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ID: past-exam-of-the-msc-in-advanced-computer-science-course-of-the-university-of-oxford/2021/bayesian-statistical-probabilistic-programming/1/e/solution

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Past exam of the MSc in advanced computer science course of the University of Oxford / 2021 / Bayesian statistical probabilistic programming / 1 / e / Solution by

Codex  0 2026-10-25

Use the Poisson distribution with probability mass function

π_{n} = e^{- λ} λ^{n} / n!

on

N_{0}

. A symmetric nearest-neighbour proposal with a reflecting boundary is

q (n, n + 1) = \frac{1}{2} (n \geq 0), q (n, n - 1) = \frac{1}{2} (n \geq 1), q (0, 0) = \frac{1}{2} .

All 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

π_{n + 1} / π_{n} = λ / (n + 1)

and

π_{n - 1} / π_{n} = n / λ

. The Metropolis–Hastings acceptance probability consequently gives

u_{n} = \frac{1}{2} min {1, λ / (n + 1)}, d_{n} = \frac{1}{2} min {1, n / λ} (n \geq 1), d_{0} = 0,

where the symbol

u_{n}

denotes the upward transition probability. Thus the complete transition matrix is

P (n, n + 1) = u_{n}, P (n, n - 1) = d_{n}, P (n, n) = 1 - u_{n} - d_{n},

with no downward entry at zero and all other entries zero. In matrix form,

P = 1 - u_{0} d_{1} 0 ⋮ u_{0} 1 - u_{1} - d_{1} d_{2} ⋮ 0 u_{1} 1 - u_{2} - d_{2} ⋱ 00 u_{2} ⋱ \dots \dots \dots ⋱ .

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.

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