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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2023

Paper 1

Condition on $N_t=n$. By <Poisson process conditional arrival times>, the jump times are the ordered version of independent $U_1,\ldots,U_n$ uniformly distributed on $[0,t]$. The sum is symmetric in the marked pairs, and the marks are independent, so
$$
\begin{aligned}
&\mathbb E\left[
\left.
\exp\left\{\theta\sum_{i=1}^{N_t}g(J_i,X_i)\right\}
\right|N_t=n\right]\\
&\qquad=
\left[
\frac1t\int_0^t
\mathbb E\left(e^{\theta g(s,X_1)}\right)ds
\right]^n.
\end{aligned}
$$
Write the bracket as $a$. Averaging over the <Poisson distribution> of $N_t$ gives
$$
\begin{aligned}
\mathbb E\exp\left\{\theta\sum_{i=1}^{N_t}g(J_i,X_i)\right\}
&=e^{-\lambda t}\sum_{n=0}^\infty
\frac{(\lambda t a)^n}{n!}\\
&=\exp\{\lambda t(a-1)\}\\
&=\exp\left\{
\lambda\int_0^t
\left(\mathbb E(e^{\theta g(s,X_1)})-1\right)ds
\right\}.
\end{aligned}
$$
This is the <exponential formula for a marked Poisson sum>.

Solution

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

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

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

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