The <inverse-square law> gives $f_s=L_0/(4\pi r_s^2)$, so a star is observed exactly when
$$
r_s\leq R=\sqrt{\frac{L_0}{4\pi f_{\min}}}.
$$
Writing $a=R/r_0$, integration of the shape-three gamma density gives
$$
\mathbb P(r_s\leq R)=1-e^{-a}\left(1+a+\frac{a^2}{2}\right).
$$
Therefore the fully normalized <truncated distribution> is
$$
p(r_s\mid I_s=1)=
\frac{r_s^2e^{-r_s/r_0}}
{2r_0^3\left[1-e^{-a}(1+a+a^2/2)\right]}
\mathbf1_{(0,R]}(r_s).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 219 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-219/1/c/solution