Solution

Let

N_{j} (A)

be the number of particles of gas

j

in a bounded Lebesgue measurable set

A

. Because Lebesgue measure is diffuse, two independent spatial Poisson point processes have no common particle almost surely. Hence the mixture count is

N (A) = N_{1} (A) + N_{2} (A) .

(347)

Writing

λ_{j} = z_{j} ∣ A ∣

, independence gives the probability generating function

E [s^{N (A)}] = E [s^{N_{1} (A)}] E [s^{N_{2} (A)}] = exp (λ_{1} (s - 1)) exp (λ_{2} (s - 1)) = exp ((z_{1} + z_{2}) ∣ A ∣ (s - 1)) .

(348)

Thus

N (A) \sim Pois ((z_{1} + z_{2}) ∣ A ∣) .

(349)

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

are pairwise disjoint, the family

{N_{j} (A_{r}) : j = 1, 2, 1 \leq r \leq m}

(350)

is independent: counts in disjoint sets are independent within each gas, and the two gases are independent. Therefore the sums

N (A_{r})

are independent. These two facts prove directly that the mixture is a Poisson point process. By the Superposition theorem for Poisson point processes, its activity is

z_{1} + z_{2} .

(351)

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 28K / a / Solution

 Ancestors (11)