 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ia/paper-2/9f/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

This is the unit-rate Erlang distribution. For completeness, use convolution. The result is clear for

n = 1

. If

f_{n} (x) = \frac{x ^{n - 1}}{( n - 1 )!} e^{- x} 1_{{x > 0}},

then independence and the unit exponential density give, for

x > 0

f_{n + 1} (x) = \int_{0}^{x} f_{n} (y) e^{- (x - y)} d y = \frac{e ^{- x}}{( n - 1 )!} \int_{0}^{x} y^{n - 1} d y = \frac{x ^{n}}{n !} e^{- x} .

Induction therefore proves

f_{X_{n}} (x) = \frac{x ^{n - 1}}{( n - 1 )!} e^{- x}, x > 0.

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