Past exam of the mathematics course of the University of Cambridge / 2005 / ii / Paper 4 / 26I / a / Solution

The embedded birth-death chain has upward probability $p_k=(k+2)/(2(k+1))$ and downward probability $q_k=k/(2(k+1))$. To calculate a return probability, solve the harmonic difference equation on $\{0,\dots ,m\}$ with boundary values one at zero and zero at $m$. Successive differences have ratio $q_k/p_k=k/(k+2)$, so their weights areTheir total infinite sum is $2$. Letting $m\to \infty$ gives the probability of ever hitting zero from $k$:In particular the chain leaves zero to one and returns with probability $1/2<1$, so it is transient. This finite-interval calculation selects the minimal nonnegative hitting-probability solution, rather than merely observing that $h_k$ is harmonic.

The continuous-time chain has the same embedded transitions and constant total exit rate $2/3$, so it is nonexplosive and has exactly the same return events. It too is transient, neither positive recurrent nor null recurrent.