 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-3/28j/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

K (t, \cdot)

t

, its transition density is

k (t, s) = q (s ∣ t) ρ (t, s) .

μ (t) k (t, s) = μ (t) q (s ∣ t) min {\frac{μ ( s ) q ( t ∣ s )}{μ ( t ) q ( s ∣ t )}, 1} = min {μ (s) q (t ∣ s), μ (t) q (s ∣ t)} = μ (s) k (s, t) .

This is detailed balance for every pair of distinct states.

Write

r (t) = 1 - \int_{Θ} k (t, s) d s

for the rejection probability, so

K (t, d s) = k (t, s) d s + r (t) δ_{t} (d s) .

For every measurable

B \subseteq Θ

\int_{Θ} K (t, B) μ (t) d t = \int_{B} \int_{Θ} μ (t) k (t, s) d t d s + \int_{B} μ (t) r (t) d t = \int_{B} μ (s) [\int_{Θ} k (s, t) d t + r (s)] d s = \int_{B} μ (s) d s .

Hence $μ$ is an invariant distribution of the chain.

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