 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2018/discrete-mathematics-continuous-mathematics-and-probability/8/b/ii/solution

c = 8/3

. Integrating the probability density from one to two gives

P (X \leq 2) = \frac{8}{3} \int_{1}^{2} x^{- 3} d x - \int_{1}^{2} x^{- 4} d x = \frac{8}{3} \frac{3}{8} - \frac{7}{24} = \frac{17}{24} .

The expected value exists, since the first-moment integrals converge, and equals

E X = \frac{8}{3} \int_{1}^{\infty} x^{- 2} d x - \int_{1}^{\infty} x^{- 3} d x = \frac{8}{3} - \frac{1}{2} = \frac{13}{6} .

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