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In a positive recurrent birth-death-immigration jump chain, let $I_n$ and $A_n$ count immigration events and upward steps before time $n$. The centered immigration indicators form bounded martingale differences; the strong law for martingales with bounded increments and the countable-state Markov chain ergodic theorem give $I_n/n\to {\sum }_k{\pi }_k\theta /(k+\theta )$ almost surely. The analogous argument for upward steps gives $A_n/n\to {\sum }_k{\pi }_k{u}_k$. Summing detailed balance gives ${\sum }_k{\pi }_k{u}_k={\sum }_k{\pi }_k{d}_k=1/2$, since every step is upward or downward. Therefore $I_n/A_n$ converges to the displayed fraction. For this chain the explicit stationary distribution reduces the fraction to $1-p/(1-p)$.

Author's authorized chapter previews. Sections 1.7, 1.8 and 1.10 develop invariant probability distributions, convergence to equilibrium and the countable-state Markov chain ergodic theorem for Markov chains. The distinction between a normalizable invariant measure and an invariant measure of infinite mass is crucial on a countably infinite state space.

Assume $0<Z={\sum }_{x}r(x)<\infty$, and work on the support $S=\{x:r(x)>0\}$. In rejection sampling, choose a normalized proposal probability mass function $g$ and a known finite constant $M$ satisfying $r(x)\le Mg(x)$ everywhere. Draw independent random variables $Y\sim g$ and $U\sim \mathrm{Unif}(0,1)$, accepting $Y$ when $U\le r(Y)/(Mg(Y))$. The joint probability of proposing and accepting $x$ is $r(x)/M$, so the overall acceptance probability is $Z/M$ and the accepted value has probability mass function $r(x)/Z$. Fresh trials therefore produce independent and identically distributed random variables with the desired probability distribution. For a fixed positive number $N$ of accepted values, the sample average $N^{-1}{\sum }_{i=1}^{N}f(Y_i)$ is an unbiased estimator of the expectation when that expectation is finite. Its variance is ${\mathrm{Var}}_p(f)/N$ if the second moment is finite. Finding a valid envelope can be difficult, and a small acceptance probability makes the method expensive.

For the Metropolis–Hastings algorithm, choose a proposal transition matrix $q$ and, at the current state $x$, propose $y\sim q(x,\cdot )$. Accept with the Metropolis–Hastings acceptance probabilityA proposal outside $S$ is rejected; an impossible proposal need not be assigned a ratio. Otherwise remain at $x$. This constructs a Markov chain satisfying detailed balance with $p$, without computing $Z$ or a global envelope. Estimate the expectation by averaging $f(X_i)$ over iterations, including repeated states after rejections. Keeping only accepted jumps generally changes the stationary distribution of the recorded process.

The Metropolis–Hastings algorithm produces dependent samples, so the variance of its average includes autocovariances; starting outside stationarity can also cause finite-run bias. With Markov chain irreducibility, suitable recurrence and an integrable $f$, the countable-state Markov chain ergodic theorem gives consistency. Aperiodicity is additionally needed for the usual convergence of one-time probability distributions. In contrast, accepted rejection sampling values need no initial transient. Neither method is universally faster: envelope rejection competes with the proposal quality and mixing time of the Markov chain. Nonnegativity of $f$ alone does not guarantee a finite expectation or variance.

For each adjacent pair, the detailed balance equations follows directly:For nonadjacent distinct states both fluxes vanish, and diagonal fluxes agree trivially. Since ${\sum }_n{\pi }_n=e^{-\lambda }{\sum }_n{\lambda }^n/n!=1$, this is a genuine stationary distribution, rather than merely an invariant measure of infinite mass.

Every upward and downward transition probability between neighbouring states is positive for $\lambda >0$. Any two states can therefore be joined by a finite path of positive probability: the Markov chain is irreducible. Moreover,State zero has period one; Markov chain irreducibility makes every state aperiodic. Finally, the standard countable-state recurrence theorem states that an irreducible Markov chain admitting a stationary distribution of total mass one is positive recurrent, with that stationary distribution unique. The countable-state Markov chain convergence theorem then states that Markov chain irreducibility, positive recurrence and aperiodicity imply $P^k(i,j)\to {\pi }_j$ for all states $i,j$. The hypotheses just verified hold on the infinite state space ${\mathbb{N}}_0$, not merely on a numerical truncation.^[ref]

Consequently, the unique stationary distribution and the limiting distribution are both $\mathrm{Poisson}(\lambda )$, from every initial state. The countable-state Markov chain ergodic theorem also gives $N^{-1}{\sum }_{k=1}^{N}f(X_k)\to {\sum }_n{\pi }_{n}f(n)$ almost surely whenever ${\sum }_n{\pi }_n|f(n)|<\infty$.^[ref]

For the countable-state Markov chain ergodic theorem, let the Markov chain be irreducible and positive recurrent, with its unique stationary distribution $\pi$. From any initial state, for every real $f$ satisfying ${\sum }_j{\pi }_j|f(j)|<\infty$,In particular, $n^{-1}{\sum }_{m=0}^{n-1}{1}_{\{X_m=j\}}\to {\pi }_j$ almost surely. This time-average conclusion requires no aperiodicity.

The countable-state Markov chain convergence theorem adds the hypothesis of an aperiodic Markov chain and concludesIt also follows that ${\mathbb{P}}_{\nu }(X_n=j)\to {\pi }_j$ for every initial probability distribution $\nu$. For an irreducible recurrent chain, the occupation-frequency theorem can also be stated as a limit $1/{\mathbb{E}}_j{T}_j^{+}$, interpreted as zero for a null recurrent state; in the positive recurrent case this equals ${\pi }_j$.

Every step changes the population by one, so $X_n$ has the parity of $n$ when $X_0=0$. Hence $\mathbb{P}(X_{2m+1}=0)=0$. A return in two steps has positive probability from every state, and all return times are even, so the period of a state in a Markov chain is exactly two.

Each transition exchanges the even and odd classes. Since $\pi$ is an invariant distribution, their masses are equal: $\pi (\text{even})=\pi (\text{odd})=1/2$. The chain sampled at times $2m$, restricted to the even states, is irreducible and positive recurrent, and it is aperiodic because its diagonal entries are positive. Its stationary distribution is $2{\pi }_k$ on even $k$. The countable-state Markov chain convergence theorem applied to this sampled chain yieldsSince ${\pi }_0>0$, the full sequence has no limit. Its time average nevertheless tends to ${\pi }_0$ by the countable-state Markov chain ergodic theorem.

Let $I_n$ count immigration events during the first $n$ transitions, and let $A_n$ count upward steps during those transitions. Write $h(k)=\theta /(k+\theta )$ for the conditional immigration probability. If $J_m$ is the indicator that transition $m$ to $m+1$ is immigration, then $J_m-h(X_m)$ is a bounded martingale difference with respect to the history that records event types as well as population sizes. The strong law for martingales with bounded increments impliesSince $h$ is bounded, the countable-state Markov chain ergodic theorem givesApplying the same argument to upward-step indicators gives $A_n/n\to {\sum }_k{\pi }_k{u}_k$ almost surely. To evaluate this denominator, sum detailed balance over all neighboring edges:Since $u_k+d_k=1$ and ${\sum }_k{\pi }_k=1$, both sums equal $1/2$. The denominator limit is positive, so division is valid and the immigration fraction among upward steps isThis is the requested answer in terms of $\pi$. Substituting part (iii) additionally gives ${\sum }_k{\pi }_k\theta /(k+\theta )=(1-r)/2$, so the fraction simplifies to $1-r=(1-2p)/(1-p)$.