 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-33/3/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-07

For a measurable

A

with

(μ + ν) (A) < \infty

, independence of the two Poisson random measures gives

P ((Φ + Ψ) (A) = k) = \sum_{j = 0}^{k} e^{- μ (A)} \frac{μ ( A ) ^{j}}{j !} e^{- ν (A)} \frac{ν ( A ) ^{k - j}}{( k - j )!} = e^{- (μ + ν) (A)} \frac{(( μ + ν ) ( A ) ) ^{k}}{k !} .

Thus its count has the Poisson distribution with the summed parameter. If

A_{1}, \dots, A_{m}

are disjoint, the variables within each family of counts are independent, and independence of

Φ

and

Ψ

makes the two families independent of each other. Hence the pairs

(Φ (A_{j}), Ψ (A_{j}))

, and therefore their sums, are independent over

j

Adding sample counting measures preserves countable additivity. The definition in (a) therefore proves

Φ + Ψ is a Poisson random measure of intensity μ + ν .

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