The inverse-square law gives $f=L_0/(4\pi r^2)$, so observation is equivalent to
$$
r\leq R,\qquad R=\sqrt{\frac{L_0}{4\pi f_{\min}}}.
$$
For $u=R/r_0$, integrating the Gamma density gives
$$
\mathbb P(r\leq R)=1-e^{-u}\left(1+u+\frac{u^2}{2}\right).
$$
Therefore
$$
\boxed{
p(r\mid I=1)=
\frac{r^2e^{-r/r_0}}
{2r_0^3\left[1-e^{-R/r_0}\{1+R/r_0+R^2/(2r_0^2)\}\right]}
\mathbf1_{\{0<r\leq R\}}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 219 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/1/d/solution