 Solution

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]