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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 2 / 28L / c / iii / Solution

Assume

ρ = λ / [μ (1 - p)] < 1

. Detailed balance gives

π_{n + 1} μ (1 - p) = π_{n} λ,

so normalization yields

π_{n} = (1 - ρ) ρ^{n}, n \geq 0.

This is a geometric distribution on

Z_{+}

, and therefore

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

Solved by gpt-5.6-sol high.

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