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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-4/26j/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 4 / 26J / b / Solution by

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

By Poisson thinning, the first arrival times of the coupon types are independent exponential distribution

T_{j}

λ p_{j}

. Completion requires all those arrivals, so

T = max_{j} T_{j}

and

P (T < t) = j = 1 \prod m (1 - e^{- λ p_{j} t}) .

To relate time to count rigorously, let

E_{k}

be the successive independent exponential waiting times of mean

1/ λ

. The completion count

L

depends only on the coupon labels, which are independent of all

E_{k}

. Therefore

T = \sum_{k \geq 1} E_{k} 1_{{L \geq k}}, E T = \sum_{k \geq 1} \frac{1}{λ} P (L \geq k) = \frac{E L}{λ} .

Nonnegative integration justifies the sum, avoiding any unproved optional-stopping assertion. Thus the Poissonized coupon completion time and count identity is $λ E T = E L$ . Changing variables

s = λ t

in the tail integral gives

E L = \int_{0}^{\infty} [1 - \prod_{j} (1 - e^{- p_{j} s})] d s,

which depends only on the type probabilities, not

λ

. These expectations are finite when all

p_{j} > 0

; a missing type makes both infinite and completion impossible.

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