= 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 $\{0,\ldots,m\}$ with boundary values one at zero and zero at $m$. Successive differences have ratio $q_k/p_k=k/(k+2)$, so their weights are
$$
r_0=1,\qquad r_j=\prod_{k=1}^j\frac{k}{k+2}
=\frac2{(j+1)(j+2)}.
$$
Their total infinite sum is $2$. Letting $m\to\infty$ gives the <probability> of ever hitting zero from $k$:
$$
h_k=\frac{\sum_{j=k}^\infty r_j}{\sum_{j=0}^\infty r_j}
=\boxed{\frac1{k+1}}.
$$
In particular the chain leaves zero to one and returns with <probability> $1/2<1$, so it is \b[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 \b[transient], neither <positive recurrent> nor <null recurrent>.
Back to article page
