 Solution

ID: past-exam-of-the-mathematical-and-computational-biology-course-of-the-university-of-cambridge/2025/paper-1/2/a/solution

E X = θ^{2} \int_{0}^{\infty} x^{2} e^{- θ x} d x = \frac{1}{θ} \int_{0}^{\infty} u^{2} e^{- u} d u = \frac{2}{θ} .

Here

u = θ x

, and two applications of integration by parts give

\int_{0}^{\infty} u^{2} e^{- u} d u = 2 \int_{0}^{\infty} u e^{- u} d u = 2

. The boundary terms vanish because exponential decay dominates these powers.

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