For equal point masses at $1/2$ and $3/2$ with $h_0=\theta$,
$$
\bar h(t)=\frac\theta2
\frac{e^{-\theta t/2}+3e^{-3\theta t/2}}
{e^{-\theta t/2}+e^{-3\theta t/2}}.
$$
Its initial value is $\theta$ because $\mathbb EU=1$, while $\bar h(t)\to\theta/2$ as the more frail survivors are progressively depleted.


Solution

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

 Solution

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