For $g\geq0$, the <Tonelli theorem> and conditioning on $N_t$ give
$$
\mathbb E[X_t\mid N_t]
=N_t\mathbb E[g(Y_1)]
=N_t\int_0^1g(y)\,dy.
$$
Since $\mathbb E[N_t]=\lambda t$,
$$
\boxed{\mathbb E[X_t]
=\lambda t\int_0^1g(y)\,dy,}
$$
with both sides allowed to be $+\infty$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 201 / 6 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-201/6/b/solution