 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/4/a/ii/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

\int_{0}^{\infty} α e^{- α t} d t = 1

E [T] = \int_{0}^{\infty} t α e^{- α t} d t = 1/ α

and

E [T^{2}] = 2/ α^{2}

. Therefore

E [T] = \frac{1}{α}, Var (T) = E [T^{2}] - E [T]^{2} = \frac{1}{α ^{2}} .

These are the mean and variance of the exponential distribution with rate

α

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