= Birth-death-immigration jump chain
{title2=$u_k=(kp+\theta)/(k+\theta),\quad d_k=k(1-p)/(k+\theta)$}
Here $0<p<1$, $\theta>0$, and the upward step is either a birth or immigration, the latter having <conditional probability> $\theta/(k+\theta)$. At zero immigration occurs with <probability> one. Put $q=1-p$, $a=\theta/p$ and $r=p/q$. The <stationary weights of a reflected birth-death chain> are
$$
w_k=\frac{k+\theta}{\theta}\frac{\Gamma(k+a)}{\Gamma(a)k!}r^k.
$$
Their successive ratio tends to $r$, so they are summable for $p<1/2$. For $p\geq1/2$, $u_k>1/2>d_{k+1}$ makes the weights strictly increasing, ruling out a <stationary distribution>. Thus <positive recurrence> holds exactly for $p<1/2$. In that case the <binomial series> gives $\sum_kw_k=2(1-r)^{-a-1}$, and
$$
\pi_k=\frac{(1-r)^{a+1}}{2\theta\Gamma(a)}\frac{(k+\theta)\Gamma(k+a)}{k!}r^k.
$$
Every step reverses parity and a two-step return has positive <probability>, so the <period of a state in a Markov chain> is two. Starting at zero, $\mathbb P(X_{2n}=0)\to2\pi_0=(1-r)^{a+1}$, whereas $\mathbb P(X_{2n+1}=0)=0$.
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