 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-4/9h/b/solution

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

π_{i} p_{ij} = π_{j} p_{ji}

for all

i, j

. Summing over

i

gives

\sum_{i} π_{i} p_{ij} = π_{j} \sum_{i} p_{ji} = π_{j},

so detailed balance implies invariance.

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