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

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- Solution c

Solution

The inverse-square law gives

f_{s} = L_{0} / (4 π r_{s}^{2})

, so a star is observed exactly when

r_{s} \leq R = \frac{L _{0}}{4 π f _{m i n}} .

a = R / r_{0}

, integration of the shape-three gamma density gives

P (r_{s} \leq R) = 1 - e^{- a} (1 + a + \frac{a ^{2}}{2}) .

Therefore the fully normalized truncated distribution is

p (r_{s} ∣ I_{s} = 1) = \frac{r _{s}^{2} e ^{- r_{s} / r_{0}}}{2 r _{0}^{3} [ 1 - e ^{- a} ( 1 + a + a ^{2} /2 ) ]} 1_{(0, R]} (r_{s}) .

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