 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2023/discrete-mathematics-probability-and-continuous-mathematics/9/b/i/solution

1 = c \int_{1}^{\infty} x^{- t} d x = c / (t - 1)

. Hence $c = t - 1$ . It is positive for the given

t > 2

, and the density is nonnegative, so this does define a probability distribution. It is a Pareto distribution with scale

1

t - 1

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