For $U\sim\operatorname{Exponential}(1)$,
$$
\bar S(t)=\int_0^\infty e^{-u\theta t}e^{-u}du
=\frac1{1+\theta t},
\qquad
\bar h(t)=\frac\theta{1+\theta t}.
$$
Again $\bar h(0)=\theta$ because $\mathbb EU=1$, but now $\bar h(t)\to0$ as $t\to\infty$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 207 / 6 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-207/6/a/ii/solution