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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ib/paper-2/18h/e/solution

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Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 2 / 18H / e / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

Adding

ε I

leaves every invariant distribution unchanged. On the unique closed communicating class it also supplies a self-loop, making the chain aperiodic; transient states do not affect the limit. Therefore

P (X_{n}^{(ε)} = 0) ⟶ π_{0} = \frac{1}{E [ T _{1} ]} = \frac{1}{\sum _{j = 1}^{N} j q _{j}},

(1)

independently of

ε \in (0, 1)

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