 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/27j/c/solution

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

For one visitor, let

T_{0} = 0, T_{j} = Z_{1} + \dots + Z_{j},

p_{j} (u) = P (T_{j - 1} \leq u < T_{j}), 1 \leq j \leq 10.

A visitor arriving at time

s \leq t

is in room

j

t

exactly when its independent service-time mark satisfies

T_{j - 1} \leq t - s < T_{j}

. Thus it receives mark

j

p_{j} (t - s)

; a further mark records that it has already left the gallery.

The Independent marking theorem for Poisson point processes makes the point processes carrying the different room marks independent. Consequently

V_{1} (t), \dots, V_{10} (t)

are independent, and Poisson thinning gives

V_{j} (t) \sim Poisson (m_{j} (t)), m_{j} (t) = \int_{0}^{t} λ (s) p_{j} (t - s) d s .

Equivalently,

m_{j} (t) = \int_{0}^{t} λ (s) P (Z_{1} + \dots + Z_{j - 1} \leq t - s < Z_{1} + \dots + Z_{j}) d s .

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