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Poisson process

Let $J_i$ be the points of a rate-$\lambda$ Poisson process and let independent identically distributed marks $X_i$ be independent of the process. Then
$$
\mathbb E\exp\left(\theta\sum_{i=1}^{N_t}g(J_i,X_i)\right)
=\exp\left\{\lambda\int_0^t
\left(\mathbb E e^{\theta g(s,X_1)}-1\right)ds\right\}.
$$
Condition on $N_t$, use uniform order statistics, and sum the resulting exponential series.


Exponential formula for a marked Poisson sum

Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 34 / 3 / Solution

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

Exponential formula for a marked Poisson sum

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Exponential formula for a marked Poisson sum

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