 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/18h/b/i/solution

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

0

and

1

gives

π_{0} p = π_{1} q^{- 1}, so π_{1} = qp π_{0} .

For

i \geq 1

i

and

i + 1

gives

π_{i} q^{- (i + 2)} = π_{i + 1} q^{- (i + 1)}, π_{i + 1} = \frac{π _{i}}{q} .

Consequently

π_{i} = qp π_{0} q^{- (i - 1)}, i \geq 1.

The expected occupation time of the positive even states during a return cycle to state

1

is therefore

\frac{π _{2} + π _{4} + \dots}{π _{1}} = \sum_{r = 1}^{\infty} q^{- (2 r - 1)} = \frac{q}{q ^{2} - 1} .

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