Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 1 / 27J / c / Solution

Fix $M$. Each time the chain is in $\{1,\dots ,M\}$ immediately before a jump, its conditional chance to jump to zero is at least the $\delta$ from part (b). Count all these opportunities, including returns to the same state after an excursion; the $T_m$ in part (b) need not count those returns. By the Strong Markov property at successive such visits, the probability of surviving $r$ opportunities without absorption is at most $(1-\delta {)}^r$. Hence infinitely many such opportunities without hitting zero have probability zero.

A positive state cannot be occupied forever: its finite positive holding rate gives an almost surely finite holding time. For a nonexplosive continuous-time Markov chain, visits at arbitrarily late times to the finite set therefore require infinitely many of the counted opportunities. On the event of no absorption, the chain consequently eventually stays above $M$, almost surely. Intersect this probability-one assertion over integers $M$. The result is $X_t\to \infty$. If zero is ever reached it remains occupied forever, and for initial state zero this is immediate. Thus $\boxed{{\mathbb{P}}_i(E\cup F)=1}$ for every initial state.

This uses the usual assumption implicit in an $I$-valued chain defined for all $t\ge 0$: no explosion with loss to a cemetery state. The rate conditions alone do not ensure it. For example rates $q_{i,i+1}=i^3$, $q_{i0}=1$ allow an infinite upward path with probability ${\prod }_i{i}^3/(i^3+1)>0$ and a finite total holding time, since ${\sum }_{i}1/(i^3+1)<\infty$. The minimal chain then explodes; with a cemetery state the printed dichotomy would require an additional explosion alternative.