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

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- Solution e

Solution

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}},

independently of

ε \in (0, 1)

.

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