 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-2/27i/d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

A

λ

and is independent of the following busy period. Thus

ϕ_{T_{k}} (θ) = \frac{λ}{λ - θ} ϕ_{B_{k}} (θ) .

Substitute

ϕ_{B_{k}} = (λ - θ) ϕ_{T_{k}} / λ

into part (c). Its service-transform argument becomes

θ + λ (ϕ_{B_{k}} - 1) = (λ - θ) (ϕ_{T_{k}} - 1)

. Therefore

\frac{λ - θ}{λ} ϕ_{T_{k}} (θ) = ϕ_{S_{k}} ((λ - θ) (ϕ_{T_{k}} (θ) - 1)) .

As before, positive arguments must lie in the domain of finite moment-generating functions.

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