 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-1/5/b/iii/solution

E [Y^{2}] = \frac{1}{2} (- 1)^{2} + \frac{1}{2} \int_{0}^{\infty} y^{2} e^{- y} d y = 1/2 + 1 = 3/2

. Since the mean is zero,

Var (Y) = \frac{3}{2} .

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