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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 areTheir successive ratio tends to $r$, so they are summable for $p<1/2$. For $p\ge 1/2$, $u_k>1/2>d_{k+1}$ 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 ${\sum }_k{w}_k=2(1-r{)}^{-a-1}$, andEvery 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, $\mathbb{P}(X_{2n}=0)\to 2{\pi }_0=(1-r{)}^{a+1}$, whereas $\mathbb{P}(X_{2n+1}=0)=0$.

We use the stationary weights of a reflected birth-death chain, also verifying why they decide the question. If a stationary distribution $\pi$ exists, its equation at zero is ${\pi }_0={\pi }_1{d}_1={\pi }_0{u}_0$. At $k\ge 1$, stationarity givesThus the edge flow is constant, and its value at the boundary is zero. Every stationary distribution must therefore satisfy detailed balance on each edge. Since irreducibility forces ${\pi }_0>0$, it must be proportional toConversely, if ${\sum }_k{w}_k<\infty$, normalizing these positive weights gives a probability distribution satisfying detailed balance, hence a stationary distribution. Existence of a stationary distribution for an irreducible Markov chain on a countable state space is equivalent to positive recurrence.

The ratio isFor $p<1/2$, the limit is less than one, so the ratio test gives summability. For $p\ge 1/2$, observe that $u_k=p+\theta q/(k+\theta )>p\ge 1/2$, whereas $d_{k+1}<q\le 1/2$. Thus $w_{k+1}/w_k>1$ for every $k$, so the weights cannot be summable. This also resolves $p=1/2$ without making an inconclusive appeal to the limiting ratio. Therefore