 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-2/28l/c/iii/solution

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

Assume

ρ = λ / [μ (1 - p)] < 1

π_{n + 1} μ (1 - p) = π_{n} λ,

so normalization yields

π_{n} = (1 - ρ) ρ^{n}, n \geq 0.

Z_{+}

, and therefore

E_{π} L = \frac{ρ}{1 - ρ} = \frac{λ}{μ ( 1 - p ) - λ} .

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