 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-45/4/e/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

The acceptance probability is the prior average of the Poisson likelihood:

A_{k} = E_{θ, T} [e^{- θ L /2} \frac{( θ L /2 ) ^{k}}{k !}] = Z_{k} = P_{prior} (S = k) .

Thus the expected number of proposals per accepted draw is

1/ Z_{k}

. A numerical rate depends on the specified prior, sample size and observed count; it is not determined by

k

alone.

For a more explicit one-dimensional expression, condition on

θ

. The probability generating function, integrating each independent exponential epoch, is

E [z^{S} ∣ θ] = E [e^{- θ (1 - z) L /2}] = \prod_{j = 2}^{n} \frac{λ _{j}}{λ _{j} + θ j ( 1 - z ) /2} = \prod_{i = 1}^{n - 1} \frac{i}{i + θ ( 1 - z )} .

Let

p_{k} (θ)

be its coefficient of

z^{k}

. Then

Z_{k} = \int π (θ) p_{k} (θ) d θ

. This identifies the acceptance rate with a fully specified marginal likelihood average.

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