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Use the Poisson distribution with probability mass function ${\pi }_n=e^{-\lambda }{\lambda }^n/n!$ on ${\mathbb{N}}_0$. A symmetric nearest-neighbour proposal with a reflecting boundary isAll other entries vanish. In particular, the boundary proposal does not force a move to one: retaining the probability $q(0,0)=1/2$ makes $q(0,1)=q(1,0)$.

The adjacent ratios of the Poisson distribution are ${\pi }_{n+1}/{\pi }_n=\lambda /(n+1)$ and ${\pi }_{n-1}/{\pi }_n=n/\lambda$. The Metropolis–Hastings acceptance probability consequently giveswhere the symbol $u_n$ denotes the upward transition probability. Thus the complete transition matrix iswith no downward entry at zero and all other entries zero. In matrix form,Every row is nonnegative and sums to one. An executable Python step of this Poisson Metropolis chain is:

def poisson_step(n, lam, rng):
    if lam <= 0 or n < 0 or int(n) != n:
        raise ValueError("require lam > 0 and an integer n >= 0")
    y = n + 1 if rng.random() < 0.5 else max(0, n - 1)
    ratio = lam / (n + 1) if y > n else n / lam if y < n else 1.0
    return y if rng.random() < min(1.0, ratio) else n

Supply a pseudorandom number generator such as random.Random(2021) and retain each returned state, including repeats. No normalizing constant or evaluation of a factorial is needed by the algorithm.

A Poisson Metropolis chain samples a Poisson distribution by proposing an upward or downward unit step with equal probability and reflecting a downward proposal at zero into a self-proposal. Its upward and downward transition probabilities are the displayed $u_n,d_n$, with $d_0=0$, and the diagonal entry is $1-u_n-d_n$. For ${\pi }_n=e^{-\lambda }{\lambda }^n/n!$, adjacent stationary fluxes equal $\frac{1}{2}\min ({\pi }_n,{\pi }_{n+1})$. Hence detailed balance holds. Positive adjacent probabilities imply an irreducible Markov chain; $P(0,0)\ge 1/2$ implies aperiodicity. The normalized invariant Poisson distribution implies positive recurrence, so the countable-state Markov chain convergence theorem applies. The boundary proposal probability is essential: forcing every proposal at zero to go upward would require a different acceptance ratio.