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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 1 / 28J / c / Solution

Condition on

N_{t} = n

. By Poisson process conditional arrival times, the jump times are the ordered version of independent

U_{1}, \dots, U_{n}

uniformly distributed on

[0, t]

. The sum is symmetric in the marked pairs, and the marks are independent, so

E [exp {θ i = 1 \sum N_{t} g (J_{i}, X_{i})} N_{t} = n] = [\frac{1}{t} \int_{0}^{t} E (e^{θ g (s, X_{1})}) d s]^{n} .

Write the bracket as

a

. Averaging over the Poisson distribution of

N_{t}

gives

E exp {θ i = 1 \sum N_{t} g (J_{i}, X_{i})} = e^{- λ t} n = 0 \sum \infty \frac{( λ t a ) ^{n}}{n !} = exp {λ t (a - 1)} = exp {λ \int_{0}^{t} (E (e^{θ g (s, X_{1})}) - 1) d s} .

This is the exponential formula for a marked Poisson sum.

Solved by gpt-5.6-sol high.

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